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h
b450g4b25gb750gt=1.54 h, 19.54 h
b450
g4
b25
gb750 2693
gt719. h ,1081. h
R|750 + 450t - 25tb0t719. and 1081.t1954.g(tank is filling or draining)
||T0.
.
11.3 a. mwabtbt0,mw750g&bt5,mw1000gmwbkg hg75050tbgBalance on methanol: Accumulation = InputOutput
Mkg CH3OH in tank
mfmw1200 kg hb75050tgkg h-170dMdt
EdM
dt45050tbkg h g
t0,M750 kg
b.
M
dM
t
b45050tgdt750
E0
M750450t25t2
EM750450t25t2
Check the solution in two ways :
(1)t0,M750 kgsatisfies the initial condition;
(2)dM
dt45050t reproduces the mass balance.
c.dM
dt
0t450 509 hM750450(9)25(9)22775 kg (maximum)
M0750450t25t2
t450
2
2b25g
d.3.40 m3103liter 0.792 kg
1 m3 1 liter2693 kg (capacity of tank)
M2693750450t25t2
t450
2
2b25gExpressions for M(t) are:
2
M(t) =S2693 (719t10.81) (tank is overflowing)(19.54t2054) (tank is empty, draining
as fast as methanol is fed to it)
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11.3 (contd)
3000
2500
2000
1500
1000
500
0
0 5 10 15 20
t(h)
11.4 a. Air initially in tank:N010.0 ft3 492R 1
lb - mole532R 359 ft3bSTPg
0.0258 lb - mole
Air in tank after 15 s:
PfV NfRT Pf 0.0258 lb - mole 114.7 psia
P0V N0RT P0 14.7 psia
.
15 s
b. Balance on air in tank: Accumulation = input
dNdt
0.0117blb - moles sg;t0 ,N0.0258 lb - mole
c. Integrate balance:
N t
dNndtN0.02580.0117tlb - mole air g0.0258 0
Check the solution in two ways :
(1)t= 0,N= 0.0258 lb - molesatisfies the initial condition
(2)dN
dt0.0117 lb - moleair / sreproduces the mass balance
d.
O2
in
tan
k
0.2
1
b143
g030 lb - mole O2t120 sN0.0258b0.0117gb120g143 lb - moles air
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M(kg)
NfN0 0.2013 lb - mole
b020130.0258glb - mole air0.0117 lb - mole air sRate of addition:n
b
. .
.
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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).
Accumulation = inputoutputdV
dt
540 m
h
31h
60 min wm3min
t0,V3.00103m3t0 corresponds to 8:00 AMV
3.00103
t t
dV9.00wdtVm33.001039.00twdt tin minutes0 0
b.
2 4 0 2 4 2 4
0
2
48
8
m
3
V
3001
03
9.00
b240g2488
2672
m3
Letwitabulated value ofwatt10bi1g i1, 2,, 2510
. . .3
c.
d.
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 Simpsons rule introduced an error.
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, VDO 10 I = 2, 25
T = H * (I1)
V = V + 9.00 * H0.5 * H * (VW(I1) + VW(I))
WRITE (6, 2) T, V
10
1
2
CONTINUE
FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)')
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
10.0020.00
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L10wdt w1w254w i2w iM P i2, 4, i3, 5,3b g b g114984 12462 113.4
.
230.00
VOLUME (CUBIC M ETERS)
3000.
2974.
2944.
2683.
240.00 2674.
Vtrapezoid2674 m3;VSimpson2672 m3;26742672
2672100%0 .07%
Simpsons rule is more accurate .
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V
out0.200Vout20.0 L minVs100 L
out60
L
.
.
.
V
F I tH K1 20.00.200V
40.0
g
26.5 minb glnb1 2 0 0
.
11.6 a. outbL mingkVbgV300
b. Balance on water: Accumulation = inputoutput (L/min).
(Balance volume directly since density is constant)dV
dt20.00200V
t0 ,V300
c.dV
dt02000200VsVs100 L
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is
2000.200(300) 40.0.As t increases, V decreases.dV/dt20.00.200Vbecomes less negative, approaches zero ast . The curve is therefore concave up.
t
d.
V t
dt30020.00.200V 0
ln0.200
0.50.005Vexpb0.200tgV100.0200.0 expb0.200tV101b100g101 Lb1% from steady stateg
101100200 exp0.200tt 0.200
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l n D2
D1 l n
b7552385
gt2t1 61
b 3000e0.230tkg week
.
30000.230t
b g45, 440 mol
b g28,920 mol min
x
n28,920
15 mol SO 2.
. . .
11.7 a.
b.
A plot ofD(log scale) vs.t(rectangular scale) yields a straight line through the points ( t1week,D2385 kg week) and (t6weeks,D755 kg week).
lnDbtlnaDaebt
b 0230
lnalnD1bt1lnb2385gb0.230gb1g8.007ae8.0073000E
D3000e0.230t
Inventory balance: Accumulation =output
g-170dI
dtt0,I18,000 kg
c.
I t
dI 3000e0.230tdtI18,00018, 000 0
t I4957 kg
0.230e
t
0I495713,043e0.230t
11.8 a. Total moles in room:N1100 m3 273 K 103mol
295 K 22.4 m3STP
Molar throughput rate:n700 m3
min
273 K 103mol
295 K 22.4 m3STP
SO2balance(t0is the instant after theSO2is released into the room):
Nbmolgbmol SO2molgmol SO2in roomAccumulation =output.
d
dtbNxg nx dx
N45,440dt 0.6364x
t0,x.
45,440 mol330105mol SO2mol
b. The plot of x vs. t begins at (t=0, x=3.3010-5). When t=0, the slope (dx/dt) is
06364330105 210105.As t increases, x decreases.dx dt 0.6364xbecomes less negative, approaches zero ast . The curveis therefore concave up.
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x
..
. .
.
. .
.
11.8 (contd)
0
t
c. Separate variables and integrate the balance equation:x
3.30105
dx
x
t
06364dtln0
x
330105
06364tx330105e0.6364t
Check the solution in two ways :
(1) t = 0, x = 3.3010-5mol SO2/ molsatisfies the initial condition;
(2)dx
dt 0.6364330105e0.6364t 0.6364xreproduces the mass balance.
d. CS O245,440 moles
1100 m3
xmol SO2
mol
1 m3
103L4131102x13632106e0.6364tmol SO2/ L
i) t2 minCSO2 382 107mol SO2
liter
ii) x10 6tlne10 63.30 105
0.6364j55. min
e. The room air composition may not be uniform, so the actual concentration of the SO 2in 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.
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mol PpMolar flow rate of entering and leaving gas:n( )
FG IJRate at which CO leaves:n( H K
F I pxd(Nx) Pp dx P
H K
dx p
p r
e jln 10035106283 hrstr
b g xmbkg mingE
dx m
11.9 a. Balance on CO : Accumulation=-output
N(mol)x( mol CO / mol) = total moles of CO in the laboratory
h RT
kmol kmol CO Pp)x = x
h kmol RT
CO balance: Accumulation = -output
x dt RT dt NRT
EPVNRT x
dt V
kmol COt0,x0.01
kmol
b.
x
0.01
dx
x
t
V0dttr
V
plnb100xg
c.
d.
11.10 a.
V350 m3350
.700
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.
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.
Total mass balance: Accumulation = inputoutput
dM
dtmmbkg ming0Mis a constant200 kg
b. Sodium nitrate balance: Accumulation = - output
x= mass fraction ofNaNO3
d xM
dt
x xdt M 200
t0,x90 2000.45
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x
x
FGH IJ
dx m m t m
.
11.10 (contd)
c.0.45
0
m50 kg / min
m100 kg / min
m200 kg / min
t(min)
dx
dt
m
200x0 , x decreases when t increases
dx
dt becomes less negative until x reaches 0;
Each curve is concave up and approaches x = 0 as t ;
mincreasesdx
dtbecomes more negativex decreases faster.
d.
x
0.45
dx
x
t
0
m
Mdtln
x
0.45
m
200tx0.45 exp
mt
200 KCheck the solution :
(1) t = 0, x = 0.45satisfies the initial condition;
(2) 0.45 exp( ) xsatisfies the mass balance.
dt 200 200 200
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
m50 kg / min
m100 kg / min
m200 kg / min
0 5 10 15 20 25
t(min)
e. m100 kg mint 2 lndxf0.45i90%xf0.045t4.6 min
99%xf0.0045t9.2 min
99.9%xf0.00045t138 min
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11.11 a. Mass of tracer in tank:Vem3jekg m3 jTracer balance : Accumulation =output. If perfectly mixed,CoutCtankC
d VC
dt
V is constant dt V
t0,C
C
m0
V
b.
C
m0V
dC
C
t
0Vdtln
F Cm0V
I
t
VC
m0
Vexp
V
c.
11.12 a.
PlotC(log scale) vst(rect. scale) on semilog paper: Data lie on straight line (verifying assumption
of perfect mixing) throughet1,C0223103j&et2,C0.050103j.
V 21
min
In tent at any time, P=14.7 psia, V=40.0 ft3, T=68F=528R
NPV
RTm(liquid)
14.7 psia
ft3psia10.73
lb - moleoR
40.0 ft3
528oR01038 lb - mole
b. Molar throughout rate:
60 ft3492R 16.0 psia 1 lb - mole
2
Balance on O2: Accumulation = inputoutput
d Nx
dt dt
c.
x
0.210.35x
t
0. b
0
.
3
5
0
21gb0.35xg
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C
b g Cbkg mingdC
z z FG IJtH KGH JK
l n b0.0500. 223g 1.495 min1
EVe30 mj e 1.495 minj201 m 3 1 3
.
.
ninn
outn
Moles of O 2in tank=N(lb - mole) F lb - mole OIHlb - moleK
.
b g 0.35nxn0.1038dx0.1695b0.35xgdxdt1.63b0.35xgt0,x0.21
z x z163dt ln . .e1.63tx0.35014e1.63t
LM FG IJOPx0.27t ln 0.343 min (or 20.6 s)N H KQ
.
.
163t
035x.
014
1 0.350.27
1.63 0.350.21
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C
FG IJ b gH KLs kC
z z F CI ktt lnbCCgC kdtln 0
t1 22.6 hrk
C0.01C0 b g17.2 hr
.
b g kC Vdt
z zdCA kdtlnCA0CA
FGCIJ ktCHCK
b g b g. .
l n b0. 018500262120.0213
.
b g kC VbVconstant; cancelsgt
z z dCA kdt CA 0CA
LM OP
11.13 a. Mass of is otope at any timeVblitersgbmg isotope literBalance on isotope: Accumulation =consumption
g
d
dtbVCg kC mg VL Cancel V
dC
dt
t0,CC0
Separate variables and integrate
C t
C0C 0 HC0K k
C0.5C0t1 2 lnb0.5g
kt 12
ln 2
k
b.ln 2
2.6 hr
t=-ln(C/C0)/k
0267 hr1
ln 0.01
t 0.267
11.14Aproducts
a. Mole balance onA: Accumulation =consumption
d CAVA
bVconstant; cancelsgt0,CACA0
CA t
0
A
A0A CA0expbktg
b. PlotCA(log scale) vs.t(rect. scale) on semilog paper. The data fall on a straight line (verifies
assumption of first-order) throught213,CA00262&t120.0,CA0.0185.
lnCA ktlnCA0
k.
g 353.10 3 min1k35103min1
11.152A2BC
a. Mole balance onA: Accumulation =consumption
d CAV 2A
t0,CACA0
CA t
0
1
CA
1
CA0 ktCA N
1
CA0
ktQ
1
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; butCA0A0
.
. . .
.
1432 satm1060209
1 01351 0683
b1015 Kgb008206 Latm molKg0582 L molsk143.2 satm
FG IJH K GH JK
s K
.
FGH IJ
b0.08206 Latm molKgb980 Kg104510. .
L OCA0.10CA0t LM OP
11.15 (contd)
b. CA0.5CA0 1
0.5CA0
1
CA0 kt1 2t1 2
1
kCA0
n
V
P0
RTt1 2
RT
kP0
nA0.5nA0
nBb0.5nA0molAreact.gb2 molB2 molAreact.g05nA0nCb0.5nA0molAreact.gb1 molC2 molAreact.g0.25nA0total moles125nA0P1 2125
nA0RT
V125P0
c. Plott1 2vs.1P0on rectangular paper. Data fall on straight line (verifying 2 ndorder
decomposition) throughdt1 21060, 1P01 0135i&dt1 2209, 1P01 0.683i
Slope:RT
k . .
.
.
.
d. t1 2RT
k0P0exp
E
RTln
Ft1 2P0RT
Iln
1
k0
E1
R T
Plott1 2P0RT(log scale) vs.1T(rect. scale) on semilog paper.
t1 2bg,P01 atm,R0.08206 Latm / (molK), TbgData fall on straight line throughdt1 2P0RT74.0, 1T1 900i&dt
1 2P0 RT0.6383, 1T1 1050i
E
R
lnb0.6383 74 .0g1 10501 900
29,940 KR=8.314 J/ (mol K)
E2 .49105J mol
ln
1
k0
lnb06383g 29 ,9401050
28.96k03.791012
L (mols)
e. T980 Kkk0expE
RT K0.204 L (mols)
CA0070b120 atmg
.2 mol L
90% conversion
1 1 1
kNCA CA0Q4222 s70.4 min
1 1
0.204N1.04510 3 1
1.045102 Q
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z dCA z dt b g b g2CACA0 tt2CA0CA
b g
FG IJ FG IJH K H K
k1 k
.
.
. . .
b g0.12035 mol gas
bCg0.60b012035 molg3.00 L0.02407 mol L COU|dCi0.40b012035 molg3.00 L0.01605 mol L ClV|W initial concentrations
.
Cl2i.
btg0.02407Cbtg U|btg0.01605Cb tg V|W Since 1 mol COCl formed requires 1 mol of each reactantC
d i 875CCOCCl 2
2.92d0.02407C pid0. 01605C p
d194124.3Ci
.
.
z d194124.3Ci
11.16AB
a. Mole balance onA: Accumulation =consumption(Vconstant)
dCA
dt
k1CA
1k2CA
t0,CACA0CA1k2CA
CA0 k1CA
t
0
1
k1
l nCA
CA0
k k
k1 k1
1
k1
l nCA
CA0
b. PlottbCACA0gvs.lnbCA/CA0 g b C A 0 CAgon rectangular paper:y
t 1 lnCACA0 k2
bCA0CAg ;CA0CA1slope intercept
y1 x1 y2 x2
Data fall on straight line through116.28,02111&130.01,0.2496
1
k1
13001116.28
0.2496b0.2111g 356.62k12 .80103L (mols)
k2
k1130.01356.62b02496g4100k20115 L mol
11.17COCl2COCl2
a.
3.00 L 273 K 1 mol
303.8 K 22.4 L STP
COi
2
CCO p
2
Cl2 p
b. Mole balance on Phosgene: Accumulation = generation
d VCp
dt
.
d1586CCl234.3Cp i2
V=3.00 LdCp
dt
. p
t0,Cp0
2
i
c. Cl2limiting; 75% conversionCp075b0.01605g0.01204 mol L
t1
2.92
0.01204
0
2
. p
d0.02407Cpid0.01605CpidC p
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.
FjGmolsIJdCdtkSVeCHsKt0,C0
*
z z e j
dCA kS kS dt lnC*ACA
CACA CA 0
kSexpb g*
11.17 (contd)
d.
20
30
40
10
1
REAL F(51), SUM1, SUM2, SIMP
INTEGER I, J, NPD(3), N, NM1, NM2
DATA NPD/5, 21, 51/
FN(C) = (1.44124.3 * C) ** 2/(0.02407C)/(0.01605C)
DO 10 I = 1, 3
N = NPD(I)NM1 = N1
NM2 = N2
DO 20 J = 1, N
C = 0.01204 * FLOAT(J1)/FLOAT(NM1)
F(J) = FN(C)
CONTINUE
SUM1 = 0.
DO 30 J = 2, NM1, 2
SUM = SUM1 + F(S)
CONTINUE
SUM2 = 0.
DO 40 J = 3, NM2, 2
SUM2 = SUM2 + F(J)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
CONTINUE
FORMAT (I4, 'POINTS', 2X, F7.1, 'MINUTES')
END
RESULTS
5 POINTS91.0 MINUTES
21 POINTS90.4 MINUTES
51 POINTS90.4 MINUTES
t904 minutes
11.18 a. Moles of CO2in liquid phase at any timeVecm3jCAemols cm3Balance on CO2in liquid phase:Accumulation = input
j
d
dtbVCAg kSeCACA
V
A
A
*A CA j
Separate variables and integrate. SincepAyAPis constant,C*ApAHis also a constant.
CA t CA
t0 0V V
lnC*ACA
C*A
Vt1
CA
C*AekSt VCACAe1ekSt V j
1CAC*A
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lnL 1 OP
a f a f d9230 atm cm moli 0.65 10 m ol c mC*AyAP H0.30 20 atm.
F0.6210I9800 s2.7 hre5000 cmj 3b002. cm sge785. cmj H 0.6510JK
v
E
z z vdtVvtdVz zdNA
CA0vkNA
F I1 CA0vkNA
H KCA0v
11.18 (contd)
b. tV
kS NMCA
C*A QPV5 L5000 cm3,k0.020 cm s ,S785 cm2,CA0.62103mol / cm3
3 3 3
t
3
2lnG1 3
(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, Smay in fact be much greater than
this value, leading to a significantly lowertthan that to be calculated)
11.19AB
a. Total Mass Balance: AccumulationinputdM d(V)dt dt
dV
dtv
t0,V0
A Balance : Accumulationinputconsumption
dNA
dtCA0v(kCA)VCA=NA/V
dNA
dtCAovkNA
t0 ,NA0
b. Steady State:dNA
dt0NA
CA0v
k
c.
V t
0 0
N A t
0 0
ln k
dt
tCA0vkNA
CA0v ekt
NA CA0v
k1expbktg t NA CA0v
k
CANA
V
CA0[1exp(kt)]kt
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(1) In our calculation, V =vtt , V .
limCAlim
QW
. .
z zo
a f
z z025 C
s.
11.19 (contd)
When the feed rate of A equals the rate at which A reacts, NAreaches a steady value.
NAwould never reach the steady value in a real reactor. The reasons are:
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.t t
CA0[1exp(kt)]
ktlim
t
CA0
kt0
From part c,t , NAa finite number, V CA NAV
0
11.20 a. MCvdT
dt
M(3.00 L)(100 kg / L) = 300 kg
CvCp(0.0754 kJ / moloC)(1 mol / 0.018 kg) = 4.184 kJ / kgoC
W0dT
dt0.0797Q(kJ / s)
t= 0,T= 18oC
b.
100oC 240 s
dT18 C 0
0.0797QdtQ10018
2400.07974.287
kJ
s4.29 kW
c. 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).
11.21 a. Energy balance:MCvdT
dtQW
M20.0 kg
CvCp( 0.0754 kJ / moloC)(1 mol / 0.0180 kg) = 4.184 kJ / (kg oC)
Q0.97 (2.50)2.425 kJ s
W0
dT
dt0.0290bC sg,t0 ,T25C
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
b.
T
t
o
dT 00290dtT25C0.0290tbg
c. T100Ctb10025g0.02902585 s43.1 minNo, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal
boiling point).
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Tb(oC)
B
eje 7.7gcm j 462gM60 cm
U0.050 J (mincmC)
dTb
T25
FGT25IJ 002635tH9525Kln .
.
11.22 a.Energy balance on the bar
MCvdTb
dtQW UAbTbTw g
Table B.1
3 3
Cv0.46 kJ (kgC),Tw25C2
A2a2fa3fa2fa10fa3fa10fcm2112 cm2
dTb
dt 0.02635bTb25gbC min g
t0,Tb95C
b.dTb
dt0 0.02635dTbf25iTbf25C
95
85
75
65
55
45
35
25
15
5
0
t
c.
Tb t
0.02635dt95b 0
b
Tbbtg2570 expb0.02635tgCheck the solution in three ways :
(1) t = 0, Tb257095oCsatisfies the initial condition;
(2)dTb
dt 700.02635e0.02635t 002635(Tb25)reproduces the mass balance;
(3) t , Tb25oCconfirms the steady state condition.
Tb30Ct100 min
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T(oC)
.
.
FG IJ. . .
. . .
.
z z
0.39470.096x( 0.015795721104x)T
L0.39470.096xe0.01579572110xj55OPlnM4
1
MM0.39470.096xe0.01579572110xj25PP400.015795.721104
N Q4
. .
..
11.23
12.0 kg/min
25oC
12.0 kg/min
T (oC)
Q (kJ/min) = UA (Tsteam-T)
a. Energy Balance :MCvdT
dtmCpb25TgUAbTsteamT g
M760 kg
m12.0 kg min
dT/dt1500.0224T(oC min),t0 ,T25oC
CvCp2.30 kJ (minC)
UA11.5 kJ (minC)
Tsteamasat'd; 7.5barsf167.8C
b. Steady State:
dT
dt 01500.0224TsTs67C
67
25
0
t
c.
T
f
25
dT
15000224T
t
dtt 0
1
00224ln H
1500.0224T
0.94 KT150094 exp(0.0224t)
0.0224
t40 min.T498C
d. Uchanged. Letx(UA)new. The differential equation becomes:
dT
dt
55
25
0.39470.096x( 0.015795.721x)T
dT
.
x.
x14.27 kJ / (minoC)
40
0
dt
.
U
Uinitial
(UA)
(UA)initial
1427115
115100%241%
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QW
b gU|VT200.0649tbsg0.0649C s|W T40Ct308 s51. min
b gTime to reachTbneglect evaporation :t 926 sb g
.
.
.
.
b gQW
b g b
11.24 a.
b.
c.
dTEnergy balance:MCv
dt
W0,Cv1.77 J gC
M350 g,Q40.2W40.2 J s
dT
dtt0,T20C
The benzene temperature will continue to rise until it reachesTb801C; thereafter the heatinput will serve to vaporize benzene isothermally.
801200.0649
Time remaining: 40 minutes60 s min926 s1474 s
Evaporation:Hvb30.765 kJ molgb1 mol 78.11 ggb1000 J kJg393 J gEvaporation rateb40.2 J sgb393 J gg0102 g sBenzene remaining350 gb0102 g sgb1474 sg200 g
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 partners homework. Fire hazard.
11.25 a.Moles of air in room:n
60 m3 273 K 1 kg - mole
283 K 22.4 m3STP 2.58 kg - moles
Energy balance on room air:nCvdT
dt
QmsHvH2O, 3bars, sat'd30.0TT0
W0
g
nCvdT
dtmsHv30.0bTT0 g
N2.58 kg - moles
Cv20.8 kJ (kg- moleC)
Hv2163 kJ kgbfrom Table B.6T00C
g
dT
dt40.3ms0.559TbC hr g
t0,T10C
(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 assumingnis constant.)
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.
t
L gO 4. 8 h r 1 13. 4 0559 b23b gPQMN. .
250 kg 4.00 kJb6020gC.
.
t
dT0001QdTT20oCQdt
bg
30
3
b g
Past 600s, Q 100 b g t600 st6
z z zT200.001Qdt200001MMQdt tP6P
MM PPN Q
GH66002JKtbsg0001Ft I.
. .
M P
11.25 (contd)
b.
c.
At steady-state ,dT dt040.3ms0.559T0ms
T24Cms0.333 kg hr
Separate variables and integrate the balance equation:
0559T
40.3
Tf
10
dT40.3ms0.559T
t
0dt
ms0.333
Tf23C
23 dT1013.40.559T
Et ln
0559 1340.559 10
11.26 a.
Q UMCvT
Integral energy balancebt0 tot20 ming
kgC40010
4 kJ
Required power input:Q 4.00104kJ 1 min
20 min 60 s
1 kW
1 kJ s333 kW
b. Differential energy balance:MCvdT
dtQ
M250 kg
dT
dt0.001Qbg
Cv4.00 kJ kgC t0,T20C
Integrate:
T t
t .
20oC 0 0
Evaluate the integral by Simpson's Rule (Appendix A.3)600 s
Qdt 334 333539445058667585950
2 34374147546270809010034830 kJ
Tb600 sg20oC +e0.001oC / kJjb34830 kJg54.8C
c.
T548 12000T248
t600t
00600
. dt
6
10 kW
60 s
2 2
LO
34830
g
T85Ct850 s14 min, 10 sexplosion at 10:1410 s
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V
C
4.00 L / s.
dMKClmi,KClmo,KClAccumulation=Input-Output
dC dV
. .
dV4dtV4004t
11.27 a.Total Mass Balance:
Accumulation=InputOutput
dMtot
dt
Emimo
d(V)
dt8004.00
=constantdV
dt
t0 , V0400 L
KCl Balance:
V C 8 4 C
dt dt
dt dt
d(CV)
dV dt4
1008.00400C
dC 88C
dt V
t0 , C00 g / L
b. (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.
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.
1
0
t
c.dV
dt4
V t
400 0
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CS1,CS2,CS3
.
ln(1C)02 ln(500.5t)0
ln(1 - C)-12 ln
ln(1
0.01t)2
E CS1v
ECS1vCS2v
ECS2vCS3v
11.27 (contd)
dC
dt
88C
V V4004tdC
dt
1C
5005tCdC
01C
t dt
050 + 0.5t
C t
500 .5t
50
1
1- C(10.01t)2C1
1
(10.01t)2
When the tank overflows,V4004t2000t400 s
C = 1-
b1+ 0.01400g1
20.96 g / L
11.28 a.Salt Balance on the 1sttank:
Accumulation=-Output
d(CS1V1) dCS1
dt dt CS1
v
V1 0.08CS1
CS1( 0 )1500 5003 g / L
Salt Balance on the 2nd tank:
Accumulation=Input-Output
d(CS2V2) dCS2
dt dt(CS1CS2)
v
V20.08(CS1CS2)
CS2( 0 )0 g / L
Salt Balance on the 3rd tank:
Accumulation=Input-Output
d(CS3V3) dCS3
dt dt(CS2CS3)
v
V30.04(CS2CS3)
CS3( 0 )0 g / L
b.
3
CS1
CS2
CS3
0
t
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CS1,CS2,CS3(g/L)
E.
E
.
11.28 (contd)
The plot of CS1vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is0.083 0.24.
As t increases, CS1decreasesdCS1/dt=-0.08CS1becomes less negative, approaches zero ast . The curve is therefore concave up.
The plot of CS2vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 0.08(30)0.24 .
As t increases, CS2increases, CS1decreases (CS2< CS1)d CS2/dt =0.08(CS1-CS2) becomes lesspositive until dCS2/dt changes to negative (CS2> CS1). Then CS2decreases with increasing t as well
as CS1. Finally dCS2/dt approaches zero as t. Therefore, CS2increases until it reaches amaximum value, then it decreases.
The plot of CS3vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 0.04(00)0 .
As t increases, CS2increases (CS3< CS2)d CS3/dt =0.04(CS2-CS3) becomes positiveCS2increases with increasing t until dCS3/dt changes to negative (CS3> CS1). Finally dCS3/dt
approaches zero as t. Therefore, CS3increases until it reaches a maximum value then itdecreases.
c.
3
2.5
2
CS1
1.5
1
0.5
0
CS2
CS3
0 20 40 60 80 100 120 140 160
t (s)
11.29 a.(i) Rate of generation of B in the 1streaction:rB12r10.2CA
(ii) Rate of consumption of B in the 2ndreaction :rB2r20.2CB2
b.Mole Balance on A:
Accumulation=-Consumption
d(CAV)
dt 01CAV
dCA
dt 01.CA
t0,CA0100 mol / L
Mole Balance on B:
Accumulation= Generation-Consumption
d(CBV)
dt0.2CAV0.2CB2V
dCB
dt0.2CA0.2CB2
t0,CB00 mol / L
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CA,CB,CC
CA,C
B
,CC
(mol/L)
. .
11.29 (contd)
c.
2
CC
1
CB
CA
0
t
The plot of CAvs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is011 01.As t increases, CAdecreasesdCA/dt=-0.1CAbecomes less negative, approaches zero as
t. CA0 as t. The curve is therefore concave up.
The plot of CBvs. t begins at (t=0, CB=0). When t=0, the slope (=dCB/dt) is 0.2(10)0.2 .
As t increases, CBincreases, CAdecreases (C2B< CA)d CB/dt =0.2(CA-C2B) becomes less positive
until dCB/dt changes to negative (C2B> CA). Then CBdecreases with increasing t as well as CA.
Finally dCB/dt approaches zero as t. Therefore, CBincreases first until it reaches a maximum
value, then it decreases. CB0 as t.
The plot of CCvs. t begins at (t=0, CC=0). When t=0, the slope (=dCC/dt) is 0.2(0)0 . As t
increases, CBincreasesdCc/dt =0.2C2Bbecomes positive also increases with increasing t
CCincreases faster until CBdecreases with increasing tdCc/dt =0.2C2Bbecomes less positive,
approaches zero as tso CCincreases more slowly. Finally CC2 as t. The curve is thereforeS-shaped.
d.
2.2
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
CB
CA
CC
0 10 20 30 40 50
t (s)
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11.30 a.When x =1, y =1.
yaxx1,y1
xb
a
1ba1b
b.Raoults Law: pC5H12yPxp*C5H12(46oC)yxp*C5H12(46oC )
P
Antoine Equation: p*C5H12(46oC ) 10(6.85221
1064 .63
46232. 00)
1053 mm Hg
y xp*C5H12(46oC )
P 0.71053
7600.970
0.70a.
TFrom part (a), a = 1+ b(2)|Tb0.078
c.Mole Balance on Residual Liquid :
Accumulation=-Output
dNL
dt nV
t0,NL100 mol
Balance on Pentane:
Accumu lation=-Output
d(NLx)
dt
dt
dN
Ldxax
V
x
dt dt xb
dNL/dt nV
t0, x = 0.70
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1
Ryaxx=0.70, y=0.970 Ra1078 (1)0970 .S| xb S0.70b
E
E
nVyx nVNL
EF IH K
dx n ax
Lxb
EQ
nVHvapQ nV
dNLt0,NL100 mol
nV L100nVt100
Q27.0nV
NL Qt
.
d.Energy Balance : Consumption=Input
H vap27.0 kJ/mol
b270 kJ / molgQt
From part (c),dt 27.0
100 -27.0
Substitute this expression into the equation for dx/dt from part (c):
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x,y
FGHn ax
JK .
GH JK
11.30 (contd)
dx
dtV
NLxb
xI Q270Qt
100 -27.0
Faxxb
xI
x(0) = 0.70
e.
1
0.9
0.8
0.7
0.6
0.5x (Q=1.5 kJ/s)
y (Q=1.5 kJ/s)
0.4
0.3
0.2
0.1
0
x (Q=3 kJ/s)y (Q=3 kJ/s)
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 fasterxandydecrease.