Since the Feed Is Equimolar, CA0 = CB0 = .1 Mols/Dm3

LEM

304S12

Fogler

CHAPTER 8

HOMEWORK SOLUTION

#8.6

P8-6

A+ B!C

Since the feed is equimolar, CA0 = CB0 = .1 mols/dm3

CA = CA0(1-X)

CB = CB0(1-X)

P8-6 (a)

PFR A

CSTR

V F

F X

For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mols/dm3

See Polymath program P8-6-a.pol.

Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 X 0 0 0.85 0.85

2 V 0 0 308.2917 308.2917

3 Ca0 0.1 0.1 0.1 0.1

4 Fa0 0.2 0.2 0.2 0.2

5 T 300. 300. 470. 470.

6 k 0.01 0.01 4.150375 4.150375

7 ra -0.0001 -0.0018941 -0.0001 -0.0009338

Differential equations

1 d(V)/d(X) = -Fa0 / ra

Explicit equations

1 Ca0 = .1

2 Fa0 = .2

3 T = 300 + 200 * X

4 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))

5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)

V = 308.2917dm3

8-24

For the CSTR,

X = .85, T = 300+(200)(85) = 470 K.

k = 4.31 (Using T = 470K in the formula).

-rA

= .000971 mol/dm3/s

0 3

-4

.1 2 .85

175 dm

9.71 10

F X

! !

= = =

" !

The reason for this difference is that the temperature and hence the rate of reaction remains

constant throughout the entire CSTR (equal to the outlet conditions), while for a PFR, the rate

increases gradually with temperature from the inlet to the outlet, so the rate of increases with

length.

P8-6 (b)

[ ]

i P

X H

T T

! C

= +

For boiling temp of 550 k,

550 = T0 + 200

T0 = 350K

P8-6 (c)

P8-6 (d)

( )

CSTR

F X

X r

" = !

For V = 500 dm3, FA0=.2

2 2 2

0 (1 ) .01 (1 ) A A

!r = k C ! X = k ! X

T = 300 + 200 X

Now use Polymath to solve the non-linear equations.

See Polymath program P8-6-d-1.pol.

8-25

Calculated values of NLE variables

Variable Value f(x) Initial Guess

1 T 484.4136 0 480.

2 X 0.9220681 -2.041E-09 0.9

Variable Value

1 k 6.072856

2 ra 0.0003688

Nonlinear equations

1 f(T) = 300 + 200 * X - T = 0

2 f(X) = 500 - .2 * X / ra = 0

Explicit equations

1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))

2 ra = 0.01 * k * (1 - X) ^ 2

Hence, X = .922 and T = 484.41 K

For the conversion in two CSTR’s of 250 dm3 each,

For the first CSTR, using the earlier program and V = 250 dm3,

Calculated values of NLE variables

Variable Value f(x) Initial Guess

1 T 476.482 1.137E-13 480.

2 X 0.88241 -5.803E-09 0.9

Variable Value

1 k 5.105278

2 ra 0.0007059

Nonlinear equations

1 f(T) = 300 + 200 * X - T = 0

2 f(X) = 250 - .2 * X / ra = 0

Explicit equations

1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))

2 ra = 0.01 * k * (1 - X) ^ 2

T = 476.48 ad X = .8824

Hence, in the second reactor,

8-26

0 1

( )

CSTR

F X X

X r X

" = ! +

See Polymath program P8-6-d-2.pol.

Calculated values of NLE variables

Variable Value f(x) Initial Guess

1 T 493.8738 0 480.

2 X 0.9693688 -1.359E-09 0.8824

Variable Value

1 k 7.415252

2 ra 6.958E-05

3 X1 0.8824

Nonlinear equations

1 f(T) = 476.48 + 200 * (X - X1) - T = 0

2 f(X) = 250 - .2 * (X - X1) / ra = 0

Explicit equations

1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))

2 ra = 0.01 * k * (1 - X) ^ 2

3 X1 = .8824

Hence, final X = .9694

P8-6 (e) Individualized solution

P8-6 (f) Individualized solution

P8-7 (a)

For reversible reaction, the rate law becomes

A A B

r k C C

! "

# = $ # %

& '

, 1 1 200( ) out CSTR

T = T + X ! X

8-27

300 200

1 1

(300) exp

300

1 1

(450) exp

450

Rxn

C C

T X

k k

R T

K K

R T

= +

! ! ""

= $ $ # %%

& & ''

)( ! "*

= + $ # %,

- & '.

Stoichiometry:

(1 )

C A

A A

B A

C C X

= !

See Polymath program P8-7-a.pol.

POLYMATH Results

No Title 03-21-2006, Rev5.1.233

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

V 0 0 10 10

X 0 0 0.0051176 0.0051176

T 300 300 301.02352 301.02352

k 0.01 0.01 0.010587 0.010587

Fa0 0.2 0.2 0.2 0.2

Ca0 0.1 0.1 0.1 0.1

Kc 286.49665 276.85758 286.49665 276.85758

ra -1.0E-04 -1.048E-04 -1.0E-04 -1.048E-04

Xe 0.8298116 0.827152 0.8298116 0.827152

ODE Report (RKF45)

Differential equations as entered by the user

[1] d(X)/d(V) = -ra / Fa0

Explicit equations as entered by the user

[1] T = 300+200*X

[2] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))

[3] Fa0 = 0.2

[4] Ca0 = 0.1

[5] Kc = 10 * exp(-6000 /1.987 * (1 / 450 - 1 / T))

[6] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)

[7] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2

P8-8 (a)

A+ B!C

Species Balance:

20 /

dX r

dW F

v dm s

P atm

= "

Stoichiometry:

, 1

A A

X T

C C where

X T

C C

X T

# " $

= % & =

' + (

# " $

) = % &

' + (

Rate Law is:

1 1

, 0.133exp

450

31400

20,000 /

A A

Rxn

r kC with k

R T

H J mol

! # $"

% = = ( & % ')

, * +-

. = %

Energy Balance:

[ ( )]

15 24 40 0

20,000

450 450 500

i P P

X H T

T T

C X C

T X

= +

+ #

# = + " =

= + = +

$ % %

See Polymath program P8-8-a.pol.

Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 X 0 0 0.8 0.8

2 W 0 0 43.13711 43.13711

3 T 450. 450. 850. 850.

4 v0 20. 20. 20. 20.

5 T0 450. 450. 450. 450.

6 k 0.133 0.133 6.904332 6.904332

Differential equations

1 d(W)/d(X) = v0 * (1 + X) * T / k / (1 - X) / T0

8-36

Explicit equations

1 T = 450 + 500 * X

2 v0 = 20

3 T0 = 450

4 k = .133 * exp(31400 / 8.314 * (1 / T0 - 1 / T))

P8-8 (b)

Species Balance for CSTR:

450 500 450 500(.8) 850

31400 1 1

.133exp 6.9

8.314 450 850

39.42

CSTR

F X

T X K

W kg

" !

= + = + =

# % &$

= ) ' " (* =

- + ,.

P8-8 (c) Individualized solution

P8-8 (d)

For pressure drop, an extra equation is added

0 ( )

0 0

2 ( / )

A A

dP T P

dW T P P

X T P

C C

X T P

# $

= % & ' +

( )

# % $

= & '

( + )

See Polymath program P8-8-d.pol.

Using POLYMATH program CRE_8_8d.pol

For ! = .019

8-37

Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 W 0 0 0.8 0.8

2 X 0 0 0.0544753 0.0544753

3 P 1.013E+06 1.002E+06 1.013E+06 1.002E+06

4 T 450. 450. 850. 850.

5 v0 20. 20. 20. 20.

6 T0 450. 450. 450. 450.

7 k 0.133 0.133 6.904332 6.904332

8 P0 1.013E+06 1.013E+06 1.013E+06 1.013E+06

9 alpha 0.019 0.019 0.019 0.019

Differential equations

1 d(X)/d(W) = k / v0 * (1 - X) / (1 + X) * T0 / T * P / P0

2 d(P)/d(W) = -alpha / 2 * (T / T0) * P0 ^ 2 / P * (1 + X)

Explicit equations

1 T = 450 + 500 * W

2 v0 = 20

3 T0 = 450

4 k = .133 * exp(31400 / 8.314 * (1 / T0 - 1 / T))

5 P0 = 1013250

6 alpha = .019