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Write out - Fuel cell system explained

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1) Stack design:
Fuel cell function at low voltage → build up the voltage to desired level by electrically
connecting cells in series to form a “stack”
The bipolar plate is used to connect multi-stack.
The Bipolar stack must be relatively impermeable to gases, sufficiently strong to withstand
stack assembly and easily mass produced.
Bipolar plates should be as thin as possible so as to minimize both the electrical resistance
between individual cells and the stack size.
2) Efficiency and Open‐Circuit Voltage
Gibbs free energy:
(2.1)
Δ𝐺𝑓 = 𝐺𝑓(π‘π‘Ÿπ‘œπ‘“π‘’π‘π‘‘) − 𝐺𝑓(π‘Ÿπ‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘‘π‘ )
Faraday constant:
(2.2)
𝐹 = 𝑁𝐴 𝑒 − = 96485 π‘π‘œπ‘’π‘™π‘œπ‘šπ‘π‘ 
Molar Gibbs free energy of formation for water, 𝐻2 ,𝑂2
1
− (𝑔𝑓 )
2
𝐻2 𝑂
𝐻2
𝑂2
2𝐻2 + 𝑂2 → 2𝐻2 𝑂
1
(Derivative from chemical reaction: {
𝐻2 + 2 𝑂2 → 𝐻2 𝑂
Δg𝑓 = (𝑔𝑓 )
− (𝑔𝑓 )
Table 2.1.
Form of water product
Liquid
Liquid
Gas
Gas
Gas
Gas
Gas
Gas
Gas
(2.3)
Temperature (°C)
Δg𝑓 (π‘˜π½π‘šπ‘œπ‘™ −1 )
25
-237.2
80
-228.2
80
-226.1
100
-225.2
200
-220.4
400
-210.3
600
-199.6
800
-188.6
1000
-177.4
1
Table 2-1: π›₯𝑔𝑓 for the reaction 𝐻2 + 2 𝑂2 → 𝐻2 𝑂 at various temperatures
For the hydrogen fuel cell, two electrons pass round the external circuit for each water molecule
produced and each hydrogen molecule used. Thus, for each mole of hydrogen consumed, 2𝑁𝐴
electrons pass round the external circuit. Given that each electron carries a unit negative charge
(𝑒 − ), the corresponding charge, in coulombs ©, that flows is
(2.4)
−2𝑁𝐴 𝑒 − = −2𝐹
Expended in moving this charge round the circuit is
πΈπ‘™π‘’π‘π‘‘π‘Ÿπ‘–π‘π‘Žπ‘™ π‘€π‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ = π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ . π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ = −2𝐹𝑉
(2.5)
If the system is thermodynamically reversible, then
Δ𝑔𝑓 = −2πΉπ‘‰π‘Ÿ π‘œπ‘Ÿ π‘‰π‘Ÿ = −
Δ𝑔𝑓
2𝐹
(2.6)
Efficiency and Its Limits
The Carnot theorem as applied to a heat engine can be expressed as:
(𝑇1 − 𝑇2 )
π‘Š
(2.7)
=
Δ𝐻
𝑇1
For a fuel cell working ideally under isothermal condition, the free change of the reaction may
be converted into electrical energy with a (Maximum) efficiency given by:
πœ‚β„Žπ‘’π‘Žπ‘‘ 𝑒𝑛𝑔𝑖𝑛𝑒 =
πœ‚π‘šπ‘Žπ‘₯ =
π‘Šπ‘šπ‘Žπ‘₯ Δ𝐺 (1 − 𝑇Δ𝑆)
=
=
ΔH
Δ𝐻
Δ𝐻
(2.8)
Efficiency and Voltage
If all the energy from the hydrogen fuel, i.e., the heating value, or enthalpy of formation were
transformed into electrical energy, the voltage would then be given by:
𝑉𝑐
(2.9)
100%(𝐻𝐻𝑉)
1.48
In practice, however, it is found that not all the fuel can be used, for reasons discussed later,
some of it usually has to pass through unreacted. A fuel utilization ecoefficiency, πœ‡π‘“ can be
defined as:
𝐢𝑒𝑙𝑙 𝑒𝑓𝑓𝑒𝑐𝑖𝑒𝑛𝑐𝑦 =
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ 𝑓𝑒𝑒𝑙 π‘Ÿπ‘’π‘Žπ‘π‘‘π‘’π‘‘ 𝑖𝑛 𝑐𝑒𝑙𝑙
(2.10)
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ 𝑓𝑒𝑒𝑙 𝑖𝑛𝑝𝑒𝑑 π‘‘π‘œ 𝑐𝑒𝑙𝑙
This parameter is equivalent to the ratio of the current delivered by the fuel cell to that which
would be obtained if all the fuel were reacted. The fuel-cell efficiency, πœ‚, is therefore given by:
πœ‡π‘“ =
πœ‚=
πœ‡π‘“ 𝑉𝑐
100%
1.48
(2.11)
Influence of Pressure and Gas Concentration
Nernst Equation
General reaction:
𝑗𝐴 + π‘˜π΅ → π‘šπΆ
(2.12)
Each of the reactants, as well as the product, has an associated ‘activitiy’, which is designated
by the symbol a. For the case of gases behaving as “ideal gases”
π‘Ž=
𝑃
π‘ƒπ‘œ
(2.13)
Where P is the pressure, or partial pressure, of the gas and π‘ƒπ‘œ is the standard pressure,
namely, 100 kPa.
The activity of a gaseous component in the system can be taken to be proportional to partial
pressure, whereas for dissolved chemicals, the activity is linked to the molarity (‘strength’) of
the solution, which is usually expressed in mol π‘‘π‘š−3. The case of the water produced in fuel
cell is somewhat difficult since this can be as either steam or liquid. For steam, the following
can be written.
𝑃𝐻2 𝑂
(2.14)
π‘ƒπ»π‘œ2 𝑂
The activities of the reactants and products modify the Gibbs free energy change of a reaction.
By using thermodynamics principles, for a chemical reaction such as the general example given
(2.15)
π‘Žπ»2 𝑂 =
𝑗
π‘Žπ΄ . π‘Žπ΅π‘˜
(2.15)
Δg𝑓 =
− 𝑅𝑇𝑙𝑛 ( π‘š )
π‘ŽπΆ
π‘œ
Where Δ𝑔𝑓 is the change in molar Gibbs free energy of formation at standard pressure.
π‘œ
Δ𝑔𝑓
For the reaction in hydrogen fuel cell, equation (2.15)becomes:
1
π‘Žπ»2 . π‘Žπ‘‚22
π‘œ
Δgf = Δ𝑔𝑓 − 𝑅𝑇𝑙𝑛 (
)
π‘Žπ»2 𝑂
(2.16)
To see how activity influences the cell voltage, Δ𝑔𝑓 can be substituted into equation (2.6) to
obtain.
π‘‰π‘Ÿ = −
π‘œ
Δ𝑔𝑓
2𝐹
+
1
π‘Žπ»2 . π‘Žπ‘‚2 2
1
π‘Žπ»2 𝑂 . π‘Žπ‘‚2 2
𝑅𝑇
𝑅𝑇
) = π‘‰π‘Ÿπ‘œ +
ln (
)
ln (
π‘Žπ»2 𝑂
2𝐹
π‘Žπ»2 𝑂
2𝐹
(2.17)
It can be assumed that the steam behaves as an ideal gas, and so:
𝑃𝐻2
𝑃𝑂2
𝑃𝐻2 𝑂
,
π‘Ž
=
,
π‘Ž
=
𝑂
𝐻
𝑂
2
2
π‘ƒπ‘œ
π‘ƒπ‘œ
π‘ƒπ‘œ
Then the Nerst equation will became:
π‘Žπ»2 =
(2.18)
1
π‘‰π‘Ÿ = π‘‰π‘Ÿπ‘œ +
𝑅𝑇
ln
2𝐹
𝑃𝐻2 𝑃𝑂2 2
π‘ƒπ‘œ . ( π‘ƒπ‘œ )
𝑃𝐻2 𝑂
π‘ƒπ‘œ
(2.19)
(
)
In nearly all cases, the pressure will be the partial pressure, if the system pressure is P that is:
𝑃𝐻2 = 𝛼𝑃, 𝑃𝑂2 = 𝛽𝑃, 𝑃𝐻𝑂 = 𝛿𝑃
Where 𝛼, 𝛽, 𝛿 are the constants that depend on the molar masses and concentrations of 𝐻2 , 𝑂2
and 𝐻2 𝑂 respectively. The Nernst equation then becomes:
1
𝑅𝑇
𝛼. 𝛽 2 1
π‘‰π‘Ÿ = π‘‰π‘Ÿ0 +
ln (
. 𝑃2)
2𝐹
𝛿
= π‘‰π‘Ÿ0 +
1
𝛼. 𝛽 2
(2.20)
𝑅𝑇
𝑅𝑇
ln (
)+
ln(𝑃)
2𝐹
𝛿
4𝐹
Hydrogen Partial Pressure:
Hydrogen can be supplied either pure or as part of a mixture. Isolation of the hydrogen pressure
term in Equation (2.20) yields:
π‘‰π‘Ÿ = π‘‰π‘Ÿ0 +
1
𝑃𝑂22
𝑅𝑇
𝑅𝑇
ln (
)+
ln(𝑃𝐻2 )
2𝐹
𝑃𝐻2 𝑂
2𝐹
(2.21)
So, if the hydrogen partial pressure changes: 𝑃1 → 𝑃2 with 𝑃𝑂2 and 𝑃𝐻2 𝑂 unchanged, then:
𝑅𝑇
𝑅𝑇
𝑅𝑇
𝑃2
(2.22)
ln(𝑃2 ) −
ln(𝑃1 ) =
ln ( )
2𝐹
2𝐹
2𝐹
𝑃1
The use of hydrogen mixture with Cox occurs particularly in phosphoric acid fuel cells (PAFCs)
that operate at about 2000 𝐢(400K). Substituting the values for R, T and F in the equation (2.22)
yields:
Δ𝑉 =
𝑃2
(2.23)
Δ𝑉 = 0.02 ln ( )
𝑃1
Fuel and Oxidant utilization
Air passes through the cathode compartment, oxygen is consumed→partial pressure is reduced.
For higher efficiency, fuel utilization should be as high as possible. OTH, (2.21) suggest that
high fuel utilization will lead to low average cell voltage or current density.
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