Solucionario Cap 11

CHAPTER ELEVEN 11.1

a.

The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is: Mp

x p

M

Therefore, the leakage rate of hydrogen peroxide is m1 M p / M b.

: Accumulation = input – output

Balance on mass

E

dM dt

 m 0 m1

t 0, M M 0 (mass in tank when leakage begins) Balance on H 2O 2: Accumulation = input – output – consumption

E

dM p dt

FG H M MKIJ kM

 m 0 x p 0 m1

p

p

t 0, M p M p0

11.2

a.

Balance on H3PO4: Accumulation = input Density of H3PO4: 1834. g / ml . Molecular weight of H3PO4: M 9800 g. / mol . Accumulation =

dn p

(kmol / min) dt 20.0 L 1000 ml 1.834 g

Input =

min

E dn p

L

ml

mol

1 kmol

98.00 g 1000 mol

 0.3743 kmol / min

 0.3743

dt t 0, n p0 150 0.05 7.5 kmol

np

b.

t

dn p 03743 . dt n p. 75 0.3743t (kmol H3PO 4 in tank ) 7.5

x p

c.

015 .

0

np n



np n0 n p n p0

7.5 03743t . 150 03743t .

7.5 0.3743t  150 0.3743t

kmol H 3PO4 kmol

 t 471. min

11-1

11.3

a.

b

g & bt 5, m h 1000g m bkg hg 750 50tbg

m w a bt  t 0, mw 750

w

w

Balance on methanol: Accumulation = Input – Output

M kg CH3OH in tank  m f mw 1200 kg h

b750 50tg kg h

-170dM dt

E

g

b

dM  450 50t kg h dt t 0, M 750 kg M

t

dM

b.

750

E

b450 50tgdt

0

M 750 450t 25t 2

E

M 750 450t 25t 2 Check the solution in two ways

:

(1) t 0, M 750 kg satisfies the initial condition; dM  450 50t  reproduces the mass balance. (2) dt c.

dM dt

 0 t 450 50 9 h M 750 450(9) 25(9)2 2775 kg (maximum)

b g b gb750g t = –1.54 h, 19.54 h

M 0 750 450t 25t 2 450  4 25 2 450 t 2 25

b g

d.

3.40 m 3 103 liter 0.792 kg 1 m3

1 liter

 2693 kg (capacity of tank)

b g b gb

g

M 2693 750 450t 25t 2 450  4 25 750 2693  t 719. h ,1081. h 450

2

b25g Expressions R|750for+ M(t) 450t - are: 25t b0 t 719. and 1081. t 1954. g (tank is filling or draining) . . 0 2693 M(t) = | TS (719 t 10.81) (tank is overflowing)

t

2

2

(19.54 t 2054)

(tank is empty, draining as fast as methanol is fed to it)

11-2

11.3 (cont’d)

3000 2500 2000 1500 1000 500 0 0

5

10

15

20

t(h)

10.0 ft 3

Air initially in tank: N 0

11.4 a.

492 R 1 lb - mole 532 R 359 ft 3 STP

b g

 0.0258 lb - mole

Air in tank after 15 s:

Pf V P0V

N f RT

Pf

N0 RT

0.0258 lb - mole 114.7 psia

P0

14.7 psia

. 15 s

b.

Balance on air in tank: Accumulation = input dN dt

 0.0117

blb - moles sg ; t 0 , N 0.0258 lb - mole N

c.

Integrate balance:

t

dN n dt N 0.0258 0.0117t lb - mole air 0.0258

g

0

Check the solution in two ways : (1) t = 0, N = 0.0258 lb - mole satisfies the initial condition dN  0.0117 lb - moleair / s reproduces the mass balance (2) dt

g 030 lb - mole O

2

b gb120g 143 lb - moles air

t 120 s N 0.0258 0.0117

d.

O2 in tan k 0.2 1

b 143

M(kg)

11-3



 N f N0

Rate of addition: n



 0.2013 lb - mole

b02013 0.0258g lb - mole air 0.0117 lb - mole air s

b

. .

.

11.5

a.

Since the temperature and pressure of the gas are constant, a volume balance on the gas is equivalent to a mole balance (conversion factors cancel). 3

540 m dV Accumulation = input output  h dt

1h 60 min



 w m3 min



t 0, V 3.00103 m3 t 0 corresponds to 8:00 AM V

t

3.0010 3

0

t

 dV9.00  dt V m  3.0010  9.00t dt t in minutes w

3

3

w

0

Let wi tabulated value of w at t 10

b. 240

bi 1g

24 24

i 1, 2,, 25 10 3

0

 2 V 4 8 3001 8 03 m 3 9.00

b240 g 2488 2672 m3

c.

Measure the height of the float roof (proportional to volume). The feed rate decreased, or the withdrawal rate increased between data points, or the storage tank has a leak, or Simpson’s rule introduced an error.

d.

REAL VW(25), T, V, V0, H INTEGER I DATA V0, H/3.0E3, 10./ READ (5, *) (VW(I), I = 1, 25) V= V0 T=0. WRITE (6, 1) WRITE (6, 2) T, V DO 10 I = 2, 25 T = H * (I – 1) V = V + 9.00 * H – 0.5 * H * (VW(I – 1) + VW(I)) WRITE (6, 2) T, V 10 CONTINUE 1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)') 2 FORMAT (F8.2, 7X, F6.0) END $DATA 11.4 11.9 12.1 11.8 11.5 11.3  Results : TIME (MIN) 0.00

10.00 20.00

. .

.

VOLUME (CUBIC M ETERS) 3000. 2974. 2944.

 230.00

 2683. 2674.

240.00

Vtrapezoid 2674 m 3 ; VSimpson 2672 m 3 ; Simpson’s rule is more accurate

2674 2672 2672

.

 100% 0 .07%



11-4







 wdt

L MP O

10  w1 w 25 4 w i 2 w i 3 i2, 4, i 3, 5,

.

bg

114 98 4 1246 2 113.4

a.

b.

b g bg

out L min  kVL

 out 0.200V out 20.0 L min Vs 100 L

V 300  out 60

Balance on water: Accumulation = input – output (L/min). (Balance volume directly since density is constant)

dV  20.0 0200V . dt t 0 , V 300

c.

dV dt

.  0 200 0200V s Vs 100 L

The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is 200 . 0.200(300)40.0. As t increases, V decreases. dV / dt 20.0 0.200V becomes less negative, approaches zero as t . The curve is therefore concave up.

V

11.6

t

V

d. 300

dV

t

 dt 20.0 0.200V 0

FHI 1

K t 0.5 0.005V expb 0.200tg V 100.0 200.0 expb0.200t V 101b. 100g 101 L b1% from steady stateg lnb120 g 26.5 min 101 100 200 exp b 0.200tg t 0.200 

ln

0.200

20.0 0.200V

 40.0

11-5

g

11.7

a.

A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points ( t 1 week, D 2385 kg week ) and ( t 6 weeks, D 755 kg week ).

ln D bt ln a D ae bt

bg

lnD21ln75238 b0230 6 1 t 2 t1

.

b2385g b0.230gb1g 8.007 a e

ln a ln D1 bt1 ln

E

8.007

3000

D 3000e0.230t

b.

Inventory balance: Accumulation = –output

g

b

3000 e0.230t kg week

-170dI dt t 0, I 18,000 kg I

30000.230t

t t

dI3000e0.230 t dt I 18,000 18, 000

c.

11.8

a.

0

e

0.230

0

 I 4957 13,043e0.230 t

t I 4957 kg

Total moles in room: N

1100 m3

273 K

b g 45, 440 mol

103 mol

295 K 22.4 m 3 STP 700 m 3

Molar throughput rate: n

min

273 K

b g 28,920 mol min

103 mol

295 K 22.4 m3 STP

SO 2 balance ( xt 0 is the instant after the SO 2 is released into the room):

bmolgbmol SO

N

2

mol

g mol SO

2

in room

Accumulation= –output. 

d dt

bNxg nx

dx

n 28,920 N45,440 dt

0.6364x 15 mol SO 2

t 0, x

. b.

.

. 45,440 mol

.  330105 mol SO2 mol

.

The plot of x vs. t begins at (t=0, x=3.30 10-5). When t=0, the slope (dx/dt) is 06364 330 105210 105. As t increases, x decreases. dx dt0.6364x becomes less negative, approaches zero as t . The curve is therefore concave up.

11-6

x

11.8 (cont’d)

0 t

c.

Separate variables and integrate the balance equation: x

dx x

3.30105

t

06364dt ln . 0

Check the solution in two ways

x 33010 .

06364t x . . 330105 e0.6364t

5

:

(1) t = 0, x = 3.3010-5 mol SO2 / mol satisfies the initial condition; dx 0.6364 33010 5 e 0.6364 t0.6364x reproduces the mass balance. (2) . dt d.

e.

CS O2

45,440 moles 1100 m 3

i)

t 2 min CSO2

ii)

x 106 t

x mol SO2 mol

1m3 103 L

 382 . 107

mol SO2 liter

e

ln 106 3.30 105 0.6364

.  413110 2 .x 13632106 e0.6364t mol SO2 / L

j 55. min

The room air composition may not be uniform, so the actual concentration of the SO 2 in parts of the room may still be higher than the safe level. Also, ―safe‖ is on the average; someone would be particularly sensitive to SO2.

11-7

11.9

a.

Balance on CO : Accumulation=-output

N (mol) x ( mol CO / mol) = total moles of CO in the laboratory Molar flow rate of entering and leaving gas: n (

kmol P p

)

h RT kmol CO P p  )x = x Rate at which CO leaves: n ( h kmol RT CO balance: Accumulation = -output kmol

d( Nx ) P p dx P  x RT dt NRT

dt

E PV NRT

dt

FHI

p

kmol t

0.01

K  x

kmol CO

dx V  p r  dt tr x  p V0

x

IJ

 dxx p V

t 0, x 0.01

b.

FHGK

b100 xg

ln

c.

V 350 m3 350 ln 100 35 106 283 . hrs tr 700

d.

The room air composition may not be uniform, so the actual concentration of CO in parts of the room may still be higher than the safe level. Also, ―safe‖ is on the average; someone could be particularly sensitive to CO.

ej

Precautionary steps

:

Purge the laboratory longer than the calculated purge time. Use a CO detector to measure the real concentration of CO in the laboratory and make sure it is lower than the safe level everywhere in the laboratory.

11.10 a.

Total mass balance: Accumulation = input – output

dM  m  m dt b.

bkg ming 0 M is a constant 200 kg

Sodium nitrate balance: Accumulation = - output x = mass fraction of NaNO3

bdt g xmbkg ming E

d xM

dt

dx m x x Mm 200 t 0, x 90 200 0.45



11-8

11.10 (cont’d) c.

0.45

m 50 kg / min m 100 kg / min x

m 200 kg / min

0 t(min)

dx m  x 0 , x decreases when t increases dt 200 dx becomes less negative until x reaches 0; dt Each curve is concave up and approaches x = 0 as t; dx becomes more negative x decreases faster. m increases dt x

d. 0.45

t

dx x

 0

m  M

x 0.45

dt ln

Check the solution



m 200

FHG

t x 0.45 exp

mt  200

IJ K

:

(1) t = 0, x = 0.45 satisfies the initial condition; dx m m t m (2)0.45 exp( ) x satisfies the mass balance. dt 200 200 200 0.45 0.4

m 50 kg / min

0.35

m 100 kg / min

0.3 0.25 x

m 200 kg / min

0.2 0.15 0.1 0.05 0 0

5

10

15

20

t(min)

e.

m  100 kg min t 2 ln

dx 0.45i f

90% x f 0.045 t 4.6 min 99% x f 0.0045 t 9.2 min 99.9% x f 0.00045 t 138 min .

11-9

25

11.11 a.

Mass of tracer in tank: V

em jekg m 3

: Accumulation = –output. If perfectly mixed, C out C tank C

Tracer balance

V is constant

d VC dt

C

b. m0 V

c.

j

3

dt

 V

C

t 0, C m0 V dC C



t

dt ln

0V

F C I m0 V

t V

m0 V

 C

exp

V

Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption of perfect mixing) through

V

et 1, C 0223 10 j & et 2, C 0.05010 j . 3

3

2 1

min

11.12 a.

In tent at any time, P=14.7 psia, V=40.0 ft 3, T=68F=528 R

 N

b.

PV RT

 m(liquid)

14.7 psia ft 3 psia 10.73 lb - moleo R

40.0 ft 3 528 o R

Molar throughout rate:

60 ft 3 492 R 16.0 psia 1 lb - mole

2

Balance on O2: Accumulation = input – output d Nx dt

dt

21 x

c. 0.21

t

0.35 x

. 0

b 0 . 3 5  

0

g

b0.35 xg

 01038 lb - mole

 163t 035 x 014

. 1

0.35 0.27

1.63

0.35 0.21

C

dC   

b g C bkg ming 11-10

z

z



GH

t FG H IJK

JK .

bg

 ln0.5 23

1.495 min1

E V e30 m j e1.495 min j 201 m

3     1 3

.

nin nout n

.

F KI

H

Moles of O2 in tank= N (lb - mole) lb - mole O lb - mole

b g 0.35n xn 0.1038 dx 0.1695b0.35 xg dxdt 1.63b0.35 xg t 0, x 0.21

z 

dx . .

z

 163dt ln

.

.

 e1.63t x 0.35 014e1.63t

x 0.27 t

LMN FGHln

IKQJOP  0.343 min (or 20.6 s)

11.13 a.

Mass of is otope at any time  V

blitersgbmgC isotope liter

g

Balance on isotope: Accumulation = –consumption

d dt

mg FG HLKIJs Vb Lg

bVCg kC

dC

Cancel V

dt

 kC

t 0, C C0

Separate variables and integrate

zz F I C

t

bg

dC C kt t ln0C C kdt ln C0 C C0

H K 

C 0.5C0 t1 2

b.

t 1 2 2.6 hr k

0

k

bg

 ln 0.5

t

k

ln 2 2.6 hr



12

ln 2 k

 0267 hr 1 .

b g 17.2 hr

 ln 0.01 C 0.01C0

t

t=-ln(C/C0)/k

0.267

11.14 A products a.

Mole balance on A: Accumulation = –consumption

b gkC V

d C AV

A

dt t 0, C A C A0 

b.

z

bV constant; cancelsg

z

CA

dC A t  kdt ln 0 CA0 C A

FG H C KC IJkt C A

A

 C A0 exp

A0

bktg

Plot C A (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies

b

gb

assumption of first-order) through t 213, C .A 00262 . & t 120.0, C A 0.0185 .

ln C Akt ln C A0 k

b

ln 0.185026 120.0 213.

g353. 10

3

min1 k 35 . 103 min1

11.15 2 A 2 B C a.

Mole balance on A: Accumulation = –consumption

bg b

g

d C AV 2 kC V V constant; cancelAs dt t 0, C A C A0 

z

CA

dC A CA0 C A

t

  kdt 0

1 CA



1 C A0

 kt C A

LM C N

1

A0

11-11

1

 kt

OP Q

g

11.15 (cont’d) b.

1

C A 0.5C A0

0.5C A0

n A 0.5n A0

b0.5n b0.5n

n B nC



1 C A0

gb2 mol B 2 mol A react.. g 05n mol A react.gb1 mol C 2 mol A react.g 0.25n

A0

P0 RT

 t1 2

kP0

n A0 RT V

A0

 125P 0 .

dt

1 2

i dt

. 1060, 1 P0 1 0135 &

1 2

209, 1 P0 1 0.683

i

RT 1060 209   1432. s atm k . .1 0683 1 0135

Slope:

 k

RT

t 1 2

RT

A0

Plot t1 2 vs. 1 P0 on rectangular paper. Data fall on straight line (verifying 2 nd order decomposition) through

d.

kC A0

n ; but C A0 A0 V

mol A react.

A0

total moles 125n . A0 P1 2 . 125

c.

1

kt1 2 t1 2

k 0 P0

b1015 Kgb. 008206 L atm mol Kg 0582 . L mol s 143.2 s atm

Ft P I ln E FG IJ  ln HRTK GH RT JK 12 0

exp

1



k0

E1 RT

Plot t1 2 P0 RT (log scale) vs. 1 T (rect. scale) on semilog paper.

bg, P  1 atm, R 0.08206 LK atm / (mol K), Tbg Data fall on straight line through dt P RT 74.0, 1 T 1 900i & d P RT 0.6383, 1 T 1 1050i t1 2 s

0

12

0

12 0

t E R ln

e.

 1

b

ln 0.6383 74 .0

g

1 1050 1 900 .

b g

 ln 06383 

 29,940 K 29 ,940

E 2 .49 105 J mol 12

28.96 k 0 3.79 10

IJ  0.204 L (mol s) T 980 K k k exp RT K . . b0.08206 L070atmbmol Kgb 120 atm g 980 Kg 104510 . C  k0

FHG

R=8.314 J/ (mol ·K)

1050

L (mol s)

E

0

2

A0

C A 0.10C A0 t

LO

90% conversion 1



1

N

1

LM



Q

1

k C A C A0  4222 s 70.4 min

11-12

1

mol L

OP 0.204 N1.045 10

3

1  1.045 102

Q

11.16 A B a.

Mole balance on A: Accumulation = –consumption(V constant) dC A dt

z

k1C A



1 k 2 C A

t 0, C A C A0 1 k 2C A

CA

k1C A

C A0

b.

dC A

z

t

A

A0

y

C A0

g bC

A / C A0

b

bg

k k  2 C A C A0t t 2 C A0 C A k1 k1

CA

ln

k1

0

bC  C g vs. lnbC

Plot t

1

 dt

g

A0

 CA

1

CA

ln

k1

C A0

g on rectangular paper:

 1 ln C A C A0 k 2   k1 C A0 C Ak C A0 C A t

b

g; slope

1

intercept

FHG

y1

IJK FGH

x1

y2

IKJ

x2

Data fall on straight line through 116.28,02111. & 130.01,0.2496

1  k1 k2

13001 116.28 .

b0.2111g 356.62 k  2 .80 10 L (mol s)  130.01 356.62b.02496 k  0115 L mol . g 4100 . 

3

1

0.2496

2

k1

11.17 CO Cl 2 COCl 2 3.00 L a.

273 K

1 mol

b g 0.12035 mol gas

303.8 K 22.4 L STP

. bC g 0.60b012035 molg 3.00 L 0.02407 mol L CO U| . dC i 0.40b012035 molg 3.00 L 0.01605 mol L Cl V|Winitial concentrations CO i

2

Cl 2 i

CCO C Cl 2 b.

btg 0.02407 C btg U| btg 0.01605 C btg V|W Since 1 mol COCl formed requires 1 mol of each reactant p

2

p

Mole balance on Phosgene: Accumulation = generation

di

. CO C Cl 2 875C

d VC p  dt

c.

d1 586C .

Cl 2 34.3C p

dC p  dt

V=3.00 L

i

2

b

z

d.1941 24.3C i d0.02407 C id0.01605 C i

g 0.01204 mol L

2

0.01204

0

p

p

d.1941 24.3C i

t 0, C p 0

Cl 2 limiting; 75% conversion C p 075 . 0.01605 1 t 2.92

di

2.9047Cp165p

dC

p

11-13

p

p

i 2

11.17 (cont’d) d.

REAL F(51), SUM1, SUM2, SIMP INTEGER I, J, NPD(3), N, NM1, NM2 DATA NPD/5, 21, 51/ FN(C) = (1.441 – 24.3 * C) ** 2/(0.02407 – C)/(0.01605 – C) DO 10 I = 1, 3 N = NPD(I) NM1 = N – 1 NM2 = N – 2 DO 20 J = 1, N C = 0.01204 * FLOAT(J – 1)/FLOAT(NM1) F(J) = FN(C) 20 CONTINUE SUM1 = 0. DO 30 J = 2, NM1, 2 SUM = SUM1 + F(S) 30 CONTINUE SUM2 = 0. DO 40 J = 3, NM2, 2 SUM2 = SUM2 + F(J) 40 CONTINUE SIMP = 0.01204/FLOAT(NM1)/3.0 * (F(1) + F(N) + 4.0 * SUM1 + 2.0 * SUM2) T = SIMP/2.92 WRITE (6, 1) N, T 10 CONTINUE 1 FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES') END RESULTS 5 POINTS — 91.0 MINUTES 21 POINTS — 90.4 MINUTES 51 POINTS — 90.4 MINUTES

t 904. minutes

11.18 a.

e j emols cm

Moles of CO2 in liquid phase at any time V cm 3 C A

j

3

Balance on CO 2 in liquid phase: Accumulation = input d dt

bVC

g kSeC FjG C molsIJ dCdt kSV eC H s K t 0, C 0 A

*

A

A

A

* A

CA

j

V

A

Separate variables and integrate. Since p A y A P is constant, C *A p A H is also a constant.

z ej

CA

0

 ln

dC A

t

C A CA

0 VC A0



C *A C A C *A

kS



CA

dt ln C *A C A

kS expb g t 1 V

CA C *A



V

kS

t

 e kSt V C A C A

1 C A C *A

11-14

e1 e *

 kSt V

j

11.18 (cont’d)

b.

V kS

t

L NM ln 1

CA C *A

OP QP

. 2 , C A 0.62 103 mol / cm3 V 5 L 5000 cm3 , k 0.020 cm s , S 785 cm

a fa f d i e5000 cm j F10.62 10 I 9800 s 2.7 hr t b002. cm sge785. cm jlnHG0.65 10 JK 33

9230atm cm mol20  0.65 10 mol cm atm C *A y A P H 0.30 3

3

3

3

2

(We assume, in the absence of more information, that the gas-liquid interfacial surface area equals the cross sectional area of the tank. If the liquid is well agitated, S may in fact be much greater than this value, leading to a significantly lower t than that to be calculated)

11.19 A B a.

Total Mass Balance: Accumulation input

dM d (V )  v dt dt

E

dV dt

 v

t 0, V 0 : Accumulation input – consumption

A Balance

dN A  C A0v  (kC A )V CA=NA/V dt

dN A  C Ao v kN A dt t 0 , N A 0

b.

Steady State:

dN A  0 N A dt

zz z

C A0 v k

V t

c.

dV vdt  V vt

z

0 0 NA t

0

dN A  dt C A0 v kN A

0

C v kN 1 C v kN e  t F I k H C v C v K C v  C v 1 expbktg t N   N  k k

 ln

A0

A

A0

A0

A

 kt

A0

A0

A0

A

C A

A

NA V



C A0[1 exp(kt )] kt

11-15

11.19 (cont’d) When the feed rate of A equals the rate at which A reacts, N A reaches a steady value. NA would never reach the steady value in a real reactor. The reasons are:

(1) In our calculation, V = vt t, V. But in a real reactor, the volume is limited by the reactor volume; (2) The steady value can only be reached at t. In a real reactor, the reaction time is finite.

d.

lim C A lim

t

t

C A0 [1 exp(kt )] kt

 lim t

C A0 kt

0 NA V

From part c, t, N A a finite number, V C A 11.20 a.

MCv

0

dT  Q W dt M (3.00 L)(100 . kg / L) . = 300 kg Cv C p (0.0754 kJ / molo C)(1 mol / 0.018 kg) = 4.184 kJ / kgo C

W 0 dT  0.0797Q (kJ / s) dt t = 0, T = 18 o C

zz 100o C

b.

dT o

18 C

c.

11.21 a.

240 s

0.0797 Q dt Q

0

100 18 240 0.0797

 4.287

kJ s

 4.29 kW

Stove output is much greater. Only a small fraction of energy goes to heat the water. Some energy heats the kettle. Some energy is lost to the surroundings (air).

Energy balance: MCv

dT dt

 Q W M 20.0 kg C v C p ( 0.0754 kJ / mol o C)(1 mol / 0.0180 kg) = 4.184 kJ / (kgo C)

af

Q 0.97 (2.50) 2.425 kJ s W 0 dT dt

 0.0290

b C sg , t 0 , T 25 C

The other 3% of the energy is used to heat the vessel or is lost to the surroundings.

zz T

b.

t

bg

dT 00290dt T 25 C 0.0290t s.

25 C

0

o

c.

T 100 C t

b100 25g 0.0290 2585 s 43.1 min

No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal boiling point).

11-16

11.22 a. Energy balance on the bar

dTb dt

MCv

g

b

  Q WUA Tb Tw

B

Table B.1

ej

3

3

M 60 7.gcm462g cm Cv 0.46 kJ (kg C), Tw 25 C 2

a2fa 3f a2fa10f a3fa10f cm  112 cm

U 0.050 J (min cm C) A 2

dTb dt

2

b

0.02635 Tb 25

2

gbC min g

t 0, Tb 95 C

d i

dTb dt

b.

 00.02635 Tbf 25  Tbf 25 C

95 85 75

Tb(oC)

65 5 45 35 25 15 5 0 t Tb

c.

t

dTb 0.02635dt 95 bT 25 0

 ln

FG . H 95T25K25IJ002635t b

btg 25 70 expb0.02635tg

 Tb

Check the solution in three ways

:

(1) t = 0, Tb 25 70 95o C satisfies the initial condition; (2)

dTb dt

70 0.02635e0.02635 t002635(T b 25) reproduces the mass balance; .

(3) t, Tb 25o C confirms the steady state condition. Tb 30 C t 100 min

11-17

11.23 12.0 kg/min 25o C

12.0 kg/min T (oC)

Q (kJ/min) = UA (Tsteam-T)

a.

dT dt

: MCv

Energy Balance

g

b g b

 mCp 25 T UA Tsteam T

M 760 kg m 12.0 kg min

dT / dt 150 0.0224T ( oC min), t 0 , T 25oC . Cv Cp 2.30 kJ (min C) UA 11.5 kJ (minC)

a sat'd; 7.5barsf 167.8 C

Tsteam

b.

Steady State:

dT dt

.  0 150 0.0224Ts Ts 67 C

T(oC)

67

25 0 t

T

c.

f

dT 150 . . 00224T

t

 dt t

2 5

0

1 00224 .

l n

. 0.0224T FG 150 IJ  T H K 0.94

150 . . 094 exp(0.0224t ) 0.0224

.

t 40 min. T 498 C d.

U changed. Let x (UA) new . The differential equation becomes:

z dT

dt

55

25

z

 0.3947 0.096x ( 0.01579 5.721x ) T

 4 0.3947 0.096x ( 0.01579 5721 104 x)T 0 dT dt 4 . 0.3947 0.096 x 0.01579 5721 10 x  55 1 ln  0 4 0.01579 5.721 104 . 0.3947 0.096x 0.01579 5721 10 x  25

ML MM N

e

x .

. o C).  x 14.27 kJ / (min U U initial

(UA)  (UA)initial

. 1427115  115

. 100% 241%

11-18

e

j OP j PP 40Q

11.24 a.

Energy balance: MCv

dT  QW dt W 0, Cv 1.77 J g C M 350 g, Q 40.2W 40.2 J s

b gU|V T 20 0.0649tbsg |W T 40 C t 308 s 51. min

dT  0.0649 C s dt t 0, T 20C b.

. The benzene temperature will continue to rise until it reaches Tb 801 C ; thereafter the heat input will serve to vaporize benzene isothermally.

bg

801 . 20  926 s 0.0649 Time remaining: 40 minutes 60 s min 926 s 1474 s Time to reach Tb neglect evaporation : t

b

g  b30.765 kJ molgb1 mol 78.11 ggb1000 J kJg 393 J g Evaporation:H  / 393 J .gg 0102 g s Evaporation rate b40.2 J sgb . Benzene remaining 350 g b0102 g sgb1474 sg 200 g v

c.

11.25 a.

1. Used a dirty flask. Chemicals remaining in the flask could react with benzene. Use a clean flask. 2. Put an open flask on the burner. Benzene vaporizes toxicity, fire hazard. Use a covered container or work under a hood. 3. Left the burner unattended. 4. Looked down into the flask with the boiling chemicals. Damage eyes. Wear goggles. 5. Rubbed his eyes with his hand. Wash with water. 6. Picked up flask with bare hands. Use lab gloves. 7. Put hot flask on partner’s homework. Fire hazard.

Moles of air in room: n

60 m3

273 K

1 kg - mole

b g  2.58 kg - moles

283 K 22.4 m 3 STP

dT  Q W dt Q msHv H 2O, 3bars, sat'd 30.0 T T0  W 0

Energy balance on room air: nCv

nCv

b

dT dt

g

b

 m sHv 30.0 T T0

b

g

g

N 2.58 kg - moles Cv 20.8 kJ (kg- mole C) Hv 2163 kJ kg

bfrom Table B.6 g

T0 0 C dT dt

 40.3ms  0.559T

b C hr g

t 0, T 10 C (Note: a real process of this type would involve air escaping from the room and a constant pressure being maintained. We simplify the analysis by assuming n is constant.)

11-19

11.25 (cont’d) b.

, dT dt 0 40.3m s 0.559T 0 m s

At steady-state

0559T . 40.3

T 24 C m s 0.333 kg hr c.

Separate variables and integrate the balance equation:

dT

Tf

t



40.3m s 0.559T

10

0

dT

23

dt

13.4 0.559T

10

m s 0.333

E

T f 23C

b b gPQ

L gO MN

1 3.4059 23 ln 4.8hr . . 0559 134 0.559 10

t

bt 0 to t 20 ming 250 kg 4.00 kJ b60 20g C .

Integral energy balance

11.26 a.

QU MCvT

 40010

kg C

4.00 104 kJ 1 min

Required power input: Q

20 min

b.

t

dT dt

Differential energy balance: MCv

1 kW 1 kJ s

60 s

 Q

.  333 kW

dT dt

M 250 kg Cv 4.00 kJ kgC

T

Integrate:

t 0, T 20 C

t

0

0

b g

Evaluate the integral by Simpson's Rule (Appendix A.3)

30  Qdt 3 33 4 33 35 39 44 50 58 66 75 85 95

600 s 0

b

g

2 34 37 41 47 54 62 70 80 90 100 34830 kJ T

b600 sg 20

o

e 0.001 C / kJjb34830 kJg 54.8 C

C+

o

bg

10 kW Past60,Q10 t 600 s t 6 60 s

c.

T 20 0.001

z

t 6 Qdt20 0001 0  .    dt 0 t

LM MM Qdt MMN  PP O

zz

F I . GH 6 6002 JK tbsg 6 34830

 T 548

. 12000 T 248

0001 t 0. 0 60 0

2

PP P Q

t 6

b

2

g

T 85C t 850 s 14 min, 10 s explosion at 10:1410 s

11-20

bg

t  0.001Q

dT 0001 Q dT T 20o C Qdt t .

20oC

4

kJ

11.27 a. Total Mass Balance: Accumulation=Input– Output

E

dM tot dt

d(V)  m i m o dt

 =constant

 800 .  4.00

dV  4.00 L / s dt t 0 , V0 400 L

KCl Balance:

d(CV ) dM KCl  m i,KCl m o,KCl dt dt

Accumulation=Input-Output

 100 8.00 . . 400C dC

V  C  8 4C

b.

dC dt

dV dt

dV dt 4

8 8C  dt V

t 0 , C 0 0 g / L

(i)The plot of V vs. t begins at (t=0, V=400). The slope (=dV/dt) is 4 (a positive constant). V increases linearly with increasing t until V reaches 2000. Then the tank begins to overflow and V stays constant at 2000.

V

2000

400

0 t

(ii) The plot of C vs. t begins at (t=0, C=0). When t=0, the slope (=dC/dt) is (8-0)/400=0.02. As t increases, C increases and V increases (or stays constant) dC/dt=(8-8C)/V becomes less positive, approaches zero as t. The curve is therefore concave down.

C

1

0 t

c.

dV dt

 4

V

t

400dV04

dt V 400 4t

11-21

11.27 (cont’d)

dC dt



C 0

dC

8 8C 

1 C

dC dt

V 400 4t

V

dt 0 50 + 0.5t t



1 C 50 05t . C

t

 ln(1 C) 0 2 ln(50 0.5t ) 0 50 0 .5t  ln(1 0.01t )2 50 1  (1 0.01t )2 C 1 (1 0.01t )2

 ln(1 - C)-1 2 ln 1



1- C

When the tank overflows, V 400 4t 2000 t 400 s

C = 1-

b1+ 0.01 400g

1 2

 0.96 g / L

11.28 a. Salt Balance on the 1st tank: Accumulation=-Output

E

d(CS1V1 ) dCS1 CS1v dt dt

v 0.08CS1 V1 CS1( 0 ) 1500 500 3 g / L CS1

Salt Balance on the 2nd tank: Accumulation=Input-Output

E

d(CS2V2 ) dCS2  CS1v CS 2v dt dt

v

 (CS1 CS 2 )

V2

 0.08(CS1 CS 2 )

CS 2( 0 ) 0 g / L Salt Balance on the 3rd tank: Accumulation=Input-Output

E

d(CS3V3 ) dCS3  CS 2v CS 3v dt dt

 (CS 2 CS 3 )

V3

v  0.04(CS 2 CS 3 )

CS3 ( 0 ) 0 g / L

b.

C S1, C S2, C S3

3

CS1

CS2 CS3

0 t

11-22

11.28 (cont’d) The plot of CS1 vs. t begins at (t=0, CS1 =3). When t=0, the slope (=dCS1 /dt) is0.08 30.24 . As t increases, CS1 decreases dCS1 /dt=-0.08CS1 becomes less negative, approaches zero as t. The curve is therefore concave up. The plot of CS2 vs. t begins at (t=0, CS2 =0). When t=0, the slope (=dCS2 /dt) is 0.08(3 0) 0.24 . As t increases, CS2 increases, CS1 decreases (CS2 < CS1 ) d CS2 /dt =0.08(CS1 -CS2 ) becomes less positive until dCS2 /dt changes to negative (CS2 > CS1 ). Then CS2 decreases with increasing t as well as CS1 . Finally dCS2 /dt approaches zero as t. Therefore, CS2 increases until it reaches a maximum value, then it decreases. The plot of CS3 vs. t begins at (t=0, CS3 =0). When t=0, the slope (=dCS3 /dt) is 0.04(0 0) 0 . As t increases, CS2 increases (CS3 < CS2 ) d CS3 /dt =0.04(CS2 -CS3 ) becomes positive CS2 increases with increasing t until dCS3 /dt changes to negative (CS3 > CS1 ). Finally dCS3 /dt approaches zero as t. Therefore, CS3 increases until it reaches a maximum value then it decreases. c. 3

CS1, CS2, CS3 (g/L)

2.5 2 CS1 1.5 CS2

1

CS3

0.5 0 0

20

40

60

80

100

120

140

160

t (s)

11.29 a. (i) Rate of generation of B in the 1st reaction: rB1 2r1 0.2CA :rB2 r2 0.2CB2

(ii) Rate of consumption of B in the 2nd reaction b. Mole Balance on A: Accumulation=-Consumption

E

d(C AV ) dt

01C . AV

dCA 01. CA dt t 0, C A0 100 . mol / L

Mole Balance on B: Accumulation= Generation-Consumption

E

d(CBV ) dt

 0.2CAV 0.2CB2V

dCB  0.2CA 0.2CB2 dt t 0, CB0 0 mol / L

11-23

11.29 (cont’d) c. 2

C A , C B , CC

CC

1 CB CA 0 t

The plot of CA vs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is01 101 . . As t increases, CA decreases dCA/dt=-0.1CA becomes less negative, approaches zero as t. CA0 as t. The curve is therefore concave up.

.

The plot of CB vs. t begins at (t=0, CB =0). When t=0, the slope (=dCB/dt) is 0.2(1 0) 0.2 . As t increases, CB increases, CA decreases ( C 2B < CA) d CB/dt =0.2(CA- C 2B ) becomes less positive until dCB/dt changes to negative ( C 2B > CA). Then CB decreases with increasing t as well as CA. Finally dCB/dt approaches zero as t. Therefore, CB increases first until it reaches a maximum value, then it decreases. CB0 as t. The plot of CC vs. t begins at (t=0, CC =0). When t=0, the slope (=dCC/dt) is 0.2(0) 0 . As t increases, CB increases dCc/dt =0.2 C 2B becomes positive also increases with increasing t  CC increases faster until CB decreases with increasing t dCc/dt =0.2 C 2B becomes less positive, approaches zero as t so CC increases more slowly. Finally CC2 as t. The curve is therefore S-shaped.

CA, CB , CC (mol/L)

d. 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0

CC

CB CA

0

10

20

30 t (s)

11-24

40

50

11.30 a.

When x =1, y =1 . y

ax x1, y1 x b

a

 a 1 b

1 b

b. Raoult’s Law:

xp *C5 H12 (46o C ) P

pC5 H12 yP xp *C5 H12 (46o C) y (6.85221

Antoine Equation:

p *C5 H12 (46o C ) 10

 y xp *C5 H12 (46o C ) P



0.7 1053 760

1064 .63 46 232. 00

)

 1053 mm Hg

 0.970

0.70a

.

TFrom part (a), a = 1+ b(2) |Tb 0.078 c. Mole Balance on Residual Liquid Accumulation=-Output

:

dN L  nV dt t 0, N L 100 mol Balance on Pentane: Accumu lation=-Output

d( N L x) dt

d Ndt

dt

x b

dN L / dtnV

L

d dx t a x

  V

      x t 0, x = 0.70

d. Energy Balance

: Consumption=Input H vap27.0 kJ/mol

From part (c),

b270 kJ / molg Qt

dt

27.0  100 27.0

expression 1 Substitute this into the equation for dx/dt from part (c):

11-25

R y ax x=0.70, y=0.970 S| x b

Ra 1078

0970

0.70 b

(1)



S

.

E

E  nV y x

 NL

nV

FHI

K

E

dx n ax N L x b

E  nVH vap Q dN L



nV

Q .

t 0, N L 100 mol

 nV

nV N L Qt

N L 100 nV t 100 Q 27.0 



11.30 (cont’d)

FHG

dx  Vn ax dt N L x b

I JK

 x 

. Q 270 Qt 100 27.0

Fax GH x b

JIK

x

x(0) = 0.70 e. 1 0.9 0.8 y (Q=1.5 kJ/s)

0.7

x, y

0.6 x (Q=1.5 kJ/s)

0.5 0.4 x (Q=3 kJ/s)

0.3

y (Q=3 kJ/s)

0.2 0.1 0 0

200

400

600

800

1000 1200 1400 1600 1800 t(s)

f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The higher the heating rate, the faster x and y decrease.

11-26