CACHE Modules on Energy in the Curriculum
Fuel Cells
Module Title: Water Gas Shift Reaction in a Palladium Membrane Reactor
Module Author: Jason Keith
Author Affiliation: Michigan Technological University
Course: Kinetics and Reaction Engineering
Text Reference: Fogler (4th edition), Section 4.9
Literature Reference: S. Uemiya et al, Ind. Eng. Chem. Res., 30, 585 (1991).
Concepts: Develop a numerical model to predict the conversion and hydrogen yield
within a membrane reactor.
Problem Motivation:
Fuel cells are a promising alternative energy conversion technology. One type of fuel
cell, a proton exchange membrane fuel cell (PEMFC) reacts hydrogen with oxygen to
produce electricity (Figure 1). Fundamental to a hydrogen economy powered by fuel cells
is the generation of high purity hydrogen.
Consider the schematic of a compressed hydrogen tank (2000 psi, regulated to 10 psi)
feeding a proton exchange membrane fuel cell, as seen in Figure 2 below. We will now
focus on hydrogen generation in a membrane reactor (to fill the compressed tank).
-
-
e
e
H2
H2O
O2
H+
H2
H2O
H2
O2
O2
+
H
H2
Computer
(Electric
Load)
Pressure
regulator
H2 feed line
Air in
H2
H2
H2
H2
H2O
H2O
+
H
H2
H+
O2
Anode
Cathode
Electrolyte
Figure 1. Reactions in the PEMFC
H2 out
H2 tank
Fuel Cell
Air / H2O out
Figure 2. Diagram for fueling a laptop.
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Background
Natural gas has been proposed as a source of hydrogen for fuel cell vehicle applications
because of the existing infrastructure. In a process known as steam reforming, natural gas
and steam are reacted into mostly carbon monoxide and hydrogen with some carbon
dioxide also produced. There is also excess water in the reformate stream.
A water gas shift reactor can be used to convert some of the remaining carbon monoxide
into hydrogen according to the reaction:
CO + H2O ↔ H2 + CO2
Figures 3 shows an axisymmetric view of an annular water gas shift reactor which is 8
cm high. In the outer (annular) region, an iron chromium oxide catalyst is present to carry
out the water gas shift reaction. A 20 m thick palladium membrane separates the
reaction (outer) zone from the separation (inner) zone.
Gas Flows In
Separation Zone
Reaction Zones
Figure 3. Annular Membrane Reactor : Top View (Left) and
Side View (Right)
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The following chemicals are present in the system: carbon monoxide (CO), water (H2O),
carbon dioxide (CO2), hydrogen (H2), and argon (Ar). Argon serves as an inert sweep gas
in the separation zone.
Performing a mass balance on each chemical in the outer (reaction) zone gives:
dFCO ,outer
 rCO
dx
dFH 2O ,outer
dx
dFH 2,outer
dx
dFCO 2,outer
dx
(1)
 rH 2O
(2)
 rH 2  j
(3)
 rCO 2
(4)
Similarly, for the inner (separation) zone we have:
dFH 2,inner
dx
dFAr ,inner
dx
 j
(5)
0
(6)
In equations 1-6 F denotes a molar flow rate in mol/min. The reaction rates are related by
 rCO  rH 2O  rCO 2  rH 2  r
(7)
where r is the reaction rate in units of mol/(min – cm reactor length) and is given in terms
of gas concentration in units of mol/cm3 as:
rk
CCO ,outerC H 2O ,outer  K 1CCO 2,outerC H 2,outer
1  2.4  10 5 C H 2O ,outer  7.2  10 5 CCO 2,outer
(8)
where k = 7.4 x 108 in units of cm6/(mol – min – cm reactor length)] and K = 11.92
(dimensionless) at the reaction conditions of 673 K. Finally, the membrane flux in units
of mol/(min – cm reactor length) is given by Equation 9 of Uemiya et al. (see the
literature reference on page 1) according to:
j

q 0.76
C H 2,outer  C H0.76
2 ,inner
t
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
(9)
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where q = 1.1 x 102 (mol0.24 – m thickness – cm2.28)/(min – cm reactor length) and t = 20
m.
Example Problem Statement: Consider a reaction zone feed of 1.1 x 10-3 mol/min CO
and 1.1 x 10-3 mol/min H2O at 2 atm pressure. In the separation zone the feed is 1.8 x 10-2
mol/min Ar at 1 atm pressure. Recall that the reactor length is 8 cm.
a) Determine the equilibrium concentrations of CO, H2O, CO2, and H2 if there is no
membrane separation of hydrogen (thus, in this problem assume j = 0).
b) Construct a numerical model to predict the molar flow rates of CO, H2O, CO2,
and H2 as a function of distance if there is no membrane separation of hydrogen
(thus, in your simulations assume j = 0). Use a step size x = 0.01 cm. Is
equilibrium reached in this reactor? What is the CO conversion in this system?
Example Problem Solution:
Part a)
Step 1) At equilibrium the forward and reverse reaction rates are equal. This corresponds
to a situation where:
CCO,eq CH 2O,eq  K 1CCO 2,eq CH 2,eq
(10)
Step 2) We note that as there is no CO2 or H2 in the feed, their initial partial
concentrations are zero. Furthermore, the equal molar feed of CO and H2O corresponds
to a partial pressure of 1 atm for each gas. The corresponding concentration is given as:
Co 
1atm
L
 1.8  10 5 mol/cm 3
3
L - atm
1000 cm
0.08206
673K
mol - K
(11)
Step 3) As the reaction occurs the concentration of CO2 and H2 will increase linearly with
conversion, whereas the concentration of CO and H2O will decrease linearly with
conversion.
If we denote C as the concentration change from the feed state to the equilibrium state,
we would have:
CCO ,eq  C H 2O ,eq  C o  C
(12)
and
C CO 2,eq  C H 2,eq  C
(13)
Step 4) Inserting Equations 12 and 13 into Equation 10 gives:
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(Co  C) 2  K 1C 2
(14)
This can be expanded and rewritten as a quadratic equation:
a(C ) 2  bC  c  0
(15)
where a = 1 – -1 = 0.916, b = –2Co = –3.6 x 10-5, and c = Co2 = 3.2 x 10-10. The solution
is C = 1.4 x 10-5 mol/cm3 such that:
C CO ,eq  C H 2O ,eq  4.0 x 10-6 mol/cm3
(16)
and
CCO 2,eq  C H 2,eq  1.4 x 10-5 mol/cm3
(17)
These can be inserted into Equation 11 to show that they satisfy the equilibrium
conditions.
Step 5) The molar flow rates of these chemicals can be determined from the change in
concentrations:
FCO ,outer  FH 2O ,outer  FCO , feed
C CO ,eq
C CO , feed
 1.1  10 3 mol/min
4.0  10 -6 mol/cm 3
 2.4  10 -4 mol/min
-5
3
1.8  10 mol/cm
 1.1  10 3 mol/min
1.4  10 -5 mol/cm 3
 8.6  10 -4 mol/min
1.8  10 -5 mol/cm 3
Also,
FCO 2,outer  FH 2,outer  FCO , feed
C CO 2,eq
C CO , feed
Part b)
Step 1) A numerical model can be made to simulate Equations 1-4 with j = 0. Using a
simple Euler discretization of the equations we have:
FCO ,outer,i 1  FCO ,outer,i  rx
(18)
FH 2O ,outer,i 1  FH 2O ,outer,i  rx
(19)
FCO 2,outer,i 1  FCO 2,outer,i  rx
(20)
FH 2,outer,i 1  FH 2,outer,i  rx
(21)
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Subject to the initial conditions FCO ,outer, 0  FH 2O ,outer, 0  1.1 x 10-3 mol/min and
FCO 2,outer,0  FH 2,outer,0  0.0 mol/min.
These equations can be solved iteratively until the end of the reactor is reached. The
procedure is:
1) Calculate the reaction rate r at feed conditions (location 0, distance x = 0).
2) Use r to calculate the chemical flow rates (location 1, distance = x) using
equations 18 – 21.
3) Calculate the reaction rate r at the updated conditions (location 1, distance x =
x).
4) Repeat steps 2-3 as you progress down the length of the reactor by increasing the
location number and distance.
The system is simulated using a step size of x =0.01 cm. For more detail please see the
MATLAB code at the end of the example problem solution.
A plot of the species molar flow rates as a function of distance is shown in Figure 4
below. It can be seen that equilibrium is reached within the first cm of the reactor. We
note that the same results were obtained with x =0.001 cm such that step size does not
influence the results.
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Figure 4. Molar flow rates as a function of distance for the water gas shift reaction
without the palladium membrane.
Step 2) The exit CO flow rate is 2.4 x 10-4 mol/min. This corresponds to a CO conversion
of:
X 
FCO , feed  FCO ,outer,801
FCO ,outer,801

1.1  10 3  2.4  10 4
 78%
1.1  10 3
(22)
Summary: Equilibrium is rapidly achieved within the first cm of reactor length. As will
be seen in the example problem, the longer reactor length will facilitate hydrogen
separation into the separation zone.
Matlab Code: Following is the Matlab code for this example problem.
%
% water-gas shift reaction
% CO + H2O <-> CO2 + H2
% without palladium membrane
%
% reaction occurs in annulus
% iron-chromium oxide catalyst
%
% catalyst is 8cm high, 12.1 g, at 673 K
%
% feed conditions
% a = CO
% b = H2O
% c = CO2
% d = H2
% e = Ar
%
clear
figure(1)
close
%
Fa0=1.1e-3; %mol/min
Fb0=1.1e-3;
Fc0=0;
Fd0=0;
%
Ca0=1.8e-5; %mol/cm^3
Cb0=1.8e-5;
Cc0=0;
Cd0=0;
%
Ftot0=Fa0+Fb0+Fc0+Fd0;
Ct0=Ca0+Cb0+Cc0+Cd0;
%
% set up numerical model
dx=0.01;
x(1)=0;
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%
Fa(1)=Fa0;
Fb(1)=Fb0;
Fc(1)=Fc0;
Fd(1)=Fd0;
%
Ca(1)=Ca0;
Cb(1)=Cb0;
Cc(1)=Cc0;
Cd(1)=Cd0;
%
for i=1:8/dx;
x(i+1)=x(i)+dx;
%
r=7.4e8*(Ca(i)*Cb(i)-Cc(i)*Cd(i)/11.92)/(1+2.4e5*Cb(i)+7.2e5*Cc(i));
%
Fa(i+1)=Fa(i)-r*dx;
Fb(i+1)=Fb(i)-r*dx;
Fc(i+1)=Fc(i)+r*dx;
Fd(i+1)=Fd(i)+r*dx;
Ftot=Fa(i+1)+Fb(i+1)+Fc(i+1)+Fd(i+1);
%
Ca(i+1) = Ct0*Fa(i+1)/Ftot0;
Cb(i+1) = Ct0*Fb(i+1)/Ftot0;
Cc(i+1) = Ct0*Fc(i+1)/Ftot0;
Cd(i+1) = Ct0*Fd(i+1)/Ftot0;
%
end
%
figure(1)
plot(x,Fa)
hold on
plot(x,Fb,'r--')
plot(x,Fc,'g-.')
plot(x,Fd,'k:')
xlabel('Distance, cm')
ylabel('Molar Flow Rate, mol/min')
legend('CO','H_2O','CO_2','H_2')
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Home Problem Statement: Consider a reaction zone feed of 1.1 x 10-3 mol/min CO and
1.1 x 10-3 mol/min H2O at 2 atm pressure. In the separation zone the feed is 1.8 x 10-2
mol/min Ar at 1 atm pressure. Recall that the reactor length is 8 cm.
a) Modify the numerical model to predict the molar flow rates of CO, H2O, CO2, and
H2 as a function of distance if hydrogen is allowed to permeate through the
palladium membrane (thus, in your simulations assume j ≠ 0). Use a step size x
= 0.01 cm. Compare your results with the example problem (when there is no
membrane). What is the CO conversion in this system?
b) Determine the equilibrium partial pressures if you change the reaction zone feed
to 7.0 x 10-4 mol/min CO and 1.5 x 10-3 mol/min H2O at 2 atm pressure if there is
no membrane separation of hydrogen (thus, in this problem assume j = 0).
c) Repeat part a) but with reaction zone feeds of 7.0 x 10-4 mol/min CO and 1.5 x
10-3 mol/min H2O (thus, in your simulations assume j ≠ 0).
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