CHE654 2012 Homework5 Solutions

8/10/2019 CHE654 2012 Homework5 Solutions

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2

53. Boiling and Draining of Ethanol in a Spherical Vessel Consider the boiling of pure ethanol in a jacketed spherical vessel with a radius R = 1

meter, as shown in the diagram. The liquid is drained at the bottom of the vessel while

some of it is boiled and escapes as vapor through the top of the vessel.

Steam TS

1

z

CV

The following data are available:

MW = 46.07 λ(TB) = 3.858x107 J/kmol

ρ = 16.575 kmol/m3 CP = 1.4682x10

5 J/kmol-°C

log10 Pvap

= 8.04494 − __1554.30__ T in °C and Pvap

in mmHg

T + 222.65

CV = 4.5 m

3

-atm

-1/2

/hr = valve constantTS = 100 °C U = 2.0x106 J/hr-m

2-°C

P0 = 1 atm z0 = z(t=0) = 1 m

(a) Model this operation and use MATLAB to determine the time it takes for the vessel

to completely empty, assuming that initially the liquid is at its boiling point. Note

that the heat transfer area AT is not constant.

(b)Repeat the calculations in Part (a), assuming that there is no draining of the liquid

at the bottom (i.e. the liquid leaves the vessel only through boiling). Determine the

solution analytically (an exact solution is possible in this case).

Useful conversion factors and formulae:

g = gravitational constant = 9.807 m/s2

1 atm = 1.01325x105 N/m

2 1 N = 1 kg-m/s

2

The volume of liquid V(z) in a spherical vessel as a function of its height z is given by

V(z) = π Rz2 − πz

3

3

while the surface area S(z) is given by

1 atm

v

P0



3

S(z) = 4π Rz − πz

Solution:

(a) First, calculate TB:

log10 (760) = 8.04494 – 1554.3/(TB + 222.65)

TB = 78.33 °C

Overall mass balance:

ρ (dV/dt) = −v − ρCv√ P – P0

dV/dt = −0.06033 v − 4.5√ P – 1

But P = P0 + φgz = 1 atm + (763.61 kg/m3) ( 1 atm ) (9.807 m/s

2)( 1 N )

1.01325x105 N/m

2 (1 kg-m/s

2)

P = 1 + 0.073908z

dV/dt = −0.06033 v − 1.22337√z

Energy balance:

v = q/ λ = (2.0x106)(100 – 78.33)AT = 1.12338(4πRz – πz

2)

3.858x107

(2πz – πz2) (dz/dt) = –0.27109πz + 0.06777πz

2– 1.22337√z

z(2 – z)(dz/dt) = –0.27109z + 0.06777z2 – 0.38941√z

dz = 0.27109 – 0.06777z + 0.38941z-1/2

dt (z – 2)

[ ]

∫ +−+

=+ 2

)ln(

b

bxaabxa

dxbxa

x



4

s.t. z(t = 0) = 1 m

Determine t at z = 0 using MATLAB t = 1.6966 hours or 101.8 minutes to empty

The MATLAB Script File:

%

% CHE654: Problem #53

%

% Boiling and Draining of Ethanol in a Spherical Vessel

%

clc

clear

format long

drain_t = 1;initZ = 1;

% Part (a)

[t, z] = ode45('spherea',drain_t,initZ);

while z(length(z),1) > 0

drain_t = drain_t + 0.01;

[t, z] = ode45('spherea',drain_t,initZ);

end

[t, z]

plot(t,z)

title('Boiling and Draining of Ethanol in a Spherical Tank')

xlabel('Time in Hours')

ylabel('Ethanol Height in Meters')

% Part (b)

drain_t = 1;

[t, z] = ode45('sphereb',drain_t,initZ);

while z(length(z),1) > 0

drain_t = drain_t + 0.01;

[t, z] = ode45('sphereb',drain_t,initZ);

end

[t, z]

The MATLAB M-File for the ODE:

function dzdt = ode(t,z)

dzdt = zeros(1,1);

dzdt(1) = (0.27109+0.06777*z(1)+0.38941*z^(-0.5))/(z(1)-2);



5

(b) Repeat the calculations in Part (a) analytically if there is no drainage.

dV/dt = −0.06033 v

(2πz – πz2) (dz/dt) = –0.27109πz + 0.06777πz

2

(2 – z)(dz/dt) = –0.27109 + 0.06777z

(z – 2)(dz/dt) = 0.06777(4 – z)

0 0 t

∫ z dz – ∫ 2 dz = ∫ 0.06777 dt

1 4 – z 1 4 – z 0

[4 – z – 4ln(4–z)] + 2ln(4–z) = 0.06777 t

0

4 – z – 2ln(4–z) = 0.06777 t

1



6

1.22741 – 0.802775 = 0.06777 t

t = 6.266 hours or 375.9 minutes

MATLAB gives an answer of about 6.27 hours.

The MATLAB M-File for the ODE:

function dzdt = ode(t,z)

dzdt = zeros(1,1);

dzdt(1) = 0.06777*(4-z(1))/(z(1)-2);



7

54. Boiling of Water in a Closed Cylindrical VesselA closed cylindrical vessel ( R = 0.5 m and H =5.0 m) contains 500 kg of water at its

boiling temperature of 100 °C (373.15 K) as shown in the figures below. Let’s assume

that initially at t = 0, the water (500 kg) is already at 100 °C and starts to boil. Initially,

there is also 5 kg of water vapor (or steam) in the gas phase of the vessel. We assume

the water vapor is an ideal gas and that the temperature in the gas and liquid phases arealways in thermal equilibrium at 100 °C.

The following data are also known about water:

MW (molecular weight) = 18.015 kg/kmol φ (mass density) = 1000 kg/m3

λ (heat of vaporization) = 2256.84 kJ/kg C P (liquid heat capacity) = 4.1813kJ/kg-K

R (universal gas constant) = 0.08206 atm-m3 /kmol-K

This problem is divided into two parts, namely Part 1 and Part 2.

Part 1: In this part, water is being boiled by a heating coil at T S = 200 °C with a

constant heat transfer area A = 2.0 m2 and an overall heat transfer coefficient U = 100

kJ/min-m2-°C as shown in the figure for Part 1.

(a) Derive an analytical expression of the liquid height z as a function of time.

(b) Derive an analytical expression of the water gas pressure PG as a function of time. Plot

your analytical solution in MATLAB and run the model until all the water has vaporized.

Hint: You may check the correctness of your analytical solution by checking PG at t = 0

and at t when all water has been vaporized.

Answer the following questions:

How long does it take for water in Part 1 to boil off completely? ____________minutes

Water V L

v

z

PG V G

Figure for Part 1 Figure for Part 2

Steam

v

PG V G

V L Water



8

What is the gas pressure in Part 1 at the following conditions?

(i) PG = ____________ atm after 30 minute

(ii)

PG = ____________ atm when water has boiled off completely

Part 2: Now, let’s study the same system in which the cylindrical vessel is being heated by

steam in a jacketed chamber as shown in the figure for Part 2. In this case, the heat

transfer area decreases with the liquid height and is no longer constant. The heating now

comes from the surrounding wall and the bottom of the vessel. We assume U and T S to be

the same as in Part 1.

(a) Derive an analytical expression of liquid volume V L as a function of time.

(b) Derive a single ODE for the gas pressure PG using the analytical expression in Part 2(a).

Then use MATLAB (ode45) to solve the ODE for PG from t = 0 to t when all liquid isgone, and make a plot of the PG profile. Note that the final value of PG when the vessel

contains only gas in Part 2(b) should be equal to that in Part 1(b).

Answer the following questions:

How long does it take for water in Part 2 to boil off completely? ____________

minutes

What is the gas pressure in Part 2 at the following conditions?

(i)

PG = ____________ atm after 30 minutes

(ii) PG = ____________ atm when water has boiled off completely

Solution:

Part 1(a): Constant heat transfer area

Mass balance for the liquid phase:

Energy balance for the liquid phase:

vdt

dV L −=φ

λ

λ φ

λ φ

)(

)(

)()()(

BS

BP BS L

BP

BP BS BP L

T T UAv

T vC vT T UAdt

dV T C

T C vT T UAdt

T C V d

−=

−−−=

+−−=



9

The mass balance equation then becomes:

When z = 0, t = 56.44 minutes

Part 1(b): Constant heat transfer area

Mass balance for the vapor phase:

mmmkg

kgt z z 6366.0

)5.0)()( / 1000(

)500()0(

230 ====π

2

2 )()(

R

T T UA

dt

dzT T UA

dt

dz R BS BS

φλπ φλ π

−−=→

−−=

t z

t t R

T T UA z z BS

01128.06366.0

)5.0)()(84.2256)(1000(

)100200)(2)(100(6366.0

)(220

−=

−−=

−−=

π φλπ

B

GG

BGG RT

V PnnRT V P =→=

)(

)(

MW

vRT

dt

V Pd BGG =

λ )( BS T T UAv

dt

dn MW −==

)( MW

vRT

dt

dPV

dt

dV P BG

G

G

G =+

)()(

)(

MW

vRT

dt

dPV V

dt

V V d P BG

LT

LT

G =−+−

)(

)(

MW

vRT

dt

dPV V

dt

dV P BG

LT L

G =−+−



11

At t = 56.44 min,

Check against the ideal gas law:

At t = 30.0 min, PG = 124.67 atm

MATLAB script to plot the derived equation:

clc

clear

time(1) = 0;

PG(1) = 2.478;

for t = 1:56

time(t+1) = t;

PG(t+1) = 1699.733-5816.493(3.427+0.008862!t);

e"#

time(58) = 56.44;

PG(58) = 1699.733-5816.493(3.427+0.008862!56.44);

$lot(time%PG)

title(&Ga' Pre''re '. *ime&)

la,el(&*ime i" i"te'&)

la,el(&Ga' Pre''re i" atm&)

atmPG 64.218

)44.56(008862.0427.3

493.5816733.1699 =

+−=

atmPG 58.218)927.3)(015.18(

)15.373)(08206.0)(5500( =+=



12

Part 2(a): Varying heat transfer area

Time for water to completely vaporize = 71.44 min when V L = 0

Part 2(b): Varying heat transfer area

φλ

π

φλ

)(2)(

2 BS

L

L BS L

T T R

V RU

dt

dV T T UA

dt

dV − +−=→

−−=

)47854.0(004431.0)84.2256)(1000(

)100200(5.0

2)5.0()100( 2

L

L

L V

V

dt

dV +−=

−

+

−=π

t V dt V

dV L

L

L 01772.07854.2

47854.0ln004431.0)47854.0(

−=

+→−=+

19635.0)01772.0exp(69635.0 −−= t V L



13

−

−=−

φ λ

G B BS G

LT

P

MW

RT T T UA

dt

dPV V

)(

)()(

−

−

+

=−φ λ

π G B

BS

L

G

LT

P

MW

RT T T

R

V RU

dt

dPV V

)(

)(2

)(

2

−

−

+

=−1000)015.18(

)15.373)(08206.0(

84.2256

)100200(5.0

2)5.0()100(

)927.3(

2

G

L

G

L

P

V

dt

dPV

π

−+=−

10006997.1)47854.0(4310.4)927.3( G

L

G

L

PV

dt

dPV

478.2)0(..1000

6997.169635.012335.4

3421.1201772.0

01772.0

=

−

−=

−

−

G

G

t

t

G Pt sP

e

e

dt

dP



14

MATLAB Script File:

clc

clear

format '/ort

'imtim = 0:0.01:71.44; i"itPG = 2.478;

t% PG = o#e45(&$&% 'imtim% i"itPG);

t% PG

$lot (t%PG)

title (&Ga' Pre''re ' *ime /e" eat *ra"fer rea ' ot o"'ta"t& )

la,el (&*ime i" i"te'&)

la,el (&Ga' Pre''re i" tm&)

f"ctio" fPG = o#e(t% PG)

fPG = ero'(1%1);

fPG(1) = 12.3421!e$(-0.01772!t)!(1.6997-PG(1)1000)(4.12335-0.69635! ...

e$(-0.01772!t));

At t = 30 min, PG = 133.71 atm

At t = 71.44 min, PG = 218.62 atm



16

Solution:

k 1 k 2f

A B C

k 2r

(a) There are 3 modeling equations which are :

dCA /dt = –k 1CA

dCB /dt = k 1CA – k 2f CB + k 2rCC

dCC /dt = k 2f CB – k 2rCC

(b) Define τ = k 1t x1 = (CA0 – CA) / CA0

x2 = CB / CA0 α = k 2f / k 1 β = k 2r / k 1

Need a 2nd

-order ODE for dimensionless concentration of B starting with:

CA = CA0 e-k

1t ⇒ 1 – x1 = e

-τ

or x1 = 1 – e-τ

d2CB /dt

2 = k 1dCA /dt – k 2f dCB /dt + k 2rdCC /dt

d2CB /dt

2 = –k 1

2CA0 e

-k 1t – k 2f dCB /dt + k 2r(k 2f CB – k 2rCC)

= –k 12CA0 e

-k 1t – k 2f dCB /dt + k 2r k 2f CB – k 2r[dCB /dt + k 2f CB – k 1 CA0 e

-k 1t]

= ( k 1k 2r – k 12) CA0 e

-k 1t – (k 2r + k 2f )dCB /dt

Introducing dimensionless variables:

d[dCB /dt]/dt = [1/(1/k 1)dτ] [CA0dx2 /(1/k 1)dτ] = k 12CA0d

2x2 /dτ

2

–(k 2r + k 2f )dCB /dt = –k 1(k 2r + k 2f )CA0dx2 /dτ

= –(α+β)k 12

CA0dx2 /dt



18

or dx2 /dτ = 1

1 = –(α+β)C2 – C3

C2 = [–1/(α+β)][1+(1–β)/(α+β–1)] = –(1/ α+β)[(α+β–1+1–β)/(α+β–1)]

= α / [(α+β)(1–α–β)]

C1 = –C2–C3 = α / [(α+β)(1–α–β)] + (β–1)/( α+β–1)

= [α+(α+β)(β–1)]/[(α+β)(α+β –1)]

∴∴∴∴ x2(ττττ) = ββββ /(αααα+ββββ) + αααα /[(αααα+ββββ)(1–αααα–ββββ)] e-(αααα+ββββ)ττττ + (1–ββββ)/(αααα+ββββ–1) e

-ττττ

(c) If k 1 = 1 min-1

, k 2f = 1.5 min-1

, k 2r = 2 min-1

CA0 = 3 mol/liter , CB = CC = 0 when t = 0

⇒ α = 1.5 and β = 2

x2(τ) = 0.57143 – 0.17143e-3.5τ

– 0.4e-τ

CB(t) = 1.71429 – 0.51429e-3.5t

– 1.2e-t

(d) Find CB,max and tmax:

dCB /dt = 1.80e-3.5t + 1.2e-t = 0

⇒ –1.5e-3.5t

= e-t

⇒ In (–1.5) – 3.5t = –t

⇒ no solution / root to the above equation!

Therefore, CB does not go through any maximum.



19

Now k 1 = 3 min-1

k 2f = 1.5 min-1

k 2r = 1 min-1

α = 0.5 β = 1/3

x2(τ) = 0.40 – 3.6e-0.8333τ

– 4e-τ

CB(t) = 1.2 + 10.8e-2.5t

–12e-3t

dCB /dt = –27e-2.5t

+ 36e-3t

= 0

⇒ –2.5t = In (1.3333) – 3t

⇒

tmax = 0.575 min

⇒ CB,max = 1.627 mol/liter

(f)

The MATLAB Script File:

%

% CHE654: Problem #58%

% Batch Reactor with a Series Reaction

%

clc

clear

simtime=[0:0.1:5];

initC=[3,0,0];

[t,c] = ode45('batchr1',simtime,initC);

[t,c]

plot(t,c)

title('Batch Reactor with a Series Reaction')xlabel('Time in Minutes')

ylabel('Concentration of A, B, and C in Mole/Liter')

legend('A','B','C')

%

[t,c] = ode45('batchr2',simtime,initC);

[t,c]

plot(t,c)

title('Batch Reactor with a Series Reaction')

xlabel('Time in Minutes')

ylabel('Concentration of A, B, and C in Mole/Liter')

legend('A','B','C')



20

The MATLAB M-Files for ODE:

function fv=batchr1(t,c)

fv = zeros(3,1);

fv(1) = -c(1);fv(2) = c(1)-1.5*c(2)+2*c(3);

fv(3) = 1.5*c(2)-2*c(3);

function fv=batchr2(t,c)

fv = zeros(3,1);

fv(1) = -3*c(1);

fv(2) = 3*c(1)-1.5*c(2)+c(3);

fv(3) = 1.5*c(2)-c(3);

The MATLAB Output:

>>

ans =

0 3.0000 0 0

0.1000 2.7145 0.2661 0.0194

0.2000 2.4562 0.4764 0.0674

0.3000 2.2225 0.6453 0.1322

0.4000 2.0110 0.7831 0.2059

0.5000 1.8196 0.8971 0.2833

0.6000 1.6464 0.9928 0.3608

0.7000 1.4898 1.0740 0.4363

0.8000 1.3480 1.1438 0.5082

0.9000 1.2197 1.2044 0.5759

1.0000 1.1036 1.2573 0.6391

1.1000 0.9986 1.3039 0.6975

1.2000 0.9036 1.3452 0.7512

1.3000 0.8176 1.3818 0.8006

1.4000 0.7398 1.4145 0.8457

1.5000 0.6694 1.4438 0.8868

1.6000 0.6057 1.4702 0.92421.7000 0.5480 1.4938 0.9582

1.8000 0.4959 1.5150 0.9891

1.9000 0.4487 1.5341 1.0172

2.0000 0.4060 1.5514 1.0426

2.1000 0.3674 1.5670 1.0656

2.2000 0.3324 1.5811 1.0865

2.3000 0.3008 1.5938 1.1054

2.4000 0.2722 1.6053 1.1226

2.5000 0.2463 1.6157 1.1381

2.6000 0.2228 1.6251 1.1521

2.7000 0.2016 1.6336 1.16482.8000 0.1824 1.6413 1.1763



22

2.4000 0.0022 1.2178 1.7799

2.5000 0.0017 1.2142 1.7841

2.6000 0.0012 1.2113 1.7874

2.7000 0.0009 1.2090 1.7901

2.8000 0.0007 1.2071 1.7922

2.9000 0.0005 1.2057 1.79383.0000 0.0004 1.2045 1.7951

3.1000 0.0003 1.2036 1.7962

3.2000 0.0002 1.2028 1.7970

3.3000 0.0002 1.2022 1.7976

3.4000 0.0001 1.2018 1.7981

3.5000 0.0001 1.2014 1.7985

3.6000 0.0001 1.2011 1.7989

3.7000 0.0000 1.2009 1.7991

3.8000 0.0000 1.2007 1.7993

3.9000 0.0000 1.2005 1.7994

4.0000 0.0000 1.2004 1.79964.1000 0.0000 1.2003 1.7997

4.2000 0.0000 1.2003 1.7997

4.3000 0.0000 1.2002 1.7998

4.4000 0.0000 1.2002 1.7998

4.5000 0.0000 1.2001 1.7999

4.6000 0.0000 1.2001 1.7999

4.7000 0.0000 1.2001 1.7999

4.8000 0.0000 1.2001 1.7999

4.9000 0.0000 1.2000 1.8000

5.0000 0.0000 1.2000 1.8000



23



24

60. Isothermal Semi-Batch Reactor (a) Consider an isothermal semi-batch reactor where a single reaction takes place in a

solvent S, which is inert. In this reaction, 2 moles of component A react with one

k 1

mole of component B to form one mole of component C: 2A + B -----> C.

The reaction rate does not conform to the stoichiometry but is 1st-order with

respect to each reactant as follows: rA = −k 1CACB

FA = 15 liter/min FB = 10 liter/min

Initially (t = 0 min), the reactor contains 100 liters of solution and 300 moles of A.

Assuming that all components have the same density of 60 mol/liter, derive 3 ODE

equations needed to compute CA, CB, and CC, the concentrations (moles/liter) of A,

B, and C, respectively. Use ode45 in MATLAB to solve for and plot (in a single

graph) the concentrations of the 3 components. Run the model for 20 minutes with

an increment of 0.5 minute.

The following experimental data have been obtained for this reaction when carried

out in a batch reactor:

Time(minute) 0 5.0 7.5 12.0 15.5 25.0 32.0 40.0

CA(mol/liter) 2.0 1.65 1.52 1.40 1.28 1.17 1.10 1.06

CB(mol/liter) 0.5

(b) At the end of 20 minutes, the 2 feeds to the reactor are suddenly shut off, and

4,500 moles of a new component called D (same density as components A, B, andC) are charged to the reactor. That is, the reactor now operates in a batch mode.

Component D reacts with component C to form A and B, and the reaction now

looks as follows:

k 1

2A + B C + D

k 2

with a reaction rate of rA = −k 1CACB + k 2CD (2nd

-order forward and 1st-order

reverse). The value of k 2 has been measured to be 1.0 min-1. Derive analyticallythe concentration of A as a function of time, and compute CA at steady state based



25

on your derived equation. Your final expression should be simplified as much as

possible and should not contain any parameters except t (time) and CA.

Useful Integrals:

where B = a + bx + cx2

γ = 4ac – b2

ω = b + 2cx

Solution:

(a) Must first determine the rate constant k using the graphical method.

dCA = 2dCB CA – CA0 = 2(CB – CB0) CB = CB0 + ½ (CA – CA0)

(dCA /dt) = –kCACB dCA = – kdt

CA[(C

B0–0.5C

A0)+0.5C

A]

Integrate from CA0 to CA on the LHS and from 0 to t on the RHS.

Integrate by partial fractions or use a table of integrals:

CA

– [1/(CB0–0.5CA0)] ln[(CB0–0.5CA0)+0.5CA)/CA] = –kt

CA0

ln(CA /CB) – ln(CA0 /CB0) = (0.5CA0– CB0)kt

∫

! +

!

−=+ x

bxa

abxa x

dx

ln

1

)(

0tanh2

0

2

0tan2

1

1

"

!

−−−

=−=

#

! +

−

−

∫

γ γ

ω

γ

γ ω

γ γ

ω

γ

if

if B

dx

if



26

So plot ln(CA /CB) vs. (0.5CA0– CB0)kt to obtain the value of k.

From MATLAB, get slope = 0.054203 k = 0.1084 liter/mole-min

dV = FA + FB = 15 + 10 = 25 V = 100 + 25t

dt

d(VCA)/dt = –kVCACB + ρAFA

because 1 dNA = dCA = –kCACB

V dt dt

d(VCA)/dt = V(dCA /dt) + CA(dV/dt) = V(dCA /dt) + 25CA

= –0.1084VCA[0 + 0.5(CA – 3.0)] + (60)(15)

= –0.1084CA[0.5CA – 1.5]V + 900

dCA /dt = 0.1084CA(1.5 – 0.5CA) + (900 – 25CA)/(100 + 25t)

s.t. CA(t = 0) = 3.0



30

CD = 4500/675 = 4.6667 mol/liter

dCA = 2dCB = –2dCD CA – CA0 = 2(CB – CB0) CB = CB0 + ½ (CA – CA0)

and CA – CA0 = 2(CD0 – CD) CD = CD0 + ½ (CA0 – CA)

(dCA /dt) = −k 1CACB + k 2CD

= −k 1CA[CB0 + ½ (CA – CA0)] + k 2[CD0 + ½ (CA0 – CA)]

= −0.1084CA[7.1794 + ½ (CA – 5.9142)] + 1.0[4.6667 + ½ (5.9142 – CA)]

= CA[−0.77825 – 0.0542CA + 0.32055 – 0.5] + 7.6238

= 7.6238 – CA[0.0542CA + 0.9577]

dCA = dt

7.6238–CA[0.0542CA+0.9577]

dCA = dt

7.6238–0.9577CA–0.0542CA

2

a = 7.6238, b = –0.9577, c = –0.0542 γ = 4ac – b2 = –2.5700

–(2/ √–γ) tanh-1

(ω / √–γ) = –(2/ √–γ) tanh-1

[(b+2cCA)/ √–γ]

–1.24756 tanh-1

[–0.62378(0.1084CA+0.9577)] = dt

–1.24756 tanh-1

[–0.62378(0.1084CA+0.9577)] – 4.1206 = t

Check: when t = 0, CA does equal to 5.9142 mol/liter.

When $$$$, CA = 5.9541 mol/liter from the above equation.

From the ODE: (dCA /dt) = 7.6238 – CA[0.0542CA + 0.9577] = 0

------> CA = 5.9541 mol/liter same answer!



31

62. Isothermal Batch Reactor with a Series Reaction

Consider an isothermal batch reactor with a series reaction where 2 moles of

component A react to form one mole of component B. Component B also reacts to

form component C. The reaction scheme can be characterized as follows:

k 1 k 2 2A B C

Initially (t = 0 hr), CA = CA0, CB = 0, and CC = 0, where CA, CB, and CC represent the

concentrations (mol/liter) of components A, B, and C, respectively.

(a) Assume constant volume and that the first reaction 2A → B is one-half order and

the second reaction B → C is first-order, i.e.

dCA /dt = −k 1CA1/2

dCB /dt = k 1 CA1/2 − k 2CB

2

Derive an analytical expression for CB as a function of time.

(b) The following experimental data were obtained for component CA:

Time(hr) 0 0.05 0.15 0.35 0.60 0.85 1.0

CA(mol/liter) 1.0 0.93 0.79 0.54 0.30 0.13 0.06

CB was also measured to be 0.11 mol/liter at t = 1 hr. Determine the values of k 1

and k 2 and the time tmax at which CB is at its maximum. Hint: Use Polyfit function

in MATLAB to help determine k 1 and k 2.

(c) Now, suppose the order of the above series reaction conforms to the stoichiometry,

derive analytically a 1st-order ODE for CB, i.e.

dCB /dt + p(t)CB = q(t)

but do not solve this ODE.

(d) When the reaction order conforms to the stoichiometry, the following experimental

data were obtained for CA:

Time(hr) 0 0.03 0.06 0.10 0.15 0.20 0.30

CA(mol/liter) 1.0 0.76 0.63 0.51 0.39 0.33 0.25

It was also observed that CB reached a maximum of 0.14 mol/liter at t = 0.1 hr.

Determine the values of k 1 and k 2.



32

Solution:

k 1 k 22A B C

(a)

If rA = dCA /dt = -k 1CA½ s.t. CA(0) = CA0

rB = dCB /dt = (k 1 /2)CA½ − k 2CB s.t. CB(0) = 0

rC = dCC /dt = k 2CB s.t. CC(0) = 0

Derive analytically CB(t):

CA t

∫ dCA /CA½

= −k 1 ∫ dt ⇒ 2(CA½ − CA0

½) = −k 1t

CA0 0

CA(t) = (√CA0 − ½k 1t)2

dCB /dt + k 2CB = (k 1 /2)(√CA0 − ½k 1t)

CB(t) = C1e-k

2t + C2t + C3

At t = 0 0 = C1 + C3

dCB /dt = −C1e-k

2t + C2

dCB /dt + k 2CB = −k 2C1e-k

2t + C2 + k 2C1e

-k 2

t + k 2C2t + k 2C3

k 2C2t + (C2 + k 2C3) = (k 1 /2)√CA0 − (k 12 /4)t

⇒ k 2C2 = −(k 12 /4) or C2 = −(k 1

2 /4k 2)

C2 + k 2C3 = k 12√CA0



33

C3 = (1/k 2)[(k 1 /2)√CA0 + (k 12 /4k 2)] = (k 1 /2k 2)√CA0 + (k 1 /2k 2)

2

= (k 1 /2k 2)(√CA0 + k 1 /2k 2)

C1 = −(k 1 /2k 2)(√CA0 + k 1 /2k 2)

∴∴∴∴ CB(t) = (k1 /2k2)(√CA0 + (k1 /2k2) [1 −−−− e-k

2t] −−−− k1

2 /4k2 t

(b) Time (hr) 0 0.05 0.15 0.35 0.60 0.85 1.0

CA (mole/liter) 1.0 0.93 0.79 0.54 0.3 0.13 0.06

Has been observed: CB (t = 1 hr) = 0.11 mole/liter

CA(t) = (1 − ½k 1t)2 = 1 − k 1t + (k 1

2 /4)t

2

Use Polyfit in MATLAB to find the CA vs. t data to a 2nd-order degree polynomial

(or plot CA1/2

vs. t)

From MATLAB : CA = 0.5729t2 – 1.5146t + 1.0023

Can obtain 2 values of k 1 as :

k 1 = 1.5146 and k 12 /4 = 0.5729 ⇒ k 1 = 1.5138

So take the average value: k 1 = 1.5142

CB(t) = (0.7571/k 2)(1+0.7571/k 2)(1 – e-k 2t) – (0.5732/k 2)t

CB(t = 1 hr) = 0.11 = (0.7571/k 2)(1+0.7571/k 2)(1 – e-k

2) – 0.5732/k 2

Use MATLAB to solve the above nonlinear equation for k 2

k 2 = 2.9818 hr-1

CB(t) = 0.3184(1 – e-2.9818t

) – 0.1922t

dCB/dt = 0.9494e-2.9818t – 0.1922 = 0



34

⇒ e-2.9818t

= 0.2024

tmax = 0.536 hr

CB,max = 0.151 mole/liter

(c) The reaction order conforms to stoichiometry

dCA /dt = –k 1CA2 s.t. same initial conditions

dCB /dt = (k 1 /2) CA2 – k 2CB

Derive 1st-order ODE for CB, i.e. dCB /dt + p (t)CB = q(t)

CA t

∫ dCA /CA2 = –k 1 ∫ dt ⇒ 1/CA – 1/CA0 = k 1t

CA0 0

⇒ CA(t) = CA0 /(CA0k 1t + 1)

∴ dCB /dt + k 2CB = (k 1CA02 /2)[CA0k 1t + 1]

–2

(d) Time (hr) 0 0.03 0.06 0.1 0.15 0.2 0.3

CA (mole/liter) 1.0 0.76 0.63 0.51 0.39 0.33 0.25

CB,max has been observed at 0.14 mol/lit at t = 0.1 hr

1/CA = k 1t + 1

Use Polyfit to regress 1/CA vs. t by fitting y = ax + b

Slope of the straight line is k 1 ⇒ y = 10.0676t + 1.0002

k 1 = 10.0676

CB,max occurs at dCB /dt = 0

⇒ k 2CB,max = (10.0676/2)(10.0676tmax + 1)–2

k2 = 8.9285 hr-1



36

But xc5 = 0.6 and P = xc5PC5VAP

+ xc6PC6VAP

0.8113[0.6*10[6.85221 – 1064.63/(T+232.0)]

+ 0.4*10[6.87776 – 1171.53/(T+224.366)]

]

= 0.6*10[6.85221 – 1064.63/(T+232.0)]

0.32452 (10[6.87776 – 1171.53/(T+224.366)]

) – 0.11322 (10[6.85221 – 1064.63/(T+232.0)]

)

= 0

Nonlinear equation in T Use MATLAB for solve for the root or T

Get T = 55.55 °°°°C

Also get, P = 845.86 + 196.69 = 1042.55 mmHg

(b) Start with mass balances:

dL/dt = –V

d( xL)/dt = – yV or x(dL/dt) + L(d x /dt) = (1.8804 x – 0.8804 x2)(dL/dt)

xdL + Ld x = (1.8804 x – 0.8804 x2)dL

Ld x = (0.8804 x

– 0.8804 x2)dL

1.13585 d x = dL Integrate from x = 0.6 to x and L = 100

to L

x( 1 – x) L

1.13585 1 + 1 = dL

x

1 – x L

ln ( x) – ln(1 – x) = 0.8804 ln(L)

x L

ln x = 0.8804 ln(L)

1 – x

0.6 100



ln 0.6667 x = ln L0.8804

1 – x 100

0.6667 x = L0.8804

1 – x 100

When 90% of the liquid L is vaporized L = 10 x = 0.165

CHE654 2012 Homework5 Solutions

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