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 m1 M p / M b.
: Accumulation = input – output
Balance on mass
E
dM dt
m 0 m1
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 m1
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
b25g 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 3.00103 m3 t 0 corresponds to 8:00 AM V
t
3.0010 3
0
t
dV9.00 dt V m 3.0010 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 3001 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 i2, 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 expb0.200t V 101b. 100g 101 L b1% from steady stateg lnb120 g 26.5 min 101 100 200 exp b 0.200tg 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 b0230 6 1 t 2 t1
.
b2385g b0.230gb1g 8.007 a e
ln a ln D1 bt1 ln
E
8.007
3000
D 3000e0.230t
b.
Inventory balance: Accumulation = –output
g
b
3000 e0.230t kg week
-170dI dt t 0, I 18,000 kg I
30000.230t
t t
dI3000e0.230 t dt I 18,000 18, 000
c.
11.8
a.
0
e
0.230
0
I 4957 13,043e0.230 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 N45,440 dt
0.6364x 15 mol SO 2
t 0, x
. b.
.
. 45,440 mol
. 330105 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 105210 105. As t increases, x decreases. dx dt0.6364x 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
3.30105
t
06364dt ln . 0
Check the solution in two ways
x 33010 .
06364t x . . 330105 e0.6364t
5
:
(1) t = 0, x = 3.3010-5 mol SO2 / mol satisfies the initial condition; dx 0.6364 33010 5 e 0.6364 t0.6364x reproduces the mass balance. (2) . dt d.
e.
CS O2
45,440 moles 1100 m 3
i)
t 2 min CSO2
ii)
x 106 t
x mol SO2 mol
1m3 103 L
382 . 107
mol SO2 liter
e
ln 106 3.30 105 0.6364
. 413110 2 .x 13632106 e0.6364t 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
dxx p V
t 0, x 0.01
b.
FHGK
b100 xg
ln
c.
V 350 m3 350 ln 100 35 106 283 . 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 xmbkg 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 0.05010 j . 3
3
2 1
min
11.12 a.
In tent at any time, P=14.7 psia, V=40.0 ft 3, T=68F=528 R
N
b.
PV RT
m(liquid)
14.7 psia ft 3 psia 10.73 lb - moleo 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 min1
E V e30 m j e1.495 min j 201 m
3 1 3
.
nin nout 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
.
.
e1.63t x 0.35 014e1.63t
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 HLKIJs 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 gkC 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 IJkt C A
A
C A0 exp
A0
bktg
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 Akt ln C A0 k
b
ln 0.185026 120.0 213.
g353. 10
3
min1 k 35 . 103 min1
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 LK 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 L070atmbmol Kgb 120 atm g 980 Kg 104510 . 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 1.045 102
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 A0t 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 .
b0.2111g 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
2.9047Cp165p
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 A0
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 0.62 103 mol / cm3 V 5 L 5000 cm3 , k 0.020 cm s , S 785 cm
a fa f d i e5000 cm j F10.62 10 I 9800 s 2.7 hr t b002. cm sge785. cm jlnHG0.65 10 JK 33
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 expbktg 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 / molo C)(1 mol / 0.018 kg) = 4.184 kJ / kgo 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 / (kgo 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 WUA Tb Tw
B
Table B.1
ej
3
3
M 60 7.gcm462g 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
gbC min g
t 0, Tb 95 C
d i
dTb dt
b.
00.02635 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 95T25K25IJ002635t b
btg 25 70 expb0.02635tg
Tb
Check the solution in three ways
:
(1) t = 0, Tb 25 70 95o C satisfies the initial condition; (2)
dTb dt
70 0.02635e0.02635 t002635(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
mCp 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 (minC)
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 104 x)T 0 dT dt 4 . 0.3947 0.096 x 0.01579 5721 10 x 55 1 ln 0 4 0.01579 5.721 104 . 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
. 1427115 115
. 100% 241%
11-18
e
j OP j PP 40Q
11.24 a.
Energy balance: MCv
dT QW 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 20C 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 msHv H 2O, 3bars, sat'd 30.0 T T0 W 0
Energy balance on room air: nCv
nCv
b
dT dt
g
b
m sHv 30.0 T T0
b
g
g
N 2.58 kg - moles Cv 20.8 kJ (kg- mole C) Hv 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 23C
b b gPQ
L gO MN
1 3.4059 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.
QU MCvT
40010
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 kgC
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,Q10 t 600 s t 6 60 s
c.
T 20 0.001
z
t 6 Qdt20 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 85C t 850 s 14 min, 10 s explosion at 10:1410 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
400dV04
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) is0.08 30.24 . 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) is01 101 . . As t increases, CA decreases dCA/dt=-0.1CA becomes less negative, approaches zero as t. CA0 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. CB0 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 CC2 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 x1, y1 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 nV 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 / dtnV
L
d dx t a x
V
x t 0, x = 0.70
d. Energy Balance
: Consumption=Input H vap27.0 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 nV y x
NL
nV
FHI
K
E
dx n ax N L x b
E nVH vap Q dN L
nV
Q .
t 0, N L 100 mol
nV
nV N L Qt
N L 100 nV 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