Termodinamika-zbirka (ciganovic)

zbirka zadataka iz termodinamike

strana 1

KVAZISTATI^KE (RAVNOTE@NE) PROMENE STAWA IDEALNIH GASOVA 2/2/ Vazduh (idealan gas), 2)q2>3!cbs-!w2>1/5416!n40lh*!kvazistati~ki (ravnote`no) mewa stawe do 3)q3>7!cbs-!w3>w2!*/ Odrediti: a) temperaturu vazduha u karakteristi~nim ta~kama procesa b) razmewenu toplotu )r23* i zapreminski rad )x23* c) promenu unutra{we energije )∆v*- entalpije )∆i* i entropije )∆t* vazduha d) skicirati proces na qw!i!Ut dijagramu a) U2 =

q2 ⋅ w 2 3 ⋅ 21 6 ⋅ 1/5416 > >411!LSh 398

U3 =

q3 ⋅ w 3 7 ⋅ 21 6 ⋅ 1/5416 > >:11!L Sh 398

b) r23 = d w ⋅ (U3 − U2 ) > 1/83 ⋅ (:11 − 411) >!543! x23!>!1!

lK lh

lK lh

c) lK lh lK = d q ⋅ (U3 − U2 ) > 2/11 ⋅ (:11 − 411) >!711! lh q3 w3 7 lK = g (q- w ) = d w mo − d q mo > 1/83 ⋅ mo >1/8:2! q2 w2 3 lhL

∆v23 = d w ⋅ (U3 − U2 ) > 1/83 ⋅ (:11 − 411) >!543! ∆i23 ∆t23

! d) q

U 3

3

2

2

w

dipl.ing. @eqko Ciganovi}

t

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strana 2

2/3/ Dva kilograma kiseonika (idealan gas) po~etnog stawa 2)q>2!cbs-!U>484!L*- usled interakcije sa toplotnim ponorom stalne temperature, mewa svoje toplotno stawe kvazistati~ki (ravnote`no) politropski )o>1/9* do stawa 3)!w3> 1/6 ⋅ w 2 */ Skicirati proces u qw!i!Ut koordinatnom sistemu i odrediti: a) mehani~ke veli~ine stawa kiseonika )q-!w-!U* u karakteristi~nim ta~kama b) koli~inu toplote )lK* koju radno telo preda toplotnom ponoru kao i zapreminski rad koji pri tom izvr{i nad radni telom )lK* c) promenu entropije izolovanog termodinami~kog sistema u najpovoqnijem slu~aju

q

U 3

2 o>1/9

o>1/9

2

3 w

UUQ t

a) w2 =

S h U2 q2

=

q2  w 3   = q 3  w 2  U3 =

371 ⋅ 484 2 ⋅ 21 6

o

⇒

>1/:7:9!

n4 lh

w q 3 = q2 ⋅  2  w3

w 3 = 1/6 ⋅ 1/:7:9 >1/595:!

n4 lh

o

  = 2 ⋅ 21 6 ⋅ 3 1/9 > 2/85 ⋅ 21 6 Qb 

q 3 ⋅ w 3 2/85 ⋅ 21 6 ⋅ 1/595: = >435/62!L Sh 371

b) 1/9 − 2/5 lK o−κ ⋅ (435/62 − 484) >−:5/66! ⋅ (U3 − U2 ) > 1/76 ⋅ 1/9 − 2 lh o −2 = n ⋅ r23 = 3 ⋅ (− :5/66) >−29:/2!lK

r23 = d w ⋅ R 23

2 2 lK ⋅ (484 − 435/62) >!−74/15! ⋅ (U2 − U3 ) > 1/37 ⋅ 1/9 − 2 o −2 lh = n ⋅ x 23 = 3 ⋅ (− 74/15 ) >237/19!lK

x23!>! S h ⋅ X23


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c) ∆Ttjtufn!>!∆Tsbeop!ufmp!,!∆Tupqmpuoj!qpops!>!///!>!−1/65!,!1/69!>!1/15!  U q  ∆Tsbeop!ufmp!>!∆T23!> n ⋅  d q mo 3 − S h mo 3  >! U2 q2   435/62 2/85  lK  > 3 ⋅  1/:2 ⋅ mo − 1/37 ⋅ mo  >−1/65! L 484 2   ∆Tupqmpuoj!qpops!>!−!

lK L !

R 23 −29:/2 lK >!−! >1/69! L 435/62 UUQ

2/4/ Kiseonik (idealan gas) n>21!lh, mewa stawe kvazistati~ki izobarski i pri tom se zagreva od temperature U2>411!L!do!U3>:11!L. Kiseonik dobija toplotu od dva toplotna izvora stalnih temperatura. Odrediti: a) promenu entropije izolovanog termodinami~kog sistema ako su temperature toplotnih izvora UUJ2>711!L!i UUJ3>:11 b) temperaturu toplotnog izvora 2!)UUJ2* tako da promena entropije sistema bude minimalna kao i minimalnu promenu entropije sitema u tom slu~aju U 3 B

UJ3 UJ2

2 t b* ∆Ttjtufn!>!∆TSU!,!∆TUJ2!,!∆TUJ2!>!///!>!21!−!5/66!−!4/14!>!3/53!  U q ∆TSU!>! n ⋅  d qmo 3 − S hmo 3 U2 q2 

 :11 lK  > 21 ⋅ 1/:2 ⋅ mo >21 411 L 

∆TUJ2!>!!−!

R 2B 3841 lK >///>!−! >−!5/66! 711 L UUJ2

∆TUJ3!>!!−!

R B3 3841 lK >///>!−! >−!4/14! L :11 UUJ3


lK L

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zbirka zadataka iz termodinamike r2B = d q ⋅ (UB − U2 ) > 1/:2 ⋅ (711 − 411) >384!

strana 4 lK lh

R 2B = n ⋅ r2B = 21 ⋅ 384 >3841!lK r B3 = d q ⋅ (U3 − UB ) > 1/:2 ⋅ (:11 − 711) >384!

lK lh

R B3 = n ⋅ r B3 = 21 ⋅ 384 >3841!lK b) dq (UB − U2 ) dq (U3 − UB )   U q − ∆Ttjtufn>!g!)!UB!*> n ⋅ dqmo 3 − Shmo 3 −  U q UB U3 2 2   U ∂)∆T tjtufn * 2  = −n ⋅ d q  23 −   ∂)UB *  UB U3  U2 ∂)∆Ttjtufn * 2 − =1 ⇔ ⇒ =1 3 ∂)UB * UB U3 UB = U2⋅U3 > :11 ⋅ 411 >62:/72!L

Pri temperaturi toplotnog izvora UB>!62:/72!L!promena entropije sistema ima minimalnu vrednost i ona iznosi: d q (62:/72 − U2 ) d q (U3 − 62:/72)   U q ∆Tnjo> n ⋅ d qmo 3 − S hmo 3 + +  U2 q2 62:/72 U3   :11 1/:2 ⋅ (62:/72 − 411) 1/:2 ⋅ (:11 − 62:/72)  lK  ∆Tnjo> 21 ⋅ 1/:2 ⋅ mo + +  >28/7: L 411 62:/72 :11  


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2/5/!Tokom kvazistati~ke (ravnote`ne) politropske ekspanzije n>3!lh idealnog gasa, do tri puta ve}e zapremine od po~etne, temperatura gasa opadne sa U2>!711!L!na!U3>444!L i izvr{i se zapreminski rad 211!lK. Da bi se proces obavio na opisani na~in, radnom telu se dovodi 31!lK toplote. Skicirati promene stawa idealnog gasa na qw!i!Ut dijagramu i odredite specifi~ne toplotne kapacitete pri stalnom pritisku (dq*! i pri stalnoj zapremini!)dw*!datog gasa. prvi zakon termodinamike za proces od 1 do 2 ⇒

R23!>!∆V23!,!X23 dw =

R 23 − X23 31 − 211 lK !> >!1/261! lhL n ⋅ (U3 − U2 ) 3 ⋅ (444 − 711)

U2  w 3   = U3  w2 

U2 711 mo U3 444 o= +2= + 2 >2/646 w3 mo 4 mo w2 mo

o −2

⇒

X23>!n!/!x23!>! n ⋅ S h ⋅

Sh!>!

!q

R23!>!n!/!dw!/!)!U3!−!U2!*!,!X23

2 ⋅ (U3 − U2 ) ⇒ o −2

− 211 ⋅ (2/646 − 2) lK = 1/211! lhL 3 ⋅ (444 − 711)

⇒

!U

!2

o>2/646

Sh =

X23 ⋅ (o − 2) = n ⋅ (U3 − U2 )

dq!>!dw!,!Sh!>!1/361!

lK lhL

!2

o>2/646

!3

!3 !w


!t

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2/6/!Dvoatomni idealan gas )o>3!lnpm* ekspandira kvazistati~ki adijabatski od U2>711!L!do!U3>411!L a zatim se od wega izobarski odvodi toplota dok mu temperatura ne dostigne U4>361!L. Odrediti koliko se zapreminskog rada dobije za vreme ekspazije )lK* i kolika se toplota odvede od gasa za vreme izobarskog hla|ewa )lK*/ X23 = n ⋅ x 23 = n ⋅ d w (U2 − U3 ) = o ⋅ )Nd w * ⋅ (U3 − U2 ) X23 = 3 ⋅ 31/9 ⋅ (411 − 711) >−23591!lK R 34 = n ⋅ r34 = n ⋅ dq (U4 − U3 ) = o ⋅ )Ndq * ⋅ (U4 − U3 ) R 34 = 3 ⋅ 3:/2 ⋅ (361 − 411) >!−3:21!lK 2/7/!Termodinami~ki sistem ~ine 21!lh kiseonika (idealan gas) kao radna materija i okolina stalne temperature Up>1pD kao toplotni ponor. Kiseonik mewa svoje stawe od 2)q>2!NQb-!U>561pD* do 3)q>2 NQb-!U>38pD* na povratan na~in (povratnim promenama stawa).Skicirati promene stawa idealnog gasa u Ut koordinatnom sistemu i odrediti razmewenu toplotu izvr{eni zapreminski rad.

!2

U !3

Up !C

!B

!t drugi zakon termodinamike za proces 1−2: ∆Ttjtufn!>!∆Tsbeop!ufmp!,!∆Tplpmjob

⇒

∆Tsbeop!ufmp!>!−∆Tplpmjob!

 U q n ⋅  d qmo 3 − S hmo 3 U2 q2 

⇒

R23!>Up/! n ⋅  dqmo

 R  = ! 23 UP 

 

U3 q  − Shmo 3  U2 q2 

411 R23!>!384! ⋅ 21 ⋅ 1/:2 ⋅ mo >!−!3296/37!lK 834


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prvi zakon termodinamike za proces 1−2 ⇒

R23!>!∆V23!,!X23!

X23!>!R23!−!n!/!dw!/)U3!.U2*

X23!>! −3296/37 − 21 ⋅ 1/76 ⋅ (411 − 834 ) >676/35!lK 2/8/ Sedam kilograma azota (idealan gas) mewa svoje stawe, na povratan na~in, od stawa 2)q>6!cbs-!u>2pD* do stawa 3, pri ~emu se dobija zapreminski rad X>2257!lK. Od okoline (toplotnog izvora) stalne temperature Up>32pD, azotu se dovodi R>2511!lK toplote. Odrediti temperaturu i pritisak radne materije (azot) na kraju procesa i skicirati promene stawa radnog tela na U−t dijagramu prvi zakon termodinamike za proces od 1 do 2

U3 = 385 +

U3 = U2 +

⇒

R23!>!∆V23!,!X23

R 23 − X23 n ⋅ dw

2511 − 2257 >!434/14!L 8 ⋅ 1/85

drugi zakon termodinamike za proces od 1 do 2 ∆Ttj!>!∆Tsu!,!∆Tp  2 q 3 = q2 ⋅ fyq  S h

⇒!

 R 23 U3 −  n ⋅ U + d q mo U P 2 

 U q  R 1!>! n ⋅  dqmo 3 − Shmo 3  − 23 U2 q2  UP    =  

 2  2511 434/19  q 3 = 6 ⋅ 21 6 ⋅ fyq + 2/15 ⋅ mo −  = !1/:!cbs 385   1/3:8  8 ⋅ 3:5

U !3

!B !C

!Up

!2 t


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zbirka zadataka iz termodinamike zadaci za ve`bawe:

strana 8

)2/9/!−!2/21/*

2/9/ 3 mola troatomnog idealnog gasa stawa )q>:!cbs-!U>484!L*!kvazistati~ki (ravnote`no) politropski ekspandira do stawa )w3>5/!w2-!q>2/:!cbs*/!Skicirati proces na qw!j!Ut dijagramu i odrediti: a) eksponent politrope, o b) promene unutra{nje energije )lK*- entalipje )lK* i entropije radnog tela )lK0L* c) koli~inu toplote koja se preda radnom telu )lK*-!u ovom procesu a) o>2/23 b) ∆V23>.6!lK-!∆I23>.7/5!lK-!∆T23>31/23!K0L c) R23!>!7/75!LK 2/:/ Idealan gas (helijum) mase n>3/6!lh izobarski (ravnote`no) mewa svoje toplotno stawe pri ~emu mu se entropija smawi za 7/6!lK0L. Po~etna temperatura gasa iznosi 311pD. Temperatura toplotnog rezervoara koji u~estvuje u ovom procesu je konstantna i jednaka je ili po~etnoj ili krajwoj temperaturi radnog tela. Odrediti promenu entropije toplotnog rezervoara. ∆TUS!>9/54!lK0L 2/21/ Termodinami~ki sistem ~ine 4!lh vazduha (idealan gas) kao radna materija i okolina stalne temperature Up>36pD kao toplotni ponor. Radna materija mewa svoje toplotno stawe od stawa 2)q>1/2 NQb-!u>61pD* do stawa 3)u>6pD* na povratan na~in (povratnim promenama stawa). Pri tome se okolini predaje 661!lK toplote. Odrediti: a) pritisak radne materije na kraju procesa b) utro{eni zapreminski rad )lK* u procesu 1−2 c) skicirati promene stawa radnog tela na Ut dijagramu a) q3>6/16!cbs b) X23>−!563/9!LK


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2/22/ Idealan gas )n>2!lh* mewa svoje toplotno stawe od 2)q>:!cbs-!w>1/2!n40lh*!do 3)q>2!cbs*/ Prvi put promena se obavwa kvazistati~ki po liniji 2B3 (vidi sliku) pri ~emu je zavisnost pritiska od zapremine linearna. Drugi put promena se obavqa kvazistati~ki linijom 2C3 po zakonu qw3>dpotu, pri ~emu se radnom telu dovodi 31!lK!toplote. Odrediti: a) dobijeni zapreminski rad )X23* du` promena 2B3!i!2C3 b) koli~inu toplote )R23* dovedenu gasu du` promena 2B3 q 2

B C 3 w a) q w 3 = w 2 ⋅  2  q3

3

3

 n4 :  = 1/2 ⋅   >!1/4! lh  2  w3

)X23 * B = n ⋅

∫

q)w*ew = n ⋅

q2 + q 3 ⋅ (w 3 − w 2 ) 3

w2

)X23 * B = 2 ⋅

: ⋅ 21 6 + 2 ⋅ 21 6 ⋅ )1/4 − 1/2* >211!/214!lK 3 w3

)X23 *C = n ⋅

∫

q)w*ew =n ⋅

w2

∫

L ⋅ w −3 ew = −n ⋅ L ⋅ w −2

w3 w2

 2 2   = −n ⋅ L ⋅  − w w 2  3

2   2 )X23 *C = −2 ⋅ : ⋅ 21 4 ⋅  −  >71!/214!lK 1/4 1/2   napomena:!

L = q2 ⋅ w 23 = : ⋅ 21 6 ⋅ 1/23 >:!/214!!


K ⋅ n4 lh3

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strana 10

b) prvi zakon termodinamike za proces 2C3:

)R 23 *C = ∆V23 + )X23 *C !!!!!)2*

prvi zakon termodinamike za proces 2B3:!

)R 23 * B = ∆V23 + )X23 * B !!!!!)3*

oduzimawem prethodne dve jedna~ine )2* i )3*!!dobija se: )R23 * B = )X23 * B − )X23 *C , )R 23 *C >211!−!71!,!31!>!71!lK 2/23/ Jedan kilogram vazduha (idealan gas) stawa 2)q>25!cbs-!U>434!L* kvazistati~ki ekspandira do stawa 2. Tokom ekspanzije zavisnost pritiska od zapremine je linearna. U toku procesa vazduhu se dovede R23>216!lK toplote i pri tom se dobije X23>211!lK zapreminskog rada. Odrediti temperaturu i pritisak vazduha stawa 2. prvi zakon termodinamike za proces 2−3:! R 23 = n ⋅ d w )U3 − U2 * + X23

w2

X23 = n ⋅

∫

q)w*ew = n ⋅

U3 = U2 +

R 23 = ∆V23 + X23

R 23 − X23 216 − 211 >441!L = 434 + n ⋅ dw 2 ⋅ 1/83

q + q3 q2 + q 3 ⋅ (w 3 − w 2 ) = n ⋅ 2 ⋅ Sh 3 3

w3

U U  ⋅  3 − 2   q 3 q2 

X   n ⋅ S h ⋅ U2 ⋅ q 33 + )3 ⋅ 23 ⋅ q2 + S h ⋅ U2 ⋅ q2 − S h ⋅ U3 ⋅ q2 * ⋅ q 3 − S h ⋅ U3 ⋅ q2 ⋅ q2  = 1 n   b ⋅ q 33 + c ⋅ q 3 + d >1

b> 2 ⋅ 398 ⋅ 434 >:3812  211 ⋅ 21 4  c> 2 ⋅  3 ⋅ ⋅ 25 ⋅ 21 6 + 398 ⋅ 434 ⋅ 25 ⋅ 21 6 − 398 ⋅ 441 ⋅ 25 ⋅ 21 6  > 3/88 ⋅ 2122   2   d>− 2 ⋅ 398 ⋅ 441 ⋅ 25 ⋅ 21 6 ⋅ 25 ⋅ 21 6 >− 2/97 ⋅ 2128

:3812⋅ q33 + 3/88 ⋅ 2122 ⋅ q 3 − 2/97 ⋅ 2128 = 1

q3 =

− 3/88 ⋅ 2122 ±

(3/88 ⋅ 21 )


22 3

⇒

+ 5 ⋅ :3812⋅ 2/97 ⋅ 2128

3 ⋅ :3812

>!6/76!cbs

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2/24/ Dvoatomni idealan gas!u koli~ini o>31!npm!sabija se ravnote`no od po~etne zapremine W2>1/: 51821 4968 + , n4 do krajwe zapremine W3>1/3!n4. Promena stawa gasa odvija se po jedna~ini: q(W ) = W W3 pri ~emu je pritisak izraèn u Qb a zapremina u n4. Odrediti: a) pritisak i temperaturu gasa na po~etku i kraju procesa b) izvr{eni nad radnim telom kao i razmewenu toplotu tokom ovog procesa a) q2 =

51821 4968 51821 4968 + >!5:46:!Qb + > 1/: W2 1/: 3 (W2 )3

q3 =

51821 4968 51821 4968 + >!3:8386!Qb + > 1/3 W3 1/3 3 (W3 )3

(

)

⇒

U2 =

q2 ⋅ W2 5:46: ⋅ 1/: >378/24!L = o ⋅ NS h 1/13 ⋅ 9426

(

)

⇒

U3 =

q 3 ⋅ W3 3:8386 ⋅ 1/3 >468/63!L = o ⋅ NS h 1/13 ⋅ 9426

q2 ⋅ W2 = o ⋅ NS h ⋅ U2 q 3 ⋅ W3 = o ⋅ NS h ⋅ U3

(

(

)

)

b) W3

X23 =

∫

W2

W3

q)W*eW >

∫

W3

4968   51821 4968   +  eW >  51281 ⋅ mo W −  3 W  W W   

W2

> W2

 W 4968 4968   1/3 4968 4968   51281 ⋅ mo 3 −  >  51281 ⋅ mo + − +  >!−86529/3!K W W W 1 /: 1/3 1/:   2 3 2   prvi zakon termodinamike za proces 2−3:

R 23 = ∆V23 + X23

R 23 = o ⋅ (N ⋅ d w ) ⋅ (U3 − U2 ) + X23 R23>! 1/13 ⋅ 31/9 ⋅ (468/63 − 378/24 ) − 86/53 >−49/36!lK


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strana 12

2/25/ Dvoatomnan idealna gas )o>3!lnpm* kvazistati~ki mewa stawe od 2)U>411!L-!q>2!cbs*!do 3)U>:11!L* po zakonu prave linije u Ut koordinatnom sistemu. Pri tome se radnom telu saop{tava 711 lK!rada. Odrediti pritisak radne materije stawa 2 i skicirati proces na Ut djagramu. R 23 = ∆V23 + X23

prvi zakon termodinamike za proces 2−3: R 23 = o ⋅ (N ⋅ d w ) ⋅ (U3 − U2 ) + X23 R23>! 3 ⋅ 31/9 ⋅ (:11 − 411) − 711 >!35471!lK R 23 =

U2 + U3 ⋅ ∆T23 3

⇒

∆T23 =

3 ⋅ R 23 3 ⋅ 35471 lK >!51/7! = U2 + U3 411 + :11 L

 U q  ⇒ ∆T23 = o ⋅  N ⋅ d q ⋅ mo 3 − NS h ⋅ mo 3  U2 q2   U3 ∆T23   :11 51/7   3:/2 ⋅ mo −  N ⋅ d q ⋅ mo U − o   411 3  2  > 2 ⋅ 21 6 ⋅ fyq q 3 = q2 ⋅ fyq    NS h 9/426         q3!>5/18!/216!Qb!>!5/18!!cbs

(

(

)

(

)

)

(

)

U 3

2

t


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strana 13

2/26/ Idealan gas sabija se kvazistati~ki od temperature U2>384!L do temperature U3>984!L po zakonu: U + D (!L>−215!lK0L>dpotu!!i!D>dpotu). Odrediti nepovratnost ove promene stawa ,!)∆TtjT = L ⋅ mo U2 lK0L*-!ako se toplota predaje izotermnom toplotnom ponoru temperature UUQ>U2 i grafi~ki je predstaviti na Ut dijagramu ∆Ttjtufn!>!∆TSU!,!∆TUQ!>!///!>!−231/:!,!339/7!>218/8! ∆TSU!>T3!−!T2>///>!−231/:! T3 = L ⋅ mo

U3 +D U2

lK L

kJ K T2 = L ⋅ mo

)3*

U2 +D U2

)2*

Oduzimawem pretnodne dve jedna~ine dobija se: U 984 lK T 3 − T2 = L ⋅ mo 3 = −215 ⋅ mo >−231/:! L U2 384 T3

∫

∆TUQ> −

U)T*eT

T R 23 =− Uuq Uuq

= /// = L ⋅

T3

napomena:

∫

U3 − U2 984 − 384 lK = −215 ⋅ >339/7 Uuq 384 L

T3

U)T*eT = /// =

T2

U2 ⋅ L

∫

U2

T −D ⋅ f L eT

= U2 ⋅ L

T −D ⋅f L

T2 T3 − D ⋅ )f L

−

T2 − D f L *

T3

= T2

= U2 ⋅ L ⋅ )f

U mo 3 U2

− 2* = L ⋅ )U3 − U2*

Postupak grafi~kiog predstavqawa promene entropije sistema zasnovan je na jednakosti povr{ina ispod:

1.

linije kojom predstavqamo promenu stawa radnog tela

2.

linije kojom predstavqamo promene stawa toplotnog ponora Obe ove povr{ine predstavqaju razmewenu toplotu izme|u radno tela i toplotnog ponora.


{fmlp@fvofu/zv


strana 14

U 3

UUQ

2

∆TSU

t

∆TUQ ∆TTJ

zadaci za ve`bawe:

)2/27/!−!2/28/*

2/27/ Vazduh (idealan gas) kvazistati~ki mewa toplotno stawe od stawa 2)U>411!L*!do stawa!3)U>711!L*!i pri tome je w2>3/!w3. Prvi put se promena vr{i po zakonu prave linije u Ut!kordinatnom sistemu, a drugi put se od stawa 1 do stawa 2 dolazi kvazisati~kom politropskom promenom stawa. Odrediti koliko se lK lK toplote )lK0lh*!dovede vazduhu u oba slu~aja . )r23 *qsbwb = 246/2 - )r23 *qpmjuspqb = 23:/7 lh lh Sh ⋅ U

b , w3 )b>82/87!On50lh3-!c> 9/138 ⋅ 21 −5 n40lh!j!Sh>79/9!K0)lhL**-!lwb{jtubuj•lj!izotermski ekspandira pri temperaturi od 1pD od w2>1/16!n40lh do w3>1/3!n40lh. Odrediti: a) po~etni i krajwi pritisak gasa kao i dobijeni zapreminski rad tokom ekspanzije b) po~etni i krajwi pritisak gasa kao i dobijeni zapreminski rad tokom ekspanzije kada bi navedeni gas posmatrali kao idealan gas iste gasne konstante )Sh* 2/28/ Neki gas koji se pona{a saglasno jedna~ini stawa:

q)w* =

w −c

−

a) q2>4/64!cbs-!q3>1/:3!cbs-!x23>36/2:!lK0lh b) q2>4/87!cbs-!q3>1/:5!cbs-!x23>37/15!lK0lh


{fmlp@fvofu/zv


strana 15

NEKVAZISTATI^KE (NERAVNOTE@NE) PROMENE STAWA IDEALNIH GASOVA 2/29/ Vazduh (idealan gas) stawa 2)q2>23!cbs-!U2>366pD* ekspandira nekvazistati~ki adijabatski sa stepenom dobrote η fy e >1/9 do stawa 3)q3>2!cbs*/!Odrediti: a) temperaturu vazduha nakon ekspanzije b) prira{taj entropije radnog tela usled mehani~ke neravnoteè c) zakon nekvazistati~ke promene stawa u obliku qwn>jefn U

q2 2

q3 B

3

3l t a) U3L

q = U2 ⋅  3L  q2

η fy e =

  

κ .2 κ

 2 = 639 ⋅    23 

1.4.2 1.4

>36:/7!L

U2 − U3 -!!!!! U3 = U2 + ηEfy ⋅ (U3L − U2 ) = 639 + 1/9 ⋅ (36:/7 − 639) >424/4!L U2 − U3l

b) qB

U = q2 ⋅  B  U2

κ

 κ .2  424/4   = 23 ⋅ 21 6 ⋅    639  

2/5 .2 2/5

= 2/:4 ⋅ 21 6 Qb

U q 2 K ∆t nfi = ∆t B3 = dqmo 3 − S hmo 3 = −398 ⋅ mo >299/82! qB 2/:4 lhL UB


{fmlp@fvofu/zv


strana 16

c) !

S h ⋅ U2

w2 =

q2

w3 =

S h ⋅ U3 q3

=

398 ⋅ 639

=

23 ⋅ 21

6

>!1/2374!

398 ⋅ 424/4 2 ⋅ 21

6

n4 lh

>!1/9::3!

n4 lh

qw n = jefn

⇒

q2 ⋅ w 2n = q 3 ⋅ w n 3

q2 23 mo q3 2 n= >2/37 = w3 1/9::3 mo mo 1/2374 w2

⇒

qw 2/37 = jefn

mo

2/2:/ Kompresor proizvo|a~a B radi izme|u pritisaka qnjo!>2!cbs!i qnby>!:!cbs. Kompresor proizvo|a~a C radi izme|u pritisaka qnjo!>2/6!cbs i!!qnby>!21!cbs. U oba slu~aja radni fluid je vazduh (idealan gas) po~etne temperature!U2>41pD. Temperature vazduha na izlazu iz oba kompresora su jednake. Odrediti koji je kompresor kvalitetniji sa termodinami~kog aspekta, predpostavqaju}i da su kompresije adijabatske Sa termodinami~kog aspekta kvalitetniji je ona kompresija kod koje je 1. na~in: 2. na~in:

ve}i stepen dobrote adijabatske kompresije mawa promena entropije sistema

1. na~in: q  U2 − U2 ⋅  3L   q   2 B B = U2 − U3LB = ηe U2 − U3 U2 − U3

q U2 − U2 ⋅  3L  q U2 − U3LC  2 C ηe = = U2 − U3 U2 − U3


κ −2 l

κ −2 l

   C

κ −2    q  l   U2 ⋅ 2 −  3L    q2   B     = U2 − U3

   q U2 ⋅ 2 −  3L  q   2   = U2 − U3

κ −2 l

   C

      

)2*

)3*

{fmlp@fvofu/zv


strana 17

Deqewem prethodne jedna~ina (1) i (2) dobija se:

B ηe ηC e

2/5 − 2 q  κ −2 :  2/5  2 −  3L  κ 2−    q   2 B  2 = 2/32 = = 2/5 − 2 q  κ −2 3 L  2−   21  2/5  q  κ 2−    2 C  2/6 

Po{to je koli~nik stepena dobrote ve}i od 1 to zna~i da je stepen dobrote kompresora proizvo|a~a B ve}i od stepena dobrote kompresora proizvo|a~a C, pa je kompresor proizvo|a~a B kvalitetniji sa termodinami~kog aspekta. Uo~iti da je zadatak mogao biti re{en i bez zadate temperature U2. 2. na~in:

(∆t tj )B

= d q ⋅ mo

U3 B q − S h ⋅ mo 3 B − ∆t p U2B q2B

)4*

(∆t tj )C

= d q ⋅ mo

U3C q − S h ⋅ mo 3C − ∆t p U2C q2C

)5*

Oduzimawem jedna~ina!)4*!j!)5*!epcjkb!tf;

(∆t tj )B − (∆t tj )C

 q q = −S h ⋅  mo 3 B − mo 3C q2C  q2B

(∆t tj )B < (∆t tj )C

⇒

 K 21   :  > − 398 ⋅  mo − mo  >−97/24! lhL 2/6   2 

kompresor proizvo|a~a A je kvalitetniji sa termodinami~kog aspekta.

U

q3 3

B 3l

q2

2 t


{fmlp@fvofu/zv


strana 18

2/31/!Pet kilograma kiseonika (idealan gas) ekspandira nekvazistati~ki politropski po zakonu qw2/2>jefn, od stawa 2)q2>8!cbs-!w2>1/23!n40lh* do stawa 3)U3>−2pD*/ Specifi~na toplota ove promene stawa iznosi d23>−761!K0lhL. Skicirati proces na Ut dijagramu i odrediti: a) prira{taj entropije radnog tela usled mehani~ke i usled toplotne neravnoteè b) promenu entropije izolovanog termodinami~kog sistema ako je temperatura toplotnog izvora 484!L U

q2 2

o

q3

n

B

3 3l

t a) q ⋅w 8 ⋅ 216 ⋅ 1/23 U2 = 2 2 = = 434 L Sh 371 [blpo!qspnfof!qwn!>jefn-!!usbotgpsnj|fnp!v!pcmjl;! U n ⋅ q2 − n = jefn n  U 2− n U2n ⋅ q22 − n = U3n ⋅ q32 − n ⇒ q 3 = q2 ⋅  2  U   3 2/2 6  434  2 − 2/2 >2/17!cbs q 3 = 8 ⋅ 21 ⋅    383  d − dw ⋅ κ o−κ ⇒ o = 23 ⇒ dolw!>!dlw d23 = d w o −2 d23 − d w o=

−761 − 831 ⋅ 2/5 >2/32 − 761 − 831


{fmlp@fvofu/zv


strana 19

o 2 / 32 o − 2 6  383  2 / 32 − 2  > 3/7 ⋅ 21 6 !Qb = 8 ⋅ 21 ⋅    434     U q  2/17  lK  ∆Tnfi/ofs/>!∆TB3>! n ⋅  d qmo 3 − S hmo 3  > 6 ⋅  − 1/398 ⋅ mo  >2/3:! UB qB  3/7  L   U q B = q2 ⋅  B  U  2



∆Tupq/ofs/>!∆T2B>! n ⋅  dqmo 

UB q − Shmo B U2 q2

 lK 383 2/17   > 6 ⋅ 2 ⋅ mo − 1/398 ⋅ mo  >1/54!  434 3 / 7 L   

∆Tsbeop!ufmp!>!∆Tnfi/ofs/!,!∆Tupq/ofs/!>2/3:,1/54>2/83!

lK L

b) ∆Ttjtufn!>!∆Tsbeop!ufmp!,!∆Tupqmpuoj!qpops!>!///!>!2/83!−!1/55!>2/39! ∆Tupqmpuoj!j{wps!>!−

lK L

R 23 277/22 lK =− = −1/55 UUJ 484 L

R 23 = R 2B + R B3 >///> R 2B = n ⋅ d w ⋅

o−κ 2/32 − 2/5 ⋅ (UB − U2 ) > 6 ⋅ 1/83 ⋅ ⋅ (383 − 434) >277/22!lK 2/32 − 2 o −2

2/32/!Termodinami~ki sistem sa~iwava n>6!lh azota (idealan gas) i okolina temperature Up>38pD. Azot nekvazistati~ki politropski mewa toplotno stawe od stawa 2)q>21!cbs-!U>566pD* do stawa 3)U>98pD*/!Specifi~ni toplotni kapacitet promene stawa 1−2 iznosi d23>481!K0lhL a nepovratnost procesa 1−2 iznosi ∆Ttj>2/56!lK0L. Skicirati proces na Ut dijagramu i odrediti stepen dobrote ove promene stawa. !U

!q2 !2

!o

q3 n

!B !3 !3l


t

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike ⇒

dolw!>!dlw o=

strana 20

d23 = d w

o−κ o −2

⇒

o=

d23 − d w ⋅ κ d23 − d w

481 − 851 ⋅ 2/5 >2/9 481 − 851

R23 = R2B + R B3 !>///> R 2B = n ⋅ d w ⋅

2/9 − 2/5 o−κ ⋅ (471 − 839) >!−!791/9!lK ⋅ (UB − U2 ) > 6 ⋅ 1/85 ⋅ 2/9 − 2 o −2

 U q  R ∆TTJ = n ⋅  dq mo 3 − S h mo 3  − 23 U2 q2  U1   2  ∆T TJ R 23 U   q 3 = q2 ⋅ fyq− + − d q mo 3  =  n ⋅ UP U2   S h  n

∆TTJ!>!∆TSU!,!∆Tp

⇒

 2  2/56 791/9 471  q 3 = 21 ⋅ 21 6 ⋅ fyq− − − 2/15 ⋅ mo   = 2/59!!cbs 6 ⋅ 411 839   1/3:8  6 q U2 =  2 U3l  q 3l q3l!>!q3

  

o−2 o

!!!!!!!!!⇒ !!!!!!!! U3l

⇒


ηFY E =

q = U2 ⋅  3l  q2

  

o−2 o

 2/59  > 839 ⋅    21 

2/9 −2 2/9

= 422/5!!L

U2 − U3 839 − 471 >1/99 = U2 − U3l 839 − 422/5

{fmlp@fvofu/zv


strana 21

2/33/! Kiseonik (idealan gas) sabija se nekvazistati~ki politropski od stawa 2)q>2!cbs-!U>384!L* do stawa 3)q>7!cbs-!U>554!L*/ U toku procesa sabijawa od kiseonika se odvodi 431!lK0lh toplote. Skicirati proces na Ut dijagramu i odrediti stepene dobrote ove promene stawa. r23 = r2B + r B3 !>!−431!

r2B = d w ⋅

lK lh

o−κ ⋅ (UB − U2 ) o −2

r2B − dw ⋅ κ UB − U2 o= r2B − dw UB − U2

⇒

−431 − 1/76 ⋅ 2/5 554 − 384 >2/2 o= − 431 − 1/76 554 − 384 q U2 =  2 U3l  q 3l ηlq E =

  

o−2 o

⇒

q = U2 ⋅  3l  q2

U3l

  

o−2 o

7 > 384 ⋅    2

2/2−2 2/2

>432/4!L

U2 − U3l 384 − 432/4 >1/39 = U2 − U3 384 − 554

q3

!U !3 !B !3l

!o

!q2

n

!2

t


{fmlp@fvofu/zv


strana 22

2/34/ Tokom nekvazistati~kog sabijawa n>4!lh butana (idealan gas) od stawa 2)q2>2!cbs-!U2>31pD* do stawa 3)q3>41!cbs-!T3>T2*- spoqa{wa mehani~ka sila izvr{i rad od 961!lK. Tokom procesa radna materija predaje toplotu toplotnom ponoru stalne temperature Uq>1pD. Skicirati proces u na Ut dijagramu i odrediti: a) promenu entropije termodinami~kog sistema tokom posmatrane promene stawa b) stepen dobrote nekvazistati~ke kompresije q3

!U !B

!3 n>κ

!3l

!q2

!o !2

t zakon nkv. promene stawa 1−2:

U2  q2   = U3  q 3 

κ −2 κ

⇒

κ −2 q  κ U3 = U2 ⋅  3  q   2

1.28 − 2  41  2/39 U3 = 3:4 ⋅   >!727/7!L  2 

prvi zakon termodinamike za proces!2−3;

R23>∆V23,X23

R23 = n ⋅ dw ⋅ (U3 − U2) + X23 > 4 ⋅ 1/6 ⋅ (727/7 − 3:4) − 961 >−475/7!LK

drugi zakon termodinamike za proces!2−3;

∆TTJ!>!∆TSU!,!∆TUQ

lK L −475/7 R23 lK >2/45!! ∆TUQ = − =− L UUQ 384 lK ∆TTJ!>!2/45!,!1!>!2/45! L ∆TSU!>!1!


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike b)

strana 23

R 23 = R 2B + R B3 !>!−475/7!lK R 2B = n ⋅ d w ⋅

o−κ ⋅ (UB − U2 ) o −2

R 2B − dw ⋅ κ n ⋅ (UB − U2 ) o= R 2B − dw n ⋅ (UB − U2 )

⇒

−475/7 − 1/6 ⋅ 2/39 4 ⋅ (727/7 − 3:4 ) >2/27 o= − 475/7 − 1/6 4 ⋅ (727/7 − 3:4) q U2 =  2 U3l  q 3l ηlq E =

  

o−2 o

⇒

U3l

q = U2 ⋅  3l  q2

  

o−2 o

 41  > 3:4 ⋅    2 

2/27 −2 2/27

>579/5!L

U2 − U3l 3:4 − 579/5 >1/65 = U2 − U3 3:4 − 727/7

2/35/ [est kilograma troatomnog idealnog gasa mewa toplotno stawe nekvazistati~ki po zakonu qwn>jefn!)n>κ* (tj. nekvazistati~ki izentropski) od stawa!2)q2>41!cbs-!U2>727/7!L* do stawa 3)q3>2 cbs*/ Tokom ove promene stawa specifi~na zapremina gasa se pove}a za 1/49!n40lh i pri tome se dobije 811!lK mehani~kog rada. Skicirati promenu stawa idealnog gasa na Ut dijagramu i odrediti: a) koli~inu razmewene toplote tokom ove promene stawa b) stepen dobrote ove nekvazistati~ke promene c) porast entropije radnog tela usled mehani~ke neravnoteè )lK0L* !q2

!U !2 n>κ !o !B

q3 !3

!3l t


{fmlp@fvofu/zv


strana 24

b* zakon nkv. promene stawa 1−2:

U2  q2   = U3  q 3 

κ −2 κ

⇒

κ −2 q  κ U3 = U2 ⋅  3  q   2

1.28 − 2  2  2/39 >!3:4!L U3 = 727/7 ⋅    41 

jedna~ina stawa idealnog gasa za kraj procesa:

q2 ⋅ w2 = Sh ⋅ U2 !!!!!!)2* q3 ⋅ w 3 = S h ⋅ U3 !!!!)3*

uslov zadatka:

w 3 − w 2 = 1/49

jedna~ina stawa idealnog gasa za po~etak procesa:!

n4 !)4* lh

Re{avawem prethodnog sistema tri jedna~ine sa 3 nepoznate dobija se: w 2 = 1/1396

N= o=

n4 n4 K - w 3 = 1/517 - S h = 249/69 lh lh lhL

SV 9426 lh = = 71 S h 249/69 lnpm n 7 = >1/2!lnpm N 71

prvi zakon termodinamike za proces 1−2:

R 23 = ∆V23 + X23

R 23 = o ⋅ )Nd w * ⋅ (U3 − U2 ) + X23 = 1/2 ⋅ 3:/2 ⋅ (3:4 − 727/7 ) + 811 b)

R 23 = R 2B + R B3 !>!−352/79!lK

R 2B

o−κ = n ⋅ dw ⋅ ⋅ (UB − U2 ) o −2

⇒

>−352/79!lK

(Nd w ) R2B − ⋅κ n ⋅ (UB − U2 ) N o= R2B (Nd w ) − n ⋅ (UB − U2 ) N

−352/79 3:/2 − ⋅ 2/39 7 ⋅ (3:4 − 727/7 ) 71 o= >2/49 3:/2 − 352/79 − 7 ⋅ (3:4 − 727/7 ) 71


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike q U2 = 2 U3l  q 3l ηEfy =

  

o−2 o

⇒

strana 25

U3l

q = U2 ⋅  3l  q2

  

o−2 o

 2  > 727/7 ⋅    41 

2/49 −2 2/49

>352/8!L

U2 − U3 727/7 − 3:4 >1/97 = U2 − U3l 727/7 − 352/8

c)  U q ∆T nfi/ofs/ = n ⋅ ∆t B3 = n ⋅  d q mo 3 − S h mo 3 UB qB 

Zakon kvazistati~ke promene stawa!2−B;

qB

U = q2 ⋅  B  U2

o

 2 lK  > 7 ⋅ (− 249/69 ) ⋅ mo >1/69! 3 L  U2  q2   = UB  q B 

o−2 o

2/49

 o−2  3:4  2/49−2  = 41 ⋅ 21 6 ⋅  >!3!cbs   727/7  

2/36/ Tri kilograma vazduha (idealan gas) stawa!2)q2>3!cbs-!U2>261pD*!mewa svoje toplotno stawe nekvazistati~ki (neravnote`no) izotermski do stawa 3)q3>9!cbs*. Promena entropije radne materije usled mehani~ke neravnoteè iznosi ∆tnfi>349!K0lhL. Odrediti: a) dovedeni rad i odvedenu toplotu tokom ove promene stawa )X23-!R23* b) stepen dobrote izotermske kompresije )ηelq* qB U

q3 q2

B 3>3l

2

t


{fmlp@fvofu/zv


strana 26

a) ∆t nfi = ∆t B3 = d qmo

U3 q − S hmo 3 UB qB

⇒

 ∆t  q B = q 3 ⋅ fyq nfi   Sh   

 349  6 q B = 9 ⋅ 21 6 ⋅ fyq  > 29/4 ⋅ 21 !Qb  398  R 23 = R 2B + R B3 = /// = −917/4/4 lK R 2B > n ⋅ U2 ⋅ S h ⋅ mo

q2 3 > 4 ⋅ 534 ⋅ 398 ⋅ mo >−917/4!lK qB 29/4 R 23 = ∆V23 + X23

prvi zakon termodinamike za proces 1−2: X23>!R23!>−917/4!lK

b) ηlq E =

X23L −615/: = /// = = 1/74 X23 − 917/4

X23L = n ⋅ S h ⋅ U ⋅ mo

q2 3 = 4 ⋅ 398 ⋅ 534 ⋅ mo >−615/:!lK q 3L 9

2/37/ Vazduh (idealan gas) stawa!2)q2>:!cbs-!U2>261pD*!mewa svoje toplotno stawe nekvazistati~ki (neravnote`no) izotermski do stawa 3)q3>2!cbs*. Promena entropije radne materije usled mehani~ke neravnoteè )∆tnfi* i promena entropije radne materije usled toplotne neravnoteè )∆tupq* su jednake. Odrediti stepen dobrote ove nekvazistati~ke izotermske ekspanzije )ηefy*/ q2

qB

U

q3

2

B

3>3l

t


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike ∆tupq!>!∆tnfi d q mo

⇒

strana 27 ∆t2B!>!∆tB3

UB q U q − S h mo B > d q mo 3 − S h mo 3 U2 UB q2 qB

⇒

q B = q2 ⋅ q 3

q B = : ⋅ 21 6 ⋅ 2 ⋅ 21 6 > 4 ⋅ 21 6 !Qb X23L = n ⋅ S h ⋅ U ⋅ mo

q2 : = 2 ⋅ 398 ⋅ 534 ⋅ mo >377/86!lK q 3L 2

R 23 = R 2B + R B3 = /// = −917/4/4 lK R 2B > n ⋅ U2 ⋅ S h ⋅ mo

q2 : > 2 ⋅ 534 ⋅ 398 ⋅ mo >244/48!lK qB 4


R 23 = ∆V23 + X23

X23>!R23!>244/48!lK η fy e =

X23 244/48 > >1/6 377/86 X23l

zadaci za ve`bawe:

)2/38/−2/39/*

2/38/!Vazduh (idealan gas) po~etnog stawa 2)q>6!cbs-!w>1/337!n40lh* ekspandira nekvazistati~ki politropski do stawa 3)q>2!cbs-!U>3:4!L*/!Tokom ove promene stawa od radne materije ka okolini se odvede 27!lK0lh toplote. Odrediti stepen dobrote ove promene stawa kao i promenu entropije vazduha lK samo usled mehani~ke neravnoteè. ! η fy e >1/73-!!∆tnfi>1/37! lhL 2/39/!Vazduh (idealan gas) stawa 2)q2>1/3!NQb-!u2* mewa svoje toplotno stawe nekvazistati~ki (neravnote`no) izotermski do stawa 3)q3?q2*/ Promena entropije radne materije usled mehani~ke neravnoteè iznosi 349!K0lhL/ Stepen dobrote ove nekvazistati~ke promene stawa iznosi ηe>1/95. Odrediti pritisak vazduha na kraju procesa )q3*/ q3 = 266/6 cbs


{fmlp@fvofu/zv


strana 28

PRVI I DRUGI ZAKON TERMODINAMIKE (ZATVOREN TERMODINAMI^KI SISTEM) 2/3:/ U vertikalno postavqenom cilindru, od okoline adijabatski izolovanom, (slika), unutra{weg pre~nika e>711 nn, nalazi se vazduh (idealan gas) temperature 31pD. Sud je zatvoren klipom zanemarqive mase, koji se moè kretati bez trewa. Na klipu se nalazi teg mase nu>3111!lh. U polaznom !∆{>411!nn poloàju ~elo klipa se nalazi na visini {>611!nn u odnosu na dowu bazu cilindra. U cilindru se nalazi elektri~ni greja~ pomo}u kojeg se vazduhu dovodi toplota. Pritisak okoline !{>611 iznosi qp>2!cbs. Odrediti: a) koli~inu toplote koju greja~ treba da preda gasu tako da se klip u procesu pomeri za ∆{>411!nn b) vreme trajawa procesa ako snaga elektri~nog greja~a ,R23 iznosi 2/77!lX c) rad koji bi izvr{io gas u cilndru ako bi se u trenutku dostizawa stawa 2 istovremeno iskqu~io greja~ i skino teg sa klipa d) skicirati sve procese sa radnim telom na qw i Ut dijagramu a) e3 ⋅ π 1/7 3 ⋅ π ⋅{ = ⋅ 1/6 = 1/2525 n 4 5 5 e3 ⋅ π 1/7 3 ⋅ π ⋅ ({ + ∆{ ) = ⋅ (1/6 + 1/4 ) = 1/3373 n 4 W3 = 5 5 W2 =

jedna~ina stati~ke ravnoteè za proizvoqan poloàj klipa: n ⋅h 3111 ⋅ :/92 q = q p + 3u = 2 ⋅ 21 6 + = 2/8 ⋅ 21 6 Qb e ⋅π 1/7 3 ⋅ π 5 5 jedna~ina stawa idealnog gasa na po~etku procesa:! n=

q ⋅ W2 2/8 ⋅ 21 ⋅ 1/2525 = = 1/3: lh S h ⋅ U2 398 ⋅ 3:4

jedna~ina stawa idealnog gasa na kraju procesa:! U3 =

q ⋅ W2 = n ⋅ Sh ⋅ U2

6

q ⋅ W3 = n ⋅ Sh ⋅ U3

q ⋅ W3 2/8 ⋅ 21 6 ⋅ 1/3373 = = 573 L n ⋅ Sh 1/3: ⋅ 398


{fmlp@fvofu/zv


strana 29 R 23 = ∆V23 + X23


R23> n ⋅ d w ⋅ (U3 − U2 ) + q ⋅ (W3 − W2 ) >

R23>! 1/3: ⋅ 1/83 ⋅ (573 − 3:4 ) + 2/8 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/3373 − 1/2525 ) >5:/9!lK b) τ=

R 23 ⋅

=

R 23

5:/9 = 41 t 2/77

c) napomena: U3  q3  =  U4  q4 

κ −2 κ

proces 2−3 je kvazistati~ki adijabatski, q4>qp>!2!cbs q  U4 = U3 ⋅  4   q3 

⇒

κ −2 κ

 2⋅ 216   = 573 ⋅   2/8 ⋅ 216   

2/5 −2 2/5

>4:8!L

R 34 = ∆V 34 + X34


X34> −n ⋅ d w ⋅ (U4 − U3 ) >− 1/3: ⋅ 1/83 ⋅ (4:8 − 573) >24/68!lK

q

U 2

3

3

4

2

w


4

t

{fmlp@fvofu/zv


strana 30

2/41/!Cilindar je napravqen prema navedenoj skici. Klip je optere}en tegom nepoznate mase i leì na osloncu A. U cilindru se nalazi azot stawa!2)q>3/6!cbs-!U>3:4!L*/!Dovo|ewem!23/6!lK!toplote zapremina azota se udvostru~i. Pritisak okoline iznosi!!qp>!2!cbs-!masa klipa je zanemarqiva a klip se kre}e bez trewa.!Odrediti: a) masu tega b) pri kojoj temperaturi azota u cilindru |e se pokrenuti klip c) promenu potencijalne energije tega E

e>291!nn E>311!nn

n

{>461!nn

{ e

a) 2−3!proces u cilindru do pokretawa klipa! 3−4!proces u cilndru nakon pokretawa klipa! W2 =

e3 ⋅ π 1/29 3 ⋅ π ⋅{ = ⋅ 1/46 = 1/119: n 4 5 5

jedna~ina stawa idealnog gasa na po~etku procesa:! n=

)w>dpotu* )q>dpotu*

q2 ⋅ W2 = n ⋅ S h ⋅ U2

q2 ⋅ W2 3/6 ⋅ 21 ⋅ 1/119: = = 1/1367 lh S h ⋅ U2 3:8 ⋅ 3:4 6

W3!>!W2>1/119:!n4 W4 = 3 ⋅ W2 > 3 ⋅ 1/119: >1/1289!n4 jedna~ina stawa idealnog gasa na kraju procesa: n ⋅ S h ⋅ U4 q 4 ⋅ W4 = n ⋅ S h ⋅ U4 ⇒ ! q4 = W4


)2*

{fmlp@fvofu/zv


strana 31

prvi zakon termodinamike za proces 1−3: R 23 + R 34 = ∆V23 + ∆V 34 + X23 + X34 > n ⋅ d w ⋅ (U4 − U2 ) + q 4 ⋅ (W4 − W3 ) !!!!!!!)3*

kada jedna~inu (1) uvrstimo u jedna~inu (2) dobija se: n ⋅ S h ⋅ U4 ⋅ (W4 − W3 ) ⇒ R 23 + R 34 > n ⋅ d w ⋅ (U4 − U2 ) + W4 R 23 + R 34 + n ⋅ d w ⋅ U2 23/6 + 1/1367 ⋅ 1/85 ⋅ 3:4 = 1/1289 − 1/119: W4 − W3 1/1367 ⋅ 1/85 + 1/1367 ⋅ 1/3:8 ⋅ n ⋅ d w + n ⋅ Sh ⋅ 1/1289 W4 U4!>8:4/7!L U4 =

q4 =

1/1367 ⋅ 3:8 ⋅ 8:4/7 > 4/5 ⋅ 21 6 Qb 1/1289

jedna~ina stati~ke ravnoteè za proizvoqan poloàj za proces 2−3 n ⋅h q − qp E3 ⋅ π ⇒ ⋅ q 4 = q p + 3u nu = 4 5 h E ⋅π 5 nu =

6

4/5 ⋅ 21 − 2 ⋅ 21 6 1/3 3 ⋅ π ⋅ >879/7!lh 5 :/92

b) jedna~ina stawa idealnog gasa za stawe 2:! U3 =

q 3 ⋅ W3 = n ⋅ S h ⋅ U3

q 3 ⋅ W3 4/5 ⋅ 21 6 ⋅ 1/119: = = 4:9 L n ⋅ Sh 1/1367 ⋅ 3:8

c) ∆Fq> n u ⋅ h ⋅ ∆{ > n u ⋅ h ⋅


W4 − W3 3

E ⋅π 5

> 879/7 ⋅ :/92 ⋅

1/1289 − 1/119: 1/3 3 ⋅ π 5

>3247!K

{fmlp@fvofu/zv


strana 32

2/42. Vertikalni cilindar zatvoren je klipom mase nl>:!lh, ~iji je hod ograni~en na kraju cilindra (slika). U cilindru se nalazi dvoatoman idealan gas stawa!2)q>2/6!cbs-!U>561pD*/ Odrediti: a) za koliko }e se spustiti klip (zanemariti trewe) dovo|ewem vazduha u mehani~ku i toplotnu ravnoteù sa okolinom stawa P)q>2!cbs-!U>31pD* b) koliko se toplote pri tome preda okolini do trenutka pokretawa klipa a koliko nakon pokrtetawa klipa do trenutka dostizawa ravnoteè sa okolinom Skicirati procese na qw i Ut dijagramu

nl e>211!nn {

{>911!nn

e

2−3!proces u cilindru do pokretawa klipa! 3−4!proces u cilndru nakon pokretawa klipa!

q

U

2

4

)w>dpotu* )q>dpotu*

2

3

3 4

w


t

{fmlp@fvofu/zv


strana 33

a) W2 =

e3 ⋅ π 1/23 ⋅ π ⋅{ = ⋅ 1/9 = 1/1174 n4 5 5

jedna~ina stawa idealnog gasa na po~etku procesa:! o=

(

)

q2 ⋅ W2 = o ⋅ NS h ⋅ U2

6

q2 ⋅ W2 2/6 ⋅ 21 ⋅ 1/1174 = = 2/68 ⋅ 21.5 lnpm NSh ⋅ U2 9426 ⋅ 834

(

)

jedna~ina stati~ke ravnoteè za poloàj klipa u stawu 2 n ⋅h : ⋅ :/92 q3 = qp + l = 2 ⋅ 21 6 + > 2/2 ⋅ 21 6 !Qb 3 3 E ⋅π 1/2 ⋅ π 5 5 W3!>!W2>1/1174!n4-

q4>q3- !

U4>Up

(

W4 =

(

)

o ⋅ NS h ⋅ U4 q4

W3 − W4 =

=

2/68 ⋅ 21 −5 ⋅ 9426 ⋅ 3:4

e3 ⋅ π ⋅ ∆{ 5

2 ⋅ 21 6 ⇒

∆{ =

>1/1149!n4

W3 − W4 3

e ⋅π 5

b) jedna~ina stawa idealnog gasa za stawe 2:! U3 =

)

q 4 ⋅ W4 = o ⋅ NS h ⋅ U4

jedna~ina stawa idealnog gasa na kraju procesa:!

=

1/1174 − 1/1149 1/23 ⋅ π 5

=0.318 m

(

)

q 3 ⋅ W3 = o ⋅ NS h ⋅ U3

q 3 ⋅ W3 2/2 ⋅ 21 6 ⋅ 1/1174 >641/96!L = o ⋅ NS h 2 ⋅ 68 ⋅ 21 .5 ⋅ 9426

(

)

prvi zakon termodinamike za proces 1−2: R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) > 2/68 ⋅ 21

−5

R 23 = ∆V23 + X23

⋅ 31/9 ⋅ (641/96 − 834) >−1/74!lK


R 34 = ∆V34 + X34

R 34 > o ⋅ (Nd w ) ⋅ (U4 − U3 ) + q 4 ⋅ (W4 − W3 ) R 34 > 2/68 ⋅ 21 −5 ⋅ 31/9 ⋅ (3:4 − 641/96) + 2/2 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/1149 − 1/1174) R 34 >!−2/16!lK


{fmlp@fvofu/zv


strana 34

2/43/!Dvoatoman idealan gas stawa 2)q>2/3!NQb-!U>411!L-!W>1/2!n4*- nalazi se u vertikalno postavqenom nepokretnom adijabatski izolovanom cilindru sa (bez trewa) pokretnim adijabatskim klipom zanemarqive mase. Preostali prostor cilindra (iznad klipa) ispuwen je nekom te~nosti (slika). Usled predaje toplote gasu (od greja~a), on se {iri do stawa 3)q>1/7!NQb-!W>1/33!n4*-!~ime izaziva prelivawe odgovaraju}e koli~ine te~nosti preko ivica cilindra. a) izvesti zakon promene stawa gasa u obliku q!>!g)W* b) prikazati promenu stawa gasa u qW koordinatnom sistemu c) odrediti zapreminski rad koji izvr{i gas pri ovoj promeni stawa kao i koli~inu toplote koja se u ovom procesu preda gasu

a) jedna~ina stati~ke ravnoteè za proizvoqan poloàj klipa u cilindru: q = qp + ρ ⋅ h ⋅ i

q = qp +

qp +

ρ ⋅ h ⋅ (Wdjmjoebs − W ) 3

e π 5

ρ ⋅ h ⋅ Wdjmjoebs 3

e π 5

q = b−c⋅W -

= b >dpotu

⇒

q = qp +

⇒

q = qp + ρ⋅h e3 π 5

ρ ⋅ h ⋅ Wuf•optu e3 π 5 ρ ⋅ h ⋅ Wdjmjoebs 3

e π 5

−

ρ⋅h e3 π 5

⋅W

>c>dpotu

zavisnost pritiska od zapremine je linearna, a konstante b i c odre|ujemo iz grani~nih uslova:

q2 = b − c ⋅ W2 q 3 = b − c ⋅ W3


)2* )3*

{fmlp@fvofu/zv


c=

strana 35

q2 − q 3 2/3 ⋅ 21 7 − 1/7 ⋅ 21 7 Qb > 6 ⋅ 21 7 = W3 − W2 1/33 − 1/2 n4

b = q2 + c ⋅ W2 = 2/3 ⋅ 21 7 + 6 ⋅ 21 7 ⋅ 1/2 > 2/8 ⋅ 21 7 Qb q = 2/8 ⋅ 21 7 − 6 ⋅ 21 7 ⋅ W !!analiti~ki oblik zavisnosti pritiska od zapremine

b) q 2

3

w c) W3

X23 =

∫

q)W*eW =

w W2

q2 + q3 2/3 ⋅ 217 + 1/7 ⋅ 217 ⋅ )W3 − W2* = ⋅ (1/33 . 1/2) >219!lK 3 3


(

)

(

)

q2 ⋅ W2 = o ⋅ NS h ⋅ U2

q2 ⋅ W2 2/3 ⋅ 21 7 ⋅ 1/33 = = 5/9 ⋅ 21 .3 lnpm NS h ⋅ U2 9426 ⋅ 411

(

)

jedna~ina stawa idealnog gasa za stawe 2:! U3 =

q 3 ⋅ W3 = o ⋅ NS h ⋅ U3

q 3 ⋅ W3 1/7 ⋅ 21 7 ⋅ 1/33 = >441/8!L o ⋅ NS h 5/9 ⋅ 21 .3 ⋅ 9426

(

)

prvi zakon termodinamike za proces 1−2: R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) + X23 > 5/9 ⋅ 21


−3

R 23 = ∆V23 + X23 ⋅ 31/9 ⋅ (441/8 − 411) + 219 >249/7!lK

{fmlp@fvofu/zv


strana 36

2/44/!Dvoatoman idealan gas, stawa 2)q2>1/7NQb-!U2>411!L-!W2>1/3!n4*- nalazi se u horizontalno postavqenom nepokretnom cilindru sa (bez trewa) pokretnim klipom. Klip je preko opruge, linearne karakteristike k, povezan sa nepokretnim zidom (slika). Predajom toplote gasu, on se dovodi do stawa 3)q3>2!NQb-!W3>1/5!n4*/ U po~etnom poloàju opruga je rastere}ena. b* izvesti zakon promene stawa gasa u obliku q>g)W* b) prikazati promenu stawa idealnog gasa na qw dijagramu c) odrediti zapreminski rad koji izvr{i gas pri ovoj promeni stawa kao i koli~inu toplote koja se u ovom procesu preda gasu

∆y

3

2

b* jedna~ina stati~ke ravnoteè za proizvoqan poloàj klipa u cilindru: q = qp +

q = qp −

k ⋅ ∆y

⇒

3

e π 5 k ⋅ W2 3

+

k

 e3 π   e3 π       5   5      k ⋅ W2 qp − = b >dpotu 3  e3 π     5    q = b+c⋅W-

3

q = qp +

k ⋅ (W − W2 )  e3 π     5   

3

⇒

⋅W

k  e3 π     5   

3

>c>dpotu

zavisnost pritiska od zapremine je linearna, a konstante b i c odre|ujemo iz grani~nih uslova:

q2 = b + c ⋅ W2 q 3 = b + c ⋅ W3


)2* )3*

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike c=

strana 37

q 3 − q2 2 ⋅ 21 7 − 1/7 ⋅ 21 7 Qb > 3 ⋅ 21 7 = W3 − W2 1/5 − 1/3 n4

b = q2 − c ⋅ W2 = 1/7 ⋅ 21 7 − 3 ⋅ 21 7 ⋅ 1/3 > 1/3 ⋅ 21 7 Qb q = 1/3 ⋅ 21 7 + 3 ⋅ 21 7 ⋅ W !!analiti~ki oblik zavisnosti pritiska od zapremine

b) q 3

2

w c) W3

X23 =

∫

q)W*eW =

w W2

q2 + q 3 1/7 ⋅ 21 7 + 2 ⋅ 21 7 ⋅ )W3 − W2 * = ⋅ (1/5 . 1/3) >271!lK 3 3


(

)

(

)

q2 ⋅ W2 = o ⋅ NS h ⋅ U2

q2 ⋅ W2 1/7 ⋅ 21 7 ⋅ 1/3 = = 5/9 ⋅ 21 .3 lnpm NS h ⋅ U2 9426 ⋅ 411

(

)


q 3 ⋅ W3 = o ⋅ NS h ⋅ U3

q 3 ⋅ W3 2 ⋅ 21 7 ⋅ 1/5 = >2113/3!L o ⋅ NS h 5/9 ⋅ 21 .3 ⋅ 9426

(

)


R 23 = ∆V23 + X23

R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) + X23 > 5/9 ⋅ 21 −3 ⋅ 31/9 ⋅ (2113/3 − 411) + 271 >972/2!lK


{fmlp@fvofu/zv


strana 38

2/45/!U vertikalnom cilindru (slika) unutra{weg pre~nika e>311!nnnalazi se o>!1/6!npm dvoatomnog idealanog gasa. Masa klipa je nl>51 lh. Klip je poduprt oprugom linearne karakteristike l. Po~etni pritisak gasa je q2>2/16!cbs, a pritisak okoline iznosi qp>2!cbs. Plin se hladi tako da u momentu rastere}ewa opruge postigne temperatura od U3>364!L, pri ~emu se od gasa odvede 2/6!lK toplote. Zanemaruju}i trewe klipa odrediti: a) po~etnu temperaturu gasa b) za koliko se podigao gas do momenta rastere}ewa opruge

R23 3 ∆{ 2

b* jedna~ina stati~ke ravnoteè za klip u trenutku rastere}ewa opruge: q3 +

nl ⋅ h 3

e ⋅π 5

= qp

⇒

nl ⋅ h

q3 = qp −

3

e ⋅π 5

= 2 ⋅ 21 6 −

(

)

o ⋅ NS h ⋅ U3 q3

1/6 ⋅ 21 −4 ⋅ 9426 ⋅ 364

=

1/98 ⋅ 21 6

> 1/98 !cbs

(

)

>1/1232!n4 R 23 = ∆V23 + X23

prvi zakon termodinamike za proces 1−2: R23 > o ⋅ (Nd w ) ⋅ (U3 − U2 ) +

1/3 3 ⋅ π 5

q 3 ⋅ W3 = o ⋅ NS h ⋅ U3

jedna~ina stawa idealnog gasa za stawe 2:! W3 =

51 ⋅ :/92

q2 + q 3 ⋅ (W3 − W2 ) 3

)2*

(

)

jedna~ina stawa idealnog gasa za stawe 1:!

q2 ⋅ W2 = o ⋅ NS h ⋅ U2

kombinovawem jedna~ina (1) i (2) dobija se:

W2>1/1259!n4-!U2>484!L

W3

napomena:! X23 =

∫

q)W*eW =

)3*

q2 + q 3 ⋅ )W3 − W2 * -!kao u prethodnom zadatku 3

w W2

b) W2 − W3 =

e3 ⋅ π ⋅ ∆{ 5


⇒

∆{ = 5 ⋅

W2 − W3 3

e ⋅π

= 5⋅

1/1259 − 1/1232 1/3 3 ⋅ π

>97!nn

{fmlp@fvofu/zv


strana 39

2/46/!U vertikalnom, toplotno izolovanom cilindru pre~nika e>311!nn sme{tena je opruga zanemarqive zapremine (slika). Na oprugu je naslowen adijabatski klip mase nl>36!lh. U cilindru se nalazi azot stawa 2)q2>2/16!cbs!U2>414!L*. U po~etnom trenutku udaqenost klipa od dna cilindra iznosi {2>611!nn. Duìna opruge (linearne karakteristike) u neoptere}enom stawu iznosi {p>711 nn. Dolivawem ìve )ρ>24711!lh0n4* iznad klipa, klip se spusti za ∆{>211!nn!(zanemariti trewe). Pritisak okoline iznosi qp>2!cbs. Odrediti: a) koliko je ìve doliveno )lh* b) za koliko se pove}ala unutra{wa energija gasa c) do koje bi temperature trebalo zagrejati azot tako da se klip vrati u po~etno stawe (pretpostaviti da ne dolazi do isticawa ìve) i koliko bi toplote pri tom trebalo dovesti e

e

∆z ∆{

{1 {2

{3

a) jedna~ina stati~ke ravnoteè za poloàj klipa u trenutku 1:     3 l ⋅ ({ p − {2 ) nl ⋅ h  q − q + nl ⋅ h  ⋅ e ⋅ π = q2 + q l = + ⇒ p 2 p  e3 ⋅ π  5 ⋅ ({ p − {2 ) e3 ⋅ π e3 ⋅ π   5  5 5      36 ⋅ :/92  1/3 3 ⋅ π O l = 2 ⋅ 21 6 − 2/16 ⋅ 21 6 + ⋅ >992/8! 3   ( ) 5 1 / 7 1 / 6 ⋅ − n 1/3 ⋅ π   5   e3 ⋅ π 1/3 3 ⋅ π ⋅ {2 = ⋅ 1/6 = 1/1268 n 4 5 5 e3 ⋅ π 1/3 3 ⋅ π ⋅ ({ 2 − ∆{ ) = ⋅ (1/6 . 1/2) = 1/1237 n 4 W3 = 5 5

W2 =

q2  W3 = q 3  W2

  

κ

⇒


W q 3 = q2 ⋅  2  W3

κ

  1/1268   = 2/16 ⋅ 21 6 ⋅    1/1237  

2/5

= 2/54 ⋅ 21 6 Qb

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strana 40

jedna~ina stati~ke ravnoteè za poloàj klipa u trenutku 2: q3 +

l ⋅ ({ p − {2 + ∆{ ) 3

e ⋅π 5

= qp +

nl ⋅ h e3 ⋅ π 5

⇒

+ ρ Ih ⋅ h ⋅ z

    l ⋅ ({ p − {2 + ∆{ ) nl ⋅ h 2 z= q q − 3 + 3 − p⋅ 3   ρ ⋅h e ⋅π e ⋅π   Ih 5 5       2483/3 ⋅ 1/3 36 ⋅ :/92 2 6 6 z>  2 / 54 21 2 / 16 21 >3:3!nn ⋅ − + ⋅ − ⋅ 3  1/3 3 ⋅ π  24711 ⋅ :/92 1/3 ⋅ π   5 5   nIh = ρ Ih ⋅

e3 ⋅ π 1/3 3 ⋅ π ⋅ z = 24711 ⋅ ⋅ 1/3:3 >235/87!lh 5 5

b) jedna~ina stawa idealnog gasa na po~etku procesa:! n=

q2 ⋅ W2 2/16 ⋅ 21 ⋅ 1/1268 = = 1/129 lh S h ⋅ U2 3:8 ⋅ 414

jedna~ina stawa idealnog gasa na kraju procesa:! U3 =

q2 ⋅ W2 = n ⋅ S h ⋅ U2

6

q 3 ⋅ W3 = n ⋅ S h ⋅ U3

q 3 ⋅ W3 2/54 ⋅ 21 ⋅ 1/1237 = = 448/2 L n ⋅ Sh 1/129 ⋅ 3:8 6

∆V23 = n ⋅ d w ⋅ (U3 − U2 ) > 1/129 ⋅ 1/85 ⋅ (448/2 − 414) >!1/56!lK c) uo~iti da je:

W4!>!W2-

q4>q3


q 4 ⋅ W4 = n ⋅ S h ⋅ U4

q 4 ⋅ W4 2/54 ⋅ 21 ⋅ 1/1268 = = 531 L n ⋅ Sh 1/129 ⋅ 3:8 6

prvi zakon termodinamike za proces 2−3: R34> n ⋅ d w ⋅ (U4 − U3 ) + q 3 ⋅ (W4 − W3 ) >

R 34 = ∆V 34 + X34

R34>! 1/129 ⋅ 1/85 ⋅ (531 − 448/2) + 2/54 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/1268 − 1/1237 ) >2/66!lK


{fmlp@fvofu/zv


strana 41

2/47/ Toplotno izolovan cilindar, sa pokretnim toplotno izolovanim klipom, podeqen je nepokretnom, toplotno propustqivom (dijatermijskom) pregradom na dva dela (slika). U delu B nalazi se troatomni idealan gas po~etnog staqa B)qB2>1/26!NQb-!WB2>1/6!n4-!UB2>911!L*- a u delu C dvoatomni idealan gas po~etnog stawa C)qC2>1/6!NQb-!WC2>1/3!n4-!UC2>411!L*/ Odrediti zapreminu u delu B i pritisak u delu C u trenutku uspostavqawa termodinami~ke ravnoteè.

B

uo~iti da je: qB2>qB3 WC2>WC3 UB3>UC3

C

jedna~ina stawa idealnog gasa A na po~etku procesa: q ⋅W o B = B2 B2 q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB ⇒ NS h ⋅ UB

(

oB =

)

(

)

1/26 ⋅ 21 7 ⋅ 1/6 > 2/24 ⋅ 21 −3 lnpm 9426 ⋅ 911

jedna~ina stawa idealnog gasa!C!ob!po~etku procesa: q ⋅W oC = C2 C2 q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC ⇒ NS h ⋅ UC

(

oC =

)

(

)

1/6 ⋅ 21 7 ⋅ 1/3 > 5/12 ⋅ 21 −3 lnpm 9426 ⋅ 411

prvi zakon za promenu stawa radnih tela B i C u celom cilindru R23!>!∆V23!,X23 1 = o B ⋅ (Nd w ) B ⋅ (UB3 − UB2 ) + oC ⋅ (Nd w )C ⋅ (UC3 − UC2 ) + q B ⋅ (WB3 − WB2 ) !!!!)2* jedna~ina stawa ideal. gasa A u trenutku uspostavqawa toplotne ravnoteè: q ⋅W )3* UB3 = B3 B3 q B3 ⋅ WB3 = o B ⋅ NS h ⋅ UB3 ⇒ o ⋅ NS h

(


)

(

)

{fmlp@fvofu/zv


strana 42

kada se jedna~ina (2) stavi u jedna~inu (1) dobija se:  q ⋅W   q ⋅W  1 = oB ⋅ (Ndw )B ⋅  B3 B3 − UB2  + oC ⋅ (Ndw )C ⋅  B3 B3 − UC2  + q B ⋅ (WB3 − WB2)  oB ⋅ NSh   oB ⋅ NSh     

(

WB 3 =

(

)

o B ⋅ (Nd w )B ⋅ UB2 + oC ⋅ (Nd w )C ⋅ UC2 + q B ⋅ WB2 (Nd w )C (Nd w )B ⋅ q B3 + ⋅ q C3 + q B NS h NS h

(

WB 3 =

)

)

(

)

2/24 ⋅ 21 −3 ⋅ 3:/2 ⋅ 21 4 ⋅ 911 + 5/12 ⋅ 21 −3 ⋅ 31/9 ⋅ 21 4 ⋅ 411 + 1/26 ⋅ 21 7 ⋅ 1/6 3:/2 ⋅ 21 4 31/9 ⋅ 21 4 ⋅ 1/26 ⋅ 21 7 + ⋅ 1/6 ⋅ 21 7 + 1/26 ⋅ 21 7 q B 9426 9426 WB3>1/416!n4

odavde se dobija:

vra}awem u jedna~inu!)3*;!

UB3 =

lK lnpmL (Nd w )C >31/9! lK lnpmL

(Nd w )B >3:/2!

napomena:

1/26 ⋅ 21 7 ⋅ 1/416 2/24 ⋅ 21 .3 ⋅ 9426

>598!L!>!UC3

troatoman idealan gas dvoatoman idealan gas

kedna~ina stawa ideal. gasa C u trenutku uspostavqawa toplotne ravnoteè; oC ⋅ NS h ⋅ UC3 q C3 ⋅ WC3 = oC ⋅ NS h ⋅ UC3 ⇒ q C3 = WC3

(

q C3 =

)

(

)

5/12 ⋅ 21 .3 ⋅ 9426 ⋅ 598 > 9/2 ⋅ 21 6 !Qb 1/3


{fmlp@fvofu/zv


strana 43

2/48/!Vertikalan, toplotno izolovan cilindar, zatvoren i sa gorwe i sa dowe strane pokretnim klipovima (toplotno izolovanim, zanemarqivih masa, koji se kre}u bez trewa), podeqen je nepropusnom, krutom i nepokretnom pregradom na deo B i deo C (slika). Pregrada je zanemarqivog toplotnog kapaciteta i pruà zanemarqiv otpor kretawu toplote. U delu B nalazi se dvoatoman idealan gas, a u delu C troatoman idealan gas. U po~etnom poloàju gas u delu B ima stawe B2)WB2>1/6!n4-!qB2>1/5!NQbUB2>411!L* gas u delu C u stawe C2)WC2>1/5!n4-!qC2>1/16!NQb-!UC2>411!L*/ Odrediti zapreminski rad koji treba obaviti pri sabijawu gasa u delu B, da bi zapremina gasa u delu C bila dva puta ve}a.

)X23*B

B

C

jedna~ina stawa idealnog gasa A na po~etku procesa: q ⋅W o B = B2 B2 q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB ⇒ NS h ⋅ UB

(

oB =

)

(

)

1/5 ⋅ 21 7 ⋅ 1/6 > 9/13 ⋅ 21 −3 lnpm 9426 ⋅ 411

jedna~ina stawa idealnog gasa C na po~etku procesa: q ⋅W oC = C2 C2 q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC ⇒ NS h ⋅ UC

(

oC =

)

(

)

7

1/16 ⋅ 21 ⋅ 1/3 > 9/13 ⋅ 21−4 lnpm 9426 ⋅ 411

uslov zadatka: dijatermijska pregrada:

WC3!>!3!/!WC2!>!1/9!n4UB3>UC3

prvi zakon termodinamike za proces u cilindru: R23!>!∆V23!,)!X23!*B!,!)!X23!*C 1 = oB ⋅ (Ndw )B ⋅ (UB3 − UB2) + oC ⋅ (Ndw )C ⋅ (UC3 − UC2) + (X23 )B + qC ⋅ (WC3 − WC2)

)2*

jedna~ina stawa idealnog gasa C na kraju procesa:

(

)

qC3 ⋅ WC3 = oC ⋅ NSh ⋅ UC3

)3*

kombinovawem jedna~ina )2* i )3*-!sistem dve jedna~ine sa dve nepoznate, dobija se: UB3>712!L-!!)X23*B>−6:1/4!lK


{fmlp@fvofu/zv


strana 44

2/49/!U izolovanom i sa obe strane zatvorenom cilindru nalaze se dva idealna gasa me|usobno odeqena bez trewa pomi~nim i toplotno propusnim klipom. Po~etni pritisak oba gasa iznosi!qB2>qB3>4!cbs/ U delu nalazi se kiseonik stawa B)UB2>3:4!L-!nB>1/2!lh*-!a u delu C nalazi se metan stawa!C)UC2>634 L-!nC>1/2!lh*/!Odrediti: a) pritisak i temperaturu oba gasa trenutku uspostavqawa termodinami~ke ravnoteè b) promenu entropije sitema koja nastaje u procesu koji po~iwe od zadatog po~etnog stawa i traje do trenutka uspostavqawa termodinami~ke ravnoteè

B

C

b* jedna~ina stawa idealnog gasa B (po~etak procesa) : q B2 ⋅ WB2 = n B ⋅ S hB ⋅ UB2 WB2 =

n B ⋅ S h ⋅ UB2 q B2

=

1/2 ⋅ 371 ⋅ 3:4 4 ⋅ 21 6

>1/1365!n4

jedna~ina stawa idealnog gasa B (kraj procesa): q C2 ⋅ WC2 = nC ⋅ S hC ⋅ UC2 WC2 =

nC ⋅ S hC ⋅ UC2 q C2

=

1/2 ⋅ 631 ⋅ 634 4 ⋅ 21 6

>1/1:17!n4 R 23 = ∆V23 + X23

prvi zakon termodinamike za proces u cilindru: ∆V23>1

⇒

V2 = V 3

V2 = n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +

U+ =

n B ⋅ d wB ⋅ UB2 + nC ⋅ d wC ⋅ UC2 1/2 ⋅ 1/76 ⋅ 3:4 + 1/2 ⋅ 2/93 ⋅ 634 = >573/6!L n B⋅ d wB + nC ⋅ d wC 1/2 ⋅ 1/76 + 1/2 ⋅ 2/93

jedna~ina stawa idealnog gasa B na kraju procesa: q B3 ⋅ WB3 = n B ⋅ S hB ⋅ U +

⇒

WB3 =

n B ⋅ S hB ⋅ U + q B3

)2*

jedna~ina stawa idealnog gasa C na kraju procesa: q C3 ⋅ WC3 = nC ⋅ S hC ⋅ U +


⇒

WC3 =

nC ⋅ S hC ⋅ U + q C3

)3*

{fmlp@fvofu/zv


strana 45

deqewem jedna~ina!)2*!i!)3*!dobija se;! WB3 = WC3

n B ⋅ S hB nC ⋅ S hC

)4*

jednake zapremine cilindra pre i posle procesa;!WB2!,WC2!>WB3!,WC3!!!!)5* kada se jedna~ina!)4*!uvrsti u jedna~inu!)5*!dobija se i re{i po!WC3!dobija se; WC3!>!

WB2 + WC2 1/1365 + 1/1:17 >1/1884!!n4 = n B ⋅ S hB 1/2 ⋅ 371 +2 +2 1/2 ⋅ 631 nC ⋅ S hC WB3 = 1/1884 ⋅

vra}awem u jedna~inu )4*!dobija se;

q B3 =

n B ⋅ S hB ⋅ U + WB3

=

1/2 ⋅ 371 >1/1498!n4 1/2 ⋅ 631

1/2 ⋅ 371 ⋅ 573/6 > 4/2 ⋅ 21 6 !Qb!>!qC3 1/1498

b) ∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>!33/:! ∆Tplpmjob>!1!

K L

K L

(adijabatski procesi u oba cilindra)

∆TSU!>! ∆T B + ∆T C >!///>!51/8!−!41/8!>!33/:! 

!∆TB!>g!)!q-!U*!> nB ⋅  dqmo 

UB3 q  − Shmo B3  > UB2 qB2 

 573/6 4/2 ⋅ 21 6 − 1/371 ⋅ mo ∆TB!> 1/2 ⋅  1/:2 ⋅ mo  3:4 4 ⋅ 21 6  

∆TC!>!g!)!q-!U*!> nC ⋅  dqmo 

K L

  >51/8! K  L 

UC3 q  − Shmo C3  > UC2 qCB2 

 573/3 4/2⋅ 216  lK >−41/89! ∆TC!>! 1/2⋅  3/45 ⋅ mo − 1/631 ⋅ mo 6   L 634 4 21 ⋅  


{fmlp@fvofu/zv


strana 46

2/4:/!U zatvorenom, delimi~no adijabatski izolovanom (vidi sliku), horizontalnom cilindru nalazi se o>3!lnpm dvoatomnog idealnog gasa. Pokretna adijabatska povr{ina (klip) deli cilindar na dva jednaka dela )WB>WC!*/ Po~etno stawe idealnog gasa (u oba dela) odre|eno je istim veli~inama stawa q>2!cbsU>399!L. Dovo|ewem toplote kroz neizolovani deo cilindra (leva ~eona povr{ina) dolazi do kretawa klipa (bez trewa) dok pritisak u delu C ne dostigne 5!cbs ( pri tome se usled kvazistati~nosti ne naru{ava mehani~ka ravnoteà tj. i pritsak u delu B iznosi 5!cbs). Odrediti: a) zapreminski rad koji izvr{i radno telo u delu B (levi deo cilindra) b) koli~inu toplote koja se preda radnom telu u istom delu cilindra

R23 B

C

a) jedna~ina stawa idealnog gasa u delu A na po~etku procesa: q ⋅W o B = B2 B2 q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB ⇒ NS h ⋅ UB

(

)

(

)

q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC

(

oC =

⇒

iz jedna~ina )2*!i!)3* se dobija: uslov zadatka:

)

q C2 ⋅ WC2 NS h ⋅ UC

(

)

oB!>!oC o>oB!,!oC

kombinovawem jedna~ina!)4*!i!)5*!dobija se:

)2* )3*

)4* )5*

oB>!2!lnpm-!oC>2!lnpm

promena stawa idealnog gasa u delu!C!je kvazistati~ka i adijabatska: UC3

q  = UC2 ⋅  C3   q C2 

κ −2 κ

 5 ⋅ 21 6 = 399 ⋅  6  2 ⋅ 21

   

2/5 −2 2/5

= 539 L

prvi zakon termodinamike za proces u delu C; !!!!! (R 23 )C = (∆V23 )C + (X23 )C

(X23 )C

= −(∆V23 )C = −oC ⋅ (Nd w ) ⋅ (UC3 − UC2 ) > −2 ⋅ 31/9 ⋅ (539 − 399) >−3:23!lK

(X23 )B

= −(X23 )C >3:23!lK


{fmlp@fvofu/zv


strana 47

b) jedna~ina stawa idealnog gasa u delu B na po~etku procesa: WB2 =

(

)

o B ⋅ NS h ⋅ UB2 q B2

(

)


(

)

q C2

(

)

o W ⋅ NS h ⋅ UC3 q C3

)

⇒

U U  WB3 − WB2 = o ⋅ NS h ⋅  B3 − B2  !)8*  q B3 q B2 

(

)

(

)

⇒

(

)

⇒

q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC2

)9* q C3 ⋅ WC3 = oC ⋅ NS h ⋅ UC3

jedna~ina stawa idealnog gasa u delu C na kraju procesa: WC3 =

(

q B3 ⋅ WB3 = o B ⋅ NS h ⋅ UB3

jedna~ina stawa idealnog gasa u delu C na po~etku procesa: oC ⋅ NS h ⋅ UC2

⇒

)7*

oduzimawem (7*!−!)6*!dobija se;!

WC2 =

)

)6*

jedna~ina stawa idealnog gasa u delu B na kraju procesa: WB3 =

(

q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB2

):* U U WC2 − WC3 = o ⋅ NS h ⋅  C2 − C3  q C2 q C3

(

oduzimawem (9*!−!):*!dobija se;

)

  !!)21* 

iz ~iwenice da su leve strane jedna~ina!)8*!i!)21*!jednake dobija se: UB3 UB2 U U > C2 − C3 − q B3 q B2 q C2 q C3

⇒

U U U UB3 = q B3 ⋅  B2 + C2 − C3  q B2 q c2 q C3

   

399 539   399 + − UB3 = 5 ⋅ 21 6 ⋅   >2987!L 6 6 2 ⋅ 21 5 ⋅ 21 6   2 ⋅ 21 prvi zakon termodinamike za proces u delu B; !!!!! (R 23 ) B = (∆V23 ) B + (X23 )B

(R23 )B

= o B ⋅ (Nd w ) ⋅ (UB3 − UB2 ) + X23 > 2 ⋅ 31/9 ⋅ (2987 − 399) + 3:23 >!46:53!lK


{fmlp@fvofu/zv


strana 48

2/51. U hermeti~ki zatvorenim i toplotno izolovanim cilindrima B i C , koji su razdvojeni slavinom (vidi sliku) nalazi se po!n>5!lh vazduha (idealan gas) stawa 2B)q>21!cbs-!U>511!L*, odnosno 2C)q>2!cbs-!U>511!L). U krajwem levom delu cilindra C nalazi se adijabatski klip koji moè da se kre}e u cilindru, ali uz savladavawe sila trewa. Otvarawem slavine, klip se usled razlike pritisaka kre}e i sa stepenom dobrote ηlq e >1/9 sabija vazduh u cilindru C dok se ne uspostavi mehani~ka ravnoteà. Skicirati procese sa radnim telom na zajedni~kom Ut dijagramu i odrediti: a) pritisak i temperaturu u cilindrima B i C u stawu mehani~ke ravnoteè b) promenu entropije izolovanog termodinami~kog sistema od zadatog po~etnog stawa do stawa mehani~ke ravnoteè izme|u vazduha u cilindrima B i C

B C

a)

prvi zakon termodinamike za proces u delu A;! (R 23 ) B = (∆V23 ) B + (X23 )B !!)2* prvi zakon termodinamike za proces u delu C;!! (R 23 )C = (∆V23 )C + (X23 )C !!!)3* sabirawem jedna~ina (1) + (2) dobija se: (∆V23 )B = −(∆V23 )C ! ⇒ UB3 − UB2 = UC2 − UC3

ili

UB2 + UC2 = UB2 + UC3

)4*

napomena: po{to su oba cilindra adijabatski izolovana od okoline (R23 )B = (R23 )C =0. zapreminski rad koji izvr{i radno telo B i zapreminski rad koji se izvr{i nad radnim telom C su jednaki, ali suprotni po znaku (X23 ) B = −(X23 )C


{fmlp@fvofu/zv


strana 49

jedna~ina stawa idealnog gasa u delu B na po~etku procesa: WB2 =

(

)


(

)


(

)

q C2

(

)

o W ⋅ NS h ⋅ UC3 q C3

)

⇒

U U  WB3 − WB2 = o ⋅ NS h ⋅  B3 − B2  !)7*  q B3 q B2 

(

)

(

)

⇒

(

)

⇒

q C2 ⋅ WC2 = oC ⋅ NS h ⋅ UC2

)8* q C3 ⋅ WC3 = oC ⋅ NS h ⋅ UC3

jedna~ina stawa idealnog gasa u delu C na kraju procesa: WC3 =

(

q B3 ⋅ WB3 = o B ⋅ NS h ⋅ UB3

jedna~ina stawa idealnog gasa u delu C na po~etku procesa: oC ⋅ NS h ⋅ UC2

⇒

)6*

oduzimawem (5*!−!)5*!dobija se;!

WC2 =

)

)5*

jedna~ina stawa idealnog gasa u delu B na kraju procesa: WB3 =

(

q B2 ⋅ WB2 = o B ⋅ NS h ⋅ UB2

)9* U U WC2 − WC3 = o ⋅ NS h ⋅  C2 − C3  q C2 q C3

(

oduzimawem )8*!−!)9*!dobija se;

)

  !!):* 

iz ~iwenice da su leve strane jedna~ina!)7*!i!):*!jednake dobija se: UB3 UB2 U U > C2 − C3 − q B3 q B2 q C2 q C3

⇒

UB3 + UC3 U U > B2 + C2 ! qy q B2 q C2

)21*

UB3 UC3 U U > B2 + C2 + q B3 q C3 q B2 q C2

⇒

kada se u jedna~inu )21* uvrsti jedna~ina )4* dobija se: UB2 + UC2 U U U + UC2 511 + 511 > B2 + C2 ⇒ qy!>! B2 = UB2 UC2 511 511 qy q B2 q C2 + + 6 q B2 q C2 21 ⋅ 21 2 ⋅ 21 6 qy!> 2/93 ⋅ 21 6 Qb napomena:!

qB3!>!qC3>!qy!(uslov mehani~ke ravnoteè na kraju procesa)


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike q UC3l = UC2 ⋅  C3l  q C2 UC3l − UC2 ηlq e = UC3 − UC2 UC3> 511 +

  

κ −2 κ

strana 50

 2/93 ⋅ 21 6 = 511 ⋅   2 ⋅ 21 6  ⇒

   

2/5 −2 2/5

UC3 = UC2 +

>585/7!L

UC3l − UC2 ηlq e

585/7 − 511 >5:4/4!L 1/9

iz jedna~ine!)4*!!!!!⇒

UB3 = UB2 + UC2 − UC3 = 511 + 511 − 5:4/4 = 417/8 L

b) ∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>!2/15! ∆Tplpmjob>!1!

lK L

lK L

(adijabatski procesi u oba cilindra)

∆TSU!>! ∆T B + ∆T C >!///>!1/9:!,!1/26!>!2/15! 

 417/8 2/93 ⋅ 216 UB3 q  − 1/398 ⋅ mo − Shmo B3  > 5 ⋅ 2⋅ mo  UB2 qB2  511 21 ⋅ 216 

  >1/9:! lK  L 

 5:4/4 2/93 ⋅ 21 6 UC3 q  − 1/398 ⋅ mo − Shmo C3  > 5 ⋅ 2 ⋅ mo  UC2 qCB2  511 2 ⋅ 21 6 

  >1/26! lK  L 

!∆TB!> nB ⋅  dqmo 



∆TC!> nC ⋅  dqmo 

lK L

qB2 3C

U 3lC

qy qC2

2B 2C

UB2>UC2

3B 3lB t


{fmlp@fvofu/zv


strana 51

2/52/!Geometrijski identi~ni, adijabatski i bez trewa poktretni klipovi hermeti~ki zaptivaju dva horizontalno postavqena, toplotno izolovana, nepokretna cilindra. Klipovi su me|usobno spregnuti preko sistema zup~astih letvi, odnosno preko fiksnog i bez trewa pokretnog zup~anika (slika). U levom cilindru )B*- nalazi se 1/9!lh sumpor dioksida (idealan gas), a u desnom cilindru )C* 1/9!lh kiseonika (idealan gas). U polaznom poloàju, sumpor-dioksid se nalazi u stawu B2)qB2>1/23!NQbUB2>411!L*- a kiseonik u stawu C2!)qC2>1/19!NQb-!UC2>411!L*. Odrediti koli~inu elektri~ne energije koju bi elektri~ni greja~ H trebao da preda sumpor-dioksidu, da bi se temperatura kiseonika snizila do UC3>393!L/

H qB

C

B

qbnc

QC

jedna~ina stawa idealnog gasa C na po~etku procesa:! q C2 ⋅ WC2 = nC ⋅ S hC ⋅ UC2 WC2 =


=

1/9 ⋅ 371 ⋅ 411 1/19 ⋅ 21 7

>1/89!n4

U q zakon kvazistati~ke adijabatske promene stawa gasa!C;!!!! C2 =  C2 q C3  UC3 q C3

U = q C2 ⋅  C3  UC2

κ

 κ −2  = 1/19 ⋅ 21 7 

κ

 κ −2  

2/5

 393  2/5 −2 ⋅ >! 1/75 ⋅ 21 6 Qb  411  

jedna~ina stawa idealnog gasa C!na kraju procesa:!!!! q C3 ⋅ WC3 = nC ⋅ S hC ⋅ UC3 WC3 =


=

1/9 ⋅ 371 ⋅ 393 1/755 ⋅ 21 6

>1/:2!n4

prvi zakon termodinamike za proces u delu C;!!!!!! (R 23 )C = (∆V23 )C + (X23 )C

(X23 )C >! − nC ⋅ d wC ⋅ (UC3 − UC2 ) > −1/9 ⋅ 1/76 ⋅ (393 − 411) >!:/47!lX


{fmlp@fvofu/zv


strana 52

(X23 )B >! (X23 )C >!:/47!lX jedna~ina stawa idealnog gasa A na po~etku procesa: q B2 ⋅ WB2 = n B ⋅ S hB ⋅ UB2 WB2 =

n B ⋅ S hB ⋅ UB2 q B2

=

1/9 ⋅ 241 ⋅ 411 1/23 ⋅ 21 7

>1/37!n4 WB3!−!WB2!>!WC3!−!WC2!

uslov jednakih promena zapremina: WB3!>!WB2!,!WC3!−!WC2

WB3>1/37!,!1/:2!−!1/89!>!1/4:!n4

jedna~ina stati~ke ravnoteè za idealan gas!C!na kraju procesa: qC3!>!qbnc!−!q{vq•bojl! ⇒

q{vq•bojl!>!qbnc!−!qC3!>!2!−!1/75!>!1/47!cbs

jedna~ina stati~ke ravnoteè za idealan gas!B!na kraju procesa: qB3!>!qbnc!,!q{vq•bojl! ⇒

qB3!>!2!,!1/47!>!2/47!cbs

jedna~ina stawa idealnog gasa!B!na kraju procesa:!! q B3 ⋅ WB3 = n B ⋅ S hB ⋅ UB3 UB 3 =

q B3 ⋅ WB3 2/47 ⋅ 21 6 ⋅ 1/4: >621!L = n B ⋅ S hB 1/9 ⋅ 241

prvi zakon termodinamike za proces u delu!B;!!!!! (R 23 ) B = (∆V23 ) B + (X23 )B

(R23 )B > n B ⋅ d wB (UB3 − UB2 ) + (X23 )B


= 1/9 ⋅ 1/56 ⋅ (621 − 411) + :/47 >95/:7!lX

{fmlp@fvofu/zv


strana 53

2/53/ U toplotno izolovanom spremniku zapremine W>1/4!n4- nalazi se idealan gas B)Sh>394!K0lhLdw>821!K0lhL-!q>2!cbs-!U>3:4!L*/ Gre{kom je u ovaj spremnik pu{tena izvesna koli~ina idealnog gasa C tako da je nastala me{avina idealni gasova stawa 2)q>2/49!cbs-!U>431!L*/ Da bi se saznalo koji je gas u{ao u spremnik izmerena je ukupna masa me{avine nB,nC>1/573!lh, a zatim je me{avina zagrejana to temperature od U3>464!L. Za ovo zagrevawe je utro{eno R23>21/4!lK toplote. Odrediti koli~inu )nC* i vrstu )Sh-!dw* dodatog gasa C. q B ⋅ WB = n B ⋅ S hB ⋅ UB

jedna~ina stawa idealnog gasa A pre me{awa: nB =

qB ⋅ W 2 ⋅ 21 6 ⋅ 1/4 = = 1/473 lh S hB ⋅ UB 394 ⋅ 3:4

nC!>)nB!!,!nC!*!−!nB!>!1/573!−!1/473!>1/2!lh

koli~ina dodatog gasa:

jedna~ina stawa me{avine idealnih gasova B,C pre zagrevawa, stawe (1):  2  q2 ⋅ W ⋅ − n B ⋅ S hB  S hC = q2 ⋅ WB = n B ⋅ S hB + nC ⋅ S hC ⋅ U2 ⇒ nC  U2 

(

S hC =

)

 K 2  2/49 ⋅ 21 6 ⋅ 1/4 ⋅ − 1/473 ⋅ 394  >37:/4!   1/2  431 lhL 

prvi zakon termodinamike za proces zagrevawa me{avine: R 23 = ∆V23 + X23 R 23 = (n B ⋅ d wB + nC ⋅ d wC ) ⋅ (U3 − U2 ) d wC =

⇒

d wC =

2 nC

 R23  ⋅  − n B ⋅ d wB   U3 − U2 

 2  21/4 ⋅ 21 4 K ⋅ − 1/473 ⋅ 821  >662  1/2  464 − 431 lhL 


{fmlp@fvofu/zv


strana 54

2/54!U adijabatski izolovanom sudu sa nepropusnim i adijabatskim pregradnim zidom odvojeno je B)O3W>8!n4-!q>5!cbs-!U>394!L* od C)DP3-!W>!5!n4-!q>9!cbs-!U>684!L*/ Izvla~ewem pregradnog zida gasovi }e se izme{ati. Odrediti: a) temperaturu )U+* i pritisak )q+* dobijene me{avine b) dokazati da je proces me{awa O3!i!DP3 nepovratan

!B

a)

!C

jedna~ina stawa idealnog gasa B pre me{awa: q B ⋅ WB = n B ⋅ S hB ⋅ UB nB =

q B ⋅ WB 5 ⋅ 21 6 ⋅ 8 = = 44/42 lh S hB ⋅ UB 3:8 ⋅ 394

jedna~ina stawa idealnog gasa C pre me{awa: q C ⋅ WC = nC ⋅ S hC ⋅ UC nC =

q C ⋅ WC 9 ⋅ 21 6 ⋅ 5 = = 3:/66 lh S hC ⋅ UC 29: ⋅ 684

prvi zakon termodinamike za proces me{awa: ∆V23>1

⇒

R 23 = ∆V23 + X23

V2 = V 3


U+ =

n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC 44/42 ⋅ 1/85 ⋅ 394 + 3:/66 ⋅ 1/77 ⋅ 684 = >522!L n B⋅ d wB + nC ⋅ d wC 44/42 ⋅ 1/85 + 3:/66 ⋅ 1/77

jedna~ina stawa me{avine idealnih gasova u trenutku uspostavqawa toplotne ravnoteè: q + ⋅ (WB + WC ) = n B ⋅ S hB + nC ⋅ S hC ⋅ U + q+ =

(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U WB + WC


(

∗

=

)

(44/42 ⋅ 3:8 + 3:/66 ⋅ 29:) ⋅ 522 8+5

> 6/89 ⋅ 21 6 Qb

{fmlp@fvofu/zv


strana 55

pritisak gasne me{avine q+ se moè odrediti i primenom Daltonovog zakona q+!> q +B + q C+ pri ~emu q +B i q C+ imaju slede}a zna~ewa:

napomena:

q +B − pritisak gasa B u gasnoj me{avini u trenutku dostizawa toplotne ravnoteè q C+ −

pritisak gasa C u gasnoj me{avini u trenutku dostizawa toplotne ravnoteè

jedna~ina stawa idealnog gasa B u trenutku uspostavqawa toplotne q +B ⋅ (WB + WC ) = n B ⋅ S hB ⋅ U + ravnoteè: q +B =

n B ⋅ S hB ⋅ U + WB + WC

=

44/42 ⋅ 3:8 ⋅ 522 = 4/81 ⋅ 21 6 !Qb 8+5

jedna~ina stawa idealnog gasa C u trenutku uspostavqawa toplotne q C+ ⋅ (WB + WC ) = nC ⋅ S hC ⋅ U + ravnoteè: q C+ =

nC ⋅ S hC ⋅ U + WB + WC

=

3:/66 ⋅ 29: ⋅ 522 = 3/19 ⋅ 21 6 !Qb 8+5

b) ∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>23/95! ∆Tplpmjob>−!

R23 lK = 1! L Up

lK !?1 L (sud izolovan od okoline)

∆TSU!>∆TB!,!∆TC>!///>!24/78!−!1/94!>23!/95!

lK L

 W + WC U∗ + S hB mo B ∆TB!> n B ⋅ g (U- w ) !> n B ⋅  d wB mo  U WB B 

 >  

522 8+5 lK  ∆TB!> 44/42 ⋅  1/85 ⋅ mo + 1/3:8 ⋅ mo  >24/78! 394 8 L    W + WC U∗ + S hC mo B ∆TC!> nC ⋅ g (U- w ) !> nC ⋅  d wC mo  UC WC 

 >  

522 8+5 lK  ∆TC!> 3:/66 ⋅  1/77 ⋅ mo + 1/29: ⋅ mo  >!−1/94! L 684 5  


{fmlp@fvofu/zv


strana 56

2/55. Toplotno izolovan sud podeqen je izolovanom pregradom na dva dela (slika). U delu B zapremine WB>2/6!n4 nalazi se vodonik (idealan gas) stawa B)qB>1/3!NQb-!UB>3:4!L*/ U delu C zapremine WC>1/5!n4, nalazi se azot stawa C)qC>1/4!NQb-!nC>2!lh*. U jednom trenutku sa pregrade se uklawa izolacioni nepropusni sloj sa pregrade, ~ime ona postaje toplotno ne izolovana polupropustqiva membrana, kroz koju mogu da prolaze samo molekuli vodonika. Odrediti a) promenu entropije sistema tokom procesa koji po~iwe uklawawem sloja pregrade i traje do uspostavqawa toplotne ravnoteè u sudu b) pritisak u delu suda B i delu suda C na kraju ovog procesa

C

B a)


q B ⋅ WB 1/3 ⋅ 21 7 ⋅ 2/6 = = 1/36 lh S hB ⋅ UB 5268 ⋅ 3:4

jedna~ina stawa idealnog gasa C pre me{awa: q C ⋅ WC = nC ⋅ S hC ⋅ UC UC =

q C ⋅ WC 1/4 ⋅ 21 7 ⋅ 1/5 = >515!L S hC ⋅ nC 3:8 ⋅ 2

prvi zakon termodinamike za proces me{awa: ∆V23>1

⇒

R 23 = ∆V23 + X23

V2 = V 3


U+ =

n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC 1/36 ⋅ 21/5 ⋅ 3:4 + 2 ⋅ 1/85 ⋅ 515 >428/7!L = n B⋅ d wB + nC ⋅ d wC 1/36 ⋅ 21/5 + 2 ⋅ 1/85

∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>1/39! ∆Tplpmjob>−!

R23 lK = 1! Up L

lK !?1 L (sud izolovan od okoline)

∆TSU!>∆TB!,!∆TC>!///>!1/57!−!1/29!>1/39!


lK L

{fmlp@fvofu/zv


strana 57

 W + WC U∗ ∆TB!> n B ⋅ g (U- w ) !> n B ⋅  d wB mo + S hB mo B  UB WB 

 >  

428/7 2/6 + 1/5  lK  + 5/268 ⋅ mo ∆TB!> 1/36 ⋅ 21/5 ⋅ mo  >1/57! L 3:4 2/6    W U∗ + S hC mo C ∆TC!> nC ⋅ g (U- w ) !> nC ⋅  d wC mo  U WC C 

 >  

428/7  lK  ∆TC!> 2 ⋅  1/85 ⋅ mo  >!−1/29! L 515   b) jedna~ina stawa idealnog gasa B u trenutku uspostavqawa toplotne ravnoteè: q +B ⋅ (WB + WC ) = n B ⋅ S hB ⋅ U + q +B =

n B ⋅ S hB ⋅ U + WB + WC

q +B −

=

1/36 ⋅ 5268 ⋅ 428/7 = 2/85 ⋅ 21 6 !Qb 2/6 + 1/5

pritisak vodonika u sudu A i istovremeno parcijalni pritisak vodonika gasa ugasnoj me{avini (vodonik +azot) u delu suda B u trenutku dostizawa toplotne ravnoteè

jedna~ina stawa idealnog gasa C u trenutku uspostavqawa toplotne q C+ ⋅ WC = nC ⋅ S hC ⋅ U + ravnoteè: q C+ =

nC ⋅ S hC ⋅ U +

q C+ −

WC

=

2 ⋅ 3:8 ⋅ 428/7 = 3/47 ⋅ 21 6 !Qb 1/5

parcijalni pritisak azota gasnoj me{avini (vodonik +azot) u delu suda C u trenutku dostizawa toplotne ravnoteè

(qC )3 > q +B + q C+ = 2/85 ⋅ 21 6 , 3/47 ⋅ 21 6 > 5/2 ⋅ 21 6 Qb (qC )3 −

pritisak u sudu C u trenutku dostizawa toplotne ravnoteè


{fmlp@fvofu/zv


strana 58

2/56/ Adijabatski izolovan termodinami~ki sistem prikazan na slici ~ine: − zatvoren rezervoar )B* stalne zapremine WB>1/4!n4, u kojem se nalazi kiseonik (idealan gas) stawa B)qB>3/7!cbs-!UB>411!L* − zatvoren vertikalni cilindar )C* sa bez trewa pokretnim klipom, u kojem se nalazi nC>!2!lh metana (idealan gas) stawa C)qC>3!cbs-!UC>511!L*/ (pokretni klip svojom teìnom obezbe|uje stalan pritisak gasa) Otvarawem ventila dolazi do me{awa gasova. Smatraju}i da pri me{awu gasova ne}e do}i do hemijske reakcije (eksplozija) odrediti: a) rad koji izvr{i klip za vreme procesa me{awa b) promenu entropije ovog adijabatski izolovanog sistema za vreme procesa me{awa b*

B

C


q B ⋅ WB 3/7 ⋅ 21 6 ⋅ 1/4 = = 2 lh S hB ⋅ UB 371 ⋅ 411

jedna~ina stawa idealnog gasa C pre me{awa: q C ⋅ WC = nC ⋅ S hC ⋅ UC WC =

nC ⋅ S hC ⋅ UC qC

=

2 ⋅ 631 ⋅ 511 3 ⋅ 21 6

>2/15!n4

prvi zakon termodinamike za proces me{awa: !!R23!>!∆V23!,!X23

[

]

1> n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U + − [n B ⋅ d wB ⋅ UB + nC ⋅ d wC ⋅ UC ] + X23 izra~unavawe zapreminskog rada:

)2*

X23!>! q C ⋅  W + − WB − WC  !! )3*  

jedna~ina stawa dobijene me{avine idealnih gasova: q C ⋅ W + = n B ⋅ S hB + nC ⋅ S hC ⋅ U + )4*

(


)

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike )4*

W+ =

⇒

strana 59

(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U +

qC ovu jedna~inu uvrstimo u jedna~inu (2):  n B ⋅ S hB + nC ⋅ S hC ⋅ U +  X23 = q C ⋅  − WB − WC  qC   ovu jedna~inu uvrstimo u jedna~inu )2* odakle se nakon sre|ivawa dobija: n B ⋅ d wB ⋅ UB + n C ⋅ d wC ⋅ UC + q C ⋅ (WB + WC ) U+ = n B ⋅ d wB + n C ⋅ d wC + n B ⋅ S hB + n C ⋅ S hC

(

U+ =

)

2 ⋅ 1/76 ⋅ 411 + 2 ⋅ 2/93 ⋅ 511 + 3 ⋅ 21 6 ⋅ 21 −4 ⋅ (1/4 + 2/15 ) >477/6!L 2 ⋅ 1/76 + 2 ⋅ 2/93 + 2 ⋅ 1/37 + 2 ⋅ 1/63

(2 ⋅ 371 + 2 ⋅ 631) ⋅ 477/6 >2/54!n4

)4*

⇒

W+ =

)3*!

⇒

X23 = 3 ⋅ 21 6 ⋅ 21 −4 ⋅ [2/54 − 1/4 − 2/15 ] >29!lK

3 ⋅ 21 6

b) ∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>1/654!

∆Tplpmjob>−!

R23 lK = 1! Up L

lK L (sud i cilidar izolovani od okoline)

∆TSU!>∆TB!,!∆TC>!///>!1/647!,!1/118!>1/654!  U+ W+ + S hB mo ∆TB> n B ⋅  d wB mo  UB WB   U∗ W+ ∆TC> nC ⋅  d wC mo + S hC mo  UC WC 


lK L

  > 2 ⋅  1/76 ⋅ mo 477/6 + 1/37 ⋅ mo 2/54  >1/647 lK  L 411 1/4   

  = 2 ⋅ 2/93 ⋅ mo 477/6 + 1/63 ⋅ mo 2/54  >1/118! lK  L 511 2/15   

{fmlp@fvofu/zv


strana 60

2/57/ Termodinami~ki sistem prikazan na slici ~ine: − zatvoren rezervoar )B* stalne zapremine WB>1/37!n4, u kojem se nalazi kiseonik (idealan gas) stawa B)qB>5!cbs-!UB>511!L* − zatvoren rezervoar )C* stalne zapremine WC>1/37!n4 u kojem vlada apsolutni vakum − okolina stalne temperature Up>399!L Otvarawem ventila kiseonik se {iri i u toku procesa {irewa okolini preda 25/5!lK toplote. a) odrediti pritisak kiseonika nakon {irewa b) dokazati da je proces {irewa kiseonika nepovratan. B

a)

C

jedna~ina stawa idealnog gasa B pre {irewa: q B ⋅ WB = n B ⋅ S h ⋅ UB2 nB =

q B ⋅ WB 5 ⋅ 21 6 ⋅ 1/37 = = 2 lh S hB ⋅ UB2 371 ⋅ 511

prvi zakon termodinamike za proces {irewa: R 23 = n ⋅ d w (UB3 − UB2 ) ⇒

UB3 = UB2 +

R 23 = ∆V23 + X23

R 23 25/5 = 511 − >491!L n ⋅ dw 2 ⋅ 1/83

jedna~ina stawa idealnog gasa B nakon {irewa: q B3 ⋅ (WB + WC ) = n B ⋅ S h ⋅ UB3 q B3 =

n B ⋅ S h ⋅ UB 3 WB + WC

=

2 ⋅ 371 ⋅ 491 >! 2/: ⋅ 21 6 Qb 1/37 + 1/37

b) ∆TTJ!>!∆TSU!,!∆Tplpmjob!>!///!>1/258,1/16!>!1/2:8!

∆Tplpmjob>−!

lK !?!1 L

R 23 − 25/5 lK =− >1/161! L Up 399

 U q ∆TSU> n B ⋅  d q mo B3 − S hmo B3 UB2 QB2 


 491 2/:  lK   > 2 ⋅  1/:2 ⋅ mo − 1/37 ⋅ mo  >1/258 L 511 5   

{fmlp@fvofu/zv


strana 61

2/58/!Adijabatski izolovan sud podeqen je nepropusnom i adijabatskom membranom na dva dela WB>1/4!n4 i WC>1/6!n4 (slika). U delu B nalazi se nB>1/6!lh kiseonika (idealan gas) temperature UB>411!L, a u delu C!nC>2!lh sumpor-dioksida (idealan gas) temperature UC>641!L. U delu A kiseonik po~iwe da se me{a ventilatorom snage 41!X/ Membrana je projektovana da pukne kada razlika pritisaka prema{i ∆q>73!lQb i u tom trenutku se iskqu~uje ventilator. Odrediti: a) vreme do pucawa membrane c* temperaturu i pritisak nastale me{avine posle pucawa membrane, a po uspostavqawu termodinami~ke ravnoteè

X23

!C

B

b* q C ⋅ WC = nC ⋅ S hC ⋅ UC

jedna~ina stawa idealnog gasa C: nC ⋅ S hC ⋅ UC

qC =

WC

=

2 ⋅ 241 ⋅ 641 > 2/49 ⋅ 21 6 Qb 1/6 ∆q = q B3 − q C

uslov pucawa membrane: 6

4

6

q B3 = q C + ∆q = 2/49 ⋅ 21 + 1/73 ⋅ 21 > 3 ⋅ 21 !Qb jedna~ina stawa idealnog gasa B neposredno pred pucawe membrane: q ⋅W 3 ⋅ 21 6 ⋅ 1/4 UB 3 = B 3 B = >572/6!L q B3 ⋅ WB = n B ⋅ S hB ⋅ UB3 ⇒ n B ⋅ S hB 1/6 ⋅ 371 prvi zakon termodinamike za proces u delu A (za vreme rada ventilatora) R 23 = ∆V23 + XU23

⇒

XU23 = −∆V23

XU23 = −n B ⋅ d wB ⋅ (UB3 − UB2 ) = −1/6 ⋅ 1/76 ⋅ (572/6 − 411) >−63/6!lK τ=

XU23 ⋅

X U23

=

− 63/6 . 41 ⋅ 21 .4


>!2861!t

{fmlp@fvofu/zv


strana 62

b) prvi zakon termodinamike za proces me{awa: ∆V23>1

⇒

R 23 = ∆V23 + X23

V2 = V 3

V2 = n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +

U+ =

n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC 1/6 ⋅ 1/76 ⋅ 572/6 + 2 ⋅ 1/56 ⋅ 641 >612/4!L = n B⋅ d wB + nC ⋅ d wC 1/6 ⋅ 1/76 + 2 ⋅ 1/56

jedna~ina stawa me{avine idealnih gasova u trenutku uspostavqawa toplotne ravnoteè: q + ⋅ (WB + WC ) = n B ⋅ S hB + nC ⋅ S hC ⋅ U + q+ =

(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U

(

∗

=

WB + WC

)

(1/6 ⋅ 371 + 2 ⋅ 241) ⋅ 612/4 > 2/74 ⋅ 21 6 Qb 1/4 + 1/6

2/59/!Adijabatski izolovan sud podeqen je nepropustqivom i adijabatskom membranom na dva dela WB>1/4 n4!i!WC>1/6!n4 (vidi sliku). U delu B nalazi se nB>1/6!lh kiseonika (idealan gas) temperature UB>411 L, a u delu C!nC>2!lh sumpor-dioksida (idealan gas) temperature UC>461!L/ Me{awe kiseonika se obavqa ventilatorom pogonske snage 41!X, sumpor-dioksida ventilatorom pogonske snage 56!X. Membrana je projektovana tako da pukne kada razlika pritisaka prema{i ∆q≥75/3!lQb i u tom trenutku se iskqu~uju oba ventilatora. Odrediti: a) vreme do pucawa membrane b) temperaturu i pritisak nastale me{avine posle pucawa membrane, a po uspostavqawu termodinami~ke ravnoteè

B )XU23*B


C )XU23*B

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strana 63

b*  ⋅  prvi zakon termodinamike za proces u delu!B;!!! (R 23 ) B = (∆V23 ) B +  X U23 ⋅ τ   B  ⋅  n B ⋅ d wB ⋅ (UB3 − UB2 ) = − X U23 ⋅ τ   B

)2*

 ⋅  prvi zakon termodinamike za proces u delu!C;! (R 23 )C = (∆V23 )C +  X U23 ⋅ τ   C  ⋅  nC ⋅ d wC ⋅ (UC3 − UC2 ) = − X U23 ⋅ τ   C

)3*

 ⋅   X U23  n ⋅ d ⋅ (UB3 − UB2 )  B > B wB !!!!!)4* deqewem jedna~ina!)2*!i!)3*!dobija se;! nC ⋅ d wC ⋅ (UC3 − UC2 )  ⋅   X U23   C jedna~ina stawa idealnog gasa!C!neposredno pred!pucawa membrane: nC ⋅ S hC ⋅ UC3 q C3 ⋅ WC = nC ⋅ S hC ⋅ UC3 ⇒ q C3 = )5* WC jedna~ina stawa idealnog gasa!B!neposredno pred!pucawa membrane: n B ⋅ S hB ⋅ UB3 q B3 = )6* q B3 ⋅ WB = n B ⋅ S hB ⋅ UB3 ⇒ WB oduzimawem jedna~ina!)6*!i )5*!dobija se: n B ⋅ S hB ⋅ UB3 nC ⋅ S hC ⋅ UC3 q B 3 − q C3 = − WB WC uslov pucawa membrane:

∆q = q B3 − q C3

)7* )8*

kombinovawem jedna~ina!)4*-!)7*!i!)8*!dobija se: UB3!>!577/26!L

UC3!>!641!L

vra}awem UB3!u jedna~inu!)2*!ili UC3!u jedna~inu )3*!dobija se: −n B ⋅ d wB ⋅ (UB3 − UB2 ) −1/6 ⋅ 1/76 ⋅ (577/26 − 411) τ= = >2911!t ⋅ − 41 X U23


{fmlp@fvofu/zv


strana 64

b) prvi zakon termodinamike za proces me{awa: ∆V23>1

⇒

R 23 = ∆V23 + X23

V2 = V 3

V2 = n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC3 V 3 = n B ⋅ d wB ⋅ U + + nC ⋅ d wC ⋅ U +

U+ =

n B ⋅ d wB ⋅ UB3 + nC ⋅ d wC ⋅ UC3 1/6 ⋅ 1/76 ⋅ 577/26 + 2 ⋅ 1/56 ⋅ 641 = >614/3!L n B⋅ d wB + nC ⋅ d wC 1/6 ⋅ 1/76 + 2 ⋅ 1/56

jedna~ina stawa me{avine idealnih gasova u trenutku uspostavqawa toplotne ravnoteè: q + ⋅ (WB + WC ) = n B ⋅ S hB + nC ⋅ S hC ⋅ U +

(

q+ =

)

(n B ⋅ S hB + nC ⋅ S hC ) ⋅ U ∗ (1/6 ⋅ 371 + 2 ⋅ 241) ⋅ 614/3 > 2/75 ⋅ 21 6 Qb = WB + WC


1/4 + 1/6

{fmlp@fvofu/zv


strana 65

)2/5:/!−2/61/*

2/5:/ Verikalni cilindar unutra{weg pre~nika e>361!nn, adijabatski izolovan od okoline, zatvoren je sa gorwe strane bez trewa pokretnim adijabatskim klipom mase nl>61!lh. Klip na sebi nosi oprugu zanemarqive teìne, linearne karakteristike l>231 O0dn3 i u po~etnom poloàju udaqen je od dna cilindra {>511!nn (slika). Pritisak okoline iznosi qp>2!cbs/ U cilindru se nalazi vazduh (idealan gas) temperature U2>3:4!L. Na oprugu se odozgo po~iwe spu{tati teg nbtf!nU>411!lh/ Od trenutka kada teg dodirne opruga , on po~inwe oprugu sa klipom potiskivati na dole, istovremeno sabijaju}i oprugu i gas u cilindru. Odrediti a) za koliko se spusti klip )∆{* a koliko sabije opruga )∆m* do trenutka kada sila u uètu postane jednaka nuli (stawe 2) b) do koje temperature bi trebalo zagrejati vazduh stawa 2 da bi klip vratili u prvobitni poloàj i koliko toplote je za to potrebno dovesti

nu m

{ e

a) ∆{>21:!nnb) U>563/9!L-

∆m>356!nn R>4/2!lK

2/61/ Cilindar je napravqen prema slici. Slobodno pomi~ni klip zanemarqive mase, optere}en tegom mase nU>311!lh, nalazi se u po~etnom poloàju na kao na slici. U cilndru se nalazi vazduh po~etne temperature U2>634!L koji se hladi predaju}i kroz zidove cilindra toplotu okolini stawa P)qp>2!cbs-!!Up>3:4!L* sve do uspostavqawa toplotne ravnoteè sa okolinom. Odrediti: a) pri kojoj temperaturi vazduha u cilindru }e klip dodirnuti oslonac (stawe 2) b) pritisak gasa na po~etku i kraju procesa c) koli~inu toplote koju vazduh preda okolini tokom procesa 1−2−3

nu ∆{

E

e>291!nn E>311!nn

{ e

{>461!nn ∆{>261!nn

a) U3!>456/7!L b) q2!>2/73!cbs-!q4!>2/48!cbs c) R24!>!−4/27!lK


{fmlp@fvofu/zv


strana 66

PRVI I DRUGI ZAKON TERMODINAMIKE (OTVOREN TERMODINAMI^KI SISTEM) 2/62/ Vazduh (idealan gas) struji stacionarno kroz vertikalnu cev visine 4/7!n, konstantnog popre~nog preseka, masenim

3

⋅

protokom od n >411!lh0i (slika). Cev je toplotno izolovana od okoline, a u cevi je instaliran greja~ koji vazduhu predaje toplotu. Stawe vazduha na ulazu u cev odre|eno je veli~inama stawa 2)q2>2/3!cbs-!U2>3:4!L-!x>5/6!n0t*, a na izlazu 3)q3>2 cbs!U3>472!L*. Odrediti: a) brzinu vazduha na izlazu iz cevi c* toplotni protok koji greja~ saop{tava vazduhu

{3−{2

a)

2

jedna~ina stawa idealnog gasa na ulazu u cev: q2 ⋅ w 2 = S h ⋅ U2 S h ⋅ U2

w2 =

q2

=

398 ⋅ 3:4 2/3 ⋅ 21 6

>1/8118!

n4 lh q 3 ⋅ w 3 = S h ⋅ U3

jedna~ina stawa idealnog gasa na izazu iz cevi: w3 =

S h ⋅ U3 q3

=

398 ⋅ 472 2 ⋅ 21 6

>2/1472!

n4 lh

x2 ⋅ jedna~ina kontinuiteta: x 3 = x2 ⋅

e23 ⋅ π e 3 ⋅π x3 ⋅ 3 5 5 = w2 w3

w3 n 2/1472 = 5/6 ⋅ >7/76! w2 1/8118 t

b) ⋅

⋅

⋅

prvi zakon termodinamike za proces u cevi:! R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 ⋅

⋅ x 33 − x 23 + n⋅ ({ 3 − {2 ) 3 411 411 7/76 3 − 5/6 3 411 = ⋅ 2 ⋅ 21 4 ⋅ (472 − 3:4) + ⋅ + ⋅ 4/7 >6779!X 4711 4711 3 4711 ⋅

⋅

R 23 = ∆ n⋅ d q ⋅ (U3 − U2 ) + n⋅ ⋅

R 23


{fmlp@fvofu/zv


strana 67 ⋅

2/63/ U toplotno izolovanoj komori me{aju se tri struje idealnih gasova: kiseonik B) n >7!lh0t-!q>1/29 ⋅

⋅

NQb-!u>361pD*- azot C) n >4!lh0t-!q>1/44!NQb-!u>6:1pD* i ugqen−monoksid D) n >3!lh0t-!q>1/49!NQbu>551pD*/ Pritisak dobijene sme{e na izlazu iz komore q+>1/2!NQb/!Zanemaruju}i promenu kineti~ke energije kao i potencijalne energije, odrediti: a) temperaturu )U+* i zapreminski protok ( W + ) dobijene sme{e c* promenu entropije sistema za proces me{awa B C

me{avina

D a) prvi zakon termodinamike za proces u me{noj komori: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23

⇒

⋅

⋅

⋅

⋅

⋅

⋅

⋅

⋅

⋅

⋅

I2 = I3

I2 = n B ⋅ d qB ⋅ UB + nC ⋅ d qC ⋅ UC + nD ⋅ d qD ⋅ UD I3 = n B ⋅ d qB ⋅ U + + nC ⋅ d qC ⋅ U + + nD ⋅ d qD ⋅ U + ⋅

+

U =

⋅

⋅

n B ⋅ d qB ⋅ UB + nC ⋅ d qC ⋅ UC + nD ⋅ d qD ⋅ UD ⋅

⋅

⋅

n B ⋅ d qB + nC ⋅ d qC + nD ⋅ d qD U+ =

7 ⋅ 1/:2 ⋅ 634 + 4 ⋅ 2/15 ⋅ 974 + 3 ⋅ 2/15 ⋅ 824 = 76:/7!L 7 ⋅ 1/:2 + 4 ⋅ 2/15 + 3 ⋅ 2/15

jedna~ina stawa idealne gasne me{avine na izlazu iz komore za me{awe: ⋅ ⋅  ⋅  q + ⋅ W + =  n B ⋅ S hB + nC ⋅ S hC + nD ⋅ S hD  ⋅ U + ⇒   ⋅ ⋅  ⋅   n B ⋅ S hB + nC ⋅ S hC + nD ⋅ S hD  ⋅ U + (7 ⋅ 371 + 4 ⋅ 3:8 + 3 ⋅ 3:8) ⋅ 76:/7   W+ = = + q 2 ⋅ 21 6 W + = 31 !

n4 t


{fmlp@fvofu/zv


strana 68

b) jedna~ina stawa idealnog gasa B u nastaloj me{avini: ⋅

q +B

=

n B ⋅ S hB ⋅ U + W

+

7 ⋅ 371 ⋅ 76:/7 >1/62 ⋅ 21 6 !Qb 31

=

jedna~ina stawa idealnog gasa C u nastaloj me{avini: ⋅

q C+ =

nC ⋅ S hC ⋅ U + W

+

⋅

⋅

=

nD ⋅ S hD ⋅ U + W ⋅

+

=

⋅

⋅

⋅

lX L

⋅

∆ T SU > ∆ T B , ∆ TC , ∆ TD >!///>4/34!,!2/44!,!2/6:!>!7/26! ⋅

⋅

q +D ⋅ W + = nD ⋅ S hD ⋅ U +

3 ⋅ 3:8 ⋅ 76:/7 >1/31 ⋅ 21 6 !Qb 31

∆ Ttj > ∆ T SU , ∆ Tp >!///!>7/26! ⋅

⋅

q C+ ⋅ W + = nC ⋅ S hC ⋅ U +

4 ⋅ 3:8 ⋅ 76:/7 >1/3: ⋅ 21 6 !Qb 31

=

jedna~ina stawa idealnog gasa D u nastaloj me{avini: q +D

⋅

q +B ⋅ W + = n B ⋅ S hB ⋅ U +

lX L

⋅

R23 lX ∆ Tp = − >1! L Up

(adijabatski izolovana komora za me{awe)

⋅ ⋅  q+ U+ ∆ T B = g (q- U ) = n B .  d qB mo − S hB mo B  UB qB 

 =  

⋅  76:/7 1/62 ⋅ 21 6  lX ∆ T B > 7 ⋅  1/:2 ⋅ mo − 1/37 ⋅ mo = 4/34 7   L 634 1/29 ⋅ 21   ⋅ ⋅  q+  U+ ∆ TC > g (q- U ) > nC /!  d qC mo − S hC mo C  >  UC q C   ⋅  76:/7 1/3: ⋅ 21 6  lX >2/44! ∆ TC > 4 ⋅ 2/15 ⋅ mo − 1/3:8 ⋅ mo 7   L 974 1/44 ⋅ 21   ⋅ ⋅  q+ U+ ∆ TD > g (q- U ) > nD /!  d qD mo − S hC mo D  UD qD 

 >  

⋅  76:/7 1/31 ⋅ 21 6 − 1/3:8 ⋅ mo ∆ TD > 3 ⋅ 2/15 ⋅ mo  824 1/49 ⋅ 21 7 


  >2/6:! lX  L 

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strana 69

2/64/ “Ludi nau~nik” tvrdi da je mogu}e, bez izmene toplote i/ili rada sa okolinom, struju vazduha stawa ⋅

2) n >2!lh0t-!q>4!cbs-!U>3:4!L* razdvojiti na dve struje. Struju 2 stawa 3)q>3/8!cbs-!U>444!L* i struju 3 stawa 4)q>3/8!cbs-!U>384!L*/ Dokazati da je “ludi nau~nik” u pravu. Zanemariti promene kineti~ke i potencijalne energije vazduha. prvi zakon termodinamike za proces u razdelnoj komori: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 ⋅

⋅

⋅

⇒

⋅

n2⋅ d q ⋅ U2 > n3 ⋅ d q ⋅ U3 + n4 ⋅ d q ⋅ U4 ⋅

jedna~ina kontinuiteta:

⋅

I2 = I3 )2*

⋅

⋅

n2 = n3 + n4

)3* ⋅

kombinovawem jedna~ina!)2*!j!)3*!dobija se:

n3 >1/44!

lh ⋅ lh -! n4 >1/78! t t

drugi zakon termodinamike za proces u razdelnoj komori: ⋅

⋅

⋅

∆ Ttj > ∆ T SU , ∆ Tp >!///!>36/7! ⋅

⋅

X L

⋅

∆ T SU > ∆ T 3 , ∆ T 4 >!///>63/85!−38/25!>!36/7! ⋅

X L

⋅

R23 X ∆ Tp = − >1! Up L

(adijabatski izolovana komora za me{awe)

⋅ ⋅  U q  ∆ T 3 = g (q- U ) = n3 .  d q mo 3 − S h mo 3  U2 q2   ⋅ 444 3/8  X  ∆ T 3 > 1/44 ⋅ 2 ⋅ mo − 1/398 ⋅ mo  = 63/85 L 3:4 4   ⋅ ⋅  U q ∆ T 4 = g (q- U ) = n4 .  d q mo 4 − S h mo 4 U2 q2 

  = 

⋅ 384 3/8  X  ∆ T 4 > 1/77 ⋅ 2 ⋅ mo − 1/398 ⋅ mo  = −38/25 3:4 L 4   ⋅

Kako je ∆ Ttj ?!1, ovaj proces je mogu} pa je “ludi nau~nik” u pravu.


{fmlp@fvofu/zv


strana 70

!2

2/65/ [ire}i se u gasnoj turbini, tok vazduha (idealan gas), mewa

R23

⋅

svoje toplotno stawe od stawa 2) W 2>1/3!n40t- q>21!cbs-!U>691!L*!, ⋅

na ulazu u turbinu, do stawa 3) W 3>2/3!n40t-!q>2!cbs*- na izlazu iz we. Tokom {irewa usled neidealnog toplotnog izolovawa turbine, toplotni protok sa vazdu{nog toka na okolni vazduh iznosi 29!lX. Zanemaruju}i promene kineti~ke i potencijalne energije vazdha , odrediti: a) snagu turbine b) dokazati da je proces u turbini nepovratan (temperatura okoline iznosi Up>3:4!L)

XU23

!3

a) ⋅

jedna~ina stawa idealnog gasa na ulazu u turbinu:

⋅

q2 ⋅ W 2 = n⋅ S h ⋅ U2

⋅

⋅

q ⋅ W 2 21 ⋅ 21 6 ⋅ 1/3 lh n= 2 = >2/3! S h ⋅ U2 398 ⋅ 691 t ⋅

jedna~ina stawa idealnog gasa na izlazu iz turbine: ⋅

U3 =

q3 ⋅ W 3

2 ⋅ 21 6 ⋅ 2/3 >459/5!L 398 ⋅ 2/3

=

⋅

⋅

q 3 ⋅ W 3 = n⋅ S h ⋅ U3

Sh ⋅ n

prvi zakon termodinamike za proces u turbini: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 ⋅

⋅

⋅

⋅

⇒

⋅

X U23 = R23 − ∆ I23 = R23 − n⋅ dq ⋅ (U3 − U2) > −29 − 2/3 ⋅ 2 ⋅ (459/5 − 691) >371!lX

b) ⋅

⋅

⋅

∆ Ttj > ∆ T SU , ∆ Tp >!///!>72/5!,!292/5!>353/9! ⋅

X ?!1 L

⋅

X − 29 R23 ∆ Tp = − =− >72/5! Up 3:4 L ⋅ ⋅  ⋅ U q ∆ T SU = ∆ T23 = g (q- U ) = n .  d q mo 3 − S h mo 3 U2 q2  ⋅



∆ T SU > 2/3 ⋅ 2 ⋅ mo 

  

X 459/5 2 − 1/398 ⋅ mo  = 292/5 691 21  L


{fmlp@fvofu/zv


strana 71 ⋅

2/66/!U dvostepenoj gasnoj turbini ekspandira n >3!lh0t vazduha (idealan gas) od po~etnog pritiska q2>71!cbs!do krajweg pritiska q5>21!cbs. Nakon ekspanzije u prvom stepenu turbine vazduh se uvodi u me|uzagreja~ u kome se izobarski zagreva do temperature U2>U4>911!L/ Ekspanzije u oba stepena su adijabatske i kvazistati~ke. Zanemariti promene potencijalne i kineti~ke energije. Skicirati proces u qw i Ut koordinatnim sistemima i odrediti: a) pritisak u me|uzagreja~u tako da snaga dvostepene turbine bude maksimalna b) snagu dvostepene turbine u tom slu~aju 2

XU23 3

4 R34

XU45

5 q

U 2

3

2

4

3

5

4

5 w

t

a) ⋅

⋅

⋅

Q = X U23 + X U 45 = n⋅ d q ⋅ (U2 − U3 + U4 − U5 ) !>!/// κ

q2  U2  κ −2 =  q3  U3 

κ

⇒

 U  κ −2 q3 = q2 ⋅  3  >qy  U2 

⇒

 U  κ −2 q4 = q5 ⋅  4  >qy  U5 

κ

q4  U4  κ −2 =  q5  U5 


)2*

κ

)3*

{fmlp@fvofu/zv


strana 72

kombinovawem jedna~ina (1) i (2) dobija se: κ

κ

 U  κ −2  U  κ −2 q2 ⋅  3  > q5 ⋅  4   U2   U5  U ⋅U  q  U3 = 4 2 ⋅  2  U5  q5 

⇒!

κ

⇒

q2  U4 ⋅ U2  κ −2  = q5  U5 ⋅ U3 

⇒

2− κ     U4 ⋅ U2  q2  κ   Q = n⋅ dq ⋅  U2 − ⋅  + U4 − U5  U5  q5     

q  U5 = U4 ⋅ U2 ⋅  2   q5 

2− κ κ

 71 ⋅ 216   > 911 ⋅ 911 ⋅   21 ⋅ 216   

911 ⋅ 911  71 ⋅ 216  U3 = ⋅ 72:/44  21 ⋅ 216 

=

U4 ⋅ U2 U5 ⋅ U3

2− κ   ⋅  U ⋅U  q  κ  ∂Q 4 2  2 2 = n⋅ dq ⋅  ⋅ −  3 q  ∂U5  U5   5  

2− κ   U ⋅U  q  κ  4 2  2 ⋅  − 2 = 1  3  U5   q5   

⇔

κ −2 κ

2− κ κ

⋅

∂Q =1 ∂U5

 q2    q   5

2−2/5 2/5

2−2/5 2/5

q  U5 = U4 ⋅ U2 ⋅  2   q5 

⇒

⇒

>72:/44!L κ

>72:/44!L-

2− κ κ

2/5

 U  κ −2  72:/44 2/5 −2 >35/6!cbs q3 = q2 ⋅  3  > 71 ⋅ 216 ⋅    911   U2 

b) Qnby = 3 ⋅ 2⋅ (911 − 72:/44 + 911 − 72:/44) >833/79!lX


{fmlp@fvofu/zv


strana 73 ⋅

2/67/!Kopresor, snage!3!lX-!usisava okolni vazduh (idealan gas) stawa!2)q>2!cbs-!U>3:4!L-! n >1/16 lh0t*!i adijabatski ga sabija do stawa!3/!Nakon toga se vazduh adijabatski prigu{uje do po~etnog pritiska )q4>q2*. Prira{taj entropije vazduha za vreme adijabatske kompresije i adijabatskog prigu{ivawa je jednak. Odrediti stepen dobrote adijabatske kompresije. Zanemariti promene kineti~ke i potencijalne energije vazduha. 2

3

4

XU23 prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ⋅

ograni~enom isprekidanom linijom: ⋅

⋅

⋅

R 24 = ∆ I24 + X U24 + ∆Fl24 + ∆Fq24 ⋅

⋅

1 = n⋅ d q ⋅ (U4 − U2 ) + X U23

⇒

U4 = U2 −

XU23 ⋅

= 3:4 +

n⋅ d q

3 >444!L 1/16 ⋅ 2

prvi zakon termodinamike za proces u prigu{nom ventilu : ⋅

⋅

⋅

R 34 = ∆ I34 + X U34 + ∆Fl34 + ∆Fq34 ∆t23 = ∆t 34 U q 3 = q2 ⋅  3  U2 q U2 =  2 U3l  q 3l ηlq e =

  

⇒

d q mo

⋅

⇒

⋅

I3 = I 4

U3>U4

U q U3 q − S h mo 3 = d q mo 4 − S h mo 4 U2 q2 U3 q3

dq

⇒

2

 3⋅Sh  444  3⋅1/398  > 2/36 ⋅ 21 6 !Qb = 2 ⋅ 21 6 ⋅   3:4    κ −2 κ

⇒

⇒

U3l

q = U2 ⋅  3l  q2

  

κ −2 κ

q3l>q3

 2/36  = 3:4 ⋅    2 

2/5 −2 2/5

>423/4!L

U2 − U3l 3:4 − 423/4 >1/59 = U2 − U3 3:4 − 444


{fmlp@fvofu/zv


strana 74

2/68/!!U turbo kompresorskoj stanici se vr{i dvostepena kvazistati~ka adijabatska kompresija. Kompresor usisava 6!lnpm0i okolnog vazduha (idelan gas) stawa!2)q>2!cbs-!U>3:4!L* i sabija ga na neki me|u pritisak pri kojem se daqe hladi do temperature okoline. Nakon toga se vazduh sabija na kona~ni pritisak 5)q>:!cbs*. Zanemaruju}i promene potencijalne i kineti~ke energije, odrediti: a) vrednost me|u pritiska )q3>q4*!pri kojem su snage potrebne za oba stepena sabijawa jednake b) u{tedu u snazi u ovom procesu u odnosu na kompresiju bez me|uhla|ewa, tj. kada bi se kvazistati~ka adijabatska kompresija od stawa 2)q>2!cbs-!U>3:4!L* do pritiska od :!cbs!vr{ila u jednom stepenu 5

XU45 3 4 R34

XU23

2 a) prvi zakon termodinamike za proces u kompresoru niskog pritiska: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23

⋅

⋅

X U23 = n⋅ d q ⋅ (U2 − U3 ) !!!!!!!)2*

⇒

prvi zakon termodinamike za proces u kompresoru visokog pritiska: ⋅

⋅

⋅

R 45 = ∆ I45 + X U 45 + ∆Fl 45 + ∆Fq45

⇒

⋅

⋅

⋅

⋅

X U 45 = n⋅ d q ⋅ (U4 − U5 ) !!!!!!!)3*

Kombinovawem uslova zadatka X U23 = X U 45 !i!U2>U4!sa jedna~inama!)2*!i!)3* dobija se!U3>U5/ κ

q2  U2  κ −2 =  q3  U3 

κ

⇒

 U  κ −2 q3 = q2 ⋅  3   U2 

⇒

 U  κ −2 q4 = q5 ⋅  4   U5 

κ

q4  U4  κ −2 =  q5  U5 


)4*

κ

)5*

{fmlp@fvofu/zv


strana 75 q U5 = U2 ⋅  5  q2

deqewem jedna~ina!)4* i!)5*!dobija se:

κ −2

 3κ  

2/5 −2

 :  3⋅2/5 U5 = 3:4 ⋅   >512/15!L!>!U3  2 2/5

 512/15  2/5 −2 >! 4 ⋅ 21 6 Qb!>!q4 q 3 = 2 ⋅ 21 ⋅   3:4   6

⇒

iz jedna~ine!)4*! b) q U5( = U2 ⋅  5  q2

  

κ −2 κ

: > 3:4 ⋅    2

2/5 −2 2/5

>659/:3!L ⋅

⋅

⋅

(

)

X U > o⋅ Nd q ⋅ (U2 − U3 + U4 − U5 ) =

⋅ -

⋅ -

⋅

(

)

⋅

6 ⋅ 3:/2 ⋅ (3:4 − 512/15 ) >!−5/48!LX 4711 ⋅

X U = X U25 (

potrebna snaga u drugom slu~aju: X U > o⋅ Nd q ⋅ (U2 − U5 ( ) =

⋅

X U = X U23 + X U 45

potrebna snaga u prvom slu~aju:

6 ⋅ 3:/2 ⋅ (3:4 − 659/:3) >!−21/45!LX 4711 ⋅

⋅ -

⋅

∆ X U = X U − X U >−5/48,21/45!>!6/:8!lX

u{teda u snazi: 5′ U

zatvorena povr{ina!!)3−4−5−5′−3*!predstavqa u{tedu u ⋅

5

snazi ( ∆ X U *!koja je ostvarena dvostepenom kompresijom (u odnosu na jednostepenu kompresiju)

3

4

2 t


{fmlp@fvofu/zv


strana 76

2/69/!U dvostepenom kompresoru sa me|uhla|ewem, pri ustaqenim uslovima, sabija se neravnote`no (nekvazistati~ki) i adijabatski!1/4!lh0t!azota (idealan gas), od polaznog stawa!2)q2>1/2!NQb-!U2>3:4!L* do stawa!5)q5>1/7!NQb-!U5>561!L*/!Stepeni dobrote prilikom sabijawa u oba stepena su jednaki i iznose ηJe = ηJJe = 1/9 /!Odrediti ukupnu snagu za pogon kompresora i toplotnu snagu koja se odvodi pri hla|ewu (izme|u dve kompresije), ako se proces me|uhla|ewa odvija pri stalnom pritisku!q3>q4!>!1/4!NQb/ Zanemariti promene kineti~ke i potencijalne energije vazduha i prikazati sve procese na!Ut!dijagramu.

5 XU45

4 3 XU23

R34

2 κ −2  κ

κ −2  κ

U2  q2  = U3l  q3l 

⇒

U3l

U2 − U3l U2 − U3

⇒

U3 = U2 −

ηJe =

q U4 =  4 U5l  q 5l ηJJe =

  

κ −2 κ

U4 − U5l U4 − U5

q = U2 ⋅  3l   q2 

U2 − U3l

q = U4 ⋅  5l  q4

  

= 3:4 −

κ −2 κ

⇒

U5l

⇒

U5l = U4 − ηJJe ⋅ (U4 − U5 )

kombinovawem jedna~ina!)2*!i!)3*!dobija se:


ηJe

4 = 3:4 ⋅    2

!

2/5 −2 2/5

>512!L

3:4 − 512 = 539!L 1/9

)2*

)3*

U4>465!L-!U5l>542!L

{fmlp@fvofu/zv


strana 77 ⋅

⋅

⋅

R 34 = n⋅ (r34 )q=dpotu

odvedena toplota u fazi me|uhla|ewa )3−4*; ⋅

R 34 = n⋅ d q ⋅ (U4 − U3 ) = 1/4 ⋅ 2/15 ⋅ (465 − 539) >!−34/2!lX

prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ⋅

⋅

⋅

⋅

⋅

⋅

⋅

R 34 = ∆ I25 + X U25 + ∆Fl25 + ∆Fq25

ograni~enom isprekidanom linijom: ⋅

X U25 = R 34 − ∆ I25 = R 34 − n⋅ d q ⋅ (U5 − U2 ) > −34/2 − 1/4 ⋅ 2/15 ⋅ (561 − 3:4) ⋅

X U25 >−83/2!lX napomena:

ukupna snaga za pogon oba kompresora jednaka je zbiru snaga potrebnih za pogon kompresora niskog pritiska i kompresora ⋅

⋅

⋅

visokog pritiska tj:! X U25 = X U23 + X U 45


{fmlp@fvofu/zv


strana 78

PRVI I DRUGI ZAKON TERMODINAMIKE (PUWEWE I PRA@WEWE REZERVOARA) 2/6:/!Vazduh (idealan gas) stawa!)q>2!cbs-!U>3:1!L) nalazi se u toplotno izolovanom rezervoaru zapremine!W>1/7!n4/!Toplotno izolovanim cevovodom u rezervoar se uvodi vazduh (idealan gas) stawa )q>21!cbs-!U>511!L). Tokom procesa u rezervoaru je stalno ukqu~en greja~ snage!3/6!lX/!Kada vazduh u rezervoaru dostigne stawe!)q>5!cbs-!U>611!L*!prekida se dotok vazduha i iskqu~uje greja~. Odrediti vreme trajawa procesa puwewa rezervoara kao i maseni protok vazduha koji se uvodi u rezervoar. q>2!cbs-!U>3:1!L q>5!cbs-!U>611!L q>21!cbs-!U>511!L

po~etak: kraj: ulaz:

jedna~ina stawa idealnog gasa za po~etak: nqpd =

q qpd ⋅ W S h ⋅ Uqpd

=

2 ⋅ 21 6 ⋅ 1/7 >1/83!lh 398 ⋅ 3:1 q ls ⋅ W = nls ⋅ S h ⋅ Uls

jedna~ina stawa idealnog gasa za kraj: nls =

q qpd ⋅ W = nqpd ⋅ S h ⋅ Uqpd

q ls ⋅ W 5 ⋅ 21 6 ⋅ 1/7 = >2/78!lh S h ⋅ Uls 398 ⋅ 611 nqp + nvm = nls + nj{

materijalni bilans procesa puwewa: nvm = nls − nqp >2/78!−!1/83!>!1/:6!lh prvi zakon termodinamike za proces puwewa: R 23 − X23 = Vls − Vqp + Ij{ − Ivm

R 23 = nls ⋅ d w ⋅ Uls − nqp ⋅ d w ⋅ Uqp − nvm ⋅ d q ⋅ Uvm

R 23 = 2/78 ⋅ 1/83 ⋅ 611 − 1/83 ⋅ 1/83 ⋅ 3:1 − 1/:6 ⋅ 2 ⋅ 511 >81/97!lK τ=

R 23 ⋅

R 23 ⋅

nvm =

=

81/97 >39/4!t 3/6

nvm 1/:6 h = >44/7! τ 39/4 t


{fmlp@fvofu/zv


strana 79

2/71/!U verikalnom toplotno izolovanom cilindru-!povr{ine popre~nog preseka B>1/2!n3-! nalazi se vazduh (idealan gas) stawa )U>291pD-!n>1/16!lh), ispod toplotno izolovanog klipa mase koja odgovara teìni od!31!lO-!a na koji spoqa deluje atmosferski pritisak od!1/2!NQb!(slika). U cilindar se, kroz toplotno izolovan cevovod, naknadno uvede vazduh stawa!)q>1/5!NQb-!U>651pD-!n>1/2!lh*!{to dovede do pomerawa klipa (bez trewa)/!Zanemaruju}i promene kineti~ke i potencijalne energije uvedenog vazduha odrediti koliki rad izvr{i vazduh nad okolinom kao i temperaturu vazduha u cilindru na kraju procesa. kraj ∆z po~etak ulaz Gufh

= 2 ⋅ 21 6 +

po~etak:

U>564!L-!q> q p +

kraj: ulaz:

q>4!cbs q>5!cbs-!U>924!L-!n>1/2!lh

B

jedna~ina stawa idealnog gasa za po~etak: Wqpd =

nqpd ⋅ S h ⋅ Uqpd

=

1/16 ⋅ 398 ⋅ 564

4 ⋅ 21 6 materijalni bilans procesa puwewa: q qpd

31 ⋅ 21 4 >4!cbs-!n>1/16!lh 1/2

q qpd ⋅ Wqpd = nqpd ⋅ S h ⋅ Uqpd

>1/1328!n4 nqp + nvm = nls + nj{

nls = nqp + nvm >1/16!,!1/2!>!1/26!lh

q ls ⋅ Wls = nls ⋅ S h ⋅ Uls !!!)2*

jedna~ina stawa idealnog gasa za kraj: prvi zakon termodinamike za proces puwewa: R 23 − X23 = Vls − Vqp + Ij{ − Ivm

−X23 = nls ⋅ d w ⋅ Uls − nqp ⋅ d w ⋅ Uqp − nvm ⋅ d q ⋅ Uvm

)3* Wls

X23 =

∫()

(

q W ⋅ eW = q qpd ⋅ Wls − Wqpd

)

!!!)4*

Wqpd


{fmlp@fvofu/zv


strana 80

kada se jedna~ine )2*!i )4* uvrste u jedna~inu )3* dobija se:

(

)

− q qpd ⋅ Wls − Wqpd = nls ⋅ d w ⋅ Wls =

Wls =

q ls ⋅ Wls − nqp ⋅ d w ⋅ Uqp − nvm ⋅ d q ⋅ Uvm nls ⋅ S h

⇒

nqp ⋅ d w ⋅ Uqp + nvm ⋅ d q ⋅ Uvm + q qpd ⋅ Wqpd q qpd +

dw ⋅ q lsbk Sh

1/16 ⋅ 1/83 ⋅ 564 + 1/2 ⋅ 2 ⋅ 924 + 4 ⋅ 21 6 ⋅ 21 −4 ⋅ 1/1328 >1/1:9:!n4 1/83 6 6 −4 ⋅ 4 ⋅ 21 4 ⋅ 21 ⋅ 21 + 398

)2*!!!⇒ ! Uls =

q ls ⋅ Wls 4 ⋅ 21 6 ⋅ 1/1:9: = >79:/3!L nls ⋅ S h 1/26 ⋅ 398

(

)

)4*!!!⇒! ! X23 = qqpd ⋅ Wls − Wqpd = 4 ⋅ 216 ⋅ 21 −4 ⋅ (1/1:9: − 1/1328) >34/27!lK 2/72/!Kroz toplotno izolovan cevovod, unutra{weg pre~nika!e>21!nn*-!biva uveden azot!)O3-!idealan gas*!stawa!)U>416!L-!q>1/7!NQb-!x>21!n0t) u toplotno izolovan rezervoar zapremine!W>1/7!n4!!u kojem se ve} nalazi ugqen−dioksid!)DP3-!idealan gas) stawa!)q>1/2!NQb-!U>3:4!L*/!Ako se proces puwewa prekida kada pritisak sme{e u rezevoaru dostigne!1/6!NQb-!odrediti: a) temperaturu me{avine idealnih gasova u rezervoaru na kraju procesa puwewa b) promenu entropije sistema za vreme procesa puwewa c) vreme trajawa procesa puwewa rezervoara po~etak: kraj: ulaz:

q>2!cbs-!U>3:4!L-!DP3 q>6!cbs-!DP3!+!O3 q>7!cbs-!U>416!L-!x>21!n0t-!!O3

a) prvi zakon termodinamike za proces puwewa: R 23 − X23 = Vls − Vqp + Ij{ − Ivm  x3  !!)2* 1 = nDP3 ⋅ dwDP3 + nO3 ⋅ dwO3 ⋅ Uls − nDP3 ⋅ dwDP3 ⋅ Uqp − nO3 ⋅  dqO3 ⋅ Uvm +  3  

(

)

jedna~ina stawa me{avine idealnih gasova za zavr{etak puwewa: q ls ⋅ W = )nDP3 ⋅ S hDP3 + nO3 ⋅ S hO3 * ⋅ Uls !!!! jedna~ina stawa idealnog gasa za )DP3*!po~etak:

!!!!!!!)3*

q qpd ⋅ W = nDP3 ⋅ S hDP3 ⋅ Uqpd

!!!!!!!)4*


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike ⇒

)4*

nDP3 =

q qpd ⋅ W S hDP3 ⋅ Uqpd

strana 81 =

2 ⋅ 21 6 ⋅ 1/7 >2/19!lh 29: ⋅ 3:4

kombinovawem jedna~ina!)2*!i!)3*!dobija se:!

Uls>494/7!L-! nO3 >2/:5!lh

b) ∆Ttjtufn!>!∆Tubeop!ufmp!,!∆Tplpmjob!>!///!>:47/4! ∆Tplpmjob>1!

K L

K ! (sud adijabatski izolovan od okoline) L

∆Tsbeop!ufmp!> ∆TDP3 + ∆TO3 >!///>!2:3!,!855/4!>!:47/4!

K L

 U W 494/7 K ∆T DP3 >g)U-!W*> nDP3 ⋅  dwDP3 mo ls + ShDP3 mo  = 2/19 ⋅ 1/77 ⋅ mo >2:3!   L Uqp W 3:4  O3  q lsbk U  ∆TO3 >g)U-!Q*> nO3 ⋅  d qO3 mo ls − S hO3 mo Uvm q vm  

   > 

494/7 4/79  K  − 1/3:8 ⋅ mo ∆TO3 > 2/:5 ⋅ 2/15 ⋅ mo  >///>!855/4! L 416 7  

jedna~ina stawa idealnog gasa )O3* za kraj: 3 qO ls =

nO3 ⋅ ShO3 ⋅ Uls W

=

O

q ls3 ⋅ W = nO3 ⋅ S hO3 ⋅ Uls

2/:5 ⋅ 3:8 ⋅ 494/7 > 4/79 ⋅ 216 Qb 1/7

c) τ=

nO3 ⋅

nO3 ⋅

>///>

2/:5 6/3 ⋅ 21−4

nO3 = ρvmb{ ⋅ x ⋅

ρ vmb{ =

>484/2!t

e3π 1/123 π lh >///> 7/73 ⋅ 21 ⋅ > 6/3 ⋅ 21−4 5 5 t

q vmb{ lh 7 ⋅ 21 6 = >7/73! 4 S hO3 ⋅ Uvmb{ 3:8 ⋅ 416 n


{fmlp@fvofu/zv


strana 82

2/73/ U rezervoaru zapremine!W>1/4!n4!nalazi se azot stawa!)q>231!cbs-!U>411!L*/!Ventil na rezervoaru se brzo otvori, ispusti se izvesna koli~ina azota u atmosferu a zatim se ventil ponovo zatvori, tako da se moè smatrati da pri takvim uslovima nema razmene toplote izme|u rezervoara i okoline. Ne posredno po zatvarawu ventila pritisak azota u rezervoaru iznosi!q>71!cbs/!Odrediti: a) koli~inu azota koja je istekla iz rezervoara )lh* kao i temperaturu azota u rezervoaru neposredno posle zatvarawa ventila c* koli~inu toplote koju bi trebalo dovesti preostalom vazduhu u sudu da bi dostigao pritisak koji je je imao pre pra`wewa )NK* b* Promena stawa azota koji se za vreme procesa pra`wewa nalazi u sudu je kvazistai~ka adijabatska promena, pa se temperatura azota u sudu na kraju procesa praèwa moè odrediti iz zakona kvazistati~ke adijabatske promene: Uls  qls  = Uqp  qqp 

κ −2 κ

q  Uls = Uqp ⋅  ls   qqp   

⇒

κ −2 κ

 71  = 411 ⋅    231 

jedna~ina stawa idealnog gasa za po~etak: nqpd =

qqpd ⋅ W Sh ⋅ Uqpd

=

>357/2!L

q qpd ⋅ W = nqpd ⋅ S h ⋅ Uqpd

231 ⋅ 216 ⋅ 1/4 >51/5!lh 3:8 ⋅ 411

jedna~ina stawa idealnog gasa za kraj: nls =

2/5 −2 2/5

q ls ⋅ W = nls ⋅ S h ⋅ Uls

q ls ⋅ W 71 ⋅ 21 6 ⋅ 1/4 = >35/74!lh S h ⋅ Uls 3:8 ⋅ 357/2

nqp + nvm = nls + nj{

materijalni bilans procesa pra`wewa: nj{ = nqp − nls >51/5!−!35/74!>!26/88!lh

b) 2!>!kraj 3 !

)n>35/74!lh-!q>71!cbs-!U>357/2!L* )n>35/74!lh-!q>231!cbs-!U>@*

jedna~ina stawa idealnog gasa za stawe 2: U3 =

q3 ⋅ W = n ⋅ Sh ⋅ U3

⇒

6

q3 ⋅ W 231 ⋅ 21 ⋅ 1/4 = >5:3/2!L n ⋅ Sh 35/74 ⋅ 3:8

R23 = n ⋅ (r23 )w =dpotu = n ⋅ dw ⋅ (U3 − U2) > 35/74 ⋅ 1/85 ⋅ (5:3/2 − 357/2) >5/59!NK


{fmlp@fvofu/zv


strana 83

ME[AVINE IDEALNIH GASOVA 2/74. Za me{avinu idealnih gasova, kiseonika )B*!i azota!)C), odredititi molsku masu!)Nn*-!gasnu konstantu!)Shn*-!specifi~ne toplotne kapacitete pri stalnom pritisku!)dqn) i pri stalnoj zapremini )dwn!*-!eksponent izentropske promene stawa!)κn*!kao i parcijalne pritiske komponenata!B!i!C!ako se me{avina nalazi na qn>2!cbs!i ako je sastav me{avine zadat na slede}i na~in: b* hB>1/7-!hC>1/5 c* sB>1/3-!sC>1/9 b* N hN =

2 hB

+

NB

Shn

hC NC

=

2 1/7 1/5 + 43 39

(NSh ) = 9426 >385/7:! = Nn

41/38

>41/38!

lh lnpm

K lhL

d Qn = h B ⋅ d QB + hC ⋅ d QC = 1/7 ⋅ 1/:2 ⋅ 21 4 + 1/5 ⋅ 2/15 ⋅ 21 4 >:73!

d wn = d qn − S hn = :73 − 385/7: >798/42 κn =

d qt d wn

qB = hB ⋅

=

K lhL

K lhL

:73 >2/5 798/42

NN 41/38 ⋅ q n = 1/7 ⋅ ⋅ 2 >1/68!cbs NB 43

q C = q n − q B = 2 − 1/68 >1/54!cbs b) Nn = sB ⋅ N B + sC ⋅ NC = 1/3 ⋅ 43 + 1/9 ⋅ 39 = 39/9!

lh lnpm

S hn =

(NS h ) Nn

=

9426 >399/83 39/9

K lhL dqn = sB ⋅

K NB N 43 39 ⋅ dqB + sC ⋅ C ⋅ dqC = 1/3 ⋅ ⋅ 1/:2 + 1/9 ⋅ ⋅ 2/15 >2122/22 NN NN 39/9 39/9 lhL

d wn = d qn − S hn = 2122/22 − 399/83 >833/4: κn =

d qt d wn

=

K lhL

2122/22 >2/5 833/4:

q B = sB ⋅ q n = 1/3 ⋅ 2 >1/3!cbs q C = q n − q B = 2 − 1/3 >1/9!cbs


{fmlp@fvofu/zv


strana 84

2/75/!Me{avina idealnih gasova-!n>2!lh-!sastoji se od azota!)B), zapreminskog udela!51& i metana!)C*zapreminskog udela!)71%). Me{avina se zagreva od temperature!U2>411!L!do temperature!U3>711!L!na dva na~ina. Prvi put je promena stawa kvazistati~ki izohorska, a drugi put se odvija kvazistati~ki po zakonu prave linije u!Ut!koordinatnom sistemu. U oba slu~aja po~etna i krajwa stawa radnog tela su jednaka. Skicirati promene stawa na Ut dijagramu i odrediti: a) zapreminski rad )lK* du` promene!2−2 koja se odvija po zakonu prave linije b) promenu entropije izolovanog sistema!)lK0L*!koji ~ine radna materija i toplotni izvor stalne temperature!UUJ>U3!za slu~aj izohorske promene stawa U 3

2

t Nn = sB ⋅ N B + sC ⋅ NC = 1/5 ⋅ 39 + 1/7 ⋅ 27 = 31/9! dwn = sB ⋅

lh lnpm

NB N 39 27 lK ⋅ dwB + sC ⋅ C ⋅ dqC = 1/5 ⋅ ⋅ 1/85 + 1/7 ⋅ ⋅ 2/93 >2/35! NN NN 31/9 31/9 lhL

a) t3

R 23 = n ⋅

∫

U (t ) ⋅ et = n ⋅

U2 + U3 U + U3 ⋅ ∆t23 = n ⋅ 2 3 3

t2

 U w ⋅  d wn ⋅ mo 3 + S hn mo 3 U w2 2 

  

711 + 411 711 R 23 = 2 ⋅ ⋅ 2/35 ⋅ mo >497/89!lK 3 411 ∆V23 = n ⋅ ∆v23 = n ⋅ d wtn ⋅ (U3 − U2 ) = 2 ⋅ 2/35 ⋅ (711 − 411) >483!lK prvi zakon termodinamike za proces 2−3 koji se odvija po pravoj liniji: R23!>!∆V23!,!X23


⇒

X23!>!R23!−!∆V23!>!497/89−483>25/89!lK

{fmlp@fvofu/zv


strana 85

b) ∆TTJ!>!∆TSU!,!∆TUJ!>!///>!96:/6!−!731!>34:/6! ∆TSU!>n!/!∆t23!>! n ⋅ d wtnmo ∆TUJ!>! −

K L

U3 711 K = 2 ⋅ 2/35 ⋅ mo >96:/6! U2 411 L

483 R23 K >!!−731! >///> − UUJ 711 L

R 23 = n ⋅ (r23 )w =dpotu = n ⋅ d wn ⋅ (U3 − U2 ) = 2 ⋅ 2/35 ⋅ (711 − 411) >483!lK

2/76/![upqa kugla zanemarqive mase unutra{weg pre~nika!e>2!n!sastavqena je od dve polovine koje su tesno priqubqene (slika). U kugli se nalazi me{avina idealnih gasova vodonika )B*ugqen−dioksida )C* i azota )D* sastava sB>1/46-!sC>1/5!j!sD>1/36!stawa 2)q>1/3!cbs-!U>3:4!L*/ Na dowoj polovini kugle obe{en je teret mase nU>5111!lh. Pritisak okoline iznosi qp>2!cbs. Odrediti koliko toplote treba dovesti me{avini idealnih gasova u kugli da bi se polovine mogle razdvojiti.

Nn = sB ⋅ N B + sC ⋅ NC + sD ⋅ ND = 1/46 ⋅ 3 + 1/5 ⋅ 55 + 1/36 ⋅ 39 = 36/4! S hn =

(NS h ) Nn

d wn = sB ⋅

=

lh lnpm

K 9426 >439/77! lhL 36/4

N NB N ⋅ d wB + sC ⋅ C ⋅ d qC + sD ⋅ D ⋅ d qD NN NN NN

d wn = 1/46 ⋅

3 55 39 lK ⋅ 21/5 + 1/5 ⋅ ⋅ 1/77 + 1/36 ⋅ ⋅ 1/85 >1/:6! 36/4 36/4 36/4 lhL

zapremina lopte:


W>

e 4 π 24 ⋅ π = >1/6347!n4 7 7

{fmlp@fvofu/zv


strana 86

jedna~ina stawa me{avine idealnih gasova stawa!)2*;! q2 ⋅ W = n ⋅ S hn ⋅ U2 n=

q2 ⋅ W 1/3 ⋅ 21 6 ⋅ 1/6347 = >1/22!lh S hn ⋅ U2 439/77 ⋅ 3:4

jedna~ina stati~ke ravnoteè neposredno pred odvajawe dowe polovine n ⋅h q 3 + 3u > q p !!! !!⇒ (stawe 2): e ⋅π 5 n ⋅h 5 ⋅ 21 4 ⋅ :/92 = 2 ⋅ 21 6 − q3!>! q p − 3u > 1/6 ⋅ 21 6 Qb e ⋅π 23 ⋅ π 5 5 jedna~ina stawa me{avine idealnih gasova stawa!)3*;!!! q 3 ⋅ W = n ⋅ S hn ⋅ U3 U3 =

q3 ⋅ W 1/6 ⋅ 21 6 ⋅ 1/6347 = >835/26!L n ⋅ S hn 1/22 ⋅ 439/77

R23 = n ⋅ (r23 )w =dpotu = 1/22⋅ 1/:6 ⋅ (U3 − U2 ) = 1/33 ⋅ 1/:6 ⋅ (835/26 − 3:4) >56/4!lK

2/77/!Za situaciju u proto~nom ure|aju za me{awe (prikazanu na slici) koji radi pri stacionarnim uslovima, odrediti da li se u sistem dovodi mehani~ka snaga ili se iz sistema mehani~ka snaga odvodi i izra~unati wenu vrednost!)lX*/ ⋅

R 23 = −21 lX

⋅

struja 1: x2>311!n0t U2>611pD n2>21!u0i

X U23 = @ struja 2: x3>1!n0t U3>531pD n3>7!u0i

molski sastav: B; DP3>61& C; O3>61&

molski sastav: D; O3>71& E; P3>51&

me{avina: x+>41!n0t U+>611pD


{fmlp@fvofu/zv


strana 87

struja 1: Nn2 = sB ⋅ NB + sC ⋅ NC = 1/6 ⋅ 55 + 1/6 ⋅ 39 = 47!

lh lnpm

⋅

⋅

o2 =

n2 21 ⋅ 21 4 lnpm(B + C) = >388/89! i Nn2 47

⋅

⋅

lnpmB i ⋅ ⋅ lnpmC oC = sC ⋅ o2 = 1/4 ⋅ 382/85 >249/9:! i o B = sB ⋅ o2 = 1/6 ⋅ 388/89 >249/9:!

struja 2: Nn3 = sD ⋅ ND + sE ⋅ NE = 1/7 ⋅ 39 + 1/5 ⋅ 43 >3:/7!

lh lnpm

⋅

n3 7 ⋅ 21 4 lnpm(D + E ) o3 = = >313/81! Nn3 3:/7 i ⋅

⋅

⋅

lnpmD i ⋅ ⋅ lnpmE oE = sE ⋅ o3 = 1/5 ⋅ 313/81 >92/19! i

o D = sD ⋅ o 3 = 1/7 ⋅ 313/81 >232/73!

prvi zakon termodinamike za proces me{awa fluidnih struja: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 ⋅

⋅

⋅

⋅

⋅

X U23 = R 23 − ∆ I23 − ∆Fl23

⇒

⋅

∆ I23 = I3 − I2 >///>435:/88!−!4229/7:!>!242/19!lX ⋅ ⋅ ⋅ ⋅  ⋅  I2 > o B ⋅ Nd q B + oC ⋅ Nd q C  ⋅ U2 + oD ⋅ Nd q D + oE ⋅ Nd q E  ⋅ U3     ⋅ 249/9: 92/19 249/9:  232/73  I2 >  ⋅ 48/5 + ⋅ 3:/2 ⋅ 884 +  ⋅ 3:/2 + ⋅ 3:/2 ⋅ 7:4 4711 4711 4711 4711    

(

)

(

)

(

)

(

)

⋅

I2 >4229/7:!lX


{fmlp@fvofu/zv


strana 88

⋅ ⋅ ⋅ ⋅  ⋅  I3 > o B ⋅ Nd q B + oC ⋅ Nd q C  ⋅ U + + oD ⋅ Nd q D + oE ⋅ Nd q E  ⋅ U +     ⋅ 249/9: 92/19 249/9:  232/73  I3 >  ⋅ 48/5 + ⋅ 3:/2 ⋅ 884 +  ⋅ 3:/2 + ⋅ 3:/2 ⋅ 884 4711 4711 4711 4711    

(

)

(

)

(

)

(

)

⋅

I3 >435:/88!lX 2 3 ∆Fl23 = ∆Fltusvkb + ∆Fltusvkb >///>−65/42!,!1/86!>−64/67!lX 23 23

( )

⋅ ⋅  x+ ∆F tusvkb2 =  o B ⋅ N B + oC ⋅ NC  ⋅ l23  

3

− (x 2 )3 3

3 3 249/9:  249/9:  41 − 311 >−65/42!lX ∆F tusvkb2 =  ⋅ 55 + ⋅ 39  ⋅ l23 4711 3  4711 

∆F tusvkb3 l23

( )

⋅ ⋅  x+ =  o D ⋅ ND + oE ⋅ NE  ⋅  

3

− (x 3 )3 3

3 3 92/19  232/73  41 − 1 ∆F tusvkb3 =  ⋅ 39 + ⋅ 43  ⋅ >,1/86!lX l23 4711 3  4711  ⋅

X U23 >!−21!−!242/19!,!65/42!>!−97/88!lX ⋅

Po{to je vrednost za X U23 negativan broj to zna~i da se u sistem dovodi

mehani~ka snaga.

⋅

2/78/ Vazduh (idealna gas) po~etne temperature!Uw2>61pD-!masenog protoka! n w>3!lh0t!zagreva se u rekuperativnom razmewiva~u toplote na ra~un hla|ewa me{avine idealnih gasova DP3!i!TP3!od ⋅

Un2>511pD!do!Un3>351pD. Maseni protok me{avine idealnih gasova je! n n>4!lh0t-!a maseni udeo!DP3!u me{avini je!91&/!Razmewiva~ toplote je toplotno izolovan od okoline. Pokazati da je proces razmene toplote u razmewiva~u toplote nepovratan. Zanemariti promene kineti~ke i potencijalne energije gasnih struja kao i padove pritiska gasnih struja pri strujawu kroz razmewiva~ toplote.

DP3!,!TP3

vazduh


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike hDP3 = 1/9

⇒

strana 89

hTP3 = 2 − hDP3 = 1/3

d qn = hDP3 ⋅ d qDP3 + hTP3 ⋅ d qTP3 = 1/9 ⋅ 1/96 + 1/3 ⋅ 1/69 >1/9!

lK lhL

prvi zakon termodinamike za proces u razmewiva~u toplote: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 ⋅

⋅

⋅

⇒

⋅

⋅

I2 > I3 ⋅

n w ⋅ d qw ⋅ Uw2 + nn ⋅ d qn ⋅ Un2 = n w ⋅ d qw ⋅ Uw3 + nn ⋅ d qn ⋅ Un3 ⋅

Uw 3 =

⋅

⋅

n w ⋅ d qw ⋅ Uw2 + nn ⋅ d qn ⋅ Un2 − nn ⋅ d qn ⋅ Un3 ⋅

n w ⋅ d qw Uw3 =

⋅

3 ⋅ 2 ⋅ 434 + 4 ⋅ 1/9 ⋅ 784 − 4 ⋅ 1/9 ⋅ 624 >626!L 3 ⋅2 ⋅

⋅

∆ T TJ = ∆ TSU + ∆ T plpmjob >//!/>1/39!

⋅

∆ T plpmjob = −

⋅

⋅

⋅ R 23 lX =1 Up L

lX L

(razmewiva~ toplote je izolovan od okoline)

⋅

lX ! L q  626 lX − S hw mo w3  > 3 ⋅ 2 ⋅ mo >1/:4! q w2  434 L

∆ TSU = ∆ T w + ∆ Tn = /// >1/:4!−!1/76!>!1/39! / /  ⋅ U ∆ T w = n w ⋅ g (q- U ) = n w ⋅  d qw mo w3 Uw2 

/ /  ⋅ U q ∆ T n = nn ⋅ g (q- U ) = nn ⋅  d qn mo n3 − S hn mo n3 Un2 q n2  napomena:

 624 lX  > 4 ⋅ 1/9 ⋅ mo >−1/76! 784 L 

po{to je promena entropije sistema pozitivan broj ⋅

( ∆ T TJ > 1 ) to zna~i da je proces razmene toplote u razmewiva~u toplote nepovratan


{fmlp@fvofu/zv


strana 90

2/79/!!Me{avina idealnih gasova (kiseonik i ugqen−dioksid), pogowena kompresorom snage!2:!lXstruji kroz kanal. Usled neidealnog izolovawa kanala i kompresora okolini se predaje!2/39!lX ⋅

toplote. Zapreminski protok i temperatura me{avine na ulazu u kanal iznose! W 2>1/26!n40t!i U2>486!L. Na izlazu iz kanala, pri pritisku!q>3!cbs-!zapreminski protok i temperatura me{avine ⋅

iznose! W 3>1/22!n40t!i!U3>586!L/!Zanemaruju}i promene kineti~ke i potencijalne energije me{avine idealnih gasova, odrediti: a) masene udele komponenata u me{avini b) promeu entropije sistema u navedenom procesu, ako temperatura okoline iznosi Up>3:4!L a) R23

XU23 2 3 jednan~ina stawa me{avine idealnih gasova na izlazu iz kanala: ⋅

(

⋅

)

⇒

q 3 ⋅ W 3 = o⋅ NS h ⋅ U3

⋅

q3 ⋅ W 3 3 ⋅ 21 6 ⋅ 1/22 lnpm o= >!6/68/21−4! = NS h ⋅ U3 9426 ⋅ 586 t ⋅

(

)

prvi zakon termodinamike za proces u kanalu: ⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23

(Ndq )n =

⋅

⋅

R 23 − X U23 ⋅

o⋅ (U3 − U2 )

=

− 2/39 + 2: 6/68 ⋅ 21

−4

⋅

(

⇒!!!!!! R 23 = o⋅ Nd q ⋅ (586 − 486)

>42/92!

lK lnpmL

(Ndq )n = sP3 ⋅ (Ndq )P3 + sDP3 ⋅ (Ndq )DP3

)2*

sP3 + sDP3 = 2

)3*

Kombinovawem jedna~ina!)2* i!)3*!dobija se:!

sP3 >1/7:-!!! sDP3 >1/42

hP3 =

sP3

⋅

)n ⋅ (U3 − U2 ) + X U23

sP3 ⋅ NP3 1/7: ⋅ 43 >1/7 = ⋅ NP3 + sDP3 ⋅ NDP3 1/7: ⋅ 43 + 1/42 ⋅ 55

hDP3 = 2 − hP3 = 2 − 1/7 >!1/5


{fmlp@fvofu/zv


strana 91

b) jedna~ina stawa me{avine idealnih gasova na ulazu u kanal: ⋅

(

⋅

⋅

)

q2 ⋅ W 2 = o⋅ NS h ⋅ U2

q2 =

⇒

(

)

o⋅ NS h ⋅ U2 ⋅

W2 q2 =

6/68 ⋅ 21

⋅

⋅

−4

⋅ 9426 ⋅ 486 > 2/27 ⋅ 21 6 !Qb 1/26 ⋅

∆ T TJ = ∆ TSU + ∆ T plpmjob >//!/>5/48!,!27/76!>32/13!

X L

⋅

R 23 − 2/39 X =− >5/48! Up 3:4 L / / ⋅ ⋅  U q ∆ T SU = ∆ T n > o⋅ g (q- U ) = o⋅  Nd qn ⋅ mo 3 − NS hn ⋅ mo 3 U2 q2  ⋅ 586 3   − 9/426 ⋅ mo ∆ T n > 6/68 ⋅ 21 −4 ⋅  42/92 ⋅ mo  > 27/76 486 2/27   ∆ T plpmjob = −

(

zadatak za ve`bawe:

)

(

)

   X L

)2/7:*

2/7:/!U ~eli~noj boci zapremine W>1/2!n4!nalazi se vazduh ) sP3 >1/32-! sO3 >1/8:* okolnog stawa P)qp>2!cbs-!Up>3:4!L*/ Boca se puni ugqen−dioksidom. Odrediti: a) koliko se lh!DP3!treba ubaciti u bocu, da bi molski udeo kiseonika u novonastaloj me{avini bio 6& i koliki je tada pritisak me{avine u boci pri temperaturi od 3:4!L b) koliko toplote treba dovesti da se me{avina u boci zagreje na 564!L

a)

nDP3 > :/6 ⋅ 21 −4 !lh-!!!q>2/16!cbs

b) R23>25/76!lK


{fmlp@fvofu/zv


strana 92

POLUIDEALNI GASOVI 2/81/!Tokom kvazistati~ke promene stawa 2−2 kiseoniku (poluidealan gas) mase!n>1/4!lh, po~etnog stawa!2)U2>484!L-!q2) predaje se toplota pri ~emu se kisonik zagreje do!U3>784!L. Specifi~ni toplotni kapacitet kiseonika tokom ove promene stawa mewa se po zakonu: d23 U = 1/: + 2 ⋅ 21 −4 [lK 0)lhL] [L ] Od stawa 2 kiseonik kvazistati~ki izotermski mewa stawe do stawa!4!)q4>q2*/!Odrediti koli~inu toplote!)lK*!koja se kisoniku preda: b* tokom procesa!2−3 c* tokom procesa!3−4 a) U3

R 23 = n ⋅

U3

∫()

d U ⋅ eU = n ⋅

U2

(

)

∫

 1/: + 2 ⋅ 21 − 4 U  ⋅ eU  

U2

(

)

R 23 = n ⋅ 1/: ⋅ U3 − U2 + 6 ⋅ 21 − 5 ⋅ U33 − U23   

(

)

R 23 = 1/4 ⋅ 1/: ⋅ (784 − 484) + 6 ⋅ 21 − 5 ⋅ 784 3 − 484 3  >239/18!lK  

b) U3

∆t23 =

∫

d(U ) ⋅ eU = U

U3

U2

∆t23 = 1/: ⋅ mo U3

dq U2

dq

U3 U2

=

−4

3

U

U2

+ 2 ⋅ 21 −4 ⋅ (U3 − U2 )

U2

784 lK + 2 ⋅ 21 −4 ⋅ (784 − 484) >1/94! 484 lhL

 2 ⋅  dq U3 − U2 

>2/129!

∫

(1/: + 2 ⋅ 21 U ) ⋅ eU = 1/: ⋅ mo U

U2  2 ⋅ (1/:757 ⋅ 784 − 1/:329 ⋅ 484) > ⋅ U3 − dq ⋅ U2  =  12 U2  784 − 484

U3

lK lhL

 U3 U q  ⇒ ∆t23 =  d q mo 3 − S h mo 3  !!  U  U q 2 2 2   q3 lK 784 S h mo = 2/129 ⋅ mo − 1/94 >−1/34 lhL q2 484


! S h mo

q3 = dq q2

U3 U2

mo

U3 − ∆t23 U2

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike t4

R 34 = n ⋅

∫

t3

strana 93

 U (t ) ⋅ et = n ⋅ U3 ⋅ ∆t 34 = n ⋅ U3 ⋅  d q  

U q mo 4 − S hmo 4 U q3 U3 3

U4

   

q3 = 1/4 ⋅ 784 ⋅ (− 1/34) >−57/55!lK q2

R 34 = n ⋅ U3 ⋅ S hmo

2/82/!Zavisnost molarnog toplotnog kapaciteta od temperature za neki poluidealan gas, pri stalnom D qN  U  pritisku, data je izrazom: = 3:/3 + 5/18 ⋅ 21 −4  − 384  [K 0 npmL]  [L ]  b* odrediti koli~inu toplote koju treba predati gasu da bi se on zagrejao od polazne temperature U2>3:1!L!do temperature!U3>684!L-!ako se predaja toplote vr{i pri stalnom pritisku!q>2/6!NQb-!a posle izobarskog {irewa gas zauzima zapreminu!W3>1/9!n4 b) odrediti koli~inu toplote koju je potrebno predati istoj koli~ini istog gasa, da bi se on zagrejao od iste polazne temperature!U2!do iste temperature!U3-!ako gas biva zagrevan pri stalnoj zapremini b*

(

o=

)

q 3 ⋅ W3 = o ⋅ NS h ⋅ U3

jedna~ina stawa idealnog gasa stawa!3;! q 3 ⋅ W3 2/6 ⋅ 21 7 ⋅ 1/9 = >1/36!lnpm NS h ⋅ U3 9426 ⋅ 684

(

)

U3

(R23 )q=dpotu

= o⋅

∫

U3

D qN (U ) ⋅ eU = o ⋅

U2

∫[

]

3:/3 + 5/18 ⋅ 21 −4 ⋅ (U − 384) ⋅ eU

U2









3

(R23 )q=dpotu = o ⋅ 3:/3 ⋅ (U3 − U2) + 5/18 ⋅ 21−4 ⋅  U3

 − U23  − 5/18 ⋅ 21−4 ⋅ 384 ⋅ (U3 − U2)  3  









3

(R23 )q=dpotu = 1/36 ⋅ 3:/3 ⋅ (684 − 3:1) + 5/18 ⋅ 21−4 ⋅  684

 − 3:13  − 2/22 ⋅ (684 − 3:1)  3  

(R23 )q=dpotu >3/22!NK


{fmlp@fvofu/zv


strana 94

b) U3

(R23 )w =dpotu

= o⋅

∫

U3

D wN (U ) ⋅ eU = o ⋅

U2

= o⋅

(

)]

D qN (U ) − NS h ⋅ eU

U2

U3

(R 23 )w =dpotu

∫[

∫[

(

)]

(

)

D qN (U ) − NS h ⋅ eU = (R 23 )q=dpotu − o ⋅ NS h ⋅ (U3 − U2 )

U2

(R 23 )w =dpotu

= 3/22 ⋅ 21 7 − 1/36 ⋅ 9426 ⋅ (684 − 3:1) = 2/63 NK

⋅

2/83/!Vazduh (poluidealan gas), masenog protoka! n w>1/3!lh0t, po~etne temperature!Uw>711pD-!pri konstantnom pritisku, struji kroz adijabatski izolovanu cev u kojoj se hladi kiseonikom (poluidealan gas) koji struji kroz cevnu zmiju, a zatim se i me{a sa jednim delom ovog kiseonika (slika). Temperatura tako nastale me{avine iznosi UN>411pD-!a maseni udeo kiseonika u toj sme{i je!hC>1/6. Temperatura kiseonika na ulazu u cev je!UL2>31pD-!a na izlazu iz cevi!UL3>311pD/!Pritisak kiseonika je stalan. Odrediti maseni protok kiseonika kojim se vr{i hla|ewe vazduha )nB,nC*. Zanemariti promene potencijalne i kineti~ke energije poluidealnih gasova.

kiseonik,!nB,nC-!3:4!L

vazduh,!nw-!984!L

me{avina,!nC!,nw-!684!L

kiseonik,!nB-!584!L


{fmlp@fvofu/zv


strana 95

⋅

nC

hC =

⋅

⇒

⋅

n w + nC

d qw

Uw2=711p D

>2/161!

1 Ul2=31p D

d ql 1

lK lhL

d qw

d ql

>1/:6!

1

Un =411p D

>2/131!

1 Ul 3 =311p D

lK >1/:222! lhL

Un =411p D

⋅

h ⋅ nw 1/6 ⋅ 1/3 lh = >1/3! nC = C 2 − hC 2 − 1/6 t ⋅

d ql

lK lhL

>1/:466!

1

lK lhL

lK lhL


⋅

⋅

⋅

⋅

⋅

⋅

! R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23

ograni~enom isprekidanom linijom: I2 = I3 I2 = n w ⋅ d qw ⋅

⋅

I3 = n w ⋅ d qw

Uw2

Ul2

1 Un

1

⋅  ⋅  ⋅ Uw2 +  n B + nC d ql   Ul 3

⋅

⋅ Un + n B ⋅ d ql

1

⋅

nB =

⋅  n w ⋅  d qw  

⋅

⋅

⋅

⋅ Ul3 + nC ⋅ d ql

Un

⋅ Un

1 Un

⋅ Un − d qw

1

 ⋅  ⋅ Uw2  + nC ⋅  d ql   

Uw2

1

1

d ql

nB =

⋅ Ul2

Ul2

Ul 3

1

1

⋅ Ul2 − d ql

 ⋅ Ul2  

Un

Ul2

1

1

⋅ Un − d ql

⋅ Ul3

lh 1/3 ⋅ (2/13 ⋅ 684 − 2/16 ⋅ 984) + 1/3 ⋅ (1/:6 ⋅ 684 − 1/:222⋅ 3:4 ) >1/17! 1/:222⋅ 3:4 − 1/:466 ⋅ 584 t ⋅

n B + nC = 1/3 + 1/17 >1/37!


lh t

{fmlp@fvofu/zv


strana 1

VELIÎNE STAWA REALNIH FLUIDA 2.1. Odrediti specifi~nu entalpiju, specifi~nu entropiju, specifi~nu zapreminu kao i specifi~nu unutra{wu energiju vode stawa (p=1 bar, t=20oC).

kJ kJ m3 , sw = 0.296 , vw = 0.001001 kg kgK kg priru~nik za termodinamiku (tabela 4.2.6. ′′iznad crte′′) strana 41−55 kJ uw = hw − p . vw = 83.9 −1.105 .10−2.0.001001 = 83.8 kg hw = 83.9

2.2. Odrediti specifif~nu entalpiju, specifi~nu entropiju, specifi~nu zapreminu i specifi~nu unutra{wu energiju pregrejane vodene pare stawa (p=25 bar, t=360oC).

kJ kJ m3 , spp = 6.870 , vpp = 0.1117 kg kgK kg priru~nik za termodinamiku (tabela 4.2.6. ′′ispod crte′′) strana 41−55 kJ upp = hpp − p . vpp = 3146– −25 .105 .10−2 .0.1117 = 2866.75 kg hpp = 3146

2.3. Odrediti specifi~nu unutra{wu energiju, specifi~nu entropiju i temperaturu a) kqu~ale vode pritiska p=10 bar b) suvozasi}ene vodene pare pritiska p=10 bar a)

kJ kJ , s’′ = 2.138 , tK= 179.88oC kg kgK priru~nik za termodinamiku (tabela 4.2.4.) strana 36−38 u′’= 761.6

b)

kJ kJ , s′′”= 6.587 , tK= 179.88oC kg kgK priru~nik za termodinamiku (tabela 4.2.4.) strana 36−38 u”′′= 2583

2.4. Odrediti specifi~nu entalpiju i temperaturu vla`ne vodene pare stawa (x=0.95, p=15 bar). kJ hx = h' +x ⋅ (h"−h' ) = ... = 844 .6 + 0.95 ⋅ (2792 − 844 .6 ) =2694 kg kJ kJ h′’ = 844.6 , h′′”= 2792 kg kg o tx = tK = 198.28 C priru~nik za termodinamiku (tabela 4.2.4.) strana 36−38


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strana 2

2.5. Odrediti specifi~nu entropiju i pritisak vla`ne vodene pare stawa (x=0.6, t=200oC). kJ s x = s ' + x ⋅ (s"−s ' ) = ... = 2.3308 + 0 .6 ⋅ (6 .4318 − 2 .3308 ) =4.7806 kgK kJ kJ s’′ = 2.3308 s′′”= 6.4318 kgK kgK px = pK = 15.551 bar priru~nik za termodinamiku (tabela 4.2.5.) strana 39−40 2.6. Primewuju}i postupak linearne interpolacijeodrediti: a) specifi~nu entalpiju pregrejane vodene pare stawa (p=1 bar, t=250oC) b) specifi~nu entropiju pregrejane vodene pare stawa (p=7 bar, t=300oC) c) specifi~nu entalpiju pregrejane vodene pare stawa (p=5 bar, t=350oC) a)

kJ tabela 4.2.6. strana 41−55, za p=1 bar i t=240oC=x1 kg kJ y2 = hpp= 2993 tabela 4.2.6. strana 41−55, za p=1 bar i t=260oC=x2 kg y − y1 2993 − 2954 kJ hpp= 2 ⋅ (x − x1) + y1 = ⋅ (250 − 240 ) + 2954 = 2973.5 x2 − x1 260 − 240 kg y1 = hpp= 2954

b)

kJ tabela 4.2.6. strana 41−55, za p=6 bar=x1 i t=300oC kgK kJ y2 = spp= 7.226 tabela 4.2.6. strana 41−55, za p=8 bar=x2 i t=300oC kgK y − y1 7.226 − 7.366 kJ spp= 2 ⋅ (x − x1) + y1 = ⋅ (7 − 6) + 7.366 = 7.296 x2 − x1 8 −6 kgK y1 = spp= 7.366

c) 1.korak

kJ kg kJ y2 = hpp= 3190 kg y1 = hpp= 3148

h1 = hpp=

tabela 4.2.6. strana 41−55, za p1 =4 bar i t=340oC=x1 . tabela 4.2.6. strana 41−55, za p1 =4 bar i t=360oC=x2 .

kJ y 2 − y1 3190 − 3148 ⋅ (x − x 1 ) + y 1 = ⋅ (350 − 340) + 3148 = 3169 x2 − x2 360 − 340 kg


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strana 3

2.korak

kJ tabela 4.2.6. strana 41−55, za p2 =6 bar i t=340oC=x1 . kg kJ y2 = hpp= 3185 tabela 4.2.6. strana 41−55, za p2 =6 bar i t=360oC=x2 . kg y − y1 3185 − 3143 kJ h2 = hpp= 2 ⋅ (x − x 1 ) + y 1 = ⋅ (350 − 340 ) + 3143 = 3164 x2 − x2 360 − 340 kg y1 = hpp= 3143

3.korak h − h1 3164 − 3169 kJ h= 2 ⋅ (p − p1 ) + h1 = ⋅ (5 - 4 ) + 3169 =3166.5 p2 − p1 6 −4 kg 2.7. Odrediti specifi~nu entalpiju i specifi~nu entropiju: a) leda temperature t=−5oC b) me{avine leda i vode (mw=2 kg, ml=3 kg) u stawu toplotne ravnoteè (t=0oC) a)

hl = cl ⋅ (Tl − 273 ) − rl = 2 ⋅ (− 5 ) − 332 .4 = −342.4 sl = cl ⋅ ln

kJ kg

Tl − 273 r − 5 332 .4 kJ − l = 2 ⋅ ln − =−1.25 273 273 273 273 kgK

b)

y=

mw 2 = =0.4 mw + ml 2 + 3

(maseni udeo vode u me{avini vode i leda)

hy = hl + y ⋅ (hw − hl ) = ... = 0 + 0.4 ⋅ (0 + 332 .4 ) =−199.4 hw= 0

kJ kg

kJ kg

hL = −332.4

kJ kg

s y = s l + y ⋅ (s w − s l ) = ... = 0 + 0.4 ⋅ (0 + 1 .22 ) =−0.732

kJ kgK

kJ kgK r 332 .4 kJ sL = − l = − −1.22 273 273 kgK sw= 0


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strana 4

2.8. Odrediti specifi~nu entalpiju i specifi~nu entropiju: a) pregrejane vodene pare stawa (t=600oC, v=1 m3 /kg) b) vla`ne vodene pare stawa (x=0.9, v=20 m3/kg) a) v=1 m3 /kg h t=600oC

h=3705 kJ/kg

s=8.46 kJ/kgK

kJ hpp = 3705 kg kJ spp = 8.46 kgK

s

(upotrebom hs dijagrama za vodenu paru) (upotrebom hs dijagrama za vodenu paru)

b) h v=20 m3 /kg

h=2325 kJ/kg

x=0.9

s=7.54 kJ/kgK

kJ kg kJ sx = 7.54 kgK hx = 2325


s

(upotrebom hs dijagrama za vodenu paru) (upotrebom hs dijagrama za vodenu paru) [email protected]


strana 5

2.9. Odrediti specifi~nu entalpiju pregrejane vodene pare stawa (u=2654 kJ/kg, v=1.08 m3 /kg). pretostavimo p= 4 bar: ⇒ 3  m  kJ  =3846.35 hpp = f p = 4 bar, v = 1.08   kg  kg  provera pretpostavke: h − u 3846 .35 − 2654 p= = =11.04 bar v 1.08 pretostavimo p= 3 bar: ⇒ 3  m  kJ  =3340.03 hpp = f p = 3 bar, v = 1 .08   kg  kg  provera pretpostavke: h − u 3340 .03 − 2654 p= = =6.35 bar v 1.08 pretostavimo p= 2 bar: ⇒ 3   m kJ  =2870 hpp = f p = 2 bar , v = 1 .08   kg  kg  provera pretpostavke: h − u 2870 − 2654 p= = =2 bar v 1 .08 kJ ta~na vrednost iznosi: hpp=2870 kg

tabela 4.2.6. strana 41−55

(pretpostavka nije ta~na)


(pretpostavka nije ta~na)


(pretpostavka je ta~na)

2.10. Odrediti specifi~nu entalpiju i specifi~nu entropiju suvozasi}ene pare amonijaka na T=300 K. kJ kJ h”′′= 2246 s′′”= 9.993 kg kgK tabela 4.4.1. strana 62, za T=300 K 2.11. Odrediti specifi~nu entalpiju i specifi~nu entropiju pregrejane pare freona 12 stawa (p=6 bar, t=200oC). kJ kJ hpp =789 spp = 1.889 kg kgK tabela 4.6.2. strana 79−81, za p=6 bar, t=200oC


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strana 6

PROMENE STAWA REALNIH FLUIDA 2.12. Vla`na para stawa 1(x=0.3, p=0.2 bar) izohorski se {iri do stawa 2(p=1.5 bar), a zatim ravnote`no izentropski ekspandira do stawa 3(p3=p1 ). Skicirati promene stawa vodene pare na Ts i pv dijagramu i odrediti razmewenu toplotu (kJ/kg) za promenu stawa 1−2 i zapreminski rad (kJ/kg) za promenu stawa 2−3. T

2

2 p

3

1

1

3

s

v

ta~ka 1: u1 = ux =u’′” + x1 . (u′′” - u′’) = ...= 251 .38 + 0.3 ⋅ (2456 − 251.38 ) =912.8 u’′ =251.38

kJ kg

u′′” =2456

kJ kg

kJ kg

v1 = vx = v’′ + x1 . (v′′” - v’′ ) = 0.0010171 + 0 .3 ⋅ (7.647 − 0.0010171) =2.29 v′’=0.0010171

m3 kg

v′′”=7.647

m3 kg

p2 =1.5 bar

m3 kg

m3 kg

ta~ka 2: v2 = v1 =2.29

m3 m3 v′′”=1.159 kg kg ta~ka 2 nalazi se u oblasti pregrejane pare

provera poloàja ta~ke 2: v2 > v”′′

⇒

h2 = hpp =3448.1

kJ , kg

v′’=0.0010527

s2 = spp=8.5

kJ kgK

u2 = upp = hpp –− p . vpp = 3448 .1 − 1 .5 ⋅ 10 5 ⋅ 10 −3 ⋅ 2.29 =3104.6


kJ kg

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strana 7

ta~ka 3: s3 = s2 =8.5

kJ kgK

p3 = p1 = 0.2 bar

kJ kJ s′′” =6.822 kgK kgK ta~ka 3 nalazi se u oblasti u pregrejane pare

provera poloàja ta~ke 3: s3 > s′′”

⇒

h3 = hpp = 2797.6

kJ , kg

s′’ =08321

v3 = vpp =14.61

m3 kg

u3 = upp = hpp –− p . vpp = 2797 .6 − 0.2 ⋅ 10 5 ⋅ 10− 3 ⋅ 14.61 =2505.4

kJ kg

(q12 )v= const = u2 − u1 = 3104 .6 − 912 .8 =2191.8 kJ (w23 )s= const

kg kJ = u2 − u3 = 3104 .6 − 2505 .4 = 599.2 kg

2.13. Pregrejana vodena para stawa 1(m=1 kg, p=0.05 MPa, t=270oC) predaje toplotu izotermnom toplotnom ponoru, usled ~ega ravnote`no mewa svoje toplotno stawe: prvo izohorski (1−2) do temperature 60oC, potom izotermski (2−3) do pritiska 0.1 MPa i kona~no izobarski (3−4) do temperature 20oC. Odrediti promenu entropije izolovanog sistema za slu~aj termodinmi~ki najpovoqnijeg temperaturskog nivoa toplotnog ponora. Skicirati proces u Ts i pv koordinatnom sistemu. 1

T

1

p

3

3

2

4

4

s


2

v

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strana 8

∆S SI =∆S RT + ∆S TP = .… .. ∆S RT =∆S14 = m ⋅ (s 4 − s1 ) =...

∆ STP = −m ⋅ ∆ STP = −m ⋅

(q12 )v =const + (q 23 ) T =const + (q34 )p=const Ttp

u2 − u1 + T2 ⋅ (s 3 − s 2 ) + h4 − h3 = ... TTP

ta~ka 1: p1 =0.5 bar,

t1 =270oC

tk=81.35oC

t1 > tk

m3 , kg kJ s1 =spp= 8.423 , kgK v1 = vpp= 5

ta~ka 1 nalazi se u oblasti pregrejane pare h1 =hpp = 3015

kJ kg

u1 = upp=h1 − p1 . v1 = 2765

kJ kg

ta~ka 2: t2 =60oC

v2 =v1 =5

v′’= 0.0010171

m3 , kg

v′ < v2 < v”′′

x2 =

m3 kg

m3 kg ta~ka 2 nalazi se u oblasti vla`ne pare

v′′”=7.678

v 2 − v' = 0.6512 v" − v'

u2 = ux =u’′” + x2 . (u′′” − u′’) =…...= 251 .1 + 0.6512 ⋅ (2456 − 251 .1) = 1684.29 u’′ =251.1

kJ kg

u′′” =2456

kJ kg

s2 = sx =s’′” + x2 . (s′′” − s′’) =…...= 0.8311 + 0.6512 ⋅ (7.9084 − 0 .8311) =5.43 s’′ =0.8311

kJ , kgK


s′′”=7.9084

kJ kg

kJ kgK

kJ kgK

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strana 9

ta~ka 3: p3 =1 bar

t3 =60oC

tk=99.64oC

t3 < tk

h3 = hw=251.1

kJ , kg

s3 = sw= 0.83

ta~ka 3 nalazi se u oblasti te~nosti

kJ kgK

ta~ka 4: p4 =1 bar

t4 =60oC

tk=99.64oC

t4 < tk

h4 =hw= 83.9

kJ , kg

s4 = sw=0.296


kJ kgK

toplotni ponor: TTP=T4 =293 K

(najpovoqniji termodinami~ki slu~aj)

∆S RT =∆S14 =1 ⋅ (0 .296 − 8.423 ) =...=−8.127

∆ STP = −1 ⋅

kJ K

kJ 1684 .29 − 2765 + 333 ⋅ (0.83 − 5.43 ) + 83 .9 − 251 .1 = 9.487 K 293

∆S SI =∆S RT + ∆S TP = … − 8.127 + 9.487 =1.36

kJ K

2.14. Vodi (m=10 kg) stawa 1(p=0.1 MPa, t=20oC) dovodi se toplota od izotermnog toplotnog izvora, usled ~ega voda mewa svoje toplotno stawe: prvo izobarski (1−2) do temperature 60oC, potom izotermski (2−3) do specifi~ne zapremine 5 m3 /kg i na kraju izohorski (3−4) do pritiska 0.05 MPa. Skicirati proces u Ts koordinatnom sistemu i odrediti promenu entropije adijabatski izolovanog sistema za slu~aj termodinmi~ki najpovoqnijeg temperaturskog nivoa toplotnog izvora.


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strana 10

∆S SI = ∆S RT + ∆S TI = …... ⋅

∆S RT = ∆S14 = m ⋅ (s 4 − s1 ) =...

∆ STI = −m ⋅ ∆ STI

(q12 )p=const + (q 23 )T =const + (q 34 ) v= const

TTI h − h1 + T2 ⋅ (s 3 − s 2 ) + u 4 − u 3 =m⋅ 2 =... TTI

ta~ka 1: t1 =20oC

p1 =1 bar, tk=99.64oC h1 =hw= 83.9

t1 < tk

kJ , kg

ta~ka 2:

s1 = sw=0.296

t2 < tklj

kJ kg

kJ kgK

t2 = 60oC

p2 =1 bar

tk=99.64oC h2 =hw = 251.1


ta~ka 2 nalazi se u oblasti te~nosti s2 =sw = 0.83

kJ kgK

ta~ka 3: t3 = 60oC v′’= 0.0010171

v3 = 5

m3 , kg

v′ < v3 < v”′′

x3 =

m3 kg m3 kg ta~ka 3 nalazi se u oblasti vla`ne pare

v′′”=7.678

v 3 − v' 5 − 0.0010171 = = 0.6512 v" −v' 7.678 − 0.0010171

kJ kg kJ s3 = sx = s '+ x 3 ⋅ (s"−s ' ) = 0.8311 + 0.6512 ⋅ (7 .9084 − 0 .8311) =5.43 kgK u3 =ux = u' + x3 ⋅ (u"−u' ) = 251 .1 + 0.6512 ⋅ (2456 − 251 .1) =1684.29


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strana 11

ta~ka 4: p4 = 0.5 bar, v′’= 0.0010299

v4 = v3 = 5

m3 kg

m3 kg

m3 kg ta~ka 4 nalazi se u oblasti pregrejane pare v′′”=3.239

v4 > v”′ ′

m3 , kg kJ s1 =spp= 8.42 , kgK v1 = vpp= 5

h1 =hpp = 3015

kJ kg

u 1 = upp=h1 − p1 . v1 = 2765

kJ kg

T4 =Tpp= 270oC = 543 K toplotni izvor: TTI =T4 = 543 K

(najpovoqniji termodinami~ki slu~aj)

kJ K kJ 251.1 − 83.9 + 333 ⋅ (5.45 − 0.83 ) + 2782 .85 − 1689 .36 = −10 ⋅ =−51.6 K 542.5

∆S RT = ∆S14 =10 ⋅ (8.42 − 0.296 ) =81.24

∆ STIi

∆S SI = 81.24 − 51.6 = 29.64

kJ K

4

T

TI

2 3 1 s


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strana 12

2.15. Jednom kilogramu leda stawa 1(p=1 bar T=−5 oC) dovodi se toplota od toplotnog izvora konstantne temperature TTI=300oC tako da se na kraju izobarske promene stawa (1−2) dobije suvozasi}ena vodena para (stawe 2). Odrediti promenu entropije izolovanog sistema pri ovoj promeni stawa i grafi~ki je predstaviti na Ts dijagramu. ∆S SI =∆S RT + ∆S TI = .… ..= 8.61 − 5.27 = 3.34 ∆S RT =m . ∆s12=m . ( s2 − s1 )=...= 8.61

(q12 )p=const

kJ K

(h2 − h1 )

kJ K

= ... = −5.27

kJ K

h1 = hl = cl ⋅ (TL − 273 ) − rl = 2 ⋅ (− 5 ) − 332.4 = −342.4

kJ kg

∆ STI = −m ⋅

= −m ⋅

Tti

Tti

ta~ka 1:

s1 = s l = c l ⋅ ln

TL 273

−

rl 273

= 2 ⋅ ln

-5 + 273 332.4 kJ − = − 1.25 kgK 273 273

ta~ka 2: h2 = h′′ = 2675

kJ , kg

s2 = s′′”=7.36

kJ kgK

T TI 2

1 ∆S RT ∆S TI

s

jednake povr{ine

∆S SI


[email protected]


strana 13

2.16. Te~an CO2, stawa 1(p=5 MPa, t=0oC), adijabatski se prigu{uje (h=idem) do stawa 2(p=0.6 MPa). Grafi~ki predstaviti po~etno i krajwe stawe CO2 u Ts i hs kJ koordinatnom sistemu i odrediti prira{taj entropije CO2 tokom procesa 1−2 ( ). kgK T

p1

h

p1

p2

p2

1

2

1 2 s

s ta~ka 1: p1 =50 bar, h1 =− 94

kJ , kg

t1 =0oC


s1 = 3.1133

kJ kgK


ta~ka 2: h2 =h1 = −94

p2 =6 bar,

kJ kJ , h′′”=142.7 kg kg h′

kJ kg

h′’= −200

x2 =

tabela 4.8.1. strana 92 ta~ka 2 nalazi se u oblasti vla`ne pare

h2 − h' −94 + 200 = =0.3093 h"−h' 142 .7 + 200

s2 = sx = s '+ x 2 ⋅ (s "−s ' ) = 2.702 + 0.3093 ⋅ (4.260 − 2 .702 ) =3.184 s’′ =2.702

kJ , kgK

s′′”=4.260

kJ kgK

∆s12 = s2 − s1 =...=3.184 − 3.1133 = 0.0707


kJ kgK

tabela 4.8.1. strana 92

kJ kgK

[email protected]


strana 14

2.17. Freon 22 stawa 1(T=−30oC, x=1) nekvazistati~ki (neravnote`no) adijabatski se komprimuje do stawa 2(p=6 bar). Prira{taj entropije freona tokom procesa J iznosi ∆s12 =51 . Predstaviti proces sa freonom na Ts i hs dijagramu i kgK odrediti stepen dobrote ove nekvazistati~ke adijabatske kompresije. ta~ka 1: h1 = h′′ = 691.92

kJ kJ , s1 = s′′= 1.7985 kg kgK


ta~ka 2k: p2k=6 bar, s′ = 1.024

kJ kgK kJ s′′ = 1.74 kgK

s2k = s1 =1.7985

kJ , kgK

s2k > s′′


ta~ka 2k nalazi se u oblasti pregrejane pare

h2k = hpp = 720.78

kJ kg

ta~ka 2: p2 =6 bar,

s2 = s1 +∆s12 =1.7895+0.051=1.8495

h2 = hpp = 739.36

kp

ηd =

kJ kgK

kJ kg

h1 − h2k 691 .92 − 720 .78 = =0.61 h1 − h 2 691 .92 − 739 .36 2

2 T

h

2k

2k

1 1 s


s

[email protected]


strana 15

(2.18. –− 2.19.)

zadaci za ve`bawe:

2.18. Kqu~ala voda temperature T1= 250oC mewa stawe ravnote`no: − izotermski (1−2) do p2 = 4.5 bar − zatim izohorski(2−3) do p= 2.2 bara − i na kraju izobarski (3−4) do stawa 4(s4 =s1 ) Skicirati promene stawa vodene pare na pv i Ts dijagramu i odrediti razmewene kJ toplote ( ) tokom procesa 1−2, 2−3 i 3−4. kg kJ kJ kJ re{ewe: q12 =2368.5 , q23 =−846.7 , q34 =−991.2 kg kg kg 2.19. Vodenoj pari stawa 1(T2 =100oC, x=0) ravnote`no se dovodi se toplota pri ~emi vodenu paru prevodimo u stawe 2(T=120oC, x=1). U procesu (1−2) temperatura pare raste linerano u Ts kordinatnom sistemu. Nakon toga se vr{i neravnote`na adijabatska ekspanzija (2−3) vodene pare (stepen dobrote nekvazistati~ke adijabatske ekspanzije: η eks d =0.9) do stawa 3(p=0.1 bar). Skicirati procese sa vodenom parom na Ts dijagramu i odrediti dovedenu toplotu za proces 1−2 i dobijeni tehni~ki rad za proces 2−3. re{ewe:

q12 =2316.2

kJ , kg


w T23 =402.4

kJ kg

[email protected]


strana 16

PRVI I DRUGI ZAKON TERMODINAMIKE (ZATVOREN TERMODINAMI^KI SISTEM) 2.20. U zatvorenom, adijabatski izolovanom, sudu zapremine V=7.264 m3 , nalazi se me{avina m′=311 kg kqu~ale vode i m’′’ ′ =? suvozasi}ene vodene pare u stawu termodinami~ke ravnoteè na p1 =0.95 bar. Vodenoj pari u sudu se dovodi toplota, od toplotnog izvora stalne temperature TTI=300oC, tako da joj pritisak poraste na p2 =68 bar. Skicirati procese sa vodenom parom na Ts i pv dijagramu i odrediti: a) koliko je toplote dovedeno u procesu (MJ) b) promenu entropije izolovanog termodinami~kog sistema za proces 1−2 (kJ/K) a) prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu

⇒

Q12 = ∆U12 + W12

Q12 = (m′ + m′′) . (u2 –− u1 )=...

ta~ka 1: p1 =0.95 bar,

x1 =?

m m3 v′′”=1.7815 kg kg V' = m'⋅v' = 311 ⋅ 0.00104205 =0.3241 m3

v’′ = 0.00104205

3

V′′” = V − V’′ =7.264 − 0.3241 = 6.9399 m3

m" =

V" 6.9399 = =3.9 kg v" 1.7815

x1 =

m" 3 .9 = =0.0124 m"+m' 3 .9 + 311

u1 = ux =u’′ + x1 .(u′′” - u′’) = 411 .23 + 0 .0124 ⋅ (2504 − 411 .23 ) =437.21

kJ u’′ = 411.23 , kg

kJ kg

kJ u”′′ = 2504 kg

v1 =vx=v’ + x1. (v”′′ − v’′ )= 0. 00104205 + 0. 0124 ⋅ (1.7815 − 0. 00104205) =0.0231 s1 = sx =s’′ + x1 .(s′′” − s′’) = 1.2861 + 0.0124 ⋅ (7 .377 − 1 .2861) =1.362

kJ s’′ = 1.2861 , kgK


m3 kg

kJ kgK

kJ s”′ ′ = 7.377 kgK

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strana 17

ta~ka 2 p2 =68 bar,

v2 = v1 = 0.0231

v′’= 0.0013445

m3 , kg

m3 kg ta~ka 2 se nalazi u oblasti vla`ne pare

v′′”=0.028382

v′’ < v2 < v”′′

x2 =

m3 kg

v 2 − v' 0 .0231 − 0.0013445 = =0.8046 v"− v' 0.028382 − 0 .0013445

u2 = ux =u’′ + x2 .(u′′” − u′’)= ...=1247 .52 + 0.8046 ⋅ (2582 − 1247 .52 ) =2321.24 u’′ = 1247.52

kJ kg

u′′” = 2582

kJ kg

kJ kg

s2 = sx =s’′ + x1 .(s′′” − s′’) = ...= 3.103 + 0.8046 ⋅ (5.829 − 3.103 ) =5.2963

kJ kgK

kJ kJ , s”′ ′ = 5.829 kgK kgK = (311 + 3.9 ) ⋅ (2321 .24 − 437.21) =593.3 MJ

s’′ = 3.103

Q 12 b)

∆S SI =∆S RT + ∆S TI = …...=216.2

kJ K

∆S RT = (m' +m' ' ) ⋅ (s 2 − s 1 ) = (311 + 3.9 ) ⋅ (5.2963 − 1.362 ) =1239.1

∆ STI = −

kJ K

Q12 593.3 ⋅ 10 3 kJ =− =1035.43 TTI 573 K

T

p

2 2

1

1 s


v

[email protected]


strana 18

2.21. Izolovan zatvoren sud zapremine V=120 litara ispuwen je kqu~alom vodom i suvozasi}enom parom u stawu termodinami~ke ravnoteè na pritisku p1 =1 bar. U posudi se nalazi greja~ snage 5 kW . Dovo|ewem toplote nivo vode u sudu raste i kada pritisak dostigne 50 bara, posuda je u celosti ispuwena te~nom fazom. Skicirati proces na pv dijagramu i odrediti koliko dugo je trajalo dovo|ewe toplote. p K 2

1 vk

v

ta~ka 2: p2 = 50 bar,

x2 =0

v2 = v′= 0.0012857

m=

m3 , kg

u2 =u′= 1148

kJ kg

V 120 ⋅ 10 −3 = =93.33 kg v 2 0.0012857

ta~ka 1: p1 =1 bar,

v1 =v2 =0.0012857

v′’= 0.0010432

m3 , kg

v′’ < v1 < v”′′

x1 =

m3 kg


v′′”=1.694

v 1 − v' 0.0012857 − 0.0010432 = =0.0001 v" −v' 1.694 − 0 .0010432

u1 = ux =u’′ + x1 .(u′′” − u′’) = 417 .3 + 0.0001 ⋅ (2506 − 417 .3 ) =417.51 u’′ = 417.3

kJ , kg


u”′′ = 2506

kJ kg

kJ kg

[email protected]


strana 19

prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu Q12 = ∆U12 + W12

⇒

Q12 =m . (u2 –− u1 )

Q12 = 93 .33 ⋅ (1148 − 417.51) =68.18 MJ

τ=

Q 12 ⋅

=

Q 12

68.18 ⋅ 10 3 =13636 s 5

Uo~iti da se u ovom zadatku pojavquje fenomen podkriti~nih zapremina , tj. izohorskim dovo|ewem toplote vla`noj pari vx1 pa zatim monotono opada do x=0. 2.22. U vertikalno postavqenom cilindru povr{ine popre~nog preseka A=0.1 m2 koji je po omota~u izolovan nalazi se m=0.92 kg vode na temperaturi od 10oC. Iznad vode je klip zanemarqive mase koji ostvaruje stalni pritisak. Pritisak okoline iznosi po=1 bar. U cilindru se nalazi greja~ toplotne snage 0.5 kJ/s. Zanemaruju}i trewe klipa o zidove cilindra odrediti vreme potrebno da se klip podigne za ?∆z=1.3 m. Predstaviti proces dovo|ewa toplote na Ts dijagramu.

τ=

Q 12 ⋅

∆z

=...

Q 12 prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu Q12 = ∆U12 + W12

⇒

Q12 =m . (u2 − u1 )+ p ⋅ (V2 − V1 ) = ...

ta~ka 1: p1 =1 bar, h1 = hw = 42

t1 =10oC

kJ , kg

v1 = vw = 0.0010005

m3 kg

u1 = uw = h1 − p 1 ⋅ v 1 = 42 − 1 ⋅ 10 5 ⋅ 10 −3 ⋅ 0 .0010005 =41.9

kJ kg

V 1 = m . v1 = 0.92 ⋅ 0.0010005 =0.0009 m3 dipl.ing. @eqko Ciganovi}

[email protected]


strana 20

ta~ka 2: p2 =1 bar

v2 =?

V 2 = V1 + A. ∆z = 0.0009 + 0.1.1.3 = 0.1309 m3

v2 =

V2 0.1309 m3 = =0.1423 m 0 .92 kg

v′’= 0.0010432

m3 , kg


v′′”=1.694

v′’ > v2 > v′′”

x2 =

v 2 − v' v" − v'

=

0.1423 - 0.0010432 =0.0834 1.694 - 0.0010432

u2 = ux =u′’ + x2 .(u′′” − u′’) = = 417 .3 + 0.0834 ⋅ (2506 − 417 .3 ) =591.5 u’′ = 417.3

kJ , kg

u”′′ = 2506

kJ kg

kJ kg

Q12 = 0.92 ⋅ (591 .5 − 41.9 ) + 1 ⋅ 10 5 ⋅ 10 −3 ⋅ (0 .1423 − 0.0009 ) =519.77 kJ

τ=

519.77 =1039.5 s 0 .5

T

p=const

1

2

s


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strana 21

2.23. Vertikalan cilindar (od okoline toplotno izolovan) unutra{weg pre~nika d=250 mm zatvoren je sa gorwe strane pomi~nim (bez trewa) i od okoline izolovanim klipom (zanemarqive mase) optere}enim sa dva tega masa: mT1=210 kg i mT2=1800 kg. Po~etna udaqenost klipa od dna cilindra je z1 =300 mm. U cilindru se nalazi 5 litara kqu~ale vode, a ostatak zapremine zauzima suvozasi}ena vodena para. Pritisak okoline iznosi po=1 bar. Dovo|ewem toplote klip se podigne za ∆z=200 mm. Zatim se istovremeno teì teg podigne dizalicom i skine sa klipa ({to dovodi do daqeg podizawa klipa) i iskqu~i greja~. Odrediti: a) koli~inu toplote dovedenu u prvom delu procesa b) izvr{eni zapreminski rad u drugom delu procesa T1

T2

T2

T1 ∆z

T1

∆z

z1

ta~ka 1:

p1 = p o +

(m T1 + m T2 ) ⋅ g = 1 ⋅ 10 5 + (210 + 1800 ) ⋅ 9.81 =5 bar d2 π 4

0.25 2 π 4

d2 π 0.25 2 π ⋅ z1 = ⋅ 0 .3 =0.0147 m3 4 4 V′′ = V − V′ =0.0147 –− 0.0050 =0.0097 m3 V1 =

v′ =0.0010927 m3 ,

v′′=0.3747 m3

V' 0.0050 V' ' 0 .0097 = =4.576 kg, m' ' = = =0.026 kg v' 0.0010927 v ' ' 0 .3747 m' ' 0.026 x1 = = =0.0056 m' +m' ' 4.576 + 0.026 m' =

u1 = ux =u’′ + x1 .(u′′” − u′’) = 639 .4 + 0.0056 ⋅ (2562 − 639 .4 ) =650.17

kJ u’′ = 639.4 , kg dipl.ing. @eqko Ciganovi}

kJ kg

kJ u”′′ = 2562 kg [email protected]


strana 22

ta~ka 2: V 2 = V1 +

p2 = p1 = 5 bar,

d2 π 0.25 2 π ⋅ ∆z = 0.0147 + ⋅ 0.2 =0.0245 m3 4 4

0 .0245 m3 =0.0053 m 4 .602 kg v′’ > v2 > v′′” v2 =

x2 =

V2

=

v 2 − v' v"− v'

ta~ka 2 se nalazi u oblasti vla`ne pare

0.0053 - 0.0010927 =0.0113 0.3747 - 0.0010927

=

u2 = ux =u’′ + x2 .(u′′” − u′’) = 639 .4 + 0.0113 ⋅ (2562 − 639.4 ) =661.13

kJ

kg kJ s2 = sx =s’′ + x2 .(s′′” − s′’) = ...= 1.86 + 0 .0113 ⋅ (6.822 − 1.86 ) =1.916 kgK kJ kJ s’′ = 1.86 , s”′ ′ = 6.822 kgK kgK ta~ka 3:

p3 = p o +

m T1 ⋅ g 2

= 1 ⋅ 10 5 +

210 ⋅ 9.81 2

d π 0.25 π 4 4 kJ kJ s’′ = 1.4184 , s”′ ′ = 7.2387 kgK kgK s′’ > s3 > s′′”

x3 =

s3 − s' s" −s'

=1.42 bar,

kJ kgK

s3 = s2 = 1.916

ta~ka 3 se nalazi u oblasti vla`ne pare

=

1.916 - 1.4184 =0.0855 7.2387 - 1.4184

u3 = ux =u’′ + x3 .(u′′” − u′’) = 461 .1 + 0.0855 ⋅ (2518 − 461 .1) =636.96

kJ u’′ = 461.1 , kg


kJ kg

kJ u”′′ = 2518 kg

[email protected]


strana 23

a) prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu Q12 = ∆U12 + W12

Q12 =m . (u2 –− u1 )+ p ⋅ (V2 − V1 )

⇒

Q 12 = 4 .602 ⋅ (661 .13 − 650 .17 ) + 5 ⋅ 10 5 ⋅ 10 − 3 ⋅ (0 .0245 − 0.0147 ) =55.34 kJ b) prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu Q23 = ∆U23 + W23

⇒

W23 =−m . (u3 –− u2 )

W23 = −4 .602 ⋅ (636 .96 − 661 .13 ) =111.23 kJ T

p1 =p2 2 1

p3 3

s 2.24. Vla`na vodena para stawa A(pA=0.11 MPa, x=0.443), koja se nalazi u toplotno izolovanom sudu A, zapremine V A=0.55 m3 , razdvojena je ventilom od suvozasi}ene vodene pare koja se pri istom pritisku (pB=pA) nalazi u toplotno izolovanom cilindru B, zapremine V B=0.31 m3 (slika). Pri zako~enom (nepokretnom) klipu K otvara se ventil i uspostavqa stawe termodinami~ke ravnoteè pare u oba suda (stawe C). Po dostizawu tog ravnote`nog stawa, pokre}e se klip K, koji pri i daqe otvorenom ventilu, kvazistati~ki sabija paru na pritisak p=2.4 MPa (stawe D). Odrediti izvr{eni zapreminski rad ( za proces C−D) i prikazati sve promene u Ts koordinatnom sistemu.

K

A B


[email protected]

zbirka zadataka iz termodinamike pA=1.1 bar

ta~ka A:

strana 24 xA=0.443

kJ kg

uA= ux =u’′ + xA.(u′′” - u′’) = ...= 428. 79 + 0. 443 ⋅ (2509 − 428.79 ) =1350.32 u’′ =428.79

kJ , kg

u”′′ = 2509

kJ kg

vA= vx= v’ + xA. (v”′′ − v′’)= ...= 0 .0010452 + 0. 443 ⋅ (1. 555 − 0 .0010452) =0.6894

VA vA

m3 kg 0 .55 = = 0.8 kg 0.6894

ta~ka B:

pB=1.1 bar

v’′ = 0.0010452

mA =

v′′”=1.555

m3 kg

m3 kg

xB=1

kJ m3 , vB = v′′”=1.555 kg kg 0.31 = 0.2 kg 1.555

uB=u”′ ′= 2509

mB =

VB vB

ta~ka C:

pC=1.1 bar

uC=?

prvi zakon termodinamike za proces me{awa u zajedni~kom sudu: Q12 = ∆U12 + W12

⇒

U1 = U2

U1 = mA. uA + mB. uB U2 = mA. uC + mB. uC

mA ⋅ u A + mB ⋅ uB 0.8 ⋅ 1350 .32 + 0 .2 ⋅ 2509 kJ = =1582.06 m A + mB 0.8 + 0.2 kg kJ kJ u’′ = 428.79 , u”′′ = 2509 kg kg u′’ < uC < u”′ ′ ta~ka 2 se nalazi u oblasti vla`ne pare uC =

xC =

u C − u' u"−u'

=

1582 .06 − 428 .79 =0.5544 2509 − 428 .79

sC= sx =s’′ + xC.(s′′” - s′’) = ...= 1.3327 + 0.5544 ⋅ (7 .238 − 1.3327 ) =4.606 s’′ = 1.3327

kJ kgK


s”′ ′ = 7.328

kJ kgK

kJ kgK [email protected]

zbirka zadataka iz termodinamike ta~ka D: s’′ = 2.534

pD=24 bar

kJ kgK

kJ kgK

sD=sC = 4.606

s”′ ′ = 6.272

s′’ < sD < s”′ ′

xD =

strana 25

kJ kgK

ta~ka D se nalazi u oblasti vla`ne pare

s D − s' 4 .606 − 2.534 = =0.55543 s"− s' 6 .272 − 2.534

uD= ux =u’′ + xD.(u′′” - u′’) = ...= 948 .9 + 0.5543 ⋅ (2602 − 948 .9) =1865.21 u’′ = 948.9

kJ , kg

u”′′ = 2602

kJ kg

kJ kg

prvi zakon termodinamike za proces adijabatske kompresije (C−D): QCD = ∆UCD + WCD

⇒

WCD= UC − UD

WCD= (m A + m B ) ⋅ (u C − uD ) = (0.8 + 0.2 ) ⋅ (1582 .06 − 1865 .21) = −283.15 kJ

T

pD D pA=pB=pC A

C

B

s


[email protected]


strana 26

2.25. U zatvorenom sudu zapremine V=2 m3 , nalazi se suvozasi}ena vodena para stawa 1(p=10 bar). Tokom hla|ewa do stawa 2 od vodene pare odvede se 14.3 MJ toplote. Odrediti promenu entropije sistema u najpovoqnijem slu~aju tokom procesa hla|ewa pare. p= 10 bar, kJ u1 = u″ = 2583 , kg ta~ka 1:

v1 = v″=0.1946

m3 , kg

x=1 s1 = s″= 6.587 m=

kJ kgK

V 2 = =10.28 kg v 1 0 .1946

m3 , u2 =? kg prvi zakon termodinamike za proces hla|ewa pare: ta~ka 2:

v2 = v1 =0.1946

Q12 = ∆U12 + W12 ⇒ Q12 = m. ( u2 –− u1 ) 3 Q 14 .3 ⋅ 10 kJ u2 = u1 + 12 = 2583 − =1191.95 kg m 10.28 pretpostavimo p2 =2.6 bar:

m3 m3 v′′”=0.6925 kg kg v 2 − v ' 0.1946 − 0 .0010685 x2 = = =0.28 v"− v ' 0.6925 − 0 .0010685 kJ kJ u’′ = 540.63 , u”′′ = 2539 kg kg v’′ = 0.0010685

kJ kg pretpostavka nije ta~na

u2 =ux = u′ + x2 . ( u″ - u′ )= 540 .63 + 0.28 ⋅ (2539 − 540 .63 ) =1100.17

pretpostavimo p2 =2.8 bar:

m3 m3 v′′”=0.6461 kg kg v − v' 0.1946 − 0.0010709 x2 = 2 = =0.3 v"− v ' 0.6461 − 0.0010709 kJ kJ u’′ = 551.1 , u”′′ = 2541 kg kg v’′ = 0.0010709

kJ kg pretpostavka nije ta~na

u2 =ux = u′ + x2 . ( u″ - u′ )= 551 .1 + 0.3 ⋅ (2541 − 551 .1) =1148.07


[email protected]


strana 27

pretpostavimo p2 =3.0 bar:

m3 m3 v′′”=0.6057 kg kg v 2 − v ' 0.1946 − 0 .0010733 x2 = = =0.32 v"− v ' 0.6057 − 0 .0010733 kJ kJ u’′ = 561.1 , u”′′ = 2543 kg kg v’′ = 0.0010733

kJ kg pretpostavka ta~na

u2 =ux = u′ + x2 . ( u″ - u′ )= 561 .1 + 0.32 ⋅ (2543 − 561 .1) = 1195.3

Obzirom da je pretpostavka ta~na to zna~i da je p2 =3 bar. Na osnovu vrednosti pritiska p2 odre|uje se temperatura T2 =133.54oC. s’′ = 1.672

kJ , kgK

s”′ ′ = 6.992

kJ kgK

s2 = sx =s’′ + x2 . (s′′” − s′’) = 1.672 + 0 .32 ⋅ (6.992 − 1.672 ) =3.374

kJ kgK

drugi zakon termodinamike za proces hla|ewa pare: ∆S SI = ∆S RT + ∆S TP = ...= 3.97

kJ K

∆S RT = m . ∆s12 = m .( s2 −– s1 ) = 10 .28 ⋅ (3.374 − 6.587 ) = − 33.03 ∆S TP = −

kJ K

Q 12 − 14.3 ⋅ 10 3 kJ =− =35.17 TTP 133 .54 + 273 K

zadatak za ve`bawe:

(2.26.)

2.26. U zatvorenom sudu nalazi se 5 kg pregrejane vodene pare stawa 1(p1 =0.1 MPa, t1 ). a) koliko iznosi temperatura pregrejane pare (t1 ) ako od we hla|ewem nastaje suva vodena para specifi~ne entalpije h=2653 kJ/kg (stawe 2) b) koliki }e biti stepen suvo}e (x3 ) vla`ne pare kada usled daqeg odvo|ewa toplote temperatura vodene pare dostigne 50oC (stawe 3) c) odrediti masu (kg) kqu~ale te~nosti (m’′’ ) i suvozasi}ene pare (m’′’ ′) stawa 3 a) t1 = 320oC b) x3 = 0.227 c) m′’=3.87 kg,

m”′ ′=1.13 kg


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strana 28

PRVI I DRUGI ZAKON TERMODINAMIKE (OTVOREN TERMODINAMI^KI SISTEM) 2.27. U adijabatski izolovanom ure|aju me{aju se suvozasi}ena vodena para stawa ⋅

1(p=0.4 MPa) i voda stawa 2(p=0.4 MPa, t=20oC, m w=1 kg/s ). Iz ure|aja izlazi voda stawa 3(p=0.4 MPa, t=80oC). Zanemaruju}i promene kineti~ke i potencijalne energije vodene pare, odrediti: a) potrebnu koli~inu pare (kg/s) b) promenu entropije sistema za proces me{awa (kW/K) 1

para

3 2

voda a)

p1 =4 bar x=1 kJ kJ h1 =h′′ = 2738 , s1 =s′′ = 6.897 kg kgK ta~ka 1:

p2 =4 bar t2 =20oC kJ kJ h2 =hw = 84.1 , s2 =sw =0.296 kg kgK

ta~ka 2:

p2 =4 bar t2 =80oC kJ kJ h3 =hw = 335.1 , s3 =sw =1.074 kg kgK

ta~ka 3:

prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⇒ H1 = H 2 ⋅ ⋅ ⋅ ⋅  mp ⋅ h1 + m w ⋅ h 2 =  mp + m w  ⋅ h3   ⋅

m w ⋅ (h3 − h2 ) 1 ⋅ (335 .1 − 84 .1) ⋅ kg mp = = = mp = 0.1 h1 − h 3 335 .1 − 2738 s ⋅


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strana 29

b) ⋅

⋅

⋅

∆ S SI = ∆ S RT + ∆ S o = ... = 0.196 + 0 = 0.196

kW K

⋅

⋅

Q12 kW ∆ S o= − =0 TO K ⋅ ⋅ ⋅ ⋅  ⋅ ∆ S RT = S izlaz − S ulaz=  mp + m w 

⋅ ⋅   ⋅ s 3 − mp ⋅ s 1 − m w ⋅ s 2  ⋅ kW ∆ S RT = (0.1 + 1) ⋅ 1.074 − 0.1 ⋅ 6.897 − 1 ⋅ 0.296 =0.196 K

⋅

t suvozasi}ene pare stawa 1(p=13 bar). Deo te pare se h koristi za potrebe nekog tehnolo{kog procesa, dok se drugi deo pare, nakon prigu{ivawa do p2 , me{a u napojnom rezervoaru sa vodom stawa 2(p=2 bar, t=20oC). Voda se iz napojnog rezervoara uvodi u toplotno izolovanu pumpu gde joj se pritisak kvazistati~i povisi do pritiska u kotlu. Ako je toplotna snaga kotla 4.56 MW, 2.28. Kotao proizvodi m =7

⋅

odrediti maseni protok pare koja se koristi u tehnolo{kom procesu ( m w) kao i snagu pumpe.

Q12 4

1

ka tehnolo{kom procesu

WT34

3

2 napojna voda


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strana 30

ta~ka 1:

p=13 bar, kJ h1 = h”″ = 2787 kg

x=1

p=2 bar, kJ h2 = hw = 84.0 kg

t=20oC

ta~ka 2:

p=13 bar,

ta~ka 4:

h4 =? ⋅

⋅

⋅

prvi zakon termodinamike za proces u kotlu: Q 12 = ∆ H12 + W T12 ⋅

⋅

⋅

Q12 = m⋅ (h1 − h 4 ) h4 = 2787 −

⇒

4.56 ⋅ 10 3 7⋅

3

10 3600

=441.86

p= 2 bar,

ta~ka 3:

h3 = hw = 440.95

Q 12 h4 = h1 − mp kJ kg

s4 = sw = 1.363

s3 = s4 = 1.363

kJ kgK

kJ kgK

kJ kg

prvi zakon termodinamike za proces me{awa: ⋅

⋅

⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⇒ H1 = H 2 ⋅ ⋅ ⋅ ⋅   mw ⋅ h2 +  mp − m w  ⋅ h1 = mp ⋅ h 3 ⇒   ⋅ t 440.95 − 2787 mw = 7 ⋅ = 6.08 h 84 − 2787

⋅

⋅

m w = mp ⋅

h3 − h1 h2 − h1

prvi zakon termodinamike za proces u pumpi: ⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⋅

⋅

⇒

W T12 = − m w ⋅ (h4 − h3 ) = 6 .08 ⋅


⋅

⋅

W T12 = − ∆ H12 10 3 ⋅ (441 .86 − 440 .95 ) =1.54 kW 3600

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strana 31

2.29. Voda stawa 1(p=2 bar, t=80oC) dostrujava kroz cev unutra{weg pre~nika d=40 mm brzinom 0.5 m/s. Prolaskom kroz delimi~no otvoren ventil se prigu{uje na p2 =0.4 bar i ulazi u odvaja~ te~nosti (od okoline toplotno izolovan). Odrediti: a) promenu entropije sistema za proces prigu{ivawa b) snagu kompresora koji izbacuje parnu fazu iz suda u okolinu pritiska p4 =1 bar c) snagu pumpe koja te~nu fazu iz suda prebacuje u parni kotao koji radi na pritisku p4 =4 bar napomena:

kompresije u kompresoru i pumpi su kvazistati~ke i adijabatske kompresor 1

2

4

3 pumpa

T

3

4

1

2′

2

2′′ s

ta~ka 1: h1 = hw =334.9 ⋅

m=

p1 =1 bar,

kJ , kg

t1 =80oC

s1 =sw=1.074

(

kJ , kgK

1 d2π 1 40 ⋅ 10 −3 ⋅w⋅ = ⋅ 0.5 ⋅ v1 4 0 .001028 4


v1 =vw = 0.001028

) π =0.61 2

m3 kg

kg s

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zbirka zadataka iz termodinamike p2 =0.4 bar,

ta~ka 2: h’”′ =317.7

x2 =

strana 32

kJ , kg

h 2 − h' h"−h'

=

h2 =h1 = 334.9

h′′ = 2636

kJ kg

kJ kg

334 .9 − 317 .7 =0.0074 2636 − 317 .7

s2 = sx = s '+ x 2 ⋅ (s ' ' −s ' ) = 1.0261 + 0.0074 ⋅ (7.67 − 1.0261) =1.075

kJ , kgK

s’′” =1.0261

s′′ = 7.67

p3 =4 bar,

ta~ka 3:

kJ kgK

s3 =s′= 1.0261

kJ kgK

kJ kg

h3 = hw= 318.5

p4 =1 bar,

ta~ka 4:

kJ kgK

h4 = hpp= 3129.7

s3 =s′= 7.67

kJ kgK

kJ kg

a) ⋅

⋅

⋅

∆ SSI = ∆ SRT + ∆ S O =0.61 ⋅

⋅

⋅

W K

⋅

∆ SRT = Sizlaz − Sulaz = m⋅ (s 2 − s 1 ) = 0 .61 ⋅ (1 .075 − 1.074 ) =0.61

W K

b) prvi zakon termodinamike za proces u kompresoru: ⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⋅

⇒

⋅

⋅

W T12 = − ∆ H12

⋅

W T 12 = − m⋅ x 2 ⋅ (h 4 − h' ' ) = −0.61 ⋅ 0 .0074 ⋅ (3129 .7 − 2636 ) =−2.23 kW c)

prvi zakon termodinamike za proces u pumpi: ⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⋅

⇒

⋅

⋅

W T12 = − ∆ H12

⋅

W T 12 = − m⋅ (1 − x 2 ) ⋅ (h3 − h' ) = − 0.61 ⋅ (1 − 0.0074 ) ⋅ (318 .5 − 317 .7 ) =−0.48 kW


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strana 33

2.30. U toplotno izolovan kompresor ulazi freon 12 (R12) stawa 1(p=1 bar, t=−20oC, ⋅

m =60 kg/h). Stawe freona 12 na izlazu iz kompresora je 2(p=8 bar), a snaga kompresora iznosi 1 kW. Nakon kompresije freon se hladi i potpuno kondenzuje u razmewiva~u toplote. Kao rashladni fluid u razmewiva~u toplote koristi se voda stawa 4(p=1bar, t=10oC) koja se prolaskom kroz razmewiva~ toplote zagreje do stawa 5(p=1 bar, t=30oC). Skicirati promene stawa freona 12 na hs dijagramu i odrediti: a) stepen dobrote adijabatske kompresije u kompresoru b) potro{wu vode u razmewiva~u toplote (kg/h) voda

4

5

3

2 WT12

1

freon

2

h 2k

1 3

s ta~ka 1: h1 =hpp

p=1 bar, kJ =647.4 kg

ta~ka 2k:

p2K=8 bar,

h2K=hpp =686.1

t=−20oC s1 =spp=1.612

kJ kgK

s2K=s1 =1.612

kJ kgK

kJ kg


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strana 34

p2 =1 bar,

ta~ka 2:

h2 =?

prvi zakon termodinamike za proces u kompresoru: ⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⋅

⋅

W T12 = − ∆ H12

⇒

W T12 h2 = h1 − mf

⋅

⋅

W T12 = − m f ⋅ (h2 − h1 ) h2 = 647 .4 −

⋅

⇒

−1 kJ =707.4 60 kg 3600

p=8 bar, kJ h1 =h′ =531.45 kg

x=0

ta~ka 3:

a) kp

ηd = b)

h1 − h2K 647.4 − 686 .1 = =0.645 h1 − h 2 647.4 − 707 .4 t=10oC

p=1 bar,

ta~ka 4: h1 =hw =42

kJ kg

(tabele za vodu)

p=1 bar, t=30oC kJ h1 =hw =125.7 (tabele za vodu) kg ta~ka 5:


⋅

⋅

Q 12 = ∆ H12 + W T12 ⋅

⋅

⇒ ⋅

⋅

m f ⋅ h 2 + m w ⋅ h4 = m f ⋅ h3 + m w ⋅ h5 ⋅

m w = 60 ⋅

⋅

⋅

H1 = H 2 ⇒

⋅

⋅

mw = m f ⋅

h2 − h3 h5 − h 4

kg 707 .4 − 531 .45 =12.61 h 125 .7 − 42


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strana 35

2.31. Pregrejana vodena para stawa 1(p=7 bar, t=450oC) ekspandira adijabatski u parnoj turbini sa stepenom dobrote η eks d =0.6 do stawa 2(p=1 bar). Po izlasku iz turbine para se u toplotno izolovanoj me{noj komori me{a sa vodom, masenog protoka ⋅

mw =2.3 kg/s stawa 3(p=1bar, t=14oC). Stawe voda na izlazu iz komore za me{awe je 4(p=1 bar, t=47oC). Skicirati procese u turbini i me{noj komori na Ts dijagramu i: a) odrediti snagu turbine (kW) b) dokazati da je proces me{awa pare i vode nepovratan para 1

WT12 3 voda

2

4 T

T 1

4 3

2 2k

s 1 –− 2 2 –− 4 3−4

2

s

promena stawa pare u turbini promena stawa pare u me{noj komori promena stawa vode u me{noj komori


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strana 36

a)

t=450oC

p=7 bar, kJ h1 = hpp = 3374.75 kg ta~ka 1:

p=1 bar,

ta~ka 2k: h′=417.4

kJ , kg

h′′= 2675

h2k>h′′ h2k

s1 =spp =7.789

kJ kgK

s2k = s1 =7.789

kJ kgK

kJ kg

ta~ka 2k se nalazi u oblasti pregrejane pare

kJ = hpp = 2854.3 kg p=1 bar,

ta~ka 2:

η eks = d

h1 − h2 h1 − h2k

⇒

η eks d 0.6 h2 = h1 − η eks d ⋅ (h1 − h 2k ) kJ kg

h2 = 3374 .75 − 0.6 ⋅ (3374 .75 − 2854 .3) = 3062.48 s2 = spp = 8.19

kJ kgK

p=1 bar, kJ h3 = hw = 58.6 kg

t=14oC

p= 1 bar, kJ h4 = hw = 196.74 kg

t=47oC

ta~ka 3:

ta~ka 4:

s3 = sw = 0.21

kJ kgK

s4 = sw = 0.66

kJ kgK

prvi zakon termodinamike za proces u me{noj komori: ⋅

⋅

⋅

⋅

⋅

Q 12 = ∆ H12 + W T12 ⇒ H1 = H 2 ⋅  ⋅  ⋅ ⋅   mw ⋅ h3 + mp h2 =  mw + m p  ⋅ h4 ⇒     ⋅ 196.74 − 58 .6 kg mp = 2.3 ⋅ = 0.11 3062 .48 − 196.74 s


⋅

⋅

mp = m w ⋅

h4 − h 3 h2 − h4

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strana 37


⋅

⋅

Q 12 = ∆ H12 + W T12

⋅

⋅

⋅

W T 12 = − ∆ H 12 = − mp ⋅ (h 2 − h1 )

⋅

W T 12 = −0.11 ⋅ (3062 .48 − 3374 .75 ) =34.35 kW b) ⋅

⋅

⋅

∆ S SI = ∆ S RT + ∆ S o = ... = 0.196 + 0 = 0.196 ⋅

kW K

⋅

Q12 kW ∆ S o= − =0 TO K ⋅ ⋅ ⋅ ⋅ ⋅ ⋅  ⋅  ∆ S RT = S izlaz − S ulaz=  mp + m w  ⋅ s 4 − mp ⋅ s 2 − m w ⋅ s 3   ⋅

∆ S RT = (0.11 + 2.3 ) ⋅ 0.66 − 0.11 ⋅ 8.19 − 2.3 ⋅ 0 .21 =0.207

kW K

2.32. Pregrejana vodena para stawa 1(p1 =70 bar, t1 =450oC) adijabatski ekspandira u parnoj turbini do stawa 2(p2 =1 bar). Snaga turbine je 200 kW. Nakon ekspanzije para se uvodi u kondenzator u kome se izobarski potpuno kondenzuje (stawe 3=kqu~ala voda). Protok vode za hla|ewe kondenzatora je mw=5 kg/s, stawe vode na ulazu u kondenzator je (p=1 bar, tw1 =20oC), a na izlazu iz kondenzatora je (p=1 bar, tw2 =45oC). Skicirati promene stawa pare (1−2−3) na hs dijagramu i odrediti: a) maseni protok vodene pare (kg/s) b) stepen suvo}e vodene pare na izlazu iz turbine c) stepen dobrote ekspanzije pare u turbini para

1

WT12 3 2 voda

4


5

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zbirka zadataka iz termodinamike a)

strana 38

ta~ka 4:

p4 =1 bar, kJ h4 = hw =83.9 kg

t4 =20oC

ta~ka 5:

p5 =1 bar, kJ h4 = hw =188.4 kg

t5 =45oC

p1 =70 bar, kJ h1 = hpp =3284.75 , kg

t1 =450oC

p= 1 bar, kJ h1 =h′ =417.4 kg

x=0

ta~ka 1:

ta~ka 3:

s1 =spp = 6.634

kJ kgK

prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ograni~enom isprekidanom linijom: ⋅

⋅

⋅

0 = H2 − H1 + W T12 ⋅

⇒

⋅

⋅

⋅

Q 12 = ∆ H12 + W T12

⋅ ⋅ ⋅ ⋅  ⋅  0 = mp ⋅ h3 + m w ⋅ h5  − mp ⋅ h1 + m w ⋅ h4  + W T 12    

⋅

m w ⋅ (h 5 − h 4 ) + W T12 5 ⋅ (188 .4 − 83 .9 ) + 200 kg mp = = =0.25 h1 − h3 3284 .75 − 417 .4 s ⋅

b) p2K =1 bar,

ta~ka 2K:

kJ , kgK s′’ > s2K > s′′” s 2k − s' s"− s'

=

kJ kgK

kJ kgK ta~ka 2K se nalazi u oblasti vla`ne pare

s′=1.3026

x 2K =

s2K=s1 = 6.634

s′′=7.36

6 .634 − 1.3026 =0.88 7.36 - 1.3026

h2K = hx = h' + x 2k ⋅ (h' '−h') = 417 .4 + 0.88 ⋅ (2675 − 417 .4 ) = 2404.09 h′=417.4

kJ , kgK


h′′=2675

kJ kg

kJ kgK

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strana 39

c) p2 =1 bar,

ta~ka 2:

h2 =?


⋅

⋅

⋅

⋅

h2 = h1 −

W T 12 ⋅

= 3284 .75 −

mp

η ex d =

⋅

⋅

W T 12 = − ∆ H 12 = − mp ⋅ (h 2 − h1 )

Q 12 = ∆ H12 + W T12

200 kJ =2484.75 0.25 kg

h1 − h2 3284 .75 − 2484 .75 = =0.91 h1 − h2k 3284 .75 − 2404 .1

h

1

2 2K 3

s


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strana 40

2.33. U sekundarnom krugu atomskog reaktora proizvodi se qm=4000 t/h suvozasi}ene vodene pare stawa 1(p=70bar). Proizvedena para se, prema skici, deli na dve struje qm1 i qm2. Para masenog protoka qm1 kvazistati~ki adijabatski ekspandira u turbini do stawa 2(p=8 bar). Vla`na para stawa 2 se u odvaja~u te~nosti (toplotno izolovan od okoline) deli na dve struje qm3 (suvozasi}ena para stawe 3) i qm4 (kqu~ala voda, stawe 4). Para stawa 3 se daqe pregreva (p=const) u razmewiva~u toplote do stawa 5(T=250 oC) na ra~un toplote koju oslobodii para qm1 kondenzacijom (p=const) do stawa 6(x=0). Skicirati stawa pare na Ts dijagramu i odrediti: a) masene protoke fluidnih struja, qm1 , qm2 , qm3 i qm4 b) snagu turbine qm

1

qm2 qm1 pregreja~ pare

WT12

qm3

2

qm2

6

3 qm4 odvaja~ te~nosti

4

p1 =70 bar, kJ h1 = h′′ =2772 , kg ta~ka 1:

ta~ka 2:

qm3

p1 =8 bar,

5 x=1 s1 =s′′ = 5.814

s2 =s1 =5.814

kJ kgK

kJ kgK

kJ kJ , s′′=6.663 kgK kgK s − s' 5 .814 − 2.046 x2 = 2 = =0.8161 s" −s' 6.663 - 2.046

s′=2.046


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strana 41

ta~ka 3:

p1 =8 bar, kJ h3 = h′′ =2769 kg

x=1

ta~ka 4:

p1 =8 bar, kJ h4 = h′ =720.9 kg

x=0

ta~ka 5:

p1 =8 bar, kJ h5 = hpp =2947.5 kg

T=250oC0

p1 =70 bar, kJ h6 = h′ =1267.4 kg

x=0

ta~ka 6:

materijalni bilans ra~ve: para koja napu{ta odvaja~ te~nosti:

qm = qm1 + qm2 qm3 = qm1 . x2

(1) (2)


⋅

⋅

Q 12 = ∆ H12 + W T12

⇒

⋅

⋅

H1 = H 2

qm 2 ⋅ h1 + qm 3 ⋅ h3 = q m2 ⋅ h6 + q m3 ⋅ h 5

(3)

Kombinovawem jedna~ina (1), (2) i (3) dobija se: t t t qm1 =3646.9 , qm2 =353.1 , qm3 = 2976.2 h h h materijalni bilans odvaja~a te~nosti: t qm4 = qm1 − qm3 = 670.7 h

qm1 = qm3 + qm4

⇒

prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ograni~enom isprekidanom linijom: ⋅

⋅

⋅

0 = H2 − H1 + W T12 ⋅

⇒

⋅

⋅

⋅

Q 12 = ∆ H12 + W T12

⋅

W T 12 = [qm ⋅ h1 ] − [qm2 ⋅ h 6 + qm 3 ⋅ h5 + q m4 ⋅ h4 ]

W T12 = [[4000 ⋅ 2772 ] − [353 .1 ⋅ 1267 .4 + 2976 .2 ⋅ 2947 .5 + 670 .7 ⋅ 720.9 ]] ⋅

10 3 3600

⋅

W T12 =384.6 MW


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strana 42

(2.34.)

zadatak za ve`bawe:

2.34. U parno-turbinskom postrojewu (slika) vodena para masenog protoka m=1.2 kg/s ekspandira u turbini visokog pritiska (TVP) od stawa 1(p=1 MPa, t=440oC) do stawa 2(p=0.5 MPa). Po izlasku iz turbine deo pare masenog protoka mA =0.4 kg/s me{a se adijabatski sa vodom stawa (p= 5 bar, tw=20oC). Stawe vode na izlasku iz komore za me{awe je (p= 5 bar, tw=45oC). Preostali deo pare se po izlasku iz turbine visokog pritiska izobarski zagreva do stawa 3(t=400oC), a zatim ekspandira u turbini niskog pritiska (TNP) do stawa 4(p=5 kPa). Ekspanzije u turbinama su adijabatske sa istim stepenom dobrote (ηdex=0.9). Odrediti: a) snagu turbina visokog i niskog pritiska (kW) b) maseni protok vode u komori za me{awe (kg/s) para

TVP

1

WT12

2 voda

5

mA

3 TNP

WT34

6

4 a) b)

⋅

WT

12 =230.5

kW, ⋅ kg mw =11.35 s

⋅

WT

34 =642.4


kW

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strana 43

PRVI I DRUGI ZAKON TERMODINAMIKE (PUWEWE I PRA@WEWE REZERVOARA) 2.35. U adijabatski izolovan rezervoar zapremine V=30 m3 , u kojem se nalazi vla`na vodena para stawa (p=1.2 bar, x=0.95), uvodi se jednim izolovanim cevovodom voda stawa (p=8 bar, t=15oC), a drugim izolovanim cevovodom suva vodena para stawa (p=30 bar). Stawe radne materije u rezervoaru na kraju procesa puwewa je (p=6 bar, x=0,1). Odrediti masu vode i masu suve pare uvedene u rezervoar. p=1.2 bar,

po~etak:

x=0.95

upo~etak= ux =u’′ + xp.(u′′” − u′’) = 439 .28 + 0 .95 ⋅ (2512 − 439 .28 ) =2408.4

kJ u’′ = 439.28 , kg

kJ kg

kJ u”′′ =2512 kg

vpo~etak= v x= v’′’ + x p. (v”′′ − v′’)= 0. 0010472 + 0. 95 ⋅ (1.429 − 0. 0010472) =1.3576 v’′ = 0.0010472 mpo~etak =

m3 , kg V =

vpo~etak

ulaz:

kraj:

v′′”=1.429

m3 kg

30 =22.1 kg 1.3576

p=30 bar,

x=1

h = h″=2804

p=8 bar,

t=15oC

h = h w = 62.8

p = 6 bar,

x=0.1

ukraj= ux =u’′ + xk.(u′′” − u′’) = 669.8 + 0.1 ⋅ (2568 − 669.8 ) =859.6 u’′ = 669.8

kJ , kg

m3 kg

u”′′ =2568

kJ kg kJ kg

kJ kg

kJ kg

vkraj= v x= v’′’ + x k. (v”′′ − v′’)= 0. 0011007 + 0. 1 ⋅ (0. 3156 − 0.0011007) =0.0326

m3 kg

m3 m3 , v′′”=0.3156 kg kg V 30 mkraj = = =920.25 kg vkraj 0.0326 v’′ = 0.0011007


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strana 44

prvi zakon termodinamike za proces puwewa suda: Q12 − W12 = Ukraj − Upo~etak − Hizlaz − Hulaz

0 = mkraj . ukraj − mpo~etak . upo~etak –− mw . hw − m″ . h″

(1)

zakon odràwa mase za proces puwewa suda: mpo~etak + mw + m″ = m kraj + mizlaz

(2)

kombinovawem jedna~ina (1) i (2) dobija se: mw =

(

)

(

mk ⋅ ukraj − h' ' − mp ⋅ up − h' '

)=

h w − h' ' 920.25 ⋅ (859. 6 − 2804) − 22. 21⋅ (2408. 4 − 2804) mw = = 649.55 kg 62.8 − 2804

m″ = mkraj − mpo~etak + mw = 920.25 − 22.21 − 649.55 = 248.49 kg 2.36. U verikalnom toplotno izolovanom cilindru, povr{ine popre~nog preseka A=0.1 m2 , nalazi se 0.05 kg vodene pare temperature 180oC, ispod toplotno izolovanog klipa mase koja odgovara teìni od 20 kN, a na koji spoqa deluje atmosferski pritisak od 0.1 MPa. U cilindar se, kroz toplotno izolovan cevovod, naknadno uvede 0.1 kg vodene pare pritiska 0.4 MPa i temperature 540oC. Zanemariti trewe klipa i odrediti: a) specifi~nu entalpiju i temperaturu vodene pare u cilindru na kraju procesa b) za koliko se podigao klip tokom ekspanzije

kraj ∆y po~etak ulaz


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strana 45

a) po~etak:

pp = patm +

mT ⋅ g 20 ⋅ 10 3 = 1 ⋅ 105 + = 3 ⋅ 10 5 Pa, A 0 .1

t=180oC

up = hp − pp . vp =…...= 2824 − 3 ⋅ 105 ⋅ 10− 3 ⋅ 0.6838 =2618.86 hp = hpp = 2824

kJ , kg

mp=0.05 kg p=4 bar, kJ hul = hpp =3572 , kg ulaz:

m3 kg V p = mp ⋅ vp = 0.05 ⋅ 0.6838 =0.03419 m3 t=540oC

(pregrejana para)

mul=0.1 kg mk = mp + mul =0.15 kg

prvi zakon termodinamike za proces puwewa:

(

)

p ⋅ Vp − Vk = mk ⋅ uk − mp ⋅ up − mul ⋅ hul

hk =

p ⋅ Vp + mp ⋅ up + mul ⋅ hul mk

hk =3322.67

kJ kg

vp=vpp = 0.6838

pk = pp = 3 bar,

kraj:

(pregrejana para)

=

Q12 − W12 = Uk − Up + Hiz − Hul

⇒ p ⋅ Vp = mk ⋅ hk − mp ⋅ up − mul ⋅ hul

3 ⋅ 10 5 ⋅ 10 − 3 ⋅ 0.03419 + 0 .05 ⋅ 2618 .86 + 0.1 ⋅ 3572 = 0.15

kJ kg

kJ , kg h′ < h k < h′′ h′=561.4

h′′ = 2725

kJ kg (stawe kraj je u pregrejanoj pari)

tk = tpp = 422.7oC,

vk = vpp =1.067

m3 kg

V k = mk ⋅ v k = 0.15 ⋅ 1.067 =0.16005 m3 b)

∆y =

Vk − Vp A

=

0 .16005 − 0.03419 =1.26 m 0 .1


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strana 46

2.37. Pomi~nim klipom sa tegom koji se kre}e bez trewa odràva se konstantan pritisak p=4 bar u vertikalnom cilindru u kojem se nalazi V=500 dm3 vode po~etne temperature t=20oC (slika kao u prethodnom zadatku). Parovodom se u cilindar postepeno uvodi mp=53 kg suvozasi}ene vodene pare pritiska p=6 bar, koja se pre me{awa prigu{uje do pritiska od p=4 bara. Temperatura me{avine (voda) na kraju procesa me{awa iznosi t=80oC. U toku me{awa usled neidealnog toplotnog izolovawa okolini se predaje 1.5 kW toplote. Odrediti vreme trajawa procesa me{awa. p= 4 bar,

po~etak:

vp = vw = 0.001001

3

m , kg

t=20oC hp = hw = 84.1

(voda) kJ

kg

up = uw = hp − pp ⋅ vp = 84. 1 − 4 ⋅ 105 ⋅ 10−3 ⋅ 0. 001001=83.7 mp =

Vp vp

=

0 .5 =499.5 kg 0.001001

p= 6 bar, kJ hu = h′′ =2757 kg ulaz:

p= 4 bar,

kraj:

kJ kg

x=1 mu= 53 kg t=80oC

uk = hk − pk . vk =…...= 335.1 − 4 ⋅ 105 ⋅ 10− 3 ⋅ 0.001028 =334.7

kJ kg

kJ m3 , vkraj = vw = 0.001028 kg kg mk = mp + mu = 499.5 + 53 = 552.5 kg, hk= hw = 335.1

V k = mk . vk = 552.5 ⋅ 0.001028 =0.568 m3 prvi zakon termodinamike za proces puwewa rezervoara:

Q12 − W12 = Uk − Up + Hiz − Hul

(

Q 12 = mk ⋅ uk − mp ⋅ up − mul ⋅ hul + p ⋅ Vk − Vp

)

Q12 = 552. 5 ⋅ 334. 7 − 499.5 ⋅ 83. 7 − 53 ⋅ 2757+ 4 ⋅105 ⋅10−3 ⋅ (0.568 − 0.5) =−2980.2 kJ

τ=

Q12 ⋅

Q12

=

−2980 .2 =1987 s − 1.5


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strana 47

2.38. U otvoren sud (slika) koji sadrì sme{u ml=15 kg leda i mw=20 kg vode u stawu termodinami~ke ravnteè, uvedeno je mp=0.8 kg pregrejane vodene pare stawa (p=3 bar, t=340oC). Okolni vazduh stawa O(p=1bar, to=7oC), tokom ovog procesa sme{i u sudu preda Q12 =320 kJ toplote. Zanemaruju}i promenu zapremine (tj. rad koji radno telo vr{i nad okolinom), odrediti promenu entropije sistema tokom ovog procesa.

po~etak:

t=0oC,

y=

mw 20 = =0.5714 m w + ml 20 + 15

up = uy ≅ hy= hl + y ⋅ (hw − hl ) = − 332 .4 + 0.5714 ⋅ 332 .4 = −142.47 sp = sl = s l + y ⋅ (s w − s l ) =

kJ kg

−332 .4 332 .4 kJ + 0.5714 ⋅ =− 0.522 kgK 273 273

mp = ml + mw =20 + 15 =35 kg p=3 bar, t=340oC (pregrejana para) kJ kJ hu = hpp = 3150 , su = spp = 7.835 , mu = 0.8 kg kgK kg ulaz:

p=1 bar,

kraj:

mk= mp + mu= 35 + 0.8 =35.8 kg

prvi zakon termodinamike za proces puwewa rezervoara:

Q12 − W12 = Uk − Up + Hiz − Hul

uk =

Q12 + mp ⋅ up + mul ⋅ h ul mk

ul < u k < uw

yk =

Q 12 = mk ⋅ uk − mp ⋅ up − mul ⋅ hul =

320 + 35 ⋅ (− 142.57 ) + 0 .8 ⋅ 3150 kJ =−60.05 35 .8 kg (stawe ″kraj″ je me{avina vode i leda)

u k − ul − 60 .05 + 332 .4 = =0.82 uw − ul 0 + 332 .4

sk = sy = s l + y ⋅ (s w − s l ) = dipl.ing. @eqko Ciganovi}

−332 .4 332.4 kJ + 0 .82 ⋅ = − 0.219 kgK 273 273 [email protected]

zbirka zadataka iz termodinamike ∆S SI = ∆S RT + ∆S O = ...=4.162 − 1.143 =3.019

strana 48

kJ K

Q 12 320 kJ =− = −1.143 To 280 K ∆S RT = mp ⋅ s k − s p + mu ⋅ (s k − s u ) = ∆S O= −

(

)

∆S RT = 35 ⋅ (− 0.219 + 0 .522 ) + 0.8 ⋅ (−0 .219 − 7.835 ) =4.162


kJ K

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strana 49

2.39. Zatvoreni rezervoar zapremine V=10 m3 sadrì kqu~alu vodu i suvu vodenu paru u stawu termodinami~ke ravnoteè na p=20 bar. Te~nost zauzima polovinu zapremine rezervoara. Iz rezervoara fluid moè isticati kroz ventil na vrhu i kroz ventil na dnu rezervoara. Dovo|ewem toplote za vreme isticawa temperatura vla`ne pare u rezervoaru se odràva stalnom. Odrediti koli~inu dovedene toplote ako je iz rezervoara isteklo 300 kg fluida kroz: a) dowi ventil b) gorwi ventil

a)

b)

a) p=20 bar,

po~etak: vp =

up=?

V 10 m3 =...= =0.0023 mp 4299 .74 kg

mp = m′ + m′′ = ... = 4249.53 + 50.21=4299.74 kg

V' 5 =…...= = 4249.53 kg v' 0 .0011766 V' ' 5 m' ' = =...= = 50.21 kg v' ' 0 .09958 m' =

v′ = 0.0011766

m3 , kg

v′′=0.09958

m3 kg

up= ux =u’′ + xp.(u′′” − u′’)=...= 906 .1 + 0.0117 ⋅ (2600 − 906 .1) =925.92 u’′ = 906.1

xp =

kJ , kg

u”′′ =2600

kJ kg

kJ kg

m' ' 50 .21 = = 0.0117 m'+ m' ' 4249 .53 + 50 .21

Tp = 212.37oC


(temperatura kqu~awa za pritisak od p=20 bar)

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strana 50

Tk=Tp=212.37oC,

kraj:

uk=?

mk =mp − miz = 4299.74 − 300 = 3999.74 kg V 10 m3 vk = = = 0.0025 mk 3999 .74 kg v′’ < vk < v”′ ′

xk =

v kj − v' v' ' − v'

(stawe kraj se nalazi u oblasti vla`ne pare)

=

0.0025 − 0.0011766 =0.0134 0.09958 − 0.0011766

uk = ux = =u’′ + xp.(u′′” − u′’)= 906 .1 + 0.0134 ⋅ (2600 − 906 .1) = 928.8

hiz= h′’ = 908.5

kJ kg

kJ kg

izlaz:

mizlaz = 300 kg,

napomena:

zbog poloàja ventila iz suda isti~e kqu~ala voda

prvi zakon termodinamike za slu~aj pra`wewa suda:

Q 12 − W12 = U k − U p + Hiz − H ul

⇒

Q 12 = mk ⋅ u k − mp ⋅ up + miz ⋅ hiz

Q12 = 3999 .74 ⋅ 928 .8 − 4299 .74 ⋅ 925 .92 + 300 ⋅ 908 .5 =6293.25 kJ b) po~etak:

nema promena u odnosu na pod a)

kraj:

nema promena u odnosu na pod a)

izlaz:

miz = 300 kg,

napomena:

kJ kg zbog poloàja ventila iz suda isti~e suva para hiz= h′′ =2799

prvi zakon termodinamike za slu~aj pra`wewa suda:

Q 12 − W12 = U k − U p + Hiz − H ul

⇒

Q 12 = mk ⋅ u k − mp ⋅ up + miz ⋅ hiz

Q12 = 3999 .74 ⋅ 928 .8 − 4299 .74 ⋅ 925 .92 + 300 ⋅ 2799 =573443.25 kJ


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strana 51

2.40. U zatvorenom, toplotno izolovanom rezervoaru, zapremine V=0.5 m3 nalazi se 30 kg vla`ne vodene pare. Kada, pri zagrevawu, pritisak pare u rezervoaru dostigne vrednost p=5 MPa, biva iskqu~en elektri~ni greja~ stalne snage i istovremeno otvoren sigurnosni ventil na rezervoaru tako da jedan deo vodene pare naglo istekne u okolinu. Po zatvarawu ventila pritisak vodene pare u rezervoaru iznosi 3 MPa. Preostala vla`na para biva potom dogrevana istim elektri~nim greja~em, stalne snage od 800 W. Skicirati promene stawa vodene pare na Ts dijagramu i odrediti: a) masu vla`ne pare u rezervoaru nakon zatvarawa sigurnosnog ventila b) vreme nakon kojeg }e se sigurnosni ventil ponovo otvoriti a) p= 50 bar,

po~etak:

xp =

v p − v' v' ' − v'

=

v′ = 0.0012857

vp =

V 0.5 m3 = 0.0167 = mp 30 kg

0 .0167 − 0.0012857 =0.404 0 .03944 − 0.0012857

m3 , kg

v′′=0.03944

m3 kg

sp= sx =s’′ + xp.(s′′” − s′’) = ...= 2.921 + 0.404 ⋅ (5.973 − 2 .921) =4.154 s’′ = 2.921

kraj:

xk =

kJ kgK

s”′ ′ = 5.973

p=30 bar,

s k − s' s' ' −s'

s’′ = 2.646

=...=

kJ kgK

kJ kgK

kJ kgK

sk =sp =4.154

kJ kgK

4.154 − 2 .646 =0.426 6.186 − 2.646 kJ s”′ ′ = 6.186 kgK

vk= vx =v’′’ + xk. (v”′′ −” v′’)= 0. 0012163 + 0. 426 ⋅ (0. 06665 − 0.0012163) =0.0291 v’′ = 0.0012163 mk =

m3 , kg

v′′”=0.06665

m3 kg

m3 kg

V 0. 5 = =17.18 kg vk 0.0291

uk= ux =u’′ + xk.(u′′” − u′’)=...= 1004.7 + 0.426 ⋅ (2604 − 1004. 7) =1686 u’′ = 1004.7

kJ , kg


u”′′ =2604

kJ kg

kJ kg [email protected]


strana 52

b)

m3 , kg

ta~ka 1=kraj

p1 =30 bar,

v1 =0.0291

ta~ka 2:

p2 =50 bar,

v2 = v1 =0.0291

v′ = 0.0012857

m3 , kg

v′ < v2 < v′′ x2 =

u2 = 1686

kJ kg

m3 kg

m3 kg (ta~ka 2 je vla`na para) v′′=0.03944

v2 − v' 0.0291 − 0. 0012857 =0.729 v"−v ' 0. 03944 − 0.0012857

u2 = ux =u’′ + x2 .(u′′” − u′’)=...= 1148 + 0. 729 ⋅ (2597 − 1148 ) =2204.32

kJ u’′ = 1148 , kg

kJ kg

kJ u”′′ =2597 kg

prvi zakon termodinamike za proces u zatvorenom termodinami~kom sistemu Q12 = ∆U12 + W12

⇒

Q12 = mk . (u2 –- u1 )

Q12 = 17. 18 ⋅ (2204. 32 − 1686) =8904.74 kW

τ=

Q12 ⋅

=

Q12

8904.74 =11131 s 0.8

≅ 3h

T 0 = po~etak 1 = kraj 0

2

1 s


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strana 53

2.41. Toplotno izolovan rezervoar zapremine V=20 m3 , sadrì vodenu paru po~etnog stawa P(p=2 MPa, T=553 K). Rezervoar je povezan sa toplotno izolovanom parnom turbinom, u kojoj se odvija ravnote`no (kvazistati~ko) {irewe pare (slika). Pritisak pare na izlazu iz turbine je stalan i iznosi piz=0.15 MPa, a proces se odvija dok pritisak pare u rezervoaru ne opadne na pk=0.3 MPa. Zanemaruju}i prigu{ewe paare u ventilu, odrediti koji izvr{i para tokom ovog procesa.

WT

piz p=20 bar,

po~etak:

t=280oC 5

−3

up=hp –− pp. vp = ...= 2972 − 20 ⋅ 10 ⋅ 10 kJ , kg

hp= hpp = 2972 mp =

p=3 bar,

kJ kgK s′ < sk < s′′

sp= spp= 6.674

kJ kgK

s k − s' s' ' −s'

kJ kgK

s”′ ′ = 6.992

=...=

vk = vx = 0.5694

sk= sp= 6.674

kJ kgK (vla`na para)

s’′ = 1.672

mk =

m3 , kg

V 20 = = 166.67 kg vp 0. 12

kraj:

xk =

vp= vpp = 0.12

(pregrejana para) kJ ⋅ 0.12 =2732 kg

6. 674 − 1. 672 =0.9402 6.992 − 1. 672

m3 kJ , uk = ux = 2424.09 kg kg

V 20 = = 35.12 kg vk 0.5694


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strana 54

p=1.5 bar,

izlaz: s’′ = 1.4336

kJ , kgK

siz = sp= 6.674

kJ kgK (vla`na para)

s”′ ′ = 7.223

s′ < siz < s′′ xiz =

kJ kgK

siz − s' 6. 674 − 1.4336 = =0.905 s' ' −s' 7. 223 − 1.4336

hiz= h′ + xiz . (h′′ − h′) =...= 467. 2 + 0.905 ⋅ (2693 − 467.2 ) =2481.55 h’′ = 467.2

kJ , kgK

h”′′ = 2693

kJ kg

kJ kgK

miz= mp −– mk = 166.67 − 35.12 = 131.55 kg prvi zakon termodinamike za slu~aj pra`wewa suda:

Q 12 − W12 = U k − U p + Hiz − H ul

⇒

W12 = −mk ⋅ uk + mp ⋅ up − miz ⋅ hiz

W12 = −35. 12 ⋅ 2424. 09 + 166. 67 ⋅ 2732 − 131. 55 ⋅ 2481. 55 =43.76 MJ 2.42. U ispariva~u zapremine V=2 m3 , u kome se odvija proces isparavawa vode na ⋅

pritisku p=1 MPa, kontinualno se uvodi mul =10 kg/s kqu~ale vode pritiska p=1 MPa, a iz wega izvodi nastala suva para istog pritiska. Greja~ima, urowenim u ⋅

kqu~alu vodu u ispariva~u, vodi se predaje Q12 = 19.26 MW toplote. Ako se u po~etnom trenutku u ispariva~u na pritisku p=1 MPa nalazila me{avina kqu~ale vode i suve pare u stawu termodinami~ke ravnoteè, a kqu~ala voda pri tom zauzimala 1/10 zapremine ispariva~a, izra~unati vreme potrebno da kqu~ala voda ispuni ceo ispariva~. Zanemariti razmenu toplote sa okolinom. suva para vla`na para

kqu~ala voda +Q12


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zbirka zadataka iz termodinamike p=10 bar,

po~etak:

strana 55 x=?

m3 m3 , v′′=0.1946 kg kg V' V' ' 0.2 1.8 m' = =… =177.42 kg, m' ' = = = 9.25 kg v' 0 .0011273 v' ' 0 .1946

v′ = 0.0011273

mp= m′ + m′′ =177.42 + 9.25 =186.67 kg

xp =

m' ' 9 .25 = =0.0496 m'+m' ' 177.42 + 9.25

up= ux =u’′ + xp.(u′′” − u′’)=...= 761.6 + 0.0496 ⋅ (2583 − 761.6 ) =851.94 u’′ = 761.6

kJ , kg

u”′′ =2583 p = 10 bar,

kraj:

vk = v’′ = 0.0011273 mk =

3

m , kg

kJ kg

kJ kg

x=0 uk = u′’ = 761.6

kJ kg

V 2 = = 1774.15 kg vk 0.0011273

p = 10 bar, x=0 ⋅ kJ hul = h′’ =762.7 , mul = mul ⋅ τ kg ulaz:

p = 10 bar, kJ hiz = h′′” = 2778 kg izlaz:

x=1

prvi zakon termodinamike za slu~aj istovremenog puwewa i pra`wewa suda:

Q 12 − W12 = U k − U p + H iz − H ul ⋅

Q12 = mk ⋅ uk − mp ⋅ up + miz ⋅ hiz − mul ⋅ hul ⋅

Q 12 ⋅ τ = mk ⋅ uk − m p ⋅ up + m iz ⋅ hiz − mul ⋅ t ⋅ hul zakon o odràwu mase: ⋅

mp + mul ⋅ τ = mk + miz


(1)

mp + mul= mk + miz (2)

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strana 56

kombinovawem jedna~ina (1) i (2) dobija se ⋅ ⋅ ⋅   Q 12 ⋅ τ = mk ⋅ uk − mp ⋅ up +  mp + m ul ⋅ τ − mk  ⋅ h iz − m ul ⋅ t ⋅ hul   mk ⋅ uk − mp ⋅ up + m p − m k ⋅ hiz τ= ⋅ ⋅ ⋅ Q 12 − m ul ⋅ h iz + m ul ⋅ hul

(

τ=

)

1774.15 ⋅ 761 .6 − 186 .67 ⋅ 851 .94 + (186 .67 − 1774 .15 ) ⋅ 2778 19 .26 ⋅ 10 3 − 10 ⋅ 2778 + 10 ⋅ 762 .7

zadatak za ve`bawe:

=3603 s

(2.43.)

2.43. Kondenzacija pare vr{i se u prostoru zapremine V=2 m3 pri pritisku od 0.1 MPa. U posudu se kontinualno uvodi 100 kg/h suvozasi}ene vodene pare, a iz we izvodi nastala kqu~ala voda istog pritiska piz=0.1 MPa. Ako se u po~etnom trenutku u posudi na pritisku pp=0.1 MPa nalazila kqu~ala voda i suvozasi}ena para u stawu termodinami~ke ravnoteè, pri ~emu je te~nost zauzimala 1/8 zapremine suda, odrediti vreme potrebno da te~nost ispuni 1/2 zapremine posude. Toplotna snaga koja se razmewuje sa hladwakom iznosi 250 kW. Zanemariti predaju toplote okolini. − Q12

suva para

vla`na para kqu~ala voda re{ewe:

τ=5.43 s


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strana 1

3. MAKSIMALAN RAD, EKSERGIJA 4/2/!U zatvorenom rezervoaru nalazi se n>21!lh vazduha (idealan gas) stawa 2)q>2/7!cbs-!U>634!L*. Stawe okoline odre|eno je sa P)q>2!cbs-!U>3:9!L*/ Odrediti koliko se najvi{e zapreminskog rada moè dobiti dovo|ewem vazduha stawa 1 u ravnoteù sa okolinom stawa O (maksimalan rad, eksergija zatvorenog termodinami~kog sistema). Dobijeni rad predstaviti na qw dijagramu. w2 =

S h ⋅ U2 q2

=

398 ⋅ 634 2/7 ⋅ 216

=1/:492!

n4 lh

wp =

S h ⋅ Up qp

=

398 ⋅ 3:9 2⋅ 216

=1/9664!

n4 lh

Xnby!>! n ⋅ [− ∆v2p + Up ⋅ ∆t2p − q p ⋅ ∆w 2p ]  Xnby!>! n ⋅ − d w ⋅ (Up − U2 ) + Up 

 U q ⋅  d q mo p − S h mo p U q2 2 

   − q p ⋅ (w p − w 2 )  

  3:9 2   3 − 1/398 ⋅ mo Xnby!>! 21 ⋅ − 1/83 ⋅ (3:9 − 634) + 3:4 ⋅ 2⋅ mo  − 2⋅ 21 ⋅ (1/9664 − 1/:492) 634 2/7     Xnby!>!2731!−!2364!,!94!>!561!lK q

q

2

P

2

P

B

⊕

−

,

w

q

B

⊕

w

q

2

=

P B

2

P B

,

w


w

{fmlp@fvofu/zv


strana 2

4/3/ Termodinami~ki sistem se sastoji od zatvorenog suda u kojem se nalazi kiseonik (idealan gas) stawa 2)q>2!cbs-!u>511pD-!n>2!lh* i okoline stawa P)q>2!cbs-!u>31pD*/ Zapreminski udeo kiseonika u okolnom vazduhu (idealan gas) iznosi sP3 >1/32. Odrediti: a) da li se navedeni termodinami~ki sistem moè upotrebiti za dobijawe X>261!lK!sbeb b) koliko bi trebalo da iznosi pritisak u sudu (q2*- uz ostale nepromewene uslove, da bi od sistema mogli dobiti X>261!lK rada povratnim promenama stawa a) w2 =

S h ⋅ U2

(q p )P3 wp =

=

371 ⋅ 784 6

=2/86!

n4 lh

2 ⋅ 21 = sP3 ⋅ q p > 1/32 ⋅ 2 !>!1/32!cbs

q2

S h ⋅ Up

(q p )P3

=

371 ⋅ 3:4 1/32 ⋅ 21 6

=4/74!

n4 lh

[

Xnby!>! n ⋅ − ∆v2p + Up ⋅ ∆t2p − (q p )P ⋅ ∆w 2p 3

 Xnby!>! n ⋅ − d w ⋅ (Up − U2 ) + Up 

]

(q p )P3  U ⋅  d q mo p − S h mo  U2 q2 

   − (q p ) ⋅ (w p − w 2 ) P 3   

  3:4 1/32   3 Xnby!>! 2⋅ − 1/76 ⋅ (3:4 − 784) + 3:4 ⋅  1/:2⋅ mo − 1/37 ⋅ mo  − 1/32⋅ 21 ⋅ (2/86 − 4/74 ) 784 2     Xnby!>!358!−!214/3!−!4:/6!>!215/4!lK X!?!Xnby

⇒!

sistem se ne moè upotrebiti za dobijawe 261!lK rada, jer najve}i mogu}i rad koji moèmo dobiti (Eksergija zatvorenog termodinami~kog sistema) iznosi 215/4!lK

b) za povratne promene stawa vaì: X>Xnby!>261!lK  2  U  Xnby  2  ⋅ d mo p − + d w ⋅ (Up − U2 ) + qp ⋅ (w p − w2 ) ⋅   q2 = (qp )P ⋅ fyq − 3  Sh  q U2  n  Up      2  3:4 261  2   q2 = 1/32⋅ fyq − ⋅ 1/:2⋅ mo − + 1/76 ⋅ (3:4 − 784) + 1/32⋅ 213 ⋅ (4/74 − 2/86) ⋅  784  2  3:4    1/37  q2!>2/92!cbs


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strana 3

3.3. U toplotno izolovanom rezervoaru zapremine W>31!n4 nalazi se vazduh (idealan gas) po~etnog stawa 2)q>31!cbs-!u>381pD*. Rezervoar je povezan sa gasnom turbinom (slika) u kojoj se vazduh {iri kvazistati~ki adijabatski. Pritisak na izlazu iz turbine je stalan i iznosi 4!cbs. Proces traje sve dok pritisak u rezervoaru ne opadne na 9!cbs. a) odrediti radnu sposobnost vazduha u rezervoaru (maksimalan rad) pre otvarawa ventila i predstaviti je grafi~ki u qw i Ut koordinatnim sistemima ako je stawe okoline P)q>2!cbsu>31pD* b) odrediti mehani~ki rad izvr{en u toku procesa (pri tome zanemariti proces prig{ivawa u ventilu)

X23

!qj{mb{ a) n2 = w2 =

q2 ⋅ W 31 ⋅ 21 6 ⋅ 31 > >367/78!lh 398 ⋅ 654 S h ⋅ U2 S h U2 q2

=

398 ⋅ 654 31 ⋅ 21 6

>!1/189!

n4 -! lh

[

wP =

Xnby!>! n ⋅ − ∆v2p + Up ⋅ ∆t2p − (q p )P ⋅ ∆w 2p 3

S h UP qP

=

398 ⋅ 3:4 2 ⋅ 21 6

>!1/952!

n4 lh

]

  Up q − Sh mo p * − qp )w2 − w p * Xnby = n2 ⋅ dw )U2 − Up * + Up )dq mo U q 2 2   

Xnby!>! 367/78− 1/83 ⋅ (3:9 − 654) + 3:4 ⋅ 2 ⋅ mo 



 3:9 2  − 1/398 ⋅ mo  − 2 ⋅ 213 ⋅ (1/952 − 1/189) 654 31  

Xnby!>!56/38!!,!2:/64!!−2:/69!>!56/33!NK


{fmlp@fvofu/zv


strana 4

q

U 2

2

B B C

P

C

P

w

t

b) Uj{mb{ = Uqpdfubl

 q ⋅  j{mb{  q qpdfubl 

   

κ −2 κ

 4  = 654 ⋅    31 

κ −2

2/5 −2 2/5

= 427 L

2/5 −2

 q lsbk  κ  9  2/5  Ulsbk = Uqpdfubl ⋅  = 654 ⋅   = 529 L  q qpdfubl   31    q lsbk ⋅ W 9 ⋅ 21 6 ⋅ 31 >244/48!lh, nj{mb{>nqp•fubl!−!nlsbk>234/4!lh nlsbk = = S h ⋅ Ulsbk 398 ⋅ 529 prvi zakon termodinamike za proces pra`wewa: R 23 − X23 = Vlsbk − Vqp•fubl + Ij{mb{ − Ivmb{ X23 = −nlsbk ⋅ d w ⋅ Ulsbk + nqp•fubl ⋅ d w ⋅ Uqp•fubl − nj{mb{ ⋅ d q ⋅ Uj{mb{ X23 = −244/48 ⋅ 1/83 ⋅ 529 + 367/78 ⋅ 1/83 ⋅ 654 − 234/4 ⋅ 2 ⋅ 427 >32/36!NK


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strana 5

4/5/!Klipni kompresor ravnote`no (kvazistati~ki) i politropski, sa eksponentom politrope o>2/4, sabija okolni vazduh (idealan gas) stawa 1)q>2!cbs-!U>3:2!L*-!na pritisak q3>5!cbs, i puni toplotno izolovan rezervoar zapremine W>21!n4. Toplotno stawe vazduha u rezervoaru na po~etku procesa puwewa isto je kao i stawe okolnog vazduha 1/ Odrediti: a) masu vazduha koju je potrebno ubaciti u rezervoar da bi pritisak vazduha u rezervoaru dostigao vrednost od 4!cbsb b) eksergiju vazduha (maksimalan rad) u rezervoaru u tom trenutku

2!⇒!4 2

3

X23 a) U2  q2   = U3  q 3  nqp•fubl =

o−2 o

q U3 = U2 ⋅  3  q2

⇒

q qp•fubl ⋅ W S h ⋅ Uqp•fubl

>

  

o−2 o

5 > 3:2 ⋅    2

2/4 −2 2/4

>511/82!L

2 ⋅ 21 6 ⋅ 21 >22/:8!lh 398 ⋅ 3:2

prvi zakon termodiamike za proces puwewa rezervoara: 1= po~etak,

2 = ulaz,

3=kraj

R 23 − X23 = Vlsbk − Vqp•fubl + Ij{mb{ − Ivmb{ 1 = nlsbk ⋅ d w ⋅ Ulsbk − nqp•fubl ⋅ d w ⋅ Uqp•fubl − nvmb{ ⋅ d q ⋅ Uvmb{

)2*

zakon odràwa mase za proces puwewa rezervoara: nqp•fubl + nvmb{ = nlsbk + nj{mb{


)3*

{fmlp@fvofu/zv


strana 6

jedna~ina stawa idealnog gasa za zavr{etak puwewa: q lsbk ⋅ W = nlsbk ⋅ S h ⋅ Ulsbk

(3)

kombinovawem jedna~ina (1) i (2) dobija se:

(

)

1 = nlsbk ⋅ d w ⋅ Ulsbk − nqp•fubl ⋅ d w ⋅ Uqp•fubl − nlsbk − nqp•fubl ⋅ d q ⋅ Uvmb{

)5*

kombinovawem jedna~ina )4*!i!)5* dobija se: q lsbk ⋅ W 1 = nlsbk ⋅ d w ⋅ − nqp•fubl ⋅ d w ⋅ Uqp•fubl − nlsbk − nqp•fubl ⋅ d q ⋅ Uvmb{ S h ⋅ nlsbk

(

dw ⋅

q lsbk ⋅ W

nlsbk = nqp•fubl +

Sh

)

− nqp•fubl ⋅ d w ⋅ Uqp•fubl d q ⋅ Uvmb{

1/83 ⋅ nlsbk = 22/:8 +

4 ⋅ 21 ⋅ 21 − 22/:8 ⋅ 1/83 ⋅ 3:2 398 >35/5:!lh 2 ⋅ 511/82 6

nvmb{>nlsbk!−!nqp•fubl!>!35/5:!−!22/:8!>23/63!lh

napomena:

Ulsbk>

q lsbk ⋅ W S h ⋅ nlsbk

>

4 ⋅ 21 6 ⋅ 21 >537/94!L 398 ⋅ 35/5:

b) okolina (ta~ka O) = ta~ka 1 w4 =

S h ⋅ U4 q2

=

398 ⋅ 537/94 4 ⋅ 21 6

kraj (ta~ka kraj) = ta~ka 3

=1/519!

S h ⋅ Up 398 ⋅ 3:2 n4 n4 -!!!!!!! w p = = =1/946! lh lh qp 2 ⋅ 21 6

Xnby!>! nlsbk ⋅ [− ∆v 4p + Up ⋅ ∆t 4p − q p ⋅ ∆w 4p ]  Xnby!>! n ⋅ − d w ⋅ (Up − U4 ) + Up 

 U q ⋅  d q mo p − S h mo p U4 q4 



Xnby!> 35/5: ⋅ − 1/83 ⋅ (3:2 − 537/94) + 3:2⋅ 2 ⋅ mo 



   − q p ⋅ (w p − w 4 )   

 3:2 2 − 1/398 ⋅ mo  − 2 ⋅ 213 ⋅ (1/946 − 1/519) 537/94 4 

Xnby!>!34:6!−!597!−53/8!>!2977/4!lK


{fmlp@fvofu/zv


strana 7

4/6/!Odrediti eksergiju struje vazduha (idealan gas) stawa 2)q2>2/7!cbs-!u2>361pD⋅

n >2!lh0t* i predstaviti je grafi~ki na qw dijagramu. Pod okolinom smatrati vazduh (idealan gas) stawa P)qp>2!cbs-!up>36pD*/ ⋅

⋅

Fy 2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p ) >!/// ⋅ ⋅  q  U Fy2 = n⋅ − dq )Up − U2* + Up )dq mo p − S h mo p * q2  U2   ⋅ ⋅ 3:9 2   Fy2 = 2⋅ − 2 ⋅ )3:9 − 634* + 3:9 ⋅ )2 ⋅ mo − 1/398 ⋅ mo * 634 2/7   ⋅

Fy2 >!336!!−!238/5!>!:8/7!lX

q

q 2 P

2

,

=

P

⊕

−

w

w

q 2 , P

w


{fmlp@fvofu/zv


strana 8 ⋅

4/7/ Eksergija toka vazduha (idealan gas), masenog protoka n >1/6!lh0t, koji struji sredwom brzinom ⋅

x>39!n0t, pri stawu vazduha u okolini P)qp>2!cbs-!Up>3:4!L), iznosi Fy 2 >94!lX. Promena entropije okoline, koja bi nastala povratnim (reverzibilnim) promenama stawa vazduha (bez promene ⋅

brzine) na pritisak i temperaturu okoline iznosila bi ∆ T plpmjof>−!1/2!lX0L. a) odredti pritisak i temperaturu vazduha stawa 1 b) grafi~ki prikazati u qw, koordinatnom sistemu eksergiju vazdu{nog toka, ne uzimaju}i u obzir deo koji se odnosi na brzinu a) drugi zakon termodinamike za proces od stawa 1 do stawa O ⋅

⋅

⋅

∆ T tjtufn = ∆ T sbeop ` ufmp + ∆ T plpmjob ⋅

∆ T2p

⋅

⋅

lX = − ∆ T plpmjob >1/2! L

⋅

⇒

∆ t2p =

⋅

1 = ∆ T2p + ∆ T plpmjob ⋅

∆ T2p ⋅

>1/3!

n

lK lhL

⋅ ⋅ ⋅  x3  Fy 2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p + f l2 ) = n⋅ − dq )Up − U2* + Up ⋅ ∆t2p +  3   ⋅

Fy 2 ⋅

U2 = UP + n

− UP ∆t2P − dq

⋅ ⋅ U ∆ T 21!>! n⋅ )dq mo p U2

x3 3

94 39 3 − 3:4 ⋅ 1/3 − ⋅ 21 −4 1 / 6 3 >511!L > 3:4 + 2

⋅  Up ∆ T2p    dq mo U − ⋅  qp  2 n  >6/:7!cbs * !!⇒!! q2 = q p ⋅ fyq − S h mo  Sh q2        

q

2 Fy2

!P

w


{fmlp@fvofu/zv


strana 9

4/8/ U horizontalnoj cevi pre~nika e>311!nn ugra|en je greja~ stalne temperature UUJ>711!L. Stawe vazduha u preseku 1 odre|eno je sa 2)q>3!cbs-!U>411!L-!x>31!n0t* a u preseku 2 sa 3)q>3!cbs-!U>511!L*. Stawe okoline odre|eno je veli~iama stawa P)qp>2!cbs-!Up>3:1!L*. Cev je toplotno izolovana od okoline. Odrediti: a) snagu ugra|enog greja~a b) eksergiju vazduha u preseku 1 i preseku 2 c) eksergijski stepen korisnosti procesa u cevi

⋅

+ R23

2

3

a) w2 = w3 =

S h ⋅ U2 q2 S h ⋅ U3 q3

>

398 ⋅ 411 3 ⋅ 21 6

>1/5416!

398 ⋅ 511

>

n4 lh

>1/685!

3 ⋅ 21 6

n4 lh

⋅

⋅

jedna~ina kontinuiteta:

n2 = n3

x2 ⋅ B2 x 3 ⋅ B 3 = w2 w3

x 3 = x2 ⋅

⋅

n=

⇒

w3 1/685 n > 31 ⋅ >!37/78! 1/5416 t w2

e3 π 1/3 3 ⋅ π 31 ⋅ lh 5 = 5 >2/57! t w2 1/5416

x2 ⋅


⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X U23 + ∆Fl23 + ∆Fq23 > n⋅ d q ⋅ (U3 − U2 ) + n⋅ ⋅

R 23 = 2/57 ⋅ 2 ⋅ (511 − 411) + 2/57 ⋅


31 3 − 37/78 3 3

x 23 − x 33 > 3

⋅ 21 −4 >256/88!lX

{fmlp@fvofu/zv


strana 10

b) ⋅

⋅

Fy 2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p + f l2 ) = ⋅ ⋅  Fy2 = n⋅ − dq )Up − U2* + Up  ⋅

 U q ⋅  d q mo p − S h mo p U q2 2 









Fy2 > 2/57 ⋅ − 2⋅ )3:1 − 411* + 3:1 ⋅ 2⋅ mo ⋅

 x 23   +  3  

 3:1 2  313 − 1/398 ⋅ mo  + ⋅ 21−4  >95/88!lX 411 3 3 

⋅

Fy 3 = n⋅ (− ∆i 3p + Up ⋅ ∆t 3p + f l3 ) = ⋅  ⋅ Fy 3 = n⋅ − dq )Up − U3 * + Up  ⋅

 U q ⋅  d q mo p − S h mo p U3 q3 









Fy3 > 2/57 ⋅ − 2⋅ )3:1 − 511* + 3:1 ⋅ 2⋅ mo

c) ⋅

⋅

Fy R = R 23 ⋅

 x 33   +  3  

 3:1 2  313 − 1/398 ⋅ mo  + ⋅ 21−4  >219/:8!lX 511 3 3 

UUJ − Up 711 − 3:1 > 256/88 ⋅ >86/42!lX 711 UUJ

drugi zakon termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

∆ T tjtufn = ∆ Tsbeop ` ufmp + ∆ T UJ =.. . ⋅ ⋅ ⋅  q U ∆ T sbeop ` ufmp = ∆ T23 > n⋅  d q mo 3 − S h mo 3  U2 q2  ⋅

∆ T UJ

  > 2/57 ⋅ 2 ⋅ mo 511 >1/53! lX  411 L 

⋅

256/88 R 23 lX >−1/35! =− >− 711 L UUJ

⋅

∆ T tjtufn = 1/53 − 1/35 >!1/29! ⋅

lX L

⋅

Fy h = Up ⋅ ∆ T tjtufn > 3:1 ⋅ 1/29 >63/3!lX ⋅

ηFy =

⋅

⋅

Fy2 + Fy R − Fy h ⋅

⋅

Fy 2 + Fy R


>

95/88 + 86/42 − 63/3 >1/78 95/88 + 86/42

{fmlp@fvofu/zv


strana 11

4/9. U neizolovanoj komori me{aju se, pri stacionarnim uslovima, dve struje idealnih gasova: kiseonika ⋅

⋅

B) n B>7!lh0t-!qB>1/29!NQb-!UB>634!L*!i azota!C) n C>4!lh0t-!qC>1/44!NQb-!UC>974!L*. U toku procesa me{awa toplotni protok u okolni vazduh stawa )qp>1/2!NQb-!Up>3:4!L-!sP3>1/32-!sO3>1/8:* iznosi 511 lX. Pritisak me{avine na izlazu iz komore je q>!1/26!NQb. Odrediti brzinu gubitka eksergije u toku procesa me{awa kao i eksergijski stepen korisnosti procesa u me{noj komori. Zanemariti promene makroskopske potencijalne i kineti~ke energije. prvi zadatak termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X U23

⋅

⋅

R 23 = I3 − I2

⋅ ⋅ ⋅ ⋅  ⋅  R23 =  nB ⋅ dqB + nC ⋅ dqC  ⋅ U+ − nB ⋅ dqB ⋅ UB − nC ⋅ dqC ⋅ UC   ⋅

+

U =

⋅

⋅

R23 + n B ⋅ d qB ⋅ UB + nC ⋅ d qC ⋅ UC ⋅

⋅

−511 + 7 ⋅ 1/:2 ⋅ 634 + 4 ⋅ 2/15 ⋅ 974 >711!L 7 ⋅ 1/:2 + 4 ⋅ 2/15

>

n B ⋅ d qB + nC ⋅ d qC

jedna~ina stawa me{avine idealnih gasova na izlazu iz me{ne komore: ⋅  ⋅   nB ⋅ ShB + nC ⋅ ShC  ⋅ U + ⋅ ⋅    W+ =  q+ ⋅ W+ =  nB ⋅ ShB + nC ⋅ ShC  ⋅ U+ ⇒ + q   ( ) 7 ⋅ 371 + 4 ⋅ 3:8 ⋅ 711 n4 W+ = >!:/915! t 2/6 ⋅ 216 jedna~ina stawa idealnog gasa za kiseonik )B* na izlazu iz me{ne komore: ⋅

⋅

q+B ⋅ W+ = nB ⋅ ShB ⋅ U +

⇒

q+B =

nB ⋅ ShB ⋅ U + W

+

>

7 ⋅ 371 ⋅ 711 > 1/:6 ⋅ 216 Qb :/915

jedna~ina stawa idealnog gasa za azot )C* na izlazu iz me{ne komore: qC+

+

⋅

⋅ W = nC ⋅ ShC ⋅ U

⋅

+

⇒

qC+

=

nC ⋅ ShC ⋅ U + W

+

>

4 ⋅ 398 ⋅ 711 :/915 ⋅ 21−3

> 1/66 ⋅ 21 6 Qb

pritisak kiseonika )B* u okolnom vazduhu: 6 6 qP B = sP3 ⋅ qp > 1/32 ⋅ 2 ⋅ 21 > 1/32 ⋅ 21 Qb pritisak azota )C* u okolnom vazduhu: 6 6 qP B = sP3 ⋅ qp > 1/8: ⋅ 2 ⋅ 21 > 1/8: ⋅ 21 Qb


{fmlp@fvofu/zv


strana 12

drugi zakon termodinamike za proces me{awa gasova B!i!C : ⋅

⋅

⋅

∆ T tjtufn = ∆ T sbeop ` ufmp + ∆ T plpmjob =.. . ⋅

⋅

⋅

∆ Tsbeop ` ufmp = ∆ T B + ∆ TC >!///!>!2/86!,!1/57!>!3/32!

lX L

⋅ ⋅  q+  711 1/:6  lX U+  − 1/37 ⋅ mo ∆ T B = n B ⋅  d qB mo − S hB mo B  > 7 ⋅  1/:2 ⋅ mo  >2/86   U q L 634 2 / 9   B  B  ⋅ ⋅  q+  711 1/66  U+ lX  ∆ TC = nC ⋅  d qC mo − S hC mo C  > 4 ⋅ 2/15 ⋅ mo − 1/3:8 ⋅ mo  >1/57   L UC 974 4/4  qC    ⋅

⋅

∆ T plpmjob

−511 R 23 lX =− >− >2/48! 3:4 L Up

⋅

∆ T tjtufn = 3/32 + 2/48 >!4/69! ⋅

lX L

⋅

Fy h = Up ⋅ ∆ T tjtufn > 3:4 ⋅ 4/69 >215:!lX ⋅

⋅

⋅

Fy 2 = Fy B + Fy C = ...  ⋅ ⋅ Fy B = n B ⋅ − d qB ⋅ (Up − UB ) + Up  

 U qp ⋅  d qB ⋅ mo p − S hB ⋅ mo B  UB qB 

   

⋅  3:4 1/32   Fy B = 7 ⋅ − 1/:2 ⋅ (3:4 − 634) + 3:4 ⋅  1/:2 ⋅ mo − 1/37 ⋅ mo  >2422!lX 634 2/9   

 ⋅ ⋅ Fy C = nC ⋅ − d qC ⋅ (Up − UC ) + Up  

 U qp ⋅  d qC ⋅ mo p − S hC ⋅ mo C  UC qC 

   

⋅  3:4 1/8:   Fy C = 4 ⋅ − 2/15 ⋅ (3:4 − 974) + 3:4 ⋅ 2/15 ⋅ mo − 1/3:8 ⋅ mo  >2275!lX 974 4/4    ⋅

Fy2 = 2422 + 2275 >!3586!lX ⋅

ηFy =

⋅

Fy2 − Fy h ⋅

Fy2

>

3586 − 215: >1/68 3586


{fmlp@fvofu/zv


strana 13

4/:. U suprotnosmernom razmewiva~u toplote pri qw>3!cbs, biva izobarski zagrevan tok vazduha (idealan gas), od temperature Uw2>524!L do temperature Uw3>654!L, a tok vrelih gasova (sme{a idealnih gasova) biva hla|ena od polaznog stawa H2)qh2>2/6!cbs-!Uh2>724!L* do stawa H3)qh3>2/4!cbs-!Uh3>@*/ Ako su maseni protoci vazduha i vrelih gasova isti, a vreli gasovi imaju iste termofizi~ke osobine kao i vazduh odrediti eksergijski stepen korisnosti procesa u ovom razmewiva~u toplote pri uslovima okoline P)qp>2!cbs-!Up>3:4!L*/ Zanemariti promene makroskopske potencijalne i kineti~ke energije kao i prisustvo hemijske neravnoteè.

Uh2

Uh3

Uw2

Uw3

prvi zadatak termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

⋅

⋅

⋅

⇒

R 23 = ∆ I23 + X U23 ⋅

⋅

I2 = I3 ⋅

n⋅ d q ⋅ Uw2 + n⋅ d q ⋅ Uh2 = n⋅ d q ⋅ Uw3 + n⋅ d q ⋅ Uh3

⇒

Uh3 = Uw2 + Uh2 − Uw3 > 524 + 724 − 654 >594!L ⋅

⋅

⋅

Fy vmb{ = Fy w2 + Fy h2 = ... ⋅ ⋅   U q Fy w2 = n⋅ − d q (Up − Uw2 ) + Up  d q mo p − S h mo p U q w w2  

  

⋅ ⋅  ⋅ 3:4 2   Fy w2 = n⋅ − 2 ⋅ (3:4 − 524) + 3:42 ⋅ mo − 1/398 ⋅ mo  = n⋅ 88/82 !lX 524 3    ⋅ ⋅   U q  Fy h2 = n⋅ − d q Up − Uh2 + Up  d q mo p − S h mo p   Uh2 q h2    

(

)

⋅ ⋅ ⋅  3:4 2   Fy h2 = n⋅ − 2 ⋅ (3:4 − 724) + 3:4 ⋅ 2 ⋅ mo − 1/398 ⋅ mo  > n⋅ 248/92!lX 724 2/6    ⋅

⋅

⋅

Fy vmb{ = n⋅ (88/82 + 248/92) > n⋅ !326/63!lX


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike ⋅

strana 14

⋅

Fy hvcjubl = Up ⋅ ∆ T tj = ... ⋅

⋅

⋅

⋅

⋅

⋅

∆ Ttj = ∆ Tsu + ∆ Tp =… ∆ Tsu = ∆ T w + ∆ Th = … ⋅

⋅

⋅ ⋅ Uw 3 q 654 lX − S h mo w * = n⋅ 2 ⋅ mo = n⋅ 1/385 L 524 Uw2 qw

⋅

⋅

Uh3

∆ T w > n /! )d q mo ∆ Th > n /! )d q mo

⋅

Uh2

− S h mo

⋅

⋅

⋅

⋅ 594 2/4  lX  − 1/398 ⋅ mo * = n⋅ 2 ⋅ mo  >− n⋅ 1/2:8 L 724 q h2 2/6  

q h3

⋅

∆ Tsu > n⋅ 1/385!− n⋅ 1/2:8> n⋅ 1/188! ⋅

⋅

⋅

⋅

∆ Ttj > ∆ Tsu , ∆ Tp > n⋅ 1/188! ⋅

lX L

lX L

⋅

Fy hvcjubl = 3:4 ⋅ n⋅ 1/188 >33/67!lX ⋅

ηFy!>

⋅

Fy vmb{ − Fy h ⋅

Fy vmb{

=

326/63 − 33/67 >1/9: 326/63 ⋅

3.10. Klipni kompresor kvazistati~ki politropski sabija n >1/6!lh0t vazduha (idealan gas) od stawa 2)q>211!lQb-!U>399!L*!do stawa 3)q>611!lQb-!U>513!L*/ Stawe okoline zadato je sa P)qp>211!lQbUp>399!lQb*. Odrediti: a) snagu kompresora b) eksergijski stepen korisnosti procesa c) ako bi se kompresor hladio vodom koja bi pri tom mewala stawe pri stalnom pritisku q>211!lQb od!Ux2>399!L!do!Ux3>414!L a) o−2  o

U2  q2  = U3  q 3  ⋅

⋅

X U23 = n⋅

mo ⇒

o> mo

q2 q3

q2 U − mo 2 q3 U3

mo > mo

211 611

211 399 − mo 611 513

>2/37

2/37 o ⋅ S h ⋅ (U2 − U3 ) > 1/6 ⋅ ⋅ 1/398 ⋅ (399 − 513) >−8:/4!lX o −2 2/37 − 2


{fmlp@fvofu/zv


strana 15

b)  U q  Fy2 = n⋅ (− ∆i2p + Up ⋅ ∆t2p ) > n⋅ − dq)Up − U2* + Up)dq mo p − Sh mo p * >!1!lX U q ⋅

⋅

⋅



2

2 

napomena:!q2>qp-!U2>Uq ⋅

⋅

Fy hvcjubl = Up ⋅ ∆ T tj = ... ⋅

⋅

⋅

∆ Ttj = ∆ Tsu + ∆ Tp =… ⋅ ⋅  U q ∆ Tsu = n⋅  d q ⋅ mo 3 − S h ⋅ mo 3 U2 q2 

 513 611  X   > 1/6 ⋅ 2 ⋅ mo − 1/398 ⋅ mo  >−75/32! 399 211 L   

⋅

−33/2 X R 23 >87/85! ∆ Tp = − >///> − 399 L Up ⋅

⋅

⋅

o−κ 2/37 − 2/5 ⋅ (U3 − U2 ) > 1/6 ⋅ 1/83 ⋅ ⋅ (513 − 399) >−33/2!lX o −2 2/37 − 2 ⋅ X ∆ Ttj >−75/32!,!87/85!>23/64! L

! R 23 = n⋅ d w ⋅

⋅

Fyhvcjubl = 399 ⋅ 23/64 >4/7!lX ⋅

⋅

⋅

Fy vmb{ = Fy 2 + X U23 >8:/4!lX ⋅

ηFy!>

⋅

Fyvmb{ − Fyh ⋅

Fyvmb{

=

8:/4 − 4/7 >1/:6 8:/4

c) ⋅

⋅

R 23 = n x ⋅ (i x2 − i x3 ) !!!!⇒ lK lh lK ix3!>!236/8! lh ix2!>!73/:6!


⋅

nx

lh −33/2 R 23 = > >!1/46! 73/:6 − 236/8 t i x2 − i x3 )voda!q>211!lQb-!U>399!L* )voda!q>211!lQb-!U>399!L*

{fmlp@fvofu/zv


strana 16

4/22/!U vertikalnom cilindru sa grani~nikom (slika) moè se trewa) klip sa tegom. U po~etnom trenutku zapremina n4 (ograni~ena klipom sa tegom), ispuwena je kqu~alom vodom makroskopski razvijenom parom u stawu termodinami~ke pritisku q>4!cbs!(vla`na para). Kqu~ala voda zauzima 1% od zapremine cilindra. Maksimalna zapremina cilindra ispod Wnby>3!n4. Odrediti termodinami~ki gubitak rada (gubitak predaji toplote, od izotermnog toplotnog izvora, temperature radnoj materiji u cilindru, ako je wena temperatura na kraju zagrevawa 684!L. Temperatura okoline iznosi Up>411!L. proces zagrevawa na Ut dijagramu. [rafirati na Ut dijagramu predstavqa gubitak eksergije.

kretati (bez cilindra W>1/7 i wenom ravnoteè na po~etne klipa iznosi eksergije) pri UUJ!>734!L, procesa Predstaviti povr{inu koja

ta~ka 1: q2>4!cbs! y2>@ n# y2 = = /// n#+n( 1/6:5 W# >///> >1/:9!lh n# = 1/7168 w# W( 1/117 >6/6:!lh n( = = /// > 1/1121844 w( n4 lh W′!>!1/12/W!>!1/117!n4 W′!>!1/::/W!>!1/6:5!n4 w′>!1/1121844!

w′′>1/7168!

n4 lh

n# 1/:9 = = 1/26 n#+n( 1/:9 + 6/6: lK lK t2>!3/58!! i2>!996/:5!! lh lhL napomena: n!>!n′!,!n′′!>!7/68!lh y2 =

ta~ka 2:

q3>q2>4!cbs- w3>wnby!>

w′!?!w3!?!w′′

Wnby n4 = 1/414 n lh

(ta~ka 2 se nalazi u oblasti vla`ne pare)

1/414 − 1/1121844 w3 − w ( > >1/6 1/7168 − 1/1121844 w #− w ( lK lK - v3!>vy!>2663/3!! -! !!!u3>!ulmk>244/65pD! i3!>iy!>!2754/3!! lh lh y3 =


{fmlp@fvofu/zv


strana 17

u>411pD!

ta~ka 3: w′!>!1/1125147! w4!?!w′′

w4!>w3>1/414!!

n4 lh

n4 n4 - !!!w′′!>!1/13275! lh lh (ta~ka 3 se nalazi u oblasti pregrejane pare)

i4!>!iqq!>!4161!!

lK lh

t4!>!tqq!>!8/29!!

lK lhL

q4!>!qqq!>!9/6!!cbs

vrednosti i4-!t4 i q4 se moraju pro~itati sa it dijagrama za vodenu paru lK v4!>!vqq>!i4!−!q4/!w4!>!4161!−!9/6/216!/!21−3/!1/414!>!38:3/6! lh

napomena:

Fyhvcjubl!>!Uplpmjob!/!∆Ttjtufn!>!///!>!411!/!:/97!>!3:69!!lK lK L lK ∆ T sbeop!ufmp> n ⋅ ∆t24 !> n ⋅ (t 4 − t2 ) > 7/68 ⋅ (8/29 − 3/58) >!41/:6! L )r23 *q=dpotu + )r34 * w =dpotu R 23 + R 34 i3 − i2 + v 4 − v 3 ∆S UJ = − −n⋅ = −n ⋅ UUJ UUJ UUJ 2754/3 − 996/:5 + 38:3/6 − 2663/3 lK >−32/1:! ∆TUJ!>! − 7/68 ⋅ L 734

∆Ttjtufn>∆Tsbeop!ufmp!,!∆Tupqmpuoj!j{wps!>!///>41/:6!−!32/1:!>!:/97!

U 4

2

3 Up

∆tsu Fyh

t

∆tUJ ∆tTJ


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 18 4/23/!U razmewiva~u toplote vr{i se atmosfersko )q>2!cbs*, potpuno isparavawe kqu~ale vode i istovremena potpuna kondenzacija suvozasi}ene vodene pare pri q>4!cbs. Ukoliko toplotna snaga razmewiva~a toplote (interno razmewena toplota izme|u pare i vode) iznosi 3/6!lX, izra~unati eksergijski stepen korisnosti procesa u ovom razmewiva~u toplote pri stawu okoline P)qp>!2!cbs-!up>31pD*/ L2 !2

!3

!4

!5

suva para, q>4!cbs

kqu~ala voda,!q>2!cbs L3

!U

!3

!2

!5

!4

!t q>!4!cbs lK i2!>!i″!>!3836! lh ta~ka 1:

q>!4!cbs lK i3!>!i′!>!672/5! lh ta~ka 2:

q>!2!cbs lK i4!>!i′!>!528/5! lh ta~ka 3:

q>!2cbs lK i5!>!i″!>!3786! lh

y>2 t2!>!t″!>!7/::3!

lK lhL

y>1 t3!>!t′!>!2/783!

lK lhL

y>1 t4!>!t′!>!2/4137!

lK lhL

ta~ka 4:


t5!>!t″!>!8/471!

lK lhL

{fmlp@fvofu/zv


strana 19 up!>!31pD

qp!>!2!cbs lK ip!>!!ix!>!94/:! lh ta~ka O:

tp!>!!tx!>!1/3:7!

lK lhL


⋅

⋅

R 23 = ∆ I23 + X U23

ograni~enom konturom 1 (K1):

⋅

⋅

R 23 = nq ⋅ (i3 − i2 )

⇒

⋅

⋅

nq =

−3/6 R 23 lh > >2/26! t i3 − i2 672/5 − 3836


⋅

⋅

R 45 = ∆ I45 + X U245

ograni~enom konturom 2 (K2):

⋅

⋅

R 45 = n x ⋅ (i 5 − i 4 )

⇒

⋅

⋅

nx =

R 45 3/6 lh > >2/22! t i 5 − i 4 3786 − 528/5 ⋅

ηFy =

⋅

Fy vmb{ − Fy hvcjubl ⋅

Fy vmb{ ⋅

>

892/16 − 286/9 >1/88 892/16

⋅

Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 1/7 >286/9!lX ⋅

⋅

⋅

⋅

⋅

∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T plpmjob!>!///>1/7! ⋅

lX L

∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!21/1:!−!:/5:>1/7! ⋅

⋅

lX L

⋅

lX L ⋅ ⋅ ⋅ lX T vmb{!>! nq ⋅ t2 + n x ⋅ t 4 > 2/26 ⋅ 7/::3 + 2/22 ⋅ 2/4137 >:/5:! L T j{mb{!>! nq ⋅ t 3 + n x ⋅ t 5 > 2/26 ⋅ 2/783 + 2/22 ⋅ 8/47 >21/1:!

⋅

⋅

⋅

Fy vmb{!>! Fy 2!, Fy 4!>!///>!892/16!,!53/92!>!934/97!lX ⋅

⋅

⋅

Fy 2> nq ⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! nq ⋅ [i2 − i p + Up ⋅ (t p − t2 )] ⋅

Fy 2> 2/26 ⋅ [3836 − 94/: + 3:4 ⋅ (1/3:7 − 7/::3)] >892/16!lX ⋅

⋅

⋅

Fy 4!> n x ⋅ (− ∆i 4p + Up ⋅ ∆t 4p ) >! n x ⋅ [i 4 − i p + Up ⋅ (t p − t 4 )] ⋅

Fy 4> 2/22 ⋅ [528/5 − 94/: + 3:4 ⋅ (1/3:7 − 2/4137)] >53/92!lX


{fmlp@fvofu/zv


strana 20 ⋅

3.13.!U ure|aj za pripremu kqu~ale vode (slika) uti~e suva para 2)q>23!cbs*, masenog protoka n 2>1/2 ⋅

lh0t i voda stawa 3)q>5!cbs-!u>71pD*!masenog protoka n 3>@/ Prolaskom kroz razmewiva~ toplote suva para se potpuno kondenzuje )stawe 3*. Nastali kondenzat se prigu{uje na pritisak q3(stawe 4), a zatim izobarski me{a sa vodom (stawe 2). Toplotni gubici razmewiva~a toplote iznose 223!lX. Prestaviti prosese u pojedina~nim ure|ajima (razmewiva~ toplote, prigu{ni ventil, me{na komora) na zasebnim Ut dijagramima i odrediti: ⋅

a) maseni protok vode ( n 2) da bi iz ure|aja isticala kqu~ala voda pritiska q3 (stawe 6) b) temperaturu vode stawa 6!)!u6!* c) eksergijski stepen korisnosti ure|aja ako se okolina defini{e kao voda stawa P!)qp>2!cbs-!u>31pD*

!U

⋅

2 n2

razmewiva~ toplote ⋅

7

n3 3

6

q>23!cbs q>5!cbs

4

2

7 4

5

6

!t !U

!U

prigu{ni ventil

me{na komora

q>23!cbs

q>23!cbs

q>5!cbs

q>5!cbs

4

5

6

5

3

!t 2!−!4 4!−!5 5!,!3!>6! 6!−!7

!t

: promena stawa pare pri proticawu kroz razmewiva~ toplote (RT) : promena stawa pare pri proticawu kroz prigu{ni ventil : proces me{awa pare i vode u me{noj komori : promena stawa me{avine pare i vode pri proticawu kroz RT


{fmlp@fvofu/zv


strana 21

a) Prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ⋅

ograni~enom isprekidanom linijom:

⋅

⋅ ⋅ ⋅ ⋅  ⋅ ⋅  ⋅ R 23 = I3 − I2 =  n2 + n3  ⋅ i 7 − n2 ⋅ i2 − n3 ⋅ i3   ⋅

⋅

R 23 = ∆ I23 + X U23 ⇒

⋅

R 23 + n2 ⋅ (i2 − i 7 ) − 223 + 1/2 ⋅ (3896 − 715/8) lh >///> n3 = >1/4! t 715/8 − 362/4 i 7 − i3 ⋅

q2>23!cbs lK i2!>i′′>3896! lh

y>2

q3>5!cbs lK i3!>ix>362/4! lh

u3>71pD

ta~ka 1:

t2>t′′!>!7/634!

ta~ka 2:

t3>tx!>!1/94!

lK lhL

lK lhL

y>1 q7>5!cbs lK lK t2>t′!>!2/888! i7!>i′>715/8! lh lhL

ta~ka 6:

b) ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 Prvi zakon termodinamike za proces me{awa: ⋅ ⋅ ⋅ ⋅ ⋅ ⋅   ⇒ I2 = I3 n2 ⋅ i 5 + n3 ⋅ i 3 =  n2 + n3  ⋅ i 6   ⋅

i6 =

⋅

n2 ⋅ i 5 + n3 ⋅ i3 ⋅

⋅

>///>

n2 + n3

1/2 ⋅ 8:9/4 + 1/4 ⋅ 362/4 lK >499/16! lh 1/2 + 1/4

q7>5!cbs lK i5!>i′>8:9/4! lh

ta~ka 4:

ta~ka 3:

i5>i4!>8:9/4!

ta~ka 5:

q>5!cbs

y>1

lK lh i>499/16!

lK lh

u6!>!ux!>:4pD


{fmlp@fvofu/zv


strana 22

c) up>31pD

qp>2!cbs lK ip!>ix>94/:! lh ta~ka O:

tp>tx!>!1/3:7!

⋅

⋅

lK lhL

⋅

Fy 2> n2 ⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! n2 ⋅ [i2 − i p + Up ⋅ (t p − t2 )] ⋅

Fy 2> 1/2 ⋅ [3896 − 94/: + 3:4 ⋅ (1/3:7 − 7/634)] >98/77!lX ⋅

⋅

⋅

Fy 3> n3 ⋅ (− ∆i3p + Up ⋅ ∆t 3p ) >! n3 ⋅ [i3 − i p + Up ⋅ (t p − t 3 )] ⋅

Fy 3> 1/4 ⋅ [362/4 − 94/: + 3:4 ⋅ (1/3:7 − 1/94 )] >4/39!lX ⋅

⋅

⋅

Fy vmb{!>! Fy 2,! Fy 3!>98/77!,!4/39!>:1/:5!lX ⋅

⋅

Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 1/2: >66/78!lX ⋅

⋅

⋅

⋅

⋅

∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T plpmjob!>!///>−1/2:!,!1/49!>!1/2:!

lX L

⋅

lX L ⋅ ⋅   ⋅ lX T j{mb{!>!  n2 + n3  ⋅ t 7 > (1/2 + 1/4 ) ⋅ 2/888 >1/82! L  

∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!1/82!−!1/:>−!1/2:!

⋅

⋅

⋅

T vmb{!>! n2 ⋅ t2 + n3 ⋅ t 3 > 1/2 ⋅ 7/634 + 1/4 ⋅ 1/94 >1/:1!

lX L

⋅

R −223 lX >1/49! ∆ T plpmjob!>!−! 23 >− L 3:4 Up ⋅

⋅

ηFy =

⋅


Fy vmb{


>

:1/:5 − 66/78 >1/4: :1/:5

{fmlp@fvofu/zv


strana 23

⋅

4/25/!U parnu turbinu ulazi n >21!lh0t vodene pare stawa 2)q>3!NQb-!u>471pD). Iz turbine se na ⋅

pritisku q3>1/5!NQb izdvaja, za potrebe nekog tehnolo{kog qspdftb-! n 3>3!lh0t!pare a preostali deo nastavqa ekspanziju do stawa 4)q>1/27!NQb-!y>2*. Stepen dobrote adijabatske ekspanzije do izdvajawa p pare iznosi η23 e >2. Pod okolinom smatrati vodu stawa P)q>1/2!NQb-!u>31 D*. Skicirati promene stawa vodene pare na Ts dijagramu i odrediti: a) snagu turbine b) eksergiju parnih tokova na ulazu u turbinu i oba izlaza iz turbine c) eksergijski stepen korisnosti procesa u turbini 2

!U 2 ⋅

3

X uvscjob

3

4l

!t

4 ta~ka 1:

q2>31!cbs

lK i2!>iqq4267! lh ta~ka 2:

4

(pregrejana para)

lK t2>tqq!>!7/:96! lhL

q3>5!cbs

i3!>iqq!>3889/27!

u>471pD

!t3!>!t2!>7/:96!

lK lhL

(pregrejana para)

lK lh

q4>2/7!cbs lK i4!>i′′!>37:7! lh

!y>2

qp>2!cbs lK ip!>ix>94/:! lh

up>31pD

ta~ka 3:

ta~ka O:


(suvo−zasi}ena para)

lK t4!>t′′!>!8/313! lhL (voda)

lK tp>tx!>!1/3:7! lhL

{fmlp@fvofu/zv


strana 24

a) ⋅

⋅

⋅

R 23 = ∆ I23 + X U23

Prvi zakon termodinamike za proces u turbini: ⋅ ⋅ ⋅ ⋅ ⋅  ⋅ ⋅  X uvscjob = I2 − I3 = n⋅ i2 − n3 ⋅ i 3 −  n− n3  ⋅ i 4   ⋅

X uvscjob > 21 ⋅ 4267 − 3 ⋅ 3889/27 − (21 − 3) ⋅ 37:7 >5/55!NX b) ⋅

⋅

⋅

Fy 2> n⋅ (− ∆i2p + Up ⋅ ∆t2p ) >! n⋅ [i2 − i p + Up ⋅ (t p − t2 )] ⋅

Fy 2> 21 ⋅ [4267 − 94/: + 3:4 ⋅ (1/3:7 − 7/:96 )] >22/23!NX ⋅

⋅

⋅

Fy 3> n3 ⋅ (− ∆i 3p + Up ⋅ ∆t 3p ) >! n3 ⋅ [i3 − i p + Up ⋅ (t p − t 3 )] ⋅

Fy 3> 3 ⋅ [3889/27 − 94/: + 3:4 ⋅ (1/3:7 − 7/:96)] >2/58!NX  ⋅ ⋅   ⋅ ⋅      Fy 4>  n− n3  ⋅ (− ∆i 4p + Up ⋅ ∆t 4p ) >!  n− n3  ⋅ [i 4 − i p + Up ⋅ (t p − t 4 )]         ⋅

⋅

Fy 4> (21 − 3) ⋅ [37:7 − 94/: + 3:4 ⋅ (1/3:7 − 8/313)] >5/82!NX c) ⋅

Bilans eksergije za proces u turbini: ⋅

⋅

⋅

⋅

⋅

⋅

⋅

⋅

Fy 2 = Fy 3 + Fy 4 + X uvscjob + Fy hvcjubl

⋅

Fy hvcjubl = Fy 2 − Fy 3 − Fy 4 − X uvscjob !>!22/23!−!2/58!−!5/82!−!5/55!>!1/6!NX ⋅

ηFy =

⋅


Fy vmb{ ⋅

>

22/23 − 1/6 >1/:7 22/23

⋅

Fy vmb{!>! Fy 2!>22/23!NX napomena:

Do gubitka eksergije se moglo do}i i na uobi~ajen na~in ⋅

⋅

primenom Hpvz!−!Tupepmjoph! zakona:! Fy h = Up ⋅ ∆ T tjtufn


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strana 25 ⋅

4/26/!Pregrejana vodena para stawa 2)q>!7!NQb-!U>844!L-! n >2!lh0t ) {iri se adijabatski u dvostepenoj turbini sa me|uprigu{ivawem (slika), do krajweg stawa 5)U>424!L- vla`na para). ⋅

Stepeni dobrote u turbinama su: ηEUWQ = 2 i ηEUOQ = 1/99 . Deo pare, masenog protoka n B>1/4!lh0t, po izlasku iz turbine visokog pritiska, pri pritisku q3>1/9!NQb odvodi se iz turbine, a preostala para prolaskom kroz prigu{ni ventil adijabatski prigu{uje na pritisak q4>1/4!NQb. Prikazati procese u it koordinatnom sistemu i odrediti snagu dobijenu na zajedni~kom vratilu kao i eksergijski stepen korisnosti procesa u ovoj dvostepenoj turbini. Pod okolinom smatrati vodu stawa P)q>1/2!NQbU>3:4!L*/

!n

UWQ

UOQ

2 X

3

4

!nB

5 2

i 3

4

5L

5

t


{fmlp@fvofu/zv


strana 26

u2!>571pD! q2>!71!cbs lK lK t2>tqq!>!7/86! i2!>!iqq!>4435!! lh lhL

ta~ka 1:

q3>!9!cbs

ta~ka 2:

i3!>!iqq!>3919/7!!

q4>!4!cbs

t4>tqq!>!8/289!

lK lhL

ta~ka 4K:

U5L!>!51pD

i4!>!i3!>3919/7!!

lK lh

lK lhL (ta~ka 4K se nalazi u oblasti vla`ne pare)

t′!?!t5L!?!t′′

 t − t(  y 5L =  5L = 1/99   t#−t(  U =51p D

ηEUWQ =

lK lhL

lK lh

ta~ka 3:

U5>51pD

ta~ka 4:

t3>t2!>!7/86!

i4 − i5 i 4 − i 5L

⇒

t5L>t4>!8/289!

i5L!>!iy!>!3396! ηEUOQ = 1/99

i 5 = i 4 − ηEUWQ ⋅ )i 4 − j 5L * =

i5 = 3919/7 − 1/99 ⋅ )3919/7 − 3396* = 3459  i − i(  y5 =  5 = 1/:17   i#−i(  U = 51p D

lK lh

lK lh

⇒

t5!>!ty!>8/65!

lK lhL

qp>2!cbs up>31pD lK lK tp>tx>1/3:7! ip>ix>94/:! lh lhL

ta~ka O:


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strana 27

Prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ⋅

⋅

⋅

ograni~enom isprekidanom linijom: R 23 = ∆ I23 + X U23 ⋅ ⋅ ⋅ ⋅ ⋅  ⋅ ⋅  X = I2 − I3 = n⋅ i2 − n B ⋅ i3 −  n− n B  ⋅ i 5   ⋅

X > 2⋅ 4435 − 1/4 ⋅ 3917/7 − (2 − 1/4 ) ⋅ 3489 >948/9!lX ⋅

⋅

⋅


Fy 2> 2 ⋅ [4435 − 94/: + 3:4 ⋅ (1/3:7 − 7/86 )] >245:/2!LX ⋅

⋅

Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 1/66 >272/26!lX ⋅

⋅

⋅

⋅

⋅

∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T plpmjob!>!///>1/57!

lX L

⋅

∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!8/41!−!7/86>1/66! ⋅

⋅

T vmb{!>! n⋅ t2 > 2⋅ 7/86 >7/86!

lX L

lX L

⋅ ⋅  ⋅ ⋅  lX T j{mb{!>! n B ⋅ t 3 +  n− n B  ⋅ t 5 > 1/4 ⋅ 7/86 + (2 − 1/4 ) ⋅ 8/65 >8/41! L   ⋅

ηFy =

⋅


Fy vmb{ ⋅

>

245:/2 − 272/26 >1/99 245:/2

⋅

Fy vmb{!>! Fy 2!>245:/2!NX


{fmlp@fvofu/zv


strana 28

4/27/ Pregrejana vodena para )1/2!lh0t* stawa 2)q>91!cbs-!u>571pD* ulazi u parno turbisnki blok gde se najpre kvazistati~ki adijabatski {iri u turbini visokog pritiska do stawa 3)q>21!cbs*. Zatim se vodenoj pari stawa 2 u dogreja~u izobarski dovodi toplota od toplotnog izvora stalne temperature UUJ>571pD sve do uspostavqawa toplotne ravnoteè (stawe 3). Nakon toga se para kvazistati~ki adijabatski {iri u turbini niskog pritiska do stawa 5)q>2!cbs*. Pod okolinom smatrati vodu stawa P)q>1/2!NQb-!U>3:4!L*/!Skicirati procese sa vodenom parom na it dijagramu i odrediti: a) mehani~ku snagu parno turbinskog bloka kao i toplotnu snagu dogreja~a pare b) ireverzibilnost procesa (gubitak eksergije) u parno turbinskom bloku c) eksergijski stepen korisnosti procesa u parnoturbinsom bloku 2

⋅

X U 23

3

4 ⋅

R 34

⋅

X U 45

5 i 2

3

4

5

t


{fmlp@fvofu/zv


strana 29

u2!>571pD! q2>!91!cbs lK lK t2>tqq!>!7/699! i2!>!iqq!>43:7!! lh lhL

ta~ka 1:

ta~ka 2:

q3>!21!cbs

i3!>!i′′!>3889!!

lK lh

t3>t2!>!7/699!

lK lhL

u4!>!571pD q4>!21!cbs lK lK t4>tqq!>!8/756! i4!>!iqq!>44:3!! lh lhL

ta~ka 3:

q5!>!2!cbs

ta~ka 4:

i5!>!iqq!>!38:3/3!

t5>t4>!8/756!

lK lhL

lK lh

up>31pD qp>2!cbs lK lK tp>tx>1/3:7! ip>ix>94/:! lh lhL ta~ka O:

a) Prvi zakon termodinamike za proces u turbini visokog pritiska: ⋅

⋅

⋅

⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X U23

⇒

⋅

X U23 = n⋅ (i2 − i3 ) ⋅

X U23 = 1/2 ⋅ (43:7 − 3889) >62/9!lX Prvi zakon termodinamike za proces u turbini niskog pritiska: ⋅

R 45 = ∆ I45 + X U 45

⇒

⋅

X U 45 = n⋅ (i 4 − i 5 ) ⋅

X U 45 = 1/2 ⋅ (44:3 − 38:3/3) >71!lX ⋅

⋅

⋅

X = X U23 + X U 45 >222/9!lX ⋅

Prvi zakon termodinamike za proces u dogreja~u pare: ⋅

⋅

⋅

R 34 = ∆ I34 + X U34

⇒

⋅

R 34 = n⋅ (i 4 − i3 ) ⋅

R 34 = 1/2 ⋅ (44:3 − 3889) >72/5!lX


{fmlp@fvofu/zv


strana 30

b) ⋅

⋅

⋅

Js = Fy hvcjubl = Up ⋅ ∆ T tj >///> 3:4 ⋅ 33 >7/56!lX ⋅

⋅

⋅

⋅

⋅

∆ T tjtufn!>!∆ T sbeop!ufmp!,!∆ T upqmpuoj!j{wps!>!///>!−94/8!,!216/8!>!33! ⋅

∆ T sbeop!ufmp> T j{mb{!−! T vmb{!>!///>!875/6!−769/9>216/8! ⋅

X L

X L

⋅

X L ⋅ ⋅ X T vmb{!>! n⋅ t2 > 1/2 ⋅ 7/699 >769/9! L T j{mb{!>! n⋅ t 5 > 1/2 ⋅ 8/756 >875/6!

⋅

R 72/5 ⋅ 21 4 X >!−94/8! ∆ T upqmpuoj!j{wps!>!−! 34 >− 844 L UUj ⋅

c) ⋅

⋅

⋅


Fy 2> 1/2 ⋅ [43:7 − 94/: + 3:4 ⋅ (1/3:7 − 7/699)] >247/96!LX ⋅

⋅

Fy R = R 34 ⋅ ⋅

ηFy =

UUJ − Up 844 − 3:4 >47/97!lX > 72/5 ⋅ 844 UUJ ⋅

⋅

Fy2 + Fy R − Fy h ⋅

⋅

Fy 2 + Fy R

zadatak za ve`bawe:

>

247/96 + 47/97 − 7/56 >1/:7 247/96 + 47/97

(3.17.) ⋅

4/28/ Kompresor usisava n =83!lh0i pare amonijaka stawa 2)q>366/:!lQb-!y>2* i sabija je adijabatski do stawa 3)q>21!cbs-!U>511!L*/ Ako za okolinu smatramo amonijak stawa P)U>331!L-!y>1* odrediti eksergijski stepen korisnosti procesa u kompresoru. re{ewe:

ηFy>1/99


⋅

⋅

⋅

)! X >7/13!lX-! Fy 2 = 5/45 lX-! Fy h >2/38!lX!*

{fmlp@fvofu/zv


strana 1

PRVI I DRUGI ZAKON TERMODINAMIKE (KOMBINOVANI PROBLEMI) 5/2/!U toplotno izolovanom rezervoaru nalazi se!211!lh!vode!)dx>5/29!lK0)lhL**!po~etne temperature Ux>3:4!L/!Komad bakra!)dDv>1/49!lK0)lhL**!mase!51!lh, po~etne temperature!UDv>469!L!i komad gvo`|a!)dGf>1/57!lK0)lhL**!mase!31!lh, po~etne temperature!UGf>454!L-!naglo se unesu u rezervoar sa vodom. U momentu uno{ewa u rezervoaru se ukqu~uje me{alica vode snage!711!X, koja radi dok se ne uspostavi stawe termi~ke ravnoteè!U+>3::!L/!Odrediti: a) vreme rada me{alice b) promenu entropije izolovanog sistema tokom navedenog procesa

Dv

Gf

I3P

a) prvi zakon termodinamike za proces u zatvorenom termodinami~kom ⇒ XU23>V2!−!V3 sistemu: R23!>!∆V23!,!XU23 V2 = n x ⋅ d x ⋅ Ux + nDv ⋅ d Dv ⋅ UDv + nGf ⋅ d Gf ⋅ UGf V2 = 211 ⋅ 5/29 ⋅ 3:4 + 51 ⋅ 1/49 ⋅ 469 + 31 ⋅ 1/57 ⋅ 454 >242182/3!lK V 3 = n x ⋅ d x ⋅ U + + nDv ⋅ d Dv ⋅ U + + nGf ⋅ d Gf ⋅ U +

V 3 = 211 ⋅ 5/29 ⋅ 3:: + 51 ⋅ 1/49 ⋅ 3:: + 31 ⋅ 1/57 ⋅ 3:: >243388/7!lK XU23!>!242182/3!−!243388/7!>!−2317/5!lK τ=

XU23 − 2317/5 = >3121/78!t ⋅ − 711 ⋅ 21 −4 X U23


{fmlp@fvofu/zv


strana 2

b) ∆TTJ!>!∆TSU!,!∆TP!>///!>5/58!

lK L

∆TSU!>!∆Tx!,!∆TDv!,!∆TGf!>///>9/58!−!3/85!−!2/37!!>!5/58! U+

∆T x = n x ⋅

∫

lK L

d x ⋅ eU U+ 3:: lK = n x ⋅ d x mo = 211 ⋅ 5/29 ⋅ mo >9/58! L U Ux 3:4

Ux U+

∆T Dv = nDv ⋅

∫

d Dv ⋅ eU U+ 3:: lK >−3/85! = nDv ⋅ d Dv mo = 51 ⋅ 1/49 ⋅ mo U UDv 469 L

∫

d Gf ⋅ eU U+ 3:: lK >−2/37! = nGf ⋅ d Gf mo = 31 ⋅ 1/57 ⋅ mo L U UGf 454

UDv U+

∆T Gf = nGf ⋅

UGf

5/3/!U kalorimetarskom sudu, zanemarqivog toplotnog kapaciteta, nalazi se te~nost polazne temperature UU>3:1!L!)stalnog toplotnog kapaciteta D>2/36!lK0L*/!U sud je unet bakarni uzorak mase nC>1/26!lh!i polazne temperature!UC>484!L/!Zavisnost specifi~nog toplotnog kapaciteta za bakar dC U od temperature data je izrazom: ! /Tokom uspostavqawa = 1/498 + 1/76: ⋅ 21 −5 ⋅ [lK 0 (lhL*)] [L ] toplotne ravnoteè u kalorimetru, okolini stalne temperature!Up>394!L-!predato je!6% od koli~ine toplote koju je predao bakarni uzorak. Odrediti: a) temperaturu u kalorimetarskom sudu u trenutku uspostavqawa toplotne ravnoteè b) promenu entropije izolovanog sistema od polaznog stawa do stawa uspostavqawa toplotne ravnoteè u kalorimetru. a) oznake koje se koriste u daqem tekstu re{ewa:

(R23 )C (R23 )p (R23 )U US

koli~ina toplote koju bakar predaje te~nosti u sudu koli~ina toplote koju bakar predaje okolini koli~ina toplote koju te~nost prima od bakra temperatura u kalorimetarskom sudu u trenutku uspostavqawa toplotne ravnoteè


{fmlp@fvofu/zv


strana 3

prvi zakon termodinamike za proces sa bakrom: US

(R23 )C , (R23 )p > nC ⋅

∫[

]

1/498 + 1/76: ⋅ 21 −5 U ⋅ eU

UC



(R23 )C , (R23 )p > nC ⋅ 1/498 ⋅ (US − UC ) + 1/76: ⋅ 21 −5 

US3 − UC3   3 

)2*

prvi zakon termodinamike za proces sa te~no{}u US

(R23 )U >

∫

D U ⋅ eU > D U ⋅ (US − UU )

)3*

UU

interno razmewena toplota izme|u bakra i te~nosti:

(R23 )C >− (R23 )U

)4*

(R23 )p > 1/16 ⋅ [(R23 )C + (R23 )p ]

uslov zadatka:

)5*

kombinovawem jedna~ina!)2*-!)3*-!)4*!j!)5*!dobija se: US!>3:4/8!L-

(R23 )U >5736!K-

(R23 )C >−5736!K-

(R23 )p >−354!K

b) K L K ∆TSU!>!∆TU!,!∆TC!>!///>!26/96!−!25/77!>!2/2:! L

∆TTJ!>!∆TSU!,!∆TP!>!///>!2/2:!,!1/97!>!3/16!

US

∆T U =

∫

U D(U ) ⋅ eU 3:4/8 K >26/96! = D U ⋅ mo S = 2/36 ⋅ mo L U UU 3:1

UU US

∆TC = nC ⋅

∫

d(U ) ⋅ eU = nC ⋅ U

UC

US

∫

(1/498 + 1/76: ⋅ 21 U ) ⋅ eU = .5

U

UC

  U K n ⋅ 1/498 ⋅ mo S + 1/76: ⋅ 21−5 ⋅ (US − UC ) = /// = −25/77 UC L   (R23 )p 354 K ∆T P = − =− >1/97! L UP 394


{fmlp@fvofu/zv


strana 4

5/4/!Meteor temperature!U>4111!L-!brzinom x>21!ln0t ule}e u ledeni breg temperature!U>384!L/ Masa meteora je!nn>21!lh-!a specifi~ni toplotni kapacitet!dn>1/9!lK0lhL/!Odrediti: a) koli~inu toplote koju meteor preda ledenom bregu c* promenu entropije izolovanog sistema koji ~ine meteor i ledeni breg d* masu otopqenog leda (toplota topqewa leda iznosi s>443/5!lK0lh)

meteor x>21!ln0t

ledeni breg

a) prvi zakon termodinamike za proces koji se de{ava sa meteorom: R23!>!∆V23!,!X23!,!∆Fl23

R23 = nn ⋅ dn ⋅ (Un3 − Un2) + n ⋅

R 23 = 21 ⋅ 1/9 ⋅ (384 − 4111) − 21 ⋅

(21 ⋅ 21 )

4 3

3

x33 − x23 3

⋅ 21 −4 >− 6/329 ⋅ 219 lK

b) ∆Ttjtufn = ∆Tnfufps + ∆Tmfe >///>−2:/286!,2:22/466>29:9/29! ∆Tnfufps = nn ⋅ dn mo ∆Tmfe =

lK L

384 Un3 lK > 21 ⋅ 1/9 ⋅ mo >!−2:/286! 4111 Un2 L

R23 6/329 ⋅ 216 lK >2:22/46! > Um 384 L

c) R23 = nm ⋅ sm

⇒


nm =

R23 6/329 ⋅ 216 >2681!lh > sm 443/5

{fmlp@fvofu/zv


strana 5

5/5/!U rezervoar sa nx>61!lh!vode!)dx>5/2:!lK0lhL*-!temperature!Ux>394!L- uroni se zatvorena boca, na~iwena od ~elika)d•>1/59!lK0lhL*-!sa!Wl>21!litara kiseonika (idealan gas). Masa ~eli~ne boce zajedno sa kiseonikom je!n•!,!nC>!31!lh/!Pre urawawa temperatura kiseonika i boce je!Ul>U•>474 L-!a pritisak kiseonika u boci je!ql>26!NQb/!Sistem koji se sastoji od vode i boce sa kiseonikom moè se smatrati izolovanim. Temperatura okolnog vazduha je!Up>3:4!L-!pritisak!qp>2!cbs-!a zapreminski (molarni) udeo kiseonika u okolnom vazduhu je!32&/!Zanemaruju}i razmenu toplote sa okolinom odrediti: a) temperaturu u sudu u trenutku uspostavqawa toplotne ravnoteè b) radnu sposobnost kiseonika u boci u trenutku postizawa toplotne ravnoteè a) prvi zakon termodinamike za proces koji po~iwe urawawem boce a zavr{ava se uspostavqawem toplotne ravnoteè; R23!>!∆V23!,!X23

⇒

V2!>!V3

V2 = n x ⋅ d x ⋅ Ux + n • ⋅ d • ⋅ U• + nl ⋅ d wL ⋅ UL V 3 = n x ⋅ d x ⋅ U + + n • ⋅ d • ⋅ U + + nl ⋅ d wL ⋅ U +

U+ =

n x ⋅ d x ⋅ Ux + n• ⋅ d • ⋅ U• + nl ⋅ d wL ⋅ UL >/// n x ⋅ d x + n • ⋅ d • + nl ⋅ d wL

jedna~ina stawa idealnog gasa za kiseonik u boci na po~etku procesa: q ⋅W 26 ⋅ 21 7 ⋅ 21 ⋅ 21 −4 = nl = l q l ⋅ Wl = nl ⋅ S hl ⋅ Ul ⇒ >2/6:!lh S hl ⋅ Ul 371 ⋅ 474 n•!>! (n • + nl ) −!nl!>!31!−!2/6:!>29/52!lh

U+ =

61 ⋅ 5/2: ⋅ 394 + 29/52 ⋅ 1/59 ⋅ 474 + 2/6: ⋅ 1/76 ⋅ 474 >397/7!L 61 ⋅ 5/2: + 29/52 ⋅ 1/59 + 2/6: ⋅ 1/76

napomena: !

U+!−

temperatura u boci u trenutku uspostavqawa toplotne ravnoteè


{fmlp@fvofu/zv


strana 6

b) jedna~ina stawa idealnog gasa za kiseonik u boci u trenutku uspostavqawa toplotne ravnoteè: q2 ⋅ Wl = nl ⋅ S hl ⋅ U2 q2 =

nl ⋅ S hL ⋅ U2 W

=

2/6: ⋅ 371 ⋅ 397/7 21 ⋅ 21 −4

>22/96!NQb

odre|ivawe pritiska kiseonika u okolnom vazduhu: q lp = sL ⋅ q p >1/132!NQb

odre|ivawe radne sposobnosti kiseonika u boci u trenutku uspostavqawa Xnby = n ⋅ (−∆v21 + Up ⋅ ∆t2p − q p ⋅ ∆w 2p ) toplotne ravnoteè:    qp  U Xnby = nL ⋅ dlw ⋅ (U2 − UP ) + UP ⋅  dq ⋅ mo P − Sh ⋅ mo L  + qLp ⋅ (w2 − w p ) = 761 lK   q2  U2   

w2 = wp =

S hl ⋅ U2 q2 S hl ⋅ Up q lp

=

371 ⋅ 397/7 22/96 ⋅ 21 7

=

371 ⋅ 3:4 1/132 ⋅ 21 7

>!1/1174!!

n4 lh

>!4/7387!!

n4 lh

napomena: U delu zadatka pod b) radi lak{e preglednosti veli~ine stawa kiseonika u boci u trenutku uspostavqawa toplotne ravnoteè obeleène su indeksom!2


{fmlp@fvofu/zv


strana 7 ⋅

5/6/!Radna materija u zatvorenom sistemu vr{i neki proces pri ~emu joj se u svakoj sekundi dovodi! R >4 ⋅

lK0t!toplote i odvodi zapreminski rad! X )lK0t*-!koji se u toku vremena mewa po zakonu: ⋅

X 12 τ = 3/5 ⋅ [lX] [i]

){b!!1! < τ ≤ 2 i *

⋅

X 23 >,3/5 [lX]

){b!!τ!?!2!i* ⋅

b* odrediti brzinu promene unutra{we energije sistema-! ∆ V23 )lX*-!u trenutku!vremena!τ>1/7!i b) odrediti promene unutra{we energije sistema-!∆V23!)lK*-!u toku prva dva ~asa ⋅

! X- [lX ]

τ-! [i] 2

3

a) ⋅

⋅

⋅

∆ V23 = R 23 − X 23 = 4 − 3/5 ⋅ τ = 4 − 3/5 ⋅ 1/7 = 2/67 lX b) 2

X12 =

2

∫

X (τ) ⋅ eτ =

∫

X (τ) ⋅ eτ = 3/5 τ

1 3

X23 =

∫

3/5 ⋅ τ ⋅ eτ = 3/5

τ3 3

2

= 2/3 lXi

1

1

3

= 3/5 lXi

2

2

Xp3!>!X12!,!X23!>!4/7!lXi!>!23:71!lK 3

R 13 =

∫

⋅

R(τ) ⋅ eτ = 4 ⋅ 3 = 7 lXi = 32711 lK

1

∆V13>!R13!−!X13!>!9751!lK


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strana 8

5/7/!Toplotne karakteristike neke radne materije zadate su zavisnostima: q!/!w!>!B!/!U v!>!C!/!U!,!D!/!U3!,!E gde je!B>!3:8!K0)lhL*-!C>7:8!K0)lhL*-!D>!1/196!K0)lhL3*-!E!>!dpotu-!a!q-!w-!U!j!u su veli~ine stawa u osnovnim jedinicama!TJ. Radna materija mewa svoje toplotno stawe kvazistati~ki adijabatski od stawa 2)q2!>!1/2!NQb-!U2>511!L*!do stawa!3!)U3>2451!L*/ b* izvesti jedna~inu kvazistati~ke adijabatske promene stawa radne materije u obliku:!q>g)U* c* odrediti pritisak radne materije u stawu!3 b* prvi zakon termodinamike za proces sa radnim telom (diferencijalni oblik) δr = ev + q ⋅ ew

)2* e(q ⋅ w ) = q ⋅ ew + w ⋅ eq )3*

diferencijal proizvoda:

kombinovawem jedna~ina!)2*!i!)3) sa toplotnim karakteristikama radne materije dobija se: 1!>!ev!, e(q ⋅ w ) − w ⋅ eq

(

⇒

ev!,! e(q ⋅ w ) >! w ⋅ eq

)

eq q eq e B ⋅ U + C ⋅ U + D ⋅ U3 + E = B ⋅ U ⋅ q

e C ⋅ U + D ⋅ U 3 + E + e(B ⋅ U ) = B ⋅ U ⋅

(

)

B +C U 3D q ⋅ mo + ⋅ (U − U2 ) = mo B U2 B q2

B + C + 3D ⋅ U eU eq ⋅ = q B U q = q2 ⋅ f

B +C U 3D ⋅mo + ⋅(U − U2 ) B U2 B

q = 1/2 ⋅ 21 7 ⋅ f

⇒

q = 1/2 ⋅ 21

7

3:8 + 7:8 U 3⋅1/196 ⋅mo + ⋅(U − 511 ) 3:8 511 3:8 ⋅f

3:8 + 7:8 U 3⋅1/196 ⋅mo + ⋅(U − 511 ) 3:8 511 3:8

b) ako u izvedenu jedna~inu stavimo!U>U3>2451!K, kao i vrednosti za navedene konstante!)B-!C!j!D*!dobija se: q 3 = 1/2 ⋅ 21 7 ⋅ f

3:8+ 7:8 2451 3⋅1/196 ⋅mo + ⋅(2451− 511 ) 511 3:8 3:8 >:/9!NQb


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strana 1

5. DESNOKRETNI KRU@NI PROCESI 6/2/ Koliko se korisnog (neto) rada moè najvi{e dobiti ako toplotni izvor temperature (UUJ=450 K) predaje toplotnom ponoru temperature (UUQ=300 K) R=800 kJ toplote, ako se izme|u toplotnog izvora i toplotnog ponora ukqu~i desnokretna toplotna ma{ina. Xepcjkfo

!Repw

!!!UJ

Radno telo

!Rpew

!!UQ

Xqplsfubokb Najvi{e korisnog rada se moè dobiti ako desnokretna tolotna ma{ina radi po Karnoovom desnokretnom ciklusu. U UUJ

UUQ

!t R EPW + R PEW U − UUQ = UJ R EPW UUJ

η =ηK

⇒

Repw = − Rpew ⋅

UUJ 561 = 911 ⋅ = 2311!lK 411 U UQ

⇒

Xofup!>!Repw!,!Rpew!>!2311!−!911!>!511!lK


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strana 2

6/3/ Radna supstanca vr{i potpuno povratni proces izme|u toplotnog izvora stalne temperature UUJ>2111!L i toplotnog ponora promenqive temperature. Toplotni kapacitet toplotnog ponora iznosi DUQ>311!lK0L, a temperatura toplotnog ponora se mewa od UUQ2>411!L do UUQ3>@ U toku obavqawa kru`nog proces toplotni izvor je radnoj supstanci predao 211!NK toplote. Odrediti: a) koristan rad kru`nog procesa b) termodinami~ki stepen korisnosti kru`nog procesa a)

(∆T tjtufn )23

= ∆T23 −

R epw UUJ

(1)

(∆T tjtufn )45

= ∆T 45 −

R pew (UUQ )mo

(2)

sabirawem jedna~ina (1) i (2) dobija se: 1=−

R epw R pew − (UUQ )mo UUJ

⇒

 2 R epw ⋅ UUQ3 = UUQ2 ⋅ fyq  D UQ UUJ

1=−

R epw D UQ ⋅ (UUQ2 − UUQ3 ) − UUQ2 − UUQ3 UUJ U mo UQ2 UUQ3

 2 211 ⋅ 21 4   > 411 ⋅ fyq ⋅  311 2111  

  >5:5/7!L  

R pew = D UQ ⋅ (UUQ2 − UUQ3 ) > 311 ⋅ (411 − 5:5/7 ) >−49/:!NX Xlps!>!Repw!,!Rpew!>!211!−!49/:!>!72/2!lX c* ηC =

Xlps 72/2 > >1/72 R EPW 211


{fmlp@fvofu/zv


strana 3

6/4/ Proveriti koji od dva prikazana kru`na procesa A i B ima ve}i stepen korisnog dejstva, ako je ∆t (vidi sliku) jednako za oba procesa. Pri kojoj temperaturi toplotnog ponora UUQ, a pri nepromeqenoj temperaturi toplotnog izvora UUJ>600 K, bi stepen korisnog dejstva kru`nog procesa B bio dva puta ve}i od stepena korisnog dejstva kru`nog procesa A. Svi procesi sa radnom telom su ravnote`ni. B U

∆t 2

600

C

U 3

3

600 ∆t03

∆t ∆t03 100

100

4 t

ciklus A:

UUj = 711 K,

2

4 t

UUQ = 211 K

R epw = R 23 = UUJ ⋅ ∆T R pew = R 34 + R 45 = ηB =

ciklus B:

R EPW + R PEW R EPW

UUj = 711 K,

R epw = R 23 + R 34 =

ηC =

UUJ + UUQ − ∆T UUJ + UUQ − ∆T UUJ + UUQ ⋅ + ⋅ = ⋅ (− ∆T) 3 3 3 3 3 U + UUQ 711 + 211 UUJ ⋅ ∆T + UJ ⋅ (− ∆T) 711 − 3 3 >1/53 > > UUJ ⋅ ∆T 711

R EPW + R PEW R EPW

UUQ = 211 K

UUJ + UUQ ∆T UUJ + UUQ ∆T UUJ + UUQ R pew = R 42 = UUQ ⋅ (−∆T) ⋅ + ⋅ = ⋅ ∆T 3 3 3 3 3 UUJ + UUQ 711 + 211 − 211 ⋅ ∆T − UUQ ⋅ ∆T + 3 3 >1/82 > > UUJ + UUQ 711 + 211 ⋅ ∆T 3 3


{fmlp@fvofu/zv


ηB =

UUJ −

UUJ + UUQ 3 UUJ

3 ⋅ η B = ηC

⇒

strana 4 UUJ + UUQ − UUQ 3 ηC = UUJ + UUQ 3 UUJ + UUQ UUJ + UUQ UUJ − − UUQ 3 3 3⋅ = UUJ + UUQ UUJ 3

⇒

UUQ >1!L zadatak za ve`bawe:

(1.4.)

6/5/ Proveriti koji od dva prikazana kru`na procesa A i B ima ve}i stepen korisnog dejstva, ako je ∆t (vidi sliku) jednako za oba procesa. U oba slu~aja temperatura toplotnog izvora iznosi UUJ>800 K, a temperatura toplotnog ponora UUQ=300 K. Svi procesi sa radnom materijom su ravnote`ni. B 1000 900 800

C

T, K

1000 900

UUJ

800

4

700

700

600

600

3

500 400 300

5

300

2

200

∆t03

100 s, J/(kgK)

η B =1/444


3

5

UUQ

2

200

∆t03

100 0

s, J/(kgK)

∆t

∆t re{ewe:

4

UUJ

500 400

UUQ

0

T, K

ηC >1/397

{fmlp@fvofu/zv


strana 5

6/6/ Dvoatomni idealan gas obavqa proces gasnoturbinskog postrojewa koji se sastoji od dve izobare i dve adijabate (Xulov ciklus). Stawe radnog tela na ulazu u kompresor je 1(q>2!cbs-!u>26pD), na izlazu iz kompresora 2(q>6!cbs) i na ulazu u turbinu 3(u4>891pD). Stepen dobrote adijabatske kompresije je fy ηlq e =0.83, a stepen dobrote adijabatske ekspanzije η e =0.85. Odrediti: a) termodinami~ki stepen korisnog dejstva ovog ciklusa (η) b) termodinami~ki stepen korisnog dejstva ovog ciklusa za slu~aj maksimalne mogu}e rekuperacije toplote (η′) a) κ −2

2/5 −2

  κ U3L!>!U2 ⋅  q 3L  > 399 ⋅  6  2/5 >!567/25!L  q   2  2  U −U 567/25 − 399 U3!>!U2!,! 3L lq 2 !>! 399 + >5:1/69!L 1/94 ηe κ −2

2/5 −2

  κ > 2164 ⋅  2  2/5 >!775/95!L U5L!>!U4 ⋅  q 5L  q  6  4  fy U5!>!U4!, η e ⋅ )U5l!−U4*!>!2164!, 1/96 ⋅ )775/95!−!2164*!>834/18!L η!>!

U − U3 + U2 − U5 R epw + R pew 2164 − 5:1/69 + 399 − 834/18 >1/34 >///> 4 !>! R epw U4 − U3 2164 − 5:1/69

Repw!>!n!/!)r34*q>dpotu!>!n!/!dq!/!)!U4!−U3!* Rpew!>!n!/!)r52*q>dpotu!>!n!/!dq!/!)!U2!−U5!* Repw !3

!4

X23

X45

!2

!5 Rpew


{fmlp@fvofu/zv


strana 6

b) Repw′ !3

X23

!2

!B

!C

R E K U P E R A T O R

!4

X45

!5

Rpew′ op{ti uslov rekuperacije:

U5!?!U3

uslov maksimalne rekuperacije:

UB!>!U3!

η′!>!

)UC!>!U5*

R epw (+R pew ( U − UC + U2 − UB 2164 − 834/18 + 399 − 5:1/69 >!1/4: >///> 4 !> R epw ( U4 − UC 2164 − 834/18

Repw′!>!n!/!)rC4*q>dpotu!>!n!/!dq!/!)!U4!−UC!* Rpew′!>!n!/!)rB2*q>dpotu!>!n!/!dq!/!)!U2!−UB!*

U

4

U

4

5

!C

3 3L

5L

t


5

3

2

bez rekuperacije

Rsfl

2

B t sa rekuperacijom

{fmlp@fvofu/zv


strana 7

6/7/ Sa vazduhom (idealan gas) kao radnim telom izvodi se idealan Xulov desnokretni ciklus (sve promene stawa su kvazistati~ke). Ekspaznzija je dvostepenom sa me|uzagrevawem radnog tela, a kompresija je dvostepena sa me|uhla|ewem (slika). Ako je qnby>27!cbs, qnjo>2!cbs!i ako je stepen q q q q kompresije u oba kompresora i stepen ekspanzije u obe turbine isti ( 8 = 2 = 3 = 5 ) i ako se q7 q 9 q4 q6 toplota dovodi od toplotnog izvora temperatura UUJ>U3>U5>711!L- a predaje toplotnom ponoru temperature UUQ>U7>U9>361!L, skicirati ciklus na Ut dijagramu i odrediti stepen korisnog dejstva ciklusa (η).

2

3

5 9

8

4

7

6

3

U

5

6 2

9


4 8

7

t

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike qnjo!>!q7>!q6!>!2!cbs

strana 8 qnby!>!q2>!q3!>!27!cbs

q3 q 5 = q4 q6

⇒

q4>q5!>!5!cbs

q 8 q2 = q7 q9

⇒

q2>q9!>!5!cbs

⇒

q U2 = U9 ⋅  2  q9

   

⇒

q U4 = U3 ⋅  4  q3

  

⇒

q U6 = U5 ⋅  6  q5

  

⇒

q U8 = U7 ⋅  8  q7

   

U2  q2  =  U9  q 9  U3  q3   = U4  q 4  U5  q 5   = U6  q 6  U7  q 7   = U8  q 8  η!>!

κ −2 κ

κ −2 κ

κ −2 κ

κ −2 κ

κ −2 κ

 27  = 361 ⋅    5 

κ −2 κ

κ −2 κ

κ −2 κ

2/5 −2 2/5

 5  = 711 ⋅    27   2 = 711 ⋅   5 5 = 361 ⋅   2

= 482/6!L

2/5 −2 2/5

2/5 −2 2/5

2/5 −2 2/5

= 514/8!L

= 514/8!L

= 482/6!L

U − U2 + U5 − U4 + U7 − U6 + U9 − U8 R epw + R pew >!///> 3 !>!1/46 R epw U3 − U2 + U5 − U4

Repw!>!n![)r23*q>dpotu!,!)r45*q>dpotu!]>!n!/!dq!/!)!U3!−!U2!,!U5!−!U4!* Rpew!>!n![)r67*q>dpotu!,!)r89*q>dpotu!]>!n!/!dq!/!)!U7!−!U6!,!U9!−!U8!* zadatak za ve`bawe:

(1.7.)

6/8/ U energetskom postrojewu za proizvodwu elektri~ne energije primewen je rekuperativni desnokretni gasnoturbinski ciklus (Xulov ciklus) sa vazduhom (idealan gas) kao radnim telom. U kompresoru se 311!lh0i radnog tela temperature 91pD adijabatski komprimuje od 1/4!NQb do 1/9!NQb, sa stepenom dobrote ηelq>1/97. Dovo|ewem toplote radno telo se zatim zagreva do 891pD i odvodi u turbinu, gde adijabatski ekspandira sa stepenom dobrote ηefy>1/:. Za vreme odvo|ewa toplote rekuperi{e se 91& od koli~ine toplote koja bi se u najpovoqnijem slu~aju mogla rekuperisati. Skicirati proces u Ut kooedinatnom sistemu i odrediti a) stepen korisnog dejstva ciklusa b) teorijsku snagu koja stoji na raspolagawu za pogon generatora, ako se turbina i kompresor nalaze na istom vratilu a) b)

η!>!1/44 Q!>!6/6!lX


{fmlp@fvofu/zv


strana 9

6/9/ Dvoatomni idealan gas obavqa teorijski (idealan) Otov kru`ni proces izme|u temperatura Unby>U4>UUJ>2111!L i Unjo>U2>UUQ>3:1!L. Odrediti stepen kompresije )ε>w20w3* tako da korisna ⋅

snaga motora bude najve}a kao i snagu motora ako je molski protok gasa kroz motor o >:!npm0t. ⋅

Xofup = R epw + R pew = ...= o⋅ (Nd w ) ⋅ (U4 − U3 + U2 − U5 ) ⋅

Repw!>!n!/!)r34*w>dpotu!>! o⋅ (Nd w ) ⋅ (U4 − U3 ) ⋅

Rpew!>!n!/!)r52*w>dpotu!>! o⋅ (Nd w ) ⋅ (U2 − U5 ) U2  w 3   = U3  w 2 

κ −2

U4  w 5   = U5  w 4 

κ −2

w  U3 = U2 ⋅  2   w3 

⇒

w  U5 = U4 ⋅  4   w5 

(

⋅

κ −2

⇒

Xofup = o⋅ (Nd w ) ⋅ U4 − U2 ⋅ ε κ −2 + U2 − U4 ⋅ ε2− κ

= U2 ⋅ ε κ −2 κ −2

= U4 ⋅ ε2− κ

)

[

]

⋅ ∂Xofup = o⋅ (Nd w ) ⋅ − U2 ⋅ (κ − 2) ⋅ ε κ −3 − U4 ⋅ (2 − κ ) ⋅ ε − κ ∂ε ∂Xofup ⇔ − U2 ⋅ (κ − 2) ⋅ ε κ −3 − U4 ⋅ (2 − κ ) ⋅ ε − κ = 1 =1 ∂ε 2

2

 U  3 κ −3  2111  3⋅2/5 −3 ε =  4  > U2 ⋅ (κ − 2) ⋅ ε = U4 ⋅ (κ − 2) ⋅ ε , >!5/8   3:1   U2  Qsj!tufqfov!lpnqsftjkf!ε>5/8!npups!ptuwbsvkf!obkwf~v!tobhv κ −3

−κ

X

Xnby

ε>5/8


ε

{fmlp@fvofu/zv


strana 10

(

)

Xnby = : ⋅ 21 −4 ⋅ 31/9 ⋅ 2111 − 3:1 ⋅ 5/82/5 −2 + 3:1 − 2111 ⋅ 5/82−2/5 >51!lX Nbltjnbmob!tobhb!npupsb!j{optj! Xnby >51!lX U

4 UJ

3−4;!w>dpotu 5−2;!w>dpotu

3 5

UQ

2

t 6/:/ Radna materija (idealan gas) obavqa idealan Xulov kru`ni ciklus izme|u temperatura U3>UUJ>2144 L i U5>UUQ>3:2!L. Odrediti temperaturu radne materije posle kvazistati~ke izentropskog sabijawa u kompresoru (U2), odnosno posle kvazistati~ke izentropske ekspanzije u turbini (U4), tako da koristan rad (rad na zajedni~kom vratilu) ima maksimalnu vrednost.

U

3 UJ 2−3;!q>dpotu 4−5;!q>dpotu 2 4

5

UQ t

Xlpsjtubo!>!Repw!,!Rpew!>!///!>n!/!dq!/!)!U3!−!U2!,!U5!−!U4!*!>!/// Repw!>!n!/!)r23*q>dpotu!>!n!/!dq!/!)!U3!−U2!* Rpew!>!n!/!)r45*q>dpotu!>!n!/!dq!/!)!U5!−U4!*


{fmlp@fvofu/zv


U2  q2   = U5  q 5  U ⋅U U2 = 3 5 U4

κ −2 κ

U3  q3   = U4  q 4 

⊕

strana 11

κ −2 κ

 q2   q5

⊕

  q3  =    q4

  

U2 U = 3 U5 U4

⇒

⇒

⇒

Xlpsjtubo!>!n!/!dq!/!)!U3!− ∂Xlpsjtop = n ⋅ dq ∂U4

U3 ⋅ U5 ,!U5!−!U4!* U4

 U ⋅U ⋅  3 3 5 − 2 -!   U 4  

U4 = U3 ⋅ U5 > 2144 ⋅ 3:2 >659/4!L

∂Xlpsjtop =1 ∂U4 !!!!!!! U2 =

U3 ⋅ U5

⇔

U43

− 2 >1

U3 ⋅ U5 2144 ⋅ 3:2 >!659/4!L > U4 659/4

X Xnby

U4

U4>659/4

6/21/ Dvoatomni idealan gas obavqa realni desnokretni kru`ni proces gasnoturbinskog bloka (Xulov) izme|u temperatura Unby>6:6pD i Unjo>26pD. Molski protok gasa kroz postrojewe iznosi 31!npm0t/!). fy Stepen dobrote adijabatske kompresije je ηlq e =0.88, a stepen dobrote adijabatske ekspanzije η e =0.88. Odrediti maksimalnu snagu gasnoturbinskog bloka pri datim uslovima. 4

U

5 3 3L

5L

2 t


{fmlp@fvofu/zv


strana 12

Korisno dobijen rad ima najve}u vrednost (vidi prethodni zadatak) kada je: U3l =

U2 ⋅ U4 >!611!L

U5l =

U2 ⋅ U4 !>!611!L U3l

Me|utim u ovom zadatku su adijabatska kompresija i adijabatska ekspanzija neravnote`ni (nekvazistati~ki) procesi. Uzimawem te ~iwenice u obzir dobija se: U3L − U2

U3!>!U2!,!

ηlq e

!>! 399 +

611 − 399 >639/:!L 1/99

U5!>!U4!, η fy e ⋅ )U5l!−U4*!>!979!, 1/99 ⋅ )611!−!979*!>655/27!L

Xlpsjtubo!>!Repw!,!Rpew!>!/// ⋅

(

)

⋅

(

)

Repw!>!n!/!)r34*q>dpotu!>!n!/!dq!/!)!U4!−U3!*!> o⋅ Nd q ⋅ (U4 − U3 ) Rpew!>!n!/!)r52*q>dpotu!>!n!/!dq!/!)!U2!−U5!*!> o⋅ Nd q ⋅ (U2 − U5 ) ⋅

(

)

Xlpsjtubo> o⋅ Ndq ⋅ (U4 − U3 + U2 − U5 ) > Xlpsjtubo!>Xnby!>! 31 ⋅ 21 −4 ⋅ (3:/2) ⋅ (979 − 639/: + 399 − 655/27 ) >59/38!lX

zadatak za ve`bawe:

(1.11.)

6/22/ Sa troatomnim idealnim gasom obavqa se Eriksonov desnokretni kru`ni proses sa izme|u temperatura Unby>UUj>711!L!i Unjo>UUQ>511!L. Odrediti stepen korisnog dejstva ovog ciklusa za slu~aj maksimalno mogu}e rekuperacije toplote i {rafirati na Ut dijagramu povr{nu koja odgovara rekuperisanoj toploti. re{ewe:

η=0.33


{fmlp@fvofu/zv


strana 13

6/23/ Nekom idealnom gasu dovodi se pri kvazistati~koj izotermskoj ekspanziji (2.3) R23>411!lK toplote od izotermnog toplotnog izvora temperature UUJ>2111!L, pri ~emu entropija idealnog gasa poraste za ∆T23>1/6!lK0L. Pri kvazistati~koj promeni stawa (3.4) entropija idealnog gasa opada linearno u Ut koordinatnom sistemu i pri tom se toplota predaje izotermnom toplotnom ponoru temperature UUQ>3:4 L sve dok se ne uspostavi stawe termodinami~ke ravnoteè. Od stawa (4) do po~etnog stawa (2) dolazi se kvazistati~kom izentropskom kompresijom. Skicirati promene stawa idealnog gasa u Ut koordinatama i odrediti: a) stepen korisnog dejstva ovog kru`nog ciklusa b) odrediti promenu entrpopije sistema (lK0L) c) {rafirati na Ut dijagramu povr{inu ekvivalentnu korisno dobijenom radu U UJ 2

3

UQ 4 t b* η=

R epw + R pew = /// = 1/37 R epw

)37&*

R epw = R23 = 411 lK T3

R epw = R 23 =

∫

U)t*eT = U3 ⋅ ∆T23

⇒ U3 =

R epw 411 = = 711 L 1/6 ∆T23

T2 T4

R pew = R 34 =

∫

U)t*eT =

U + U4 U3 + U4 ⋅ ∆T 34 = 3 ⋅ )−∆T23 * 3 3

T3

R pew

711 + 3:4 = ⋅ )−1/6* = −334/4 lK 3


{fmlp@fvofu/zv


strana 14

c* ∆T tjtufn = ∆Tsbeopufmp + ∆T UJ + ∆T UQ = /// = −1/4 + 1/87 = 1/57 ∆T UJ = −

lK lhL

R epw 411 lK =− = −1/41 UUJ 2111 L

∆T UQ = −

R pew lK − 334/4 =− = 1/87 UUQ 3:4 L

d* Xlpsjtubo!>!Repw!,!Rpew!>!Repw!−! R pew Princip {rafirawja korisnog rada na Ut dijagramu je princip oduzimawa povr{ina koje predstavqaju dovedenu (Repw* i odvedenu )Rpew*!toplotu. U

U 2

3

3

4 t

t −

Repw

R pew

U 2

3

>

4

t Xlpsjtubo


{fmlp@fvofu/zv


strana 15

6/24/ Radna susptancija (idealan gas) obavqa desnokretni kru`ni proces koji se sastoji iz izobarskog dovo|ewa toplote, kvazistati~ke (ravnote`ne) adijabatske ekspanzije, izotermskog odvo|ewa toplote i kvazistati~ke (ravnote`ne) adijabatske kompresije. Toplota se radnoj supstanciji predaje od dimnih gasova (idealan gas), koji se pri tome izobarski hlade (Cp=1.06 kJ/K), od po~etne temperature!uh2>:11pD. Od radne susptancije, okolini stalne temperature up>29pD- predaje se 411!lK!toplote na povratan na~in. Odrediti stepen korisnosti ovog kru`nog procesa i skicirati ga u Ut i qw koordinatnom sistemu. U

q 2

3

3

2 5 4 5

4 t

w

Rpew!>!R45!>!−!411!lK T5

R pew = R 45 =

∫

U)t*eT = U4 ⋅ ∆T 45

⇒ ∆T 45 =

R epw − 411 lK = = −2/14 U4 3:2 L

T4

∆T23!>!−!∆T45!>!2/14!

(

(∆T tjtufn )23

(∆T

ejn o hbt

lK L

= ∆T23 + ∆T ejno hbt

)

23

= n eh ⋅ S h ⋅ mo

)

q eh3 q eh2

 ∆T ejnoj hbt UEH3 = UEH2 ⋅ fyq  D qEH 

23

(

!!!!!!!⇒!!!!!!! ∆T ejno hbt

− D qEH ⋅ mo

UEH3 UEH2

)

23

= −∆T23 >2/14!

lK L

⇒

  > 2284 ⋅ fyq − 2/14  >554/:!L   2/17  

Repw!>!R23!>−!REH!>! − D qEH ⋅ (UEH3 − UEH2 ) > − 2/17 ⋅ (554/: − 2284 ) >883/96!lK η!>!

R epw + R pew 883/96 . 411 >1/72 > R epw 883/96

zadatak za ve`bawe (1.14.) 6/25/ Vazduh (idealan gas) vr{i slede}i kru`ni proces. Od po~etnog stawa (U2>411!L) vr{i se kvazistati~ka promena stawa po zakonu prave linije u Ut koordinatnom sistemu pri ~emu se radnom telu dovodi toplota od toplotnog izvora stalne temperature UUJ>U3?U2, pri ~emu je w2>w3. Nakon toga vr{i se kvazistati~ka izentropska ekspanzija do po~etne temperature. Kru`ni proces se zatvara kvazistati~kom izotermom. Stepen korisnog dejstva ovog ciklusa iznosi η>1/36. Skicirati ciklus na Ut dijagramu i odrediti promenu entropije izolovanog sistema za najpovoqniji poloàj temperatura toplotnog izvora i toplotnog ponora. re{ewe:

∆ttjtufn!>!85!


K lhL {fmlp@fvofu/zv


strana 16

6/26/ Vazduh (idealan gas) u prvom slu~aju obavqa desnokretni Xulov kru`ni proces. U drugom slu~aju pri istom odnosu pritisaka qnby0qnjo>4, istoj dovedenoj koli~ini toplote i istoj Unby>:84!L, izentropska kompresija zamewuje se izotermskom kompresijom, pri temperaturi Unjo>4:6!L. Sve promene stawa vazduha su ravnote`ne. a) odrediti termodinami~ke stepene korisnosti kru`nih procesa za oba slu~aja b) odrediti termodinami~ke stepena korisnosti, ako se u oba prethodna slu~aja obavqaju kru`ni procesi sa potpunim regenerativnim zagrevawem radne materije b* U

4

U

4

Unby

Unby

3

3

5

5

Unjo 2′

2 t q3!>!q4!>!qnby!

!

κ −2  κ

q U5!>!U4! ⋅  5   q4  q  U2!>!U3!  2   q3 

κ −2 κ

q2!>!q5!>!qnjo 2/5 −2  2  2/5

= ! :84 ⋅   4

4 = 4:6 ⋅    2


t

2/5 −2 2/5

= 821/:!L

= !399/7!L

{fmlp@fvofu/zv


strana 17

prvi slu~aj: ηJ!>!

r epw + r pew 689 − 553/4 = !1/38 = ///!> 689 r epw

repw!>!r34!>!dq!/!)!U4!−!U3!*!>! 2 ⋅ (:84 − 4:6 ) >689!!

lK lh

rpew!>!r52!>!dq!/!)!U2!−!U5!*!> 2 ⋅ (399/7 − 821/: ) >−!533/4!

lK lh

drugi slu~aj: ηJJ!>!

689 − 551/5 repw + rpew = !1/35 = ///!> 689 repw

lK lh lK rpew!>!r52′!,!r2′3!>!///!>!−426/:!−!235/6!>−!551/5!! lh lK r52′!>!dq!/!)!U2′!−!U5!*!>! 2 ⋅ (4:6 − 821/: ) >−!426/:! lh q 2 lK r2′3!>!U2′!/!Sh!/!mo! 2( = 4:6 ⋅ 1/398 ⋅ mo >−235/6! q3 4 lh repw!>!r34!>!dq!/!)!U4!−!U3!*!> 2 ⋅ (:84 − 4:6 ) >!689!!

c* u oba slu~aja maksimalna rekuperisana toplota je jednaka i iznosi: lK rsfl!>!r52′!>!−!427! lh

prvi slu~aj: r epw (+r pew ( = ///!>!1/6: r epw ( lK repw′!>!repw!.! q rek !>!373! lh

ηJ′!>!

lK rpew′!>!rpew!,! q rek !>!−!217! lh

drugi slu~aj: r epw (+r pew ( = ///!>!1/63 r epw ( lK repw′!>!repw!.! q rek !>!373! lh ηJJ′!>!


rpew′!>!rpew!,! q rek !>!−!217!

lK lh

{fmlp@fvofu/zv


strana 18

6/27/ Ciklus gasne turbine koji radi sa troatomnim idealnim gasom kao radnim telom sastoji se iz izotermskog kvazistati~kog sabijawa (2.3), izohorskog dovo|ewa toplote (3.4), adijabatske ekspanzije (4. 5) i izobarskog odvo|ewa toplote (5.2). Ako je odnos pritisaka )q40q5*>9/5 i )q30q2*>4/6- odrediti termodinami~ki stepen korisnosti kru`nog procesa (η) za slu~aj da je adijabatska ekspanzija (4.5): a) kvazistati~ka b) nekvazistati~ka sa stepenom dobrote ekspanzije ηefy>1/:6 a)

!4

!U

!5l

!2

!3

!t q3!>!4/6/q2

U3!>!U2 U3 q3 = U4 q 4

U4 = U3 ⋅

⇒

q U5l!>!U4!  5l  q4

  

κ −2 κ

q4 9/5 ⋅ q2 = U2 ⋅ = 3/5 ⋅ U2 q3 4/6 ⋅ q2

 q2   = 3/5 ⋅ U2 ⋅   9/5 ⋅ q2 

κ −2 κ

 2  = 3/5 ⋅ U2 ⋅    9/5 

2/39 −2 2/39

= 2/6!/U2

R epw = n ⋅ (r34 )w =dpotu = o ⋅ [(Nd w ) ⋅ (U4 − U3 )] !>!)Ndw*!/!2/5!/!U2

[

]

 q  R pew = n ⋅ (r 5l2 )q = dpotu + (r23 )U = dpotu > o ⋅  Nd q ⋅ (U2 − U5l ) + NS h ⋅ U2 ⋅ mo 2  q3  

(

(

)

(

)

)

2   R pew = o ⋅  Nd q ⋅ (− 1/6 ⋅ U2 ) + NS h ⋅ U2 ⋅ mo ! 4/6   + R pew R η!>! epw =! R epw

)

⇒

(Nd w ) ⋅ 2/5 ⋅ U2 − (Ndq ) ⋅ 1/6 ⋅ U2 + (NSh ) ⋅ U2 ⋅ mo 2 4/6 (Nd w ) ⋅ 2/5 ⋅ U2

3:/2 ⋅ 2/5 − 48/5 ⋅ 1/6 + 9/426 ⋅ mo η>!

(

3:/2 ⋅ 2/5


!>

2 4/6 >1/396

{fmlp@fvofu/zv


strana 19

b) !4

!U

!5 !5l !3

!2 !t

ηfy e =

U4 − U5 !⇒ U4 − U5l

U5 = U4 − ηfy e ⋅ (U4 − U5l ) ⇒

U5 = 3/5 ⋅ U2 − 1/:6 ⋅ (3/5 U2 − 2/6 ⋅ U2 ) !>!2/66!/U2 R epw = n ⋅ (r34 )w =dpotu = o ⋅ [(Nd w ) ⋅ (U4 − U3 )] !>!)Ndw*!/!2/5!/!U2

[

] (

 q  R pew = n ⋅ (r 52 )q = dpotu + (r23 )U = dpotu >  Nd q ⋅ (U2 − U5 ) + NS h ⋅ U2 mo 2  ! q 3 

(

(

)

)

(

)

)

2   R pew = o ⋅  Nd q ⋅ (− 1/66 ⋅ U2 ) + NS h ⋅ U2 mo ! 4 /6   + R pew R η!>! epw =! R epw

(Nd w ) ⋅ 2/5 ⋅ U2 − (Ndq )⋅ 1/66 ⋅ U2 + (NSh )⋅ U2 ⋅ mo 2 4/6 (Nd w ) ⋅ 2/5 ⋅ U2

3:/2 ⋅ 2/5 − 48/5 ⋅ 1/66 + 9/426 ⋅ mo η>!

3:/2 ⋅ 2/5


!>

2 4/6 >1/35

{fmlp@fvofu/zv


strana 20

6/28/ Vazduh (idealan gas) obavqa desnokretni kru`ni proces koji se sastoji od dve kvazistati~ke (ravnote`ne) izentrope i dve nekvazistati~ke (neravnote`ne) izoterme, izme|u temperatura Unjo>411!L i Unby>911!L. Pritisak vazduha na kraju izotermske kompresije iznosi q2>1/2!NQb, a na kraju izotermske ekspanzije iznosi q4>1/9!NQb. Temperature toplotnog izvora i toplotnog ponora su stalne i iznose UUJ>961!L i UUQ>391!L. Odrediti: a) termodinami~ki stepen korisnosti kru`nog procesa, ako promena entropije izolovanog sistema za proces dovo|ewa toplote iznosi 71!K0)lhL*- odnosno promena entropije izolovanog sistema za proces odvo|ewa toplote iznosi 211!K0lhL b) promenu entropije izolovanog sistema (koji sa~iwavaju toplotni izvor, toplotni ponor, i radno telo) za slu~aj da se sve promene stawa odvijaju kvazistati~ki (ravnote`no) a) U UJ 3 U nby

4

Unjo 2

5

UQ t

repw + rpew = /// repw repw!>!r34!>!///

η!>!

rpew!>!r52!>!///  q3!>!q2 ⋅  U3 U  2

κ

2/5

 κ −2 > 2 ⋅ 21 6 ⋅  911  2/5 −2 >!42/216!Qb   411  

proces 2-3: ∆t34!>! dq mo

U4 q 9 K = 49:! − S h mo 4 = − 398 mo 42 U3 q3 lhL

(∆T tjtufn )34 !>!∆t34!−!

r 34 UUJ

r34> 961 ⋅ (49: − 71) >38:/76!


⇒

r34!>!UUJ ⋅ (∆t 34 − (∆t tjtufn ) ) 34

lK lh

{fmlp@fvofu/zv


 q5!>!q4 ⋅  U5 U  4

strana 21

κ

2/5

 κ −2 > 9 ⋅ 21 6 ⋅  411  2/5 −2 >!1/37!/216!Qb    911  

proces 4-1: ∆t52!>! d q mo

U2 q 2 K = −49:! − S h mo 2 = − 398 mo !)uo~iti da je!∆t52>−∆t23* 1/37 U5 q5 lhL

(∆T tjtufn )52 >!∆t52!−! r52

⇒

Uuq

r52>! 391 ⋅ (−49: − 211) >−!247/:3!

r52!>!UUQ ⋅ (∆t 52 − (∆t tjtufn )52 )

kJ kg

lK lh lK rpew!>!r52!>!−!247/:3! lh repw!>!r34!>!38:/76!

η!>!

repw + rpew 38:/76 − 247/:3 !>!1/62 = repw 38:/76

b) repw′!>! (r 34 )U =dpotu > U3 ⋅ S h mo rpew′!>! (r 52 )U =dpotu > U5 ⋅ Sh mo

q3 42 lK = 911 ⋅ 398 ⋅ mo = 422! 9 lh q4

2 lK q5 = −227! = 391 ⋅ 398 ⋅ mo 1/37 lh q2

∆t tjtufn = ∆t sbeopufmp + ∆t UJ + ∆t UQ = /// = −476/: + 525/4 = 59/5 ∆t UJ = −

repw 422 K =− = −476/: UUJ 961 lhL

∆t UQ = −

−227 rpew K =− = 525/4 UUQ 391 lhL


K lhL

{fmlp@fvofu/zv


strana 22

6/29/ Parno turbinsko postrojewe radi po idealnom Rankin−Klauzijus–ovom kru`nom procesu izme|u qnjo>1/2!cbs!i qnby>21!cbs. U kondenzatoru, se rashladnoj vodi predaje toplota i pri tom se rashladna ⋅

voda n x >4!lh0t, zagreje od stawa B)q>2!cbs-!u>21pD* do stawa B!)q>2!cbs-!u>31pD*/ Snaga napojne pumpe iznosi 0.2!lX. Skicirati kru`ni proces na Ut dijagramu i odrediti: a) termodinami~ki stepen korisnog dejstva ciklusa b) snagu turbine

2

U

3

3

2 5

4 C

5

4

B

t

a) q>1/17!cbs-

ta~ka 4: lK i5!>!2:2/:! lh

i2!>!2:4/72!!!

(kqu~ala te~nost)

lK t5!>!1/75:3! lhL q!>21!cbs-

ta~ka 1:

y>1

t>1/75:3!

lK )te~nost* lhL

lK lh q!>2!cbs

ta~ka 3:

⋅

⋅

⋅

⋅

⋅

X 23 = − ∆ I23 > − nq ⋅ (i2 − i 5 ) > X q ⋅

nq =

⋅

⋅

R 23 = ∆ I23 + X 23

Prvi zakon termodinamike za proces u pumpi: ⇒

⋅

lh −1/2 XQ > >6/: ⋅ 21 −3 t i 5 − i2 2:2/: − 2:4/72


{fmlp@fvofu/zv


strana 23 ⋅

Prvi zakon termodinamike za proces u kondenzatoru: ⋅

⋅

I2 = I3

⋅

⋅

⋅

nq ⋅ i 5 + n x ⋅ (iC − i B ) ⋅

⋅

>///>!

6/: ⋅ 21 −3 ⋅ 2:2/: + 4 ⋅ (95 − 53)

>3422/4!

6/: ⋅ 21 −3

nq lK t4!>8/3:6! lh

)q>1/2!cbs-!i>3422/4!

ta~ka A:

q!>21!cbs-

u>21pD

⇒

ta~ka C:

q!>21!cbs-

u>31pD

⇒

ta~ka 2:

q!>21!cbs-

t>8/3:6!

i3!>!4266/6!!!

⋅

nq ⋅ i 4 + n x ⋅ i B > n q ⋅ i 5 + n x ⋅ i C

⇒

⋅

i4!>!

⋅

⋅

R 23 = ∆ I23 + X 23

lK lh

lK * lh

lK lhL

lK lh lK iC!>!95! lh iB!>!53!

)pregrejana para*

lK lh

a) ⋅

η=

⋅

R epw + R pew ⋅

>!///>!

R epw ⋅

⋅

⋅

⋅

⋅

⋅

285/8 − 236 >1/39 285/8

R epw = R 23 = n q ⋅ (i 3 − i2 ) > 6/: ⋅ 21 −3 ⋅ (4266/6 − 2:4/72) >285/8!lX R epw = R 45 = n q ⋅ (i 5 − i 4 ) > 6/: ⋅ 21 −3 ⋅ (2:2/: − 3422/4 ) >−236!lX b) ⋅

Prvi zakon termodinamike za proces u turbini: ⋅

⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X 23 ⋅

X 23 = − ∆ I23 > − nq ⋅ (i 4 − i 3 ) > − 6/: ⋅ 21 −3 ⋅ (3422/4 − 4266/6 ) >61!lX> X uvs


{fmlp@fvofu/zv


strana 24

6/2:/ U parnom kotlu uz konstantan pritisak od q>51!cbs od vode temperature 41pD proizvodi se vodena para temperature u>611pD. Ta para izentropski (ravnote`no) ekspandira u turbini do pritiska od q>1/17!cbs, a zatim se odvodi u kondenzator. Napojna pumpa vra}a u kotao pothla|en kondenzat. Toplota potrebna za proizvodwu pare u parnom kotlu obezbe|uje se hla|ewem dimnih gasova (idealan gas) od po~etne temperature 2711pD do temperature od 311pD. Koli~ina dimnih gasova je 6611!lnpm0i- a wihov zapreminski sastav 29&!DP3-!:&P3-!84&O3 . Skicirati promene stawa vodene pare na Ut dijagramu i odrediti: a) termodinami~ki stepen korisnog dejstva kru`nog procesa c* snagu turbine 3 U

2 4

5

t ta~ka 1:

q!>51!cbsu>41pD lK lK - !t2>1/544! i2!>!23:/4! lhL lh

)te~nost*

q>51!cbsu>611pD lK lK - t3!>!8/198! i3!>!4556! lhL lh

(pregrejana para)

ta~ka 2:

ta~ka 3:

q!>1/17!cbs-

y4!>!1/95-

i4!>!3291/6!!!

ta~ka 4:

q!>1/17!cbs-

i5!>!236/:!!!

lK lhL

(vla`na para)

lK lhL

(te~nost)

t4>!t3!>!8/198! lK lh t5>!t2>!1/544!

lK lh


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike η!>!

strana 25

repw + rpew 4426/8 − 3165/7 = ///!> = !1/49 4426/8 repw

lK lh lK rpew!>!r45!>i5!−!i4!>!236/:!−!3291/6>−3165/7! lh repw!>!r23!>!i3!!−!i2!>!4556!−!23:/4!>!4426/8!

c* analiza dimnih gasova: N eh = sDP3 ⋅ NDP3 + sP3 ⋅ NP3 + sO3 ⋅ NO3 > 1/29 ⋅ 55 + 1/1: ⋅ 43 + 1/84 ⋅ 39 >

Neh!>42/35! d qeh =

lh lnpm

(

)

2 ⋅ sDP3 ⋅ NDP3 ⋅ d qDP3 + sP3 ⋅ NP3 ⋅ d qP3 + sO3 ⋅ NO3 ⋅ d qO3 > N eh

2 lK ⋅ (1/29 ⋅ 55 ⋅ 1/96 + 1/1: ⋅ 43 ⋅ 1/:2 + 1/84 ⋅ 39 ⋅ 2/15 ) >1/:9! 42/35 lhL ⋅ 6611 lh ⋅ 42/35 >58/8! = o eh ⋅ N eh > 4711 t

d qeh = ⋅

n eh

⋅

prvi zakon termodinamike za proces u parnom kotlu: ⋅

⋅

⋅

⋅

nq ⋅ i2 + n eh ⋅ d qeh ⋅ Uh2 = nq ⋅ i 3 + n eh ⋅ d qeh ⋅ Uh3 ⋅

nq =

⋅

n eh ⋅ d qeh ⋅ )Uh2 − Uh3 * i3 − i2

>

⋅

⋅

⋅

⇒

⋅

⋅

⋅

58/8 ⋅ 1/:9 ⋅ )2711 − 311* lh >2:/85! 4556 − 23:/4 t

Prvi zakon termodinamike za proces u turbini: ⋅

⋅

R 23 = ∆ I23 + X 23

R 23 = ∆ I23 + X 23 ⋅

X 23 = − ∆ I23 > − nq ⋅ (i 4 − i 3 ) > −2:/85 ⋅ (3291/6 − 4556 ) >36!NX> X uvs


{fmlp@fvofu/zv


strana 26

6/31/ Parnoturbinsko postrojewe radi po Rankin-ovom kru`nom procesu. Stepen dobrote adijabatske ekspanzije u turbini iznosi ηefy>1/9, a stepen dobrote adijabatske kompresije u pumpi iznosi ηelq>1/:7. Stawe vodene pare na ulazu u turbinu je q>51!cbs i u>431pD, a pritisak u kondenzatoru kf!q>1/13!cbs. Skicirati promene stawa vodene pare na Ut i it dijagramu i odrediti: a) termodinami~ki stepen korisnog dejstva ciklusa (η) b) termodinami~ki stepen korisnog dejstva Karnoovog ciklusa koji radi izme|u istih temperatura toplotnog izvora i toplotnog ponora )ηL* i

U

3

3

2L

2 2

4L

2L 5

t

4L 4

4 t

5

b* q>51!cbs-

ta~ka 2: i3>!4121!

lK lh

t3>!7/557!

ta~ka 3k:

q!>1/13!cbs-

y4l!>!1/84-

i4l!>!2979/:!!!

ta~ka 3:

q!>1/13!cbs-

ηfy e =

i3 − i 4 i3 − i4l

u>431pD

⇒

(pregrejana para)

lK lhL

t>7/557!

lK lhL

lK lh ηefy>1/9

i4!>!i3!−! ηfy e (i3 − i 4l ) =

i4!>!4121!−! 1/9 ⋅ (4121 3 − 2979/: ) = 31:8/2! ta~ka 4:

q>1/13!cbs-

lK i5!>!84/63! lh


(vla`na para)

y>1

lK ! (vla`na para) lh (kqu~ala te~nost)

lK t5!>!1/3:1:! lhL

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike q!>51!cbs-

ta~ka 1k: i2l!>!88/:7!!! ta~ka 1: i 5 − i2l ηlq e = i 5 − i2

strana 27 t>1/3:1:!

lK )te~nost* lhL

lK lh ηelq>1/:7 i −i lK i2!>!i5!.− 5 lq 2l = 89/25! lh ηe

q!>51!cbs-

⇒

84/63 − 89/25 lK = 89/25! (te~nost) 1/:7 lh + rpew r 3:42/: − 3134/7 = ///!> = !1/42 η!>! epw 3:42/: repw lK lK rpew!>!r45!>i5!−!i4!>!−!3134/7! repw!>!r23!>!i3!!−!i2!>!3:42/:! lh lh

i2!>!84/63!.−

c* UUJ − UUQ 6:4 − 3:1/6 = 1/62 = ///!> 6:4 UUJ UUJ!>!U3!>!6:4!L UUQ!>!U4!>!U5!>!)Ulmk*q>1/13!cbs!>!3:1/6!L

ηL!>!

6/32/ Idealni Rankin-Klauzijusov ciklus obavqa se sa vodenom parom izme|u pritisaka qnjo>1/15!cbs i qnby>51!cbs, sa pregrejanom vodenom parom (u>571pD*!na ulazu u turbinu. Za rekuperativno zagrevawe napojne vode (u zagreja~u me{nog tipa), do temperature od uC>215/9pD, iz turbine se pri pritisku pe ⋅

q4>2/3!cbs oduzima deo qbsf!) n 4>291!lh0i*!(slika). Zanemaruju}i radove napojnih pumpi, skicirati proces na Ut dijagramu i odrediti: a) termodinami~ki stepen korisnosti ovog kru`nog procesa b) snagu parne turbine Repw !2

!3

!C

!4

Xuvs

!B !6

!5

Rpew


{fmlp@fvofu/zv


strana 28

U

3

2 B

4

C 6

5 t

b* u>571pD

q>51!cbs-

ta~ka 2:

(pregrejana para)

lK lK - t3!>!7/:76! i3!>!4464! lhL lh ta~ka 3:

q!>2/3!cbs-

y4!>!1/:5-

i4!>!3659/5!!!

ta~ka 4:

q>1/15!cbs-

y5!>!1/92-

i5!>!31:2/9!!!

t>7/:76!

lK lhL

(vla`na para)

lK lhL

(vla`na para)

lK lh t>7/:76! lK lh

q>1/15!cbsy>1 lK lK -t6!>!1/5336! i6!>!232/53! lhL lh ta~ka 5:

ta~ka A = ta~ka 5

q>!2/3!cbs

ta~ka B: iC!>!54:/5!

(kqu~ala te~nost)

(jer se zanemaruje rad pumpe) u>215/9pD

)kqu~ala te~nost)

lK lh

ta~ka 1 = ta~ka B!


(jer se zanemaruje rad pumpe)

{fmlp@fvofu/zv


η=

⋅

R epw + R pew ⋅

>!///>!

R epw ⋅

⋅

strana 29

2218/28 − 761/34 >1/52 2218/28

⋅

R epw = R 23 = n⋅ (i 3 − i2 ) > 1/49 ⋅ (4464 − 54:/5 ) >2218/28!lX ⋅ ⋅  ⋅ ⋅  Rpew = R 56 =  n− n4  ⋅ (i6 − i5 ) > (1/49 − 1/16 ) ⋅ (232/53 − 31:2/9 ) >−761/34lX  

prvi zakon termodinamike za proces u me{nom zagreja~u vode: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅

⋅

I2 = I3

⋅ ⋅ ⋅ ⋅ i − iB  ⋅ ⋅   n− n 4  ⋅ i B + n 4 ⋅ i 4 = n⋅ iC ⇒ n = n4 ⋅ 4 i   C − iB ⋅ ⋅ i − iB 3659/5 − 232/53 lh >1/49! > 1/16 ⋅ n = n4 ⋅ 4 54:/5 − 232/53 t iC − i B

c* ⋅

⋅

⋅

prvi zakon termodinamike za proces u parnoj turbini: R 23 = ∆ I23 + X 23 ⋅ ⋅ ⋅ ⋅ ⋅  ⋅  ⋅  ⋅ ⋅  X 23 = − ∆ I23 > I2 − I3 > n⋅ i 3  − n 4 ⋅ i 4 +  n− n 4  ⋅ i 5  >567/:5!lX> X uvs      

6/33/ Vodena para obavqa Rankin-Klauzijusov ciklus (slika kao u prethodnom zadatku) izme|u pritisaka qnjo>1/2!cbs i qnby>2!cbs. U kotlu se voda zagreva i isparava. Suvozasi}ena vodena para ulazi u turbinu gde ekspandira kvazistati~ki adijabatski. Pri ekspanziji se iz turbine oduzima jedan deo pare na pritisku od q>1/4!cbs i koristi za rekuperativno zagrevawe napojne vode u me{nom zagreja~u od temperature koja vlada u kondenzatoru do temperature od 7:/23pD. Zanemaruju}i radove napojnih pumpi odrediti snagu turbine ako kotao proizvodi 2!lh0t pare i skicirati procese na Ut dijagramu. U

2 B

3 4

C 6

5 t


{fmlp@fvofu/zv


strana 30

q>2!cbsy>2 lK lK - t3!>!8/47! i3!>!3786! lhL lh ta~ka 2:

ta~ka 3:

q!>1/4!cbs-

y4!>!1/:5-

i4!>!3595/:!!!

ta~ka 4:

q>1/2!cbs-

y5!>!1/9:-

i5!>!3431/:!!!

ta~ka 5:

q>1/2!cbs-

(suva para)

t>8/47!

lK lhL

(vla`na para)

lK lhL

(vla`na para)

lK lh t>8/47! lK lh y>1

(kqu~ala te~nost)

lK i6!>!2:2/:! lh ta~ka A = ta~ka 5!


q>!1/4!cbs-

ta~ka B: iC!>!39:/4!

u>7:/23pD

(kqu~ala te~nost)

lK lh

ta~ka 1 = ta~ka B!


prvi zakon termodinamike za proces u me{nom zagreja~u vode: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅

⋅

I2 = I3

⋅ ⋅  ⋅ ⋅   n− n 4  ⋅ i B + n 4 ⋅ i 4 = n⋅ iC   ⋅ 39:/4 − 2:2/3 lh n4 > 2 ⋅ >1/154! 3595/: − 2:2/3 t

⇒

⋅

⋅

n 4 = n⋅

iC − i B i4 − i B

⋅

⋅

⋅

prvi zakon termodinamike za proces u parnoj turbini: R 23 = ∆ I23 + X 23 ⋅ ⋅ ⋅ ⋅ ⋅  ⋅  ⋅  ⋅ ⋅  X 23 = − ∆ I23 > I2 − I3 > n⋅ i 3  − n 4 ⋅ i 4 +  n− n 4  ⋅ i 5  >!453!lX!> X uvs      


{fmlp@fvofu/zv


strana 31

6/34/ Parno turbinsko postrojewe (slika), radi po Rankinovom kru`nom procesu. U parnoj turbini nekvazistati~ki adijabatski {iri se pregrejana vodena para stawa 3)q>36!cbs-!u>571pD* do pritiska q5 >1/15!cbs. Deo pare pri pritisku q4>4!cbs!se oduzima iz turbine radi regenerativnog zagrevawa napojne vode (u zagreja~u vode povr{inskog tipa) od temperature!)uLMK*Q5!do temperature )uLK*Q4. Ako prvi deo turbine (do oduzimawa pare) radi sa stepenom dobrote ηefy>1/:!i masenim protokom ⋅

⋅

n >23/6!lh0t!i ako je korisna snaga turbine X uvs >22!NX, zanemaruju}i rad napojnih pumpi odrediti: a) toplotni protok koji para predaje okolini u kondenzatoru b) stepen dobrote adijabatske ekspanzije u drugom delu turbine (nakon oduzimawa pare) c) termodinami~ki stepen korisnog dejstva ciklusa d) skicirati procese sa vodenom parom na Ut dijagramu Repw !2

!3

!C

⋅

⋅

n

!4

n4

Xuvs ⋅

⋅

n− n 4

!B !6

!5

Rpew U

!3

2

B

4 C

6


!4l 5l

!5

!t

{fmlp@fvofu/zv


strana 32

u>571pD lK t3!>!8/313! lhL

q>36!cbs-

ta~ka 2: lK i3!>!4484! lh

q!>4!cbs-

ta~ka 3k: i4l!>!3927/:!!!

lK lhL

(pregrejana para)

lK lh q!>4!cbs-

ta~ka 3: i4!>!3982/8!!!

t>8/313!

(pregrejana para)

lK lh

t4!>8/433!

q>1/15!cbs-

ta~ka 4k: y5l!>!1/97-

ηfy e =

i3 − i 4 >1/: i3 − i4l

lK lhL

t>8/433!

i5l!>!3324/5!!! q>1/15!cbs-

ta~ka 5:

lK lhL

(vla`na para)

lK lh

y>1

(kqu~ala te~nost)

lK i6!>!232/53! lh ta~ka A = ta~ka 5!


q>!1/15!cbs-

ta~ka B:

y>1

(kqu~ala te~nost)

lK iC!>!672/5! lh ta~ka 1 = ta~ka B!


prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ograni~enom isprekidanom konturom: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅

⋅ ⋅  ⋅ ⋅   n− n 4  ⋅ i B + n 4 ⋅ i 4 = n⋅ iC   ⋅ 672/5 − 232/53 lh >3! n 4 > 23/6 ⋅ 3982/8 − 232/53 t


⋅

I2 = I3 ⇒

⋅

⋅

n 4 = n⋅

iC − i B i4 − i B

{fmlp@fvofu/zv


strana 33 ⋅

⋅

⋅

prvi zakon termodinamike za proces u parnoj turbini: R 23 = ∆ I23 + X 23 ⋅ ⋅ ⋅ ⋅ ⋅   ⋅ ⋅  ⋅  ⋅ X 23 = − ∆ I23 > I2 − I3 > n⋅ i 3  − n 4 ⋅ i 4 +  n− n 4  ⋅ i 5  >! X uvs       ⋅

i5!>!

⋅

⋅

n⋅ i 3 − X uvs − n 4 ⋅ i 4 ⋅

⋅

n− n4 i′!=!i5!!=!i″

>

23/6 ⋅ 4484 − 22 ⋅ 21 4 − 3 ⋅ 3982/8 lK >3531/:! 23/6 − 3 lh

ta~ka 4 je u vla`noj pari

b* ⋅ ⋅  ⋅ ⋅  R lpoe!>! R 56>  n− n4  ⋅ (i 6 − i 5 ) !>! 2(23/6 − 3 ) ⋅ (232/53 − 3531/: ) >−35/2!NX  

c* ηfy e =

i4 − i 5 3982/8 − 3531/: > >1/79 i4 − i5l 3982/8 − 3324/5

c) ⋅

⋅

R epw + R pew

η=

⋅

>!///>!

R epw ⋅

⋅

46/2 − 35/2 >1/42 46/2

⋅

R epw = R 23 = n⋅ (i 3 − i2 ) > 23/6 ⋅ (4484 − 672/5 ) >46/2!NX ⋅ ⋅  ⋅ ⋅  R epw = R 56 =  n− n 4  ⋅ (i 6 − i 5 ) > (23/6 − 3 ) ⋅ (232/53 − 3531/: ) >−!35/2!NX  

6/35/ Sa vodenom parom kao radnim telom, izvr{ava se Rankin−Klauzijus−ov kru`ni proces sa maksimalnom regeneracijom toplote. Regeneracija toplote, koja se odvija sa beskona~no mnogo predajnika toplote povr{inskog tipa, naizmeni~no povezanih sa beskona~no mnog toplotno izolovanih turbina, vr{i se sa ciqem predgrevawa napojne vode pre ulaza u parni kotao. Kru`ni proces se odvija izme|u pritisaka qnjo>1/16!cbs i qnby>61!cbs i najve}om temperaturom u tokou procesa od unby>511pD. Kotao ⋅

proizvodi n >1/2!lh0t!pare, a procesi u turbinama su ravnote`ni (kvazistati~ki). Skicirati proces na Ts dijagramu i zanemaruju}i rad napojne pumpe, odrediti: a) termodinami~ki stepen korisnosti kru`nog procesa b) snagu turbine visokog pritiska, UWQ c) snagu turbine niskog pritiska, UOQ, (sve turbine osim prve) d) relativno pove}awe stepena korisnog dejstva (%) u odnosu na ciklus bez regeneracije toplote(sa samo jednom turbinom)


{fmlp@fvofu/zv


strana 34

U

U 3 Unby 2

4 B

B

4

2

qnby

5

qnjo 6

5 t

t

napomena: {rafiran povr{ina (ispod linije A−1) i povr{ina ispod stepenaste linije 3−4 je jednaka i predstavqa maksimalno mogu}u regenerisanu (rekuperisanu) toplotu u ovom ciklusu

q>1/16!cbsy>1 lK lK -! t6>!1/5872 i6!>!248/94! lhL lh

ta~ka 5:

ta~ka A = ta~ka 5!


q>61!cbsy>1 lK lK t2>3/:32! i2!>!2265/5! lhL lh ta~ka 1:

u>511pD lK t3!>!7/75! lhL

q>61!cbs-

ta~ka 2: lK i3!>!42:4! lh

(kqu~ala te~nost) !u2>374/:2pD (pregrejana para)

u4>!u2>374/:2pD-!!!!!t4>t3!>!7/75!

ta~ka 3: i4!>!3:51!

(kqu~ala te~nost)

lK lh

lK lhL

(pregrejana para)

(ova vrednost se ~ita sa it dijagrama za vodenu paru)


{fmlp@fvofu/zv


strana 35 ∆t45!>!−∆tB2!(uslov ekvidistantnosti)

q>1/16!cbs-

ta~ka 4:

t5!−!t4!>!tB!−!t2!!!!!⇒!!!!!t5!>!tB!−!t2!,!t4!>1/5872!−!3/:32!,7/75!>5/2:6! y5>1/58

!i5>2387/8

lK lhL

lK lh

a) η!>!

repw + rpew 3349/7 − 2249/: = ///!> = !1/55 3349/7 repw

lK lh lK rpew!>!r56!>i6!−!i5!>!248/94!−!2387/8>−2249/:! lh repw!>!r23!>!i3!!−!i2!>!42:4!−!2265/5!>!3149/7!

c* prvi zakon termodinamike za proces u UWQ: ⋅

⋅

⋅

⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⋅

⋅

X 23 = − ∆ I23 > I2 − I3 > n⋅ (i3 − i 4 ) >! 1/2⋅ (42:4 − 3:51) >36/4!lX!> X UWQ d* prvi zakon termodinamike za proces u UOQ: ⋅

⋅

⋅

⋅

⋅

⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X 23 ⋅

⋅

X 23 = R 23 − ∆ I23 > R sfl − ∆ I23 > n⋅ (i B − i2 ) − n⋅ (i 5 − i 4 ) >! X UOQ ⋅

⋅

X UOQ > 1/2 ⋅ (248/94 − 2265/5 ) − 1/2 ⋅ (2387/8 − 3:51) >75/8!lX!> X UOQ e* U 3

B 6

C t


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike η′!>!

strana 36

r epw (+r pew ( 4166/28 − 29:1/18 = ///!> = !1/49 r epw ( 4366/28

lK lh lK rpew′>!rC6!>i6!!−!iC!>!248/94!−!3138/:>−29:1/18! lh repw′!>!rB3!>!i3!!−!iB!>!42:4!−!248/94!>!4166/28!

ta~ka C

qC>!1/16!cbs-!!!!!tC!>t3!>!7/75!

y4!>!1/89-

i4!>!3138/:!

η′!;!211!>!)!η!−!η′!*!;!y

lK lhL

(pregrejana para)

lK lh

⇒

y=

η − η( 1/55 − 1/49 ⋅ 211 >26/9& > η( 1/49

zadatak za ve`bawe (1.25.) 6/36/ Parni kotao proizvodi paru stawa 3)q>31!cbs-!u>471pD*/!Para se po izlasku iz kotla deli: jedan deo ide u turbinu, a drugi deo se prigu{uje. Prigu{ena para se zatim me{a sa onom koja je kvazistati~ki adijabatski ekspandirala u turbini, a dobijena me{avina odvodi u kondenzator u kojoj se kondezuje na 231pD. Dobijeni kondenzat se pumpom vra}a u kotao. Snaga turbine iznosi 2!NX, a toplota predana okolini u kondenzatoru iznosi 6/:!NX. Skicirati procese sa paroma na Ut koordinatnom sistemu i odrediti: a) koliko pare proizvodi kotao, koliko se prigu{uje a koliko ide u turbinu b) termodinami~ki stepen korisnosti ovog postrojewa ,R23 2

3

XQ

XUVS 4 7

5

6

−R67 ⋅

re{ewe:

nlpubp >3/7

lh ⋅ lh ⋅ lh -! n uvscjob >2/:5 - n wfoujm >1/77 -!η>1/25 t t t


{fmlp@fvofu/zv


strana 37

6/37/ Vodena para stawa 2)u>511pD-!q>91!cbs*!kvazistati~ki izentropski ekspandira u turbini visokog pritiska do stawa!4)q>5!cbs*-!posle ~ega joj se izobarski dovodi r45>599!lK0lh toplote. Nakon dovo|ewa toplote para kvazistati~ki izentropski ekspandira u turbini niskog pritiska do stawa )q>1/19!cbs*/ Proces se daqe nastavqa po idealnom Rankinovom ciklusu (slika). Skicirati ciklus na Ut dijagramu i odrediti termodinami~ki stepen korisnosti ovog kru`nog procesa. 2

3

,R23

XUWQ 5 4

XQ ,R45

XUOQ

7

6 −R67

U 3 5 2 4 7

6 t

ta~ka 2:

u>511pD lK t3!>!7/469! lhL

q>91!cbs-

lK i3!>!4246! lh


(pregrejana para)

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike ta~ka 3:

q!>5!cbs-

y4!>!1/9:-

i4!>!3625!!!

ta~ka 4:

q!>5!cbs-

i5!>!4113!!!

lK lh

y6!>!1/:1-

q>1/19!cbs-

lK -! i7!>!284/:! lh

r45>599!

lK lh

(vla`na para)

lK lh

lK lhL

t6!>t5>8/558!

(pregrejana para) lK lhL

(vla`na para)

lK lh

y>1

(kqu~ala te~nost)

lK t7>!1/6:38 lhL q!>91!cbs-

ta~ka 1:

η!>!

lK lhL

i6!>!3446/9!!!

ta~ka 6:

i2!>!293/7!!!

t>7/469!

t5>8/558!

q>1/19!cbs-

ta~ka 5:

strana 38

t2>!t7>!1/6:38!

lK lhL

(te~nost)

lK lh

repw + rpew 4551/5 − 3268/2 = ///!> = !1/48 4551/5 repw

repw!>!r23!,!r45!>!i3!!−!i2!,!r45!>!4246!−!293/7!,!599!>!4551/5! rpew!>!r67!>!i7!−!i6!>!284/:!−!3446/9>−3268/2!


lK lh

lK lh

{fmlp@fvofu/zv


strana 39

6/38/ Parno turbinsko postrojewe radi po Rankin-Klauzijus-ovom kru`nom procesu sa dvostepenim adijabatskim {irewem vodene pare. Pregrejana vodena para stawa 2)q2>211!cbs-!u2>551pD* {iri se u turbini visokog pritiska nekvazistati~ki, sa stepenom dobrote ηEUWQ>1/:, do pritiska!q3>6!cbs. Potom se para izobarski zagreva do temperature u4>411pD, nakon ~ega se, u turbini niskog pritiska, nekvazistati~ki {iri, sa stepenom dobrote ηEUOQ>1/9, do pritiska q5>1/16!cbs, koji vlada u kondenzatoru. a) da li je termodinami~ki stepen korisnosti ovog kru`nog procesa mogu}e dosti}i u RankinKlauzijus-ovom kru`nom procesu sa jednostepenim adijabatskim {irewem vodene pare stawa 2 do pritiska q5, uz maksimalno pove}awe stepena dobrote procesa u turbini b) koliko mimimalno mora da iznosi stepen dobrote jednostepene adijabatske ekspanzije da bi dostigli stepen korisnog dejstva koji ima navedeni ciklus sa dvostepenom adijabatskom ekspanzijom U svim slu~ajevima zanemariti rad napojne pumpe. 2 !i 4

3l

3 5l

5

1 6 !t ta~ka 1:

u>551pD

q>211!cbs-

(pregrejana para)

lK lK - t2!>!7/488! i2!>!4322! lhL lh ta~ka 2k:

q!>6!cbs-

y3l>1/:2-

i3l!>!366:/3!


t3l>t2>7/488!

lK lhL

(vla`na para)

lK lh

{fmlp@fvofu/zv


strana 40 ηeuwq =

q!>6!cbs-

i3!>! i2 − η euwq ⋅ (i2 − i 3l ) >3735/5!!! ta~ka 3:

i2 − i3 >1/: i2 − i3l

lK lh

(vla`na para)

u>411pD

q>6!cbs-

(pregrejana para)

lK lK - t4!>!8/574! i4!>!4173! lhL lh ta~ka 4k:

q>1/16!cbs-

y5l!>!1/99-

i5l!>!3381/3!

ta~ka 4:

q!>1/16!cbs-

t>8/574!

q>1/16!cbs-

(vla`na para)

lK lh ηeuoq =

i5!>! i 4 − η euoq ⋅ (i 4 − i 5l ) >3544/6! ta~ka 5:

lK lhL

i2 − i3 >1/9 i2 − i3l

lK lh

y>1

(vla`na para) (kqu~ala te~nost)

lK i6!>!248/94! lh ta~ka 0 = ta~ka 5: η!>!

(jer se zanemaruje rad pumpi)

repw + rpew 4621/9 − 33:6/67 = ///!> = 1/46 repw 4621/9

repw!>!r12!,!r34>!i2!−!i1!,!i4!−!i3!>!4322−248/94!,!4173!−3735/5>4621/9! rpew!>!r56!>!i6!−!i5!>!248/94!−!3544/6!>−!33:6/67!


lK lh

lK lh

{fmlp@fvofu/zv


strana 41

ciklus sa jednostepenom ekspanzijom sa maksimalnim stepenom dobrote!ηe>2 2 !i

B 1 6 !t ta~ka A:!

q!>1/16!cbs-

yB>1/85-

iB!>!2:41/:!!!

η′!>!

( ( + rpew repw ( repw

= ///!>

t>7/488!

kJ (vla`na para) kgK

lK lh

4184/3 − 28:4/26 = !1/53 4184/3

( repw !>!r12!>!i2!−!i1!>!4322−248/94!!>!4184/3!

lK lh

( rpew !>!rB6!>i6!−!iB!>!248/94!−!2:41/:!>−!28:4/26!

lK lh

kako je η′!?!η- u ciklusu sa jednostepenom adijabatskom ekspanzijom sa maksimalnim pove}awem stepena dobrote ekspanzije ( ηfy e = 1) moè se dosti}i stepen korisnog dejstva navedenog ciklusa sa dvostepenom adijabatskom ekspanzijom.


{fmlp@fvofu/zv


strana 42

c*! ciklus sa jednostepenom adijabatskom ekspanzijom sa stepenom dobrote ekspanzije ηnjo e 2 !i

B

C

1 6 !t η>1/42 (( r epw !>!r12!>!i2!−!i1 (( r pew >!rC6!>i6!−!iC

!η!>!

(( (( r epw + r pew (( r epw

=

i2 − i1 + i6 − iC i2 − i1

⇒

iC> i2 − i 1 + i 6 − η ⋅ (i2 − i 1 )

iC!> 4322 − 248/94 + 248/94 − 1/42 ⋅ (4322 − 248/94 ) >!3369/4!

ηnjo = e

lK lh

i2 − iC 4322 − 3369/4 > >1/85 4322 − 2:41/: i2 − i B


{fmlp@fvofu/zv


strana 43

6/39/ Parno turbinsko postrojewe radi sa dvostepenim {irewem i me|uzagrevawem pare uz jednostepeno regenerativno zagrevawe napojne vode od temperature koja vlada u kondenzatoru do temperature od uC>323pD (slika). Zanemaruju}i radove napojnih pumpi i ako je: − pritisak pare u kondenzatoru 7!lQb − pritisak pare u kotlu 23!NQb − pritisak pare na izlazu iz turbine visokog pritiska q>5!NQb − temperatura pare na ulazu u turbinu visokog pritiska u>641pD − temperatura pare na ulazu u turbinu niskog pritiska u>641pD ⋅

− protok pare kroz turbinu visokog pritiska n =1/5!lh0t ⋅

⋅

− protok pare kroz turbinu niskog pritiska n− n4 =1/4!lh0t − stepen dobrote adijabatske ekspanzije u turbini niskog pritiska ηeuoq>1/93 a) odrediti stepen dobrote adijabatske ekspanzije u turbini visogog pritiska, ηeuwq b) odrditi termodinami~ki stepen korisnog dejstva kru`nog procesa c) skicirati promene stawa vodene pare na hs dijagramu 2

3 ⋅

n R23

XUWQ

C ⋅

n4

4

5

R45

B

7

XUOQ

6

R67


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike a) η euwq =

strana 44

i3 − i4 4537 − 4286/6 = ///> = 1/87 4537 − 41:8/7 i 3 − i 4l

lK i3!>!4537! lh

u>!641pD lK t3!>!7/6996! lhL

ta~ka 3k:

q!>51!cbs-

q>231!cbs-

ta~ka 2:

i4l!>!41:8/7!!!

t>7/6996!

(pregrejana para)

lK (pregrejana para) lhL

lK lh q!>51!cbs

ta~ka 3:

prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ograni~enom isprekidanom konturom: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅

⋅ ⋅  ⋅ ⋅   n− n 4  ⋅ i B + n 4 ⋅ i 4 = n⋅ iC  

ta~ka 6:

q>!1/17!cbs-

⋅

I2 = I3 ⇒

i4 =

⋅  ⋅ ⋅  n⋅ iC −  n− n 4  ⋅ i B   ⋅

>///

n4 y>1

(kqu~ala te~nost)

lK i7!>!262/6! lh ta~ka A = ta~ka 6 ta~ka B:

q!>!51!cbs-

(jer se zanemaruje rad pumpi) u>!323pD

(voda)

lK iC!>!:18/6! lh i4 =

1/5 ⋅ :18/6 − 1/4 ⋅ 262/6 lK >!4286/6! 1/2 lh


{fmlp@fvofu/zv


strana 45

b) ⋅

η=

⋅

R epw + R pew ⋅

>!///>!

R epw

2219/46 − 797/8 >1/49 2219/46

⋅ ⋅ ⋅  ⋅ ⋅  R epw = R 23 + R 45 = n⋅ (i 3 − i2 ) +  n− n 4  ⋅ (i 5 − i 4 ) >   ⋅

⋅

R epw > 1/5 ⋅ (4537 − :18/6 ) + 1/4 ⋅ (4625 − 4288/6 ) >2219/46!lX ⋅ ⋅  ⋅ ⋅  R pew = R 67 =  n− n 4  ⋅ (i 7 − i 6 ) > 1/4 ⋅ (262/6 − 3551/5 ) >−!797/8!lX  

u>641pD

ta~ka 4:

q>51!cbs-

lK i5!>!4625! lh

lK t5!>!8/2856! lhL

ta~ka 5k:

q!>1/17!cbs-

y6l!>!1/96-

i6l!>!3315/8!!!

ta~ka 5:

q>1/17!cbs-

t>8/2856!

(pregrejana para)

lK lhL

(vla`na para)

lK lh ηeuoq =

i 5 − i6 i5 − i6l

i6!>!i5!−! η euoq ⋅ (i 5 − i 6l ) = 4625!−! 1/93 ⋅ (4625 − 3315/8 ) = 3551/5! ta~ka 1 = ta~ka B

lK lh

(jer se zanemaruje rad pumpi) 5

3 i 4 4l

6 6l 2 C B 7


t

{fmlp@fvofu/zv


strana 46

⋅

6/3:/ Parni kotao proizvodi n =31!u0i pare stawa 2)q3>27!cbs-!u3>511pD*/ Para u turbini ekspandira ⋅

u dva stepena. Nakon prvog stepena, deo pare ( n4 )se odvodi za potrebe nekog spoljnog predajnika toplote u kojem se vr{i potpuna kondenzacija pare na u>291pD pri ~emu se od pare odvodi 3!NX toplote. Tako nastalo kondenzat se ne vra}a u kotao, nego ispu{ta u okolinu, a umesto wega se u kotao dodaje ista koli~ina vode iz okoline stawa )q>2!cbs-!u>26pD*. Ostatak pare ekspandira u drugom stepenu turbine a zatim odvodi u kondenzator, u kome vlada temperatura od 41pD. Ekspanzije u turbinama su ravnote`ne (kvazistati~ke) i adijabatske. Zanemaruju}i snage napojnih pumpi, skicirati proces na Ts dijagramu i odrediti: ⋅

a) maseni protok sveè vode, n4 b) ukupnu snagu koja se dobije u turbinama c) termodinami~ki stepen korisnog dejstva ciklusa

3 5

4 ,R23

−R47

−R56

2 7

1

6 U

3

4 2 6

7 5 t


{fmlp@fvofu/zv


strana 47 u>!511pD

ta~ka 2:

q>27!cbs-

lK i3!>!4364! lh

lK t3!>!8/344! lhL

ta~ka 3:

q!> (q)u =291p D >21!cbs- t4>!t3!>!8/344!

i4!>!4228/4!!!

lK lhL

(pregrej. para)

lK lh

ta~ka 4:

u>!41pD

y5!>1/96

i5!>!32:2/5!

ta~ka 5:

u>!41pD-

lK i6!>!236/82! lh ta~ka 1: i2!>!239/2!!!

(pregrejana para)

t5!>!t4>!t3!>!8/344!

lK (vla`na para) lhL

lK lh y>1

(kqu~ala te~nost)

lK t6!>1/5477! lhL

q!>27!cbs-

t2>!t6!>!1/5477!

u>!291pD-

y>1

lK lhL

(te~nost)

lK lh

ta~ka 6:

(kqu~ala te~nost)

lK i7!>!874/2! lh b* prvi zakon termodinamike za proces u spoqnom predajniku toplote: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅

⋅

R qsfebkojlb = n4 ⋅ (i 7 − i 4 )

⋅

lh − 3 ⋅ 21 4 R qsfebkojlb >1/96 > n4 = 874/2 − 4228/4 t i7 − i4 ⋅

b) prvi zakon termodinamike za proces u turbini visokog pritiska: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅

⋅

X uwq = n⋅ (i 3 − i 4 ) >

31 ⋅ 21 4 ⋅ (4364 − 4228/4 ) 4711

⋅

X uwq >1/86!NX


{fmlp@fvofu/zv


strana 48

prvi zakon termodinamike za proces u turbini niskog pritiska: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⇒

⋅  31 ⋅ 21 4   ⋅ ⋅  X uoq =  n− n4  ⋅ (i 4 − i 5 ) >  − 1/96  ⋅ (4228/4 − 32:2/5 ) >  4711     

⋅

X uwq >5/47!NX c) prvi zakon termodinamike za proces u parnom kotlu: ⋅

⋅

⋅

⇒

R 23 = ∆ I23 + X 23

⋅

⋅

R lpumb = n⋅ (i3 − i2 ) >

31 ⋅ 21 4 ⋅ (4364 − 239/2) 4711

⋅

R lpumb >28/47!NX ⋅

η=

⋅

X uwq + X uoq ⋅

>

R lpumb

1/86 + 5/47 >1/3: 28/47

zadatak za ve`bawe (1.30.) 6/41/ Parno turbinsko postrojewe radi po Rankin−Klauzijus−ovom kru`nom procesu sa dvostepenim adijabatskim {irewem vodene pare(slika kao u zadatku 1.26) . Pregrejana vodena para stawa 3)q>21 NQb-!u>551pD* {iri se u turbini visokog pritiska nekvazistati~ki, sa stepenom dobrote η euwq >1/:, do pritiska od q4>1/6!NQb. Potom se izobarski zagreva do temperature od u5>411pD, nakon ~ega se, u turbini niskog pritiska, {iri nekvazistati~ki, sa stepenom dobrote η euoq >1/9 do pritiska od q6>1/116!NQb, koji vlada u kondenzatoru. Skicirati proces na it dijagramu i zanemaruju}i snage napojnih pumpi odredit stepen korisnosti posmatranog kru`nog procesa. re{ewe:

η>1/476


{fmlp@fvofu/zv


strana 1

6. LEVOKRETNI KRU@NI PROCESI 7/2/ Odrediti minimalan rad koji treba uloìti da se od nekog tela, konstantne temperature u>!−24pD, oduzme 21!lK!toplote i preda okolnom vazduhu, konstantne temperature od 48pD. Koliko se toplote u tom slu~aju predaje okolnom vazduhu. Najmawe rada se mora uloìti ako levokretna tolotna ma{ina radi po Karnoovom levokretnom ciklusu (sve promene stawa radne materije su povratne). U 4

3

5

2

UUQ

UUJ

!t UUJ>−24pD!>!371!LR epw Xofup

=

UUQ>48pD>421!L-

Repw!>21!lK

Uuq − Uuj Uuj 421 − 371 !> 21 ⋅ !!!!!! ⇒ !!!!!! Xofup = R epw ⋅ >2/:3!lK Uuq − Uuj 371 Uuj

R pew >!Repw!,! Xofup !>!21!,!2/:3!>!22/:3!lK

7/3/ Rashladni ure|aj (slika) koristi kao radnu materiju vazduh (idealan gas) i radi po levokretnom kru`nom Xulovom procesu. Stawe vazduha na ulazu u izentropski kompresor je 2)u2>−21pD-!q2>2!cbs*- a na izlazu iz kompresora 3)q3>5!cbs*/!Temperatura vazduha na ulazu u izentropsku turbinu je u4>31pD. Maseni protok vazduha kroz rashladni ure|aj 2311!lh0i a sve promene stawa radne materije su ravnote`ne (kvazistati~ke). Skicirati promene stawa vazduha na Ut dijagramu i odrediti: a) rashladni efekat instalacije )lX* b) koeficijent hla|ewa instalacije, εi c) ako je svrha rashladnog ure|aja proizvodwa leda temperature um>−4pD od vode temperature ux>21pD, odrediti masu proizvedenog leda za vreme od τ>2!i


{fmlp@fvofu/zv


strana 2

U

4

3

3

3 2 5

5

2

!!!ledomat t a)  U3!>!U2 ⋅  q 3 q  2

  

 U5!>!U4 ⋅  q 5 q  4

   

⋅

⋅

κ −2 κ

κ −2 κ

> 374 ⋅  5   2 > 3:4 ⋅  2  5

⋅

2/5 −2 2/5

2/5 −2 2/5

R epw = R 52 = n⋅ d q ⋅ (U2 − U5 ) >

>!4:1/9!L

>2:8/3!L

2311 ⋅ 2 ⋅ (374 − 2:8/3 ) >32/:4!lX 4711

b) ⋅

εi!>!

R epw ⋅

X ofup ⋅

= ///!>

32/:4 >!3/16 21/78

⋅

⋅

⋅

X ofup > X lpnqsftps!, X uvscjob!> n⋅ d q ⋅ (U2 − U3 + U4 − U5 ) ⋅

X ofup >

2311 ⋅ 2 ⋅ (374 − 4:1/9 + 3:4 − 2:8/3 ) !>!−21/78!lX 4711

c) ⋅

32/:4 ⋅ 4711 R epw ⋅ τ nmfe = = /// = >318/6!lh 53 + 449/5 i x − im lK )q>!2!cbs-!u>21pD* ix!>!53! lh im = d m ⋅ (Um − 384 ) − sm > 3 ⋅ (−4 ) − 443/5 >−449/5!


lK lh

{fmlp@fvofu/zv


strana 3

7/4/ Helijum (idalan gas) obavqa realan levokretni Xulov proces sa potpunim rekuperativnim (regenegrativnim) zagrevawem radne materije. Rashladna snaga ovog postrojewa je 33!lX. Temperatura u rashladnoj komori je stala i jednaka je temperaturi na ulazu u gasnu turbinu UUJ>U4!>356!L>dpotu. Temperatura okoline je stalna i jednaka temperaturi na ulazu u kompresor UUQ>U2>431!L. Odnos q pritiska na ulazu i izlazu iz gasne turbine iznosi 4 >3/2. Stepeni dobrote u adijabatskom q5 fy kompresoru i adijabatskoj turbini su jednaki i iznose ηlq e = η e >1/93. Prikazati ovaj proces u Ut

koordinatnom sistemu i odrediti: a) neto snagu potrebnu za pogon ovog postrojewa b) faktor hla|ewa ovog postrojewa

4

!C

5

R E K U P E R A T O R

!B

3

Rpew

2

Repw

3 U

3l

B 2 !Rsfl

C

4

5l

!UQ !UJ

5 t


{fmlp@fvofu/zv


strana 4

a) κ −2

2/78 −2   κ >! 431 ⋅ (3/2) 2/78 >541/:!L U3L!>!U2 ⋅  q 3L   q   2  U −U 541/:6 − 431 U3!>!U2!,! 3L lq 2 !>! 431 + >566/4!L 1/93 ηe κ −2

2/78 −2

  κ > 356 ⋅  2  2/78 >293/2!L U5L!>!U4 ⋅  q 5L   q   3/2   4  U U5!>!U4!, η e ⋅ (U5l − U4 ) !>! 356 + 1/93 ⋅ (293/2 − 356 ) >2:4/5!L ⋅

⋅

⋅

⋅

R epw = R 5C = n⋅ d q ⋅ (UC − U5 ) ⇒ ⋅

n=

⋅

R epw n= > d q ⋅ (UC − U5 )

33 lh >9/3!/21.3! 6/3 ⋅ (356 − 2:4/5 ) t ⋅

⋅

⋅

⋅

X ofup > X lpnqsftps!, X uvscjob!> n⋅ d q ⋅ (U2 − U3 + U4 − U5 ) ⋅

X ofup > 9/3 ⋅ 21 −3 ⋅ 6/3 ⋅ (431 − 566/4 + 356 − 2:4/5 ) !>!−46/8!lX UC>U3, uslov potpune (maksimalne) regeneracije (rekuperacije) toplote za Xulov ciklus

napomena:

b) ⋅

εi!>!

R epw ⋅

>

⋅

R pew − R epw ⋅

⋅

33 >!1/73 68/8 − 33

⋅

R pew = R 3 B = n⋅ d q ⋅ (UB − U3 ) > 9/3 ⋅ 21 −3 ⋅ 6/3 ⋅ (431 − 566/4 ) >−68/8!lX


{fmlp@fvofu/zv


strana 5

7/5. Rashladno postrojewe (slika) koristi kao radni fluid freon 12 )S23*/ Temperatura isparavawa je 354!L, a teperatura kondenzacije 426!L. Snaga kompresora u kojem se vr{i kvazistati~ko adijabatsko sabijawe freona iznosi 1/94!lX/ Skicirati promene stawa S23 u Ut i it koordiantnom sistemu i odrediti: ⋅

a) rashladni kapacitet ) R epw* i koeficijent hla|ewa ovog postrojewa )εi) b) ako bi se kqu~ala te~nost S23 pre prigu{ivawa podhladila za ∆U>21!L koliko bi tada iznosio ⋅ -

⋅-

rashladni kapacitet ) R epw * i koeficijent hla|ewa ovog postrojewa ) ε i ) koeficijent hla|ewa c) na Ut dijagramu {rafirati povr{inu koja predstavqa pove}awe rashladnog kapaciteta postrojewa usled pothla|ivawem kondenzata pre prigu{ivawa 4

3 ⋅

⋅

R pew

n

⋅

X lq ⋅

5

2

n ⋅

n⋅ y 5

⋅

n ⋅ (2 − y 5

)

⋅

R epw

3

i U 3

4 2 4 5

2 t


5

t

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike !

strana 6

U>354!L>!−!41pD

ta~ka 1:

lK i2!>!752/92! lh

y>2 lK t2!>!2/6993! lhL

q!>!)q*Ul>426L!>!21/28!cbs

ta~ka 2: i3!>!794/1:!

t!>!t2!>!2/6993!

lK lhL

lK lh U>426!L>53pD

ta~ka 3:

y>1

lK i4!>!652/58! lh i5!>!i4!>!652/58!

ta~ka 4:

lK lh

a) ⋅

⋅

⋅

⋅

⋅

R epw = R 52 = n⋅ (i2 − i 5 ) >///> 3 ⋅ 21 −3 ⋅ (752/92 − 652/58 ) >3!lX X lq = X u23 ⋅

εi!>!

R epw ⋅

X lq

=

⋅

−1/94 lh X lq > 3 ⋅ 21 −3 = n⋅ (i2 − i 3 ) !!!!!!⇒!!!!!! n = > 725/92 − 794/1: t i2 − i3 ⋅

⋅

3 >3/52 1/94

b) 3

i U 3

4 2

4

B C

5

B

2 t


C

5 t

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strana 7

ta~ka A:

iB! ≅ !)i′*Ub>416!L!>!642/22!

ta~ka B:

iC!>!iB!>!642/22!

⋅ -

⋅

lK lh

lK lh

⋅

R epw = R C2 = n⋅ (i2 − iC ) > 3 ⋅ 21 −3 ⋅ (752/92 − 642/22) >3/32!lX ⋅ -

εi!>!

R epw ⋅

X lq

=

3/32 >3/78 1/94

U 3

4 B 2 5

C

t ⋅

∆ R epw ⋅

6.5. Levokretni kru`ni proces obavqa se sa n =711!lh0i amonijaka )OI4* , izme|u Unjo>−24pD i qnby>2!NQb. U toplotno izolovan kompresor ulazi suva para koja se nekvazistati~ki sabija do stawa 3)U3>211pD*/ Po izlasku iz kondenzatora vr{i se pothla|ivawe do temperature od 26pD. Skicirati proces na Ut dijagramu i odrediti: a) koeficijent hla|ewa b) za koliko bi se pove}ala vrednost koeficijenta hla|ewa )&*, ako bi sabijawe u kompresoru bilo kvazistati~ko c) u{tedi u snazi za pogon kompresora u slu~aju kvazistai~kog sabijawa u odnosu na stvarno nekvazistati~ko sabijawe )lX* i u Ut koordinatnom sistemu {rafirati povr{inu ekvivalentnu toj u{tedi


{fmlp@fvofu/zv


strana 8

U

3 3l

4 2 5 t a) u!>!−24pD

ta~ka 1: lK i2!>!3318! lh

t2!>!21/5:3! q!>!21!cbs

ta~ka 2: i3!>!3552/6!

q>21!cbs lK i4! ≅ !)i′*u>26pD!>!2141/2! lh

u!>211pD

u>26pD

i5!>!i4!>!2141/2!

ta~ka 4: ⋅ -

R epw ⋅

= ///!>

X lq ⋅

⋅

⋅

⋅

lK lhL

lK lh

ta~ka 3:

εi!>!

y>2

lK lh

2:7/26 >6 4:/19

⋅

711 ⋅ (3318 − 2141/2) >2:7/26!lX 4711 ⋅ 711 = n⋅ (i2 − i 3 ) > ⋅ (3318 − 3552/6 ) >−4:/19!lX 4711

R epw = R 52 = n⋅ (i2 − i 5 ) > X lq = X u23

U 3

4

5


2

t

{fmlp@fvofu/zv


strana 9

b)

i3l>3513/:!

⋅ -

R epw

= ///!>

⋅ X lq

⋅

lK lhL

lK lh

⋅

εi!>!

t!>!t2>21/5:3!

q>21!cbs

ta~ka 2k:

2:7/26 >7 43/76

⋅

X lq = X u23l = n⋅ (i2 − i 3l ) >

711 ⋅ (3318 − 3513/: ) >−43/76!lX 4711

 ε 7  ∆ε i (&) =  i − 2 ⋅ 211& !>  − 2 ⋅ 211& >31&  εi  6    d* ⋅

⋅ -

⋅

∆ X lq !>! X lq !.! X lq !>!7/54!lX

U

3 3l

∆Xlq

4

5

2 t


{fmlp@fvofu/zv


strana 10

7/7/ Amonija~ni kompresorski rashladni ure|aj sa prigu{nim ventilom radi izme|u pritisaka qnjo>4/92:!cbs!i!qnby>26!cbs. Kompresor usisava suvozasi}enu paru amonijaka i kvazistati~ki adijabatski je sabija. Ure|aj je projektovan tako da iz prostorije koju hladi oduzima 61!lX toplote. Odrediti: a) snagu kompresora c* koliko bi trebalo da iznosi stepen suvo}e vla`ne pare koja napu{ta kondenzator da bi koeficijent hla|ewa iznosio εh=0 a) q>4/92:!cbs

ta~ka 1: lK i2!>!332:! lh ta~ka 2: i3!>!3528/6!

y>2

t2!>!21/465!

lK lhL

q!>26!cbs

t!>!t2!>!21/465!

q>26!cbs

y>1

lK lhL

lK lh

ta~ka 3: lK i4!>!2255/2! lh

i5!>!i4!>!2255/2!

ta~ka 4: ⋅

⋅

⋅

⋅

R epw = n⋅ (i2 − i 5 )

⇒

lK lh ⋅

⋅

61 lh R epw n= > > 5/76 ⋅ 21 −3 t i2 − i 5 332: − 2255/2

⋅

X lq = X u23 = n⋅ (i2 − i 3 ) > 5/67 ⋅ 21 −3 ⋅ (332: − 3528/6 ) >−:/16!lX b) i 4( = i2

⇒ i

332: − 2255/2  i − i(  y 4 =  2 >1/:8 >   i( (−i(  q=26cbs 3363 − 2255/2 3

4′ 2 4

5

t


{fmlp@fvofu/zv


strana 11

7/8/!Levokretni kru`ni proces sa pothla|ivawem, prigu{ivawem, isparavawem i pregrevawem pare pre ulaska u kompresor obavqa se sa freonom 12 )S23*- kao radnim telom (slika). Rashladni kapacitet postrojewa je 6/9!lX, a snaga kompresora, koji vr{i nekvazistati~ko sabijawe pare freona, je 3!lX. Radna materija obavqa ciklus izme|u pritisaka qnjo>1/2!NQb i qnby>1/7!NQb i pri tom dostiè maksimalnu temperaturu od 71pD. Temperatura pothla|ivawa je 27pD. Skicirati promene stawa radnog tela na qw i Ut dijagramu i odrediti: a) stepen dobrote adijabatske kompresije b) {rafirati na Ut dijagramu potrebnu snagu kompresora

LE!,!QI 4

3

5

JT!,!QH

2

3

U

3l

4

5

2

t


{fmlp@fvofu/zv


strana 12

q!>!7!cbs

u!>!71pD

q>7!cbs

u>27pD

lK i3!>!7:1/6! lh ta~ka 3:

lK i4! ≅ !)i′*u>27pD!>!626/3:! lh ta~ka 4:

i5!>!i4!>!626/3:!

ta~ka 1:

q2!>!2!cbs

⋅

⋅

R epw >6/9!lX ⋅

⋅

⋅

⋅

lK lh

X lq >−3!lX ⋅

R epw = R 52 = n⋅ (i2 − i 5 ) ⋅

X lq = X u23 = n⋅ (i2 − i 3 )

)2* )3*

Kombinovawem jedna~ina (1) i (2) dobija se:!i2!>!756/7! t2!>!2/716!

lK -! lh

n>55/6!

h t

lK lhL q!>!7!cbs

ta~ka 2k: i3l!>!789/8!

t!>!t2!>!2/716!

lK lhL

lK lh

b* ηlq e =

i2 − i3l 756/7 − 789/8 !>! >1/85 i2 − i3 756/7 − 7:1/6


{fmlp@fvofu/zv


strana 13

b) 3 U

3 U

3l

3l

4

4 2

2

5

5 t

t

⋅

⋅

R pew

R epw ⋅

⋅

⋅

X lq = R pew − R epw 3

U

3l

4 2

2

5

t ⋅

X lq


{fmlp@fvofu/zv


strana 14

7/9/!Levokretna instalacija radi sa amonijakom kao radnim telom. Rashladni kapacitet postrojewa je 51!lX. Temperatura isparavawa je −44pD, pritisak u kondenzatoru 6!cbs-!a temperatura prehla|ivawa −4pD. Toplota oslobo|ena prehla|ivawem kondenzata koristi se za pregrevawe suve pare amonijaka, tako da u kompresor ulazi pregrejana para koja se sabija nekvazistati~ki adijabatski sa stepenom dobrote ηLQ e >1/9. Predstaviti kru`ni proces na Ut dijagramu i odrediti: a) maseni protok amonijaka kroz instalaciju )lh0t* b) snagu kompresora )lX* 5

4 3

6

7

2

U 3 3L

4 5

6

2 7 t


{fmlp@fvofu/zv

zbirka zadataka iz termodinamike ta~ka 3: i4>:88/:6!

strana 15

q>6!cbs-!

y>1

q>!6!cbs-!

U>381!L

lK lh

ta~ka 4: lK i5>:56/8!! lh ta~ka 5:

i6!>!i5>:56/8!!

ta~ka 6:

U>351!L-

lK lh

y>2

lK i7!>!3288! lh

i4!−!i5!>!i2!−!i7!

i2>!i4!−!i5!,!i7

i2>!:88/:6!−!:56/8!,!3288!>!331:/36

q3L>6!cbs-!

ta~ka 2K: i3l>3537!

lK lh

t3>!t2>21/:!

t2>21/:!

lK lhL

lK lhL

lK lh ηLQ e >1/9

q3>!6!cbs-!

ta~ka 2: ηLQ e >

−∆i45!>!∆i72

q2!>!q7!>2/7647!cbs

ta~ka 1:

i2 − i 3L i −i 331:/36 − 3537 lK i 3 = i2 − 2 LQ 3L > 331:/36 − >3591/2:! i2 − i 3 1 / 9 lh ηe

a) ⋅

⋅

R epw = n⋅ (i 7 − i 6 )

⇒

⋅

⋅

lh R epw 51 > 4/36 ⋅ 21 −3 n= = t i7 − i6 3288 − :56/8

b) ⋅

⋅

M lpnq = n⋅ (i2 − i3 ) = 4/36 ⋅ 21 −3 ⋅ (331:/36 − 3591/2: ) >!−9/9!lX


{fmlp@fvofu/zv


strana 16

7/:/!Rashladno postrojewe (slika) radi, sa freonom 12 kao radnim telom, izme|u pritisaka q2>366!lQb i q3>2!NQb. U kompresoru snage 2/4!lX nekvazistati~ki adijabatski sabija se pare freona pri ~emu specifi~na entropija freona (usled mehani~ke neravnoteè) poraste za ∆t23>4/3!K0lhL. Odrediti: a) rashladnu snagu postrojewa (lX) b) koeficijent hla|ewa postrojewa c) {rafirati na Ut dijagramu povr{inu koja koja je evivalentna rashladnoj snazi postrojewa !kondenzator 4

3

2

5

ispariva~ 6

7

b* q>3/66!cbs y>2 lK lK t2!>!2/6829! i2!>!764/12! lhL lh ta~ka 1:

ta~ka 2: i3!>!789/6!

q!>!21!cbs

t>t2!,!∆t23!>!2/686!

q>21!cbs

y>1

lK lhL

lK lh

ta~ka 3: lK i4!>!651/84! lh


{fmlp@fvofu/zv


strana 17

Prvi zakon termodinamike za proces u odvaja~u te~nosti: ⋅

⋅

⋅

/

⋅

⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X 23 ⋅

⋅

n⋅ i 4 + n⋅ i 7 = n⋅ i 5 + n⋅ i2

X lq = X u23

⋅

⋅

⇒

I2 = I3

⇒

i7!−!i5!>!i2!−!i4 ⋅

X lq lh −2/4 = n⋅ (i2 − i 3 ) !!!!!!⇒!!!!!! n = > > 6/2 ⋅ 21 −3 764/12 − 789/6 t i2 − i3 ⋅

⋅

⋅

⋅

⋅

R epw = R 67 = n⋅ (i 7 − i 6 ) = n⋅ (i 7 − i 5 ) = n⋅ (i2 − i 4 )

⇒

⋅

R epw > 6/2 ⋅ 21 −3 ⋅ (764/12 − 651/84 ) >6/84!lX b) ⋅

εi!>!

R epw ⋅

X lq

=!

6/84 >!5/5 2/4

c) 3 3l

U 4 5 ⋅

7

R sfl

!!2

6

t ⋅

R epw


{fmlp@fvofu/zv


strana 18 ⋅

7/21/!Levokretni kru`ni proces sa amonijakom kao radnim telom ) n >1/12!lh0t* odvija se izme|u qnjo>2 cbs i qnby>36!cbs. Kompresija je ravnote`na izentropska i dvostepena. Stepen povi{ewa pritiska u oba q q stepena je jednak ( 3 = 5 ) . Na ulazu u kompresor niskog pritiska (stawe 1) para amonijaka je q2 q 4 suvozasi}ena. Nakon prvog stepena kompresije para amonijaka se hladi do temperature od T>431!L/ Skicirati proces na Ut dijagramu i odrediti: a) rashladni efekat instalacije koja radi po ovom ciklusu (lX) b) snagu kompresora niskog pritiska i snagu kompresora visokog pritiska )lX* c) koeficijent hla|ewa )εi* d) procentualno pove}awe koeficijenta hla|ewa koje je ostvareno dvostepenom kompresijom (u odnosu na jednostepenu kompresiju izme|u istih pritisaka i istog stawa na ulazu u kompresor) e) na Ut dijagramu {rafirati povr{inu koja predstavqa u{tedu u snazi kompresora usled dvostepene kompresijeϕ 6

5 ⋅

X lwq

⋅

R 56

⋅

n

3 4 ⋅

n ⋅

⋅

⋅

X loq

R 34 2

n⋅ y 7 7

⋅

n⋅ (2 − y 7 )

⋅

Repw


{fmlp@fvofu/zv


strana 19

U 5

6 3 4 2

7

t q>2!cbs

ta~ka 1: lK i2!>!3272! lh

t2!>!22/14!

t>t2>22/14!

lK lhL

lK lh q!>6!cbs

ta~ka 3:

U>431!L lK t4!>!21/74! lhL

lK i4!>!3444! lh ta~ka 4: i5!>!3739!

lK lhL

q!>! q2 ⋅ q 5 >6!cbs

ta~ka 2: i3!>!3584!

y>2

lK lhL

q!>36!cbs

t!>21/74!

q>36!cbs

y>1

q>2!cbs

i>!i6!>!2353!

lK lh

ta~ka 5: lK i6!>!2353! lh ta~ka 6:


lK lh

{fmlp@fvofu/zv


strana 20

b* ⋅

⋅

⋅

R epw = R 72 = n⋅ (i2 − i 7 ) = 1/12 ⋅ (3272 − 2353) >:/2:!lX c* ⋅

⋅

⋅

⋅

⋅

⋅

X loq = X u23 = n⋅ (i2 − i3 ) > 1/12 ⋅ (3272 − 3584) >−4/23!lX X lwq = X u45 = n⋅ (i 4 − i 5 ) > 1/12 ⋅ (3444 − 3739 ) >−3/:6!lX d* ⋅

R epw

εi!>!

⋅

=!

⋅

X loq + X lwq

:/2: >!2/6 4/23 + 3/:6

e* U B

6

2

7

t q!>36!cbs

ta~ka A: iB!>!3915/3! ⋅

t!>22/14!

lK lhL

lK lh

⋅

⋅

X lq = X u2B = n⋅ (i2 − i B ) > 1/12 ⋅ (3272 − 3915 ) >−7/54!lX ⋅

ε i-

!>!

R epw ⋅

X lq

=!

:/2: >!2/54 7/54

ε   2/6  ∆ε i (&) =  i- − 2 ⋅ 211& !>  − 2 ⋅ 211& >8/25& ε   2/54   i 


{fmlp@fvofu/zv


strana 21

f* U

B 5 ⋅

∆X

6 3 4 2

7

t zadaci za ve`bawe:

(2.11. − 2.12.)

7/22/!U komori za hla|ewe potrebno je odràvati stalnu temperaturu od −26pD, pri ~emu temperatura spoqa{weg (okolnog) vazduha iznosi 41/4pD. Toplotni dobici kroz zidove komore iznose 91!lK0t. Za hla|ewe komore primeweno je kompresiono rashladno postrojewe bez pothla|ivawa kondenzata i sa wegovim prigu{ivawem. Pri tome kompresor usisava suvu paru freona 22 )S33*!i sabija je adijabatski. Odrediti minimalnu snagu za pogon rashladnog postrojewa kao i faktor hla|ewa. Skicirati promene stawa freona na Ut i it dijagramu. ⋅

re{ewe:

X l >28/5!lX-

εi>5/7

7/23/ U postrojewe koje radi po levokretnom kru`nom procesu, kondenzuje se i pothla|uje amonijak pri pritisku od q>2!NQb. Te~ni rashladni fluid ulazi u prigu{ni ventil pri temperaturi od 28pD, gde se prigu{uje do temperature isparavawa u>−34pD. Kompresor, u kojem se obavqa adijabatsko sabijawe radi lq

sa stepenom dobrote η e >1/9, usisava suvu paru, koja od stawa vla`ne pare prelazi u stawe suve pare na ra~un pothla|ivawa te~ne faze. Snaga kompresora je 67!lX. Odrediti rashladnu snagu ovog postrojewa i u Ut koordinatnom sistemu {rafirati povr{inu ekvivalentnu snazi za pogon kompresora. ⋅

re{ewe:

R epw!>!29:/7!lX


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strana 22

7/24/ Kaskadna rashladna instalacija (slika), sastoji se iz me|usobno spregnutog “kola visoke temperature” i “kola niske temperature”. “Kola” su spregnuta preko toplotno izolovanog predajnika toplote, u kome rashladni fluid kola niske temperature (preko kondenzatora kola niske temperature) u potpunosti predaje toplotu rashladnom fluidu kola visoke temperature (preko ispariva~a kola visoke temperature). Kolo visoke temperature radi sa freonom 11 )S22*, izme|u pritisaka qnjo>q{)−45pD* i ⋅

qnby>1/3!NQb i masenim protokom n 2>1/45!lh0t. Kolo niske temperature radi sa freonom 22 )S33*izme|u pritisak qnjo>q{)−:1pD* i qnby>1/3!NQb. Izra~unati stepen (koeficijent) hla|ewa ovog postrojejwa, ako oba kola rade bez pothla|ivawa kondenzata i sa kvazistati~kom adijabatskom kompresijom suvozasi}ene pare.

3

4 kolo visoke temperature

⋅

X lwu

S22

5 4′

2 3′ kolo niske temperature

⋅

X lou

S33 5′

2′ ⋅

lh n S22!>!1/45! kolo visoke temperature: t y>2 ta~ka 1: u!>!−45pD lK lK t2!>!2/8415! i2!>!783/86! lhL lh ta~ka 2: i3!>!841!

q!>!3!cbs

t!>!t2!>2/8415!

q!>!3!cbs

y>1

lK lhL

lK lh

ta~ka 3: lK i4!>!651! lh ta~ka 4:

i5!>!i4!>!651!


lK lh

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strana 23

kolo niske temperature: u!>!.!:1pD

ta~ka 1′: lK i2′!>!774/:7! lh

t2′!>!2/::74!

q>3!cbs

ta~ka 2′: i3′!>!863!

y>2 lK lhL

t!>!t2′!>!2/::74!

lK lhL

lK lh q!>!3!cbs

ta~ka 3′:

y>1

lK i4′!>!582/3! lh i5′!>!i4′!>!582/3!

ta~ka 4′:

lK lh

prvi zakon termodinamike za toplotno izolovani predajnik toplote ⋅

⋅

⋅

R 23 = ∆ I23 + X 23 ⋅

⋅

⇒ ⋅

/

⋅

nS33 ⋅ i 3( + nS22 ⋅ i 5 = nS33 ⋅ i 4( + nS22 ⋅ i2 ⋅

nS33 = 1/45 ⋅

⋅

⋅

I2 = I3 ⇒

⋅

⋅

nS33 = nS22⋅

i2 − i 5 i 3( − i 4 (

kg 783/6 − 651 >1/27! 863 − 582/3 s

⋅

⋅

R epw = R 5 (2( = nS33 ⋅ (i2( − i 5 ( ) = 1/27 ⋅ (774/:7 − 582/3 ) = 41/9!lX ⋅

⋅

⋅

X loq = X u2(3( = nS33 ⋅ (i2( − i3( ) > 1/27 ⋅ (774/:7 − 863) >−25/2!lX ⋅

⋅

⋅

X lwu = X u23 = nS22 ⋅ (i2 − i3 ) > 1/45 ⋅ (783/86 − 841) >−2:/6!lX ⋅

εi!>!

R epw ⋅

⋅

=!

X lou + X lwu


41/9 >!1/:3 25/2 + 2:/6

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strana 24

7/25/!Rashladno postrojewe sa dva prigu{na ventila, dva odvaja~a te~nosti, dva kompresora i jednim ⋅

ispariva~em prikazano je na slici. Ako iz kondenzatora rashladnog postrojewa (stawe 1) izlazi n =1/2 lh0t kqu~alog freona 12 temperature 41pD, i ako se prvim prigu{nim ventilom sniàva pritisak freona na q>281!lQb, a drugim na q>31!lQb, skicirati proces u it koordinatnom sistemu i odrediti rashladni kapacitet postrojewa. 1 7

5

n 6 4

2 3

ispariva~

5 7 i

4

6 2 1 3 t


{fmlp@fvofu/zv


ta~ka 0:

strana 25

u>41pD

y>1

q!>!2/8!cbs

i!>!i1!>!63:/19!

q!>1/3!cbs

i3!>!)!i′!*q>2/8!cbs!>!595/6!

lK i1!>!63:/19! lh ta~ka 1:

lK lh

y2!>!1/38 ta~ka 2: y3!>!1/32

s!>!291/44! q!>1/3!cbs

ta~ka 3:

lK lh

lK lh

y>1

lK i4!>!737/6! lh 1. na~in: ⋅

⋅

R epw = n⋅ (2 − y 2 ) ⋅ (2 − y 3 ) ⋅ (s )q=1/3cbs > 1/2 ⋅ (2 − 1/38) ⋅ (2 − 1/32) ⋅ 291/34 >21/5!lX 2. na~in: ⋅

⋅

R epw = n⋅ (2 − y2 ) ⋅ (i 4 − i 3 ) > 1/2 ⋅ (2 − 1/38) ⋅ (737/6 − 595/6 ) >21/5!lX


{fmlp@fvofu/zv


strana 26

7/26/!Freon 12 (R12) kao rashladni fluid obavqa levokretni ciklus sa dvostepenim isparavawem (slika). U nisko−temperaturskom ispariva~u vlada pritisak 2!cbs, u visoko-temperaturskom ispariva~u 4!cbs, a u kondenzatoru 9!cbs. Kondenzovani fluid (stawa 6) se razdvaja na dve struje i svaka od wih se adijabatski prigu{uje u odgovaraju}em prigu{nom ventilu (do stawa 7 odnosno stawa 8). Suva para (stawa 2) iz visoko−temperaturskog ispariva~a (VTI) se adijabatski prigu{uje do pritiska 1 bar (stawe 3) i zatim izobarski me{a sa suvom parom (stawa 1) iz nisko−temperaturskog ispariva~a (NTI). Dobijena me{avina (stawa 4) se kvazistati~ki izentropski sabija u kompresoru do stawa 5. Ako je rashladni kapacitet visokotemperaturskog ispariva~a 25!lX, a niskotemperaturskog 8!lX, skicirati promene stawa freona 12 na Ut dijagramu i odrediti: a) masene protoke rashladnog fluida kroz oba ispariva~a b) snagu kompresora c) faktor hla|ewa rashladnog postrojewa 7

6

VTI 8

3

NTI 9

4

2

5

6

U

7 3

8

4 9

2

5 t


{fmlp@fvofu/zv


strana 27

a) ta~ka 6:

q!>!9!cbs

y>1

ta~ka 7:

i8!>!i7!>!642/6!

ta~ka 8:

i9!>!i8!>!i7!>!642/6!

ta~ka 2:

q!>!4!cbs

y>2

ta~ka 1:

q!>!2!cbs

y>2

i>642/6!

lK lh

lK lh lK lh lK lh lK i2!>!752/2! lh

i3!>!766/69!

⋅

⋅

8 R ouj lh n ouj!>! >! >1/17! 752/2 − 642/6 t i2 − i 9 ⋅

⋅

25 R wuj lh n wuj!>! > >!1/22! t i 3 − i 8 766/69 − 642/6 b) lK lh i!>@

ta~ka 3:

i4!>!i3!>!766/69!

ta~ka 4:

q>!4!cbs

prvi zakon termodinamike za me{awe fluidnih struja: ⋅

⋅

⋅

R 23 = ∆ I23 + X 23

⋅  ⋅  n wuj ⋅ i 4 + nouj ⋅ i2 =  n wuj + nouj  ⋅ i 5   ⋅

i5 =

⋅

⇒

⋅

⋅

⇒

1/22 ⋅ 766/69 + 1/17 ⋅ 752/2 lK >761/5! 1/22 + 1/17 lh q!>!9!cbs

ta~ka 5:

⋅

I2 = I3

t!>!t5!>!2/734!

i5 =

⋅

n wuj ⋅ i 4 + nouj ⋅ i2 ⋅

⋅

n wuj + nouj lK t5!>!2/734! lhL lK lhL

i6!>!799/2!

lK lh

⋅ ⋅  ⋅  X lq =  nouj + nouj  ⋅ (i 5 − i 6 ) > (1/17 + 1/22) ⋅ (761/5 − 799/2) >−7/5!lX  

c) ⋅

εi!>!

R epw ⋅

X lq

⋅

=!

⋅

R ouj + R wuj ⋅

X lq


!!>

8 + 25 >!4/4 7/5

{fmlp@fvofu/zv

zbirka zadataka iz termodinamike zadatak za ve`bawe:

strana 28

(2.16.)

7/27/ Za potrebe hla|ewa dve odvojene rashladne komore koristi se levokretni kru`ni proces sa zajedni~kim kondenzatorom (KD) ,pri ~emu je rashladni fluid freon 12. U nisko temperaturskom ispariva~u (NTI) vlada temperatura od −41PD, u visoko temperaturskom ispariva~u (VTI) temperartura −2PD, dok je pritisak u kondenzatoru 1/8:42!NQb. Kondezovani fluid (stawa 6) razdvaja se na dve struje i svaka od wih se adijabatski prigu{uje u odgovaraju}em ventilu (do stawa 7 odnosno stawa 8). Suva para (stawa 1) iz nisko temperaturskog ispariva~a se kvazistati~ki adijabatski sabija u prvom kompresoru do pritiska koji vlada u visoko temperaturskom ispariva~u (stawe 3) i zatim izobarski me{a sa suvom parom (stawa 2) iz visoko temperaturskog ispariva~a. Dobijena me{avina se kvazistati~ki adijabatski sabija u drugom kompresoru do temperature od 51PD (stawe 5). Ako je maseni protok rashladnog fluida kroz nisko temperaturski ispariva~ 1/174!lh0t, a kroz visoko temperaturski ispariva~ 1/224!lh0t, odrediti: a) rashladne snage oba ispariva~a b) snage oba kompresora c) koeficijent hla|ewa rashladnog postrojewa KD 7

6

VTI

8

3

NTI 9

⋅

a)

R ouj!>8!lX⋅

b) X u24 >!−2/29!lX c) εi!>!6

2

4

5

⋅

! R wuj>25!lX ⋅

X u 56 >!−4!lX


{fmlp@fvofu/zv


strana 29

7/28. U toplotnoj pumpi, radna materija obavqa levokretni kru`ni proces koji se sastoji od ravnote`nog (kvazistati~kog) adijabatskog sabijawa, izotermskog ravnote`nog (kvazistati~og) sabijawa, ravnote`nog (kvazistati~kog) adijabatskog {irewa i ravnote`nog (kvazistai~kog) izotermnog {irewa. Maksimalna odnosno minimalna temperatura radne materije iznose: Unby>!431!L i Unjo>!391!L, a temperature toplotnog izvora odnosno toplotnog ponora su stalne i iznose Uuj>!3:1!L i Uuq>!421!L. Nepovratnost predaje toplote radnoj materiji iznosi ∆TJ!>!6!K0L. Predstaviti proces u Ut koordinatnom sistemu i odrediti: a) nepovratnost predaje toplote toplotnom ponoru b) koeficijent grejawa toplotne pumpe U Unby 4

3

!Uuq !Uuj Unjo

5

2 t

a) ∆TJ!>! (∆T tj )52 = (∆Tsu )52 −

(∆T tj )52

Repw!>!

2 Unjo

−

2 Uuj

R epw R R !>! epw − epw !!!!!!! Uuj Unjo Uuj 6

!>

2 2 − 391 3:1

⇒

>!51/7!lK

Rpew!>!R34!> Unby ⋅ (∆T SU )34 >///!> 431 ⋅ (−1/256 ) >!−57/5!lK

(∆TSU )34

= −(∆T SU )52 = −

(∆Ttj )34 = (∆Tsu )34 − R pew Uuq

R epw 51/7 K >−256 =− L Unjo 391 !>! − 256 −

.57511 K >!5/79! 421 L

b) εh!>!

R pew R pew − R epw

>

57/5 >!9 57/5 − 51/7


{fmlp@fvofu/zv


strana 30

7/29/!Dve toplotne pumpe me|usobno spojene redno (slika) rade po idealnom Karnoovom kru`nom procesu. Toplotna pumpa 2 uzima 511!lK toplote od toplotnog izvora stalne temperature UUJ>411!L. Toplotu odvedenu od toplotne pumpe 2 preuzima toplotna pumpa 3, koja predaje toplotu toplotnom ponoru stalne temperature UUQ>2311!L. Ako obe toplotne pumpe rade sa istim faktorom grejawa odrediti: a) temperaturu radne materije pri kojoj se vr{i razmena toplote izme|u dve toplotne pumpe, UY b) neto mehani~ke radove koji se dovode radnoj materiji u toplotnoj pumpi, 2(UQ2) i toplotnoj pumpi, 3(UQ3) TOPLOTNI Repw IZVOR

Ry

Rpew

UQ2

UQ3

X2

X3

TOPLOTNI PONOR

a) ε h2 =

UY UY − UUJ

ε h2 = ε h3 UY =

ε h3 =

⇒

UUQ UUQ − UY

UY UUQ > UY − UUJ UUQ − UY

⇒

UUJ ⋅ UUQ > 411 ⋅ 2311 >711!L

b) ε h2 = εh2!>!

UY 711 > >3 711 − 411 UY − UUJ RY R Y − R epw

⇒

ε h3 = RY =

ε h2 ε h2 − 2

UUQ 2311 > >3 2311 − 711 UUQ − UY ⋅ R epw >911!lK

X2> R Y − R epw >911!−!511!>!511!lK εh3!>!

R pew R pew − R Y

⇒

R pew =

ε h3 ε h3 − 2

⋅ R Y >2711!lK

X3> R pew − R Y >2711!−!911!>!911!lK


{fmlp@fvofu/zv


strana 31

7/2:/!Izobarskim odvo|ewem toplote od 4!lh vodene pare stawa (y>1/8-!q>31!lQb) sve dok se ne postigne stawe kqu~ale te~nosti, pomo}u levokretnog kru`nog procesa u postrojewu sa toplotnom pumpom, toplota se predaje vodoniku (idealan gas). Masa vodonika je 29!lh, a po~etna temperatura :6pD. Vodonik se nalazi u zatvorenom sudu. Koeficijent grejawa toplotne pumpe je εh>2/9. Odrediti krajwu temperaturu vodonika u sudu kao i snagu kompresora ako toplotna pumpa radi jedan sat. Koli~ina toplote koja se oduzme od vodene pare u procesu kondenzacije )Rqbsb* istovremeno predstavqa dovedenu toplotu za toplotnu pumpu )Repw* Repw!>!Rqbsb!>!nqbsb!/!)iy!−!i′*!>!///!> 4 ⋅ (2:12/83 − 362/5 ) >!5:62!lK iy!>!i′!,!y/!)i′′!−!i′*!>!!///!> 362/5 + 1/8 ⋅ (371: − 362/5 ) >2:12/83! i′!>!362/5!

lK -! lh

i′′!>!371:!

lK -! lh

lK lh

)q!>!1/3!cbs*

Koli~ina toplote koju primi vodonik )RW*!istovremeno predstavqa odvedenu toplotu za toplotnu pumpu!)Rpew* εh!>!


R pew =

⇒

R pew =

εh εh − 2

⋅ R epw

2/9 ⋅ 5:62>!2224:/86!lK!>!RW 2/9 − 2

RW!>!nw!/!dw!)!Uw3!.!Uw2* !!!!!! UW3> 479 +

⇒

!!!!!!Uw3!> UW2 +

RW > nw ⋅ d W

2224:/86 >!538/62!L 29 ⋅ 21/5

XL!>! R pew − R epw = 7299/86!lK •

XL =

XL 7299/86 = 2/83!lX > 4711 τ


{fmlp@fvofu/zv


strana 32

7/31. Instalacija toplotne pumpe (slika) radi sa ugqendioksidom (idealan gas) kao radnim fluidom po Xulovom kru`nom procesu izme|u qnjo>1/2!NQb i qnby>1/5!NQb. Stepen dobrote adijabatske kompresije je ηlq>!1/:7, a stepen dobrote adijabatske ekspanzije ηfy>!1/:3. Svrha toplotne pumpe je da se u prostoriji odràva temperatura UUQ>28pD pri temperaturi okolnog vazduha UUJ>1pD. Pri tome se iz okoline radnom telu dovodi 311!lK0t toplote. Usvojiti da se ugqendioksid ispred kompresora zagreva do temperature koja vlada u okolini, a ispred ekspanzionog ure|aja hladi do temperature koja vlada u prostoriji. Odrediti faktor grejawa toplotne pumpe kao i snagu kompresora i snagu ekspanzionog ure|aja (turbine).

U

grejana prostorija

3 3l

4

3

4 2 5

5l

t κ −2

2

5 2/39 −2

  κ >!384 ⋅  5  2/39 >!47:/82!L U3L!>!U2 ⋅  q 3L   q   2  2  U −U 47:/82 − 384 !>484/85!L U3!>!U2!,! 3L lq 2 !>!384!,! 1/:7 η  U5L!>!U4 ⋅  q 5L  q  4

   

κ −2 κ

>!3:1 ⋅  2  5

2/39 −2 2/39

>!325/25!L

U5!>!U4!,!ηfy!/)U5l!.!U4*!>!3:1!,!1/:3 ⋅(325/25 − 3:1) !>331/32!L ⋅

⋅

⋅

R epw = R 52 = n⋅ d q ⋅ (U2 − U5 ) ⋅

n= ⋅

⇒

⋅

R epw 311 lh > >5/57! d q ⋅ (U2 − U5 ) 1/96 ⋅ (384 − 331/32) t ⋅

⋅

R pew = R 34 = n⋅ d q ⋅ (U4 − U3 ) > 5/57 ⋅ 1/96 ⋅ (3:1 − 484/85 ) >−428/57!lX


{fmlp@fvofu/zv


strana 33

⋅

R pew εh!>!

⋅

>!

⋅

R pew − R epw ⋅

⋅

428/57 >3/8 428/57 − 311

⋅

X lpnqsftps!>! X u23 > n ⋅ d q ⋅ (U2 − U3 ) > 5/57 ⋅ 1/96 ⋅ (384 − 484/85 ) >−492/:!lX ⋅

⋅

⋅

X uvscjob!>! X u45 > n ⋅ d q ⋅ (U4 − U5 ) > 5/57 ⋅ 1/96 ⋅ (3:1 − 331/32) >!375/68!lX 7/32/!Toplotna pumpa koja se koristi za zagrevawe vazduha (idealan gas) od Uw2>66pD!do Uw3>71pD! na ra~un hla|ewa vode od (q>2!cbs- Ux2>29pD*!do!)q>2!cbs-!Ux3>25pD*, radi izme|u pritisaka qnjo>5/66!cbs i!qnby>27/83!cbs, sa freonom 12 )S23* kao radnim telom, po idealnom Rankin−Klauzijusovom kru`nom procesu sa prigu{ivawem te~ne faze,. Protok vode kroz ispariva~ toplotne pumpe je 2/6!lh0t/ Skicirati promene stawa freona 12 na Ut!i!it dijagramu i odrediti faktor grejawa toplotne pumpe, snagu kompresora kao i maseni protok vazduha kroz kondenzator toplotne pumpe. vazduh 4

3

S23

5

2 3

voda 3

U

i

4

2 4 5 5

2 t


t

{fmlp@fvofu/zv


strana 34

q>5/66!cbs

ta~ka 1: lK i2!>!771/95! lh

t2!>!2/675!

ta~ka 2: i3!>!795/13!

y>2 lK lhL

q>27/83!cbs

t!>!t2!>2/675!

q>27/83!cbs

y>1

lK lhL

lK lh

ta~ka 3: lK i4!>!677/2! lh

i5!>!i4!>!677/2!

ta~ka 4: ⋅

⋅

⋅

⋅

⋅

⋅

lK lh

R epw = R 52 = n g ⋅ (i2 − i 5 ) >!/// R pew = R 34 = n g ⋅ (i 4 − i3 ) >!/// ⋅

R pew εh =

⋅

>

⋅

R pew − R epw

i4 − i3 i 4 − i 3 − i2 − i 5

>

677/2 − 795/13 677/2 − 795/13 − 771/95 − 677/2

>6/19

prvi zakon termodinamike za proces u ispariva~u toplotne pumpe: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 ⋅

⋅

⋅

⇒ ⋅

⋅

n x ⋅ i x2 + n g ⋅ i 5 = n x ⋅ i x3 + n g ⋅ i2 ⋅

⋅

ng = nx ⋅

⋅

I2 = I3 ⇒

i x2 − i x3 86/46 − 69/71 lh >1/376! > 2/6 ⋅ 771/95 − 677/2 t i2 − i 5

napomena: lK lh lK ix3!>!69/71! lh ix2!>!86/46!


)q!>!2!cbs-!u>29pD* )q!>!2!cbs-!u>25pD*

{fmlp@fvofu/zv


strana 35

prvi zakon termodinamike za proces u kompresoru toplotne pumpe: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23

⇒

⋅

⋅

⋅

X U23 = −∆ I23 = − n g ⋅ (i3 − i2 )

⋅

X U23 = −1/376 ⋅ (795/13 − 771/95 ) >−7/25!lX

prvi zakon termodinamike za proces u kondenzatoru toplotne pumpe: ⋅

⋅

⋅

R 23 = ∆ I23 + X U23 ⋅

⇒

⋅

⋅

⋅

I2 = I3

⋅

⋅

n w ⋅ d q ⋅ Uw2 + n g ⋅ i 3 = n w ⋅ d q ⋅ Uw3 + n g ⋅ i 4 ⋅

⋅

nw = ng ⋅

⇒

i3 − i 4 795/13 − 677/2 lh >7/36! > 1/376 ⋅ d q ⋅ (Uw3 − Uw2 ) 2 ⋅ (71 − 66 ) t

7/33/!Termodinami~ki stepen korisnosti Xulovog kru`nog procesa (2−3−4−5−2), koji obavqa vazduh (idealan gas) iznosi η =0.3. Koliki bi bio faktor grejawa toplotne pumpe, kada bi metan (idealan gas) obavqao levokretni Xulov kru`ni procec izme|u istih stawa )2−5−4−3−2*. desnokretni

levokretni U

U 4

4

5

5

3

3

2

2 t


t

{fmlp@fvofu/zv


strana 36

desnokretni kru`ni proces: R epw = R 34 = n ⋅ d q ⋅ (U4 − U3 ) R pew = R 52 = n ⋅ d q ⋅ (U2 − U5 )

η!>!

U − U3 + U2 − U5 R epw + R pew ! >! 4 R epw U4 − U3

)2*

levokretni kru`ni proces: R epw = R 25 = n ⋅ d q ⋅ (U5 − U2 )

R pwe = R 43 = n ⋅ d q ⋅ (U3 − U4 )

!!!εh!>!


>

U3 − U4 U3 − U4 − U5 + U2

pore|em izraza (1) i (2) uo~ava se:

zadatak za ve`bawe:

>

U4 − U3 U4 − U3 − U5 + U2

εh =

)3*

2 2 >4/44 > η 1/4

(2.23.)

7/34. Toplotna pumpa radi, sa vodenom parom kao radnim fluidom, po realnom levokretnom Rankin−Klauzijusovom kru`nom procesu bez podhla|ivawa kondenzata izme|u pritisaka qnjo>9!lQb!i qnby>1/7!NQb. U kompresor ulazi suvozasi}ena vodena para, a na izlazu iz kompresora temperatura pare je 811pD. Kao izvor toplote koristi se otpadna voda nekog industrijskog procesa, temperature 61pD>dpotu, koja predaje vodenoj pari 2311!lX toplote. Potro{a~ toplote (toplotni ponor) nalazi se na temperaturi 261pD>dpotu. Odrediti: a) koli~inu toplote koja se predaje potro{a~u (lX) b) koeficijent grejawa toplotne pumpe )εh* c) promenu entorpije sistema koji sa~iwavaju radna materija, toplotni izvor i toplotni ponor ⋅

a)

R pew >3161/4!lX

b)

εh!>!3/5

c)

∆ T tj!>!2/24!

⋅

lX L


{fmlp@fvofu/zv


strana 1

VLA@AN VAZDUH 8/2/!U sudu zapremine!W>1/96!n4!nalazi se!nww>2/12!lh!nezasi}enog vla`nog vazduha stawa!)q>2!cbsu>31pD*/!Odrediti: a) masu suvog vazduha u sudu i masu vodene pare u sudu b) pritisak suvog vazduha i pritisak vodene pare u sudu c) gustinu suvog vazduha, gustinu vodene pare i gustinu vla`nog vazduha u sudu lhI3 P d) sadràj vlage (apsolutnu vla`nost) vla`nog vazduha, lhTW e) relativnu vla`nost vla`nog vazduha f) specifi~nu entalpiju vla`nog vazduha a) nww!>!ntw!,!nI3P

)2*

(

)

q ⋅ W = nTW ⋅ S hTW + nI3PS hI3P ⋅ U

)3*

re{avawem sistema jedna~ina (1) i (2) dobija se;  2⋅ 216 ⋅ 1/96  2 2 q⋅W  n tw =  > >2!lh − 2/12⋅ 573  ⋅ − n ww ⋅ S hI3P  ⋅   − U S S 3:4 398 573 −   htw hI3P  

nI3P!>!nww!−!ntw!>!2/12!−!2!>1/12!lh b) q TW =

nTW ⋅ S hTW U

q I3P =

=

W nI3P ⋅ S hI3P U W

2 ⋅ 398 ⋅ 3:4 = :9:41!Qb 1/96 =

1/12 ⋅ 573 ⋅ 3:4 = 26:3!Qb 1/96

c) ρtw> !!

nTW q TW :9:41 lhTW = = >2/287! W S hTW U 398 ⋅ 3:4 n4

!ρI3P>

nI3P W

=

q I3P S hI3P U

=

lhI3 P 26:3 >1/119! 573 ⋅ 3:4 n4

ρww!>!ρtw!,!ρI3P!>2/287!,!1/119!>2/195!!

lhWW n4

d) y>

nI3P nTW

=

NI3P NTW

⋅

q I3P q − q I3P


=

lhI3 P 29 26:3 ⋅ >1/12! 6 3: 2 ⋅ 21 − 26:3 lhTW

{fmlp@fvofu/zv


strana 2

e) ϕ>

q I3P

(qqt )U =31p D

=

26:3 >1/79 3448

f) i = d qTW ⋅ u + y ⋅ )2/97 ⋅ u + 3611* = 2 ⋅ 31 + 1/12 ⋅ (2/97 ⋅ 31 + 3611) >56/48!

lK lhTW

8/3/!Odrediti temperaturu!vla`nog vazduha ~ije je stawe pri!q>2!cbs!zadato na na~in: lhI3 P a) y>1/13! !)apsolutna vla`nost*-!ϕ>1/9!(relativna vla`nost) lhTW b) Uwu>31pD!)temperatura vla`nog termometra*-!Us>21pD!)ta~ka rose* a) q I3P =

y NI3P NTW

q qt =

q I3P

ϕ u4!>!39/6pD b)

=

⋅q = +y

1/13 ⋅ 2 ⋅ 21 6 >4232/75!Qb 29 + 1/13 3:

4232/75 >4:13!Qb 1/9 )temperatura kqu~awa vode na!q>4:13!Qb*

(qI3P )S = ϕS ⋅ (qqt )US =21 > 2⋅ 2338 >2338!Qb NI3P

yS>

NTW

⋅

q I3P q − q I3P

=

lhI3 P 29 2338 ⋅ >1/1188! 6 3: 2 ⋅ 21 − 2338 lhTW

y!>!yS

(qI3P )wu = ϕ wu ⋅ (qqt )Uwu =31 > 2⋅ 3448 >3448!Qb ywu>

NI3P NTW

⋅

q I3P q − q I3P

=

lhI3 P 29 3448 ⋅ >1/1259! 6 3: 2 ⋅ 21 − 3448 lhTW

iwu = dq ⋅ u + y wu ⋅ )2/97 ⋅ u wu + 3611* = 2 ⋅ 31 + 1/1259 ⋅ (2/97 ⋅ 31 + 3611) >56/48

lK lhTW

i!>!iwu

u!>!

i − y ⋅ 3611 68/66 − 1/1188 ⋅ 3611 !> >48/87pD 2 + 1/1188 ⋅ 2/97 dq + y ⋅ 2/97


{fmlp@fvofu/zv


strana 3

8/4/!Vla`nom vazduhu stawa!2)q2>2!cbs-!u2>31pD-!ϕ2>1/9-!nww>31!lh0i) dovodi se toplota u zagreja~u vazduha dok vazduh ne dostigne stawe!3)q3>2!cbs-!u3>91pD), a zatim se tako zagrejan vazduh u adijabatski izolovanoj komori vlaì pregrejanom vodenom parom stawa!Q)q>2!cbs-!u>231pD-!nqq>2!lh0i*!do stawa 4)q>2!cbs). Skicirati promene stawa vla`nog vazduha na Molijerovom i!−y!dijagramu i odrediti: a) toplotnu snagu zagreja~a vazduha!)lX* b) entalpiju!)i*-!apsolutnu vla`nost!)y*!i temperaturu!)u*!vla`nog vazduha stawa!4

i 4 3

3828

u3 ϕ>2 u2

2 ϕ2

y

ta~ka 1: q qt >3448!Qb

)napon pare ~iste vode na!u>31pD*

q I3P = ϕ ⋅ q qt > 1/9 ⋅ 3448 >297:/7!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 297:/7 ⋅ >1/1229! 6 lhTW 3: 2 ⋅ 21 − 297:/7

i2> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1229 ⋅ )2/97 ⋅ 31 + 3611* >61/13! ntw!>!

lK lhTW

n ww 31 lhTW > >2:/88! 2 + y 2 2 + 1/1229 i

ta~ka 2: y3!>!y2>1/1229!

lhI3 P lhTW

i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 91 + 1/1229 ⋅ )2/97 ⋅ 91 + 3611* >222/44!


lK lhTW

{fmlp@fvofu/zv


strana 4 ⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

2:/88 ⋅ (222/44 − 61/13) >1/45!lX 4711

R 23 = n tw ⋅ (i3 − i2 ) = ta~ka 3:

⋅

⋅

⋅

prvi zakon termodinamike za proces vlaèwa vazduha:!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

⋅

n tw ⋅ i3 + nqq ⋅ iqq = n tw ⋅ i 4

⋅

⇒!

i4 =

⋅

ntw ⋅ i3 + nqq ⋅ iqq ⋅

ntw 2 2:/88 ⋅ 222/44 + ⋅ 3828 lK 4711 4711 i4 = >359/87! 2:/88 lhTW 4711 materijalni bilans vlage za proces vlaèwa vazduha: ⋅

⋅

⋅

n tw ⋅ y 3 + nqq = n tw ⋅ y 4

⋅

⇒!

y4 =

⋅

n tw ⋅ y 3 + nqq ⋅

n tw y4 =

lhI3 P 2:/88 ⋅ 1/1229 + 2 >1/1735! 2:/88 lhTW

u4!>!

i 4 − y 4 ⋅ 3611 359/87 − 1/1735 ⋅ 3611 >94/22pD !> d q + y 4 ⋅ 2/97 2 + 1/1735 ⋅ 2/97

napomena: iqq>!3828!

lK -!entalpija pregrejane vodene pare stawa!Q!)q>2!cbs-!u>231pD* lh


{fmlp@fvofu/zv


strana 5

8/5/!Za pripremu vla`nog vazduha stawa!5)q>2!cbs-!u>47pD-!ϕ>1/4) koristi se sve` vazduh stawa!2)q>2 cbs-!u>21pD-!ϕ>1/9). Sve` vazduh se najpre zagreva u zagreja~u do stawa!3)q>2!cbs*-!a onda adijabatski vlaì ubrizgavawem vode!X)q>2!cbs-!ux>61pD*!dok ne postane zasi}en!)q>2!cbs-!ϕ>2*/!Na kraju se vazduh dogreva u dogreja~u. Potro{wa vode u fazi vlaèwa vazduha iznosi!71!lh0i. Skicirati promene stawa vla`nog vazduha na Molijerovom!i!−y!dijagramu i odrediti: a) veli~ine stawe vla`nog vazduha na ulazu u dogreja~!!4)i-!y-!u* b) toplotne snage zagreja~a i dogreja~a!)lX* i

5

u5

ϕ5

3 ϕ>2

u2

4

2 ϕ2

y 31: ta~ka 1: q qt >2338!Qb


q I3P = ϕ ⋅ q qt > 1/9 ⋅ 2338 >:92/7!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

lhI3 P 29 :92/7 ⋅ >1/1173! 6 3: 2 ⋅ 21 − :92/7 lhTW

>

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 21 + 1/1173 ⋅ )2/97 ⋅ 21 + 3611* >36/73!

lK lhTW

ta~ka 4; q qt >6:51!Qb


q I3P = ϕ ⋅ q qt > 1/4 ⋅ 6:51 >2893!Qb

y5!>

NI3P NTW

⋅

q I3P q − q I3P

lhI3 P 29 2893 ⋅ >1/1224! 6 3: 2 ⋅ 21 − 2893 lhTW

>

i5> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 47 + 1/1224 ⋅ )2/97 ⋅ 47 + 3611* >76/12!

lK lhTW

ta~ka 3: y4!>y5!>1/1224! q I3P =

y NI3P NTW

q qt =

q I3P

ϕ p u4!>!27 D

=

lhI3 P lhTW ⋅q =

+y

1/1224 ⋅ 2 ⋅ 21 6 >2899!Qb 29 + 1/1224 3:

2899 >2899!Qb 2 )temperatura kqu~awa vode na!q>2899!Qb*

i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 27 + 1/1224 ⋅ )2/97 ⋅ 27 + 3611* >55/6:!


lK lhTW {fmlp@fvofu/zv


strana 6

ta~ka 2: y3!>y2!>1/1173!

lhI3 P lhTW

materijalni bilans vlage za proces vlaèwa vazduha!)3−4*; ⋅

⋅

⋅

⋅

ntw ⋅ y3 + nX = ntw ⋅ y4 !!!!!!!!⇒!!!!!!!! n tw

71 ⋅ nX lh 4711 >4/38! = = t 1/1224 − 1/1173 y4 − y3 ⋅

⋅

⋅


⋅

⋅


⋅

⇒

i3 =

⋅

ntw ⋅ i4 − nX ⋅ iX ⋅

ntw i3 =

71 ⋅ 31: lK 4711 >54/63! 4/38 lhTW

4/38 ⋅ 55/6: −

⋅

⋅

⋅

⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

R 23 = n tw ⋅ (i3 − i2 ) > 4/38 ⋅ (54/63 − 36/73) >69/64!lX prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!! R 45 = ∆ I45 + X u 45 ⋅

⋅

R 45 = n tw ⋅ (i 5 − i 4 ) > 4/38 ⋅ (76/12 − 55/6:) >77/88!lX

8/6/!Vlaàn vazduh, pri konstantnom pritisku!)q>2/37!cbs*-!struji kroz adijabatski izolovan kanal i pri tome se najpre zagreva a potom i vlaì suvozasi}enom vodenom parom!)q>2/4!cbs*!)slika*/!Jedan deo vodene parekoristi se za zagrevawe vazduha (ulazi u cevnu zmiju i iz we izlazi potpuno kondenzovan tj. kao kqu~ala te~nost), a drugi deo pare (istog po~etnog stawa) koristi se za vlaèwe vla`nog vazduha (isti~e kroz mlaznicu i me{a se sa vla`nim vazduhom stawa 2). Zapreminski protok vla`nog vazduha na ulazu u kanal iznosi!1/65!n40t-!a wegovo stawe je definisano temperaturom suvog termometra i temperaturom vla`nog termometra!2)utu>33pD-!uwu>23pD*/!Odrediti potrebne masene protoke vodene pare posebno kroz cevnu zmiju i posebno kroz mlaznicu, da bi se ostvarilo stawe!4)u>71pD-!ϕ>1/4*!vla`nog vazduha na izlazu iz kanala. Skicirati promene stawa vla`nog vazduha na!Molijerovom!i−y!dijagramu. nB

nC

vlaàn vazduh 2


3

4

{fmlp@fvofu/zv


strana 7

ta~ka WU:

(qI3P )wu = ϕ wu ⋅ (qqt )Uwu =23 > 2⋅ 2512 >2512!Qb ywu>

NI3P

⋅

NTW

q I3P

lhI3 P 29 2512 ⋅ >1/1181! 6 3: 2/37 ⋅ 21 − 2512 lhTW

=

q − q I3P

iwu = dq ⋅ u + y wu ⋅ )2/97 ⋅ u wu + 3611* = 2⋅ 23 + 1/1181 ⋅ (2/97 ⋅ 23 + 3611) >3:/77

lK lhTW

ta~ka 1: i2>iwu!>3:/72! y2 =

lK lhTW

i2 − d q ⋅ u 2

=

2/97 ⋅ u 2 + 3611

(qI3P )2 = N

y2 I3P

lhI3 P 3:/77 − 2 ⋅ 33 >1/1141! lhTW 2/97 ⋅ 33 + 3611

⋅q = + y2

NTW

1/114 ⋅ 2/37 ⋅ 21 6 >717!Qb 29 + 1/114 3:

(q tw )2 = q − (q I3P )2 = 2/37 ⋅ 21 6 (ρ tw )2 =

(q TW )2 S hTW ⋅ U2

⋅

=

− 717 > 2/36 ⋅ 21 6 !Qb

2/36 ⋅ 21 6 lhTW >2/59! 398 ⋅ 3:6 n4

⋅

n tw = (ρ tw )2 ⋅ W 2 = 2/59 ⋅ 1/65 >1/9!

lhTW t

ta~ka 3: q qt >2::21!Qb )napon pare ~iste vode na!u>71pD* q I3P = ϕ ⋅ q qt > 1/4 ⋅ 2::21 >6:84!Qb

y4!>

NI3P

⋅

NTW

q I3P q − q I3P

>

lhI3 P 29 6:84 ⋅ >1/141:! 6 lhTW 3: 2/37 ⋅ 21 − 6:84

i4> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 71 + 1/141: ⋅ )2/97 ⋅ 71 + 3611* >251/8

lK lhTW

ta~ka 2: y3>y2>1/1141!

lhI3 P lhTW

i3>@

materijalni bilans vlage za proces vlaèwa vazduha!)3−4*; ⋅

⋅

⋅

ntw ⋅ y3 + nB = ntw ⋅ y4 !!!!!!!!⇒ ⋅

n B = 1/9 ⋅ (1/141: − 1/1141) > 3/34 ⋅ 21 −3


⋅

⋅

n B = ntw ⋅ (y 4 − y 3 ) lh t

{fmlp@fvofu/zv


strana 8

prvi zakon termodinamike za proces u otvorenom sistemu ograni~enom isprekidanom konturom: ⋅

⋅

⋅

R 23 = ∆ I23 + X u23 ⋅

⋅

⋅

⋅

⋅

n tw ⋅ i2 + n B ⋅ i B2 + nC ⋅ iC = n tw ⋅ i 4 + nC ⋅ i B3 ⋅

⋅

n tw ⋅ (i 4 − i2 ) − nC ⋅ iC 1/9 ⋅ (251/8 − 3:/77 ) − 3/34 ⋅ 21 −3 ⋅ 3798 > nC = = i B2 − i B3 3798 − 55:/3 ⋅

⋅

lh t lK iB2>iC2!>!3798! lh iB3>55:/3 nC > 2/3: ⋅ 21 −3

)suva para!q>2/4!cbs* )kqu~ala voda!q>2/4!cbs*

i 4 3 ϕ>2 3798

2 WU

y

)8/7/*

zadatak za ve`bawe:

8/7/!21!)2,y*!lh0t!vla`nog vazduha stawa!2)q>2!cbs-!u>71pD-!qI3P>1/12!cbs) vlaì se vodenom parom stawa!Q)q>2!cbs-!u>271!pD). Parcijalni pritisak vodene pare u vla`nom vazduhu nakon vlaèwa iznosi 3)qI3P>1/16!cbs*/!Dobijeni vlaàn vazduh stawa 2 hladi se do zasi}ewa (stawe 3). Svi procesi sa vla`nim vazduhom su izobarski. Skicirati procese sa vla`nim vazduhom na Molijerovom iy!dijagramu i odrediti: a) temperaturu )u* i apsolutnu vla`nost!)y*!vla`nog vazduha stawa!3!i stawa!4 b) koliko se toplote odvede od vla`nog vazduha u procesu hla|ewa!)3−4*-!)lX* re{ewe: a) u3>75/79pD-!y3>1/1438!

lhI3 P lhI3 P -!u4>43/:pD-!y4>1/1438! lhTW lhTW

⋅

b)

R 34!>!−447!lX


{fmlp@fvofu/zv


strana 9

8/8/!U adijabatski izolovanom rashladnom torwu, za potrebe hla|ewa neke prostorije, hladi se voda X2)q>2!cbs-!ux2>68pD) isparavawem u struji vazduha, ~ije je stawe na ulazu u toraw 2)q>2!cbs-!u>33pDϕ>1/3) a na izlazu iz torwa 3)q>2!cbs-!u>38pD-!ϕ!>1/:*/!Protok suvog vazduha kroz toraw iznosi!9/6!lh0t. Ohla|ena voda iz torwa!X3)q>2!cbs-!ux3>33!D*-!se me{a sa sveòm vodom!Xp)q>2cbs-!uxp>27pD*!da bi se nadoknadila isparena koli~ina vode i ponovo odvodi u prostoriju koju treba ohladiti. Odrediti: a) potro{wu sveè vode!)X1* b) razmewenu toplotu u torwu!)lX* c) protoke tople )X2* i ohla|ene vode!)X3* d) koli~inu toplote koju prostorija koja se hladi predaje vodi-!R′!!)lX* X2 2)u3-!ϕ3* Rups R′ X1

X3 2)u2-!ϕ2* Xp vlaàn vazduh: ta~ka 1: q qt >3754!Qb


q I3P = ϕ ⋅ q qt > 1/3 ⋅ 3754 >639/7!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 639/7 ⋅ >1/1144! 6 lhTW 3: 2 ⋅ 21 − 639/7

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/1144 ⋅ )2/97 ⋅ 33 + 3611* >41/48!


lK lhTW

{fmlp@fvofu/zv


strana 10

ub•lb!3; )napon pare ~iste vode na!u>38pD*

q qt >4675!Qb

q I3P = ϕ ⋅ q qt > 1/: ⋅ 4675 >4318/7!Qb

y3!>

NI3P NTW

⋅

q I3P

>

q − q I3P

lhI3 P 29 4318/7 ⋅ >1/1317! 6 lhTW 3: 2 ⋅ 21 − 4318/7

i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 38 + 1/1317 ⋅ )2/97 ⋅ 38 + 3611* >8:/64

lK lhTW

voda: lK lh lK ix3>:2/:7! lh lK ix1>77/99! lh

entalpija vode!q>2!cbs-!u>68pD

ix2>349/37!

entalpija vode!q>2!cbs-!u>33pD entalpija vode!q>2!cbs-!u>27pD

materijalni bilans vlage za proces vlaèwa vazduha u torwu: ⋅

⋅

X2 + ntw ⋅ y2 = X3 + ntw ⋅ y3

⇒

⋅

X2 − X3 = ntw ⋅ (y 3 − y2 ) > Xp

⋅

Xp > n tw ⋅ (y 3 − y 2 ) > 9/6 ⋅ (1/1317 − 1/1144 ) >1/258!

lh t

⋅

⋅

⋅

prvi zakon termodinamike za proces u torwu:!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

X2 ⋅ i x2 + n tw ⋅ i2 = X3 ⋅ i x3 + n tw ⋅ i3 ⋅

R ups > X2 ⋅ i x2 − X3 ⋅ i x3 = n tw ⋅ (i 3 − i2 ) ⋅

R ups > n tw ⋅ (i 3 − i2 ) > 9/6 ⋅ (8:/64 − 41/68) >528/97!lX Xp > X2 − X3 R ups > X2 ⋅ i x2 − X3 ⋅ i x3

)2* )3*

Kombinovawem jedna~ina!)2*!i!)3*!dobija se:

X 2>3/869!

lh lh -! X 3>3/722! t t

prvi zakon termodinamike za proces u hladwaku prostorije koju treba hladiti: ⋅

⋅

⋅

R 23 = ∆ I23 + X u23

R ( = X2 ⋅ ix2 − X3 ⋅ ix3 − Xp ⋅ ixp

R ( = 3/869 ⋅ 349/37 − 3/722 ⋅ :2/:7 − 1/258 ⋅ 77/99 >518/29!lX


{fmlp@fvofu/zv


strana 11

8/9/!U nekom procesu izobarski se hladi vlaàn vazduh, po~etnog stawa!)q2>2!cbs-!u2>41pD-!ϕ2>1/9nww>211!)2,y*!lh0i*/!Odrediti: a) koli~inu toplote koja se odvodi od vla`nog vazduha kao i koli~inu izdvojenog kondenzata ako se hla|ewe vazduha vr{i do!u3>21pD b) koli~inu toplote koja se odvodi od vla`nog vazduha kao i koli~inu izdvojenog leda ako se hla|ewe vazduha vr{i do!u4>−21pD c) koli~inu toplote koja se odvodi od vla`nog vazduha kao i koli~inu izdvojenog leda i kondenzata ako se hla|ewe vazduha vr{i do!u4>1pD!i pri tome nastaje jednaka koli~ina leda i kondenzata Sve procese predstaviti na Molijerovom i−y dijagramu za vlaàn vazduh ⋅

− R 23 2

3′

3

izdvojeni kondenzat i/ili led ta~ka 1: q qt >5352!Qb


q I3P = ϕ ⋅ q qt > 1/9 ⋅ 5352>44:3/9!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 44:3/9 ⋅ >1/1329! 6 lhTW 3: 2 ⋅ 21 − 44:3/9

i2> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 41 + 1/1329 ⋅ )2/97 ⋅ 41 + 3611* >96/83!


lK lhTW

{fmlp@fvofu/zv


strana 12

a) i 2

ϕ2

u2

ϕ>2 3 u3

3′ y



q I3P = ϕ ⋅ q qt > 2⋅ 2338 >2338!Qb

y3!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 2338 >1/1188! ⋅ 6 3: 2 ⋅ 21 − 2338 lhTW

i3> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 21 + 1/1188 ⋅ )2/97 ⋅ 21 + 3611* >3:/4:!

lK lhTW

koli~ina izdvojenog kondenzata: ⋅

⋅

nlpoe = n tw ⋅ (y2 − y 3 ) = 211 ⋅ (1/1329 − 1/1188) >2/52!

lh i ⋅

⋅

⋅

prvi zakon termodinamike za proces u hladwaku vazduha:!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

⋅

R23 = ntw ⋅ (i3 − i2) + nlpoe ⋅ ix >

ix!>!53!

lK lh


211 2/52 ⋅ (3:/4: − 96/83) + ⋅ 53 >−2/66!lX 4711 4711

entalpija kondenzata (vode)!q>2!cbs-!u>21pD

{fmlp@fvofu/zv


strana 13

b) i 2

ϕ2

u2

ϕ>2

y 3 u3 3′ ta~ka 2: q qt >36:/5!Qb )napon pare ~iste vode na!u>−21pD* q I3P = ϕ ⋅ q qt > 2⋅ 36:/5 >36:/5!Qb

y3!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 36:/5 ⋅ >1/1127! 6 lhTW 3: 2 ⋅ 21 − 36:/5

i3> dqtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ (−21) + 1/1127 ⋅ )2/97 ⋅ (−21) + 3611* > !!!>−7/14

lK lhTW

koli~ina izdvojenog leda: ⋅

⋅

nmfe = ntw ⋅ (y2 − y3 ) = 211 ⋅ (1/1329 − 1/1127) >3/13!

lh i ⋅

⋅

⋅


⋅

⋅

R23 = ntw ⋅ (i3 − i2) + nlpoe ⋅ im >

211 3/13 ⋅ (− 7/14 − 96/83) + ⋅ (− 463/5 ) >−3/86!lX 4711 4711

im!>! d m ⋅ (Um − 384) − sm = 3 ⋅ (−21) − 443/5 >−463/5!


lK !!!!!!entalpija leda-!u>−21pD lh

{fmlp@fvofu/zv


strana 14

c) i 2

ϕ2

u2

ϕ>2

3

u3

y 3′

ta~ka 2: q qt >721/9!Qb )napon pare ~iste vode na!u>1pD* q I3P = ϕ ⋅ q qt > 2⋅ 721/9 >721/9!Qb

y3!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 721/9 ⋅ >1/1149! 6 3: 2 ⋅ 21 − 721/9 lhTW

i3> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 1 + 1/1149 ⋅ )2/97 ⋅ 1 + 3611* >:/6

lK lhTW

koli~ina izdvojenog kondenzata i leda: ⋅

⋅

⋅

nlpoe + nmfe = n tw ⋅ (y2 − y 3 ) = 211 ⋅ (1/1329 − 1/1149) >2/9! ⋅

⋅

nlpoe = nmfe >1/:!

lh i

lh i ⋅

⋅

⋅


⋅

⋅

⋅

R 23 = ntw ⋅ (i3 − i2 ) + nlpoe ⋅ i x + nm ⋅ im > ⋅

211 1/: ⋅ (:/6 − 96/83) + ⋅ (− 443/5 ) >−3/3!lX 4711 4711 lK im!>! d m ⋅ (Um − 384) − sm = 3 ⋅ 1 − 443/5 >−443/5! !!!!!!entalpija leda-!u>1pD lh lK ix!>1! entalpija kondenzata (vode)!q>2!cbs-!u>1pD lh R 23 =


{fmlp@fvofu/zv


strana 15

8/:/!Iz!21!)2,y*!lh0t!vla`nog vazduha stawa!2)q>2!cbs-!u>31pD-!y>1/13!lhI3P0!lhTW*!izdvaja se vlaga u te~nom stawu, a zatim se preostali vazduh zagreva izobarski dok se ne postigne relativna vla`nost od ϕ>1/4/!Odrediti maseni protok izdvojene vlage )lh0t* kao i temperaturu vla`nog vazduha nakon zagrevawa. Prikazati procese sa vla`nim vazduhom na Molijerovom!i−y!dijagramu. ta~ka 1: i2>g)u2-!y2*!>!69! ta~ka 2:

lK !)pro~itano sa Molijerovog!i−y!dijagrama* lhTW

(qI3P )3 = ϕ3 ⋅ (qqt )U3 =31 > 2⋅ 3448 >3448!Qb y3>

NI3P q I3P lhI3 P 29 3448 >1/125:! ⋅ = ⋅ 6 lhTW NTW q − q I3P 3: 2 ⋅ 21 − 3448

ta~ka 3: lhI3 P lhTW y4 1/125: = ⋅q = ⋅ 2 ⋅ 21 6 >1/1345!cbs NTW 29 + 1/125: + y4 3: NI3P

y4>y3>1/125:!

(qI3P )4 (qqt )4 =

(qI3P )4 ϕ4

=

1/1345 >1/189!cbs 1/4

⇒

u4!>!)ul*Q>1/189!cbs>52/6pD

koli~ina odstrawene vlage: ⋅

⋅

n x = n tw ⋅ (y 2 − y 3 ) = 21 ⋅ (1/13 − 1/125:) >1/162!

lh t

i ϕ4 i2 ϕ>2 3 u2

2

y y2


{fmlp@fvofu/zv


strana 16

8/21/!Vlaàn vazduh stawa!2)q>286!lQb-!utu>39pD-!uwu>33pD*!i zapreminskog protoka!W>1/6!n40t izobarski se vlaì u adijabatski izolovanoj komori sa!1/13!lh0s pregrejane vodene pare stawa!)q>286 lQb-!u>511pD*!do stawa 2. Odrediti: a) temperaturu vla`nog vazduha stawa!3 b) koli~inu toplote koju bi trebalo odvesti od vla`nog vazduha stawa!2 da bi ga izobarski ohladili do temperature od −21pD!)stawe!4*-!kao i masu leda u jedinici vremena!)lh0i*!koja se tom prilikom izdvoji iz vla`nog vazduha c) skicirati sve procese sa vla`nim vazduhom na Molijerovom i−y dijagramu

ta~ka WU: q qt >3784!Qb


q I3P = ϕ ⋅ q qt > 2⋅ 3784 >3784!Qb

ywu!>

NI3P NTW

⋅

q I3P

>

q − q I3P

lhI3 P 29 3784 ⋅ >1/11:7! 6 3: 2/86 ⋅ 21 − 3784 lhTW

iwu> d qtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/11:7 ⋅ )2/97 ⋅ 33 + 3611* >57/4:

lK lhTW

ta~ka 1: lK lhTW i2 − d q ⋅ u 2

i2>iwu!>57/25! y2 =

2/97 ⋅ u 2 + 3611

(qI3P )2 = N

y2 I3P

NTW

=

lhI3 P 57/25 − 2 ⋅ 39 >1/1182! lhTW 2/97 ⋅ 39 + 3611

⋅q = + y2

1/1182 ⋅ 2/86 ⋅ 21 6 >2:8:!Qb 29 + 1/1182 3:

(q tw )2 = q − (q I3P )2 = 2/86 ⋅ 21 6 − 2:8: > 2/84 ⋅ 21 6 !Qb (ρ tw )2 = ⋅

(q TW )2 S hTW ⋅ U2

=

2/84 ⋅ 21 6 lhTW >3! 398 ⋅ 412 n4

⋅

n tw = ρ tw ⋅ W = 3 ⋅ 1/6 >2!


lhTW t

{fmlp@fvofu/zv


strana 17

ta~ka 2: materijalni bilans vlage za proces vlaèwa vazduha: ⋅

⋅

⋅

⋅

n tw ⋅ y2 + nqq = ntw ⋅ y 3

⇒!

y3 =

⋅

n tw ⋅ y2 + nqq ⋅

n tw y3 =

lhI3 P 2 ⋅ 1/1182 + 1/13 >1/1382! 2 lhTW ⋅

⋅

⋅


⋅

⋅

⋅


⇒!

i3 =

⋅

ntw ⋅ i2 + nqq ⋅ iqq ⋅

ntw lK 2 ⋅ 57/25 + 1/13 ⋅ 4387/6 >222/78! lhTW 2 i 3 − y 3 ⋅ 3611 222/78 − 1/1382 ⋅ 3611 u3!>! !> >52/9pD 2 + 1/1382 ⋅ 2/97 d q + y 3 ⋅ 2/97 i3 =

iqq>!4387/6!

lK -! lh

entalpija pregrejane vodene pare, q>2/86!cbs-!u>511pD

ta~ka 3: q qt >36:/5!Qb )napon pare ~iste vode na!u>−21pD* q I3P = ϕ ⋅ q qt > 2⋅ 36:/5 >36:/5!Qb

y4!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 36:/5 ⋅ >1/111:! 6 lhTW 3: 2/86 ⋅ 21 − 36:/5

i3> dqtw ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ (−21) + 1/111: ⋅ )2/97 ⋅ (−21) + 3611* > !!!>−8/88

lK lhTW

koli~ina izdvojenog leda: ⋅

⋅

nmfe = n tw ⋅ (y 3 − y 4 ) = 2 ⋅ (1/1382 − 1/111:) >1/1373!

lh lh >!:5/43! t i ⋅

⋅

⋅


⋅

⋅

R 34 = n tw ⋅ (i 4 − i 3 ) + nm ⋅ im > 2 ⋅ (− 8/88 − 222/78) + im!>! d m ⋅ (Um − 384) − sm = 3 ⋅ (−21) − 443/5 >−463/5!


:5/43 ⋅ (− 463/5 ) >−239/8!lX 4711

lK !!!!!!entalpija leda-!u>−21pD lh

{fmlp@fvofu/zv


strana 18

i

4387/6 3 2

u2

ϕ>2

uwu

WU y 4

u4 4′

zadatak za ve`bawe:

)8/22/* ⋅

8/22/!Pri izobarskom hla|ewu W >91!n40i vla`nog vazduha stawa!2)q>2!cbs-!u>31pD-!ϕ>1/7* do stawa 3)u>11D) od vla`nog vazduha odvede se 9:1!X toplote. Rashladna povr{ina sastoji se iz 23 plo~a dimenzija 31!Y!41!dn zanemarqive debqine. Odrediti vreme potrebno da se na rashladnim plo~ama stvori sloj leda debqine δ=5!dn. Pretpostaviti ravnomernost debqine leda. )ρM>:11!lh0n4* re{ewe:

τ>351111!t


{fmlp@fvofu/zv


strana 19 ⋅

8/23/!Struja vla`nog vazduha stawa 2)q>2!cbs-!u>41!pD-!ϕ>31&- W >26

n4 ) me{a se sa strujom vla`nog njo

n4 */!Skicirati proces me{awa na Molijerovom i−y njo dijagramu i odrediti temperaturu!)u*!-!apsolutnu vla`nost!)y) i entalpiju!)i*!novonastale me{avine ako se me{awe vr{i: a) adijabatski b) neadijabatski, pri semu se okolini predaje!!4!lX!toplote ⋅

vazduha stawa!3)q>2!cbs-!u>51!pD-!ϕ>91&- W >31

ϕ3

i 3 N u3 O u2

ϕ>2

2 4

ϕ2

y



q I3P = ϕ ⋅ q qt > 1/3 ⋅ 5352>959/3!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 959/3 ⋅ >1/1164! 6 lhTW 3: 2 ⋅ 21 − 959/3

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 41 + 1/1164 ⋅ )2/97 ⋅ 41 + 3611* >54/66!

lK lhTW

q tw = q − q I3P > 2 ⋅ 21 6 − 959/3 >::262/9!Qb

ρtw>

⋅ ⋅ q TW ::262/9 26 lhTW lh = 2/25! - !!! n tw2> ρ tw ⋅ W 2 > 2/25 ⋅ > >1/396! 4 S hTW ⋅ U 398 ⋅ 414 71 t n


{fmlp@fvofu/zv


strana 20



q I3P = ϕ ⋅ q qt > 1/9 ⋅ 8486 >6:11!Qb

y3!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 6:11 ⋅ >1/149:! 6 lhTW 3: 2 ⋅ 21 − 6:11

i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/149: ⋅ )2/97 ⋅ 51 + 3611* >251/25

lK lhTW

q tw = q − q I3P > 2 ⋅ 21 6 − 6:11 >:5211!Qb

ρtw>

⋅ ⋅ q TW :5211 31 lh lhTW n > W !!! = 2/16! ρ ⋅ >1/46! > 3 > 2/16 ⋅ tw3 tw 4 S hTW ⋅ U 398 ⋅ 424 71 t n

ta~ka!N; materijalni bilans vlage za proces me{awa dva vla`na vazduha: ⋅  ⋅  n tw2 ⋅ y 2 + n tw3 ⋅ y 3 =  ntw2 + n tw3  ⋅ y n ⇒!   ⋅

⋅

yn =

⋅

yn =

⋅

n tw2 ⋅ y 2 + ntw3 ⋅ y 3 ⋅

⋅

n tw2 + n tw3

lhI3 P 1/396 ⋅ 1/1164 + 1/46 ⋅ 1/149: >1/1349! 1/396 + 1/46 lhTW ⋅

⋅  ⋅  n tw2 ⋅ i2 + ntw3 ⋅ i 3 =  n tw2 + n tw3  ⋅ in ⇒!   ⋅

⋅

⋅

⋅

R 23 = ∆ I23 + X u23


in =

⋅

ntw2 ⋅ i2 + n tw3 ⋅ i 3 ⋅

⋅

n tw2 + n tw3

1/396 ⋅ 54/66 + 1/46 ⋅ 251/25 lK >:7/8:! 1/396 + 1/46 lhTW in − y n ⋅ 3611 :7/8: − 1/1349 ⋅ 3611 un!>! >46/82pD !> d q + y n ⋅ 2/97 2 + 1/1349 ⋅ 2/97 in =

ta~ka!O; ⋅

⋅

R 23

⋅

⋅

R 23 = ∆ I23 + X u23


⋅

⋅

⋅ ⋅ ⋅ n tw2 ⋅ i2 + n tw3 ⋅ i 3 − R 23  ⋅  =  ntw2 + n tw3  ⋅ io − ntw2 ⋅ i2 − n tw3 ⋅ i3 -!! io = ⋅ ⋅   n tw2 + ntw3

1/396 ⋅ 54/66 + 1/46 ⋅ 251/25 − 4 lK >:3/18! 1/396 + 1/46 lhTW lhI3 P :3/18 − 1/1349 ⋅ 3611 ! uo!>! yo!>yn!>1/1349! >!42/26pD 2 + 1/1349 ⋅ 2/97 lhTW io =


{fmlp@fvofu/zv


strana 21 ⋅

8/24/!Za prostoriju u kojoj se gaje {ampiwoni (slika) priprema se! n 4>6111!lh0i!vla`nog vazduha na slede}i na~in: sve` vazduh stawa!2)q>2!cbs-!u>−21pD-!ϕ>1/9*!!adijabatski se me{a se sa delom iskori{}enog vazduha stawa!5)q>2cbs-!u>33!pD-!ϕ>1/:*!u odnosu!2;3/!Dobijeni vlaàn vazduh stawa N)q>2!cbs) se zagreva u zagreja~u do stawa!3)q>2cbs-!u>36pD) a zatim adijabatski vlaì uvo|ewem suvozasi}ene vodene pare stawa!Q)u>211pD*!do stawa!4)q>2!cbs*!kada vazduh dostiè apsolutnu vla`nost otpadnog vazduha. Tako dobijen vazduh se u komori sa {ampiwonima hladi. Skicirati promene stawa vla`nog vazduha na Molijerovom!i!−y!dijagramu i odrediti: a) temperaturu vla`nog vazduha stawa!N b) temperaturu vla`nog vazduha stawa!4 c) toplotnu snagu zagreja~a vazduha!)lX* d) potro{wu vodene pare u fazi vlaèwa!)lh0t* n′′ 4 komora za vlaèwe

3

5

prostorija sa {ampiwonima

N

5

5

recirkulacioni vazduh

2 sve` vazduh

otpadni vazduh

i 4 3 3786 5 ϕ>2 N

2

y


{fmlp@fvofu/zv


strana 22

ta~ka 1: q qt >36:/5!Qb )napon pare ~iste vode na!u>−21pD* q I3P = ϕ ⋅ q qt > 1/9 ⋅ 36:/5 >318/6!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 318/6 ⋅ >1/1124! 6 lhTW 3: 2 ⋅ 21 − 318/6

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2⋅ (−21) + 1/1124 ⋅ )2/97 ⋅ (−21) + 3611* >−7/88!

lK lhTW



q I3P = ϕ ⋅ q qt > 1/: ⋅ 3754 >3489/8!Qb

y5!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 3489/8 >1/1262! ⋅ 3: 2⋅ 216 − 3489/8 lhTW

i5> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/1262 ⋅ )2/97 ⋅ 33 + 3611* >71/48!

lK lhTW

ta~ka!N; materijalni bilans vlage za proces me{awa dva vla`na vazduha: ⋅

n tw2 ⋅  ⋅  n tw2 ⋅ y2 + n tw 5 ⋅ y 5 =  n tw2 + n tw 5  ⋅ y n   ⋅

⋅

⋅

⇒!

yn =

⋅ y2 + y 5

ntw 5 ⋅

n tw2 ⋅

+2

n tw 5 yn

2 ⋅ 1/1124 + 1/1262 lhI3 P >1/1216 = 3 2 lhTW +2 3 ⋅

⋅

⋅

R 23 = ∆ I23 + X u23

prvi zakon termodinamike za proces me{awa:

⋅

n tw2 ⋅  ⋅  n tw2 ⋅ i2 + n tw 5 ⋅ i 5 =  n tw2 + n tw 5  ⋅ in   ⋅

⋅

⋅

⇒!

in =

⋅ i2 + i 5

n tw 5 ⋅

ntw2 ⋅

+2

n tw 5 in

2 ⋅ (− 7/88) + 71/48 lK 3 = >48/::! 2 lhTW +2 3

un!>!

in − y n ⋅ 3611 48/:: − 1/1216 ⋅ 3611 !> >22/63pD d q + y n ⋅ 2/97 2 + 1/1216 ⋅ 2/97 ⋅

napomena:

n tw5!je oznaka za maseni protok samo recirkulacionog vazduha !!


{fmlp@fvofu/zv


strana 23

ta~ka 2: y3!>!yn>1/1216!

lhI3 P lhTW

i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 36 + 1/1216 ⋅ )2/97 ⋅ 36 + 3611* >62/85

lK lhTW

ta~ka 3: lhI3 P lhTW 6111 ⋅ H4 lhTW n tw4!>! > 4711 >2/49! 2 + y 4 2 + 1/1262 t

y4!>!y5>1/1262!

⋅

⋅

⋅

⋅

⋅

n tw4!>! n tw3!>! n twn> n tw2!,! n tw5 ⋅

n tw2 =

)2*

⋅

ntw 5 3

)3* ⋅

Kombinovawem jedna~ina!)2*!i!)3*!dobija se;! n tw2>1/57!

lh ⋅ lh -! n tw5>1/:3! t t

materijalni bilans vlage za proces vlaèwa vazduha; ⋅ ⋅ ⋅  ⋅   ⋅   ⋅   n tw2 + ntw 5  ⋅ y 3 + n( ( =  ntw2 + ntw 5  ⋅ y 4 !!!!⇒ n( ( =  n tw2 + n tw 5  ⋅ (y 4 − y 3 )      

! n( ( = 2/49 ⋅ (1/1262 − 1/1216) >7/46!/21−4!


lh t

{fmlp@fvofu/zv


strana 24 ⋅

⋅

⋅

prvi zakon termodinamike za proces vlaèwa vazduha:! R 23 = ∆ I23 + X u23 ⋅  ⋅   ntw2+ ntw 5  ⋅ i3 + n( (⋅i# ⋅ ⋅  ⋅   ⋅    ntw2+ ntw 5  ⋅ i3 + n#⋅i# =  ntw2+ ntw 5  ⋅ i4 !!!!⇒!!! i4 =  ⋅ ⋅     ntw2+ ntw 5

i4 =

2/49 ⋅ 62/85 + 7/46 ⋅ 21 −4 ⋅ 3786 lK >75/16! 2/49 lhTW

u4!>!

i 4 − y 4 ⋅ 3611 75/16 − 1/1262 ⋅ 3611 >36/69pD !> d q + y 4 ⋅ 2/97 2 + 1/1262 ⋅ 2/97

napomena: i′′>!3786!

lK -!entalpija suvozasi}ene vodene pare stawa!Q)u>211pD* lh ⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!!! R 23 = ∆ I23 + X u23 R {bh = (Htw2 + Htw 5 ) ⋅ (i3 − in ) > 2/49 ⋅ (62/85 − 48/:: ) >29/:9!lX

zadatak za ve`bawe:

)8/25/*

8/25/ n2>3!)2,y*!lh0t!vla`nog vazduha stawa!2)q>2!cbs-!y>1/116!lh0lhTW) adijabatski se me{a sa n3>4 )2,y*!lh0t!vla`nog vazduha stawa!3)q>2!cbs-!y>1/17!lh0lhTW-!u>61!pD*/!Ne koriste}i i−y dijagrama odrediti: a) temperaturu vla`nog vazduha 1 tako da vazduh dobijen me{awem vazduha 1 i 2 bude zasi}en b) temperaturu dobijenog zasi}enog vla`nog vazduha c) temperaturu vla`nog vazduha 1 tako da vazduh dobijen me{awem vazduha 1 i 2 bude zasi}en za slu~aj da je me{awe neadijabatsko uz toplotne gubitke u okolinu od Rp>5!lX d) skicirati sve procese na i−y dijagramu a) u2>23/6pD b) uN>46/5pD c) u2′>25/8pD


{fmlp@fvofu/zv


strana 25

8/26/ Za klimatizaciju nekog objekta potrebno je obezbediti vlaàn vazduh stawa ⋅

4)q>1/23!NQb-!u>33pD-!ϕ>61&-! W >1/5!n40t*/!U tu svrhu koristi se ure|aj koji se sastoji iz filtera, hladwaka, zagreja~a vazduha i ventilatora-duvaqke, (slika). Snaga ventilatora koji adijabatski sabija vazduh sa pritiska!q3)>q2>qp*!na pritisak!q4!je 2/5!lX/!Stawe okolnog nezasi}enog vla`nog vazduha je P)qp>1/2!NQb-!up>41pD-!ϕ>61&-*/!Prikazati proces pripreme vla`nog vazduha na Molijerovom!i!−y dijagramu i odrediti: b* koli~inu izdvojenog kondenzata!)lh0i* c* toplotnu snagu hladwaka vazduha,!Rimb!)lX* d* toplotnu snagu zagreja~a vazduha,!R{bh!)lX* 4 W4 Rimb h l a d w a k

f i l t e r

1

,R{bh

2

z a g r e j a ~

v e n t i l a t o r

3

X

kondenzat i ϕ>2-!q>2/3!cbs

4

i

1

y

3

ϕ>2-!q>2!cbs

2

y


{fmlp@fvofu/zv


strana 26

ta~ka 3: )napon pare ~iste vode na!u>33pD*

q qt >3754!Qb

q I3P = ϕ ⋅ q qt > 1/6 ⋅ 3754 >2432/6!Qb

y4!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 2432/6 ⋅ >1/117:! 6 lhTW 3: 2/3 ⋅ 21 − 2432/6

i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 33 + 1/117: ⋅ )2/97 ⋅ 33 + 3611* >4:/64!

lK lhTW

q tw = q − q I3P > 2/3 ⋅ 21 6 − 2432/6 >229789/6!Qb

ρtw>

⋅ q TW 229789/6 lh lhTW > > 2/51 ⋅ 1/5 >1/67! !!!H = 2/51! ρ ⋅ W > tw tw 4 S hTW ⋅ U 398 ⋅ 3:6 t n

ta~ka 2: y3>y4>1/117:!

lhI3 P lhTW ⋅

⋅

⋅

prvi zakon termodinamike za proces u ventilatoru:!!!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

⋅

X u23 = ntw ⋅ (i 3 − i 4 )

⇒

i3 = i4 +

X u23 ⋅

> 4:/64 −

n tw

2/5 lK >48/14! 1/67 lhTW

ta~ka 1: y2>y3>1/117:! q I3P =

q qt =

lhI3 P lhTW

y NTW +y NI3P

q I3P

=

ϕ p u2!>9/6 D

⋅ q2 >

1/117: ⋅ 2 ⋅ 21 6 = 21::/5!Qb 29 + 1/117: 3:

21::/5 = 21::/5!Qb 2 )temperatura kqu~awa vode na!q>21::/5!cbs*

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 9/6 + 1/117: ⋅ )2/97 ⋅ 9/6 + 3611* >36/97

lK lhTW



q I3P = ϕ ⋅ q qt > 1/6 ⋅ 5352>3231/6!Qb

yp!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 3231/6 >1/1245! ⋅ 3: 2 ⋅ 21 6 − 3231/6 lhTW

ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 41 + 1/1245 ⋅ )2/97 ⋅ 41 + 3611* >75/36!


lK lhTW

{fmlp@fvofu/zv


strana 27 ⋅

koli~ina izdvojenog kondenzata:

⋅

X l = n tw ⋅ (y 2 − y 3 )

⋅

X l = 1/67 ⋅ (1/1245 − 1/117:) ⋅ 4711 >24/21!

lh i ⋅

⋅

⋅

prvi zakon termodinamike za proces u hladwaku vazduha:!!!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

⋅

R imb = n tw ⋅ (i2 − i 1 ) + X l ⋅ il > 1/67 ⋅ (36/97 − 75/36 ) + napomena:

24/21 ⋅ 46/64 >−32/48!lX 4711

il!−!entalpija kondenzata (voda!q>2!cbs-!u>9/6pD* ⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

R {bh = n tw ⋅ (i 3 − i2 ) > 1/67 ⋅ (48/14 − 36/97 ) >7/37!lX 8/27/ Postrojewe za delimi~no su{ewe vazduha sastoji se od vodom hla|enog klipnog kompresora i ⋅

hladwaka za vlaàn vazduh (slika). U klipnom kompresoru se sabija!! n ww>1/38!)2,y*!lh0t!vla`nog vazduha stawa!2)q2>1/2!NQb-!u2>31pD-!ϕ2>1/9*!do stawa!3)q3?q2-!u3>56pD-!ϕ3>2), a potom se uz izdvajawe te~ne faze vlaàn vazduh stawa!3!izobarski hladi do stawa!4)u4>u2). Ukupan toplotni fluks sa vla`nog vazduha na rashladnu vodu u toku procesa sabijawa i izobarskog hla|ewa vla`nog vazduha iznosi!R>RI2,RI3>24!lX/!Odrediti pritisak vla`nog vazduha na kraju procesa sabijawa, koli~inu izdvojenog kondenzata kao i pogonsku snagu za pogon klipnog kompresora. RI3 4

3

X

2 vlaàn vazduh

vlaàn vazduh

kondenzat

RI2


{fmlp@fvofu/zv


strana 28


q qt >3448!Qb

q I3P = ϕ ⋅ q qt > 1/9 ⋅ 3448 >297:/7!Qb

y2!>

NI3P NTW

⋅

q I3P q2 − q I3P

>

lhI3 P 29 297:/7 ⋅ >1/1229! lhTW 3: 2 ⋅ 21 6 − 297:/7

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1229 ⋅ )2/97 ⋅ 31 + 3611* >61/13!

lK lhTW

ta~ka 2: lhI3 P lhTW q qt >:695!Qb )napon pare ~iste vode na!u>56pD*

y3>y2>1/1229!

q I3P = ϕ ⋅ q qt > 2⋅ :695 >:695!Qb

NI3P N tw q3!>! y3

+ y3 ⋅ q I3P

29 + 1/1229 = 3: ⋅ :695 >!624821!Qb!>!6/248!cbs 1/1229

i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 56 + 1/1229 ⋅ )2/97 ⋅ 56 + 3611* >86/5:!

lK lhTW



q I3P = ϕ ⋅ q qt > 2⋅ 3448 >3448!Qb

y4!>

NI3P NTW

⋅

q I3P q 4 − q I3P

>

lhI3 P 29 3448 ⋅ >1/1139! lhTW 3: 6/248 ⋅ 21 6 − 3448

i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1139 ⋅ )2/97 ⋅ 31 + 3611* >38/21!

lK lhTW

koli~ina izdvojenog kondenzata: ⋅

⋅

X = ntw ⋅ (y 3 − y 4 ) = 1/38 ⋅ (1/1229 − 1/1139) > 3/54 ⋅ 21 −4 !


lh t

{fmlp@fvofu/zv


strana 29


⋅

⋅

⋅

⋅

⋅

⋅

! R 23 = ∆ I23 + X u23

ograni~enom isprekidanom konturom: ⋅

⋅

X !>! − ntw ⋅ i 4 − X x ⋅ i x + n tw ⋅ i2 − R I2 − R I3 ⋅

X >! − 1/38 ⋅ 38/21 − 3/54 ⋅ 21 −4 ⋅ 299/5 + 1/38 ⋅ 61/13 − 24 >−8/38!lX ix>299/5!

napomena:

lK lh

entalpija vode!!q>2!cbs-!u>56pD

)8/28/*

zadatak za ve`bawe:

8/28/ 611!lh0i!vla`nog vazduha stawa!2)q>2!cbs-!u>3pD!ϕ>1/9*!me{a se izobarski sa!611!lh0i!vla`nog vazduha stawa!3)q>2!cbs-!u>57pD-!ϕ>1/8). Zatim se kondenzat koji je nastao me{awem izdvaja, a preostali vazduh zagreva do!81pD. Nakon zagrevawa vazduhu se dodaje vodena para ~ija entalpija iznosi 3111!lK0lh!i vlaèwe se obavqa do postizawa stawa zasi}ewa. Skicirati procese sa vla`nim vazduhom na Molijerovom i!−y! dijagramu i odrediti: a) apsolutnu vla`nost me{avine )y* kada kondenzat jo{ nije izdvojen (ra~unskim putem) b) maseni protok odvedenog kondenzata!)lh0i* c) toplotnu snagu greja~a!)lX* d) maseni protok vodene pare koja se dodaje u ciqu vlaèwa!)lh0i* za stavke b), c) i d)!moè se koristiti Molijerov dijagram za vlaàn vazduh lhI3 P a) yn!>!1/1355! lhTW lh b) nlpoefo{bu!>!2/5! i c) R45!>!23/2!lX lh d) nwpefob!qbsb!>!54/8! i

i 5

3

6

ϕ>2

4 N

2

y

ix>3111!lK


{fmlp@fvofu/zv


strana 30

DRUGI VLA@NI GASOVI 8/29/!Me{avina vodonika (idealan gas) i vodene pare (idealan gas) ima temperaturu!u>41pD-!relativnu vla`nost!ϕ>:1&!i pritisak!q>311!lQb/!Za navedenu gasnu me{avinu odrediti: a) apsolutnu vla`nost!)y*!i specifi~nu entalpiju!)i*!vla`nog vodonika b) masene udele vodonika i vodene pare u vla`nom vodoniku a) )napon pare ~iste vode na!u>41pD*

q qt >5352!Qb

q I3P = ϕ ⋅ q qt > 1/: ⋅ 5352>4927/:!Qb

y2!>

NI3P NI3

⋅

q I3P q2 − q I3P

>

lhI3 P 29 4927/: ⋅ >1/2862! 6 lhTW 3 3 ⋅ 21 − 4927/:

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 25/66 ⋅ 41 + 1/2862 ⋅ )2/97 ⋅ 41 + 3611* >995/13

lK lhI3

b) nI3P hI3P =

nI3P nI3P + nI3

=

nI3 nI3P nI3

+

nI3

=

y 1/2862 = >1/26 y + 2 1/2862 + 2

nI3

nI3 hI3 =

nI3 nI3P + nI3

=

nI3 nI3P nI3


+

nI3

=

2 2 = >1/96 y + 2 1/2862 + 2

nI3

{fmlp@fvofu/zv


strana 31

8/2:/!U vertikalnom cilindru sa klipom, po~etne zapremine!W>1/2!n4 nalazi se, pri stalnom pritisku q>3!cbs!sme{a ugqen−dioksida (idealan gas) i pregrejane vodene pare. Maseni udeo vodene pare u sme{i je! hI3P >1/2-!a po~etna temperatura!sme{e!:1pD/!Odrediti koli~inu toplote koju treba odvesti od vla`nog ugqen-dioksida da bi zapo~ela kondenzacija vodene pare. ta~ka 1: y2!>!

hI3P 2 − hI3P

=

lhI3 P 1/2 >1/2222! 2 − 1/2 lhDP3

i2> dqDP3 u + y ⋅ )2/97 ⋅ u + 3611* > 1/96 ⋅ :1 + 1/2222 ⋅ )2/97 ⋅ :1 + 3611* >483/96 qI3P =

y2 NI3P NDP3

!

⋅q = + y2

lK lhI3

1/2222 ⋅ 3 ⋅ 216 > 1/4 ⋅ 216 Qb!>1/41!cbs 29 + 1/2222 55

qDP3 = q − qI3P >3!−!1/4!>2/8!cbs

ρDP3 =

qDP3 ShDP3 ⋅U

=

lhDP3 2/8 ⋅ 216 >3/59 29: ⋅ 474 n4

nDP3 = ρDP3 ⋅ W = 3/59 ⋅ 1/2 >1/359!lh

ta~ka 2: y3!>!y2!>!1/2222! qqt3 =

qI3P ϕ3

=

lhI3 P !! lhDP3

qI3P = dpotu > 1/4 ⋅ 216 Qb!>1/41!cbs

1/4 >1/4!cbs 2

u3!>!)ulr*q>1/4!cbs!≈!7:pD

i3> dqDP3 u + y ⋅ )2/97 ⋅ u + 3611* > 1/96 ⋅ 7: + 1/2222 ⋅ )2/97 ⋅ 7: + 3611* >461/77

lK lhI3

prvi zakon termodinamike za proces u cilindru: R23!>!∆V23!,!X23

⇒

R23!>!V3!−!V2!,!q!/)W3!.−!W2*!

R23!>!I3!−!I2!

⇒

R23!>! nDP3 ⋅ (i3 − i2)

R23!>! 1/359 ⋅ (461/77 − 483/96 ) >−6/6!lX


{fmlp@fvofu/zv


strana 32

8/31/!U toplotno izolovanoj komori me{aju se dva toka razli~itih vla`nih gasova zadatih ⋅

termodinami~kih stawa : zasi}en vlaàn kiseonik!)P3*!stawa!2)q>1/6!NQb-!U>464!L-! n2 >3!)2,y*!lh0t*!i ⋅

vlaàn metan!)DI5) stawa!3)q>1/4!NQb-!U>3:4!L-!ϕ>1/5-! n3 >4!)2,y*!lh0t*/!Promene kineti~ke i potencijalne energije gasnih tokova su zanemarqive. Odrediti temperaturu vla`ne gasne sme{e koja izlazi iz komore. 1. vlaàn kiseonik M. me{avina vla`nog kiseonika i vla`nog metana 2. vlaàn metan

ta~ka 1: q qt >58471!Qb )napon pare ~iste vode na!u>91pD* q I3P = ϕ ⋅ q qt > 2⋅ 58471 >58471!Qb

y2!>

NI3P NP3

⋅

qI3P q2 − qI3P

>

lhI3P 29 58471 ⋅ >1/169:! 43 6 ⋅ 216 − 58471 lhP3

i2> dqP3 ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 1/:2 ⋅ 91 + 1/169: ⋅ )2/97 ⋅ 91 + 3611* >339/92! ⋅

nP3 >3

lK lhP3

lh t



q I3P = ϕ ⋅ q qt > 1/5 ⋅ 3448 >:45/9!Qb

y3!>

NI3P NDI5

⋅

q I3P q 3 − q I3P

>

lhI3 P 29 :45/9 ⋅ >1/1146! lhDI5 27 4 ⋅ 21 6 − :45/9

i3> dqDI5 ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 3/45 ⋅ 31 + 1/1146 ⋅ )2/97 ⋅ 31 + 3611* >66/79! ⋅

nDI5 >4

lK lhDI5

lh t


{fmlp@fvofu/zv


strana 33

ta~ka!N; materijalni bilans vlage za proces me{awa dva vla`na gasa: ⋅ ⋅ ⋅  ⋅  nP3 ⋅ y2 + nDI5 ⋅ y3 =  nP3 + nDI5  ⋅ yn  

⋅

⇒!

yn =

⋅

nP3 ⋅ y2 + nDI5 ⋅ y3 ⋅

⋅

nP3 + nDI5

lhI3 P 3 ⋅ 1/169: + 4 ⋅ 1/1146 >1/1368! 3+4 lh(P 3 + DI 5 )

yn =

⋅

⋅  ⋅ nP3 ⋅ i2 + nDI5 ⋅ i 3 =  nP3 + nDI5  ⋅

in =

⋅

⋅

⋅

⋅

nP3 ⋅ i2 + nDI5 ⋅ i 3   ⋅ in !!!!!!⇒!!!!! in = ⋅ ⋅  nP3 + nDI5

3 ⋅ 339/92 + 4 ⋅ 66/79 lK >235/:4! 3+4 lh(P 3 + DI 5 ) ⋅

hP3 =

nP3 ⋅

⋅

nP3 + nDI5

⋅

3 >1/5 = 3+4

hDI5 =

nDI5 ⋅

⋅

nP3 + nDI5

d qn = h P3 ⋅ d qP3 + h DI5 ⋅ d qDI5 = 1/5 ⋅ 1/:2 + 1/7 ⋅ 3/45 >2/88!

un!>!

⋅

R 23 = ∆ I23 + X u23

prvi zakon termodinamike za proces me{awa:

=

4 >1/7 3+4

lK lh(P 3 + DI 5 )

in − y n ⋅ 3611 235/:4 − 1/1368 ⋅ 3611 !> >44/5pD>417/5!L d qn + y n ⋅ 2/97 2/88 + 1/1368 ⋅ 2/97


{fmlp@fvofu/zv


strana 34

8/32/!U toplotno izolovanom kanalu izobarski se me{aju tok kiseonika stawa!2)q>1/3!NQb-!U>391!L⋅

⋅

W >1/657!n40t*!i tok pregrejane vodene pare stawa!Q)q>1/3!NQb-!U>664!L-! nq >1/17!lh0t*/!Nastali vlaàn kiseonik stawa!3, biva potom u vodom hla|enom klipnom kompresoru, pogonske snage!Q>76!lXsabijan do stawa!4)q>1/4!NQb-!ϕ>1/83*/!Odrediti toplotni protok sa vla`nog kiseonika na vodu za hla|ewe kompresora i prikazati sve procese u!!i−y!koordinatnom sistemu. nqq 2

3

4

ta~ka!Q; iqq!>!4141!!

lK lh

)q>!3!cbs-!u>391pD*

ta~ka 1: y2!>1!

lhI3 P lhP3

i2!>! d qP3 ⋅ u 2 + y2 ⋅ )2/97 ⋅ u 2 + 3611* > 1/:2 ⋅ 8 + 1 ⋅ 2/97 ⋅ 8 + 3611* >!7/48! ρ P3 =

lK lhP3

⋅ ⋅ kg q P3 lhP3 3 ⋅ 21 6 = >3/86! !!!!! n P3 = ρ P3 ⋅ W = 3/86 ⋅ 1/657 >2/6 4 S hP3 U2 371 ⋅ 391 n s

ta~ka 2: materijalni bilans vlage za proces vlaèwa kiseonika!)2−3*; ⋅

⋅

⋅

⋅

nP3 ⋅ y2 + nqq = nP3 ⋅ y3 !!!!⇒

y3 =

⋅

nP3 ⋅ y2 + nqq ⋅

=

nP3

lhI3 P 1/17 >1/15! lhP3 2/6 ⋅

⋅

⋅

prvi zakon termodinamike za proces vlaèwa kiseonika:!!!!! R 23 = ∆ I23 + X u23 ⋅

⋅

⋅

nP3 ⋅ i2 + nqq ⋅ iqq = nP3 ⋅ i3

⋅

⇒!

i3 =

⋅

nP3 ⋅ i2 + nqq ⋅ iqq ⋅

nP3 i3 =

2/6 ⋅ 7/48 + 1/17 ⋅ 4141 lK >238/68! 2/6 lhP 3


{fmlp@fvofu/zv


strana 35

ta~ka 3: y4!>!y3!>1/15!

y4 1/15 ⋅ q4 = ⋅ 4 ⋅ 21 6 !>!2::28!Qb NI3P 29 + 1/15 + y4 43 NP3

q I3P =

q qt =

lhI3 P lhP3

q I3P ϕ

=

2::28 >38774Qb!≈1/39!cbs 1/83

u4!>!)ulr*Q>1/39!cbs>78/6pD

i4!>! d qP3 ⋅ u 4 + y 4 ⋅ )2/97 ⋅ u 4 + 3611* > 1/:2 ⋅ 78/6 + 1/15 ⋅ )2/97 ⋅ 78/6 + 3611* !> >277/56! ⋅

prvi zakon termodinamike za proces u kompresoru:! ⋅

⋅

⋅

lK lhP3

⋅

R 34 = ∆ I34 + X u34

⋅

R 34 = nP3 ⋅ (i 4 − i 3 ) + X u34 = 2/6 ⋅ (277/56 − 238/68) − 76 >−7/79!lX i i

ϕ>2-!q>4!cbs

4

y 3

4141

ϕ>2-!q>3!cbs

2

y

zadatak za ve`bawe:

)8/33/*

8/33/!Vlaàn azot, masenog protoka!1/5!lh0t-!stawa!2)q2>4!cbs-!u2>55pD-!ϕ2>1/:*!izobarski se ohladi do temperature od!1pD!)stawe 2*-!pri ~emu se od azota odvede!47!lX!toplote. Odrediti masuformiranog kondenzata i masu formiranog leda ako je proces trajao!2!sat. re{ewe:!

nlpoefo{bu!>!23/:7!lh


nmfe!>!21/9!lh

{fmlp@fvofu/zv


strana 36

8/34/!âsovni kapacitet teorijske tunelske teorijske su{are iznosi!261!lh suvih banana. Vla`nost sirovih banana (maseni udeo vlage) je!z2>81!nbt&!a suvih!z3>23!nbt&/!Temperatura vazduha na izlazu iz su{are kf!51pD a maksimalna temperatura vazduha u su{ari!96pD/!Atmosferski vazduh ima temperaturu od!29pD!i ta~ku rose!23pD/!Skicirati promene stawa vla`nog vazduha na Molijerovom!i −y!dijagramu i odrediti potro{wu grejne pare u zagreja~u vazduha (suvozasi}ena vodena para) ako joj je temperatura za!31!L!vi{a od maksimalne temperature vazduha u su{ari (smatrati da je kondenzat grejne pare na izlazu iz zagreja~a vazduha neprehla|en). Sve promene stawa vla`nog vazduha su izobarske na q>2!cbs/ nwn

1

vazduh

2

zagreja~ vazduha

npn komora za su{ewe materijala

3

grejna para

i 2 u2 3 u3

ϕ>2

1

up us

S

y napomena: Teorijski uslovi su{ewa (adijabatska su{ara) podrazumevaju: 1−2; y>dpotu 2−3; i>dpotu


{fmlp@fvofu/zv


strana 37

ub•lb!S; yS>!g)uS-!ϕS>2*!>!1/1199!

lhI3 P lhTW

ub•lb!1; yp!>yS!>!1/1199!

lhI3 P lhTW

ip!>!g)up-!yp*!>!51/4!

lK lhTW

ub•lb!2; y2!>!y1!>!yS!>!1/1199!

lhI3 P lhTW

i2!>!g)u2-!y2*!>!219/5!

lK lhTW

ub•lb!3; lhI3 P lhTW

i3!>!i2>219/5!

lK lhTW

napomena:

Sve vrednosti pro~itane sa Molijerovog!i!−!y!dijagrama

y3!>!g)u3-!i3*!>!1/1376!

materijalni bilans vlage za proces su{ewa banana: ⋅

n tw ⋅ (y 3 − y 2 ) > n pn ⋅ ⋅

n tw =

z2 − z 3 2 − z2

⇒

⋅

n tw = npn ⋅

z2 − z 3 2 ⋅ 2 − z 2 y 3 − y2

261 1/8 − 1/23 2 lh ⋅ ⋅ >5/66 4711 2 − 1/8 1/1376 − 1/1199 t

⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!! R 12 = ∆ I12 + X u12 ⋅

⋅

⋅

⋅

n tw ⋅ i p + nq ⋅ i( ( = n tw ⋅ i2 + nq ⋅ i( ⋅

nq =

⇒

⋅

n tw ⋅ (i2 − i p ) nq = i( (−i( ⋅

5/66 ⋅ (219/5 − 51/4 ) lh >1/25 3354 t

i′′!−!i′!>!s!>3354!

lK lh


toplota kondenzacije vodene pare na!u>216pD

{fmlp@fvofu/zv


strana 38

8/35/!U teorijskoj su{ari su{i se, pri!q>2!cbs-!61!lh0i!kva{~eve biomase koja sadrì!z2>81!nbt& vlage, pri ~emu se dobija suvi kvasac sa!z3>8!nbt&!vlage. Na ulazu u zagreja~ stawe vazduha odre|eno je temperaturom suvog termometra i temperaturom vla`nog termometra!1)utu>27pD-!uwu>21pD*/!Stawe otpadnog vazduha odre|eno je entalpijom i relativnom vla`nosti vazduha!3)i>:1!lK0lhTW-!ϕ>1/7). Skicirati promene stawa vla`nog vazduha na Molijerovom!i!−y!dijagramu i odrediti: b* potro{wu suvog vazduha u su{ari!) no4 0t* c* toplotnu snagu zagreja~a vazduha!)lX* d* koliko bi se toplote moglo u{tedeti hla|ewem otpadnog vazduha do stawa zasi}ewa i rekuperativnim kori{}ewem oslobo|ene toplote za zagrevawe sveèg vazduha u predgreja~u!)lX* i

i3

2 ϕ3

ϕ>2

3 up uwu

1 WU

y ta~ka 0: ip>3:/6

lK lhTW

yp>1/1168

lhI3 P lhTW

y3>1/1326

lhI3 P lhTW

ta~ka 2: i3>:1

lK lhTW

ta~ka 1: i2>i3>:1 napomena:

lK lhTW

y2>yp>1/1168

lhI3 P lhTW

Sve vrednosti pro~itane sa Molijerovog!i!−!y!dijagrama


{fmlp@fvofu/zv


strana 39

b* materijalni bilans vlage za proces su{ewa kva{~eve biomase: ⋅

n tw ⋅ (y 3 − y 2 ) > n wn ⋅ ⋅

ntw =

z2 − z 3 2− z3

z2 − z 3 2 ⋅ 2 − z 3 y 3 − y2

⋅

⇒

n tw = n wn ⋅

61 1/8 − 1/18 2 lh >1/7 ⋅ ⋅ 4711 2 − 1/18 1/1326 − 1/1168 t

⋅

⋅

W o tw = n tw ⋅

n4 33/5 33/5 >1/57! o > 1/7 ⋅ 3: t N tw

c* ⋅

⋅

⋅


⋅

R {bh = n tw ⋅ (i2 − i p ) = 1/7 ⋅ (:1 − 3:/6 ) >47/3:!lX d*

n wn

n pn X

Htw

1

C

2

3

Rsfl B

prvi zakon termodinamike za proces u otvorenom sistemu ograni~enom ⋅

⋅

⋅

isprekidanom konturom:!! R 23 = ∆ I23 + X u23 ⋅

⋅

R sfl > − ntw ⋅ (i3 − i B ) > −1/7 ⋅ (:1 − 92/3) >−6/39!lX napomena:

iB>g)yB>y3-!ϕ>2*>92/3!


lK )!Molijerov!i−y!dijagram) lhTW

{fmlp@fvofu/zv


strana 40

8/36/!Jednostepena, teorijska su{ara, radi sa vazduhom kao agensom za su{ewe po zatvorenom ciklusu (slika) na pritisku!q>:1!lQb>jefn/!Nakon zagrevawa vazduha )2−3*- wegovog prolaska kroz komoru za su{ewe )3−4*-!te hla|ewa )4−5*- u predajniku toplote, u kome se kondenzuje vodena para, ulazi zasi}en vlaàn vazduh stawa!5)U>424!L*-!a napu{ta ga ohla|eni zasi}en vlaàn vazduh i izdvojeni kondenzat temperature!U2>3:4!L. Maseni protok odvedenog kondenzata je!X>1/14!lh0t. Toplotna snaga zagreja~a vazduha je!R{bh>:6!lX/!Skicirati promene stawa vla`nog vazduha na Molijerovom i!−!y!dijagramu i odrediti potreban maseni protok suvog vazduha i relativnu vla`nost!)ϕ4*!do koje se, su{ewem vla`nog materijala, ovlaì vazduh. vlaàn materijal zagreja~ 2

3

komora za su{ewe

4

osu{en materijal

5 !!!L hladwak kondenzat i 3 4

5

ϕ>2

2

y


{fmlp@fvofu/zv


strana 41


q qt >3448!Qb

q I3P = ϕ ⋅ q qt > 2⋅ 3448 >3448!Qb

y2!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 3448 ⋅ >1/1276! 6 lhTW 3: 1/: ⋅ 21 − 3448

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1276 ⋅ )2/97 ⋅ 31 + 3611* >72/97!

lK lhTW


q qt >8486!Qb

q I3P = ϕ ⋅ q qt > 2⋅ 8486 >8486!Qb

y5!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 8486 ⋅ >1/1665! lhTW 3: 1/: ⋅ 21 6 − 8486

i5> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/1665 ⋅ )2/97 ⋅ 51 + 3611* >293/73! ⋅

X

⋅

⋅

!>! n tw/!)y4!−!y3*!>! n tw/!)y5!−!y2*!

⋅

⇒!

⋅

lK lhTW

⋅

X n tw!> y 5 − y2

1/14 lh >!1/88! 1/1665 − 1/1276 t

n tw!> ta~ka 2:

y3!>!y2!>!1/1276!

lhI3 P lhTW

i3!>!@ ⋅

⋅

⋅


⋅

⋅

R {bh = n tw ⋅ (i 3 − i2 )

i 3 = i2 +

R {bh ⋅

> 72/97 +

n tw

:6 lK >296/35! lhTW 1/88

ta~ka 3: lhI3 P lK i4>i3>296/35! lhTW lhTW i 4 − y 4 ⋅ 3611 296/35 − 1/1665 ⋅ 3611 >53/48pD u4!>! !> d q + y 4 ⋅ 2/97 2 + 1/1665 ⋅ 2/97

y4>y5>!1/1665!

q I3P =

y4 NI3P N tw

ϕ4!>!

q I3P

(qqt )U4

⋅q> + y4

!>

1/1665 ⋅ 1/: ⋅ 21 6 >8485/9!Qb 29 + 1/1665 3:

8485/9 >!1/99 9472


{fmlp@fvofu/zv


strana 42

8/37/!U dvostepenu teorijsku su{nicu uvodi se vlaàn vazduh zapreminskog protoka!Wp>1/94!n40t!i stawa ⋅

1)q>1/2!NQb-!u>25pD-!ϕ>1/5). Nakon zagrevawa vazduha u zagreja~u toplotne snage! R J>62/:!lX!)do stawa 2) vazduh se uvodi u prvi stepen su{are odakle izlazi sa temperaturom!u>41pD!)stawe!3). Ovaj vazduh se ⋅

zatim zagreva u drugom zagreja~u toplotne snage! R JJ>35!lX!)do stawa!4), te uvodi u drugi stepen su{are koji napu{ta sa relativnom vla`no{}u!ϕ>1/9!)stawe!5*/!Ako se zanemare padovi pritiska odrediti masu vlage uklowenu iz vla`nog materijala u prvom i drugom stepenu su{ewa (posebno za svaki stepen) za vreme od τ=1 sat. Skicirati promene stawa vla`nog vazduha na!i!−y!dijagramu. 4 i 2 5 3

ϕ>2

1

y ta~ka 0: q qt >26:8!Qb


q I3P = ϕ ⋅ q qt > 1/5 ⋅ 26:8 >749/9!Qb

yp!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 749/9 ⋅ >1/1151! 6 lhTW 3: 2 ⋅ 21 − 749/9

ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 25 + 1/1151 ⋅ )2/97 ⋅ 25 + 3611* >35/2!

lK lhTW

q tw = q − q I3P > 2 ⋅ 21 6 − 749/9 >::472/3!Qb

ρtw>

⋅ ⋅ q TW ::472/3 lh lhTW = 2/32! !!! n > tw = ρ tw ⋅ W > 2/32 ⋅ 1/94 >2! 4 S hTW ⋅ U 398 ⋅ 398 t n


{fmlp@fvofu/zv


strana 43

ta~ka 1: y2!>!yp>1/1151!

lhI3 P lhTW

i3!>!@ ⋅

⋅

⋅

⋅

⋅

prvi zakon termodinamike za proces u 1. zagreja~u vazduha:! R 12 = ∆ I12 + X u12 R J = n tw ⋅ (i2 − i 1 ) ⋅

i2 = i p +

RJ ⋅

> 35/2 +

n tw

62/: lK >87! lhTW 2

ta~ka 2: lK lhTW i . dq ⋅ u

i3!>!i2!>!87! y3!>!

2/97 ⋅ u + 3611

>!

lhI3 P 87 . 2 ⋅ 41 >!1/129! 2/97 ⋅ 41 + 3611 lhTW

ta~ka 3: y4!>!y3!>!1/129!

lhI3 P lhTW

i4!>!@ ⋅

⋅

⋅

prvi zakon termodinamike za proces u 2. zagreja~u vazduha: R 34 = ∆ I34 + X u34 ⋅

⋅

⋅

R JJ = n tw ⋅ (i 4 − i 3 )

i 4 = i3 +

R JJ ⋅

n tw

> 87 +

35 lK >211! lhTW 2

ta~ka 4: i5!>!i4!>!211!

lK lhTW

y5!>!g)ϕ5-!i5*!>!1/1374!

lhI3 P ! )i!−!y!dijagram* lhTW

⋅

X2!>! n tw!)y3!−!y2* ⋅ τ >! 2 ⋅ (1/129 − 1/115 ) ⋅ 4711 >61/5!lh ⋅

X3!>! n tw!)y5!−y4* ⋅ τ >! 2 ⋅ (1/1374 − 1/129 ) ⋅ 4711 >3:/:!lh


{fmlp@fvofu/zv


strana 44

8/38/!U teorijskoj konvektivnoj su{ari su{i se neki materijal koji ne sme biti izloèn temperaturi vi{oj od!91pD/!Maksimalna relativna vla`nost, koju dostiè vazduh pri svakom prolasku preko vla`nog materijala, iznosi!ϕnby>:1%. Odrediti koli~inu vlage, koja se u toku jednog sata odstrani iz materijala, ⋅

ako je stawe vla`nog vazduha na ulazu u su{aru odre|eno sa!P)q>2cbs-!u>31pD-!ϕ>1/6-! n ww>1/6!lh0t*; a) u slu~aju dvostepene teorijske su{are b) u slu~aju teorijske su{are sa beskona~no mnogo stepeni su{ewa (naizmeni~no povezanih komora za su{ewe i zagreja~a vazduha) Smatrati da se tokom svih proces pritisak vazduha u su{ari ne mewa. a) i

4

2 u2>u4

ϕ3>ϕ5>ϕnby 5 3

ϕ>2

1

y ta~ka 0: q qt >3448!Qb

)napon pare ~iste vode na u>31pD*

q I3P = ϕ ⋅ q qt > 1/6 ⋅ 3448 >2279/6!Qb

yp!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 2279/6 >!1/1184! ⋅ lhTW 3: 2 ⋅ 21 6 − 2279/6

ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1184 ⋅ )2/97 ⋅ 31 + 3611* >49/63! ⋅

n tw =

lK lhTW

⋅

1/6 n ww lh > >1/5:7! t 2 + y 1 2 + 1/1184


{fmlp@fvofu/zv


strana 45

ta~ka 1: y2>yp>!1/1184!

lhI3 P lhTW

i2> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 91 + 1/1184 ⋅ )2/97 ⋅ 91 + 3611* >::/45!

lK lhTW

ta~ka 2: i3>i2!>::/45

lK lhTW

y3>!g)ϕ3-!i3*>1/1376!

lhI3 P ! lhTW

)i!−!y!dijagram*

ta~ka 3: y4>y3>!1/1376!

lhI3 P lhTW

i4> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 91 + 1/1376 ⋅ )2/97 ⋅ 91 + 3611* >261/2:!

lK lhTW

ta~ka 4: lK lhTW lhI3 P y5>!g)ϕ5-!i5*>1/154! ! lhTW i5>i4!>261/2:

)i!−!y!dijagram*

⋅

X!>! n tw!)y5!−!yp* ⋅ τ >! 1/5:7 ⋅ (1/154 − 1/1184) ⋅ 4711 >74/86!lh b) ta~ka!3o; q qt >58471!Qb )napon pare ~iste vode na!u>91pD* q I3P = ϕ nby ⋅ q qt > 1/: ⋅ 58471 >53735!Qb

y3o!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 53735 ⋅ >!1/5722! 6 3: 2 ⋅ 21 − 53735 lhTW

⋅

X′!>! n tw!)y3o!−!yp* ⋅ τ >! 1/5:7 ⋅ (1/5722 − 1/1184) ⋅ 4711 >921/4!lh


{fmlp@fvofu/zv


strana 46

8/39/!U teorijskoj su{ari sa recirkulacijom, jednog dela iskori{}enog vazduha, protok atmosferskog ⋅

vla`nog vazduha, stawa!1)i>61!lK0lhTW-!y>1/12!lhI3P0lhTW*-!iznosi! n p>7!u0i. Stawe me{avine sveèg i opticajnog vazduha na ulazu zagreja~ vazduha je N)u>51pD-!y>1/145!lhI3P0lhTW*/!Me{avina se u kaloriferu zagreva do stawa!2)u>99pD*/!Po~etna vla`nost materijala je! Z2 >81&!ra~unato na suvu materiju, a krajwa! Z3 >9&!tako|e ra~unato na suvu materiju. Skicirati promene stawa vla`nog vazduha na!i!−!y!dijagramu i odrediti: a) masene protoke: odstrawene vlage i osu{enog materijala!)lh0i* b) maseni udeo sveèg i opticajnog vazduha u me{avini c) potrebnu koli~inu toplote za zagrevawe vla`nog vazduha!)lK0t* d) kolika bi bila potro{wa toplote da se su{ewe izvodi samo sveìm vazduhom tj. da nema recirkulacije i kolika bi bila temperaturu vla`nog vazduha na ulazu u komoru za su{ewe u tom slu~aju i 2 u2 3 ϕ>2

un ip

N 1

y y2

yN

ta~ka 0: y1>1/12! ⋅

n twp!>!

lhI3 P lhTW ⋅

n1 2+ y1

ip!>!!61!

lK lhTW

21 4 4711 = 2/76! lh = 2 + 1/12 t 7⋅

ta~ka M: yn>1/145!

lhI3 P lhTW

in!> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/145 ⋅ )2/97 ⋅ 51 + 3611* >238/64


lK lhTW

{fmlp@fvofu/zv


strana 47

ta~ka 1: y2>yn>1/145!

lhI3 P lhTW

i2!> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 99 + 1/145 ⋅ )2/97 ⋅ 99 + 3611* >289/67

lK lhTW

ta~ka 2: i3>i2>289/67!

lK lhTW

y3!>!@

prvi zakon termodinamike za proces me{awa dva vla`na vazduha: ⋅ ⋅ ⋅ ⋅ ⋅ ⋅  ⋅  R 23 = ∆ I23 + X u23 n twp ⋅ i p + n tw3 ⋅ i 3 =  ntwp + n tw3  ⋅ in   ⋅

n twp ⋅ (in − i p ) 2/76 ⋅ (238/64 − 61) lh > = >3/62 i 3 − in 289/67 − 238/64 t

⋅

n tw3

materijalni bilans vlage za proces me{awa dva vla`na vazduha: ⋅ ⋅  ⋅   n twp + n tw3  ⋅ y n − n twp ⋅ y p ⋅ ⋅ ⋅ ⋅     n twp ⋅ y p + n tw3 ⋅ y 3 =  n twp + n tw3  ⋅ y n !!! y 3 = ⋅   n tw3 (2/76 + 3/62) ⋅ 1/145 − 2/76 ⋅ 1/12 >1/15:9! lhI3 P y3 = 3/62 lhTW a) ⋅ ⋅  ⋅  lh X =  ntwp + n tw3  ⋅ (y 3 − y 2 ) > (2/76 + 3/62) ⋅ 4711 ⋅ (1/15:9 − 1/145 ) >347/73! i   Z2 Z 1/8 1/19 3 z2 = > z3 = > >1/52>1/18 2 + Z2 2 + 1/8 2 + Z3 2 + 1/19 ⋅

X = npn ⋅

z2 − z 3 2 − z2

⋅

n pn = X⋅

2 − z2 2 − 1/52 lh >521/72! > 347/73 ⋅ z2 − z 3 1/52 − 1/18 i

b) ⋅

hp>

n TWp ⋅

⋅

nTWp + nTW3

2/76 > >1/52/76 + 3/62

⋅

h3>

nTW3 ⋅

⋅

nTWp + nTW3

>

3/62 >1/7 2/76 + 3/62

c) ⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!! R 23 = ∆ I23 + X u23 ⋅ ⋅  ⋅  R {bh =  !n twp + n tw3  ⋅ (i2 − in ) (2/76 + 3/62) ⋅ (289/67 − 238/64 ) >323/39!lX  


{fmlp@fvofu/zv


strana 48

d) i

2′

2 u2 3 ϕ>2 un ip

N 1

y y1

yN

lhI3 P lK i2′>!i2>!i3!>289/67! lhTW lhTW 289/67 − 1/12 ⋅ 3611 i − y ⋅ 3611 u2′!>! !> >261/87pD 2 + 1/12 ⋅ 2/97 d q + y ⋅ 2/97

y2′!>!yp>1/12!

8/3:/!U teorijskoj su{ari se obavqa proces izdvajawa vlage iz koncentrata paradajza. Maseni protok koncentrata paradajza na ulazu u su{aru je!1/237!lh0t. Na ulazu u su{aru koncentrat paradajza sadrì z 2 =31!nbt&!vode, a prah na izlazu! z 3 =6!nbt%. Parcijalni pritisak vodene pare u okolnom (sveèm) vazduhu je! q I3P 1 >2/44!lQb, dok na izlazu iz su{are ne sme biti vi{i od! q I3P 3 >37/8!lQb/!Da bi se taj

(

)

(

)

uslov ispunio potrebno je me{awe dela iskori{}enog i okolnog sveèg vazduha tako da parcijalni pritisak vodene pare u vla`nom vazduh na ulazu u zagreja~ iznosi! q I3P >7/8!lQb. Pritisak vazduha za

(

)n

vreme su{ewa je konstantan i iznosi!q>212/4!lQb/!Odrediti: a) maseni protok sveèg i recirkulacionog vazduha (ra~unato na suv vazduh) c* specifi~nu potro{wu toplote u su{ari!)lK0lh!odstrawene vlage) ako se u fazi zagrevawa vazduh zagreje za u 2 − u n >41pD


{fmlp@fvofu/zv


strana 49

ta~ka 0: y1!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 2/44 ⋅ >1/1194! 3: 212/4 − 2/44 lhTW

>

lhI3 P 29 37/8 ⋅ >1/3333! 3: 212/4 − 37/8 lhTW

ta~ka 2: y3!>

NI3P NTW

⋅

q I3P q − q I3P

ta~ka M: yn!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 7/8 ⋅ >1/155! 3: 212/4 − 7/8 lhTW

ta~ka 1: y2>yn!>1/155!

lhI3 P lhTW

a) materijalni bilans vlage za proces su{ewa: ⋅ z − z3  ⋅   n tw p + n tw3  ⋅ (y 3 − y 2 ) > n wn ⋅ 2 2− z3   ⋅

⋅

n tw p + ntw3 = 1/237 ⋅

⇒

⋅

z2 − z 3 2 ⋅ 2 − z 3 y 3 − y2

⋅

n tw p + ntw3 = n wn ⋅

1/3 − 1/16 2 lh ⋅ >1/22 2 − 1/16 1/3333 − 1/155 t

materijalni bilans vlage za proces me{awa dva vla`na vazduha: ⋅ ⋅ ⋅  ⋅  n twp ⋅ y p + n tw3 ⋅ y 3 =  n twp + n tw3  ⋅ y n   ⋅

)2*

⋅

n tw p + n tw3 >1/22

)3* ⋅

Kombinovawem jedna~ina!)2*!i!)3*!dobija se;! n tw p >1/1:3

lh ⋅ lh -! ntw 3 >1/129 t t

c* i2!> d q ⋅ u 2 + y2 ⋅ )2/97 ⋅ u 2 + 3611*

)2*

in!> d q ⋅ u n + y n ⋅ )2/97 ⋅ u n + 3611*

)3*

(

Oduzimawem jedna~ina!)2*!i!)3*!dobija se;!!!!!!i2!−in!> (u 2 − u n ) ⋅ d q + 2/97 ⋅ y2

⋅  ⋅   n twp + n tw3  ⋅ (i2 − in ) (u 2 − u n ) ⋅ d q + 2/97 ⋅ y2 R n2  lK  >293/2! = ⋅ = = ⋅ ⋅ lhX y 3 − y2   X  n twp + n tw3  ⋅ (y 3 − y2 )   ⋅

rx

)


(

)

{fmlp@fvofu/zv


strana 50

8/41/!Jabuke koje ne podnose temperaturu vi{u od!81pD su{e se u teorijskoj su{ari sa recirkulacijom dela iskori{}enog vazduha. Stawe sveèg vazduha odre|eno je sa!1)u>7pD-!y>6/42!hI3P0lhTW*/ Apsolutna vla`nost iskori{}enog vazduha je y3>45!hI3P0lhTW-a specifi~na potro{wa toplote u su{ari iznosi!rx>4761!lK0lh!odstrawene vlage. Skicirati promene stawa vla`nog vazduha na!i!−!y!dijagramu i odrediti: b* masene udele sveèg i recirkulacionog vazduha u me{avini b) minimalnu temperaturu do koje se mora zagrejati sve` vazduh pre me{awa da bi se izbeglo stvarawe magle za vreme procesa me{awa a) ta~ka 0: yp>1/11642!

lhI3 P lhTW

ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 7 + 1/11642 ⋅ )2/97 ⋅ 7 + 3611* >2:/44!

lK lhTW

ta~ka 2: y3>1/145!

lhI3 P lhTW

⋅

rx =

R n2 ⋅

=

i − h p ⋅ i p − h3 ⋅ i3 i2 − in i − in = 3 = 3 y 3 − y2 y 3 − y n y 3 − h p ⋅ y p − h3 ⋅ y 3

X i3 ⋅ (2 − h3 ) − h p ⋅ i p i − ip rx> > 3 y 3 ⋅ (2 − h3 ) − y p ⋅ i p y3 − yp

⇒

i3>2:/44!, 4761 ⋅ (1/145 − 1/11642) >235/16

⇒

i3>ip!, r x ⋅ (y 3 − y p ) ⇒ lK lhTW

ta~ka 1: lK lhTW i2 − d q ⋅ u 2 lhI3 P 235/16 − 2 ⋅ 81 y2 = > >1/1316! 2/97 ⋅ u 2 + 3611 2/97 ⋅ 81 + 3611 lhTW ta~ka M: lhI3 P yn>y2>1/1316! lhTW y − yn 1/145 − 1/1316 > hp> 3 >1/58 1/145 − 1/11642 y3 − yp i2>i3!>235/16!

h3>

yn − yp 1/1316 − 1/11642 > >1/64 1/145 − 1/11642 y3 − yp

in = h p ⋅ i p + h3 ⋅ i3 > 1/58 ⋅ 2:/44 + 1/64 ⋅ 235/16 >85/94


lK lhTW

{fmlp@fvofu/zv


strana 51

b)

i 2 u2 4761 3

N′ N

1′ up

ϕ>2

1

y y3

y2

P−N−3; P−N′−3;

pravac me{awa pre zagrevawa okolnog vazduha pravac me{awa nakon zagrevawa okolnog vazduha

ta~ka 0′:

grafi~ki postupak: Konstrui{e se prava kroz ta~ke!3!i!N′)!yN′>yN). Presek ove prave sa linijom!yp>dpotu!!defini{e poloàj ta~ke O′. Iz dijagrama se o~itava!uP′! /

ra~unski postupak: lhI3 P y p( >yp>1/11642! lhTW lK in′!>88/68 lhTW in = h p ⋅ i p( + h 3 ⋅ i 3

i p( >@

⇒

i p( =

in − h3 ⋅ i 3 hp

88/68 − 1/58 ⋅ 235/16 lK >44/46 lhTW 1/64 47/46 − 1/11642 ⋅ 3611 i − y ⋅ 3611 up′!>! >33/96pD !> d q + y ⋅ 2/97 2 + 1/11642 ⋅ 2/97 i p( >


{fmlp@fvofu/zv


strana 52

8/42/!U teorijskoj su{ari sa recirkulacijom jednog dela iskori{}enog vazduha su{i se vlaàn lhX materijal po~etne vla`nosti!!411&!ra~unato na suvu materiju!)Z2>4! */!U su{ari se odstrani lhTN 91% od vlage koju sa sobom u su{aru unosi vlaàn materijal i pri tom dobijamo!43!lh0i!osu{enog materijala. Stawe sveèg vazduha odre|eno je sa!)u>31pD-!ϕ>1/7*!a stawe otpadnog vazduha odre|eno je sa!)u>51pD-!ϕ>1/9*/!Temperatura vazduha nakon faze zagrevawa iznosi!u>87pD/!Odrediti: b* toplotnu snagu zagreja~a vazduha!R{bh!)lX* b) koliko bi se toplote moglo u{tedeti (u zagreja~u) hla|ewem otpadnog vazduha do stawa zasi}ewa i rekuperativnim kori{}ewem tako oslobo|ene toplote za zagrevawe vazduha nastalog me{awem sveèg i recirkulacionog vazduha (slika) ta~ka 0: )napon pare ~iste vode na!u>31pD*

q qt >3448!Qb

q I3P = ϕ ⋅ q qt > 1/7 ⋅ 3448 >2513/3!Qb

yp!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 2513/3 ⋅ >!1/1199! 6 lhTW 3: 2 ⋅ 21 − 2513/3

ip> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 31 + 1/1199 ⋅ )2/97 ⋅ 31 + 3611* >53/44!

lK lhTW


q qt >8486!Qb

q I3P = ϕ ⋅ q qt > 1/9 ⋅ 8486 >6:11!Qb

y3!>

NI3P NTW

⋅

q I3P q − q I3P

>

lhI3 P 29 6:11 ⋅ >!1/149:! 3: 2 ⋅ 21 6 − 6:11 lhTW

i3> d q ⋅ u + y ⋅ )2/97 ⋅ u + 3611* > 2 ⋅ 51 + 1/149: ⋅ )2/97 ⋅ 51 + 3611* >251/25!

lK lhTW

ta~ka 1: i2>!i3>!251/25! y2>!

lK lhTW

i . dq ⋅ u 2/97 ⋅ u + 3611

>!

lhI3 P 251/25 . 2 ⋅ 87 >1/1354! 2/97 ⋅ 87 + 3611 lhTW

ta~ka M: lhI3 P lhTW 1/149: − 1/1354 > >1/596 1/149: − 1/1199

yn>y2>1/1354! hp>

y3 − yn y3 − yp

h3>2−h2>1/626

in = h p ⋅ i p + h3 ⋅ i3 > 1/596 ⋅ 53/44 + 1/626 ⋅ 251/25 >:3/81


lK lhTW

{fmlp@fvofu/zv


strana 53

a) z2 =

Z2 4 lhX > >1/86! lh)X + TN* 2 + Z2 4 + 2

materijalni bilans komore za su{ewe materijla: nwn>npn,!X

!!!!)2* ⋅

bilans vlage komore za su{ewe materijala:

n wn ⋅ z 2 = n pn ⋅ z 3 + X !!)3*

uslov zadatka:

1/9 ⋅ n wn ⋅ z 2 = X !!

⋅

!!!!)4*

⋅

kada se odstrawena vlaga!) X *!iz jedna~ine )4*!uvrsti u jedna~ine!)2*!i!)3*!⇒ n wn ⋅ z 2 = n pn ⋅ z 3 + 1/9 ⋅ n wn ⋅ z 2

tj.

n wn = n pn + 1/9 ⋅ n wn ⋅ z 2

tj.

1/3 ⋅ n wn ⋅ z 2 = npn ⋅ z 3 !!!)5* npn n wn = !!!!)6* 2 − 1/9 ⋅ z 2

kada se jedna~ina!)6) uvrsti u jedna~inu!)5*!dobija se: z3 >

1/3 ⋅ z 2 1/3 ⋅ 1/86 lhX >1/486 > lh)X + TN* 2 − 1/9 ⋅ z 2 2 − 1/9 ⋅ 1/86 n pn ⋅ z 3 43 ⋅ 1/486 lh > >91 1/3 ⋅ 1/86 1/3 ⋅ z 2 i

)5*

⇒

n wn =

)2*

⇒

X = n wn − n pn !>91!−!43!>59

⋅

⋅

⋅

lh i

⋅

59 2 lh X ⋅ n tw1!,! n tw3!> > >1/:2 t y 3 − y2 1/149: − 1/1354 4711 ⋅

⋅

⋅

prvi zakon termodinamike za proces u zagreja~u vazduha:!!!!!! R 23 = ∆ I23 + X u23 ⋅ ⋅  ⋅  R {bh =  !n twp + n tw3  ⋅ (i2 − in ) > 1/:2 ⋅ (251/25 − :3/8) >54/28!lX  


{fmlp@fvofu/zv


strana 54

b) nwn zagreja~ vazduha

C predgreja~ vazduha

B

komora za su{ewe materijala

2

npn 3

otpadni vazduh

3

recirkulacioni vazduh

N 3

1 sve` vazduh i 2 u2 3 C

u3

ϕ>2 N

up

ϕ3

B

1 ϕp

⋅ ⋅  ⋅  lh n tw1!>! h p ⋅  n tw1 !+!n tw3  > 1/596 ⋅ 1/:2 >1/55 t   ⋅ ⋅ ⋅ ⋅   lh n tw3!>! h3 ⋅  n tw1 !+!n n tw3  > 1/626 ⋅ 1/:2 >1/58 t   ta~ka A: lK y B = y3 iB>g (y B - ϕ = 2) >247/28 lhTW ⋅

⋅

R qsfe!>! n twp ⋅ (i3 − i B ) >!1/55 ⋅(251/25 − 247/28) >2/86!lX


{fmlp@fvofu/zv


strana 55

8/43/!U dvostepenoj teorijskoj su{ari za!7!sati osu{i se!2111!lh!vla`nog materijala.!Maseni odnos vlage lhX prema suvoj materiji u materijalu koji ulazi u prvi stepen su{ewa je 1/54! !a maseni odnos vlage lhTN lhX prema suvoj materiji u materijalu koji napu{ta drugi!stepen su{ewa je!1/25! . Sve` ulazni vazduh lhTN stawa!1)q>2!cbs-!u>27pD-!ϕ>1/6*!me{a se sa recirkulacionim vazduhom stawa!6)q>2!cbs-!u>57pD-!ϕ>1/7*!u odnosu!3;2, a zatim se predgreja~u vazduha (razmewiva~ toplote) pomo}u dela vla`nog vazduha oduzetog iz prvog stepena su{are. U greja~ima vazduha!H2!i!H3 vlaàn vazduh se zagreva do temperature od!91pD/ Temperatura vla`nog vazduha na izlazu iz postrojewa je!41pD/!Skicirati promene stawa vla`nog vazduha na Molijerovom!i!−!y!dijagramu i odrediti: a) veli~ine stawa vla`nog vazduha!)i-!y-!u*!u karakteristi~nim ta~kama b) toplotne snage greja~a vazduha-!H2!i!H3 c) vla`nost materijala (maseni udeo vlage) na kraju prvog stepena su{ewa sve` vazduh 1

nwn N

p r e d g r e j a ~

6

7

2

npn prvi stepen su{ewa

3

drugi stepen su{ewa

4

H2

H3

6

5

otpadni vazduh recirkulacioni vazduh

i

5 3 6 4 2 7

ϕ>2

N 1

y


{fmlp@fvofu/zv


strana 56

b* ta~ka 0: lhI3 P lhTW

yp!>!g)up-!ϕp*>1/1168!

ip>!g)up-!yp*>41/53!

lK lhTW

ta~ka 5: y6!>!g)u6-!ϕ6*!>1/14::7!

lhI3 P lhTW

i6!>!g)u6-!y6*!>25:/43!

lK lhTW

ta~ka M: prvi zakon termodinamike za proces me{awa dva vla`na vazduha: ⋅ ⋅ ⋅ ⋅ ⋅ ⋅  ⋅  R 23 = ∆ I23 + X u23 n twp ⋅ i p + n tw 6 ⋅ i 6 =  n twp + n tw 6  ⋅ in   ⋅

n twp ⋅

in =

⋅ ip + i6

n tw 6 ⋅

n twp ⋅

+2

3 ⋅ (41/53) + 25:/43 lK >81/16! >2 3 lhTW +2 2

n tw6 materijalni bilans vlage za proces me{awa dva vla`na vazduha: ⋅

ntwp ⋅  ⋅  n tw 1 ⋅ y p + n tw 6 ⋅ y 6 =  n twp + n tw 6  ⋅ y n   ⋅

⋅

⋅

yn =

⇒!

⋅ yp + y6

ntw 6 ⋅

ntwp ⋅

+2

ntw 6 3 ⋅ 1/1168 + 1/15 lhI3 P yn = 2 >1/1282 3 lhTW +2 2 i − y n ⋅ 3611 81/13 − 1/1282 ⋅ 3611 !> un!>! n >37/54pD d q + y n ⋅ 2/97 2 + 1/1282 ⋅ 2/97 ta~ka 2: y3>yn!>1/1282!

lhI3 P lhTW

i5!>i6!>25:/43!

lK lhTW

y5>!g)u5-!i5*>1/1373!

lK lhTW

y4>!y5>1/1373!

i3!>!g)u3-!y3*>236/46!

lK lhTW

ta~ka 4: lhI3 P lhTW

ta~ka 3: i4!>i3>236/46!


lhI3 P lhTW

{fmlp@fvofu/zv


strana 57

ta~ka 6: y7!>!y4!>!y5>1/1373!

lhI3 P lhTW

i7!>!g)u7-!y7*!>!:7/9:!

lK lhTW

ta~ka 1: y2!>!yn!>!1/1282!

lhI3 P lhTW

i2!>!@ ⋅

⋅

⋅

prvi zakon termodinamike za proces u predgreja~u vazduha:!!! R 23 = ∆ I23 + X u23 ⋅ ⋅  ⋅  n twp ⋅ (i 4 − i 7 ) =  ntwp + ntw 7  ⋅ (i2 − iN )  

⇒

⋅

n twp ⋅

i2 = in −

⋅ (i 7 − i 4 )

n tw 7 ⋅

n twp

3 ⋅ (:7/9: − 236/46 ) lK >9:/13 > 81/16 − 2 3 lhTW +2 2

+2 ⋅ n tw 7 i − y2 ⋅ 3611 9:/13 − 1/1282 ⋅ 3611 !> u2!>! 2 >88/95pD d q + y2 ⋅ 2/97 2 + 1/1282 ⋅ 2/97 c* z2 =

Z2 1/54 lhX > >1/41! 2 + Z2 1/54 + 2 lh)X + TN*

z3 =

Z3 1/25 lhX > >1/23! 1/25 + 2 2 + Z3 lh)X + TN*

materijalni bilans vlage za oba stepena su{ewa zajedno: z 2 − z 3 2111 1/41 − 1/23 lh > ⋅ >45/1: 7 2 − 1/23 2− z3 i

⋅

X = n wn ⋅

⋅ ⋅ ⋅  ⋅  X =  ntwp + n tw 6  ⋅ (y 4 − y 3 ) + n tw 7 ⋅ (y 6 − y 5 )   ⋅

n tw 6 ⋅

⇒

45/1: ⋅ lh X 4711 = > >1/34 t 4 ⋅ (y 4 − y 3 ) + (y 6 − y 5 ) 4 ⋅ (1/1373 − 1/1282) + 1/15 − 1/1373 ⋅

n twp = 3 ⋅ n tw 6 > 3 ⋅ 1/34 >1/57


lh t

{fmlp@fvofu/zv


strana 58 ⋅

⋅

⋅

⋅

⋅

prvi zakon termodinamike za proces u greja~u vazduha H2 :!!! R 23 = ∆ I23 + X u23 ⋅ ⋅  ⋅  R 23 =  ntwp + ntw3  ⋅ (i3 − i2 ) > 1/7: ⋅ (236/46 − 9:/13) >36/18!lX   ⋅

prvi zakon termodinamike za proces u greja~u vazduha H3 :!!! R 45 = ∆ I45 + X u 45 ⋅

⋅

R 45 = n tw6 ⋅ (i 5 − i 4 ) > 1/34 ⋅ (25:/43 − 236/46 ) >6/62!lX d* ⋅

X 2!>! n wn ⋅

z 2 − z( 2 − z(

⋅ ⋅  ⋅  X 2 =  n twp + n tw 6  ⋅ (y 4 − y 3 )  

)2* )3* ⋅

Kombinovawem jedna~ina!)2*!i!)3*!dobija se! X 2>!33/7!

zadatak za ve`bawe:!

lh !!j!z′>1/3 i

)8/44/*

8/44/!U dvostepenoj teorijskoj su{ari su{i se!2911!lh0i nekog proizvoda koji sadrì!4:!nbt&!vlage. Nakon su{ewa proizvod sadrì!:3!nbt% suve materije. Vazduh izlazi iz su{are na temperaturi od 56pD/!Temperatura okoline je!31pD/!Vazduh se pred svakim stepenom zagreva do!91pD!a na izlazu iz svakog stepena ima relativnu vla`nost!81&/!Sve promene stawa vla`nog vazduah u su{ari se doga|aju pri!q>2!cbs>dpotu/!Skicirati promene stawa vla`nog vazduha na!i!−y!!dijagramu i odrediti: a) ukupnu potro{wu toplote u su{ari!)lX* b) izra~unati vla`nost materijala (maseni udeo vlage) na izlazu iz prvog stepena su{ewa re{ewe: a) R>684!lX b) z′>1/36


{fmlp@fvofu/zv


strana 1

KRETAWE TOPLOTE 9/2/!Sa jedne strane ravnog zida povr{ine B>4!n3 nalazi se suva vodena para U2>247pD, a sa druge ravnog zida nalazi se vazduh U6>36pD. Zid je sastavqen od dva sloja: ~eli~nog lima (1) )λ2>61-!X0nLδ2>21!nn* i izolacionog materijala (2) )λ3>1/17!X0nL-!δ3>31!nn*/ Koeficijent prelaza toplote sa pare na zid iznosi α2>6111!X0n3L, a sa zida na okolni vazduh α3>2111!X0n3L. Odrediti: b* toplotni protok sa pare na vazduh, kroz zid )X* b) temperaturu izolacije (do vazduha) i temperaturu lima (do pare) δ3

δ2 α2 para

λ2

λ3

vazduh

α3 U6

U2

U4

U3

U5 a) ⋅

R=

U2 . U6 δ δ 2 2 + 2 + 3 + α2 λ2 λ 3 α 3

B=

247 . 36 ⋅ 4 >2111!X 2 1/12 1/13 2 + + + 6111 61 1/17 2111

b) U − U3 R= 2 ⋅B ⇒ 2 α2 ⋅

⋅

U .U R= 5 6 2 α3

⋅

2111 R >246/:4pD U3!>!U2!−! >247!−! 6111 4 α2 ⋅ B ⋅

B

⇒

2111 R >36/69pD U5!>!U6!, >36,! 2111 4 α3 ⋅ B

dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv


strana 2

9/3/!Za izgradwu privremenog skloni{ta u polarnim oblastima istraìva~i mogu upotrebiti ponetu {per plo~u, debqine!δ2>6!nn!)λ2>1/218!X0)nL**-!vla`nu zemqu!)λ3>1/767!X0)nL**-!i nabijen sneg )λ4>1/218!X0)nL**/!Na unutra{woj povr{i zida skloni{ta ustali se temperatura U2>3:4!L-!a koeficijent prelaza toplote, sa spoqa{we povr{i zida na okolni vazduh temperature!U6>341!L, iznosi!α>:/7!X0)n3L*/!′′Povr{inski toplotni protok′′ (toplotni fluks) pri tome treba da iznosi r>69!X0n3. Odrediti: a) najmawu debqinu sloja vla`ne zemqe u konstrukciji zida, tako da ne do|e do topqewa snega b) potrebnu debqinu sloja nabijenog snega u konstrukciji zida δ2

{per plo~a

U2

δ3

δ4

vla`na zemqa

U3

U4

nabijen sneg

U5

U6

a) napomena:

!r!>!

Uo~iti da je U4!>384!L (uslov ne topqewa snega na grani~noj povr{ini vla`na zemqa sneg).

U2 − U4 δ2 δ3 + λ2 λ3

⇒

U −U δ  δ3!>!λ3!  2 4 − 2  λ2   r

 3:4 − 384 6 ⋅ 21 −4 − δ3!>! 1/767 ⋅   69 1/218 

b)

r!>!

U2 − U6 δ2 δ3 δ4 2 + + + λ2 λ3 λ 4 α

⇒

⇒

  >2:7!nn  

U −U δ δ 2 δ4!>!λ4!  2 6 − 2 − 3 −  λ2 λ3 α   r

 3:4 − 341 6 ⋅ 21 −4 79 ⋅ 21 −4 2  >!79!nn δ4!>!λ4! ⋅  − − −  69 1/218 1/767 :/7  



strana 3

9/4/![upqi cilindar od stiropora!)λ>1/138!X0nL) unutra{weg pre~nika!ev>1/3!n, spoqa{weg pre~nika!et>1/4!n!i visine!I>2/6!n!napuwen je ledom sredwe!temperature!u2>1pD/!Temperatura okolnog vazduha je!u4>41pD, a koeficijent prelaza toplote na stiropor sa okolnog vazduha iznosi!α>9 X0n3L/!Temperatura unutra{we povr{ine cilindra je!1pD/!Zanemaruju}i razmenu toplote kroz baze cilindra, odrediti: a) toplotne dobitke cilindra )lX* b) temperaturu spoqa{we povr{ine zida cilindra d* vreme za koje }e se sav led otopiti ako toplota topqewa leda iznosi sm>443/5!lK0lh-!a gustina leda ρm>:11!lh0n4

α led vazduh

u2

u3

u4

a) ⋅

!R =

u2 − u 4 36 − 1 ⋅I = ⋅ 2/6 = 28/95!X et 2 2 1/4 2 2 mo + mo + 1/4π ⋅ 9 3π ⋅ 1/138 1/3 e t π ⋅ α 3π ⋅ λ e v

b) u − u3 R= 2 ⋅I ⇒! 2 et π ⋅ α ⋅

⋅

28/95 R u3!>!u2!−! >!36!− >39/5pD 1/4π ⋅ 9 ⋅ 2/6 ev π ⋅ α ⋅ I

c) τ!>!

Rm ⋅

R

= ///!>

251:4/87 28/95 ⋅ 21 −4

= 8:1119!t!

(9 dana 3 sata 27 min)

Rm!>!nm!/!sm!>!///!>53/5/!443/5!>251:4/87!lK nm!>!ρm!/!Wm!>!ρm!/

1/3 3 π ev3 π ⋅ M >!:11 ⋅ ⋅ 2/6 >!53/5!lh 5 5



strana 4

9/5/!Odrediti kolika se maksimalna debqina leda!)λmfe!>3/67!X0)nL**!moè obrazovati na spoqnoj povr{ini aluminijumske cevi!)λBm>33:/2!X0)nL**-!pre~nika!∅>:6094!nn-!duìne!M>2!n-!koju obliva voda, ako je temperatura na wenoj unutra{woj povr{ini U4>374!L-!pri ~emu je toplotni protok sa vode na cev!2661!X/

e4

e3

e2

U4 napomena:

!U2!>!384!L

U3

U2

(uslov stvarawa leda)

e4 − e3 1/216 − 1/1:6 >1/116!n>!6!nn >!///>! 3 3 ⋅ U2 − U4 R= ⋅M ⇒ e 2 e4 2 mo mo 3 + 3π ⋅ λmfe e3 3π ⋅ λ Bm e2 δ mfe !>!

 3π ⋅ λ ⋅ M λ mfe e 3  mfe ( ) e4!> e 3 ⋅ fyq U U mo  ⋅ − − 2 4 ⋅ λ Bm e2   R   3/67 :6   3π ⋅ 3/67 ⋅ 2 ⋅ (384 − 374 ) − mo >1/216!n e4!> 1/1:6 ⋅ fyq 33:/2 94   2661



strana 5

)9/6/!−!9/7/*

9/6/!Sa jedne strane staklene!)λ>1/9!X0nL*!plo~e, ukupne povr{ine!B>21!n3-!nalazi se vlaàn vazduh temperature!71pD!dok je sa druge strane voda temperature!31pD/!Pad temperature kroz staklenu plo~u iznosi!6pD. Koeficijent prelaza toplote sa vla`nog vaduha na plo~u iznosi!α2>31!X0n3L-!a sa plo~e na vodu α3>211!X0n3L/!Odrediti: a) temperature staklene povr{ine u dodiru sa vla`nim vazduhom i vodom b) debqinu staklene plo~e c) toplotni protok sa vla`nog vazduha na vodu d) toplotni protok sa vla`nog vazduha na vodu ako bi se sa strane vode formirao sloj kamenca toplotnog otpora S>1/2!n3L0X re{ewe: b* u3!>!41/94pD-!u4>36/94pD b) δ>7/97!nn ⋅

c) R >6945!X ⋅

d)

R( >3484!X

9/7/!U ~eli~noj cevi!)λ>57/6!X0)n3L**-!pre~nika!26:y5/6!nn, po celoj duìni deonice ime|u dva ventila, usled duèg prekida rada u ma{inskoj hali, u zimskom periodu, obrzaovao se ledeni ~ep sredwe temperature!1pD/!êli~na cev je toplotno izolovana slojem stiropora debqine!δ>61!!nn/ Naglim zagrevawem, temperatura vazduha u ma{inskoj hali povisi se do!41pD i potom ostaje nepromewena. Koeficijent prelaza toplote sa okolnog vazduha na spoqa{wu povr{ stiropora tako|e je stalan i iznosi!α>26!X0)n3L*/!Uz predpostavku da je temperatura na unutra{woj povr{i cevi stalna i da iznosi!1pD-!odrediti vreme (u danima) za koje }e se ledeni ~ep potpuno otopiti. re{ewe:

τ!≈!7 dana



strana 6

9/8/!U zidanom kanalu od hrapave crvene opeke!)ε3>1/:4*-!duìne!M>2!n, kvadratnog popre~nog preseka stranice!b>511!nn!postavqena je ~eli~na cev!)ε2>1/9*!spoqa{weg pre~nika!e>211!nn/ Temperatura spoqa{we povr{i cevi je!U2>684!L-!a unutra{wih povr{i zidova kanala!U3>434!L/ Prostor izme|u cevi i kanala je vakumiran. Odrediti: a) toplotni protok koji zra~ewem razmene cev i zidani kanal b) toplotni protok koji zra~ewem razmene cev i zidani kanal (pri istim temperaturama U2 i U3 ) ako se izme|u cevi i zidova kanala postavi cilindri~ni toplotni ekran koeficijenta emisije εF>1/96 i pre~nika eF>311!nn c) temperaturu tako postavqenog ekrana (zanemariti debqinu ekrana ) ε3 ε2

U2

U3

a) 5

⋅   R {   23

5

5 5  U2   U   684   434    −  3  −      211   211   211   211  = ⋅ M >///> ⋅ 2 >2496!X 2 2 1/2 ⋅ π ⋅ 5/56 eπ ⋅ D23

D23!>!Dd/!ε23!>//!!/!>!6/78/!1/8:!>!5/56!

X

n3L 5 2 2 ε23!>! >///>! >!1/8: 2 1/425  2   2 B2  2  − 2 + − 2  +  1/9 2/7  1/:4 ε2 B3  ε3   B2!>!e π /!M!> 1/2 ⋅ π ⋅ 2 >1/425!n3 B3!>! 5 ⋅ b ⋅ M > 5 ⋅ 1/5 ⋅ 2 >2/7!n3



strana 7

b) ε3

ε2

U εF

U2 5

⋅   R {   2F3

U3

5

5 5  U   U2   684   434    − 3    −   211   211   211   211  ⋅ M >///> ⋅ 2>!99:/77!X = 2 2 2 2 + + eπ ⋅ D2F eF π ⋅ D F3 1/2 ⋅ π ⋅ 5/35 1/3 ⋅ π ⋅ 5/8

D2F!>!Dd!ε2F!>5/35!

X 3 5

nL

BF!>!eF π /!M!> 1/3 ⋅ π ⋅ 2 >!1/739!n3 X DF3!>!Dd/!εF3!>5/8! n3L 5

ε2F!>!

2 >!1/86  2 B2  2  + − 2 ε2 BF  εF 

εF3!>!

2 >!1/94  2 BF  2  − 2 +  εF B3  ε3 

c) U trenutku uspostavqawa stacionarnog reìma kretawa toplote ⋅  ⋅  postavqamo toplotni bilans za toplotni ekran:  R [  =  R [  odakle  2F   F3 sledi da je:

UF!>! 5

D2F ⋅ e ⋅ U25 + D F3 ⋅ eF ⋅ U35 D2F ⋅ e + D F3⋅ eF

>

5

5/35 ⋅ 1/2 ⋅ 684 5 + 5/8 ⋅ 1/3 ⋅ 434 5 5/35 ⋅ 1/2 + 5/8 ⋅ 1/3

= >561!L



strana 8

9/9/!U industrijskoj hali nalazi se pe} od vaqanog ~eli~nog lima!)ε2>1/68*!ukupne povr{ine!B>3/6 n3/!Temperatura spoqa{we povr{i pe}i je!u2>271pD, a okolnog vazduha i unutra{wih povr{i zidova hale!u3>u4>26pD. Koeficijent prelaza toplote sa spoqa{we povr{i pe}i na vazduh u hali je!α>24/24 X0)n3L*/!Odrediti: a) ukupan toplotni protok (zra~ewe + prelaz) koji odaje spoqa{wa povr{ina pe}i b) temperaturu unutra{we povr{i pe}i ako je debqina zida pe}i!δ>31!nn-!a koeficijent toplotne provodqivost zida pe}i!λ>61!X0)n3L*

α ε2

U1

ε3

U2 U3

U4

a) ⋅ ⋅  ⋅   R  >!  R QSFMB[  ,!  R [SB•FOKF  >!///>!5871!,!3351!>!8111!X  ∑  23  23 ⋅  U −U 271 − 26  R QSFMB[  >! 2 3 ⋅ B2 >! ⋅ 3/6 >5871!X 2 2  23 α 24/24 5

5

5

5

 544   399   U2   U     −   − 4  ⋅  211 211    211   211  ⋅ B >!   R [SB•FOKF  >  ⋅ 3/6 >3351!X 2 2 2  23 D24 4/34 ε24!>!ε2!>!1/68!)B2!==!B4* D24!>!Dd/!ε24!>! 6/78 ⋅ 1/68 >!4/34!

X n3L 5

b) ⋅ ⋅  U − U2  R  >!  R QSPWPEKFOKF  >! 1 ⋅ B !!! ⇒ δ  ∑   12 λ −4 8111 31 ⋅ 21 ⋅ !>!271/4pD U1!>!271!, 3/6 61

⋅  R   Σ δ ⋅ !>!271/4pD U1!>!U2!, B λ



strana 9

9/:/!Temperatura vrelih gasova, koji se kre}u kroz kanal, meri se temperaturskom sondom!)ε2>1/9*/ Pri stacionarnim uslovima sonda pokazuje temperaturu!u2>411pD/!Temperatura povr{i zidova kanala je!u4>311pD/!Koeficijent prelaza toplote sa vrelih gasova na povr{ sonde iznosi!α>69!X0)n3L). Odrediti stvarnu temperaturu vrelih gasova u kanalu!)!u3>@*/ u4 u3 u2

ε α

(r{sb•fokf )24 = (rqsfmb{ )32

toplotni bilans temperaturske sonde: 5

5

5

 U2   U    − 4   211   211  = U3 − U2 2 2 D24 α 5

 684   584    −   211   211  U3!>!684!,! 2 5/65 ε24!>!ε2!>!1/9!)B2!==!B4*

⇒

5

 U2   U    − 4  211   211  !!⇒ U3!>!U2!,!  2 D24

5

>729!L!)!456pD!*

D24!>!Dd/!ε24!> 6/78 ⋅ 1/9 >!5/65!

X n3L 5



strana 10

9/21/!U prostoru izme|u dve, koncentri~no postavqene posrebrene cevi ostvaren je potpuni vakum. Temperatura na spoqa{woj povr{i unutra{we cevi, spoqa{weg pre~nika!e2>261!nn-!iznosi u2>711pD-!a temperatura na unutra{woj povr{i spoqa{we cevi, unutra{weg pre~nika!e3>311!nniznosi u3>311pD/!Emisivnost svake posrebrene cevi je!ε>1/16/!Odrediti “ekvivalentnu” toplotnu provodqivost materijala!)λ*-!~ijim bi se postavqewem u prostor izme|u cevi, pri nepromewenim temperaturama i pre~nicima cevi ostvarila ista linijska gustina toplotnog protoka!)!toplotni fluks-!X0n* e2 e3

e3

e2

(r{sb•fokf )23 = (rqspwp} fokb )23 5

5

 U2   U    − 3  211    211  = U2 − U3 2 e 2 mo 3 3π ⋅ λ e2 e2π ⋅ D23 5

⇒

5

e  U2   U  mo 3   − 3  e2 211   211  ⋅ !>/// λ!>!  2 3π ⋅ (U2 − U3 ) e2π ⋅ D23 D23!>!Dd!ε23!>///!>! 6/78 ⋅ 1/14 >1/276! ε23!>!

X n3L 5

2 2 2 >!1/14 > >!  2 1/26  2   2 e2  2 2 B2  2 + − 2    − 2  +  ε + e  ε − 2 1/16 1/3  1/16 ε2 B3  ε3  2 3  3   5

5

 984   584  1/3   −  mo 211 211 X     1/26 λ!> ⋅ >!1/158! 2 3π ⋅ (984 − 584 ) nL 1/26 ⋅ π ⋅ 1/276



strana 11

9/22/!Cilindri~ni kolektor za vodenu paru, spoqa{weg pre~nika!e2>386!nn-!nalazi se u velikoj prostoriji. Koeficijent emisije kolektora iznosi!ε2>1/:2/!Radi smawewa toplotnih gubitaka zra~ewem postavqa se osno postavqen toplotni {tit (ekran), zanemarqive debqine, koeficijenta emisije!εF>1/66/!Predpostavqaju}i da se postavqawem toplotnog {tita ne mewa temperatura na spoqa{woj povr{i kolektora i unutra{woj povr{i zidova prostorije odrediti pre~nik toplotnog {tita!)eF*-!tako da je u odnosu na neza{ti}eni kolektor smawewe toplotnih gubitaka zra~ewem 61&!/

(r{sb•fokf )23 = (r{sb•fokf )2F3 5

5

 U2   U    − 3   211   211  = 3 ⋅ 2 e2π ⋅ D23 2 3 = 2 2 2 + e2ε23 e2ε2F eFεF3

5

5

 U2   U    − 3   211   211  2 2 + e2π ⋅ D2F eFπ ⋅ DF3 ⇒

2 3 = 2  2 e2  2  − 2 + e2ε23 ε2 eF  εF  e2

 3 2 2  2 2  = + − 2 + ⇒  e2ε2 e2ε2 eF  εF  eFεF

⇒

+

2 eFεF3

 3  eF!>! e2 ⋅ ε2 ⋅  − 2  εF 

 3  − 2 !>!1/76:!n!>!76:!nn eF!>! 1/386 ⋅ 1/:2 ⋅    1/66 napomena: ε23!>!ε2 )!B2!==!B3!* ε2F!>!

εF3!>!εF)!BF!==!B3!*

2 2 >    2 e2  2 2 B2 2  − 2  − 2 + +     ε2 BF  εF   ε2 eF  εF



strana 12

9/23/!Oko duga~kog cilindra, pre~nika!e2>361!nn-!koncentri~no je postavqen ekran pre~nika eF>461!nn-!zanemarqive debqine. Ukupan koeficijent emisije povr{i cilindra i povr{i ekrana su jednaki i iznose!!ε!>1/9/!U stacionarnim uslovima, temperatura ekrana je!2:1pD-!temperatura okolnog vazduha i okolnih povr{i iznosi!61pD-!a sredwi koeficijent prelaza toplote sa spoqa{we povr{i ekrana na okolni vazduh!α>46!X0)n3L). Zanemaruju}i konvektivnu predaju toplote izme|u cilindra i ekrana odrediti temperaturu povr{i cilindra.

α ε

ε

U2

5

5

Uw

U3

(r{sb•fokf )23 = (r{sb•fokf )F3 + (rqsfmb{ )FW

toplotni bilans ekrana:  U2   U    − F  211    211  = 2 e2π ⋅ D2F

UF

5

5

 UF   U    − 3  211    211  + UF − UW 2 2 eFπ ⋅ α eFπ ⋅ DF3



strana 13

D2F!>!Dd!ε2F!>///> 6/78 1/8 >4/:8!

X n3L 5

2 2 > >!>!1/8 2 B2  2  2 e2  2  + +  − 2  − 2 ε BF  ε  ε eF  ε  X DF3!>!Dd!εF3!>//!!/!>!5/65! n3L 5 εF3!>!εF!>!1/8

ε2F!>!

5

2  U  U2 = 211 ⋅ 5  3  + e2 ⋅ D2F  211 

5  U 5     3  −  U2    211   U − U     211   + 3 4  >!835!L  2  2   eF ⋅ α  eF ⋅ DF3  

9/24/!U kanalu kvadratnog popre~nog preseka!)b>711!nn*!nalazi se ~eli~na cev!∅>3310311!nnλ>57!X0nL/!Kroz kanal proti~e suv vazduh. Temperatura spoqa{we povr{i cevi je!U2>711!L-!a unutra{we povr{i zida je!U4>411!L/!Koeficijenti emisije zra~ewa su!ε2>1/92!(za cev) i!ε4>1/97!(za zidove kanala). Koeficijent prelaza toplote sa cevi na vazduh je!α2>41!X0n3L/!Odrediti: a) temperaturu vazduha u kanalu )U3>@* ako su toplotni gubici spoqa{we povr{i cevi, radijacijom i konvekcijom jednaki c* temperaturu unutra{we povr{i cevi )Up>@* c) koeficijent prelaza toplote )α3*!sa vazduha na zidove kanala

α3 U4

α2

U3 U2 Up

ε2 ε4


zbirka zadataka iz termodinamike a)

strana 14

(rqsfmb{b )23 = (r{sb•fokf )24

uslov zadatka: 5

 U   U2    − 4  U2 − U3 211   211  =  2 2 et π ⋅ α et π ⋅ D24

5

5

⇒

D24!>!Dd/!ε24!>///> 6/78 ⋅ 1/9 >!5/53!

5

 U   U2    − 4  211   211  >!/// U3!>!U2! −  α D24 X n3L 5

2 2 2 >1/89 > >!! 2 1/33 ⋅ π  2    2 et π  2 2 B2  2 + 2 −       + − 2 + − 2  1/92 5 ⋅ 1/7  1/97 ε2 B 4  ε4    ε2 5 ⋅ b  ε3 B2!>! et π ⋅ M B4!>!5b!/!M ε24!>!

5

5

 711   411    −   211   211  U3!>!711! − >!532!L 41 5/53 b) toplotni bilans spoqa{we povr{ine cevi:

(rqspwp} fokf )12 = (rqsfmb{b )23 + (r{sb•fokf )24 !!⇒ (rqspwp} fokf )12 = 3 ⋅ (rqsfmb{b )23 U − U3 3⋅ 2 = 2 et π ⋅ α

⇒

e mo t ev U2 − U3 ⋅ U1!>!U2!.! 3 ⋅ 2 3π ⋅ λ et π ⋅ α

U1 − U2 !> e 2 mo t 3π ⋅ λ ev

⇒

1/33 mo 711 − 532 1/3 !>!713/5!L Up!>!711!−! 3 ⋅ ⋅ 2 3π ⋅ 57 1/33 ⋅ π ⋅ 41 c) toplotni bilans vazduha u kanalu: U − U4 U2 − U3 ⋅M = 3 ⋅b⋅M ⋅ 5 2 2 et π ⋅ α2 α3 α3!>!

⇒

•  •   Rqsfmb{  =  Rqsfmb{       23  34 U − U3 2 ⋅ α3!>! 2 2 5b ⋅ (U3 − U4 ) et π ⋅ α2

⇒

X 711 − 532 2 >23/9! ⋅ 2 5 ⋅ 1/7 ⋅ (532 − 411) n3L 1/33π ⋅ 41



strana 15

9/25/!!Unutar metalnog!cilindri~nog rezervoara, unutra{weg pre~nika!e>1/6!n! i visine i>2!n, ostvaren je potpuni vakum. Temperatura na unutra{woj povr{i doweg dna je stalna i iznosi U2>511!L, dok su temperature na preostalim unutra{wim povr{ima tako|e stalne i iznose U3>411!L/ Koeficijent emisije svih unutra{wih povr{i je jednak i iznosi ε>1/9/ Odrediti debwinu izolacionog materijala )δj{* toplotne provodnosti λj>1/4!X0)nL* koeficijenta emisije εj{>1/:6, kojim treba izolovati dowe dno, da bi pri stacionarnim uslovima i pri nepromewenim temperaturama (na dodirnoj povr{i doweg dna i izolacionog sloja U2 i na ostalim unutra{wim povr{inama U3) toplotni protok sa doweg dna bio smawen za 31&/ e

U3

vakum

i

δj{

Uj{

U2 5

⋅   R {   23

5

5 5  U2   U   511   411    −  3     −  211   211   211   211  = ⋅ B 2 >///> ⋅ 1/2:7 >262/72!X 2 2 D23 5/53

D23!>!Dd/!ε23!>//!!/!>!6/78/!1/89!>!5/53!

X

n3L 5 2 2 >!1/89 ε23!>! >///>! 2 1/2:7  2   2 B2  2 + − 2  − 2   +  1/9 2/878  1/9 ε2 B3  ε3   e 3 π 1/6 3 π !>1/2:7!n3 = 5 5 e3 π 1/6 3 π > 1/6 ⋅ π ⋅ 2 + >2/878!n3 B3!>! e ⋅ π ⋅ i + 5 5 B2!>



strana 16

⋅ ⋅  R( = 1/9 ⋅  R {  >232/3:!X  23 5

5

 Uj{   U    − 3  ⋅  211   211  ⋅ B2 R( = 2

⋅

⇒

R(  U  Uj{ = 211 ⋅ 5 + 3  D J{3 ⋅ B 2  211 

5

D J{3 DJ{3>!Dd/!εJ{3′!>//!!/!>!6/78/!1/:4!>!6/38!

X

n3L 5 2 2 >!1/:4 εJ{3!>! >///>! 2 1/2:7  2   B2  2 2 + − 2    + − 2 1/:6 2/878  1/9  ε j{ B 3  ε 3  5

Uj{ = 211 ⋅ 5

232/3:  411  +  >486/43!L 6/38 ⋅ 1/2:7  211 

toplotni bilans gorwe povr{i izolacije:

U − Uj{ ⋅   R qspwp} fokf  = 2 ⋅ B2 δ {j  2J{ λ j{ δ j{ =

⇒

δ j{ =

(511 − 486/43) ⋅ 1/4 ⋅ 1/2:7 >22/:7!nn

napomena:

⋅ ⋅   R qspwp} fokf  = R(  2J{

(U2 − Uj{ ) ⋅ λ j{ ⋅ B 2 ⋅   R qspwp} fokf   2J{

>

232/3:

Pri izra~unavawu A2 za slu~aju sa izolaciju zanemaruje se smawewe povr{ine A2 zbog male debqine izolacije



strana 17

9/26/!Ravan zid debqine!6!dn!sa jedne svoje strane izloèn je dejstvu toplotnog zra~ewa, ~iji intenzitet u pravcu normale na zid iznosi!r{>2111!X0)n3L*/!Usled na ovaj na~in prenete koli~ine toplote ozra~ena povr{ zida odràva se na temperaturi od!u2>62/2pD/!Ukupan koeficijent emisije povr{i zida je!ε>1/8-!a toplotna provodqivost materijala od kojeg je zid na~iwen je!λ>1/86!X0)nL*/ Temperatura okolnog vazduha (sa obe strane zida) je 31pD/!Zanemaruju}i sopstveno zra~ewe zida i smatraju}i da je koeficijent prelaza toplote sa obe strane zida na okolni vazduh (α) isti odrediti temperaturu neozra~ene povr{i zida!)u3* r{ rsfg rqsfmb{2

rqspw

rqsfmb{3

Uw

U2

U3

Uw

toplotni bilans ozra~ene povr{i zida (povr{ 1) r{!>!rsfg!,!rqsfmb{2!,!rqspw

r{!>!)2.!ε!*!r{!,!

U2 − Uw U −U !,! 2 3 !!!)2* 2 δ α λ

toplotni bilans ne ozra~ene povr{i zida (povr{ 2) U2 − U3 U − Uw > 3 δ 2 λ α

rqspw!>!rqsfmb{3

!!!

!!!!!!!!)3*

re{avawem sistema dve jedna~ine )2*!i )3* sa dve nepoznate )U3!i!α* dobija se: U3!>419/7!L-!!α!>26!

X n3L



strana 18

9/27/!Ravan zid od mermera!)λ>3/9!X0)nL*-!ε>1/66*-!debqine!1/167!n!izloèn je sa obe strane dejstvu toplotnog zra~ewa ~iji intenziteti u pravcu normala na povr{i iznose!r{2>253:!X0n3!i!r{3>2:4 X0n3/!Hla|ewe mermera sa obe strane zida obavqa se iskqu~ivo konvektivnim putem (zanemaruje se sopstveno zra~ewe mermera). Temperatura vazduha sa jedne strane zida je!UW2>61pD-!a sa druge UW3>51pD-!a odgovaraju}i koeficijenti prelaza toplote!α2>9!X0)n3L*!i!α3>31!X0)n3L*/!Odrediti temperature obe povr{i mermera!)U2!j!U3*/ r{2

r{3 rsfg3

rsfg rqsfmb{2

rqspw

rqsfmb{3

Uw2

U2

U3

Uw3

toplotni bilans ozra~ene povr{i zida (povr{ 1) r{2>!rsfg!,!rqsfmb{2!,!rqspw

r{2>)2−ε!*!r{2,

U2 − Uw2 U −U !,! 2 3 !!)2* δ 2 λ α

toplotni bilans ozra~ene povr{i zida 2 (povr{ 2) r{3!,!rqspw!>!rsfg3!,!rqsfmb{

r{3,

U − Uw3 U2 − U3 !)3* >!)2−ε!*!r{3!,! 3 δ 2 λ α3

re{avawem sistema dve jedna~ine )2* i )3* sa dve nepoznate )U3!i U2*!dobija se: U2>91pD-!U3>81pD



strana 19

9/28/!Toplotna provodnost materijala od kojeg je na~iwen ravan zid, moè da se izrazi u funkciji temperature zida u obliku:! λ = 1/5 + 7 ⋅ 21 −4 u -!!pri ~emu je toplotna provodnost, λ, izraèna!v!X0)nL*a temperatura, u, u!pD/!Debqina zida je!δ>41!cm a temperature sa jedne i druge strane zida su!u2>231pD i!u3>41pD/!Pdrediti toplotni fluks (povr{insku gustinu toplotnog protoka) kroz zid i predstaviti grafi~ki raspored temperatura u zidu. 1. na~in: diferencijalna jedna~ina provo|ewa toplote kroz ravan zid pri λ>g)u*: r = −λ(u ) ⋅ δ

eu ey u3

∫(

∫

(

)

r ⋅ δ = −1/5 ⋅ (u 3 − u 2 ) − 7 ⋅ 21 −4

r ⋅ ey = − 1/5 + 7 ⋅ 21 −4 u ⋅ eu ⇒ 1

u2

(

)

− 1/5 ⋅ (41 − 231) − 4 ⋅ 21 − 1/5 ⋅ (u 3 − u 2 ) − 4 ⋅ 21 −4 u 33 − u 23 > δ 41 ⋅ 21 −3 X r>366! 3 n r=

)

r ⋅ ey = − 1/5 + 7 ⋅ 21 −4 u ⋅ eu

⇒

−4

(41

3

u 33 − u 23 3

− 231 3

)

2. na~in: Jedna~ina za toplotni fluks kroz ravan zid pri λ>g)u*; 2 r!>! − δ r!>! −

U3

∫

U2

2 λ)u* ⋅ eu = − δ

U3

∫ (1/5 + 1/117 ⋅ u ) ⋅ eu

⇒

U2

2 u3 − u3  1/5 ⋅ (u3 − u2 ) + 1/117 ⋅ 3 2  δ  3 

 231 3 − 41 3  2   >!366! X r=− 1/5 ⋅ (231 − 41) + 1/117 ⋅  1/14  3 n3  


zbirka zadataka iz termodinamike e3 y

1. na~in:

eu 3

eu r =− ey 1/5 + 7 ⋅ 21 −4 u e3 y

=−

eu 3

strana 20

>@ ⇒

ey 1/5 + 7 ⋅ 21 −4 u =− eu r

7 ⋅ 21 −4 =1 r

Kako je drugi izvod funkcije y>g)u* negativan to zna~i da je y>g)u*!konkavna (ispup~ena na gore). δ>g)!u!*>@

2. na~in:

Do istog zakqu~ka se moè do}i posmatrawem funkcije δ>g)u*, kada koristimo 2. na~in za izra~unavawe toplotnog fluksa. δ!> −

2 u3 − u23  1/5 ⋅ (u − u2 ) + 1/117 ⋅  r  3 

∂δ 2 = − ⋅ [1/5 + 1/117 ⋅ u ] ∂u r 3

∂ δ ∂u3

!=!1

⇒

⇒

∂ 3δ ∂u

3

=−

2 ⋅ 1/117 r

funkcija!δ!>!g)!u!*!je konkavna (ispup~ena na gore)

u

u2 isprekidana linija predstavqa temperaturni profil pri λ>dpotu puna linija predstavqa temperaturni profil pri λ≠dpotu, tj. pri! λ = 1/5 + 7 ⋅ 21 −4 u

u3 δ

y



strana 21

9/29/!Cev od {amotne opeke!∅>4310391!nn!)λ>1/5,1/113/!u) obloèna je sa spoqa{we strane slojem izolacije!)δj>6!nn-!λj>1/16,1/1112/!u*-!gde je!u!temperatura zida u!pD-!b!λ!v!X0)n/L). Temperatura unutra{we povr{i cevi od {amotne opeke je!241pD-!a spoqa{we povr{i izolacionog materijala 41pD/!Odrediti topotni fluks (gustinu toplotnog protoka) po du`nom metru cevi. u4 u3 u2

3π r23!!>− e mo 3 e2

U3

∫ (1/5 + 1/113 ⋅ u ) ⋅eu !>! mo e3 [1/5(u3 − u2) + 1/112⋅ (u3 − u2 )] 3π

U2

3

3

e2

U4

∫(

)

[

(

)]

[

(

)]

3π 3π 1/15 + 21 ⋅ 21−6 u ⋅eu !> 1/15 ⋅ (u 4 − u3 ) + 6 ⋅ 21−6 u34 − u33 e4 e4 mo mo U e3 3 e3 toplotni bilans spoqa{we povr{i {amotne opeke (unutra{we povr{i izolacije): r23!>!r34 r34!>−

[

(

)]

3π 3π 1/5 ⋅ (u3 − u2) + 1/112⋅ u33 − u23 !>! 1/15 ⋅ (u 4 − u3 ) + 6 ⋅ 21−6 u34 − u33 e3 e4 mo mo e2 e3 4/856 ⋅ 21 −6 ⋅ u 33 + 2/876 ⋅ 21 −3 ⋅ u 3 − 3/397 >1 u3 =

− 2/876 ⋅ 21 −3 ±

(2/876 ⋅ 21 )

−3 3

− 5 ⋅ 4/856 ⋅ 21 −6 ⋅ (− 3/397)

3 ⋅ 4/856 ⋅ 21 −6

>216/9pD

u3!>!216/9pD-!(pozitivno re{ewe) r24!>!r23!>!r34!>!−

r24!>−

[

[

(

3π 1/5 ⋅ (u3 − u2) + 1/112⋅ u33 − u23 e3 mo e2

(

)]

)]

3π X 1/5 ⋅ (216/9 − 241) + 1/112 ⋅ 216/9 3 − 241 3 !>!835! 1/43 n mo 1/39



strana 22

9/2:/!Preko gorwe povr{i horizontalne ravne plo~e postavqena je toplotna izolacija!debqine!δ>39 nn-!toplotne provodqivosti!λ>1/4!,21−5!/!u!,!3/21−8/!u3!-!gde je!λ!)X0nL*!b!u!)pD*-!i emisivnosti hrapave povr{ine u pravcu normale!εo>1/:/!Na dodirnoj povr{i plo~e i izolacije je stalna temperatura!u2>511pD. Temperatura gorwe povr{i izolacije je tako|e stalna i iznosi!u3>211pD. Temperatura zidova velike prostorije u kojoj se nalazi izolovana plo~a iznosi!u5>31pD/!Koeficijent prelaza toplote sa gorwe povr{i izolacije na vazduh iznosi α>48/:!X0)n3L*/!Odrediti temperaturu vazduha u prostoriji )u4*/

u5 u4 ε3

α u3 u2

izolacija plo~a toplotni bilans gorwe povr{i izolacije:

(rqspwp} fokf )23 = (rqsfmb{ )34 + (r{sb•fokf )35 5

−

2 δ

u3

 U   U3    − 5  u − u 211   211  1/4 + 2⋅ 21− 5 ⋅ u + 3 ⋅ 21−8 ⋅ u3 ⋅ eu !>! 3 4 ,  2 2 D35 α

∫(

u2

5

)

⇒

5 5   U3   U    u3   − 5   2 2 211   211   u4!>!u3! − − 1/4 + 2⋅ 21− 5 ⋅ u + 3 ⋅ 21−8 ⋅ u3 ⋅ eu −   2 α δ   u2 D35  

∫(

u4>u3!−

[(

)

)

(

2 ⋅ rqspwp } fokb 23 − r{sb•fokb α

)35 ] >///>211!− 482/: ⋅ [4743 − 711] >31pD



strana 23

napomena: u3

(rqspwp} fokf )23 !>! − 2δ ∫ (1/4 + 2⋅ 21−5 ⋅ u + 3 ⋅ 21−8 ⋅ u3 )⋅ eu ! u2



3

3

4

4

X



n3

(rqspwp} fokf )23 !>! − 2δ 1/4 ⋅ (u3 −u2) + 2⋅ 21−5 u3 3− u2 + 3 ⋅ 21−8 u3 4− u2  !>!4743! 

5

5

5 5  U3   U   484   3:4    − 5    −  (r{sb•fokf )35 >  211  2  211  >  211  2  211  >711! X3 n 6 D35

D35!>!Dd/!ε35!>!///>6/78!/1/993>6

X n3L 5

ε35!>!ε3!)!kfs!kf!B3!==!B5!* ε3!>!L/εo!>!///>!1/:91/:>!1/993 L!>!1/:9-!(hrapava povr{ izolacije) )9/31/*

zadatak za ve`bawe:

9/31/!Izme|u homogenog i izotropnog izolacionog materijala debqine δj>51!nn!i cevi, spoqa{weg pre~nika e>91!nn, ostvaren je idealan dodir. Zavisnost toplotne provodqivosti materijala od temperature data je izrazom: 2  U  λ = 1/59 + 1/27 ⋅ mo   − 384  . Termoelementima, pri ustaqenim uslovima, izmerene su [X 0 (nL )]   31  [L ] slede}e temperature: − na spoqa{woj povr{i cevi U2!>!534!L U3!>!429!L − na spoqa{woj povr{i izolacionog materijala U4!>!384!L − za okolni fluid. Odrediti koeficijent prelaza toplote sa spoqa{we povr{i izolacionog materijala na okolni fluid. re{ewe: α!>!41/6!

X n3L



strana 24

9/32.!Tanka plo~a visine!i>1/3!n-!{irine b>1/6!n-!potopqena je, vertikalno, u veliki rezervoar sa vodom temperature ug!>!31pD. Odrediti snagu greja~a, ugra|enog u plo~u, potrebnu za odràvawe temperature povr{i na!u{>!71pD/ ⋅ u − ug R= { ⋅ b ⋅ i ⋅ 3 !>!/// 2 α 1. korak: fizi~ki parametri za vodu na temperaturi!ug!>!31pD X n3 2 λg!>!6:/:!/21.3! -!βg!>!2/93!/21.5! -!νg!>!2/117!/21.7! t L nL 2. korak:

karakteristi~na duìna ~vrste povr{i

ml!>!i!>!1/3!n 3. korak:

potrebni kriterijumi sli~nosti

Hsg!>!

β g ⋅ h ⋅ mL4 ⋅ )U{ − Ug * υ 3g

Qsg!>!8/13

!>!

2/93 ⋅ 21 −5 ⋅ :/92 ⋅ 1/3 4 ⋅ )71 − 31*

(2/117 ⋅ 21 )

−7 3

>6/76!/219

Qs{!>!3/:9 1/36

 Qsg   4. korak: konstante u kriterijalnoj jedna~ini  Qs{  Hsg!/Qsg!>!4/:8!/21: turbulentno strujawe fluida u grani~nom sloju Ovg!>!D!)!Hsg!/Qsg!*o! 

D>!1/26 5. korak:

o>1/44

izra~unavawe Nuseltovog broja

 8/13  Ovg!>!1/26!)!4/:8!/21:!*1/44!    3/:9 

1/36

>384/46

6. korak:

izra~unavawe koeficijenta prelaza toplote!)α* λ 6:/: ⋅ 21 −3 X α!>!Ovg! ⋅ g >!384/46 ⋅ >929/8! 1/3 mL n3L ⋅

R=

71 − 31 ⋅ 1/6 ⋅ 1/3 ⋅ 3 >765:/7!X 2 929/8



strana 25

9/33/!Odrediti povr{insku gustinu toplotnog protoka (toplotni fluks) konvekcijom, sa spoqa{we povr{i vertikalnog zida neke pe}i na okolni prividno miran vazduh stalne temperature!ug>31pD/ Temperatura spoqa{we povr{ine pe}i je!u{>!91pD/!Smatrati!da je strujawe vazduha u grani~nom sloju turbulentno po celoj visini zida. U − Ug r!>! { >/// 2 α α!>!Ovg!

 Qs λg >!D!/!)!Hsg!/!Qsg!*o!/  G mL  Qs{

  

1/36

λg >/// mL

/

fizi~ki parametri za vazduh na temperaturi tf = 20oC λg!>!3/6:!/21.3!

n3 X !!!!! νg!>!26/17!/21.7! nL t

βg!>!

2 2 2 > >!4/52!/21.4! UG 3:4 L

potrebni kritrijumi sli~nosti: Qsg!>!1/814

Qs{!>!1/7:3

konstante u kriterijalnoj jedna~ina za prirodnu konvekciju D>1/26

o>!1/44!!

(turbulentno strujawe u grani~nom sloju)

 β g ⋅ h ⋅ ml4 ⋅ (u { − u g )  ⋅ Qsg  3   υg  

1/44

α!>!1/26/ 

 β g ⋅ h ⋅ (u { − u g )  ⋅ Qsg  3   υg  

1/44

α!>!1/26/  

4/52⋅ 21 α!>!1/26/   

r!>!

−4

⋅ :/92⋅ (91 − 31)

(26/17 ⋅21 )

−7 3

 Qsg  Qs  {

/! 

 Qsg  Qs  {

/! 

 ⋅ 1/814   

  

  

1/36 /

λg mL

⇒

1/36

1/44

/λ

g>

/  1/814 

 1/7:3 

1/36

⋅ 3/6: ⋅ 21−3 !>7/74!

X n3 L

X 91 − 31 >4:8/9! 2 n3 7/74



strana 26

9/34/!Ukupan toplotni protok koji odaje horizontalna ~eli~na cev!)ε>1/86) spoqa{weg pre~nika!et>91 nn!iznosi 611!X. Ako sredwa temperatura spoqa{we povr{i cevi iznosi!U2>91pD-!temperatura mirnog okolnog vazduha!U3>31pD!i temperatura unutra{we povr{i zidova velike prostorije u kojoj se cev nalazi U4>26pD!-!odrediti: a) duìnu cevi b) ukupan toplotni protok koji odaje ista ova cev kada bi je postavili vertikalno (pretpostaviti da je strujawe u grani~nom sloju turbulentno po celoj visini cevi) a) 5

5

 U   U2   −  4     ⋅ ⋅ ⋅ 211 U − U3          211   R  >!  R qsfmb{  ,!  R {sb•fokf  > 2 ⋅ M !,!! ⋅M 2 2 23  23  ∑  e t πα e t π ⋅ D24

M!>!

⋅  R   Σ   e t πα ⋅ (U2 − U3 ) + D24 

 U  5  U  5   ⋅  2  −  4     211   211   

ε24!>!ε2!>!1/86!)B2!==!B4*

= ///

D24!>!Dd!ε24!>!6/78!/1/86>!5/37!

X n3 L 5

fizi~ki parametri za vazduh na temperaturi!ug!>!31pD 2 n3 2 2 X > λg!>!3/6:!/21.3! !!!!!!βg!>! >!4/52!/21.4! νg!>!26/17!/21.7! nL t UG 3:4 L

1. korak:

2. korak:


ml!>!et!>!91!nn 3. korak: Hsg!>!

potrebni kriterijumi sli~nosti β g ⋅ h ⋅ mL4 ⋅ )U{ − Ug * υ 3g

Qsg!>1/814

4. korak:

4/52 ⋅ 21 −4 ⋅ :/92 ⋅ 1/19 4 ⋅ )91 − 31*

(26/17 ⋅ 21 )

−7 3

>!5/65!/217

Qs{!>!1/7:3  Qs Ovg!>!D!)!Hsg!/Qsg!*o!  g  Qs{ laminarno strujawe fluida u grani~nom sloju

konstante u kriterijalnoj jedna~ini

Hsg!/Qsg!>!4/2:!/217 D>!1/6

!>

  

1/36

o>1/36


zbirka zadataka iz termodinamike 5. korak:

strana 27


 1/814  Ovg!>!1/6!)!4/2:!/217!*1/36!    1/7:3 

1/36

>32/32

6. korak:

izra~unavawe koeficijenta prelaza toplote!)α* λ X 3/6: ⋅ 21 −3 α!>!Ovg! g >32/32! >!7/98! . 4 mL 91 ⋅ 21 n3L

M!>!

611   464  5  399  5   1/19π7/98 ⋅ (91 − 31) + 5/37 ⋅   −    211     211  

>!3/66!n

b) 5

5

 U   U2    − 4   ⋅   211  ⋅ M >///>591!X  R (  > U2 − U3 ⋅ M !,!!  211    2 2  ∑ eT π ⋅ α ( eT π ⋅ D24 α′>@ 2. korak:

ml( >M!>!3/66!n

3. korak:

Hsg′!>!

4.korak:

Hsg′!/!Qsg!>!2/14!/2122

4/52 ⋅ 21 −4 ⋅ :/92 ⋅ 3/66 4 ⋅ )91 − 31*

(3/6: ⋅ 21 )

−3 3

⇒

>!2/58!/2122

D>!1/26

o>1/44

)turbulentno strujawe fluida u grani~nom sloju du` cele visine cevi)

5. korak:

 1/814  Ovg′!>!1/26!)!2/14!/2122!*1/44!    1/7:3 

6. korak:

α′!>!75:/32!

1/36

>75:/32

3/6: ⋅ 21 −3 X >!7/6:! 3/66 n3L 5

5

 464   399    −     211   R (  > 91 − 31 ⋅ 3/66 !,!!  211  ⋅ 3/66 >59:!X   2 2  ∑ 1/19π ⋅ 7/6: 1/19π ⋅ 5/37 ⋅



strana 28

9/35/!Vertikalna cev, visine!M>1/9!n, nalazi se u prividno mirnom vazduhu stalne temperature Ug>31pD!j!pritiska!q>2!cbs/!Temperatura na grani~noj povr{i cevi je stalna i iznosi U{>91pD/ Odrediti toplotni protok sa cevi na okolni vazduh u laminarnom delu strujawa, turbulentnom delu strujawa i du` cele cevi. fizi~ki parametri za vazduh na temperaturi!ug!>!31pD 2 X n3 2 2 >!4/52!/21−4! !!!!!!βg!>! νg!>!26/17!/21−7! > λg!>!3/6:!/21−3! nL t UG 3:4 L

1. korak:

2. korak:


ml!>!M!>!1/9!n 3. korak:

potrebni kriterijumi sli~nosti β ⋅ h ⋅ ml4 ⋅ )U{ − Ug * 4/52 ⋅ 21 −4 ⋅ :/92 ⋅ 1/9 4 ⋅ )91 − 31* >!5/64!/21: !> Hsg!>! g −7 3 υ 3g 26/17 ⋅ 21 Qsg!>1/814 Qs{!>!1/7:3

(

4. korak:

)

 Qs Ovg!>!D!)!Hsg!/Qsg!*o!  g  Qs{ )kriti~na vrednost proizvoda!Hsg!/Qsg!*


(Hsg ⋅ Qsg )ls >2/21:

Hsg!/Qsg!>!4/2:!/21:!?! (Hsg ⋅ Qsg )ls

  

1/36

prirodno strujawe fluida u grani~nom sloju do visine M>mls je laminarno a nakon toga turbulentno

D>1/87-!

o>1/36

mls

5. korak:

2 4

⋅ υ 3g

)

izra~unavawe Nuseltovog broja za laminarnu oblast strujawa

(Ov g )mbn >!1/87!)!2!/21:!*1/36!  1/814   1/7:3 

6. korak:

(

2

3  4  2 ⋅ 21 2 ⋅ 21 : ⋅ 26/17 ⋅ 21 −7  >1/65!n  > =  4/52 ⋅ 21 −4 ⋅ :/92 ⋅ )91 − 31* ⋅ 1/814   β g ⋅ h ⋅ )U{ − Ug * ⋅ Qsg      :

1/36

>246/26

izra~unavawe koeficijenta prelaza toplote!)αmbn) u laminarnom delu strujawa.

αmbn!>! (Ov g )mbn ⋅

λg 3/6: ⋅ 21 −3 X >!135.15 ⋅ >7/59! 1/65 mls n3L



R mbn =

strana 29

91 − 31 ⋅ 1/65 >76/5!X 2 1/2 ⋅ π ⋅ 7/5

oblast turbulentnog strujawa u grani~nom sloju:

2. korak:


ml!>!M!−!mls!>!1/:!−!1/65!>1/47!n 3. korak:


(Hsg ⋅ Qsg )M >4/2:!/21:- (Hsg ⋅ Qsg )ls >2!/21: 4. korak:


Hsg!/Qsg!>!4/2:!/21:!?! (Hsg ⋅ Qsg )ls 5. korak:

⇒

 Qsg    Qs{ 

1/36

Ovg!>!D!)!Hsg!/Qsg!*o!  D>1/26-!

o>1/44

izra~unavawe Nuseltovog broja za turbulentnu oblast strujawa

(Ov g )uvs >! D ⋅ [(Hsg ⋅ Qsg )oM − (Hsg ⋅ Qsg )omls ]⋅ 

 Qsg    Qs{ 

(Ov g )uvs > 1/26 ⋅ (4/2: ⋅ 21 : )

1/44



6. korak:

(

− 2 ⋅ 21 :

)

1/36

 1/814  ⋅   1/7:3 

1/44 

1/36

>76/3:

izra~unavawe koeficijenta prelaza toplote!)αmbn) u turbulentnom delu strujawa.

αuvs!>! (Ov g )uvs ⋅ ⋅

R uvs =

λg 3/6: ⋅ 21 −3 X >!76/3: ⋅ >5/81! ml 1/47 n3L

91 − 31 ⋅ 1/47 >42/:!X 2 1/2 ⋅ π ⋅ 5/8

⋅ ⋅ ⋅  R  = R mbn !,! R uvs >76/5!,!42/:!>:8/4!X  Σ



strana 30

)9/36/!−!9/37/*

9/36/!Toplotni gubici prostorije, u kojoj je potrebno odràvati temperaturu od!29pD-!iznose!2!lX/ Grejawe vazduha u prostoriji se ostvaruje grejnim telima (konvektorima). Ova tela imaju oblik kvadra!)2111!y!291!y!711!nn*-!i toplotno su izolovana na gorwoj i dowoj osnovi ({rafirani deo na slici). Temperatura na spoqa{wim povr{ima konvektora je!55pD/!Odrediti potreban broj konvektora za nadokna|ivawe toplotnih gubitaka prostorije. Zanemariti razmenu toplote zra~ewem.

re{ewe:

o>7 291!nn

!711!nn

2111!nn ⋅

9/37/!U horizontalnoj cevi spoqa{weg pre~nika e>231!nn!vr{i se potpuna kondenzacija n >36!lh0i suvozasi}ene vodene pare pritiska q>211!lQb. Cev se nalazi u velikoj prostoriji i okruèna je mirnim okolnim vazduhom stalne temperature 31pD. Sredwa temperatura na povr{i zidova prostorije iznosi 28pD, a sredwa temperatura na spoqa{woj povr{i cevi :8pD. Koeficijent emisije zra~ewa sa cevi iznosi ε>1/:. Odrediti duìnu cevi. re{ewe:

M>47/23!n



strana 31

9/38/!Kroz cev unutra{weg pre~nika!e>91!nn!struji transformatorsko uqe sredwe temperature ug>61pD-!sredwom brzinom!x>1/3!n0t. Temperatura unutra{we povr{i zidova cevi je!u{>36pD/!Ukupno ⋅

razmewen toplotni protok izme|u uqa i unutra{we povr{i cevi iznosi! R >896!X/!Odrediti duìnu cevi. M ug

ug

u{ ⋅

⋅

u − u{ R= g ⋅M 2 eπ ⋅ α

R M!>! = /// eπ ⋅ α ⋅ (u g − u { )

⇒

fizi~ki parametri za transformatorsko uqe temperaturi ug!>!61pD 2 lh X -! βg>7:/6/21−6! λg>1/233! -! ρg>956! 4 nL L n lK µg>!:/:!/21.4! Qb ⋅ t -! dqg>3/154! lhL

1. korak:

2. korak:

karakteristi~na duìna ~vrste povr{i e3 π B ml!>! 5 ⋅ = 5 ⋅ 5 >e>!91!nn P eπ

3. korak:


Sfg!>! Hsg> dqg ⋅ µ g λg

ρ g ⋅ x ⋅ mL 956 ⋅ 1/3 ⋅ 1/19 > >2476/7 µg :/: ⋅ 21 −4

β g ⋅ h ⋅ mL4 ⋅ )Ug − U{ * ⋅ ρ 3g µ 3g 4

>

3/154 ⋅ 21 ⋅ :/: ⋅ 21 1/233 d q{ ⋅ µ {

>

)Sf=3411-!laminarno strujawe*

7:/6 ⋅ 21 −6 ⋅ :/92 ⋅ 1/19 4 ⋅ )61 − 36* ⋅ 956 3

(:/: ⋅ 21 )

−4 3

>7/47!/216 Qsg!>

−4

>276/9

2/:29 ⋅ 21 4 ⋅ 35 ⋅ 21 −4 >!485/3 1/234 λ{ fizi~ki parametri za transformatorsko uqe temperaturi u{!>!36pD lK X λ{>1/234! dq{>2/:29! -! µ{>!35!/21.4! Qb ⋅ t -! lhL nL Qs{!>!

>///>



4. korak:

o q konstante u kriterijalnoj jedna~ini:!!!!!!Ovg> D ⋅ Sfn g ⋅ Qsg ⋅ Hsg ⋅ ε U

 Qs n>1/44-!!!o>1/54-!!!q>1/2-!!!εU>  g  Qs{ M pretpostavimo: > 61 ml 5. korak:

strana 32

  

1/2:

 276/9  >   485/3 

⇒

εM>2

>1/97-!!!D> 1/26 ⋅ ε M ⇒!

D>1/26


(

Ov g = 1/26 ⋅ (2476/7 )1/44 ⋅ (276/9 )1/54 ⋅ 7/47 ⋅ 21 6 6. korak:

1/2:

)

1/2

⋅ 1/97 1/36 >56/5

izra~unavawe koeficijenta prelaza toplote!)α*

α!>!Ovg! ⋅

λg X 1/233 >56/5! >!7:/3! . 4 mL 91 ⋅ 21 n3L ⋅

R 896 M!>! > >2/9!n eπ ⋅ α ⋅ (u g − u { ) 1/19π ⋅ 7:/3 ⋅ (61 − 36 ) provera pretpostavke:

2/9 M > >33/6!=!61! mL 1/19

pretpostavimo:

M( >33/6 ml

M( = M ⋅

εM ε M(

> 2/9 ⋅

⇒

pretpostavka nije ta~na

ε M( >2/22

2 >2/73!n 2/22

provera pretpostavke:

M( 2/73 > >31/36!≠!33/6! 1/19 ml

pretpostavimo:

M( ( >31/36 ml

M(( = M ⋅

εM ε M((

> 2/9 ⋅

⇒

pretpostavka nije ta~na

ε M(( >2/24

2 >2/6:!n 2/24


M( ( 2/6: > >2:/99!≈!31/36! ml 1/19

pretpostavka ta~na !!

stvarna duìna cevi iznosi M>2/6:!n/



strana 33

9/39/ Kroz prav kanal, prstenastog popre~nog preseka dimenzija ∅2>910:1!nn, ∅3>3110321!nn, proti~e voda sredwom brzinom od x>2/3!n0t. Ulazna temperatura vode je Ux2>91pD, a sredwa temperatura zidova kanala iznosi U{>31pD. Odrediti duìnu cevi na kojoj }e temperatura vode pasti na Ux3>71pD. Zanemariti prelaz toplote sa vode na spoqa{wu cev. Ux2

Ux3

U{ M prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

⇒

R 23!>!∆ I 23!,! X U23 lK lh lK ix3!>!362/2! lh

⋅

⋅

R 23!>! n x /!)!ix3!−!ix2!*!>!///

)q>2!cbs-!u>91pD*

!ix2!>!445/:!

)q>2!cbs-!u>71pD* ⋅

⋅

 1/3 3 ⋅ π 1/1: 3 ⋅ π   e3 ⋅ π e33 ⋅ π   >!3:/5!lh0t  !> :88/9 ⋅ 2/3 ⋅  nx = ρ x ⋅ x ⋅  4 − −  5   5  5 5     lh napomena:! ρx>:88/9! 4 -!je gustina vode odre|ena za sredwu n U + Ux 3 91 + 71 temperaturu vode u cevi;! Uxts = x2 >81pD = 3 3 ⋅

R 23 = 3:/5 ⋅ (362/2 − 445/: ) >−3574/8!lX

∆Uts R= ⋅M 2 e3 π ⋅ α ⋅

⇒

⋅

⋅

R !>! R23 >3574/8!lX

⋅

⇒

R M!>! = /// e3 π ⋅ α ⋅ ∆Uts



strana 34

U ∆Unby!>!91!−!31!>71pD ∆Unjo!>!71!−!31!>51pD

91pD

wpeb 71pD

∆Uts!>!

71 − 51 = 5:/4pD 71 mo 51

dfw

31pD

31pD

M

1. korak:

fizi~ki parametri za vodu odre|eni za sredwu temperaturu vode u U + Ux 3 91 + 71 = >81pD cevi: Uxts = x2 3 3 n3 X λg>77/9!/21−3!! ! ! υg>1/526/21−7! t nL

2. korak:

karakteristi~na duìna ~vrste povr{i e34 ⋅ π e33 ⋅ π − B 5 = e − e = 311 − :1 >221!nn ml!>! 5 ⋅ = 5 ⋅ 5 4 3 P e 4 ⋅ π − e3 ⋅ π

3. korak:


Qsg!> (Qs )Ug =81p D >3/66Sfg!>! 4. korak:

Qs{!> (Qs )U{ =31p D >8/13

x ⋅ mL 2/3 ⋅ 1/22 !>!4/29!/216! (turbulentno strujawe) = νg 1/526 ⋅ 21 −7 konstante u kriterijalnoj jedna~ini 1/36

1/36

 Qs   3/66  > >1/89-!!!!D>1/132!/!εM n>1/9-!!!o>1/54-!!!q>1-!!!εU!>!  g   Qs  8/13   { M D>1/132 > 61 ⇒ εM>2 ⇒! pretpostavimo: ml 5. korak: izra~unavawe Nuseltovog broja Ovg!>!1/132!/!)!4/29!/216!*1/9!/!)!3/66!*1/54!/!1/89!>!726/5



strana 35


λg X 77/9 ⋅ 21 .3 >!4848! 3 > 726/5 ⋅ .4 mL 221 ⋅ 21 nL 3574/8 M!>! >!58/4!n :1 ⋅ 21 −4 π ⋅ 4/848 ⋅ 5:/4

α!> Ov g ⋅


58/4 M >541!?!61! > mL 1/22

pretpostavka ta~na

stvarna duìna cevi iznosi M>58/4!n/

9/3:/!Predajnik toplote se sastoji od cilindri~nog, toplotno izolovanog omota~a, unutra{weg pre~nika!E>1/5!n!i snopa od!o>66!pravih cevi, spoqa{weg pre~nika!e>41!nn/!Podu`no, kroz prostor izme|u omota~a i cevi, struji suv vazduh i pritom se izobarno, pri!q>6!NQb!-!hladi od temperature!Ug2>447!L!ep!Ug3>424!L/ Maseni protok vazduha iznosi n>1/5!lh0t/!Uemperatura na spoqa{woj povr{i cevi pre~nika e je stalna i iznosi!U{>3:4!L. Odrediti duìnu predajnika toplote.

E

e

M

⋅

∆Uts R= ⋅o⋅M 2 e⋅ π⋅α ⋅

R M= >!/// e ⋅ π ⋅ α ⋅ ∆Uts ⋅ o

⇒


⋅

⋅

R 23!>!∆ I 23!,! X U23

!

⇒

⋅

⋅

R 23!>! n w ⋅ d qg ⋅ (Ug 3 − Ug2 )

⋅

R 23!>! 1/5 ⋅ 2/196 ⋅ (424 − 447) >−:/:9!lX

⇒

⋅

⋅

R !>! R23 >:/:9!lX

U ∆Unby!>!447!−!3:4!>!54!L

447!L

∆Unjo!>!424!−!3:4!>!31!L ∆Uts =

54 − 31 !>!41!L 54 mo 31

wb{evi 424!L

3:4!L

dfw

3:4!L

M dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv


strana 36

fizi~ki parametri za vazduh (q>61!cbs- Ugts =

1. korak

λg>4/14/21−3!

X nL

ρg>64/84!

lh 4

n

Ug2 + Ug 3 >435/6!L* 3

µg!>31/74/21−7!Qb/t

-

2. korak

karakteristi~na duìna ~vrste povr{i E3 π e3 π −o⋅ 3 3 3 3 B 5 > E − o ⋅ e = 1/5 − 66 ⋅ 1/14 >1/165!n ml = 5 ⋅ = 5 ⋅ 5 E +o⋅e 1/5 + 66 ⋅ 1/14 P Eπ + o ⋅ e π

3. korak


Qsg!> (Qs )Ug =435/6L -q=61 cbs >1/849Sfg!>!

ρ g ⋅ x ⋅ ml 64/84 ⋅ 1/197 ⋅ 1/165 >!///>! >2/32/215!!!!(turbulentno strujawe) µg 31/74 ⋅ 21 −7 ⋅

 E3 π e 3 π  ng = ρ g ⋅ x ⋅  o⋅ −  5 5   ⋅

x=

Qs{!> (Qs )U{ =3:4L -q=61 cbs >1/887

⇒

1/5  1/5 π 1/14 π  64/84 ⋅  − 66 ⋅  5 5   3

3

x=

>1/197!

ng  E3 π e 3 π  ρg ⋅  −o⋅  5 5  

n t



strana 37


 Qs n>1/9-!!!o>1/54-!!!q>1-!!!εU!>!  G  Qs{ M pretpostavimo: > 61 ml 5. korak:

  

1/36

 1/849  >   1/887 

⇒

1/36

εM>2

≈2-!!!!D>1/132!/!εM ⇒!

D>1/132


(

Ov g = 1/132 ⋅ 2/32 ⋅ 21 5

)

1/9

⋅ (1/849)1/54 >45

6. korak:

izra~unavawe koeficijenta prelaza toplote!)α* λ X 4/14 ⋅ 21 −3 >2:/2!! 3 α = Ov g ⋅ g !>! 45 ⋅ 1/165 ml nL

M=

R :/:9 >4/47!n >! M = e ⋅ π ⋅ α ⋅ ∆Uts ⋅ o 1/14 ⋅ π ⋅ 2:/2 ⋅ 21 −4 ⋅ 41 ⋅ 66


4/47 M >73!?!61! > mL 1/165


stvarna duìna cevi iznosi M>4/47!n/

9/41/ Kroz prav kanal pravougaonog popre~nog preseka proti~e voda brzinom x>2n0t/!Dimenzije unutra{wih stranica!pravougaonika iznose b>21!nn!i!c>31!nn/!Temperatura vode na ulazu u kanal je ux2>21pD-!a na izlazu!ux3>81pD/!Temperatura zidova kanala je!u{>211pD>dpotu/!Odrediti: a) toplotni protok sa zidova kanala na vodu )lX* b) duìnu kanala c M α

U{

voda

b

Ux>g)!M!*

voda



strana 38

b* prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

⇒

R 23!>!∆ I 23!,! X U23 lK lh lK ix3!>3:4! lh

⋅

⋅

R 23!>! n x /!)!ix3!−!ix2!*!>!///

)q>2!cbs-!u>21pD*

!ix2!>53!

)q>2!cbs-!u>81pD* ⋅

⋅

n x = ρ x ⋅ x ⋅ b ⋅ c !> ::3/3 ⋅ 2 ⋅ 21 ⋅ 21 −4 ⋅ 31 ⋅ 21 −4 >1/3! napomena:!

ρx>::3/3!

lh n4

lh t

-!je gustina vode odre|ena za sredwu

temperaturu vode u cevi;! Uxts = ⋅

Ux2 + Ux3 21 + 81 = >51pD 3 3 ⋅

⋅

R 23 = 1/3 ⋅ (3:4 − 53) >61/3!lX

⇒

∆U R >! ts ⋅ (3b + 3c) ⋅ M 2 α

R M!>! >!/// α ⋅ ∆Uts ⋅ 3 ⋅ (b + c)

R !>! R23 >61/3!lX

b) ⋅

⋅

⇒

U ∆Unby!>!:1pD ∆Unjo!>!41pD ∆Uts!>!

:1 − 41 = 65/7pD :1 mo 41

211pD

{je

211pD 81pD wpeb

21pD M

1. korak:

fizi~ki parametri za vodu odre|eni za sredwu temperaturu vode u U + Ux3 21 + 81 = >51pD cevi: Uxts = x2 3 3 n3 X lh υg>1/76:/21−7! λg>74/6!/21−3! !-! ρg!>::3/3! t nL n4

2. korak: ml!>! 5 ⋅ 3. korak:

karakteristi~na duìna ~vrste povr{i 1/12 ⋅ 1/13 B b ⋅c = 5⋅ !> 5 ⋅ >24/44!nn P 3 ⋅ (b + c) 3 ⋅ (1/12 + 1/13) potrebni kriterijumi sli~nosti

x ⋅ mL 2 ⋅ 24/24 ⋅ 21 −4 > >3!/215 (turbulentno strujawe) νg 1/76: ⋅ 21 −7 Qs{!> (Qs )U{ =211p D >2/86 Qsg!> (Qs )Ug =51p D >5/42Sfg!>!

4. korak: konstante u kriterijalnoj jedna~ini: dipl.ing. @eqko Ciganovi}!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!{fmlp@fvofu/zv

zbirka zadataka iz termodinamike  Qs n>1/9-!!!o>1/54-!!!q>1-!!!εU!>  G  Qs{ M > 61 pretpostavimo: ml 5. korak:

(

1/36

 5/42  >   2/86  ⇒

1/36

>2/36-!!!!D>1/132!/!εM

εM>2

⇒!

D>1/132

)

1/9

⋅ (5/42)1/54 ⋅ 2/36 >246/9


α!>!Ovg! ⋅

M!>!

  


Ov g = 1/132 ⋅ 3 ⋅ 21 5 6. korak:

strana 39

λg 74/6 ⋅ 21 .3 X >246/9! ⋅ >!757:/2! . 4 mL 24/44 ⋅ 21 n3L 61/3

757:/2 ⋅ 21

−4

⋅ 65/7 ⋅ 3 ⋅ (1/12 + 1/13)


>!3/48!n

3/48 M >288/9?!61! pretpostavka ta~na !! > mL 24/44 ⋅ 21 −4

stvarna duìna kanala iznosi M>3/48!n/



strana 40

9/42/ Dve cevi spoqa{wih pre~nika!e>1/229!n-!od kojih je jedna od wih toplotno izolovana nalaze se u toplotno izolovanom kanalu kvadratnog popre~nog preseka, unutra{we stranice!b>1/4!m (slika). Kroz slobodan prostor, pri stalnom pritisku!q>212/436!lQb-!proti~e voda, sredwom brzinom!x>1/3 n0t/!Temperatura vode na ulazu u kanal je!Ug2>3::!L, a iz kanala izlazi voda sredwne temperature Ug3>418!L/ Temperatura povr{i neizolovane cevi je stalna i iznosi U{>474!L/!Odrediti toplotni protok koji razmeni voda ca neizolovanom cevi kao i duìnu kanala,

α

e

b e voda prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu: ⋅

⋅

⋅

R 23!>!∆ I 23!,! X U23 lK lh lK !ix3!>!253/5! lh

!ix2!>!21:/1

⇒

⋅

⋅

R 23!>! n x /!)!ix3!−!ix2!*!>!///

)q>2!cbs-!u>37pD* )q>2!cbs-!u>45pD*

⋅

⋅

  lh e 3 π  1/229 3 π  n x = ρ x ⋅ x ⋅  b3 − 3 ⋅ !> ::6/8 ⋅ 1/3 ⋅  1/4 3 − 3 ⋅ >24/68     5 5 t     lh napomena:! ρx>::6/8! 4 -!je gustina vode odre|ena za sredwu n U + Ux3 37 + 45 temperaturu vode u cevi;! Uxts = x2 = >41pD 3 3 ⋅

R 23 = 24/68 ⋅ (253/5 − 21:/1) >564/3!lX

⇒

⋅

⋅

R !>! R23 >564/3!lX


zbirka zadataka iz termodinamike ∆Uts R= ⋅M 2 eπ ⋅ α ⋅

strana 41 ⋅

⇒

R !>!/// M!>! eπ ⋅ α ⋅ ∆Uts U

∆Unby!>474!−!3::!>75pD ∆Unjo!>474!−!418!>!67pD

cev

474!L 75 − 67 = 6:/:pD ∆Uts!> 75 mo 67

474!L 418!L voda

3::!L M 1. korak:

fizi~ki parametri za vodu odre|eni za sredwu temperaturu vode u U + Ux3 37 + 45 = >41pD cevi: Uxts = x2 3 3 n3 X lh υg>1/916/21−7! λg>72/9!/21−3! !-! ρg!>::6/8! t nL n4

2. korak:

karakteristi~na duìna ~vrste povr{i B 1/179 >1/25!n ml!>! 5 ⋅ >///> 5 ⋅ P 2/:52 e3π 1/229 3 π >1/179!n3 !> 1/4 3 − 3 ⋅ 5 5 P!>! 5 ⋅ b + 3 ⋅ eπ !>! 5 ⋅ 1/4 + 3 ⋅ 1/229 ⋅ π >2/:52!n B = b3 − 3 ⋅

3. korak:


x ⋅ mL 1/3 ⋅ 1/25 > >4/59/215 (turbulentno strujawe) νg 1/916 ⋅ 21 −7 Qs{!> (Qs )U{ =:1p D >2/:6 Qsg!> (Qs )Ug =41p D >6/53Sfg!>!

4. korak:

konstante u kriterijalnoj jedna~ini:

 Qs n>1/9-!!!o>1/54-!!!q>1-!!!εU!>  G  Qs{ M > 61 pretpostavimo: ml

  

1/36

 6/53  >   2/:6  ⇒

1/36

>2/3:-!!!!D>1/132!/!εM

εM>2

⇒!

D>1/132




(

Ov g = 1/132 ⋅ 4/59 ⋅ 21 5 6. korak:

strana 42

)

1/9

⋅ (6/53)1/54 ⋅ 2/3: >351/9


α!>!Ovg! ⋅

λg 72/9 ⋅ 21 .3 X >351/9! ⋅ >!2174! mL 1/25 n3L ⋅

R 564/3 M!> > >!2:/3!n eπ ⋅ α ⋅ ∆Uts 1/229 ⋅ π ⋅ 2174 ⋅ 21 −4 ⋅ 6:/:


M 2:/3 >248/2?!61! > mL 1/25


stvarna duìna cevi iznosi M>2:/3!n/ 9/43/!U tolplotno izolovanom kanalu kvadratnog popre~og preseka stranice!b>1/6!n!postavqena je cev spoqa{weg pre~nika!e>1/3!n/!Kroz kanal struji suv vazduh temperature!ug>41pD-!brzinom!x>9 n0t/!Temperatura zidova kanala iznosi U3>321pD-!a temperatura spoqa{we povr{i cev!U2?U3>@/ Koeficijent emisije zra~ewa cevi iznosi!ε2>1/:6-!a zidova kanala!ε3>1/9/!Smatraju}i da koeficijenti prelaza toplote )α*!sa obe povr{i (sa cevi na vazduh i sa kanala na vazduh) imaju istu vrednost, odrediti: b* temperaturu spoqa{we povr{i cevi b) ukupan toplotni fluks koji odaje spoqa{wa povr{ cevi Pri odre|ivawu Nuseltovog broja smatrati da εU!i!εM!iznose!εU>εM>2

α ε2

ε3

α U3 Ug U2



strana 43

a) ⋅  ⋅  toplotni bilans unutra{wih povr{i zidova kanala:  R {sb•okf  =  R qsfmb{   23  3g 5

5

 U2   U    −  3  U − Ug  211   211  ⋅M = 3 ⋅b⋅M ⋅ 5 2 2 α eπ ⋅ D23

⇒

5

5b ⋅ α  U  U2 = 211 ⋅ 5  3  + (U3 − Ug ) ⋅ >!/// eπ ⋅ D23  211  D23!>!Dd!ε23!>//!!/!>!6/78/!1/995!>!6! ε23!>!

X n3L 5

2 2 2 >!1/995 > >! 2 1/3π  2    2 B2  2 2 eπ ⋅ M  2 + 2 −       + − 2 + − 2  1/:6 5 ⋅ 1/6  1/9 ε2 B3  ε3   ε 2 5b ⋅ M  ε 3 

fizi~ki parametri za vazduh na temperaturi!ug!>41pD lh X νg!>!27!/21−7! Qb ⋅ t λg!>!3/78!/21−3!! -! !ρg!>!::3/3! 4 nL n

1. korak:

2. korak:

karakteristi~na duìna ~vrste povr{i 1/3 3 π e3π 1/6 3 − b3 − B 5 >1/444!n 5 !> 5 ⋅ ml!>! 5 ⋅ = 5 ⋅ 5 ⋅ 1/6 + 1/3π P 5 ⋅ b + eπ

3. korak:


x ⋅ mL 9 ⋅ 1/444 >2/77!/216 > νg 27 ⋅ 21 −7 Qsg!> (Qs )Ug =41p D >1/812

Sfg!>!

4. korak:

(turbulentno strujawe)


n>1/9-!!!o>1/54-!!!q>1-!!!εU!>2-!!!!D>1/132!/!εM>1/132



strana 44


(

Ov g = 1/132 ⋅ 2/77 ⋅ 21 6 6. korak:

)

1/9

⋅ (1/812)1/54 >381/5


α!>Ovg! ⋅

λg 3/78 ⋅ 21 .3 X >32/8! >381/5! ⋅ 1/444 mL n3L 5

5 ⋅ 1/6 ⋅ 32/8  594  U2 = 211 ⋅ 5  >853!L  + (321 − 41) ⋅ 211 1/3 ⋅ π ⋅ 6   b) 5

(

r Σ = r {sb•okf

5

 U2   U    −  3  U − Ug  211   211  + 2 > 2g 2 2 eπ ⋅ α eπ ⋅ D23

)23 + (rqsfmb{ )

5

5

 853   594    −  211 X 853 − 414    211  + >8924/15!,!6:96/66!>!248:9/7! rΣ = 2 2 n 1/3π ⋅ 32/8 1/3π ⋅ 6

9/44/!Horizontalna cev, spoqa{weg pre~nika!e>41!nn!i duìne!M>6!n, se hladi popre~nom strujom vode sredwe temperature!ug>21pD/!Voda struji brzinom!x>3!n0t-!pod napadnim uglom od!β>71p/ Temperatura spoqa{we povr{i cevi iznosi!u{>91pD/!Odrediti toplotni protok konvekcijom sa cevi na vodu. u{

voda

ug α β



R=

strana 45

u{ − ug ⋅ M >!/// 2 e⋅π⋅α

fizi~ki parametri za vodu na temperaturi!ug>21pD X n3 λg!>!68/5!/21.3!! ! !υg!>!2/417!/21−7! t nL

1. korak:

2. korak:


ml!>!et!>!41!nn 3. korak:


x ⋅ mL 3 ⋅ 1/14 > >5/6:!/215 (prelazni reìm strujawa) νg 2/417 ⋅ 21 −7 Qs{!> (Qs )U{ =91p D >3/32 Qsg!> (Qs )Ug =21p D >:/63-

Sfg!>!

4. korak:


 Qs n>1/7-!!!o>1/48-!!!q>1-!!!εU!>  G  Qs{ /! /! D>1/37! εβ>1/37! 1/:4>1/35 5. korak:

  

1/36

1/36

 :/63  > >2/55  3/32  )β>71pD! ⇒

εβ>!1/:4*


Ovg!>!1/35/!)!5/6:!/215!*1/7!/!)!:/63!*1/48!/!2/55!>!5:9/7 6. korak:


α!>!Ovg! ⋅ ⋅

R=

λg 68/5 ⋅ 21 .3 X >5:9/7! ⋅ >!:46:/:! mL 1/14 n3L 91 − 21 2

⋅ 6 >419/86!lX

1/14 ⋅ π ⋅ :46:/: ⋅ 21 −4



strana 46

9/45/!Suvozasi}ena vodena para!)u>291pD*!transportuje se kroz parovod na rastojawe od!M>5!ln/!Parovod je napravqen od ~eli~nih cevi!)λ2>61!X0nL*-!pre~nika!)∅>211091!nn*!i izolovan je slojem staklene vune!)λ3>1/15!X0nL*!debqine!δ>:1!nn/!Pra}ewe atmosferskih uslova pokazalo je da: - maksimalna brzina vetra koji duva normalno na parovod je x>21!n0t - minimalna temperatura okolnog vazduha je!−21pD U parovod treba ugraditi kondenzacione lonce na drugom!)3/*!i ~etvrtom!)5/*!kilometru. Ukoliko gubici zra~ewem iznose!71&!od gubitaka konvekcijom, a koeficijent prelaza toplote sa strane pare koja se kondenzuje du` celog cevovoda iznosi α2>:111!X0n3L!-!odrediti potreban kapacitet kondenzacionih lonaca!)lh0t*/!Pri izra~unavawu koeficijenta prelaza toplote sa strane vazduha zanemariti popravku!εUtj. smatrati da je!εU>2 razmewen toplotni protok na prva dva kilometra!)M2>3111!n*; ⋅ ⋅ ⋅ ⋅ R 2 =  R  +  R  > 2/7 ⋅  R    qsfmb{   {sb•fokf   qsfmb{

Uqbsb − Uwb{evi

⋅

R 2 = 2/7 ⋅

e e 2 2 2 2 + mo 3 + mo 4 + e2π ⋅ α 2 3π ⋅ λ 2 e2 3π ⋅ λ 3 e 3 e 4 π ⋅ α 3

⋅ M2 >!///

fizi~ki parametri za vazduh!ob!ufnqfsbuvsj!ug>−21pD n3 X λg!>!3/47!/21−3! νg!>!23/54/21−7! nL t

1. korak:

2. korak:


mfl>e4>!1/39!n 3. korak:

izra~unavawe potrebnih kriterijuma sli~nosti

x ⋅ mL 21 ⋅ 1/39 >3/36!/216 !> υg 23/54 ⋅ 21 −7 Qsg!> (Qs )Ug = −21p D >1/823

Sfg!>!

4. korak:

(turbulentno strujawe)


D>1/134-

n>1/9-

o>1/5-

q>1



strana 47


(

Ov g = 1/132 ⋅ 3/36 ⋅ 21 6 6. korak

)

1/9

⋅ (1/823)1/5 ⋅ (Hsg )1 ⋅ 2 >495/2

izra~unavawe koeficijenta prelaza toplote!)α* α = Ov g ⋅

λg 3/47 ⋅ 21 −3 X >43/5!! !>! 495/2 ⋅ mfl 1/39 n3L

⋅

R 2 = 2/7 ⋅

⋅

⋅

R 2 = n2 ⋅ s napomena:

291 + 21 ⋅ 3111 >! 2 2 211 2 391 2 mo mo + + + 1/19π ⋅ :111 3π ⋅ 61 91 3π ⋅ 1/15 211 1/39π ⋅ 58/6

⇒

⋅

⋅

R 2 >!258!lX

⋅

258 lh R2 >1/184! > n2 = 3126 t s

Razmewena toplota na drugom delu cevovoda!)duìne!M3>3!ln) je identi~na kao na prvom delu cevovoda!)duìne!M2>3!ln*-!pa je i kapacitet drugog kondenzazcionog lonca jednak kapacitetu prvog ⋅ lh kondenzacionog lonca-! n3 >1/184! t

9/46/!Upravno na cev, spoqa{weg pre~nika!e>311!nn!i duìne!M>9!n-!struji suv vazduh temperature ug>−31pD-!pri!q>212/4!lQb/!Temperatura na spoqa{woj povr{i cevi je konstantna i iznosi u{>291pD/ Odrediti brzinu strujawa vazduha pri kojoj toplotni protok sa cevi na vazduh iznosi 31!lX/ ⋅

u − ug X R R= { !>!31! 3 ⋅M ⇒ α!>! 2 e ⋅ π ⋅ M ⋅ (u { − u g ) nL e⋅π⋅α X 31 α!>! !>!31! 3 1/3 ⋅ π ⋅ 9 ⋅ (291 + 31) nL ⋅



strana 48

fizi~ki parametri za vazduh na!ug>−31pD λg!>!3/39/21−3!!

X ! nL

ml>1/3!n-!

!Qsg!>!1/827-!

α!>! Ovg ⋅

νg!>!23/8:/21−7!

!

Qs{!>!1/792-

λ λ = D ⋅ Sfn ⋅ Qso ⋅ Hsq ⋅ ε U ⋅ ml ml

 m 2 Sf!>!  α ⋅ l ⋅ o  λ D ⋅ Qsg ⋅ Hsgq ⋅ ε U 

lh n4  Qs εU>  G  QsB

   

1/36

>2

⇒

2

n  >!///  

predpostavimo da je strujawe vazduha oko cevi turbulentno tj. da vaì: 3/216!=!Sfg!=!2/218

⇒

D!>1/134/!εβ!>1/134-

n!>1/9-!!!o!>1/5-!!!q>1 2

  1/9 1/3 2  >9/53/215 ⋅ Sf>  31 ⋅ −3 1/5   3 / 39 ⋅ 21 1 / 134 ⋅ 1 / 827 ⋅ 2  

pretpostavka neta~na!"

predpostavimo da je strujawe vazduha oko cevi preobraàjno tj. da vaì: 2/214!=!Sfg!=!3/216

⇒

D!>1/37/!εβ!>1/37-!!!n!>1/7-!!!o!>1/48-!!!q>1 2

  1/7 1/3 2  >7/49/215 ⋅ Sf>  31 ⋅ −3 1/48   3 / 39 ⋅ 21 1 / 37 ⋅ 1 / 827 ⋅ 2   x!>!

pretpostavka ta~na!"

Sf g ⋅ υ g n 7/49 ⋅ 21 5 ⋅ 23/8: ⋅ 21 −7 >5/19! >! 1/3 ml t


zbirka zadataka iz termodinamike zadatak za ve`bawe:

strana 49

)9/47/*

9/47/!Kroz cev od ner|aju}eg ~elika!)!λ>28!X0nL!*-!pre~nika!∅>62y3/:!nn!i duìne!M>3/6!n-!struji voda!ux!>91pD>dpotu-!!sredwom brzinom!xx>2!n0t. Upravno na cev struji vazduh sredwe temperature ug>31pD>dpotu-!sredwom brzinom!xg>3!n0t/!Odrediti toplotni protok sa vode na vazduh kao i temperaturu spoqa{we povr{i cevi ( smatrati da je!εUv!>!εUt!>!2*/ ⋅

re{ewe:

u!>!8:/6pD

R >655/3!X-

9/48/!Dve kvadratne plo~e stranica duìne!b>2!n obrzauju ravnu povr{ (zanemarqive debqine) du` koje brzinom!x>2!n0t!struji suv vazduh!)u>21pD-!q>2!cbs*/!Odrediti koliko se toplote preda vazduhu za slede}a tri slu~aja: a) obe plo~e su stalne temperature u{>231pD b) prva plo~a je stalne temperature u{>231pD, a druga je adijabatski izolovana d* prva plo~a je adijabatski izolovana, a druga plo~a je stalne temperature u{>231pD b

b

b

b* ml!>!

Sfls ⋅ ν g 6 ⋅ 21 6 ⋅ 25/27 ⋅ 21 −7 > >7/:!n x 2

kako je M>b,b>!3!n! ≤ !mls strujawe vazduha du` cele plo~e je laminarno napomena:

υg!>25/27!/21−7

n3 t

(vazduh na temperaturi!ug!>21pD)



strana 50

b* ⋅

R 1−3b >!

u{ − ug ⋅ b ⋅ b ⋅ 3 >/// 2 α 1−3b

fizi~ki parametri za vazduh na temperaturi!ug!>21pD n3 X λg!>!3/62!/21−3! υg!>25/27!/21−7 t nL

1. korak:

2. korak:


ml!>3b!>3!n 3. korak:


x ⋅ ml 2⋅ 3 >2/52!/216 = υg 25/27 ⋅ 21 −7 Qsg!> (Qs )Ug =21p D >1/816Qs{!> (Qs )U{ =231p D >1/797

Sfg!>

4. korak:


 Qs n>1/6-!!!o>1/44-!!!q>1 -!!!εU!>!  G  Qs{ 5. korak:

  

1/36

 1/816  =   1/797 

1/36

>2-!!!D>1/775


Ovg!>!1/775/!)!2/52!/216!*1/6!/!)!1/816!*1/44!/!2!>333/28 6. korak:


α1−3b!>Ovg!/ ⋅

R 1−3b >!

λg X 3/62 ⋅ 21 .3 >3/8:! 3 >333/28/! 3 mL nL

231 − 21 ⋅ 2 ⋅ 2 ⋅ 3 >!724/9!X 2 3/8:



strana 51

b) !

⋅

R 1−b >!

u{ − ug ⋅ b ⋅ b >/// 2 α 1 −b

2. korak: karakteristi~na duìna ~vrste povr{i ml!>b!>2!n 3. korak:

potrebni kriterijumi sli~nosti x ⋅ ml 2⋅ 2 >1/8!/216 = Sfg!> −7 υg 25/27 ⋅ 21

5. korak:


Ovg!>!1/775/!)!1/8!/216!*1/6!/!)!1/816!*1/44!/!2!>!267/65 6. korak:


λg 3/62⋅ 21.3 X >!4/:4! 3 >267/65/! mL 2 nL ⋅ 231 − 21 ⋅ 2 ⋅ 2>!543/4!X R 1−b >! 2 4/:4

α1−b!>!Ovg!/

napomena:

prikazani su samo oni koraci koji nisu isti kao pod a)

c) ⋅

⋅

⋅

R b−3b !>! R 1−3b !−! R 1−b !>!724/9!−!534/4!>292/6!X napomena: ⋅

R b−3b >!

zadatak pod c) se moè re{iti i na slede}i na~in:

u{ − ug 231 − 21 ⋅ 2 ⋅ 2>292/6!X ⋅ b ⋅ (M − b) >///> 2 2 α b−M 2/76

(Ov g )b−M > D ⋅ [(Sf g )n1−M − (Sf g )n1−b ]⋅ Qsgo ⋅ ε u (Ov g )b−M >!1/775/! (2/52 ⋅ 21 6 )

1/6



αb−M!>! (Ov g )b−M /

(

− 1/8 ⋅ 21 6

)

1/6 



/!)!1/816!*1/44!/!2>76/74

λg X 3/62⋅ 21.3 >!2/76! 3 >76/74/! 2 mL nL



strana 52

9/49/ Vazduh temperature!ug>31pD-!struji sredwom brzinom!3/6!n0t preko ravne plo~e duìne!b>6!n!i {irine!c>2/6!n. Povr{i plo~e se odràvaju na stalnoj temperaturi od u{>:1pD/!Odrediti toplotni protok sa plo~e na vazduh u laminarnom delu strujawa, turbulentnom delu strujawa kao i ukupni du` cele plo~e.

mls

vazduh

c

b Sfls ⋅ ν g 6 ⋅ 21 6 ⋅ 26/17 ⋅ 21 −7 >4!n > 3/6 x kako je M>b>!6!n!?!mls strujawe vazduha na duìni mls je laminarno a na duìni M−mls turbulentno n3 napomena: υg!>26/17!/21−7 (vazduh na temperaturi!ug!>31pD) t mls!>!

u{ − ug ⋅ mls ⋅ c >/// 2 α mbn 1. korak: fizi~ki parametri zavazduh na temperaturi!ug!>!31pD n3 X λg!>!3/6:!/21−3! -!!υg!>26/17!/21−7! t nL ⋅

R mbn!>!

2. korak: karakteristi~na duìna ~vrste povr{i ml!>!mls!>4!n 3. korak:


Sfg!>Sfls>6!/216 Qsg!> (Qs )Ug =31p D >1/814-

Qs{!> (Qs )U{ =:1p D >1/7:



strana 53



  

1/36

 1/814  =   1/7: 

1/36

>2-!!!D>1/775


Ovg!>!1/775/!)!6!/216!*1/6!/!)!1/814!*1/44!/!2!>!528/:4 izra~unavawe koeficijenta prelaza toplote!)αmbn*

6. korak:

3/6: ⋅ 21.3 λg X >!4/72! 3 >528/:4! 4 mL nL ⋅ :1 − 31 R mbn!>! ⋅ 4 ⋅ 2/6 >2248/3!X 2 4/72

αmbn!>!Ovg!/

⋅

R uvs!>!

u{ − ug ⋅ (b − mls ) ⋅ c >/// 2 α uvs

2. korak: karakteristi~na duìna ~vrste povr{i ml!>!b!−!mls!>3!n 3. korak:


(Sf g )mls >Sfls>6!/216-! (Sf g ) M > x ⋅ M = υg

4. korak:

26/17 ⋅ 21 −7

>9/4!/216



3/6 ⋅ 6

  

1/36

 1/814  =   1/7: 

1/36

>2-!!!D>1/148


(Ov g )mls −M > D ⋅ [(Sf g )nM − (Sf g )nls ]⋅ Qsgo ⋅ ε u (Ov g )mls −M >!1/148/! (9/4 ⋅ 21 6 )

1/9



(

− 6 ⋅ 21 6

)

1/9 



/!)!1/814!*1/54!/!2>687/24



strana 54

izra~unavawe koeficijenta prelaza toplote!)αuvs*

6. korak:

αuvs> (Ov g )m

ls −M

⋅

R uvs!>!

/

λg X 3/6: ⋅ 21 .3 >!8/57! 3 >687/24/! 3 mL nL

:1 − 31 ⋅ (6 − 4 ) ⋅ 2/6 >2677/9!X 2 8/57

⋅ ⋅  ⋅  +  R uvs  >2248/3!,!2677/9!>3815!X R 1−m =  R mbn    1−mls   mls −M

9/4:/!Vertikalnu plo~u (zanemarqive debqine), visine!i>3/5!n!i {irine!b>1/9!n!sa obe strane (u pravcu kra}e strane) opstrujava vazduh temperature!ug>31pD. Toplotni protok konvekcijom sa plo~e na vazduh iznosi!6!lX/!Odrediti sredwu brzinu strujawa vazduha, tako da se temperatura na povr{ima plo~e odràva konstantnom i iznosi u{>81pD, smatraju}i da je strujawe vazduha turbulentno po celoj

plo~i. b

i

⋅

u − ug R >! { ⋅b⋅i⋅3 ⇒ 2 α X 6111 α!>! !>37! 3 (81 − 31) ⋅ 1/9 ⋅ 3/5 ⋅ 3 nL

vazduh

⋅

R α!>! !> (u { − u g ) ⋅ b ⋅ i ⋅ 3

fizi~ki parametri zavazduh na temperaturi!ug!>!31pD n3 X λg!>!3/6:!/21−3! -!!υg!>26/17!/21−7! t nL

1. korak:



strana 55

2. korak: karakteristi~na duìna ~vrste povr{i ml!>!b!>1/9!n 3. korak:


Qsg!> (Qs )Ug =31p D >1/8144. korak:

Qs{!> (Qs )U{ =81p D >1/7:5



1/36

 1/814  =   1/7:5 

1/36

>2-!!!D>1/148


Ov g = α ⋅ 5. korak:

  

ml 1/9 = 37 ⋅ >914/1: λg 3/6: ⋅ 21 −3

izra~unavawe Rejnoldsovog broja

o q Ovg> D ⋅ Sfn g ⋅ Qsg ⋅ Hsg ⋅ ε U

⇒

 Ov g Sfg!>!   D ⋅ Qs o ⋅ Hs q ⋅ ε U g g 

2

n   

2

  1/9 914/1:  > 4/295 ⋅ 21 6 Sfg!>!   1/148 ⋅ 1/814 1/54 ⋅ 2 ⋅ 2    x!>!

Sf g ⋅ υ g n 4/295 ⋅ 21 6 ⋅ 26/17 ⋅ 21 −7 !>!7! = ml 1/9 t

zadatak za ve`bawe:

)9/51/*

9/51/!Vaqanu, vertikalnu bakarnu plo~u, visine!3/3!n!i {irine!1/:!m sa obe strane opstrujava vazduh sredwe temperature!31pD-!sredwom brzinom!7!n0t. Sredwe temperature obe povr{i plo~e iznose 81pD-!a zidova velike prostorije u kojoj se plo~a nalazi!31pD/!Odrediti toplotni protok koji se odvodi sa plo~e (debqinu plo~e zanemariti) ako se strujawe vr{i u pravcu kra}e strane. ⋅ ⋅  ⋅  re{ewe:!!!!!!!  R  >  R qsfmb{  ,  R {sb•fokf  >!3126/7!,!:23/4!>!3:83/:!X  ∑  23  24



strana 56

9/52/!Iznad horizontalne ravne betonske plo~e, duìne M>3!n, toplotno izolovane sa dowe strane, suv vazduh stawa (q>2!cbs-!Ug>394!L) proti~e brzinom x>4!n0t. Ako se pod dejstvom toplotnog zra~ewa, na gorwoj povr{i plo~e ustali temperatura U{>434!L, odrediti povr{inski toplotni protok (toplotni fluks) tog zra~ewa i grafi~ki predstaviti raspored temperatura u betonskoj plo~i i okolnom vazduhu. rep{sb•fop Ug

U{

vazduh, x>4!n0t

rtpqtuw/{sb•fokf

α

rsfgmflupwbop

U{>dpotu M

toplotni bilans ozra~ene povr{i: r ep{sb•fop = rsfgmflupwbop + r bqtpscpwbop r ep{sb•fop = r ep{sb•fop ⋅ (2 − ε ) + 5   U{     2  U − Ug  211  r ep{sb•fop > ⋅  { + 2 ε  2  α ε ⋅ Dd 

 U{    U{ − U3 U2 − Ug  211  + + 2 2 δ ε ⋅ Dd λ α

5

    >///   

ε!>!L!/!εo!>!1/:9!/!1/:5!>!1/:3 α!>!@



strana 57

fizi~ki parametri za vazduh na temperaturi!ug!>21pD n3 X λg!>!3/62!/21−3! υg!>25/27!/21−7 t nL

1. korak:

2. korak:


ml!>M!>3!n 3. korak: Sfg!>

potrebni kriterijumi sli~nosti x ⋅ ml 4⋅3 >5/35!/216 = υg 25/27 ⋅ 21 −7

Qsg!> (Qs )Ug =21p D >1/8164. korak:

Qs{!> (Qs )U{ =231p D >1/7:9



  

1/36

 1/816  =   1/7:9 

1/36

>2-!!!D>1/775


Ovg!>!1/775/!)!5/35!/216!*1/6!/!)!1/816!*1/44!/!2!>496/37 6. korak:


α!>Ovg!/

λg X 3/62 ⋅ 21 .3 >5/95! 3 >496/37/! 3 mL nL

5   U{     2  U − Ug  211  r ep{sb•fop > ⋅  { + 2 ε  2  α ε ⋅ Dd 

5    434        2  434 − 394 X  211   ⋅ + >  >938/7! 3 2 2 n  1/:3     5/95 1/:3 ⋅ 6/78    



strana 58 ⋅

9/53/!U istosmernom razmewiva~u toplote tipa cev u cevi!zagreva se! n f>2611!lh0i!etanola!od U2>21pD!to!U3>41pD/!Grejni fluid je suvozasi}ena vodena para!q>1/23!cbs!koja se u procesu razmene toplte sa etanolom potpuno kondenzuje. Etanol proti~e kroz cev a para se kondenzuje u anularnom prostoru. Unutra{wi pre~nik unutra{we cevi iznosi!e>211!nn/!Zanemaruju}i toplotni otpor prelaza sa pare na cev, toplotni otpor provo|ewa kroz cev kao i toplotne gubitke u okolinu odrediti: a) maseni protok grejne pare b) duìnu cevi b* prvi zakon termodinamike za proces u razmewiva~u toplote: ⋅

⋅

⋅

⋅

⋅

⋅

⇒

R 23!>!∆ I 23!,! X U23 ⋅

⋅

I 2!>! I 3

⋅

nq ⋅ i2 + nf ⋅ d qf ⋅ U2 = nq ⋅ i3 + nf ⋅ d qf ⋅ U3

⋅

⋅

nq =

nf ⋅ d qf ⋅ (U3 − U2 ) i2 − i3

lK lh lK !i3>318! lh

!i2>36:2

dqf>3/56!

lK lhL

2611 ⋅ 3/56 ⋅ (41 − 21) lh >9/67!/21−3! > 4711 3495 t )i′′-!!q>1/23!cbs* )i′-!!!q>1/23!cbs*

specifi~ni toplotni kapacitet etanola odre|en za sredwu temperaturu etanola: Uf!>!

21 + 41 = 31pD 3

b) ⋅

R sb{

∆UTS = ⋅M 2 l

⋅

⇒

R sb{ M!>! !>/// l ⋅ ∆Uts

⋅

R sb{!−!interno razmewena toplota u razmewiva~u izme|u pare i etanola l!−!koeficijent prolaza toplote sa pare na etanol ∆Uts!−!sredwa logaritamska razlika temperatura izme|u pare i etanola



strana 59

L

para-!i2

para-!i3

⋅

R sb{

etanom-!U2

etanol-!U3

prvi zakon termodinamike za proces u otvorenom termodinami~kom sistemu ograni~enom ⋅

konturom K: ⋅

⋅

⋅

R sb{!>!∆ I 23!,! X U23

⋅

R sb{!> nq ⋅ (i2 − i3 ) = 9/67 ⋅ 21 −3 ⋅ (36:2 − 318) >315/18!lX

U

∆Unby!>5:/26!−!21>!4:/26pD ∆Unjo!>5:/26!−!41!>!2:/26pD 4:/26 − 2:/26 = 39/4pD ∆Uts!> 4:/26 mo 2:/26

para

5:/26

5:/26 41

etanom 21 M

e 2 2 2 2 mo 3 + = + l e3 π ⋅ α Q 3π ⋅ λ e3 e2π ⋅ α f

⇒

l> e2 ⋅ π ⋅ α >///

2 !!!!−!!toplotni otpor prelaza sa strane pare, zanemaren uzadatku e3 π ⋅ α Q e 2 ⋅ mo 3 !−!toplptni otpor provo|ewa kroz cev, zanemaren u zadatku 3π ⋅ λ e3 2 !!!!−!!toplotni otpor prelaza sa strane etanola e2π ⋅ α f αf!>!@



strana 60

1. korak:

fizi~ki parametri za etanol odre|eni za sredwu temperaturu U + Uf3 21 + 41 = >31pD etanola u cevi: Ufts = f2 3 3 lh X λg!>!1/294!! ! ! ρg!>!89:! 4 nL n lK dqg!>!3/56! µg!>!2/2:!/21−4! Qb ⋅ t lhL

2. korak:

karakteristi~na duìna ~vrste povr{i e23 ⋅ π B ml!>! 5 ⋅ = 5 ⋅ 5 >e2>211!nn P e2 ⋅ π

3. korak: Qsg!> Qs{!>

potrebni kriterijumi sli~nosti dqg ⋅ µ g λg d q{ ⋅ µ { λ{

>

3/56 ⋅ 21 4 ⋅ 2/2: ⋅ 21 −4 >26/:4 1/294

>///>

3/92 ⋅ 21 4 ⋅ 1/7:6 ⋅ 21 −4 >21/:8 1/289

µ{-!λ{-!dq{

fizi~ki parametri etanola na temperaturi!U{>5:/26pD lK X µ{!>!1/7:6!/21−4! Qb ⋅ t ! λ{!>!1/289! ! dq{!>!3/92! lhL nL

Sfg!>

ρ g ⋅ x ⋅ mL 89: ⋅ 7/8 ⋅ 21 −3 ⋅ 1/2 >!5553/4!!!!!prelazni reìm strujawa >!///> µg 2/2: ⋅ 21 −4

2611 4711 >7/8!/21−3! n x!>! > 3 t ρ g ⋅ e2 ⋅ π 89: ⋅ 1/23 ⋅ π ⋅

5 ⋅ nF

4. korak:

5⋅


 Qs n>1-!!!o>1/54-!!!q>1-!!!εU!>!  g  Qs{ L1!>g!)!Sf!*>!24/9 5. korak:

  

1/36

 26/:4  >   21/:8 

1/36

>2/2-!!!!D>L1>//!/>24/9


Ovg!>24/9!/)!26/:4!*1/54!/!2/2!>!61



strana 61


αf!>!Ovg! ⋅

λg X 1/294 >61! ⋅ >!:2/6 3 . 4 mL 211 ⋅ 21 nL

l> e2 ⋅ π ⋅ α f> 1/2 ⋅ π ⋅ :2/4 >39/86!

X nL

⋅

R sb{ 315/18 >!362!n M!>! !> l ⋅ ∆Uts 39/86 ⋅ 21 .4 ⋅ 39/4 9/54/!U kqu~alu vodu pritiska!q>21!cbs-!koja pri konstantnom pritisku isparava u proto~nom kotlu, potopqeno je 31 pravih cevi, unutra{weg pre~nika!e>:6!nn/!Kroz cev !)pri konstantnom pritisku, q>2!cbs*-!brzinom!xg>7!n0t-!struji dimni gas!)sDP3>1/24-!sI3P>1/22-!sO3>1/87*/!Temperatura gasa na ulazu u cevi je!571pD-!a na izlazu iz wih!371pD/!Ako se zanemari toplotni otpor provo|ewa kroz zidove cevi kao i toplotni otpor prelaza sa cevi na kqu~alu vodu odrediti: a) koli~inu vode koja ispari u kotlu )lh0t* b) !potrebnu duìnu cevi b* prvi zakon termodinamike za proces u razmewiva~u toplote: ⋅

⋅

⋅

⇒

R 23!>!∆ I 23!,! X U23 ⋅

⋅

⋅

⋅

⋅

I 2!>! I 3

⋅

n x ⋅ i2 + neh ⋅ d qeh ⋅ U2 = n x ⋅ i3 + neh ⋅ d qeh ⋅ U3 ⋅

nx =

⋅

neh ⋅ d qeh ⋅ (U2 − U3 )

⋅

i3 − i2

n eh = ρ eh ⋅ x ⋅

>///>

lh 1/59 ⋅ 2/25 ⋅ (571 − 371) >6/54!/21−3! 3126/4 t

lh e3 π 1/1:6 3 ⋅ π ⋅ o = 1/673 ⋅ 7 ⋅ ⋅ 31 >1/59! 5 5 t

lh lK ! specifi~ni toplotni kapacitet i gustina dimnnog ρeh>1/673! 4 lhL n gasa odre|eni za sredwu temperaturu dimnog gasa: Uf!> 571 + 371 = 471pD 3 lK !i2>!873/8 )i′-!!q>21!cbs* lh lK !i3>3889! )i′′-!!!q>21!cbs* lh

dqeh>2/25!



strana 62

b) ⋅

R sb{ =

∆UTS ⋅M 2 l

⋅

⇒

M!>!

R sb{ !>/// l ⋅ ∆Uts

⋅

R sb{!−!interno razmewena toplota u razmewiva~u izme|u dimnog gasa i vode l!−!koeficijent prolaza toplote sa dimnog gasa na vodu ∆Uts!−!sredwa logaritamska razlika temperatura izme|u dimnog gasa i vode L dimni gas gas-!U2

dimni gas-!U3

⋅

R sb{ voda-!i2

para-!i3


ograni~enom konturom K: ⋅

⋅

⋅

R sb{!>!∆ I 23!,! X U23

⋅

R sb{!> n eh ⋅ d qeh (U2 − U3 ) = 1/59 ⋅ 2/25 ⋅ (571 − 371) >21:/55!lX

∆Unby!>571!−291>!391pD ∆Unjo!>371!−!291!>!91pD

U 571 dimni gas

391 − 91 = 26:/76pD ∆Uts!> 391 mo 91 291

voda

371

291 M



strana 63

e 2 2 2 2 = + mo 3 + l e3 π ⋅ α x 3π ⋅ λ e3 e2π ⋅ α eh

⇒

l> e2 ⋅ π ⋅ α eh>///

2 !!!!−!!toplotni otpor prelaza sa strane vode, zanemaren uzadatku e3 π ⋅ α x e 2 ⋅ mo 3 !−!toplptni otpor provo|ewa kroz cev, zanemaren u zadatku 3π ⋅ λ e3 2 !!!!−!!toplotni otpor prelaza sa strane dimnog gasa e2π ⋅ α eh αeh!>!@ 1. korak:

fizi~ki parametri za dimni gas odre|eni za sredwu temperaturu Ueh2 + Ueh3 571 + 371 = >471pD dimnog gasa: Uehts = 3 3 lh X λg!>!6/47!/21−3!! ! ! ρg!>!1/673! 4 nL n lK dqg!>!2/24:! µg!>!41/4!/21−7! Qb ⋅ t lhL

2. korak:

karakteristi~na duìna ~vrste povr{i e23 ⋅ π B ml!>! 5 ⋅ = 5 ⋅ 5 >e2>:6!nn P e2 ⋅ π

3. korak:


Qsg!> (Qs )Ug =471p D >1/75Sfg!>

Qs{!> (Qs )U{ =291p D >1/78

ρ g ⋅ x ⋅ mL 1/673 ⋅ 7 ⋅ 1/1:6 >21683 !> µg 41/4 ⋅ 21 −7

4. korak:

konstante u kriterijalnoj jedna~ini U 564 n>1-!!!o>1/54-!!!q>1-!εU!>2/38!−!1/38 ⋅ { >2/38!−!1/38 ⋅ >2/188Ug 744 predpostavimo !

M ?61! mL

⇒

εM>!2

⇒

D>1/132

D>1/132!/!εM!>1/132



strana 64


Ovg!>1/132/!)21683!*1/9!/)!1/75!*1/54!/2/188>41/:4 6. korak:


αeh!>!Ovg! ⋅

λg X 6/47 ⋅ 21 .3 >41/:4 ⋅ >28/56! 3 .4 mL :6 ⋅ 21 nL

l> e2 ⋅ π ⋅ α eh> 1/1:6 ⋅ π ⋅ 28/56 >!6/32!

X nL

⋅

R 21:/55 = M!>! >!7/7!n l ⋅ ∆Uts ⋅ o 6/32 ⋅ 21 −4 ⋅ 26:/76 ⋅ 31 provera pretpostavke iz 4. koraka:!

M ?61 mL

M 7/7 >7:/6!?61!! > 1/1:6 mL

pretpostavka je ta~na

⇒

⋅

9/55/!U razmewiva~u toplote sa suprotnosmerim tokom fluida zagreva W >7111!n40i!se )!pri!q>212/4 lQb-!u>1pD*!vazduha (ideala gas) od po~ete temperature!U2>!51pD!do krajwe temperature!U3>91pDpomo}u vode temperature!Ux2>:1pD. Procewena vrednost koeficijenta prolaza toplote iznosi!l>611 X0)n3L). Ukupna povr{ina za razmenu toplote iznosi!B>29!n3/!Odrediti maseni protok vode!)lh0t*/

voda-!Ux2

voda-!Ux3

⋅

R sb{ vazduh-!U2

vazduh-!U3 L

⋅

n wb{evi

⋅

q⋅W = > Sh ⋅ U

7111 4711 >3/27! lh 398 ⋅ 384 t

212/4 ⋅ 214 ⋅



strana 65


⋅

⋅

⋅

R sb{!>!∆ I 23!,! X U23

ograni~enom konturom K: ⋅

R sb{!> n w ⋅ d qw (U3 − U2 ) = 3/27 ⋅ 2 ⋅ (91 − 51) >97/5!lX ⋅

R sb{

⋅

∆UTS = ⋅B 2 l

∆Uts =

97/5 R sb{ ∆Uts !>! > !>!:/7pD l ⋅ B 1/6 ⋅ 29

⇒

(:1 − 91) − (Ux3 − 51) :1 − 91 mo Ux3 − 51

:1pD voda 91pD

Ux3!>!@

@ vazduh 51pD

pretpostavimo!ux3>71pD

⇒

∆Uts >25/5pD

(nije ta~no!)

pretpostavimo!ux3>59pD

⇒

∆Uts >9/:pD

(nije ta~no!)

pretpostavimo!ux3>5:pD

⇒

∆Uts >:/6pD

(ta~no!)


⋅

⋅

⋅

⋅

⋅

⇒

R 23!>!∆ I 23!,! X U23 ⋅

⋅

I 2!>! I 3 ⋅

n x ⋅ i x2 + n w ⋅ d qw ⋅ U2 = n x ⋅ i x3 + n w ⋅ d qw ⋅ U3 ⋅

nx =

⋅

n w ⋅ d qw ⋅ (U3 − U2 ) i x2 − i x3

lK lh lK ix3>31:/4! lh ix2>488/1!

>

lh 3/27 ⋅ 2 ⋅ (91 − 51) >1/6! 488 − 31:/4 t )q>2!cbs-!Ux2>:1pD* )q>2!cbs-!Ux3>5:pD*


Termodinamika-zbirka (ciganovic)

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