Page 168 - Chemical engineering design

Solution
Basis: 100 mol/h dry feed gas. FLOW-SHEETING 145
H 2 O in feed stream D 3.0 ð 11.0 D 33 mol.

2

1
500°K
Shift converter

Let fractional conversion of CO to H 2 be C. Then mols of CO reacted D 11.0 ð C.From
the stoichiometric equation and feed composition, the exit gas composition will be:
CO D 11.0 1 C

CO 2 D 8.5 C 11.0 ð C
H 2 O D 33 11.0 ð C
H 2 D 76.5 C 11.0 ð C
P CO ð P H 2 O
At equilibrium K p D
P CO 2 ð P H 2
The temperature is high enough for the gases to be considered ideal, so the equilibrium
constant is written in terms of partial pressure rather than fugacity, and the constant will
not be affected by pressure. Mol fraction can be substituted for partial pressure. As the
total mols in and out is constant, the equilibrium relationship can be written directly in
mols of the components.
11 1 C 33 11C
K p D
8.5 C 11C 76.5 C 11C
Expanding and rearranging
2
K p 121 121 C C K p 935 C 484 C C K p 650 363 D 0 1
K p is a function of temperature.
Ž
For illustration, take T out D 700 K,atwhich K p D 1.11 ð 10 1
2
107.6C C 587.8C 290.85 D 0
C D 0.57
The reaction is exothermic and the operation can be taken as adiabatic, as no cooling is
provided and the heat losses will be small.
The gas exit temperature will be a function of the conversion. The exit temperature
must satisfy the adiabatic heat balance and the equilibrium relationship.
A heat balance was carried over a range of values for the conversion C,using the
program Energy 1, Chapter 3. The value for which the program gives zero heat input or