Mench M.M. Fuel Cell Engines

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c03 JWPR067-Mench December 18, 2007 1:59 Char Count=

80 Thermodynamics of Fuel Cell Systems

Here, P is the total mixture pressure and y

is the mole fraction of constituent i, deﬁned as

total

(3.38)

where n

is the number of moles of species i and n

total

is the total number of moles in the

mixture. The

and

total

terms represent the molar ﬂow rates of species i and the mixture,

respectively. The sum of the partial pressures equals the total mixture pressure and the sum

of the mole fractions equals 1:



= P and



= 1 (3.39)

Because it is so common, thermodynamic tables have been developed to describe most of

the thermodynamic properties for air. However, for other gas mixtures, one usually has to

calculate the mixture properties. For molar speciﬁc properties, the mixture value is based

on an average value among the constituents, weighted by the relative mole fractions:

mix



i=1

(3.40)

Here,

x is the molar intensive thermodynamic parameter of interest (e.g., enthalpy and

internal energy) and

is the molar intensive parameter for species i.

Example 3.5 Determination of Nonreacting Ideal Gas Mixture Properties Given a

mixture of air (21% oxygen and 79% nitrogen by volume) at 2 atm pressure and 350 K,

ﬁnd

(a) the partial pressures of each species and

(b) the mixture molar speciﬁc heat.

SOLUTION (a) Since we know the mole fractions are y

= 0.21 and y

= 0.79, we

can easily determine the partial pressures of each constituent:

= 0.21 × 2atm= 0.42 atm

= 0.79 × 2atm= 1.58 atm



= 2.0atm= P

(b) The mixture molar speciﬁc heat is calculated by ﬁrst looking up the constituent

speciﬁc heats at 350 K or calculated from Eqs. (3.25) and (3.26) and averaging them on a

molar basis:

p,mix



i=1

p,i

= 0.21

p,O

+ 0.79

p,N

= 0.21 × 29.7kJ/(kmol O

· K) + 0.79 × 29.2kJ/(kmol N

· K)

= 29.3kJ/(kmol mix ·K)

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3.3 Determination of Change in Enthalpy for Nonreacting Species and Mixtures 81

COMMENTS: Since air is commonly tabulated, we usually do not have to calculate

its mixture properties. However, you can use air tables as a reference to make sure you

understand the concepts here.

In some cases, property values are tabulated in mass-speciﬁc units (e.g., enthalpy in

kilojoules per kilogram). In this case, we can evaluate mixture properties on a mass basis:

mix



i=1

(3.41)

where the mass fraction of each species is deﬁned as

ﬁ

total

(3.42)

and the sum of the individual mass fractions equals 1:



ﬁ

= 1 (3.43)

The mixture property on a mass basis can be converted to a molar basis by

mix

· MW

mix

(3.44)

where the molecular weight of the mixture, MW

mix

, can also be calculated based on an

averaging technique as in Eq. (3.40).

For a nonreacting mixture, determination of the change in enthalpy (or other thermody-

namic properties) is the same as for a single ideal gas species, but follows ideal gas mixture

property relations discussed in this section and exempliﬁed in the following example.

Example 3.6 Change in Properties for a Nonreacting Ideal Gas Mixture Find the

change in absolute enthalpy for a 3-kg mixture of air (21% O

, 79% N

by volume) heated

from 400 to 600 K.

SOLUTION We are dealing with an ideal gas mixture. The mixture molecular weight

(MW

= 32 kg/kmol, MW

= 28 kg/kmol) is ﬁrst determined:

mix



i=1

= 0.79 · MW

+ 0.21 · MW

= 28.85 kg/kmol

To determine the change in enthalpy, we use a molar average approach. Since the mixture

is not reacting, the heats of formation cancel out and we are left with a simple calculation

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82 Thermodynamics of Fuel Cell Systems

of the sensible enthalpy change of the mixture: From Eqs. (3.36) and (3.40) the change in

enthalpy can be shown as



mix





s,i



−





s,i



= 0.21









ref

p,O

dT −



ref

p,O







+0.79









ref

p,N

dT −



ref

p,N







To solve for the molar speciﬁc enthalpy change, we can integrate the expressions using the

polynomials given in this chapter. If we assume constant speciﬁc heat, then this reduces to

mix

= 0.21

p,O

(

− T

ref

)

−

p,O

(

− T

ref

)

+ 0.79

p,N

(

− T

ref

)

−

p,N

(

− T

ref

)

and



mix

= 0.21

p,O

,ave

(

− T

)

+ 0.79

p,N

,ave

(

− T

)

Evaluating the speciﬁc heats of oxygen and nitrogen from Eqs. (3.25) and (3.26) at an

average temperature of 500 K, we can show



mix

= 0.21 × 31.09 kJ/(kmol ·K)

(

200 K

)

+ 0.79 × 29.56 kJ/(kmol · K)

(

200 K

)

= 5976 kJ/kmol

To ﬁnd the absolute enthalpy change, we need to multiply the mixture molar speciﬁc

enthalpy change by the total number of moles in the mixture:

mix

3kg

mix

3kg

28.85 kg/kmol

= 0.104 kmol

Finally,

H

mix

= n

mix



mix

= (0.104 kmol) (5976 kJ/kmol) = 621.5kJ

COMMENTS: Note that the reference temperature enthalpy cancels out of the sensible

enthalpy expression. In some thermodynamic tables, the reference point is chosen differ-

ently; therefore the absolute enthalpy calculated may be different. However, the enthalpy

change between two states will always be the same. Also, there are several ways the air

mixture properties could have been calculated. The mixture speciﬁc heat for air could have

been used and

p,mix

(

− T

)

would have given the molar speciﬁc enthalpy change. Also,

since tables abound for air, we could have read the enthalpy changes directly from an ideal

gas air table.

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3.4 Determination of Change in Enthalpy for Reacting Species and Mixtures 83

3.4 DETERMINATION OF CHANGE IN ENTHALPY FOR

REACTING SPECIES AND MIXTURES

Consider a generic chemical or electrochemical reaction:

A + ν



 

Reactants

→ ν

C + ν



 

Products

(3.45)

where the ν’s represent the coefﬁcients of the balanced electrochemical reaction. In order

to determine the change in enthalpy for the reaction, we use the following relationship to

account for the change in molar speciﬁc enthalpy between product and reactant states:



h =

−



i,P





−



i,R







j,P



◦

+ LH



−



i,R



◦

+ LH



(3.46)

where R represents reactants and P represents products. To determine the total enthalpy

change, simply multiply by the number of moles reacted of each species:

H

P−R





i=1



−







j=1





(3.47)

Here we have assumed there is no latent heat term in the enthalpy expression, although it

should be added as needed if not already accounted for in the heat of formation. Similar

expressions can be written for the change in any intensive thermodynamic parameter of

interest. For example, consider generic parameter x, where x can represent s, u,org, that

is,



x =

−



i,P

(

)

−



j,R

(

)

(3.48)

and

X

P−R





i=1



−







j=1





(3.49)

Example 3.7 Change in Enthalpy for Reacting Gas-Phase Mixture Find the change in

enthalpy per mole of hydrogen for combustion of a 50% hydrogen–50% oxygen mixture

(by mole) at 298 K. The ﬁnal product is water vapor and oxygen at 1000 K.

SOLUTION The chemical reaction can be written as

+ O

→ H

O +

where the initial mixture was 50% oxygen, 50% hydrogen by volume (mole), so there is

unreacted oxygen in the products by a simple species balance. The enthalpy per mole of

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84 Thermodynamics of Fuel Cell Systems

hydrogen is simply the molar speciﬁc enthalpy change of the reaction, since there is one

mole of hydrogen in the balanced chemical reaction.



h =



i, p



◦



−



j,R



◦



h =







1.0





1000

ref





O,v

+ 0.5





1000

ref











00 00

−











1.0





298

ref





+ 1.0





298

ref















h =







1.0





1000

298





O,v

+ 0.5





1000

298











where the sensible enthalpy of the reactants is zero, based on a 298 K reference temperature,

and the formation enthalpies of oxygen and hydrogen are zero by convention. Also, we

will use the heat of formation for gas-phase water vapor to eliminate the need to include

a separate latent-heat-of-vaporization term. We can integrate the sensible enthalpy terms

using the polynomial expressions given in this chapter. If we evaluate the speciﬁc heat

integrals using the polynomials given in Eqs. (3.25) and (3.28), after evaluation, this

reduces to



h =

1.0

(

−241,820 +29,745

)

O,v

+ 0.5

(

0 + 22,712

)

=−200,718 kJ/kmol H

COMMENTS: We could have also come close using an average speciﬁc heat value, but

in this case the polynomial approximating the speciﬁc heat behavior with temperature

was integrated exactly. For reacting mixtures, since the products at state “P” are different

from the reactants at state “R”, the mixture property mole fractions change for products

and reactants and the reference temperature enthalpy in the sensible enthalpy term does

not cancel out. In this case it is important that the reference temperature for the heat of

formation is the same as the sensible enthalpy. Also, it is interesting to see the dominance

of the enthalpy of formation compared to the sensible enthalpy. The energy released from

an exothermic chemical or electrochemical reaction is typically an order of magnitude or

more larger than the sensible enthalpy change.

Entropy (S) The concept of entropy is typically more difﬁcult to grasp than most other

thermodynamic parameters, yet it is an important part of the science of fuel cells to

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3.4 Determination of Change in Enthalpy for Reacting Species and Mixtures 85

understand. Often, students describe the entropy of a system as some nebulous measure of

“randomness.” While a dorm room may indeed be a high-entropy environment, a more fun-

damental physical representation of the meaning of entropy can be obtained from statistical

thermodynamics. From a statistical thermodynamic viewpoint, entropy is proportional to

the number of quantum microstates available, consistent with the given system constraints

and thermodynamic parameters [5]:

S = k

ln  (3.50)

where k

is Boltzmann’s constant (1.3807 × 10

−23

J/K) and  is the number of thermody-

namically available quantum microstates consistent within the macroscopic thermodynamic

constraints of the system. One can heuristically envision the number of possible microstates

available for the molecules to occupy as a set of available boxes in which to put marbles, as

shown in Figure 3.9. The more boxes available, the more possible combinations the marbles

could be distributed among, hence the randomness of the distribution is increased. From

the second law of thermodynamics, a system in equilibrium will maximize the number of

available microstates within a given set of macroscopic constraints and occupy the available

microstates with an equal randomness. This representation also satisﬁes the fundamental

principle of entropic maximization for a given system.

Practically, we need an expression for calculating entropy in terms of regularly mea-

surable thermodynamic quantities. We cannot inspect a system to determine the number of

available quantum microstates or buy an entropy meter at the store like we can purchase

(a)

ABCDEFGH

OIK

(b)

JMNL

Figure 3.9 Heuristic representation of available microstates. (a) lower entropy; (b) higher entropy.

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86 Thermodynamics of Fuel Cell Systems

a temperature or pressure sensor. Therefore, we seek to derive an expression relating the

change in entropy to things we can easily measure like temperature and pressure. Consider

the ﬁrst law of thermodynamics for a simple compressible system in an internally reversible

process with no changes in potential or kinetic energy:

dU = δ Q − δW (3.51)

From the second law of thermodynamics and the deﬁnition of compression work for this

system, we can show that

(

δ Q

)

rev

= TdS (3.52)

(

δW

)

rev

= PdV (3.53)

Plugging Eqs. (3.52) and (3.53) into Eq. (3.51) yields

dU = TdS− PdV (3.54)

From the deﬁnition of enthalpy we can show that

H = U + PV (3.55)

dH = dU + PdV+ VdP (3.56)

dH = TdS+ VdP (3.57)

Rearranging these equations and dividing by temperature and mass, we can show that

ds =

−

vdP

(3.58)

ds =

Pdv

(3.59)

These are the entropy equations we seek, which relate the entropy to properties such as

enthalpy, temperature, and pressure, which we can already measure or solve for. This was

derived considering a reversible process but is applicable for any process, since entropy is

a thermodynamic state property which does not depend on the path taken to arrive at the

ﬁnal equilibrium state.

Determination of Entropy for Ideal Gas For a nonreacting ideal gas, enthalpy and internal

energy are functions of temperature only, and the ideal gas EOS can be used to relate the P,

V, and T variables together. Using the ideal gas law along with the deﬁnitions of speciﬁc

heat, we can simplify Eq. (3.58) on a molar intensive basis as follows:

s =

(T ) dT

− R

(3.60)

Note that an analogous expression could be derived unsing the constant-volume speciﬁc

heat and Eq. (3.59), but working with pressure is often more intuitive than volumes, so we

will use this relationship from here.

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3.4 Determination of Change in Enthalpy for Reacting Species and Mixtures 87

For a change from a reference state 1 to some other state 2 for a nonreacting system,

we can integrate:

−

ref



ref

(T ) dT

− R

ref

(3.61)

ref

At 0 K

ref

(T ) dT

T departure

− R

ref

P departure

(3.62)

This is analogous to Eq. (3.34) for the absolute enthalpy except the reference temperature

chosen is different and there is a temperature and pressure dependence. Because entropy

is a measure of the number of possible microstates available, it is only zero at absolute

zero, where there is no molecular motion. Thus, the reference point of the entropy of every

substance is 0 K, 1 atm. The absolute entropy at some state 2 is the entropy caused by

departure from 0 K and 1 atm.

Evaluation of the integral in Eq. (3.61) has been done for many ideal gas substances

and has been tabulated as a function of temperature:

◦

(T ) =



(T ) dT

(3.63)

For gases that have a variable speciﬁc heat, the appropriate speciﬁc heat polynomial can be

integrated. If we are to determine the entropy change for a nonreacting ideal gas from state

1 to state 2, we can show

, P

) −

, P

) =

◦

) −

◦

)



 

f (T )

− R

  

f (P)

(3.64)

Thus, the entropy change can be broken down into temperature-dependent and pressure-

dependent portions. Tabulated values for the temperature-dependent

◦

term for various

common fuel cell species are given in the Appendix. Because this is a tabulation of the

more precise integration of the speciﬁc heat function, it is more accurate than assuming

constant speciﬁc heats.

Determination of Entropy for Nonreacting Liquids and Solids For liquids and solids,

one can safely assume incompressibility over any pressure range common to fuel cells so

that the pressure dependence of entropy is negligible. Additionally, the speciﬁc heat of

liquids and solids varies little over a wide temperature range. Using these simpliﬁcations,

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88 Thermodynamics of Fuel Cell Systems

the entropy change for a liquid or a solid between two states can be determined to be

−

c ln

(3.65)

Determination of Change in Entropy for Nonreacting Gas Mixture For a species in a

mixture we have

s(T )

◦

(

T, P

ref

)

− R

ref

(3.66)

and the reference partial pressure is still 1 atm. Note that species i now has a mixture partial

pressure, calculated with Eq. (3.37). The entropy of a mixture can be evaluated using a

molar or mass averaging among the constituents in the mixture, as in Eq. (3.40).

Determination of Change in Entropy for Reacting Gas Mixture The change in entropy

for a reacting mixture can be determined in an analogous fashion to the change in enthalpy

(of any other intensive parameter of interest) for a reacting mixture.

For the generic chemical or electrochemical reaction

A + ν



 

Reactants

→ ν

C + ν



 

Products

(3.67)

The entropy change between products and reactants can be evaluated as



s =

−



i,P

(

)

−



j,R

(

)



i,P



◦

(

T, P

ref

)

− R

ref



−



j,R



◦

(

T, P

ref

)

− R

ref



(3.68)

For the total entropy change, we can write

S

P−R



i=1

(

)

−



j=1





(3.69)

Gibbs Function (G) The Gibbs function (G) is deﬁned from other thermodynamic prop-

erties, enthalpy and entropy, so that it is itself a thermodynamic property:

G ≡ H − TS (3.70)

The Gibbs function is a measure of the maximum work possible at a given state from a

constant temperature and pressure reversible process. It can be shown from a combination

of the ﬁrst and second laws of thermodynamics for a simple compressible system that the

Gibbs function of a system will always decrease or remain the same for a spontaneous

process. Consider the differential of the Gibbs function:

dG = dH − TdS− SdT (3.71)

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3.4 Determination of Change in Enthalpy for Reacting Species and Mixtures 89

From the ﬁrst law for a simple compressible system at a given constant temperature (T)

and pressure (P) with only compression work

δ Q − δW = dU = δ Q − PdV (3.72)

where the inexact differential of work and heat transfer are used because they are not

properties but path functions. From the second law of thermodynamics, we know that

dS ≥

δ Q

(3.73)

Combining these we can show that

dU + PdV− TdS≤ 0 (3.74)

Considering the differential of absolute enthalpy

dH = dU + PdV+ VdP (3.75)

Rearranging, we can show that

VdP TdS≤ 0

(3.76)

where the pressure differential is zero for this constant pressure and temperature process:

dG = dH

TdS SdT

(3.77)

Plugging Eq. (3.76) into Eq. (3.77), we can show that

T, p

≤ 0 (3.78)

In other words, the Gibbs function of a system will always be minimized in a spontaneous

process (dG < 0). When the Gibbs energy reaches a local minima, where the change is zero

(dG = 0), the reaction stops and local thermodynamic equilibrium is achieved, as illustrated

in Figure 3.10.

The term Gibbs free energy is often used synonymously with Gibbs function.Theterm

free energy is often confusing to people but makes perfect sense in the context of a natural

tendency to reduce the chemical energy of a system through a spontaneous reaction. For

the Gibbs free energy to increase, external work must be applied to the system to increase

the energy state of the system.

The Gibbs free energy can also be thought of as the maximum energy available for

conversion to useful work. The change in Gibbs energy for a reaction is calculated in the

same fashion as other thermodynamic intensive parameters. The absolute molar intensive

Gibbs function at a given temperature and pressure is written as

g =

◦

h − T

s (3.79)