CACHE Modules on Energy in the Curriculum
Fuel Cells
Module Title: Simulation of a Methane Steam Reforming Reactor
Module Author: Jason Keith
Author Affiliation: Michigan Technological University
Course: Kinetics and Reaction Engineering
Text Reference: Fogler (4th edition), Sections 4.7, 4.8, 4.9
Literature References: J. Xu and G. Froment, AIChE J., 35, 88 (1989), J. Xu and G.
Froment, AIChE J., 35, 97 (1989), F. A. N. Fernandes and A. B. Soares, Lat. Am. Appl.
Res., 36(3) 155 (2006).
Concepts: Develop a numerical model to predict the conversion and hydrogen yield
within a steam reforming 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. The focus of
this module is hydrogen generation by steam reforming of methane 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
H2
H2O
H2O
+
H
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.
1st Draft
2nd Draft
J.M. Keith
Page 1
October 14, 2008
March 20, 2009
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 can also be excess water in the reformate stream.
The steam reforming reaction is given as:
CH4 + H2O ↔ 3 H2 + CO
(1)
In the steam reformer, the water gas shift reaction also takes place as:
CO + H2O ↔ H2 + CO2
(2)
Adding together the steam reforming and water gas shift reactions gives the overall
reaction:
CH4 + 2 H2O ↔ 4 H2 + CO2
(3)
The equilibrium constants can be expressed in terms of partial pressures (in atm) and
temperature in degrees Kelvin as [J. R. Rostrup-Nielsen and K. Aasberg-Petersen,
“Steam Reforming, ATR, Partial Oxidation; Catalysts and Reaction Engineering,” Ch. 14
of Handbook of Fuel Cells: Fundamentals, Technology, and Applications, Vol 3., W.
Vielstich, A. Lamm, H. A. Gasteiger, eds., Wiley, 2003]. The subscript on the following
equilibrium constants refers to the equation number given above:
K1 
PH3 2 PCO
 exp( 30.42  27106 / T )
PCH 4 PH 2O
(4)
K2 
PH 2 PCO 2
 exp( 3.798  4160 / T )
PCO PH 2O
(5)
PH4 2 PCO 2
K3 
 exp( 34.218  31266 / T )
PCH 4 PH2 2O
(6)
In the reactor, methane (CH4) and water (H2O) are fed as reactants and carbon dioxide
(CO2), carbon monoxide (CO), and hydrogen (H2) are produced over a nickel catalyst on
an alumina support.
In laboratory experiments, a nonreacting inert gas such as helium (He) may also be
present. In the most general form, the governing conservation equations for each of these
species is given below, where Fi denotes the molar flow rate of species i in mol/h, W
1st Draft
2nd Draft
J.M. Keith
Page 2
October 14, 2008
March 20, 2009
denotes the catalyst weight in g, and Ri denotes the reaction rate of equation i in units of
mol/(g-h):
dFCH 4
0
 ( R1  R3 ) with FCH 4 (W  0)  FCH
4
dW
(7)
dFH 2O
 ( R1  R2  2 R3 ) with FH 2O (W  0)  FH0 2O
dW
(8)
dFH 2
 (3R1  R2  4R3 ) with FH 2 (W  0)  FH0 2
dW
(9)
dFCO
0
 ( R1  R2 ) with FCO (W  0)  FCO
dW
(10)
dFCO 2
0
 ( R2  R3 ) with FCO 2 (W  0)  FCO
2
dW
(11)
dFHe
0
 0 with FHe (W  0)  FHe
dW
(12)
The reaction rates are given by:
R1 
k1
PH2.25

PH3 2 PCO 
P
P

 CH 4 H 2O

K1 

DEN 2
(13)
PH 2 PCO 2 
k2 
 PCO PH 2O 

K2 
PH 2 
R2 
DEN 2
R3 
k3
PH3.25
(14)

PH4 2 PCO 2 
2
P
P

 CH 4 H 2O

K3 

DEN 2
(15)
DEN  1  K CH 4 PCH 4  K CO PCO  K H 2 PH 2 
K H 2O PH 2O
PH 2
(16)
Furthermore, the coefficients in Equations 13-16 are given by the Arrhenius relationships
as:
1st Draft
2nd Draft
J.M. Keith
Page 3
October 14, 2008
March 20, 2009
k1  4.22  1015 exp( 240100 / RT )
(17)
k 2  1.96  10 6 exp( 67130 / RT )
(18)
k 3  1.02  1015 exp( 243900 / RT )
(19)
K CH 4  6.65  10 4 exp( 38280 / RT )
(20)
K H 2O  1.77  10 5 exp( 88680 / RT )
(21)
K H 2  6.12  10 9 exp(82900 / RT )
(22)
K CO  8.23  10 5 exp( 70650 / RT )
(23)
Note that in the above expressions, R = 8.314 J/(mol-K) is the gas constant.
The reaction stoichiometry suggests that the number of moles will increase with the
distance down the reactor. Thus, for a negligible pressure drop in the reactor, the gas
expands by increasing the volumetric flow rate. The partial pressure of a chemical species
is calculated from the total pressure and the number of moles of that species.
Pi  P
Fi
Ftot
1st Draft
2nd Draft
(24)
J.M. Keith
Page 4
October 14, 2008
March 20, 2009
Example Problem Statement: Consider a feed of 10000 mol/h CH4, 10000 mol/h H2O,
and 100 mol/h H2 to a steam reforming reactor that operates at 1000 K and a 1 atm feed
pressure. Determine the molar flow rates of CH4, H2O, CO2, CO, and H2 as a function of
catalyst weight up to 382 g. Also, determine the overall methane conversion.
Example Problem Solution:
Step 1) A numerical model can be made to simulate Equations 7 – 12 with the
coefficients determined in equations 4 – 6 and 13 – 23. Using a simple Euler
discretization of the equations we have:
FCH 4,i 1  FCH 4,i  W ( R1  R3 )
(25)
FH 2O ,i 1  FH 2O 4,i  W ( R1  R2  2 R3 )
(26)
FH 2,i 1  FH 2,i  W (3R1  R2  4 R3 )
(27)
FCO ,i 1  FCO ,i  W ( R1  R2 )
(28)
FCO 2,i 1  FCO 2,i  W ( R2  R2 )
(29)
FHe,i 1  FHe,i
(30)
subject to the initial conditions FCH 4,0  FH 2O ,0  10000 mol/h and FH 2, 0  100 mol/h.
The other chemicals have zero initial molar flows: FCO 2,0  FCO ,0  FHe,0  0 mol/h.
These equations can be solved iteratively until the end of the reactor is reached. The
procedure is as follows:
1) Calculate the rate constants k1, k2, k3, KCH4, KH2O, KCO, KH2 and use them to
compute the reaction rates R1, R2, R3 at the feed conditions (location 0, total
catalyst weight W = 0).
2) Calculate chemical flow rates (location 1, catalyst weight = W) using Equations
25 – 30.
3) Calculate the total and partial pressures using Equation 24.
4) Calculate the rate constants k1, k2, k3, KCH4, KH2O, KCO, KH2 and use them to
compute the reaction rates R1, R2, R3 at the feed conditions (location 1, total
catalyst weight W = W).
5) Repeat steps 2-4 as you progress down the length of the reactor.
The system is simulated using a step size of W = 0.1 g. 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 catalyst weight is shown in Figure 3 below. There are some
observations to be made from this plot. First of all, as there is no helium present in the
1st Draft
2nd Draft
J.M. Keith
Page 5
October 14, 2008
March 20, 2009
feed the molar flow rate is zero everywhere in the reactor. Secondly, both the CO and
CH4 are shown as solid lines. The molar flow rate of CH4 decreases with catalyst weight,
while the molar flow rate of CO increases with catalyst weight. We also note that if W
=0.01 g the results are nearly identical.
Figure 3. Species molar flow rates as a function of catalyst weight.
Step 2) The exit CH4 molar flow rate is about 2400 mol/h. This corresponds to a CH4
conversion of:
X 
FCH 4, 0  FCH 4,exit
FCH 4, 0

10000  2400
 76%
10000
(21)
Summary: After a quick change in the first g of catalyst (see the H2O molar flow rate),
there is a slow approach towards equilibrium in the reactor.
Matlab Code: Following is the Matlab code for this example problem.
%
% steam reforming plug flow model
% includes water-gas shift reaction and overall reaction
%
% this is an isothermal model with no pressure drop!
1st Draft
2nd Draft
J.M. Keith
Page 6
October 14, 2008
March 20, 2009
%
% CH4 + H2O <-> CO + 3H2 rxn 1
% CO + H2O <-> CO2 + H2 rxn 2
% CH4 + 2 H2O <-> CO2 + 4 H2 rxn 3
%
% feed conditions
% a = CO
% b = H2O
% c = CO2
% d = H2
% e = He
% f = CH4
%
clear
figure(1)
close
%
Fa0=0; %mol/hr
Fb0=10000;
Fc0=0;
Fd0=100;
Fe0=0;
Ff0=10000;
%
Ftot0=Fa0+Fb0+Fc0+Fd0+Fe0+Ff0;
%
% partial pressures in atm
Ptot0 = 1;
Pa0 = Ptot0*Fa0/Ftot0;
Pb0 = Ptot0*Fb0/Ftot0;
Pc0 = Ptot0*Fc0/Ftot0;
Pd0 = Ptot0*Fd0/Ftot0;
Pe0 = Ptot0*Fe0/Ftot0;
Pf0 = Ptot0*Ff0/Ftot0;
%
% temperatures in K
T0 = 1000;
%
% set up numerical model
dw=1e-1;
w(1)=0;
%
Fa(1)=Fa0;
Fb(1)=Fb0;
Fc(1)=Fc0;
Fd(1)=Fd0;
Fe(1)=Fe0;
Ff(1)=Ff0;
%
Pa(1)=Pa0;
Pb(1)=Pb0;
Pc(1)=Pc0;
Pd(1)=Pd0;
Pe(1)=Pe0;
Pf(1)=Pf0;
%
1st Draft
2nd Draft
J.M. Keith
Page 7
October 14, 2008
March 20, 2009
for i=1:382/dw+1;
w(i+1)=w(i)+dw;
%
K1=exp(30.420-27106/T(i));
K2=exp(-3.798+4160/T(i));
K3=exp(34.218-31266/T(i));
%
KCH4=6.65e-4*exp(38280/8.314/T(i));
KCO=8.23e-5*exp(70650/8.314/T(i));
KH2=6.12e-9*exp(82900/8.314/T(i));
KH2O=1.77e5*exp(-88680/8.314/T(i));
%
kin1=4.2248e15*exp(-240100/8.314/T(i));
kin2=1.955e6*exp(-67130/8.314/T(i));
kin3=1.0202e15*exp(-243900/8.314/T(i));
%
DEN=1+KCH4*Pf(i)+KCO*Pc(i)+KH2*Pd(i)+KH2O*Pb(i)/Pd(i);
r1=kin1/Pd(i)^2.5/DEN^2*(Pf(i)*Pb(i)-Pd(i)^3*Pa(i)/K1);
r2=kin2/Pd(i)/DEN^2*(Pa(i)*Pb(i)-Pd(i)*Pc(i)/K2);
r3=kin3/Pd(i)^3.5/DEN^2*(Pf(i)*Pb(i)^2-Pd(i)^4*Pc(i)/K3);
%
Fa(i+1)=Fa(i)+(r1-r2)*dw;
Fb(i+1)=Fb(i)-(Fb0/Ff0)*(r1+r2+2*r3)*dw;
Fc(i+1)=Fc(i)+(r2+r3)*dw;
Fd(i+1)=Fd(i)+(3*r1+r2+4*r3)*dw;
Fe(i+1)=Fe(i);
Ff(i+1)=Ff(i)-(r1+r3)*dw;
Ftot=Fa(i+1)+Fb(i+1)+Fc(i+1)+Fd(i+1)+Fe(i+1)+Ff(i+1);
%
Pa(i+1) = Ptot0*Fa(i+1)/Ftot;
Pb(i+1) = Ptot0*Fb(i+1)/Ftot;
Pc(i+1) = Ptot0*Fc(i+1)/Ftot;
Pd(i+1) = Ptot0*Fd(i+1)/Ftot;
Pe(i+1) = Ptot0*Fe(i+1)/Ftot;
Pf(i+1) = Ptot0*Ff(i+1)/Ftot;
%
end
%
figure(1)
plot(w,Fa)
hold on
plot(w,Fb,'r--')
plot(w,Fc,'g-.')
plot(w,Fd,'k:')
plot(w,Fe,'c')
plot(w,Ff,'b')
xlabel('Catalyst Weight, g')
ylabel('Molar Flow Rate mol/hr')
legend('CO','H_2O','CO_2','H_2','He','CH_4')
1st Draft
2nd Draft
J.M. Keith
Page 8
October 14, 2008
March 20, 2009
Home Problem Statement: Consider a feed of 10000 mol/h CH4 and 100 mol/h H2 to a
steam reforming reactor that operates at 900 K and a 2 atm feed pressure.
a) Determine the molar flow rates of CH4, H2O, CO2, CO, and H2 as a function of
catalyst weight up to 382 g for H2O feed flow rates of 20000 mol/h, 30000 mol/h, 40000
mol.h. For each water molar flow feed rate, determine the methane conversion and the
exit hydrogen molar flow rate.
b) If the water feed flow rate is 20000 mol/h, determine the best choice for reactor
pressure and temperature to give a minimum of 90% methane conversion.
1st Draft
2nd Draft
J.M. Keith
Page 9
October 14, 2008
March 20, 2009