Download Efficiency and Open Circuit Voltage (OCV)

Efficiency and Open Circuit Voltage (OCV)
Why we have to study efficiency of fuel cell?
From the definition and calculation of efficiency of fuel cell, the limitation of Fuel Cell
can be discussed. Also, it can give us
• OCV
• Pressure effect
• Gas concentration effect
• Temperature effect
The electrical power/energy can be defined as following:
Power = VI (w) or Energy = VIt (J)
However, the energy of chemical input and output is not so easily defined. It can be said
as “chemical energy”, but the “chemical energy” is NOT simple defined.
There are several terms to describe the “chemical energy”:
• Internal energy
• Gibbs free energy
• Enthalpy
• Helmhotz function
These terms are defined in the thermodynamics. So, let’s review it before discuss more
further.
Thermodynamics Review
Internal energy
A fuel cell converts energy stored within a fuel into other, more useful forms of energy.
The total intrinsic energy of a fuel is quantified by a property known as internal energy
(U).
First Law
The first law of thermodynamics is also known as the law of conservation of energy.
That means the energy can never be created or destroyed. It can be expressed as
following equation:
dEuniv = dEsystem + dEsurroundings = 0
or
dEsystem = -dEsurroundings
The equation states the change of energy of a (closed) system must be equal to the
energy transfer to the surroundings. There are two ways that energy can be transferred
between a (closed) system and its surroundings: via heat (Q) or work (W). So, the above
equation can be rewritten as:
dU = dQ - dW
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dWmech = pdV and
dW = dWmech + dWelectrical
The expression for the internal energy change of a system is:
dU = dQ – pdV – dWelectrical
Second Law
The second law of thermodynamics introduces the concept of entropy. Entropy is
determined by the number of possible ways of configuring a system. Entropy can be
thought of as a measure of “disorder,” since an increasing entropy indicates an increasing
number of ways of configuring a system.
Except for extremely simple system, it is impossible to calculate entropy exactly. Instead,
a system’s entropy is usually inferred based on how heat transfer causes the entropy of
the system to change.
Q dS = T rev
and
Q dS > T irr
so,
dS Q
T
From above equation, it can be rewrite as following:
dS =
Q
+ S gen
T
It can be concluded that
Sgen 0
Enthalpy
The enthalpy (H) is a description of thermodynamic potential of a system, which can be
used to calculate the "useful" work obtainable from a closed thermodynamic system
under constant pressure. The energy needed to create a system plus the work needed to
make room for it.
H = U + pV
dH = dU + Vdp + pdV
and dU = TdS – pdV
dH = TdS + Vdp
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Thermodynamic Potentials
The thermodynamic potentials indicate how energy can be transferred from one form to
another based on the first and second laws of thermodynamics.
From the first and second law of thermodynamics and ignore the electrical work,
dU = dQ – pdV
and
Q dS = T rev
the equation can be combined as
dU = TdS – pdV
Two useful relations are driven which show the dependent variables (T and P) are related
to variables in the independent variables (S and V):
dU =T
dS V
and
dU = p
dV S
Unfortunately, S and V are not easily measurable in most experiments (there is no such
thing as an “entropy meter.”). Therefore, a new thermodynamic potential is needed
equivalent to U but depending on quantities that are more readily measured than S and V.
Temperature T and pressure P fall into this category.
Helmhotz function
Helmholtz energy (A) is a thermodynamic potential which measures the “useful” work
obtainable from a closed thermodynamic system at a constant temperature. The energy
needed to created a system and make room for it minus the energy that you can get from
system’s environment due to heat transfer.
dF = dU – SdT – TdS
and dU = TdS – pdV
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dF = -SdT – pdV
Gibbs free energy
The Gibbs free energy (G) can be considered to be the net energy required to created a
system and make room for it minus the energy received from the environment due to
heat transfer. In the other words, G represents the net energy cost for a system created at
a constant environmental temperature T from a negligible initial volume after subtracting
what the environment automatically supplied.
and
dG = dU – SdT – TdS + Vdp + pdV
where dU = TdS – pdV
dG = -SdT + Vdp
If G represents the net energy you had to transfer to create system, then G should also
represent the maximum energy that you ever get back out of the system. For Fuel cell,
the Gibbs free energy is important.
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In fuel cell, the “Gibbs free energy of formation”, Gf, is used instead of the “Gibbs
free energy” since Gf gives us the energy released. This change is the different between
the Gibbs free energy of products and the Gibbs free energy of the inputs or reactants.
Gf = G of products – G of reactants
Standard State
Because most thermodynamic quantities depend on temperature and pressure, it is
convenient to reference everything to a standard set of conditions. This set of conditions
is called the standard-state. Standard-state conditions are taken as room temperature
and atmospheric pressure. (298.15K and 1atm = 101.325kPa ~ 1bar)
The standard-state conditions are also known as standard temperature and pressure, or
STP. Standard-state condition are designated by a superscript zero, i.e., Po, To, Ho, Go, or
So .
Molar Quantities
Typical notation distinguishes between intensive and extensive variables. Intensive
quantities such as temperature and pressure are independent of system size. In the other
hand, the extensive quantities, i.e. internal energy or entropy, are dependent of system
size.
It is conventional to denote intensive quantities with a lowercase letter (p, pressure) and
extensive quantities with an uppercase letter (U, internal energy).
Molar quantities are intensive. It can be defined as physical quantities per mole and
denoted as lowercase letter with bar. For instant,
•
•
•
•
Internal energy, u (kJ/mol)
Enthalpy, h (kJ/mol)
Entropy, s (kJ/mol)
Gibb free energy, g (kJ/mol)
For the quantities at STP,
•
•
•
•
Internal energy, u o (kJ/mol)
Enthalpy, h o (kJ/mol)
Entropy, s o (kJ/mol)
Gibb free energy, g o (kJ/mol)
For Gibbs free energy of formation, the denotation is g f .
Reversibility
From the below figure, it shows a ball has potential energy (PE=mgh) but no kinetic
energy (KE) in position A. When this ball is released at point A, the PE starts to convert
to the KE. Until the ball reaches point B, the PE has been converted to KE completely
if there is no friction loss, wind resistance. This process without energy loss is
reversible which means the ball can roll up the other side and recover its PE.
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Simple Reversible Process
The energy loss involved in the process is called irreversible process. Once the
irreversible process happens, the system cannot get back to the original status.
Calculating Reaction Enthalpies
For a reaction aA + bB mM + nN, the formation of enthalpy (reaction of enthalpy)
also can be defined at STP.
h of = mh o M + nh o N a h o A + b h o B An expression analogous to above equation may be written for the standard-state
entropy of a reaction.
s of = ms o M + ns o N a s o A + b s o B Ex:
Calculate the h of and s of for the methanol combustion.
CH3OH + 3/2 O2 CO2 + 2H2O (liquid)
Solution:
Chemical Species
h of (kJ/mole)
s of (J/mole-K)
CH3OH
-200.95
239.83
O2
0
205.14
CO2
-393.51
213.80
H2O (liquid)
-285.83
69.95
h o f = h o CO2 + 2 h o H 2O h o CH 3OH + 1.5 h o O2 =-719.19 kJ/mole
o
o
o
o
o
s f = s CO2 + 2 s H 2O s CH 3OH + 1.5 s O2 =-193.84 J/mole-K
Temperature dependence of enthalpy
hf = ho f +
T
To
C p (T )dT
where T0 = 298.15 K; Cp is the constant-pressure heat capacity of the substance.
Temperature dependence of entropy
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s = so +
T
C p (T )
To
T
dT
where T0 = 298.15 K; Cp is the constant-pressure heat capacity of the substance.
Calculating Gibbs Free Energy of Formation
Recall the equation of Gibbs free energy of formation:
Gf = G of products – G of reactants
It can be re-written in molar quantities form as following:
g f = g prod g react
Consider a reaction
aA + bB mM + nN
The Gibbs free energy of formation can be calculated as
[
] [
g f = mg M + ng N a g A + b g B
]
Ex:
Calculate the Gibbs free energy of formation of the reaction H2 + O2 H2O at STP.
Solution:
From the reaction equation, the Gibbs free energy can be expressed as:
g f = g H 2O g H 2 0.5 g O2
From thermodynamics properties table, the Gibbs free energy at STP (1atm, 25oC) can
be found.
g H 2O = 306.69 kJ mole
g H 2 = 38.96 kJ
mole
= 61.12 kJ
mole
g O2
So,
g f = (-306.69)-(-38.96+0.5*(-61.12)) = -237.13 (kJ/mole)
The other way to determined the Gibbs free energy when the thermodynamic properties
of the Gibbs free energy are lacked:
Recalling the definition of G and H, where
G = U TS + PV and H = U + PV
It can be concluded as:
G = H - TS
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Differentiating the expression gives
dG = dH – TdS – SdT
At the isothermal process (constant temperature), this relationship in terms of molar
quantities gives
g f = h f T s f
Ex:
Determine the Gibbs free energy of formation of the reaction H2 + O2 H2O at
STP.
Solution:
The enthalpy and entropy properties can be found from the thermodynamic table as
following:
h H 2O = 285.83 kJ mole
h H 2 = 0 kJ mole
s H 2O = 69.95 J mole K
and
h O2 = 0 kJ mole
s H 2 = 130.68 J mole K
sO2 = 205.0 J mole K
h f = (-285.83)-(0+0.5*0) = -285.83 (kJ/mole)
s f = (69.95)-(130.68+0.5*205.0) = -163.23 (J/mole-K)
g f = h f T s f =-285830 – (163.23 * 298.15) = -237.16 kJ/mole
2.1 Reversible Open Circuit Voltage (Reversible OCV)
Again, from the definition of Gibbs free energy,
dG = dU – SdT – TdS + Vdp + pdV
and, the internal energy is
dU = dQ - dW
dU = TdS – pdV – dWelectrical
which yields dG as
dG = – SdT + Vdp – dWelectrical
For a constant-temperature, constant-pressure process (dT, dP = 0) , the equation
reduces to
dG = – dWelectrical
For a reaction using molar quantities, the equation can be written as:
Welectrical = g f
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Remember that the constant temperature, constant pressure assumption used here is not
really as restrictive as it seems. The only limitation is that the temperature and pressure
do not vary during the reaction process. Since fuel cells usually operate at constant
temperature and pressure, this assumption is reasonable. It is important to realize that
the expression derived above is valid for different values of temperature and pressure as
long as these values are not changing during the reaction. We could apply this equation
for T = 200 K and p = 1 atm, or just as validly apply it for T = 400 K and p = 5 atm.
For the electrical work,
Welectrical = charge (C) * voltage (J/C) = zFE (J)
Where F is Faraday constant which means the charge of a mole of electrons. z is the
number of electrons produced per mole of the fuel. N is Avogadro’s number, e is the
charge on one electron.
F = Ne = 6.022 10231.602 10-19 = 96485 C
So,
zFE = g f
The reversible OCV of a fuel cell is,
E=
g f
zF
Ex:
For PEMFC, please determine its reversible open circuit voltage at STP.
Solution:
The PEMFC’s overall reaction is H2 + O2 H2O which produced 2 mole electrons
per mole fuels. So that,
z = 2.
From the reaction equation, the Gibbs free energy can be expressed as:
g f = g H 2O g H 2 0.5 g O2
From thermodynamics properties table, the Gibbs free energy at STP (1atm, 25oC) can
be found.
g H 2O = 306.69 kJ
mole
g H2
= 38.96 kJ
mole
g O2
= 61.12 kJ
mole
So,
g f = (-306.69)-(-38.96+0.5*(-61.12)) = -237.13 (kJ/mole)
The reversible open circuit voltage can be calculated:
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E=
g f
zF
= - (-273130) / (2 * 96485) = 1.23 (V)
The Effect of Temperature, Pressure, and Gas Concentration
The reversible open circuit voltage of a fuel cell is determined at the standard-state.
However, a fuel cell usually operates at the conditions that vary greatly from the STP.
For instant, the SOFC operate at 700~1000 oC. In that case, the Gibbs free energy
changes in a chemical reaction vary with temperature, Pressure, and gas concentration.
Thus, the reversible OCV will be changed.
Reversible OCV Variation with Temperature
To understand how the reversible voltage varies with temperature, we need to go back to
the original differential expression for Gibbs free energy:
dG = -SdT + VdP
At the constant pressure,
dG = S
dT p
Use the molar quantities of Gibbs free energy of formation, the equation becomes,
d g
dT
( ) p
= s
from the equation,
zFE = g f
The variation of reversible OCV with temperature can be expressed as following.
dE s
=
dT p nF
Therefore, the voltage varies as a function of temperature can be derived.
E =
s
(T T o )
zF
So, the reversible OCV varied with temperature (ET) are
ET = E o + E = E o +
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s
(T T o )
zF
s is assumed to be independent of temperature. If more accurate value of ET is needed,
it may be calculated by integrating the heat capacity related temperature dependence of
s .
From the equation,
• If s for a chemical reaction is positive, then ET will increase with temperature.
• If s for a chemical reaction is negative, then ET will decrease with temperature.
For most fuel cell reactions, s is negative; therefore reversible fuel cell voltages tend to
decrease with increasing temperature.
For a PEMFC, s =-44.43 J/mole-K. The variation of cell voltage with temperature is
approximated as
ET = E o +
s
44.43
(T T o ) = E o +
(T T o ) = E o 2.3 104 (T T o )
zF
2 96485
Thus, for every 100 degrees increase in cell temperature, there is an approximate 23 mV
decrease in cell voltage.
Should we operate the fuel cell at low temperature since the reversible OCV
decreases by increasing the operation temperature? NO! Activation losses tend to
decrease with increasing temperature. Therefore real fuel cell performance typically
increases with increasing temperature even though the thermodynamically reversible
OCV decreases.
The other way to determine the reversible OCV reversible OCV varied with temperature
(ET) is,
[
] [
g f (T ) = mg M (T ) + ng N (T ) a g A (T ) + b g B (T )
E=
]
g f (T )
zF
For PEMFC at 400 K,
[
] [
]
g f (T ) = g H 2O (T ) g H 2 (T ) + 0.5 g O2 (T )
=(-317.89)-(-52.73)-0.5(82.51) = 221.405 kJ/mol
Thus,
E=
g f (T )
zF
=
221405
= 1.147V
2 96485
Reversible OCV Variation with Pressure and Gas Concentration
Consider a general reaction such as
jJ + kK mM
In the case of gases behaving as “ideal gases”, the activity, a, can be defined as
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ai =
Pi
Po
where
P is the pressure or partial pressure of the gas
PO is standard pressure, 0.1 MPa.
The activity is proportional to partial pressure. For above reaction equation, the activities
can be shown.
PJ
P
P
a = Ko ; a M = Mo
o ; K
P
P
P
aJ =
The activities of the reactants and products modify the Gibbs free energy change of a
reaction. It can be expressed that in above chemical reaction.
a j ak
o
g f = g f RT ln J m K
aM
For PEMFC, H2 + O2 H2O
g f =
o
g f
1 2
a H 2 aO2 RT ln
a H 2O If the activity of reactant increases, g f becomes more negative, that means more
energy is released.
Plug into the equation
zFE = g f
The equation of reversible OCV becomes
E=
g of
zF
+
k
RT a Jj a K
ln m
zF a M
k
RT a Jj a K
= E o +
ln m
zF a M
The equation in terms of product and reactant activity is called Nernst equations.
From the definition of activity,
aJ =
PJ
P
P
a = Ko ; a M = Mo
o ; K
P
P
P
and Po = 1 bar with all pressure are given in bar, the equation is simplifies to
E = Eo +
RT PJ j PKk ln
zF PMm 30
All pressures in the equation are partial pressure. If the system pressure is P, and , , are constants depending on the molar masses and concentrations of the species.
PJ = P; PK = P; PM = P
The equation becomes
E = Eo +
k
k
RT Jj K ( j +km) RT Jj K
= E o +
ln
m P
ln
zF M
zF Mm
RT
+
( j + k m) ln( P)
zF
The second term of equation can be seen as the effect is occurred by the concentration
of fuels. The third is the effect of the system pressure.
For PEMFC, H2 + O2 H2O
1 2
RT PH 2 PO2 E=E +
ln
zF PH 2O o
1 2 RT
RT
E = Eo +
ln
ln( P)
+
zF 4F
Partial Pressure
In a mixture of gases, the total pressure is the sum of all the “partial pressure” of the
components of the mixtures.
Partial Pressures of Atmospheric Gases
Gas
Partial Pressure (MPa)
Nitrogen
0.07809
Oxygen
0.02095
Argon
0.00093
Others (including CO2)
0.00003
Total
0.10000
System pressure
The system pressure effect can be described by the last term of Nerst equation.
For a PEMFC,
RT
( j + k m) ln( P)
zF
RT
ln( P)
4F
If the pressure changes from P1 to P2, the voltage difference is,
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V =
P RT
ln 2 4F
P1 A similar effect occurs when studying the change from air to oxygen. Recall the Nerst
equation of a PEMFC,
1 2 RT
RT
E = Eo +
ln
ln( P)
+
2F 4F
E = Eo +
RT 1 RT
RT
ln +
ln( ) +
ln( P)
2F 2 2F
4F
So, the voltage change caused by using oxygen instead of air is,
V =
RT
1 8.314 (80 + 273) 1 RT
ln 2 =
ln
ln
=
= 0.012V
0.21 0.21 4 96485
4F
1 4F
In fact, there are many reports show the change of voltage will more significant than the
prediction of Nerst equation that is because the pressure effect will reduce the losses.
Efficiency and Efficiency Limit
The efficiency of a fuel cell is not straightforward to be defined. From the previous
section in this chapter, the formation of Gibbs free energy is used to define the useful
energy in a fuel cell. If it were not for irreversibilities, all this energy would be converted
into electrical energy, and the efficiency could be said to be 100%. So, the fuel cell
efficiency definition can be expressed as:
electrical energy produced
Gibbs free energy change
However, it is not very useful, and is rarely done, as whatever conditions are used the
efficiency limit is always 100%.
In contrast to a fuel cell, the maximum theoretical efficiency of a conventional
heat/expansion engine is described by the Carnot cycle. This efficiency may be derived
from classical thermodynamics. It can be expressed as:
Carnot =
T H Tc
T
= 1 C
TH
TH
In the expression, TH is the maximum temperature of heat engine and TL is the rejection
temperature of the heat engine. For a heat engine that operates at 400 oC (673 K) and
reject heat at 50 oC (323 K), the efficiency is 52%.
Let’s consider a hydrogen-oxygen reaction, no wonder that occurs in a heat engine or a
fuel cell, the total input energy, the energy to produce water (said formation of enthalpy,
h f ), is the same. So, the suitable definition of a fuel cell is the ration of useful energy to
total energy.
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However, there are two different values that we can use for h f . For the reaction which
produces steam, h f = -243.83 kJ/mole. For the product water is condensed back to
liquid, h f =-285.84 kJ/mole.
The higher figure is called the higher heating value (HHV) and the lower one is called the
lower heating value (LHV). Which of these values should be used in computing a fuel
cell’s efficiency? The most equitable calculations of fuel cell efficiency use the HHV.
Using the HHV instead of the LHV is appropriate because it acknowledge the true total
heat that could theoretically be recovered from hydrogen combustion reaction. Use of
the LHV will result in higher, but perhaps misleading, efficiency numbers.
From the above discussion, it can be concluded that the efficiency limit (or reversible
efficiency, maximum efficiency) can be written as:
=
g f
h f
100%
If the equation of efficiency can be expressed in voltage, the calculation can be more
convenient to perform.
Eh =
h f
zF
=1.48 V if using the HHV
So that the efficiency becomes,
=
E g f
V
=
=
100%
Eh h f 1.48
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