Download Module 28: Hydrogen Production by Natural Gas Assisted Steam

CACHE Modules on Energy in the Curriculum
Fuel Cells
Module X (Draft): H2 Production by Natural Gas Assisted Steam Electrolysis
Module Author: Michael D. Gross
Module Affiliation: Bucknell University
Course:
Membrane Separations
Text Reference:
Geankoplis 4th ed., Chapter 13
Concept Illustrated:
Problem Motivation: Fuel cells are a promising alternative energy conversion
technology. Hydrogen is currently the preferred fuel for fuel cells. The generation of pure
hydrogen is important, particularly for low temperature fuel cells such as PEMFC,
because the Pt catalyst is sensitive to impurities. One way to generate hydrogen is by
electrolysis, a process that uses electricity to decompose water into hydrogen and oxygen.
One could accomplish this by running a fuel cell backwards as discussed in Module x.
Another route would be to supply methane at the anode and water at the cathode. This
method, called Natural Gas Assisted Steam Electrolysis (NGASE), produces hydrogen
without supplying electricity from an external source. In this configuration, the
electrolyzer operates as a fuel cell and the electricity is generated by the electrochemical
reaction.
The NGASE reactions are:
Anode:
Cathode:
Overall:
Electron
Flow
(Current)
-
-
e
e
CH4
H2O
H2
O2-
CO2
H2O
CH4
CH4
O
H2O
CO2 CH4
CH4 H2O
CH4 + 4O-2  CO2 + 2 H2O + 8 e4 H2O + 8 e-  4 O-2+ 4 H2
CH4 + 2 H2O  4 H2 + CO2
O2O
2-
H2
H
Air
2O
In
Anode
Gas
Chamber
Cathode
Gas
Chamber
Fuel Cell
H2O
H
&
CO
2 2&
H22O
Out
H2
Anode
Cathode
Electrolyte
Figure 1: Reactions within NGASE
Draft 1
Cell Voltage
CH
H24
In
H2
H2
2-
Electric Load
H2
Air
Out
Figure 2: Flow Diagram for NGASE
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September 18, 2008
For each mole of methane consumed, 4 moles of H2 are generated and 8 moles of
electrons are passed through the external circuit. To convert electron flow (moles of
electrons/s) to electrical current (coulombs/s or amps), one would use Faraday’s
constant: F  96,485 coulombs / mole of electrons. The rate of hydrogen production is
directly related to the current as follows:
i  nF
where
dN H2
or
dt
dN H2
i

dt
nF
i  current density 
n
amperes coulombs

cm 2
s  cm 2
2 moles of e 1 mole of H 2
coulombs
mole of e moles of H 2

cm 2  s
F  96,485
dN H2
t
Figure 3 shows the relationship between current density, i, and electrolyzer voltage.
There are several things to note here.
0.3
Voltage, V
EOCV = 0.25 V
0.2
0.1
0.0
0.0
0.1
0.2
0.3
Current Density, A•cm-2
Figure 3. A performance curve for NGASE.
Draft 1
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September 18, 2008






The theoretical maximum voltage of this electrolyzer is 0.25 V. This is called
the “open circuit voltage” EOCV.
Any drop in voltage from EOCV is termed “overvoltage.” It is desired to
minimize the overvoltage so that the electrolyzer can operate as efficiently as
possible.
The hydrogen reaction rate is directly proportional to the current, since for
each hydrogen molecule that reacts, two electrons are formed.
For operating voltages (Eop) 0 ≤ Eop ≤ EOCV, hydrogen is produced without an
external power source. This corresponds to a possible current range of 0.25
A·cm2 to 0 A·cm2.
At current densities between 0 A/cm2 and 0.25 A/cm2, there is a linear fall in
voltage as the current density increases. Assuming fast kinetics, this voltage
drop is caused by a resistance to oxygen ion flow across the electrolyte
membrane. In this case the operating voltage can be described as Eop = EOCV –
iRA, where i is current density, R is resistance, and A is cross sectional area of
the electrolyzer. In physics and electrical engineering, this effect is referred to
as Ohm’s law.
The maximum rate of hydrogen generation corresponds to the maximum
current, 0.25 A/cm2 and Eop=0, which is called short circuit. It is always
desired to operate at short circuit since this is the highest and most
efficient rate of H2 generation.
While the NGASE method uses an electrochemical device for separation of water into
hydrogen and oxygen, it draws many parallels to typical membrane separation processes
across a dense membrane. Our primary interest is the rate of H2 production, however, O2is the species transported across the membrane and so the flux across the membrane is
written with respect to O2-. For every one mole of H2 produced, 2 moles of e- are
collected and one mole of O2- is produced at the cathode. The flux equations for an
electrolyzer and a separation with a dense membrane are compared below.
Draft 1
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September 18, 2008
cio
Ni
ciL
L
Figure 4. Concentration profile for a membrane process
Table 1. Flux equations for a membrane process and NGASE.
Equations
Units
 mol O

2
 cm  s
2-
Membrane
Separation
D
N i  i c iO  ciL 
L
N O2 
i
nF
i
i
V
E OCV  E op  E OCV  E op 
V


RA
RA
L
E
OCV
N O2 



 mol O 2 -   C
mol O 2 - mol e- 

   2 


2
C 
 cm  s   cm  s mol e
V  i  RA
NGASE
  cm 2 mol O 2  

s

cm
cm 3
 
C
C 
 cm 2 
cm 2  s
s
C
1
1
V
 V


cm 2  s
  cm cm   cm 2
 E op 
L

L  nF
E
OCV
 mol O 2 -   V
mol O 2 - mol e- 







2
2
 
C 
 cm  s     cm mol e
 E op 
N O 2 -  flux
i  current density
E OCV  open circuit vo ltage
E op  operating voltage
RA  area specific resistance
  O 2 - conductivi ty (S/cm or 1/  cm)
L  thickness of the membrane
D  diffusivit y
ciO and ciL  concentrat ions at the membrane interfaces
Draft 1
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September 18, 2008
The only difference between the flux equations for the electrolyzer and membrane
separation is that the electrolyzer takes into account electrical units. For the electrolyzer,
voltage directly corresponds to the concentration of oxygen at each membrane interface
via the Nernst equation. The concentration of oxygen is typically expressed as a partial
pressure of oxygen, PO2. For more information regarding the Nernst equation, refer to
Module 9, under Thermodynamics.
Draft 1
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September 18, 2008
Example Problem
A company is developing a new car powered by a fuel cell system that runs on H2. You
have been asked to consider generating the H2 by NGASE. The H2 tank to be used is 10
liters in volume and a fill-up requires a pressure of 500 psi.
a. The electrolyzer should be designed for maximum performance. Plot the
maximum rate of H2 generation as a function of electrolyte membrane thickness
(1 μm to 1 mm) given the following data. What is the ideal membrane thickness?
EOCV = 0.25 V
σ = 0.02 S/cm (S = 1/Ω)
b. Given the following data, how long would it take to fill the tank? Assume an ideal
gas and 25ºC.
Electrolyzer stack consists of 100 cells in series
Each cell has an area of 10 cm2
Electrolyte membrane thickness = 10 μm
Example Problem Solution
Part a
Step 1. Calculate the flux of O2-, NO2-, across the electrolyte membrane.
N O2 

L  nF
E
OCV
 E op 
The conductivity, σ, and open circuit voltage, EOCV, were given in the problem statement.
n
2 mol e mol O 2 
F  96485
C
mol e -
The maximum rate of H2 generation will occur at the maximum current density. The
maximum current density corresponds to Eop = 0 V.
For an electrolyte membrane thickness of 10 μm:
Draft 1
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September 18, 2008
N O2 

L  nF
E
1


cm
0.25 V - 0 V 
 E op  
2 mol e
C
10 10  4 cm 

96485
1 mol O 2mol e 0.02
OCV
N O 2   2.59 10 5
mol O 2cm 2  s
The oxygen ion flux as a function of electrolyte membrane thickness is summarized in
the following table:
l (cm)
1.00E-04
1.00E-03
1.00E-02
1.00E-01
NO22.59E-04
2.59E-05
2.59E-06
2.59E-07
Step 2: Calculate the rate of hydrogen generation with the oxygen flux.
Every one mole of H2O reacted, generates one mole of H2 and one mole of O2-.
Therefore, the rate of H2 generation is equal to the flux of oxygen ions.
1.E-03
dN H 2  mol H 2 


dt  cm 2  s 
1.E-04
l (cm)
1.00E-04
1.00E-03
1.00E-02
1.00E-01
1.E-05
dNH2/dt
2.59E-04
2.59E-05
2.59E-06
2.59E-07
1.E-06
1.E-07
1.E-05
1.E-04
1.E-03
1.E-02
1.E-01
1.E+00
l (cm)
Ideally, the smallest electrolyte membrane thickness possible would be used. However, in
real systems the electrolyte is typically 10 μm thick or more. Membranes thinner than 10
μm can leak, causing the reactants on either side of the membrane to mix. Once reactants
from either side of the electrolyte mix, the concentration of oxygen is equal on both sides
and EOCV = 0.
Part b
From part a, we know for a 10 μm thick electrolyte membrane
Draft 1
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September 18, 2008
dN H 2
dt
 2.59E - 05
mol H 2
cm 2  s
Total area of electrolyzer system:
# of cells in stack area of one cell   100 cells  10 cm
2
cell
 1000 cm 2
The moles of H2 required can be calculated with the ideal gas law, PV=nRT.
 1 atm 
  34 atm
P  500 psi 
 14.696 psi 
n
34 atm 10 L
 13.9 mol H 2
L  atm
0.08206
 298 K
mol  K
fill up time 
Draft 1
13.9 mol H 2
n

 540 sec onds ~ 9 min
dN H 2
mol H 2
2
1000 cm  2.59 E  04
area 
cm 2  s
dt
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September 18, 2008
Home Problem
The example problem demonstrated how the flux across an electrolyte membrane
changes with thickness. This problem addresses how the flux changes with temperature.
The conductivity of an electrolyte membrane as a function of temperature (in Kelvin) is
as follows:
 T  
3.6  10  e
T
5
-8104
RT
A company is developing a new truck powered by a fuel cell system that runs on H2, and
you have been asked to consider generating the H 2 by NGASE. The H2 tank to be used is
20 liters in volume and a fill-up requires a pressure of 750 psi.
a. The electrolyzer should be designed for maximum performance. Plot the
maximum rate of H2 generation as a function of electrolyzer temperature (500ºC
to 1000ºC) given the following data. What is the ideal operating temperature of
the electrolyzer?
Electrolyte membrane thickness = 10 μm
EOCV = 0.25 V
b. Given the following data, how long would it take to fill the tank? Assume an ideal
gas and a tank temperature of 25ºC.
Electrolyzer stack consists of 50 cells in series
Each cell has an area of 5 cm2
Electrolyzer T = 800ºC
Draft 1
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September 18, 2008