CACHE Modules on Energy in the Curriculum
Fuel Cells
Module Title: Equation of State for Hydrogen Fuel
Module Author: Jason Keith
Author Affiliation: Michigan Technological University
Course: Thermodynamics
Text Reference: Sandler (2nd ed.), section 4.4
Smith, Van Ness, and Abbott (7th ed.), section 3.5
Concepts: Gas law, equation of state
Problem Motivation:
Fuel cells are a promising alternative energy technology. One type of fuel cell, a proton
exchange membrane fuel cell reacts hydrogen and oxygen together to produce electricity.
Fundamental to the design of fuel cells is an understanding of nonideal equations of state,
as hydrogen fuel is often stored at high pressure. After completing this module, you will
be able to size compressed fuel tanks for fuel cell applications.
Consider the schematic of a compressed hydrogen tank feeding a proton exchange
membrane fuel cell, as seen in figure 1 below. The electricity generated by the fuel cell is
used here to power a laptop computer. We are interested in determining the impact of the
nonideal behavior of the compressed hydrogen gas on the fuel cell performance.
Computer
(Electric Load)
H2 feed line
Air in
Anode
Gas
Chamber
Cathode
Gas
Chamber
Air / H2O out
H2 out
H2 tank
Fuel Cell
Problem Information
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Example Problem Statement: In this example problem we will determine the number of
moles of hydrogen gas in a tank at 2000 psi and 298 K. The tank volume is 49.9 L. We
will utilize the Peng-Robinson equation of state. Then, compare with the ideal gas law
result which is 278 mol.
The Peng-Robinson equation of state is given by:
P
RT
a 2

1  b 1  2b  b 2  2
where P is the pressure,  is the density, R is the gas constant, T is the absolute
temperature, and a and b are coefficients for the Peng-Robinson equation of state. They
are given by the following relationships (which are defined in standard chemical
engineering thermodynamic textbooks) and can be calculated knowing the critical
temperature Tc and pressure Pc of the gas under pressure. Here a = ac where ac =
0.45723553 R2Tc2/Pc with  = [1+(1-Tr1/2)] 2, with Tr = T/Tc, b = 0.07779607 RTc/Pc,
and  = 0.37464 + 1.54226  – 0.26993 2. Note that is the acentric factor and relates
to the compressibility of the gas.
Additional information:
Critical properties of hydrogen gas: Tc = 33.3 K, Pc = 1.297 MPa,  = -0.215.
Example Problem Solution:
For simplicity all terms in the Peng-Robinson equation of state will be in atmospheres.
The tank pressure is thus 136.1 atm. Since the pressure and temperature are known we
will solve for the density. Note that the Peng-Robinson equation of state is a cubic
equation of state. This means there are three answers for the gas density. We may have to
decide which answer is appropriate to use.
a) We will calculate the a and b parameters for use in the equation of state.
For the given value of we calculate  = 0.0306.
The reduced temperature is T/Tc = 298.15/33.3 = 8.95. Thus, we can calculate the
parameter  = 0.882. To determine the value of a in the Peng-Robinson equation of state,
we need to calculate ac.
2
L2 atm
0.45723553  0.08206 L atm  33.3 2 K 2 0.101325 MPa


 0.2667
We have a c 
mol K
atm
mol 2

 1.297 MPa
ac = 0.2667 L2atm/mol2,
from which we can calculate
a = ac = 0.2667 (0.882)
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a = 0.2352 L2atm/mol2.
Furthermore, we can calculate
0.07779607  0.08206 L atm  33.3 K 0.101325 MPa
L


b
 0.0166
mol K
atm
mol

 1.297 MPa
b) The a and b parameters can be inserted into the Peng-Robinson equation to solve for
the density. This can be done by graphing
RT
a 2

P
1  b 1  2b  b 2  2
as a function of the density . A single real root occurs when  = 5.286 mol/L, as seen in
the figure below.
Alternatively one can use a root finding program using the ideal gas density as an initial
guess.
c) Multiplying the gas density by the tank volume gives approximately 264 mol. The
relative error when compared with the ideal gas law is given as 100(278 – 264)/278 =
5%. If we had used the ideal gas law, and predicted for 278 moles of hydrogen that the
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fuel cell would operate for 1 hour, in reality, with 264 moles the fuel cell would operate
for 57 minutes.
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Home Problem Statement: A compressed gas cylinder is full of hydrogen gas at room
temperature and 2000 psig pressure at 298 K. The volume of the cylinder is 49.9 L.
Using the Van der Waal’s equation of state
P
RT
 a 2
1  b
where P is the pressure,  is the density, R is the gas constant, T is the absolute
temperature, and a and b are coefficients for the Van der Waal’s equation of state. They
are given by the following relationships (which are defined in standard chemical
engineering thermodynamic textbooks) and can be calculated knowing the critical
temperature Tc and pressure Pc of the gas under pressure. Here a = 27/64 R2Tc2/Pc and b =
0.125 RTc/Pc where Tc is the critical temperature and Pc is the critical pressure.
Determine the number of moles of hydrogen and compare with the value using the ideal
gas law, 278 mol.
Additional information:
Critical properties of hydrogen gas: Tc = 33.3 K, Pc = 1.297 MPa,  = -0.215.
Home Problem Solution:
Step a. The Van der Waal’s equation of state is a cubic equation of state that accounts for
the volume of the molecules and the fact that the molecules collide with each other. There
are correction factors to the ideal gas law that reflect these physical phenomena within
this equation of state. For simplicity all terms in the equation:
P
RT
 a 2
1  b
will be computed in units of atmospheres. Thus, P = 2000 psi = 136.1 atm. Since the
pressure and temperature are known we will solve for the density  . With Tc = 33.3 K
and Pc = 1.297 MPa we have
27 R 2Tc2 27 0.08206 2 L2 atm 2
a

64 Pc
64
mol 2 K 2
33.3 2 K 2
 0.2461 L2 atm/mol 2
6
10 Pa
atm
1.297MPa
MPa 1.01325  10 5 Pa
And
b  0.125
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RTc
0.08206L atm
 0.125
Pc
mol K
33.3K
 0.02668 L/mol
10 Pa
atm
1.297 MPa
MPa 1.01325  10 5 Pa
6
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Step b. Noting that RT = 24.45 L atm/mol, we can substitute into the equation of state to
give the expression:
136.1atm 
24.45
 0.2461 2
1  0.02668
The plot below illustrates the dependence of the expression
1
24.45
 0.2461 2  136.1  0
1  0.02668
on the gas density . There is only one real root at  = 5.038 mol/L.
Step c. From this we can multiply by the tank volume of 49.9 L to give the number of
moles = 250 mol, a difference of 9.5% when compared to the ideal gas law. If we had
used the ideal gas law, and predicted for 278 moles of hydrogen that the fuel cell would
operate for 1 hour, in reality, with 250 moles the fuel cell would operate for 54 minutes.
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