Masters G.M. Renewable and Efficient Electric Power Systems

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208 DISTRIBUTED GENERATION

if fuel cells are powered by hydrogen obtained by electrolysis of water using

renewable energy sources such as wind, hydroelectric, or photovoltaics, they

have no greenhouse gas emissions at all. Fuel cells are easily modulated to track

short-term changes in electrical demand, and they do so with modest compro-

mises in efﬁciency. Finally, they are inherently modular in nature, so that small

amounts of generation capacity can be added as loads grow rather than the con-

ventional approach of building large, central power stations in anticipation of

load growth.

4.6.1 Historical Development

While fuel cells are now seen as a potentially dominant distributed generation

technology for the twenty-ﬁrst century, it is worth noting that they were ﬁrst

developed more than 160 years ago. Sir William Grove, the English scientist cred-

ited with the invention of the original galvanic cell battery, published his original

experiments on what he called a “gaseous voltaic battery” in 1839 (Grove, 1839).

He described the effects caused by his battery as follows: “A shock was given

which could be felt by ﬁve persons joining hands, and which when taken by a sin-

gle person was painful.” Interestingly, this same phenomenon is responsible for

the way that the organs and muscles of an electric eel supply their electric shock.

Grove’s battery depended upon a continuous supply of rare and expensive

gases, and corrosion problems were expected to result in a short cell lifetime,

so the concept was not pursued. Fifty years later, Mond and Langer picked

up on Grove’s work and developed a 1.5-W cell with 50% efﬁciency, which

they named a “fuel cell” (Mond and Langer, 1890). After another half century of

little progress, Francis T. Bacon, a descendent of the famous seventeenth-century

scientist, began work in 1932 that eventually resulted in what is usually thought

to be the ﬁrst practical fuel cell. By 1952, Bacon was able to demonstrate a 5-

kW alkaline fuel cell (AFC) that powered, among other things, a 2-ton capacity

fork-lift truck. In the same year, Allis Chalmers demonstrated a 20-horsepower

fuel-cell-powered tractor.

Fuel cell development was greatly stimulated by NASA’s need for on-board

electrical power for spacecraft. The Gemini series of earth-orbiting missions

used fuel cells that relied on permeable membrane technology, while the later

Apollo manned lunar explorations and subsequent Space Shuttle ﬂights have used

advanced versions of the alkaline fuel cells originally developed by Bacon. Fuel

cells not only provide electrical power, their byproduct is pure water, which is

used by astronauts as a drinking water supply. For longer missions, however,

photovoltaic arrays, which convert sunlight into electric power, have become the

preferred technology.

Fuel cells for cars, buildings, central power stations, and spacecraft were

the subject of intense development efforts in the last part of the twentieth

century. Companies with major efforts in these applications include: Ballard

Power Systems, Inc. (Canada), General Electric Company, the International

Fuel Cells division of United Technologies Corp. and its ONSI subsidiary,

FUEL CELLS 209

Plug Power, Analytic Power, General Motors, H-Power, Allison Chalmers,

Siemens, ELENCO (Belgium), Union Carbide, Exxon/Ashthom, Toyota, Mazda,

Honda, Toshiba, Hitachi Ltd., Ishikawajima-Harima Heavy Industries, Deutsche

Aerospace, Fuji Electric, Mitsubishi Electric Corp. (MELCO), Daimler Chrysler,

Ford, Energy Research Corporation, M-C Power Corp., Siemens-Westinghouse,

CGE, DenNora, and Ansaldo. Clearly, there is an explosion of activity on the

fuel cell front.

4.6.2 Basic Operation of Fuel Cells

There are many variations on the basic fuel cell concept, but a common conﬁg-

uration looks something like Fig. 4.26. As shown there, a single cell consists of

two porous gas diffusion electrodes separated by an electrolyte. It is the choice

of electrolyte that distinguishes one fuel cell type from another.

The electrolyte in Fig. 4.26 consists of a thin membrane that is capable of

conducting positive ions but not electrons or neutral gases. Guided by the ﬂow

ﬁeld plates, fuel (hydrogen) is introduced on one side of the cell while an oxidizer

(oxygen) enters from the opposite side. The entering hydrogen gas has a slight

tendency to dissociate into protons and electrons as follows:

↔ 2H

+ 2e

−

(4.17)

This dissociation can be encouraged by coating the electrodes or membrane with

catalysts to help drive the reaction to the right. Since the hydrogen gas releases

Proton

Transport

2 H

Fuel H

Unused Fuel

Recirculates

from Air

O Air + Water Vapor

Electrons 2e

−

Current

Electrical Load

(40% - 60% Efficiency)

Electrolyte

(PEM)

Gas Diffusion Electrode

(Anode)

> 2H

+ 2e

−

Catalyst

Gas Diffusion Electrode

(Cathode)

1/2 O

+ 2H

+ 2e

−

> H

−

Heat (≈ 85°C)

Flow Field Plate

Figure 4.26 Basic conﬁguration of a proton-exchange membrane (PEM) fuel cell.

210 DISTRIBUTED GENERATION

protons in the vicinity of the electrode on the left (the anode), there will be a

concentration gradient across the membrane between the two electrodes. This

gradient will cause protons to diffuse through the membrane leaving electrons

behind. As a result, the cathode takes on a positive charge with respect to the

anode. Those electrons that had been left behind are drawn toward the positively

charged cathode; but since they can’t pass through the membrane, they must

ﬁnd some other route. If an external circuit is created between the electrodes,

the electrons will take that path to get to the cathode. The resulting ﬂow of

electrons through the external circuit delivers energy to the load (remember that

conventional current ﬂow is opposite to the direction that electrons move, so

current I “ﬂows” from cathode to anode).

A single cell, such as that shown in Fig. 4.26, typically produces on the order

of 1 V or less under open circuit conditions and produces about 0.5 V under

normal operating conditions. To build up the voltage, cells can be stacked in

series. To do so, the gas ﬂow plates inside the stack are designed to be bipolar;

that is, they carry both the oxygen and hydrogen used by adjacent cells as is

suggested in Fig. 4.27.

4.6.3 Fuel Cell Thermodynamics: Enthalpy

The fuel cell shown in Fig. 4.26 is described by the following pair of half-

cell reactions:

Anode: H

→ 2H

+ 2e

−

(4.18)

Cathode:

+ 2H

+ 2e

−

→ H

O (4.19)

When combined, (4.18) and (4.19) result in the same equation that we would

write for ordinary combustion of hydrogen:

→ H

O (4.20)

Gas Flow Bipolar Plates

Electrodes/Catalysts

Single Cell

Multi-Cell Stack

Electrolyte

Figure 4.27 A multicell stack made up of multiple cells increases the voltage. After

Srinivasan et al. (1999).

FUEL CELLS 211

The reaction described in (4.20) is exothermic; that is, it liberates heat (as

opposed to endothermic reactions, which need heat to be added to make them

occur). Since (4.20) is exothermic, it will occur spontaneously—the hydrogen

and oxygen want to combine to form water. Their eagerness to do so provides

the energy that the fuel cell uses to deliver electrical energy to its load. The

questions, of course, are how much energy is liberated in reaction (4.20) and how

much of that can be converted to electrical energy. To answer those questions,

we need to describe three quantities from thermodynamics: enthalpy, free energy,

and entropy. Unfortunately, the precise deﬁnitions of each of these quantities tend

not to lend themselves to easy, intuitive interpretation. Moreover, they have very

subtle properties that are beyond our scope here, and there are risks in trying to

present a simpliﬁed introduction.

The enthalpy of a substance is deﬁned as the sum of its internal energy U and

the product of its volume V and pressure P :

Enthalpy H = U + PV (4.21)

The internal energy U of a substance refers to its microscopic properties, includ-

ing the kinetic energies of molecules and the energies associated with the forces

acting between molecules, between atoms within molecules, and within atoms.

The total energy of that substance is the sum of its internal energy plus the

observable, macroscopic forms such as its kinetic and potential energies. The

units of enthalpy are usually kJ of energy per mole of substance.

Molecules in a system possess energy in various forms such as sensible and

latent energy, which depends on temperature and state (solid, liquid, gaseous),

chemical energy (associated with the molecular structure), and nuclear energy

(associated with the atomic structure). For a discussion of fuel cells, it is changes

in chemical energy that are of interest, and those changes are best described in

terms of enthalpy changes. For example, we can talk about the potential energy of

an object as being its weight times its height above some reference elevation. Our

choice of reference elevation does not matter as long as we are only interested

in the change in potential energy as an object is raised against gravity from one

elevation to another. The same concept applies for enthalpy. We need to describe

it with respect to some arbitrary reference conditions.

In the case of enthalpy a reference temperature of 25

◦

C and a reference pres-

sure of 1 atmosphere (standard temperature and pressure, STP) are assumed. It

is also assumed that the reference condition for the chemically stable form of

an element at 1 atmosphere and 25

◦

C is zero. For example, the stable form of

oxygen at STP is gaseous O

, so the enthalpy for O

(g)iszero,where(g)just

means it is in the gaseous state. On the other hand, since atomic oxygen (O) is

not stable, its enthalpy is not zero (it is, in fact, 247.5 kJ/mol). Notice that the

state of a substance at STP matters. Mercury, for example, at 1 atmosphere and

◦

C is a liquid, so the standard enthalpy for Hg(l) is zero, where (l)meansthe

liquid state.

One way to think about enthalpy is that it is a measure of the energy that

it takes to form that substance out of its constituent elements. The difference

212 DISTRIBUTED GENERATION

between the enthalpy of the substance and the enthalpies of its elements is called

the enthalpy of formation. It is in essence the energy stored in that substance

due to its chemical composition. A short list of enthalpies of formation at STP

conditions appears in Table 4.6. To remind us that a particular value of enthalpy

(or other thermodynamic properties such as entropy and free energy) is at STP

conditions, a superscript “

◦

”isused(e.g.H

◦

Table 4.6 also includes two other quantities, the absolute entropy S

◦

and the

Gibbs free energy G

◦

, which will be useful when we try to determine the maxi-

mum possible fuel cell efﬁciency. Notice when a substance has negative enthalpy

of formation, it means that the chemical energy in that substance is less than that

of the constituents from which it was formed. That is, during its formation,

some of the energy in the reactants didn’t end up as chemical energy in the

ﬁnal substance.

In chemical reactions, the difference between the enthalpy of the products and

the enthalpy of the reactants tells us how much energy is released or absorbed

in the reaction. When there is less enthalpy in the ﬁnal products than in the

reactants, heat is liberated—that is, the reaction is exothermic. When it is the

other way around, heat is absorbed and the reaction is endothermic.

If we analyze the reaction in (4.20), the enthalpies of H

and O

are zero so

the enthalpy of formation is simply the enthalpy of the resulting H

O. Notice

in Table 4.6 that the enthalpy of H

O depends on whether it is liquid water or

gaseous water vapor. When the result is liquid water:

→ H

O(l) H =−285.8kJ (4.22)

When the resulting product is water vapor:

→ H

O(g) H =−241.8kJ (4.23)

TA BLE 4.6 Enthalpy of Formation H

◦

, Absolute Entropy S

◦

and Gibbs Free Energy G

◦

at 1 atm, 25

◦

Cfor

Selected Substances

Substance State H

◦

(kJ./mol) S

◦

(kJ/mol-K) G

◦

(kJ/mol)

H Gas 217.9 0.114 203.3

Gas 0 0.130 0

O Gas 247.5 0.161 231.8

Gas 0 0.205 0

O Liquid −285.8 0.0699 −237.2

OGas−241.8 0.1888 −228.6

C Solid 0 0.006 0

Gas −74.9 0.186 −50.8

CO Gas −110.5 0.197 −137.2

Gas −393.5 0.213 −394.4

OH Liquid −238.7 0.1268 −166.4

FUEL CELLS 213

The negative signs for the enthalpy changes in (4.22) and (4.23) tell us these

reactions are exothermic; that is, heat is released. The difference between the

enthalpy of liquid water and gaseous water vapor is 44.0 kJ/mol. Therefore, that

amount is the familiar latent heat of vaporization of water. Recall that latent heat

is what distinguishes the higher heating valuez (HHV) of a hydrogen-containing

fuel and the lower heating value LHV. The HHV includes the 44.0 kJ/mol of

latent heat in the water vapor formed during combustion, while the LHV does not.

Example 4.8 The High Heating Value (HHV) for Methane. Find the HHV

of methane CH

in kJ/mol and kJ/kg when it is oxidized to CO

and liquid H

Solution. The reaction is written below, and beneath it are enthalpies taken from

Table 4.6. Notice that we must balance the equation so that we know how many

moles of each constituent are involved.

(g) + 2O

(g) → CO

(g) + 2H

O(l)

(−74.9) 2 × (0)(−393.5) 2 × (−285.8)

Notice, too, that we have used the enthalpy of liquid water to ﬁnd the HHV.

The difference between the total enthalpy of the reaction products and the

reactants is

H = [(−393.5) + 2 × (−285.8)] −[(−74.9) + 2 × (0)]

=−890.2kJ/mol of CH

Since the result is negative, heat is released during combustion; that is, it is

exothermic. The HHV is the absolute value of H , which is 890.2 kJ/mol.

Since there are 12.011 + 4 × 1.008 = 16.043 g/mol of CH

, the HHV can

also be written as

HHV =

890.2kJ/mol

16.043 g/mol

× 1000 g/kg = 55,490 kJ/kg

4.6.4 Entropy and the Theoretical Efﬁciency of Fuel Cells

While the enthalpy change tells us how much energy is released in a fuel cell,

it doesn’t tell us how much of that energy can be converted directly into elec-

tricity. To ﬁgure that out, we need to review another thermodynamic quantity,

entropy. Entropy has already been introduced in the context of heat engines in

Section 3.4.2, where it was used to help develop the Carnot efﬁciency limit. In

a similar fashion, the concept of entropy will help us develop the maximum

efﬁciency of a fuel cell.

214 DISTRIBUTED GENERATION

To begin, let us note that all energy is not created equal. That is, for example,

1 joule of energy in the form of electricity or mechanical work is much more

useful than a joule of heat. We can convert that joule of electricity or work

into heat with 100% conversion efﬁciency, but we cannot get back the joule

of electricity or work from just a single joule of heat. What this suggests is

that there is a hierarchy of energy forms, with some being “better” than others.

Electricity and mechanical energy (doing work) are of the highest quality. In

theory we could go back and forth between electricity and mechanical work with

100% conversion efﬁciency. Heat energy is of much lower quality, with low-

temperature heat being of lower quality than high-temperature heat. So, where

does chemical energy ﬁt in this scheme? It is better than thermal, but worse than

mechanical and electrical. Entropy will help us ﬁgure out just where it stands.

Recall that when an amount of heat Q is removed from a thermal reservoir

large enough that its temperature T does not change during the process (i.e., the

process is isothermal), the loss of entropy S from the reservoir is deﬁned to be

S =

(4.24)

With Q measured in kilojoules (kJ) and T in kelvins (K =

◦

C + 273.15), the

units of entropy are kJ/K. Recall, too, that entropy is only associated with heat

transfer and that electrical or mechanical work is perfect so that these forms have

zero entropy. And, ﬁnally, remember that in any real system, if we carefully

add up all of the entropy changes, the second law of thermodynamics requires

that there be an overall increase in entropy. Now let us apply these ideas to a

fuel cell.

Consider Fig. 4.28, which shows a fuel cell converting chemical energy into

electricity and waste heat. The fuel cell reactions (4.22) and (4.23) are exother-

mic, which means that their enthalpy changes H are negative. Working with

negative quantities leads to awkward nomenclature, which we can avoid by say-

ing that those reactions act as a source of enthalpy H that can be converted to

heat and work as Fig. 4.28 implies.

Fuel

Cell

Enthalpy in

Enthalpy output

Rejected heat

Figure 4.28 The energy balance for a fuel cell.

FUEL CELLS 215

The cell generates an amount of electricity W

and rejects an amount of thermal

energy Q to its environment. Since there is heat transfer and it is a real system,

there must be an increase in entropy. We can use that requirement to determine

the minimum amount of rejected heat and therefore the maximum amount of

electric power that the fuel cell will generate. To do so, we need to carefully

tabulate the entropy changes occurring in the cell:

→ H

O + Q(4.25)

where we have included the fact that heat Q will be released. The entropy of the

reactants H

and O

will disappear, but new entropy will appear in the H

Othat

is formed plus the entropy that appears in the form of heat Q. As long as the

process is isothermal, which is a reasonable assumption for a fuel cell, we can

write the entropy appearing in the rejected heat as

S =

(4.26)

What about the entropy associated with the work done, W

? Since there is no

heat transfer in electrical (or mechanical) work, that entropy is zero.

To make the necessary tabulation, we need values of the entropy of the reac-

tants and products. And, as usual, we need to deﬁne reference conditions. It is

conventional practice to declare that the entropy of a pure crystalline substance

at zero absolute temperature is zero (the “third law of thermodynamics”). The

entropy of a substance under other conditions, referenced to the zero base con-

ditions, is called the absolute entropy of that substance, and those values are

tabulated in a number of places. Table 4.6 gives absolute entropy values, S

◦

,for

several substances under STP conditions (25

◦

C, 1 atm).

The second law of thermodynamics requires that in a real fuel cell there be a

net increase in entropy (an ideal cell will release just enough heat to make the

increase in entropy be zero). Therefore, we can write that the entropy that shows

up in the rejected heat and the product water (liquid water) must be greater than

the entropy contained in the reactants (H

and O

Entropy gain ≥ Entropy loss (4.27)



products

≥



reactants

(4.28)

which leads to

Q ≥ T





reactants

−



products



(4.29)

Equation (4.29) tells us the minimum amount of heat that must appear in the

fuel cell. That is, we cannot convert all of the fuel’s energy into electricity—we

are stuck with some thermal losses. At least our thermal losses are going to be

less than if we tried to generate electricity with a heat engine.

216 DISTRIBUTED GENERATION

We can now easily determine the maximum efﬁciency of the fuel cell. From

Fig. 4.28, the enthalpy supplied by the chemical reaction H equals the electricity

produced W

plus the heat rejected Q:

H = W

+ Q(4.30)

Since it is the electrical output that we want, we can write the fuel cell’s efﬁ-

ciency as

η =

H − Q

= 1 −

(4.31)

To ﬁnd the maximum efﬁciency, all we need to do is plug in the theoretical

minimum amount of heat Q from (4.29).

Example 4.9 Minimum Heat Released from a Fuel Cell. Suppose a fuel cell

that operates at 25

◦

C (298 K) and 1 atm forms liquid water (that is, we are

working with the HHV of the hydrogen fuel):

→ H

O(l) H =−285.8kJ/mol of H

a. Find the minimum amount of heat rejected per mole of H

b. What is the maximum efﬁciency of the fuel cell?

Solution

a. From the reaction, 1 mole of H

reacts with 1/2 mole of O

to produce

1 mole of liquid H

O. The loss of entropy by the reactants per mole of H

is found using values given in Table 4.6:



reactants

= 0.130 kJ/mol-K × 1molH

+ 0.205 kJ/mol-K × 0.5molO

= 0.2325 kJ/K

The gain in entropy in the product water is



product

= 0.0699 kJ/mol-K ×1molH

O(l) = 0.0699 kJ/K

From (4.29), the minimum amount of heat released during the reaction

is therefore

min

= T





reactants

−



products



= 298 K (0.2325 − 0.0699) kJ/K = 48.45 kJ per mole H

FUEL CELLS 217

b. From (4.22), the enthalpy made available during the formation of liquid

water from H

and O

is H = 285.8kJ/molofH

The maximum efﬁciency possible occurs when Q is a minimum; thus

from (4.31)

max

= 1 −

min

= 1 −

48.45

285.8

= 0.830 = 83.0%

4.6.5 Gibbs Free Energy and Fuel Cell Efﬁciency

The chemical energy released in a reaction can be thought of as consisting of two

parts: an entropy-free part, called free energy G, that can be converted directly

into electrical or mechanical work, plus a part that must appear as heat Q.The

“G” in free energy is in honor of Josiah Willard Gibbs (1839–1903), who ﬁrst

described its usefulness, and the quantity is usually referred to as Gibbs free

energy. The free energy G is the enthalpy H created by the chemical reaction,

minus the heat that must be liberated, Q = TS, to satisfy the second law.

The Gibbs free energy G corresponds to the maximum possible, entropy-

free, electrical (or mechanical) output from a chemical reaction. It can be found

at STP using Table 4.6 by taking the difference between the sum of the Gibbs

energies of the reactants and the products:

G =



products

−



reactants

(4.32)

This means that the maximum possible efﬁciency is just the ratio of the Gibbs

free energy to the enthalpy change H in the chemical reaction

max

G

H

(4.33)

Example 4.10 Maximum Fuel Cell Efﬁciency Using Gibbs Free Energy.

What is the maximum efﬁciency at STP of a proton-exchange-membrane (PEM)

fuel cell based on the higher heating value (HHV) of hydrogen?

Solution. The HHV corresponds to liberated water in the liquid state so that the

appropriate reaction is

→ H

O(l) H =−285.8kJ/mol of H

From Table 4.6 the Gibbs free energy of the reactants H

and of O

are both

zero, and that of the product, liquid water, is −237.2 kJ. Therefore, from 4.32)

G =−237.2 − (0 + 0) =−237.2kJ/mol