Solution Manual For Chemical Reactor Analysis and Design Fundamentals 2nd Edition by James B. Rawlings

SOLUTION MANUAL FOR
Chemical Reactor Analysis and Design Fundamentals
by James B. Rawlings
2nd Edition

, Chapter 1

1.1. For The Thermal Cracking Of Ethane In A Tubular Reactor, The Following
Data Were Obtained For The Rate Coefficient At Different Reference
Temperatures:

T(°C) 702 725 734 754 773 789 803 810 827 837
K(S-1) 0.15 0.273 0.333 0.595 0.923 1.492 2.138 2.718 4.137 4.665



Solution

The Arrhenius Expression

K =A Exp − 
E

 Rt 

Is Transformed Logarithmically Into:

E
Ln K = Ln A −
Rt

For Each Data Point Ln K And 1/T Is Calculated:

X = 1/T 10³ Y = Ln K

1.025 –1.897
1.002 –1.298
0.993 –1.100
0.974 –0.519
0.956 –0.080
0.941 0.400
0.929 0.760
0.923 1.000
0.909 1.420
0.901 1.540

The Slope And The Intercept (Ln A) Are Calculated By Linear Regression:

XY
E  xy
−
− = N
R (X)
2


X 2 − N

Ln A =
 Y − MX
N

With X = 1/T And Y = Ln K.

, So:

E
− = −28497. E = 56623 Kcal/Kmol Or 2.37 105 Kj/Kmol.
Or
R

Ln A = 27.245 Or A = 6.800 1011 S-1

1.2. Derive The Result Given In Table 1.2.4.2-1 For The Reaction
.S
A
B

Q




+ → +

Solution
A
B

Q
S




+ → +

The Continuity Equation For Species A Reads:

Dca
= −K Ca Cb (1)
Dt

To Integrate (1), Cb Has To Be Expressed As A Function Of Ca:

Cb = Cb 0 − (Ca 0− Ca )

Hence,
Ca T
Dc A
 = −K  Dt
Ca0
Ca (Cb − C A + Ca ) 0
0 0

Or
−1 C − Ca
Kt = Ln b0
CB − CA C A − CB
0 0 0




Expressing The Concentrations Ca And Cb As A Function Of The Conversion Of The
Reactant, Xa:

Ca = Ca (0 1 − Xa )


( )
C B = C B0 − CA 0 − C A = C B0 − C A 0 + CA 0 (1 − XA )

= Cb 0 − Ca 0Xa

Hence

Kt Ln
−1 C B − C A (1 − Xa )
0 0
=
CB − CA C A (C B − Ca 0Xa )
0 0 0 0

, Or
1 M(1 − Xa )
C A 0 Kt = − Ln
M −1 M − Xa
With
Cb0
M=
Ca
0




1.3. Derive The Solutions To The Rate Equation For The First Order Reversible
Reaction Given In Section 1.2.3.

Solution
1
For A Q
2


Dca
RA = − = K1 Ca − K 2 Cq
Dt
(
= K1ca − K 2 C A 0 + Cq0 − Ca )
Or
)Ca = K2 ( CA 0 +C Q0 )
Dca
+ (K + K
Dt 1 2


This Is A Standard Form, With Integrating Factor

Exp[ (K1 + K2 ) Dt ] = Exp(K1 + K2 ) T

Thus

D
(E( K1 +K 2 )T
A
)= E ( − K1 +K 2 )T
(
2 C A0 + Cq0
K )
And C Dt
C = E−(K1 +K 2 )T \
K2 (Ca 0 + Cq0 E(K1 +K 2 )Tdt + K) )
(
A
B


=
K2
(
C A + Cq0 + Ke−(K1 +K 2 )T )
K1 + K2 0

Now, At T = 0, Ca = Ca0, Leading To

C = C ( +C
K2
) +
K1ca − K2cq
0 0
E−(K1 +K 2 )T
A A0 Q0
k1 + K K1 + K2
2


Which Is The Solution Given In Section 1.2.3.

The Alternate Approach In Terms Of Conversions Is Somewhat Simpler: