VARIABLE LOADS 1. A Di Dies esel el power power plant plant has a ma maxi ximum mum deman demand d of 12 !" wi with th a lo load ad fa#to fa#torr of .$ .$ an and d #apa#it% of .&. Estimate the plant #apa#it%. A. 1 !" B. 160 MW '. 1$ !" D. 2 !" Solution( Average load Loadfactor = Peak load 0.80
=
Average load 120
Average load= 96 kw
Usefactor =
0.60
=
Average load Plant capacit 96
Plant capacit
= 160 kw Plant capacity 2. 'al#ul 'al#ulate ate the !" pow power er #apa#it #apa#it% % of a )eo )eother thermal mal plant plant with a loa load d fa# fa#tor tor of .$2 and 12 !" pea* load. +he operation is limited to $, hours a %ear with a use fa#tor of .-. A. 1-, B. 1 '. 2// D. 145 Solution( Average load Loadfactor = Peak load 0.82
=
Average load 120
Average load= 98.4 kw
Usefactor =
0.70
=
Average load Plant capacit
98.4
Plant capacit
= 144.9 kw Plant capacity /. A 0ien euipment euipment #onsumes #onsumes , *w3hr4month *w3hr4month at 256 rated plant plant #apa#it% #apa#it%.. It operates at 25 hours7 / da%s4month. "hat is the rated #apa#it%8 A. 1-., *w 1
B. 2/. *w C. 28.90 kw D. $2. 2. 5 *w Solution(
Average load= 5000
kw− hr 1 month 1 day month 30 days 24 hrs
Average load= 6.94 kw Plant factor =
0.24
=
Average load Ratingof equipmen 6.94
Ratingof equipme
= 28.93 kw Ratingof equipment 5. A #entral station station is suppl%in0 suppl%in0 ener0% ener0% to a #ommun #ommunit% it% throu0h throu0h two su9stations. su9stations. One su9statio su9station n feeds four distri9ution #ir#uits: the other7 six. +he maximum dail% re#orded demandsare( ;ower station 127 *w Su9station A &7 *w
=
600
+ 1500 + 1000 + 2900 + 2200 + 3000 9000
= 1.247 ,. +he annual pea* load on a 1,7 1,73*w 3*w power power plant is 17, 17, *". *". +wo +wo su9stations su9stations are 9ein0 supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a pea* at &7&, *w. =e0le#tin0 line losses7 >nd the #apa#it% fa#tor of the power plant. A. ./,/ B. .,// 2
B. 2/. *w C. 28.90 kw D. $2. 2. 5 *w Solution(
Average load= 5000
kw− hr 1 month 1 day month 30 days 24 hrs
Average load= 6.94 kw Plant factor =
0.24
=
Average load Ratingof equipmen 6.94
Ratingof equipme
= 28.93 kw Ratingof equipment 5. A #entral station station is suppl%in0 suppl%in0 ener0% ener0% to a #ommun #ommunit% it% throu0h throu0h two su9stations. su9stations. One su9statio su9station n feeds four distri9ution #ir#uits: the other7 six. +he maximum dail% re#orded demandsare( ;ower station 127 *w Su9station A &7 *w
=
600
+ 1500 + 1000 + 2900 + 2200 + 3000 9000
= 1.247 ,. +he annual pea* load on a 1,7 1,73*w 3*w power power plant is 17, 17, *". *". +wo +wo su9stations su9stations are 9ein0 supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a pea* at &7&, *w. =e0le#tin0 line losses7 >nd the #apa#it% fa#tor of the power plant. A. ./,/ B. .,// 2
C. 0.335 D. .,/, Solution( actualenergy produced produced max imum possibl possible energy for thesame perio perio
Plant capacity factor =
Plant capacity factor =
277500 500 7000
+ 1675007000 kw− hr
(157000 000 kw) ( 8760 hr )
335 Plant capacity factor = 0.335
&. A power plant plant is said to hae hae had a use fa#tor fa#tor of 5$.,6 and a #apa#it #apa#it% % fa#tor of 52.56. 52.56. ?ow man% hours did it operate durin0 the %ear8 A. 7660 B. $-& '. $2 D. 172 Solution( actualkwhr produced produced Plant usefactor = × operatinghour kw plant plant capacity Plant capacity factor =
actualkwhr produced produced × 8760 hours kw plant plant capacity
+herefore 8760 hrs Plant usefactor = Plant capacity factor operating hour 48.5 42.4
=
hrs operating hour 8760
= 7660 hr operating hours
-. A distri9ution distri9ution transforme transformerr supplies supplies a 0roup 0roup of 0eneral power power #ustomers #ustomers hain0 a #onne#ted #onne#ted load of 1$& *w. If ener0% sells at /., #ents per *"3hr7 what will 9e the monthl% in#ome from ener0% deliered throu0h this transformer hain0 an aera0e motor e@#ien#% of -,6. )eneral data( Demand fa#tor .-, Diersit%
Aera0e Aera0e load on transformer .5, x 125 ,,.$ *w w3hrs deliered monthl% C,,.$ *wC/ da%s4mo.C25 hrs4da% w3hrs deliered monthl% 571-& *w3hrs4mo. !onthl% in#ome from ener0% deliered C./,4*w3hrC571C./,4*w3hrC571-& & *w3hr4mo. !onthl% in#ome from ener0% deliered ;175&.1& $. A dail% load load #ure whi#h whi#h exhi9it exhi9ited ed a 1,3minute 1,3minute pea* of &7,2 &7,2 *" is drawn drawn to s#ales of 1 #m 2 hrs and 1 #m , *w. *w. +he area under the #ure is measured 9% a planimeter and found to 2 9e 5$.-2 #m . Determine the load fa#tor 9ased on the 1,3minute pea*. A. .211 B. .511 C. 0.311 D. .,11 Solution( 1 #m2 ,C2 1 *w3hrs Aera0e load C5$.-2 #m 2C1 *w3hrs4#m 2C 1 da%425 hours Aera0e load 2/ *w Load fa#tor Aera0e load 4 ;ea* load Load fa#tor 2/ 4 &,2 Load fa#tor ./11 . A unit unit is rate rated d at $7 $7 *" *".. It oper operat ates es for for &, &, hrs hrs duri durin0 n0 the the %ear %ear and and 0ene 0enera rate tes s /,77 *"3hr. "ith a pea* load of -,7, *"7 #al#ulate the #apa#it% fa#tor. A. 50% B. ,/6 '. -6 D. &,6 Solution( Capacity factor =
35070007000
( 8760 )() ( 807000)
= 0.50 = 506
1.A 2,7 *" plant has a utiliation fa#tor of -16 and a load fa#tor of /.&6. "hat is the aera0e load on the plant in *"8 A. 257$, B. 1/7$& C. 9,841 D. 17,21 Solution( Aera0e Aera0e load C./&C2,7 C./&C2,7 *" 7 *" 11.A steam #%#le produ#es 5 !" of power7 , !" of pro#ess heat and reFe#ts & !" of heat. "hat is the utiliation fa#tor for this #%#le8 A. ,6 B. 60% '. -6 D. $6 Solution(
=
+ 50 = 0.60 = 60% 40 + 50 + 60 40
Gtiliation
D. .5& Solution( Plant Capacity "actor =
!nergy produced per year ( Plant capacity )( ours per year )
Energy produced = ( Average Load )( Hours per year )
= Load"actor
Average Loa Peak Load
− Reserve#ver Pea Peak Load= Plant Capacity Peak Load= 42 − 5 = 37 $% Average Load= ( .5$) ( /-) = 1-.-&$% Plant Capacity "actor =
= Plant Capacity "actor
!nergy produced per year ( Plant capacity )( ours per year )
(17.76 $% )( 8760 hrs year ) = 0.423 = 42.36 ( 42 $% )( 8760 hrs year )
1/.+he resere oer and a9oe the pea* load of a power station is 2 !".
( A!P)( 8760 ) A!P = ( PC)( 8760 ) PC
= Plant Use"actor
!nergy produced per year ( Plant capacity )( &o. of hoursnot in operation ) per year
,
PU" =
( A!P)( $-& )
( PC) ( oursnot in operation per year )
A! Euate
( PC" )( PC) =
( per year ) ( PU" )( PC) ( oursnot in operation 8760
oursnot in operation per year =
( PC" )( 8760 ) PU"
=
( 0.405)( 8760 ) 0.465
= 7629 .677 hrs
+ 13( 3.76)( 28) = 15.38 8(12 ) + 20(1)
13( 32)
% aa = 1.20% ta = 1.20(15.38) = 18.456
1&.A fuel 0as has the followin0 olumetri# anal%sis( '?5 &$6 '2?& /26 Assume #omplete #om9ustion with /6 ex#ess air at 11./2, *;a7 21 ' wet 9ul9 and 2- ' dr% 9ul9. "hat is the partial pressure of the water apor in *;a8 A. $.&2 B. 1.2$ '. 1&.5 D. 14.24 Solution( +heoreti#al Air 0.68C4 + 0.32C26 + a#2 + a( 3.76 ) &2 → bC#2 + c2#+ a( 3.76) &2
+ 0.32C + 2.48# + 2.48( 3.76) & → 1.32C# + 2.32 #+ 2.48( 3.76 ) & 0.68C4
2
2
2
6
2
2
2
"ith /6 Ex#ess air 0.68C + 0.32C + 2.48(1.3) # + 2.48( 1.3)( 3.76) & → 1.32C# + 2.32 # + 0.3( 2.48) # + 2.48( 1.3)( 3.76) & 4
2
2
2
6
2
2
2
2
&
+otal moles in produ#ts 1./2 2./2 ./C2.5$ 1./C2.5$C/.-& +otal moles in produ#ts 1&.,& ;artial pressure C2./2 4 1&.,&C11./2, 15.25 *;a 1-.A diesel ele#tri# plant supplies ener0% for !E'O. Durin0 a 25hr period7 the plant #onsumed / 0allons of fuel at 2$ ' and produ#ed 5 *w3hr. Industrial fuel used is /2 A;I and was pur#hased at ;/2. per liter at 1,.& '. "hat should the #ost of fuel to produ#e one *w3hr8 A. ;-., B. P7.35 '. ;-.& D. ;&. Solution( 'omputin0 for pri#e per *0 from per liter at 1,.& ' ()15.6 = ()15.6 =
141.5
AP'+ 131.5 141.5 32
+ 131.5
= 0.8654
Densit% at 1,.& ' .$&,5 C1 *04L .$&,5 *04L ;ri#e per *0 ;/2 4 .$&,5 ; /&.-- per *0 'onertin0 / 0al per *w3hr to *0 per *w3hr at 2$ ' ()t = ()15.6 [1 − 0.0007 ( t − 15.6 ) ] ()28
= 0.8654 [1 − 0.0007 ( 28 − 15.6 ) ]
()28 = 0.8579 300 gal 4900
×
kw− hr
3.7854 L
gal
×
0.8579
kg
L
= 0.1988
kg kw− hr
'ost per *w3hr .1$$ x ; /&.-- ;-./, per *w3hr 1$.'al#ulate the mass of the #om9ustion produ#ts of a diesel fuel oil C' 1&?/ with /6 ex#ess air if the mass of fuel is / *0. A. 1$.- *0 B. 1&.1 *0 C. 599.1 kg D. ,&.1 *0 Solution( C1630 + a#2 + a( 3.76 ) &2 → bC#2 + c2# + a( 3.76 ) &2 Solin0 for a7 b and c C1&/ + 2/.,#2 + 2/.,( /.-&) &2 → 1&C#2 + 1,2# + 2/.,( /.-&) &2 % ta =
+ 23.5( 3.76)( 28) = 14.53 16(12 ) + 30( 1)
23.5( 32 )
% aa = 1.30% ta = 1.30(14.53) = 18.889
mf / *0 !ass of produ#t of #om9ustion
-
C1 % tamf C11$.$$C/ ,&.&- *0 1.A mixture of 11 *;a and 2/ ' that is /6 9% wei0ht 'O 2 and -6 9% wei0ht = 27 determine the partial pressure of 'O 2. A. /.1$2 *;a B. 21.65 kPa '. 1 *;a D. 2, *;a Solution( nC# pC# = n+ P+ 2
2
2
0 682 = (101) = 21.65 kPa 3.182
.
2.
Solin0 for a7 b and c C820 + 13#2 + 13( 3.76) &2 → 8C#2
% ta =
+ 10 #+ 13( 3.76) & 2
2
+ 13( 3.76)( 28) = 15.38 8( 12) + 20( 1)
13( 32 )
% aa = 1.30% ta = 1.30( 15.38) = 20
21.At 2$ A;I spe#i># 0rait%7 what is the heatin0 alue of oil in *J4*08 A. 5&7- B. 5-71/ C. 45,039 D. 5$7/ Solution( ,h = 417130 + 139 .6( AP') ,h = 417130 + 139 .6( 28) ,h = 457038.80 k- kg
$
22.Dode#ane C'12?2& is 9urned at #onstant pressure with 156 ex#ess air. Determine the num9er of moles of ox%0en in the produ#ts of #om9ustion8 A. 22. B. 25.9 '. 55., D. ,,., Solution( +heoreti#al Air C122& + a#2 + a( /.-&) &2 → bC#2 + c2#+ a( /.-&) &2 C122& + 1$.,#2 + 1$.,( /.-&) &2 → 12C#2 + 1/2#+ 1$.,( /.-&) &2 "ith 156 Ex#ess air C122& + 1$.,( 1+ 1.5) #2 + 1$.,( 1+ 1.5)( /.-&) &2 → 12C#2 + 1/2#+ 1.5( 1$.,) #2 + 1$.,( 1+ 1.5)( /.-&) &2 !ols of O2 1.5 x 1$., 2,. mols 2/.A mixture of 15.- psia and &$ < that is /6 9% wei0ht 'O 2 and -6 9% wei0ht = 2 has a partial pressure of 'O2 in psia that is nearest to( A. 2.15 B. $.$/ '. -.$$ D. 3.15 Solution( nC# pC# = n+ P+ 2
2
pC#
2
0.682 = (14.7 ) = 3.15 psi 3.182
25.)ien 0aseous fuel ' $?1$ and the olumetri# produ#ts of #om9ustion( 'O2 1.167 'O .$,67 = 2 $,67 O2 156. Determine the air3fuel ratio. A. 1$. B. 20.1 '. 1,. D. 25. Solution( aC$1$ + b( #2 + /.-&&2 ) → 1.1C#2 + .$,C#+ .15#2 + c2#+ $,&2 'ar9on Balan#e( $a 1.1 .$,7 a 1./,-, =itro0en Balan#e( 2C9C/.-& 2C$,7 9 22.&& =o need to determine #. 22.606 [ 32 + 3.76( 28) ] kg % aa = = 20.1 1.3575 [ 8( 12 ) + 18( 1) ] kg C9
2,.A manufa#turin0 #ompan% in Bi#ol operates a Diesel Ele#tri# ;lant to suppl% its ele#tri# ener0% reuirements. Durin0 a 25 hr period7 the plant #onsumed 2, 0allons of fuel at $ < and produ#ed 2- *w3hrs. Industrial fuel used is / A;I and was pur#hased at ;/. per liter at & <. Determine the oer3all e@#ien#% of the plant. A. 2,.&6 B. 2.-6 C. 26.08% D. /.16 Solution( outpu e= input 2700
e=
kw− hr
24 hr
m,h ,h
Solin0 for ( .&( /) = 5,7/1$k- kg ,h = 5171/+ 1/.&( AP') = 5171/+ 1/
m Solin0 for
(
(.).60 " = (.).15.6 C = (.).60 " = (.).15.6 C =
141.5 131.5
+ AP'
141.5 131.5
+ 30
= 0.876
(.).80 " = (.).26.6 C = 0.876[1 − 0.0007 ( 26.67 − 15.6) ] = 0.869 m= ρ .
2,gal 1hr ( ./-$, m/ gal) 25 /& hr s
m= ( .$&) (1kg m/ ) m= 0.00952 kg s
+hen7 k% − hr 24 hr e= ( 0.00952 kg s)( 457318 k- kg) 2700
.2&$2&.$6 2&.A #ertain #oal has the followin0 ultimate anal%sis( A ,6 ' -6 ?2 2.,6 ! 16 =2 56 S ,6 K2 /.,6
'. 1,.5 D. 1/.-, Solution(
A F = 11.5C + 34.5 H 2 −
O2
+ 4.32 S 0.035 A F = 11.5( 0.70 ) + 34.5 0.025 − + 4.32( 0.05) 8 A F = 8.977 kg air kg coal 8
nd the theoreti#al olume of air at 2 ' and 1 *;a a9solute pressure to 9urn one *ilo0ram #oal. +he ultimate anal%sis of #oal as3>red is as follows( ' &,.&,6 O 1$.&16 S 1.,16 !oist. /6 ? ,.$-6 = 1./6 Ash ,./&6 A. .5/ m/air4*0#oal B. 7.43 m3air/kgcoa '. $.5/ m/air4*0#oal D. &.5/ m/air4*0#oal Solution( +he theoreti#al olume of air reuired: mR . = p
m Solin0 for
(
m= ( A " ) t = 11.,C + /5., −
#
+ 5./(
$
m= 11.,( .&,&, ) + /5., .,$-−
.1$&1
+ 5./( .1 ,)
$
m= $.$5kgair kgcoa
( $.$5kgair kgcoal) ( .2$-k- kgair / )(2+ 2-/ / )
. =
2
1k& m
. = -.5/m/ air kgcoa
2$.A steam 9oiler 9urns fuel oil with /6 ex#ess air and represented as '1&?/2. +he fuel 0as leaes the preheater at 22 *;a.
B%
+ 0.25 × 32) = 14.708 12 ×16 + 32
137.28( 16
% aa = (1 + e)% ta = (1.30 )( 14.708) = 19.12
2.In 9urnin0 a t%pi#al #oal7 the theoreti#al air3fuel ratio was found to 9e 1, *04*0. +he approximate heatin0 alue of the #oal in *J4*0 is7 A. 57&, B. 45,610 '. 5&7, D. 517,1 Solution( 7.65. % ta = 107000
. =
(15)(107000) 7.65
Btu lb = 197607.84 Btu lb
= 457613 k- kg I=+ER=AL 'O!BGS+IO= E=)I=E /.Determine the thermal e@#ien#% of an air standard Otto #%#le with 16 #learan#e. A. ,/.$6 B. 61.68% '. ,$.//6 D. -2.//6 Solution(
= −
η 1
1 k −1
r
r = ratioof compressio n=
c=
. 1 . 2
. 2 . 1 − . 2
c. − c. 2 = . 2 1 . 1 . 2 η
=
c + 1 0.1 + 1 = = 11 c 0.1
=1 −
1
−
1.4 1
11
= 0.6168 = 61.686
/1.A two3stro*e Diesel en0ine has a fri#tion power of 1/6 of the heat 0enerated. Determine the 9ra*e thermal e@#ien#% if the indi#ated thermal e@#ien#% of the en0ine is ,6. A. //6 12
B. 37% '. /56 D. /&6 Solution( % f = 0.13, A % i = 0.5, A
% b = % i − % f = 0.5, A − 0.13, A = 0.37, A e=
% b , A
= 0.37
or /-6 /2.Durin0 a !orse test on a four3#%linder automotie en0ine7 the speed was *ept #onstant at 25., rps 9% the 9ra*e and the followin0 9ra*e torue readin0s were o9tained( All #%linders >rin0 1/.$ =3m =o. 1 #%linder #ut3out 1/.$ =3m =o. 2 #%linder #ut3out 1/.2 =3m =o. / #%linder #ut3out 12. =3m =o. 5 #%linder #ut3out 1/1.1 =3m A##ount for the fri#tional power7 in *". A. $.-2 B. 9.13 '. 1.1, D. 12.Solution( BP= 2π +&=
2π (193 .8)( 24.5) 1000
= 29.83 k%
=o. 1 #%linder #ut3out( BP1 =
2π (130.8)( 24.5) 1000
= 20.14 k%
'P1 = 29.83 − 20.14 = 9.69 k%
=o. 2 #%linder #ut3out( BP2 =
2π ( 130 .2)( 24.5) 1000
= 20.04 k%
'P2 = 29.83 − 20.04 = 9.79 k%
=o. / #%linder #ut3out( BP3 =
2π ( 129.9)( 24.5) 1000
= 20.00 k%
'P3 = 29.83 − 20.00 = 9.83 k%
=o. 5 #%linder #ut3out( BP4 =
2π (131.1)( 24.5) 1000
= 20.18 k%
1/
'P4
= 29.83 − 20.18 = 9.65 k
'P = 'P1 + 'P2 + 'P3 + 'P4 = 38.96 k%
"P= 'P − BP= 38.96 − 29.83 = 9.13 k%
//.An en0ine 9urns a liter of fuel ea#h 12 min. +he fuel has a spe#i># 0rait% of .$ and a heatin0 alue of 5, !J4*0. +he thermal e@#ien#% of the en0ine is 2,6. "hat is the 9ra*e horsepower of the en0ine8 A. 11 B. 17 '. 1/ D. 1, Solution( , A = m
1 Liter min 12
m= ( (P.)R.) ( ρ w )
1 Liter = 0.06667 kg mi 12 min
m= ( 0.8) (1.0 kg L )
1 min = 50 k% 60 se#
, A = ( 0.06667 kg min)( 457000 k- kg) % = e,A = 0.25( 50 k% ) = 12.5 k% hp=
12.5 k% 0.746 k% hp
= 17 hp
/5.A Diesel en0ine used 2- tons of fuel per da% when deelopin0 5& *" indi#ated power and 5& *" 9ra*e power. Of the total heat supplied to the en0ine7 /1.-6 is #arried awa% 9% #oolin0 water and radiation7 and /.$6 in the exhaust 0ases. 'al#ulate the indi#ated thermal e@#ien#%. A. 37.5% B. /,.-6 '. /1.2,6 D. /2.-,6 Solution( ei = 100 − 31.7 − 30.8 = 37.56 /,.+he #om9ined me#hani#al ele#tri#al e@#ien#% of an en0ine30enerator set is $26. If the #om9ined spe#i># fuel #onsumption is .2, *04*"3hr7 the indi#ated spe#i># fuel #onsumption is7 A. .2/, *04*"3hr B. .2, *04*"3hr C. 0.168 kg/kW!"r D. .1-, *04*"3hr Solution( )( .$2) kg k% − hr ind. s. f .c. = ( comb . s. f .c.)η comb= ( .2,
15
.1&$ *04*"3hr /&.+he thermal e@#ien#% of an en0ine is /6 and the fuel used hae a heatin0 alue of 57, *J4*0. +he en0ines spe#i># fuel #onsumption is7 A. /./-, *04*"3hr B. .2$, *04*"3hr C. 0.296 kg/kW!"r D. ./-, *04*"3hr Solution( 3600 k- k% ⋅ hr = 0.296 kg k% ⋅ hr s. f .c = 0.30( 407500 k- kg) /-.A #ar en0ine produ#es an aera0e of 2, *" of power in a period of 1 min. durin0 whi#h 1.2, *0 of 0asoline is #onsumed. If the heatin0 alue of 0asoline is 5/7 *J4*07 the e@#ien#% of this en0ine durin0 the period is7 A. 256 B. 28% '. /26 D. /&6 Solution( 1.25 kg 1 min , A = 60 s ( 437000 k- kg) = 89.58 k% 10 min % = 25 k% e=
% 25 = (1006) = 286 , A 89.58
/$.An en0ine 9urns a liter of fuel ea#h 12 minutes. +he fuel has a relatie densit% .$ and a heatin0 alue of 5, !J4*0. +he en0ine has a 9ra*e thermal e@#ien#% of 2,6. "hat is the 9ra*e horsepower of the en0ine8 A. 16.76 B. 12.&$ '. 1/.,& D. 2$./5 Solution( 1 Liter .1 m / , = mf .
0.001 m 1 min ( 457000 k- kg) 60 se# 12 min
, = 0.80(9000 kg m3 )
3
, = 50 k%
1 hp = 16.76 hp 0.746 k%
bhp= 0.25( 50 k% )
/.A , *" Diesel en0ine deelops torue of 2 *=3m. +he en0ine dries an alternatin0 #urrent 0enerator with $ poles produ#in0 #urrent at & ?. "hat is the speed ratio of redu#tion 0ear8 A. /.&(1 B. 1.&(1 C. 2.6#1 1,
D. 5.&(1 Solution( 2+& P= 60
500
=
2( 2 ) & 60
& = 2387 .32 rpm &s =
120 f
p
=
120 ( 60) 8
= 900 rpm
+hus7 the speed ratio is & 2387 .32 2.6 R= = = 900 1 &s 5.A 1, *" diesel en0ine operates at an altitude of 12 meters eleation.
p pL Bra*e power at a hi0her eleation Bra*e power at a lower eleation p pressure at a hi0her eleation pL pressure at sea leel Cstandard + temperature at a hi0her eleation + L temperature at a lower eleation Cstandard
1 in. g (1200 m)( 3.28 ft m) = 25.984 in. g ft 1000
p = 29.92 in. g−
3.566 R (1200 m)( 3.28 ft m) = 515 .964 R ft 1000
+ = ( 70 + 460 ) R −
1&
+ + L
25.984 29.92
BP = (1500 )
515.964 70
+ 460
= 1285 k%
51.Determine the power output of a Diesel power plant if the en0ine and 0enerator e@#ien#% is $/6 and ,6 respe#tiel%. +he en0ine uses 2, o A;I fuel and has a fuel #onsumption of 22$ *04hr. A. 2815 kw B. 2, *w '. 2&, *w D. 2-, *w Solution( = ( eat generatedfromfuel)η mη g Plant #utput
= ( mf . )η mη g Plant #utput . = 417130 + 139.6( AP') = 417130 + 139.6( 25) = 447620 k- kg
mf = 288 kg hr η m
en0ine me#hani#al e@#ien#% $/6 η g
0enerator e@#ien#% ,6
1 hr ( 447620 k- kg) ( 0.83)( 0.95) 3600 s
Plant #utput = ( 288 kg s)
Plant #utput = 2815 k%
52.An en0ine deelopin0 / hp7 is transferrin0 1 hp more to the #oolin0 tower. If the top water and 9ottom water of the radiator are 2 < and 1 <7 respe#tiel%7 #al#ulate the reuired water Now rate in 0allons per minute. A. 20.34 B. 1,.2, '. ,.$ D. 1.1Solution( , = ( 30 + 10 hp)( 2544 Btu hp− hr ) = 1017760 Btu hr
∆+ = 200 − 190 = 10 " c p = 1.0 Btu lb − " , = mcp∆+ 1017760
= m(1.0 )(10)
m= 107176 lb hr
In 0pm( 1-
1 hr 1 gal 8.34 = 20.34 gpm 60 min
(107176 lb hr )
5/.+he e@#ien#% of an Otto en0ine is ,,6 and the stro*e olume is /, liters. If the heat added at the 9e0innin0 of the #om9ustion is 12.& *J7 #al#ulate the mean eMe#tie pressure. A. 15 *;a B. 1$$ *;a C. 198 kPa D. 1$5 *;a Solution( % pm = net . 0 "here: . 0 = 35 liters= 0.035 m3
W net Solin0 for % e = net , A 0.55
(
% net
=
12.6
% net = 6.93 k-
+hus7 pm =
6.93 0.035
= 198 kP
55.A test 9ed7 1&3#%linder7 V3t%pe7 tur9o3#har0ed Diesel en0ine deeloped 7&1 metri# horsepower at ,15 rpm. ?eat reFe#tion to lu9ri#atin0 oil is -& *'al per metri# hp3hr. Spe#i># 0rait% of lu9ri#atin0 oil used is . and spe#i># heat of oil is ., Btu4l93<. 'ompute the uantit% of lu9ri#atin0 oil #ir#ulated in m /4hr if the temperature of lu9ri#atin0 oil enterin0 the en0ine is ,, ' and &, ' leain0. A. 2,$./5 m/4hr B. 12/.5, m/4hr C. 168.23 m3 /"r D. /25.,& m/4hr Solution( m . = ρ
Solin0 for mass Now rate of lu9ri#atin0 oil( , heat #arried awa% 9% lu9ri#atin0 oil C&1C-& -,-7/& *'al4hr c p = 0.5 Btu lbο " = 0.5 kCalkgο C
1$
, = mcp∆t 757 7036
= m( 0.5) ( 65 − 55)
m= 1517407.2 kg hr ρ
= 0.90(1000 ) = 900 kg m
3
+hen: . =
1517407.2 kg hr 900 kg
3
m
= 168.23 m
3
hr
5,.In an ideal Otto #%#le7 the initial pressure and temperature of air are 1 *;a and 1$ de0ree '. Determine the maximum pressure in the #%#le if the maximum temperature in the #%#le is & de0ree '7 and the #ompression ratio is $. A. 1$ *;a B. 225 *;a C. 2400 kPa D. 1, *;a Solution(
p1 1 *;a + 1 1$ 2-/ 21 + 2 & 2-/ $-/ r k . 14. 3 $
p3 + 3
=
p2 + 2
+ 2 = + 1 ( r k ) k −1 p2 = p1 ( r k )
p3 + 3
k
p1 ( r k ) k p1 ( r k ) = = + 1 + 1 ( r k ) k −1
p3 873
=
100( 8) 291
p3 = 2400 kP
1
5&.Air enters the #%linder of an internal #om9ustion en0ine at an initial pressure and temperature of , *;a and 2$ de0 '. A four #%linder7 sin0le3a#tin07 5 mm x ,, mm7 four stro*e en0ine runs at 5, rpm. +he mass Now rate of air per se#onds is .1 *0. "hat is the olumetri# e@#ien#% of the en0ine8 A. -26 B. $16 C. 87% D. -.-6 Solution( mR . 1′ = p1 . 1′ =
( 0.01)( 0.287)( 28 + 273) 95
π
. 0 =
4
. 0 =
η v
=
π 4
3
s
02L& p&c
4500 2
& p =
= 0.00909 m
= 2250 rpm 2250 ( 4) = 0.01037 m 60
( 0.040) 3 ( 0.055)
. 1′ 0.00909 = . 0 0.01037
3
s
= 0.877 = 87.76
5-.A 2 #m x /& #m7 twin #%linder7 two stro*e #%#le diesel en0ine runnin0 at &, rpm. +he en0ine rate is 112 *w. Determine the en0ine displa#ement in #u9i# meter4se# per 9ra*e horsepower. A. .1&/ B. .1&/ C. 0.00163 D. .&/1 Solution( . 0 =
π 4
02L& p&c
& p = 650 rpm π
. 0 = BP
4
BP π
. 0 = BP
02L& p&c
4
( 0.20) 2 ( 0.36)
650
60
( 2)
112 0.746
. 0 = 0.001632 m3 se#− bhp BP
2
5$.A 1 *w3hr diesel 0eneratin0 set has a 0enerator e@#ien#% of $,6. If the mass of the fuel is 1, *07 #ompute the en0ine fuel rate. A. .1, *04*w3hr B. .11-, *04*w3hr C. 0.127 kg/kw!"r D. 22.- *04*w3hr Solution( mf 150 = = 0.1275 kg k%h BP 1000 0.85 5.In an air standard diesel #%#le7 #ompression starts at 1 *;a and // . +he #ompression ratio is 2 to 1. +he maximum #%#le temperature is 22 . Determine the thermal e@#ien#%. A. 5&6 B. 64% '. &6 D. -6 Solution(
e=1−
1
r k k −1
r ck − 1 k ( r − 1) c
r k =
. 1 = 20 . 2
r c =
. 3 . 2
+ 3 + 2
=
k −1
. + = + = ( 303)( 20 ) . − .
14 1
1
2
1
2
1004 / + 2 =
+ 3 = 2200 /
r c =
2200 1004
e=1−
= 2.19
1
r k k −1
r ck − 1 k ( r − 1) c 21
e=1−
( 2.19) . − 1 1.4( 2.19 − 1) 14
1
( 20) 1.4−1
e = 0.6384 = 63.846
,.If the pressure at the 9e0innin0 and end of #ompression in an Otto #%#le en0ine are 1 *;a and , *;a7 respe#tiel%7 the maximum #%#le temperature is 1, . Determine the air standard e@#ien#%. A. /,.-6 B. /.,6 C. 47.5% D. ,-.-6 Solution(
e= 1−
1
r k k −1 1
. p r k = = . p 1
2
2
1
k
1
950 . = 4.993 r k = 100 14
e=1−
1
−
1.4 1
4.993
= 0.4744 = 47.446
C# ,1.An Otto en0ine has a #learan#e olume of $6. It produ#es , *w power. 'ompression starts at 1 *;a and 2, '. "hat is the heat reFe#ted in *w8 A. 273 kw B. /-2 *w '. /2- *w D. -/2 *w Solution( e= 1−
r k =
1
1
r k k −1
+c c
=
1
+ 0.08 0.08
= 13.5
22
1
e=1−
e=
(13.5) 1.4−1
= 0.6469
% , A
0.6469
=
500
, A
, A = 772 .92 k% , A = % + ,R 772 .92
= 500 + ,R
,R = 272 .92 k% ,2.+he followin0 data are the results on a test of a two #%linder7 four stro*e #%#le Otto en0ine( torue 2 =3m: indi#ated mean eMe#tie pressure - *;a: fuel #onsumption .& *04se#: fuel heatin0 alue 5/7 *J4*0: 9ore x stro*e / #m x 5 xm: speed & rpm. 'al#ulate the 9ra*e mean eMe#tie pressure. A. /-, *;a B. /2, *;a C. 445 kPa D. 2$, *;a Solution(
600 = 125.66 k% 60
BP = 2π +&= 2π ( 2.0) π
. 0 =
4
& p = . 0 =
02L& p&c
600 2
π 4
pmb =
= 300 rpm 300 ( 2) = 0.2827 m 60
( 0.30) 3 ( 0.40)
BP 125.66 = . 0 0.2827
3
s
= 444 .49 kPa
,/. A four3#%linder7 four3#%#le en0ine with 13#m diameter pistons and a 1&3#m stro*e operates at a speed of & rpm and %ields an indi#ator dia0ram. +he area under the #ure CpV dia0ram is eual to 12. #m 2. +he len0th of the dia0ram is . #m7 and the sprin0 #onstant of the indi#ator sprin0 is & *;a4#m. Determine the mean eMe#tie pressure. A. 800 kPa B. 5 *;a '. & *;a D. , *;a Solution( 2/
pm =
pm =
Areaunder curve × springscal stroke 12 9
× 600 = 800 kPa
,5.+he #ompression ratio of an air3standard 0asoline en0ine is -.,. +he heat added is 1$ *J4*0. If the initial pressure and temperature are 1 9ar and 2- ' respe#tiel%7 determine the temperature in ' at the end of the isentropi# expansion. A. 1206 B. 2&$ '. 1-&, D. 1/5Solution(
+ 1 2- 2-/ / r k -., , A4m 1$ *J4*0 1.4 1 + 2 = + 1r k k −1 = 300( 7.5) − = 671.7 /
, A = cv ( + 3 − + 2 ) m 1890
= 0.716( + − 671.7 ) 3
+ 3 = 3311 .4 /
+ 3 + 4
= r k k −
1
3311 .4
+ 4
= ( 7.5) . −
14 1
+ 4 = 1479 / t 4
= 1479 − 273 = 1206 C
,,.'al#ulate the thermal e@#ien#% of an air3standard Diesel #%#le operatin0 with a #ompression ratio of 21 and #ut3oM ratio of /. A. ,,.6 B. 61.4% '. &&.6 D. -1.6 25
Solution(
r ck − 1 k ( r − 1) c
1
e=1−
r k k −1
e=1−
( 3) . − 1 1.4( 3 − 1) 14
1
( 21) 1.4−1
e = 0.6137 = 61.376
)AS +GRBI=E ;O"ER ;LA=+ ,&.A 0as tur9ine plant wor*in0 on a #losed air3standard Bra%ton #%#le has a pressure ratio of ,. +he heat reFe#ted in the heat sin* is 2 *". If one3>fth of the expansion wor* is supplied for #ompression ideall%7 determine the theoreti#al power of the 0as tur9ine. A. 1460 kW B. -5 *" '. 12& *" D. 1&5 *" Solution(
r p = 5 η
η
= 1−
=
1 k −1 r p k
( )
1
= 1− ( 5)
−
1.4 1
= 0.3686
1.4
% net , A −,R = , A , A
, A =
,R
−
1 η
=
2000 k% 1 0.3686
−
= 3168 k%
% net = η ,A = ( 0.3686 )( 3168 ) = 1168 k 1
5
,
,
% net = % /−5 − % 1−2 = % /−5 − % /−5 = % /−5 % 3−4
5
) = 1460 k% = (1168 4
,-.A 0as tur9ine power plant operatin0 in a Bra%ton #%#le7 the #ompressor power needed is 2, *" whi#h is drien 9% a 1 *" tur9ine. +he #ompressor e@#ien#% is 6 and that of the tur9ine is 6. Determine the 9a#* wor* ratio. A. /6 B. /56 2,
C. 25% D. 56 Solution( Ba#* wor* ratio #ompressor power 4 tur9ine power 2, *w 4 1 *w .2, 2,6 ,$.A 0as tur9ine plant utiliin0 a #losed #%#le has the followin0 temperatures at ea#h state point( + 1 <7 + 3 , <7 + 2 155 <7 and + 4 ,/ <. Based on these alues7 #al#ulate the thermal e@#ien#% of the #%#le. A. 48% B. 5,6 '. 5&6 D. 5-6 Solution( + − + 530 − 90 = 0.48 e= 1− 4 1 = 1− 1440 − 590 + 3 − + 2 or 5$6 ,.In a simple 0as tur9ine plant wor*in0 on the ideal #onstant pressure #%#le7 air is ta*en into the #ompressor at 1 9ar7 1& ' and deliered at ,.5 9ar. If the temperature at tur9ine inlet is - '7 #al#ulate the ideal thermal e@#ien#%. +a*e * 1.5. A. /,.2$6 B. 52.&16 C. 38.23% D. 5,.&&6 Solution( + −+ et = 1 − 4 1 + 3 − + 2
+ 4 + 2 + 3 Solin0 for 7 7 and ( + 2 + 1
p = p
2
1
k −1 k
+ 1 = 16 + 273 = 289 / −
1.4 1
= 289 1 + 2
5.4
1.4
+ 1 = 467 .90 /
+ 4 + 3
p = p
4 3
k −1
k
+ 3 = 700 + 273 = 973 /
2&
−
1.4 1
5.4 = 973 1
+ 4
1.4
+ 4 = 600.98 /
+hen: et = 1 −
600.98 973
− 289
− 467.9
= 0.3823 = 38.236
&.+he net power output of an air3standard Bra%ton #%#le is 2 *". Air enters the #ompressors at /2 ' and leaes the hi0h3temperature heat ex#han0er at $ '. "hat is the mass Now rate of air if it leaes the tur9ines at /, '. A. .,- *04s B. .&- *04s C. 0.87 kg/$ D. .-- *04s Solution( % n = % t − % c % t
% c
Solin0 for and + 1 = 32 + 273 = 305 /
.
+ 3 = 800 + 273 = 1073 / + 4 = 350 + 273 = 623 /
+ 2 + 1
=
+ 2 305
+ 3 + 4
= 1073 623
+ 2 = 525.30 /
% c = mcp ( + 2 − + 1 )
% c = m(1) ( 525 .30 − 305 ) = 220 .3m
% t = mcp ( + 3 − + 4 )
% t = m(1) ( 1073 − 623 ) = 450 m +hen: 200
= 450m− 220 .3m 2-
m= 0.87 kg s &1.Air enters the #ompressor of a 0as tur9ine at 11 *;a and 2- ' with a olume Now rate of $., m/4se#. +he #ompressor pressure ratio is 1 and its isentropi# e@#ien#% is $26. At the inlet to the tur9ine7 the pressure is , *;a and the temperature is 15 . +he tur9ine has an isentropi# e@#ien#% of $,6 and the exit pressure is 1 *;a. On the 9asis of an air standard anal%sis7 what is the thermal e@#ien#% of the #%#le in per#ent8 A. 27.3 B. 22. '. 21.$ D. 2.2 Solution( p1 11 *;a7 + 1 2- 2-/ / . 1 $., m/4se# p34 p1 17 η c .$2 p2 , *;a7 p4 1 *;a7 η t .$, + 2 15
kp. p % c = k − 1 p 1 1
2
1
% c =
k − k − 1 1
1.4(110 )( 8.5) 1.4
(10)
−1
−
1.4 1 1.4
− 1
% c = 3045 .7 kw
% c′ =
% c η c
=
3045 .7 kw 0.82
kp. p % t = 1− k − 1 p 3 3
= 3714 .3 kw
k − k 1
4 3
kp. + p 1 − % t = k − 1 + p 1 1
3
4
1
3
k −1 k
100 . .− 1 − 300 950
14 1
% t =
1.4(110 )( 8.5) 1.4
−1
1400
14
% t = 7245 .0 kw % t ′ = η t % t = ( 0.85) ( 7245 .0 kw) = 6158 .3 kw
′ = % t ′ − % c′ = 6158 % net .3 − 3714 .3 = 2444 kw , A = mc p ( + 3 − + 2 )
2$
, A = mcp ( + 3 − + 2 )
p p
+ 2 = + 1
2
1
k −1 k
p kp. , A = + − + ( k − 1) + p 1 1
2
3
1
1
, A =
1
k −1 k
1.4(110)( 8.5)
1400 − 300(10) (1.4 − 1)( 300)
−
1 .4 1 1.4
, A = 8953 .5 kw
e=
′ 2444 kw % net (1006) = 27.306 = , A 8953 .5 kw
&2.Air enters a 0as tur9ine power plant at 1 9ar and / ' with a olume Now rate of , m /4s. +he #ompressor pressure is 1 and its isentropi# e@#ien#% is $6. +he tur9ine inlet pressure and temperature are 1 *;a and 1/ '7 respe#tiel%. +he tur9ine isentropi# e@#ien#% is $,6 and the exit pressure is 1 *;a. Determine the thermal e@#ien#% of the #%#le in per#ent8 A. ,2.6 B. /.,6 C. 29.5% D. 2.,6 Solution( p1 1 9ar 1 *;a + 1 / 2-/ // . 1 , m/4s p34 p1 1 η c .$ p2 1 *;a + 2 1/ 2-/ 1,-/ ht .$, p4 1 *;a k −1
kp. p k % c = 1 1 2 k − 1 p1
% c =
1.4(100 )( 5) 1.4
−1
− 1
(10)
−
1.4 1 1.4
− 1
% c = 1628 .7 kw
% c′ =
% c η c
=
1628 .7 kw 0.80
= 2035 .9 kw
2
kp. p % t = 1− k − 1 p 3
3
4 3
k − k 1
kp. + p 1 − % t = k − 1 + p 1 1
3
4
1
3
k −1 k
. − 100 1 − . 303 1000
14 1
% t =
1.4( 100)( 5) 1.4
−1
1573
14
% t = 4379 .4 kw % t ′ = η t % t = ( 0.85) ( 4379 .4 kw) = 3722 .5 kw
′ = % t ′ − % c′ = 3722 .5 − 2035 .9 = 1686 .6 kw % net , A = mcp ( + 3 − + 2 ) , A = mcp ( + 3 − + 2 )
p p
+ 2 = + 1
2
1
k −1 k
k −1
p k kp1. 1 , A = + 3 − + 1 2 ( k − 1)+ 1 p1
, A =
1.4(100 )( 5)
1573 − 303(10) (1.4 − 1)( 303 )
−
1.4 1 1.4
, A = 5706 .3 kw
e=
′ 1686 .6 kw % net = (1006) = 29.566 , A 5706 .3 kw
&/.A -, !" stationar% 0as tur9ine power plant has air enters the #ompressor at 1 *;a7 / temperature and pressure ratio of 1. +he tur9ine inlet temperature is 1&, . Gsin0 the standard Bra%ton #%#le7 determine the 9a#*3wor* ratio of the 0as tur9ine power plant. A. .2,, B. .$- '. 1.&1 D. 0.351 Solution( p1 1 *;a + 1 / + 2 1&, p34 p1 p24 p4 1 /
p + = + p
2
2
1
1
k −1
k = 300(10)
−
1.4 1 1.4
+ 2 = 579 .2 / + 3 + 4
p = p
1650
+ 4
3 4
k −1 k
= (10)
−
1.4 1 1.4
+ 4 = 854.6 /
− work ratio= Back
+ 2 − + 1 + 3 − + 4
− work Back ratio=
− 300 1650 − 854.6 579.2
Back ratio= 0.351 − work
&5.A 27 !" 0as tur9ine operatin0 in the simple open #%#le has an exhaust 0as Now of 1 *0 per se#ond. +he 0as enters a waste heat re#oer% 9oiler at 5, ' and leaes at 1$ '7 spe#i># heat of the 0as is .$, *J4*037 heat reuired to produ#ed 1 *0 of steam at 5., !;a and /2 ' from feedwater at 11 ' is ,., *J. 'al#ulate the uantit% of steam that #an 9e produ#e in *0 per se#ond. A. 4173 kg/$ B. 1-2/ *04s '. 21-/ *04s D. /1-/ *04s Solution( ?eat reFe#ted heat a9sor9ed mc p ( t 4 − t 5 ) = ms∆h 100( 0.85)( 450
− 180) = ms( 5.5)
.-2kg se ms = 51-2
&,.A simple Bra%ton #%#le whi#h uses helium as the wor*in0 Nuid7 has a maximum temperature of 11, 7 and a pressure ratio of 2.. At the start of the #ompression7 the helium pressure and temperature are *;a and / . Based upon #old air3standard anal%sis assumptions7 determine the thermal e@#ien#% of the #%#le in per#ent8 A. 34.7% B. 1-.,6 '. 2/.-6 D. 2$.-6 Solution( k 1.&&- for helium p1 *;a /1
+ 1 / + 2 11, p34 p1 p24 p4 2.
p p
+ 2 = + 1
2
1
k −1 k
= 300( 2.9)
−
1.667 1 1.667
+ 2 = 459 .34 / + p = + p 3
3
4
4
1150
+ 4
k −1
k
= ( 2.9)
−
1.667 1 1.667
+ 4 = 751 .07 / % c = mcp ( + 2 − + 1 )
% t = mc p( + 3 − + 4 ) % net = % t − % c , A = mcp ( + 3 − + 2 ) e=
e=
% net ( + 3 − + 4 ) − ( + 2 − + 1 ) = , A + 3 − + 2
(1150 − 751.07 ) − ( 459.34 − 300) 1150
− 459.34
(1006) = 34.696
&&.+he mass Now rate of the 0as in a 0as tur9ine is / *04se#. +he spe#i># enthalp% and elo#it% in the inlet are 1/ *J4*0 and 2 m4se# respe#tiel% while in the outlet are / *J4*0 and , m4s7 respe#tiel%. 'al#ulate the power output in *w of the tur9ine if there is a heat loss of / !". A. /27&-$., *w B. 27,562.50 kw '. /-712,., *w D. 5,7,., *w Solution( m(. 22 − . 12 ) − ,L % t = m( h2 − h1 ) + 2( 1000 )
[ % t = 30(1300 − 300) +
30 ( 200)
2
− ( 50)
2(1000 )
2
] − 3000
% t = 277562 .5 k%
/2
&-.An ideal 0as tur9ine operates with a pressure ratio of -., and temperature limits of 2, ' and 11 '. +he ener0% input in the hi0h temperature heat ex#han0er is , *w. Determine the air Now rate in *04hr. A. ,./ B. 2135 '. 2/, D. /5, Solution( p34 p1 -., + 1 2, 2-/ 2$ + 2 11 2-/ 1/-/ + p = + p 2
2
1
1
+ 2 298
k −1
k
= ( 7.5)
−
1.4 1 1.4
+ 2 = 530.0 /
, A = mcp ( + 3 − + 2 ) 500
= m(1.0)(1373 − 530)
m= 0.593 kg s = 2135 kg hr
&$.Air at 5 *;a and , is extra#ted from a Fet en0ine #ompressor to 9e used for the 0eneration of auxiliar% power for the #a9in. +he extra#ted air is #ooled in a #onstant pressure heat ex#han0er down to 5& . It then enters an isentropi# tur9ine and expands to 1 *;a 9efore 9ein0 reFe#ted into the #a9in. If the mass Now is 1, *04min7 determine the power deeloped 9% the tur9ine in *w. A. &/.- *w B. -/.& *w C. 37.6 kw D. -&./ *w Solution( + 3 , 7 + 2 5& p4 1 *;a7 p2 5 *;a m 1, *04min + 5 + /
k −1 k
p = 5 p/
1.5−1
1 1.5 = /.,&/ + 5 = 5& 5 ;ower deeloped 9% the tur9ine mc pC+ 2 H + 4 C1,4&C1.C5& H /.,& /-.&1 *" &.+he mass Now rate of the 0as in 0as tur9ine is 5 *04se#. +he spe#i># enthalp% and elo#it% in the inlet are 1& *J4*0 and 2 m4se# respe#tiel% while the outlet are , *J4*0 and , m4se#7 respe#tiel%. 'al#ulate the power output in *w of the tur9ine if there is a heat loss of 1., !". A. /-75&2 //
B. 57$-2 C. 43,250 D. /57&, Solution( % t = m( h2 − h1 ) +
m(. 22 − . 12 ) 2( 1000 )
% t = 40(1600 − 500) +
[
− ,L
40 ( 200)
2
− ( 50)
2( 1000 )
2
] −1500
% t = 437250 k%
-.An ideal 0as tur9ine operates with a pressure ratio of and temperature limits of 2, ' and 11 '. +he ener0% input in the hi0h temperature heat ex#han0er is / *w. Determine the air Now rate in *04hr. A. 1&, B. 1$, C. 1325 D. 1-/2 Solution( p34 p1 + 1 2, 2-/ 2$ + 2 11 2-/ 1/-/ + p = + p 2
2
1
1
+ 2 298
= ( 9)
k −1
k −
1.4 1 1.4
+ 2 = 558 .3 /
, A = mcp ( + 3 − + 2 ) 300
= m(1.0) (1373 − 558.3)
m= 0.3682 kg s = 1325 .5 kg hr
S+EA! ;O"ER ;LA=+ -1.+he ener0% input to the tur9ine is 2$ *J4*0 and the ener0% at the exit is 15 *J4*07 e@#ien#% of the en0ine is -,6 and power output at full load is & *". "hat is the tur9ine Now rate at full load in *04*"3hr8 A. ,.B. 5.// C. 3.43 D. .2/ Solution(
/5
+urbine5owrate=
+urbine5owrate=
+urbine5owrate=
3600 m
% t 3600 m
m( h1 − h2 )η t 3600
( h − h )η t 1
+urbine5owrate=
2
3600
( 2800 − 1400 )( 0.75)
+urbine5owrate= 3.43 kg k%h
-2.A water3tu9e #ondenser has a total of tu9es. If these t%pe of #ondenser are two passes7 then #ompute the num9er of tu9es per pass. A. 45 &'($ B. / tu9es '. & tu9es D. 1$ tu9es Solution(
− 160 ) = mw ( 4.187 )( 24 − 13)
mw = 523.3 kg s
-5.In a steam Ran*ine #%#le7 saturated liuid at 15.- psia Cf .1&-2 #u. ft. per l97 is pump to a hi0h pressure liuid at 1 psia. ?ow mu#h wor* is reuired to pump one l9m of water8 A. 11$ Btu B. 1$., Btu '. .12 Btu D. 0.264 B& Solution( % p = mv f pf − pi
(1 lb) ( 0.01672 ft lb)(100 − 14.7 psia) (144 in ft ) 3
% p =
2
778
2
ft − lb Btu
% p = 0.264 Btu
/,
-,.A steam power plant has a tur9ine wor* of /2 *J4*0 and the pump wor* is 2 *J4*0. If the mass Now rate is 1 *0 per se#ond7 #ompute the power produ#ed 9% this plant8 A. /2 *w B. /22 *w C. 3180 kw D. 2$ *w Solution( % = % t − % p % = 10( 320 − 2 ) = 3180 k%
-&.A tur9ine has an e@#ien#% of -,6 and has a steam Now rate of 227 *04hr. If the aaila9le ener0% is 12 *J4*07 determine the *ilowatt output at full load. A. 5500 B. /, '. 5, D. 5 Solution( % t = m( h1 − h2 )η t
227000 (1200 )( 0.75) = 5500 k% 3600
% t =
--.+he aaila9le ener0% of a tur9ine is 15, *J4*07 e@#ien#% of the en0ine is -6 and the power output at full load is ,., !". "hat is the en0ine Now rate at full load in *04*"3hr8 A. 2., B. 3.5 '. /. D. 2.-, Solution( 3600 m !ngine5owrate= % t !ngine5owrate=
!ngine5owrate=
!ngine5owrate=
3600 m
m( h1 − h2 )η t 3600
( h1 − h2 )η t 3600
(1450 )( 0.70)
!ngine5owrate= 3.54 kg k%h
-$.A steam plant operates with an initial pressure of 1.- !;a and /- ' temperature and to a #ondenser at 1- *;a. If the tur9ine e@#ien#% and 9oiler e@#ien#% are -,6 and $,67 respe#tiel%7 determine the #%#le thermal e@#ien#%. Steam properties( At 1.- !;a /- '( h /1$-.1 *J4*0 s -.1$1 *J4*0 At 1- *;a( /&
hf 5$/.2 hfg 221&. sf 1.5-,2 sfg ,.-&2 A. 2,.&6 B. ,2.&6 C. 12.1% D. /,.&6 Solution(
h1
= 3187 .1 k- kg
s2
= s = 7.1081 k- kg− / 1
s2 = sf + xsfg 7.1081
= 1.4752 + x ( 5.7062 )
x = 0.987
h2
= hf + xhfg
h2 = 483.2 + 0.987( 2216 .0 ) h2 = 2670 .6 k- kg % t = ( h1 − h2 )η t % t = ( 3187 .1 − 2670 .6 ) ( 0.75) % t = 387.36 k- kg
% p = v 3 ( p4 − p3 )
Assume water densit% at 1 *04m /. % p =
1700
−170
1000
= 1.53 k- kg
% p = h4 − h3 1.53
= h − 483.2 4
/-
h4 = 484.73 k- kg
, A =
, A =
h1 − h4 η b 3187 .1
− 484 .73
0.85
, A = 3179 .25 k- kg
e=
e=
% t − % p , A 387.36
− 1.53
3179 .25
e = 0.121 = 12.16
-.=a0a priate power plant is usin0 water as the wor*in0 Nuid that operates on an ideal Ran*ine #%#le. Superheated apor enters the tur9ine at 17 *;a and ,2 ' and is exhausted into the #ondenser at $ *;a. +he net output of the Ran*ine #%#le is 17 *". Determine the mass Now rate of steam in *0 per se#ond. Steam properties( At 17 *;a C1 9ar and ,2 '( h /52,.17 s &.&22 At $ *;a( hf 1-/.$$7 sf .,2& hg 2,--7 sg $.22$v f .1$5 A. &- *04se# B. 2&- *04se# C. 74.4 kg/$(c D. 2.1 *04se# Solution(
h1 = 3425 .1 k- kg s2
= s = 6.622 k- kg− / 1
s2 = sf + xsfg 6.622
= 0.5926 + x ( 8.2287 − 0.5926 ) /$
x = 0.7896
h2
= hf + xhfg
h2 = 173.88 + 0.7896 ( 2577 − 173.88) h2 = 2071 .38 k- kg
% t = h1 − h2 m
% t = 3425 .1 − 2071 .38 = 1353 .72 k- kg m % p m
% p m
= v ( p − p ) 3
4
3
= ( 0.0010084 )(107000 − 8) = 10.08 k- kg
% % t % p = − m m m 1007000
m
= 1353 .72 − 10.08
m= 74.4 kg se
$.A 1,7 *" steam tur9ine30enerator power plant has full load steam rate of -., *04*"3hr. =o load steam #onsumption is around 16 of full load steam #onsumption. 'al#ulate the hourl% steam #onsumption at three3fourths load in *04hr. A. 1127, *04hr B. 1,7, *04hr C. 87,187.50 kg/"r D. &7-, *04hr Solution( ms = kx + b At full load ms = 7.5(157000 ) = 1127500 lb hr At no load b = 0.10(1127500 ) = 117250 lb hr x = 157000
+hen7 ms = kx + b 112 7500
= k (157000) + 117250
k = 6.75
/
x = (157000 )( 0.75) = 117250 k%
At P load7 ms = 6.75(117250 ) + 117250 ms = 87 7187.5 kg hr
$1.A t%pi#al steam 0enerator with an e@#ien#% of $,6 is produ#in0 1$ *04se# of steam. +he enthalp% of the enterin0 water is 15 *J4*0 and is superheated to an enthalp% of //., *J4*0. +he fuel used has a heatin0 alue of 557, *J4*0. +he A4< ratio 9% wei0ht is 2. Determine the amount of air needed in *04hr. A. 1&$7/1B. 1-$7/2$ '. &712/ D. 108,288 Solution( ms ( h2 − h1 ) η b = mf . 0.85
=
18( 3600 )( 3300 .5
− 140)
mf ( 447500)
mf = 5414 .42 kg hr ma = r famf = 20( 5414 .42) = 108 7288 kg hr
$2.A 127 *" steam tur9ine30enerator power plant has full load steam rate of ,., *04*"3hr. =o load steam #onsumption is around 16 of full load steam #onsumption. 'al#ulate the hourl% steam #onsumption at two3third of load in *04hr. A. 572, B. /&7$/ C. 46,200 D. /,7$2 Solution( ms = kx + b At full load ms = 5.5(127000 ) = 667000 lb hr At no load b = 0.10( 667000 ) = 67600 lb hr x = 127000
+hen7 ms = kx + b 66 7000
= k ( 6600 ) + 127000
k = 4.95
x = (127000 ) ( 2 3) = 87000 k%
At P load7 5
ms = 4.95( 87000 ) + 67600 ms = 467200 kg hr
$/.A #ondenser 0au0e reads 2,., in ?0 a#uum when the 9arometer stands at 1./2 m of ?2O. Determine the a9solute pressure in the #ondenser in mm ?0. A. -$. mm ?0 B. 2/,.& mm ?0 '. 1,-.& mm ?0 D. 111.6 mm )g Solution(
760 mmg . . 29 92 in g
pg = −25.5 in. g
pg = −647 .73 mmg patm = (10.32 m2#) (9.81 k& m3 ) patm = 101.24 kPa
760 mmg . 101 325 kPa
patm = 101.24 kPa
760 mmg . 101 325 kPa
patm = 101.24 kPa pabs = pg + patm
pabs = −647.73 + 759.4 pabs = 111.67 mmg
$5.Steam at a pressure of 1 !;a C hf -&$.$1 *J4*07 hfg 21,./ *J4*0 in the main steam line is passed throu0h a throttlin0 #alorimeter with an a9solute pressure of 11 *;a and a temperature of 1, ' C h 2&$,.1, *J4*0. 'al#ulate the steam ualit%. A. 95.39% B. /.,6 '. .,/6 D. 1./,6 Solution( h − hf 2685 .15 − 768 .81 = = 95.096 x = 2015 .3 hfg $,.A steam tur9ine is powered 9% a set of >xed ori>#e noles. Ea#h has an isentropi# e@#ien#% of 6. 1-&.&- ' sat. steam C hg 2--/.1 *J*0 at $,.5 m4s enters the noles. +he steam expands adia9ati#all% to ,//., m4s. "hat is the enthalp% of the steam as it leaes the noles8 A. 2&1 *J4*0 B. 2&5/ *J4*0 C. 2634 k*/kg D. 2&1 *J4*0 51
Solution( v 12 h1 +
(
2 1000
)
=h + 2
v 22
(
2 1000
)
( 85.4) 2 ( 533.5) 2 2773 .10 + = h2 + 2(1000 ) 2( 1000 ) h2 = 2634 k- kg
$&.An industrial plant operates a , *" tur9ine hain0 an en0ine e@#ien#% of -/6. +he initial steam #onditions are at /2.5 *J4*0 enthalp% and &.52 *J4*03 entrop%. +he 9a#* pressure is /.5 *;a Csf ./$ *J4*037 sg $.,25& *J4*037 hf 1./2 *J4*07 hg 2,5&.$2 *J4*0. +he tur9ine steam rate in *04*w3hr is7 A. /./2 B. 4.54 '. /.2/ D. 5.5, Solution( s1 = 6 .4200 k- kg− / s2
= s = 6.4200 k- kg− / 1
x =
− sf 6.4200 − 0.3818 = = 0.7415 sg − sf 8.5246 − 0.3818 s2
h2
= hf + x hg − hf = 109.32 + 0.7415( 2546 .82 − 109.32)
h2
= 1916 .73 k- kg
h1 = 3002 .4 k- kg
steamrate=
( 3600 s hr ) (
0.73 3002 .4
− 1916 .73 k- kg)
= 4.54 kg k% − hr
$-.A water tu9e 9oiler has a heatin0 surfa#e area of , m 2.
D. 1.5, Solution( 3 19 litersmin= 0.019 m mi mw
=
3
min
−3
m3 kg
0.019 m 170305
× 10
= 18.44 kg min
ms∆h = mwcp∆t
At $ 6 ualit% h = hf + x ( hg − hf ) = 529.65 + 0.80( 2715 .0 − 529.65) = 2277 .93 k- kg ms ( 2277 .93 − 529 .65) = (18.44 )( 4.187 )( 82 − 16 ) mw = 2.9 kg min
$.After expandin0 2., L of superheated steam at 2., !;aa and 5 '7 its pressure was de#reased to .1 !;aa. If its dr%ness fra#tion is 67 what is the >nal olume of the steam in L8 Q 2., !;aa 5 '7 v 12,.2 x 1 3/ m/4*0. Q .1 !;aa7 v f 1.12 x 1 3/ m/4*07 v g 157&-5 x 1 3/ m/4*0. A. 264 B. 1&5 '. /&5 D. 5&5 Solution( v 2 = v f + x ( v g − v f ) = 1.01022 ×10 −3
+ 0.90(147674 ×10− − 1.01022 ×10− ) = 137206 ×10− m kg 3
3
. 1
m=
v 1
=
. 2 =
3
. 2 v 2
2.5 L 125 .2
3
× 10 −
3
=
. 2 137206
× 10 −
3
264 L
.+he euialent eaporation of a 9oiler7 from and at 1 ' is 1, *0 steam per *0 fuel and the #alori># alue of the fuel 9urned is 51. !J4*0. "hat is the 9oiler e@#ien#%8 Q 2., !;aa 5 ': hf0 22,- *J4*0. A. $$.6 B. 80.8% '. $5.$6 D. $2.6 Solution( 2257 k- kg η b = (15 kgsteamkg fuel) 417900 k- kg (100 6) = 80.86
5/
1.In a steam power plant7 the #apa#it% is 2 !" and ,6 is for auxiliaries. If the ener0% #har0ed to the station is 1 !" and the 9oiler e@#ien#% is -,67 the 0ross station heat rate in *J4*"3 hr is . A. 15721 B. 1/7, '. 1$7, D. 18,000 Solution( (100 $% )( 3600 k- k% − hr ) grossstationheat rate= = 18000 k- k% − hr 20 $% 2.+he power deeloped 9% the steam tur9ine is 2$ *" while the power reuired 9% the feedwater pump is 12 *". If the heat supplied to the 9oiler is /, *J4*0 and the heat reFe#ted from the #ondenser is 22 *J4*07 >nd the mass Now rate of the steam in *04s. A. 2.1 B. 1.'. 2., D. /.1 Solution( m=
2800 3500
− 12
− 2200
= 2.145 kg s
/.Determine the wor* of a pump in *J reuired in raisin0 the pressure of 1 *0 of water from 2 *;aa to / !;aa7 assumin0 that the spe#i># olume of saturated water at 2 *;aa is 1.1- x 13/ m/4*0. A. 30.3 B. 2.&, '. /./ D. 2&., Solution( % = mv ∆ p = ( 1) (1.1 -× 1−/ m/ kg)( /− 2kPa) = /./k5.)ien the steam pressure of l94ft 27 temperature of / <7 spe#i># olume of ,.$ ft /4l9. If the spe#i># enthalp% is , ft3l94l9. "hat is the internal ener0% per pound of the s%stem8 A. 55 ft3l9 B. / ft3l9 C. 4280 +!' D. /5 ft3l9 Solution( u = h− pv = 9500 − ( 900) ( 5.8) = 4280 ft − lb lb ,.Determine the heat transferred to the #oolin0 Nuid in a #ondenser operatin0 under stead% Now #onditions with steam enterin0 with an enthalp% of 2/ *J4*0 and a elo#it% of /, m4s. +he #ondensate leaes with an enthalp% of 1& *J4*0 and elo#it% of - m4s. A. 312 *J4*0 B. 21 *J4*0 C. !2199 k*/kg D. 21 *J4*0 Solution( . 22 − . 12 , = h2 − h1 + 2( 1000 )
55
( 70) 2 − ( 350) 2 , = 160 − 2300 + 2(1000 ) ( 70) 2 − ( 350) 2 , = 160 − 2300 + 2(1000 ) , = −2199 k- kg
&.Steam at the rate of , *04hr is produ#ed 9% a 9oiler from 5 ' feedwater. If the enthalp% of the steam is 2,12 *J4*07 >nd the hourl% rate of heat reuired. A. 171,17/- B. 171/572- C. 1,172,264 D. 1712172-5 Solution( hf = cpw( ∆t ) = ( 4.187 )( 40 − 0) = 167.48 k- kg h1 = 2512 k- kg m s = 500 kg hr
=m (h − h ) , A s 1 f , A = ( 500 )( 2512 − 167 .48) = 17172 7260 k- hr
-.A simple Ran*ine #%#le operates 9etween superheated steam at & !;a. & ' enterin0 the tur9ine7 and 1 *;a enterin0 the pump. Determine the maximum possi9le #%#le thermal e@#ien#%. Steam properties( at & !;a7 & '7 h /&,$7 s -.1&$,: at 1 *;a7 hf 127 hfg 2//7 sf .&57 sfg -.,27 f .1 m /4*0. A. /,. B. /-., C. 40.0 D. 52., Solution( h1 = 3658 s2
= s = 7.1685 = 0.649 + x ( 7.502)
x 2
= 0.869
h2
= hf + x
1
2
2
hfg = 192 + 0.869( 2393 ) = 2271 .52
h3 = hf at 10kPa= 192 h4
= h + v ( p − p ) = 192 + 0.001( 6000 − 10 ) = 197.99 3
3
4
3
% p = v 3 ( p4 − p3 ) = 0.001( 6000 − 10) = 5.93 e=
h1 − h2 − % p h1 −h4
=
− 2271 .52 − 5.93 = 0.39894 = 39.8946 3658 − 197.99
3658
5,
$.An open feedwater heater utilies saturated steam at 1, ' extra#ted from the steam tur9ine.
At & ' h5 = 251.13 me =
ms ( h6 − h5 ) h2 − h5
=
187000 ( 632.2 2746 .5
− 251.13)
− 251.13
= 2748 .795 kg hr
.A steam 0eneratin0 unit produ#es steam at the rate of 1 *04s at , !;a7 5, ' whi#h is #ontinuousl% 9lown down at .2, *04s. red is 9urned at the rate of 1 *04s. "hat is the oer3all e@#ien#% of the steam 9oiler8 Steam properties at , !;a7 5, '7 h //1&.2: at , !;a7 1 '7 h 522.-2: at , !;a7 hf 11,5.2/7 hg 2-5./. A. 69.33 B. &/./ '. &&./& D. -2./ Solution( hbo = hf at 5$Pa= 1154 .23 Oer3all 9oiler e@#ien#%7 eo. ms hs − hfw + mbo hbo − hfw eo = mf ( . ) eo =
10( 3316 .2
− 422.72) + 0.25(1154 .23 − 422.72) (1)( 427000)
×100 6
eo = 69.336 C#. 1. A horiontal return tu9ular 9oiler with a heatin0 surfa#e of 2 m 2 0enerates steam at the rate of 5., *04s.
5&
A. &2.5& B. &&.-$ C. 64.62 D. &$.5 Solution( re tu9e 9oiler totalheatingsurface 2000 m2 ratedbohp= = 2 k 1.1 m
= 1818 .18
Deeloped 9oiler hp( ms hs − hfw 4.5( 3600 )( 3068 .4 − 507.09) = dev bohp= 357316 k- hr 357316
= 1174 .91
;er#ent ratin0 dev bohp 1174 .91 6rating= ×100 = ×100 = 64.626 1818 .18 ratedbohp 11. A ,3!" steam tur9ine 0enerator power plant has a full3load steam rate of &. *04*"3hr. Assumin0 that no3load steam #onsumption as 16 of full3load steam #onsumption7 #ompute for the hourl% steam #onsumption at &6 load7 in *04he. A. 127 B. 2,7, C. 19,200 D. /17$ Solution( At full load Steam #onsumption7 ms3 ms3 C, *wC&. *04*"3hr /7 *04hr At no3load Steam #onsumption7 ms1 ms1 C.1C/7 *04hr /7 *04hr Gsin0 two3point form( y − y y − y 1 = 2 1 ( x − x 1 ) − x x 2 1 "here( ;1Cx17%1 ;1C7/ ;2Cx27%2 ;2C,7/ ;Cx7% ;Cms7L +hus ms ,.5L / At & 6 load7 L .&C, / ms ,.5C/ / 172 *04hr 12. +he a#uum in the surfa#e #ondenser of a small #ondensin0 steam power plant is &5 mm ?0 as referred to a -& mm ?0 9arometer. If the temperature in the #ondenser is /, ' C#orrespondin0 pressure of ,.&2$ *;a7 the a#uum e@#ien#% is approximatel%( A. $1.& B. $&.1 C. 89.16 D. /./ Solution( 5-
η vac
=
patm− pcond patm− psat
patm -& mm ?0 pcond &5 mm ?0 a# 12 mm ?0 psat ,.&2$ *;a C-& mm ?0 4 11./2, *;a 52.21 mm ?0 η vac
=
− 120 ×100 = 89.166 760 − 42.21 760
1/. A 9oiler feedpump re#eies water at 2 ' Centhalp%7 h $, *J4*0 from a surfa#e #ondenser at the rate of 1 L4s. It operates a0ainst a total head of $, meters. Determine the enthalp% leain0 the pump7 in *J4*0. A. $-$., B. $.2 C. 857.8 D. 1. Solution( ;ump wor*7 "p % p mCh3 H h1 mgh41 h3 H $, C.$1C$41 h3 $,-.$ *J4*0 15. A 9inar% mer#ur%3steam #%#le produ#es a net power output of / !". Steam enters the tur9ine at 5 !;a7 5 ' where it is 9ein0 exhausted to a #ondenser pressure of 1 *;a. Assumin0 an oerall thermal e@#ien#% of $6 for the #%#le and 16 heat transfer e@#ien#% in the heat ex#han0er7 determine the enthalp% diMeren#e in the 9oiler per *ilo0ram of mer#ur% per se#ond. Steam properties at 5 !;a7 5 '7 h /21/.&7 s &.-&: at 1 *;a7 hf 11.$/7 hfg 2/2.$7 sf .&/57 sfg -.,. A. 37,500 B. /7 '. /27, D. /,7 Solution( % eb = net , A , A =
% net 307000 = = 377500 k% 0.80 eb
, A = mg( ha − hd )
( ha − hd ) =
, A 377500 = = 377500 k- kg 1 mg
1,. In a #o30eneration steam power plant7 steam enters the tur9ine at , !;a7 5 '. A uarter of the steam is extra#ted from the tur9ine at 1- *;a while the remainin0 steam is allowed to expand to 1 *;a. +he extra#ted steam is then #ondensed and mixed with feedwater at #onstant pressure. +he mixture is then pumped to a 9oiler pressure of , !;a. !ass Nowrate of steam throu0h the 9oiler is 2, *04s. Assumin0 no pressure drops and heat losses in the pipin0 s%stem7 #al#ulate for the pro#ess heat reuired in *". Steam properties at , !;a7 5 '7 h /1,.-7 s &.&5,: at .1- !;a7 hf 5$/.27 hfg 221&7 sf 1.5-,27 sfg ,.-&2: at .1 !;a7 hf 11.$/7 hfg 2/2.$7 sf .&5/7 sfg -.,. A. 12,551 5$
