Solucionario Cap 11

8/11/2019 Solucionario Cap 11

1/30


2/30

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)

11-2


3/30

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


(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


4/30

M(kg)

NfN0 0.2013 lb - mole

b020130.0258glb - mole air0.0117 lb - mole air sRate of addition:n

b

. .

.


5/30

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


6/30

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 .

11-4


7/30

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

11-5

g


8/30

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.

11-6


9/30

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


(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.

11-7


10/30

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

11-8


11/30

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

11-9


12/30

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


13/30

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

11-10


14/30

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

11-11


15/30

; 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

11-12


16/30

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

11-13


17/30

.

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



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

11-14


18/30

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

11-15


19/30

(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).

11-16


20/30

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

11-17


21/30

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%

11-18


22/30

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.)

11-19


23/30

.

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

11-20


24/30

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

11-21


25/30

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

11-22


26/30

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

11-23


27/30

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)

11-24


28/30

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


29/30

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):

11-25


30/30

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.

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Solucionario Cap 11