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Lecture 8 Thermodynamics Relevant for Fuel Cells

Thermodynamics : study of conversion of energy from one form to another in fuel cells, chemical energy (fuel) is converted into electrical energy

and heat in combustion engines, chemical energy is converted into

mechanical energy and heat

It is important to realize that in modern physics today, we have no knowledge of what energy is

R.P. Feynman, R. B. Leighton, and M. Sands, The Feynberg Lectures on Physics, Section 4-1, Addison-Wesley, Reading, MA 1963

Other thermodynamic terms such as entropy and enthalpy are even harder to understand

Thermodynamics is just a very elaborate bookkeeping scheme for tracking the properties of a system in a self-consistent manner, based on several assumptions or laws.

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First Law of Thermodynamics Energy of a system is conserved Energy crosses system boundaries in form of heat and work

dE = Q - W

E is a point function, depending only on initial and final states Q and W are path functions (values are dependent on path)

When the above equation is integrated, the First Law relation becomes

E = Q - W

Piston cylinder

Steam turbine

Closed system (no mass flow across system boundaries)

E = U + KE +PE

Open system (mass flow across system boundaries)

H = Q W H= U + pV

H.enthalpy

p= pressure of fluid V = volume of fluid

PV represents the work that is exerted

on the fluid to keep if flowing

U. internal energy (total intrinsic energy of a substance, associated with microscopic movement, i.e. kinetic energy and interactions between particles on atomic scale, i.e. chemical and potential energy)

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Fuel cell is an open system Instead of the shaft rotation as in a turbine, work is in form of electron transport across a potential difference

-E

+E

Second Law of Thermodynamics

Entropy S measure of disorder in a system A process that does not generate entropy ( i.e. more disorder) is called reversible. In a reversible process no net exchange of heat or work occurs in system or surroundings

both return to their original states

An irreversible process does generate entropy ( e.g. heat loss from friction, or heat transfer through a finite temperature difference)

If the temperature gradient is made infinitesimally small

(at the expense of the heat transfer rate!), the process approaches a reversible one.

Entropy, a property or point function, is based on this reversible heat transfer:

dS = (Q/T)rev After integration: S = Qrev/T0

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Increase in entropy principle Isolated system . no heat, work, or mass flow across system boundary

S 0

Note: all systems can become isolated by extending the system boundary to a large enough size

Stotal =Ssystem + Ssurroundings 0

For a fuel cell where a chemical reaction takes place, the system is a control volume where mass crosses the system boundaries. Entropy change of system is the difference between Sproducts and Sreactants

Ssystem = Sproducts - Sreactants

Heat generated or consumed in reaction is included in the expression for the surroundings at T=T0

Ssurroundings = Qsurroundings / T0

Stotal = (Sproducts Sreactants)system + (Qsurroundings / T0)

Thermodynamic Potentials First and second law can be used to set up a number of rules governing how energy can be transferred from one form to another. These rules are called thermodynamic potentials.

dU =TdS -pdV

reversible heat transfer mechanical work

(dU/dS)v = T (dU/dV)S = -p

U, the internal energy of a system, is a function of entropy and volume U = U(S, V) but S and V tend to be difficult to measure, while T and p are easier to measure. A new thermodynamic potential equivalent to U is needed that can be determined form T and p

Lets define a new thermodynamic potential, the Gibbs free energy, G(T,p)

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Gibbs free energy, G

Conversion from U(S,V) to G(T,p) can be accomplished via Legendre transform G = U (dU/dS)v S - (dU/dV)s V G = U TS + pV dG = dU TdS SdT +pdV +Vdp since dU = TdS pdV, it follows that dG = -SdT + Vdp

Enthalpy H

Thermodynamic potential that is a function of S and p can be obtained by transforming U with respect to V by definition, we call this new thermodynamic potential H, Enthalpy H = U (dU/dV)S V Since (dU/dV)s = -p, one gets H = U + pV Through differentiation, one obtains dH = dU +pdV + Vdp Knowing that dU = TdS pdV gives dH = TdS + Vdp

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Helmholtz free energy F

Thermodynamic potential that depends on T and V F = U TS dF = -SdT - pdV

How the four thermodynamic potentials are related by expansion work term pV and energy from the environment term TS

Internal energy, U U = energy required to create a system

Helmholtz free energy, F = U-TS

F = energy required to create a system minus the energy supplied by the environment

Enthalpy, H = U + pV

H = energy required to create a system plus the work needed to make room for it

Gibbs free energy, G = U+pV-TS

G= energy required to create a system plus the work needed to make room for it minus the energy supplied by the environment

-TS

+pV

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Definition of Standard State

Many thermodynamic quantities depend on both temperature and pressure. Therefore, we need a way to reference everything to a standard set of conditions. When dealing with chemical reactions, the zero energy point is normally defined as pure elements, in the normal state, at standard T and P:

T = 298.15 K P = 1 bar = 100 kPa (note: atmospheric pressure is actually 101.325 kPa. This slight difference is usually ignored)

Standard-state conditions also specify that all reactant and product species are present at unit activity. For a typical hydrogen fuel cell operating at standard conditions, the Gibbs free energy of formation of the input is zero- a very useful simplification.

Heat engines

1. Heat input from external high-T source 2. Partial conversion of heat to work ( e.g. in turbine) 3. Rejection of waste heat to low-T sink (e.g. condenser with cooling water)

Operate in closed system (no mass flow) on cycle:

Typical application: steam engine In contrast, combustion engines and gas turbines are open system ( working fluid is continually replaced) and therefore they do not follow this type of cycle

initial and final states are identical, thus: Q - W = E = 0 Wnet (net work) = net heat = Qin- Qout

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Carnot cycle Thermal efficiency, th = Wnet/Qin= (Qin-Qout)/Qin = 1-Qout/Qin Maximum work is achieved when all processes are conducted reversibly as proposed for idealized cycle by Carnot:

adapted from Fuel Cell Technology Handbook, CRC Press 2003

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The Carnot cycle for steady-state flow devices. All processes are reversible.

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The Carnot Cycle

Wnet

Q in

Q out

TH

TL

1 2

3 4 S

T

S1 =S4 S2 =S3

Qrev = T0S th, Carnot= 1- TL/TH

maximum possible conversion efficiency for any heat engine

The concept of Exergy

Potential to do work .. property called exergy Exergy is the term used to quantify the work potential of a system from its initial state to the dead state ( i.e. the state of the surroundings, usually at standard T and P conditions ( STP, 25 C, 1 atm) Besides the work potential of temperature, exergy also considers pressure Pressure could be relieved through a turbine converting the pressure into shaft work

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Steady-flow device with heat and work outputs Heat output can be used by Carnot cycle heat engine to produce more work

Work potential is the amount of useful work that the steady flow device can perform as it changes from initial T and P to dead state T0, P0 Since it is an open system with mass flow, we must account for both internal energy and flow work of the mass by using enthalpy, H dH = -Q W ( note: minus signs because system rejects heat and does work) WCarnot = (1- T0/T) Qin, Carnot = Qin, Carnot T0dSin, Carnot Qin, Carnot = WCarnot + T0dSin, Carnot

adapted from Fuel Cell Technology Handbook, CRC Press 2003

But. equation should be written in terms of the system, not just the Carnot cycle heat engine Heat leaving the system, Qout, system , = heat entering the heat engine, Qin, Carnot

Qout, system = - Qin, Carnot Assuming the ideal case where heat transfer between system and engine occurs isothermally ( in reality that would lead to zero rate of heat transfer !), we can use the relationship dS = Q/T System loses entropy ( because it loses heat, -Q) Heat engine gains entropy (because it gains heat, +Q)

Therefore dSout, system = -dSin, Carnot Shaft work done by steady-state device: W = Wshaft Substituting into original expression dH = -Q W gives. - (WCarnot T0dS) Wshaft = dH or Wshaft + WCarnot = - dH + T0dS

Integration form initial to dead state gives.. Wtotal, useful = (H H0) =T0(S-S0)

EXERGY

Exergy destroyed = irreversibility, I, ( difference between reversible and actual work )

I = Wrev Wactual = T0Sgenerated

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Efficiencies Thermal efficiency of heat engine, thermal = Wnet / Qin thermal, Carnot = 1 TL / TH

note: TL is usually fixed at ambient condition; efficiency is thus determined by TH However, to get to high T, the fuel loses some of its chemical energy to irreversible processes:

for example, in adiabatic combustion of methane with excess air, ~35% of the work potential of the fuel is used up to change the temperature of the molecules involved Cengel and Boles, Thermodynamics: An Engineering Approach, 3rd ed, WCB/McGraw-Hill, Boston, 1998

Example:

Steam turbine operating at TH = 700 K

Water leaving at 350 K thermal, Carnot = 1 TL / TH = 1- 350/700 = 0.5 ( 50 % efficiency)

Fuel cells operate at constant T. Therefore, more of the chemical energy of the reactants can be converted to electrical energy, since we are not using up energy to change the temperature of the reaction products. Fuel cells are not combustion-cycle limited. Many textbooks and scientific articles get this wrong by stating fuel cells are more efficient because they are not Carnot-cycle limited The Carnot cycle efficiency, which limits the maximum work to the highest T of the cycle, is irrelevant for fuel cells. The maximum work for a fuel cell is equal to the change in the Gibbs energy, G

Wmax, cell = -G Work is done by electrons moving though a potential difference, E

Wcell = nFE

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Comparison between heat engines and fuel cells thermal,cell = Wcell/ HHVfuel = (nFE)/HHVfuel Maximum theoretical efficiency of fuel cell is achieved at open-circuit voltage, E0 ( but of course, since there is no current drawn, no power is available!) thermal,cell, max = (nFE0)/HHVfuel Value for E0 can be obtained from tabulated Gibbs energy data

( 1.23 V for H2- O2 fuel cell ) thermal,cell, max = (2 x 96485 C/mol x 1.23 V)/ 285840 J/mol = 0.83

H0f of H2O (J/mol)

The 17 % inefficiency is due to entropy changes during the chemical reaction

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Reversible work produced by H2/O2 fuel cell and by Carnot engine

Fuel cell

Carnot heat engine

In fuel cell, as T increases, G decreases Since Wmax,cell = G, decreasing values of G lead to decrease in reversible work ( Gibbs energy of formation of water vapor =

-228.582 kJ/mol @ STP (298 K, 1 atm) -164.429 kJ/mol @ 1500 K

Fuel Cell Technology Handbook, CRC Press 2003

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Thermal efficiency of combustion engine is usually determined in terms of power. Heat input is expressed via flow rate of fuel and completeness of fuel combustion. Parameter for fuel cell equivalent to completeness of fuel combustion :

fuel utilization . a measure of fuel consumed to produce an electric current current efficiency, i = I / (-n F Nfuel )

I = current (in A) Nfuel = flow rate of fuel ( in mol/s)

e.g. Fuel cell current efficiency of 70 % means that 70 % of the hydrogen fed to the cell is converted into electricity. The remaining 30 % leave the cell without reaction or reacts via processes ( such as combustion) that do not contribute the electrons to the cell current. In practice,excess hydrogen is fed to assure that sections at the end of the flow channels get enough hydrogen. So, fuel utilization is always less, as the excess hydrogen does not react and leaves the cell.

Energy (E) = total power consumed (or produced) Units of energy: 1 J = 1 N m = 107 erg 1 ft lbf = 1.356 J 1 cal = 4.186 J 1 Btu = 1055 J = 252 cal = 778 ft lbf 1 eV = 1.602 x 10-19 J 1 kWh = 3.6 x 106 J

Power (P) = rate at which energy is made available Units of power: 1W = 1 J/s 1 hp = 746 W = 550 ft lbf s-1 = 0.707 Btu/s 1Btu/hr = 0.293 W

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For fuel cell: Chemical energy = thermal energy + electrical energy H2 + O2 H2O (this reaction is exothermic) exothermic reactions give off heat

G = H - TS

change in Gibbs Free Energy

change in Enthalpy (heat)

change in Entropy (disorder)

H2 + O2 H2O

Lets determine how much heat energy and electric energy can be generated in a hydrogen fuel cell under ideal conditions:

Use Gibbs free energy equation to calculate ________ = _-285840_ - ______ _________

For 298 K and 1 atm:

Entropy S0 (j/mol K)

H2 130.547 O2 205.033 H2O (l) 69.92

S = 69.92-0.5*205.033-130.547= -163.14

G = H S

T -

for liquid water (higher heating value)

H = -241830 kJ/mol for H2O(g) lower heating value H = -285840 kJ/mol for H2O(l) higher heating value

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H2 + O2 H2O

Lets determine how much heat energy and electric energy can be generated in a hydrogen fuel cell under ideal conditions:

Use Gibbs free energy equation to calculate G = -285840 - (298 * -163.14 ) = -237224 kJ/mol

For 298 K and 1 atm:

Entropy S0 (kJ/mol K)

H2 130.547 O2 205.033 H2O (l) 69.92

S = 69.92-0.5*205.033-130.547= -163.14

H S

T -

for liquid water (higher heating value)

H = -241830 kJ/mol for H2O(g) lower heating value H = -285840 kJ/mol for H2O(l) higher heating value

To calculate maximum possible fuel cell voltage (at open circuit) G = -nFE0 where F = Faradays constant = ___________ coulombs (C)

1 C = 1 J/V E0 = -G/nF where n = ______ electrons E0 = -( _______ J) / ( ____ ) ( ___________ J/V ) = ________ V But. PEM fuel cells do not operate at 25 C. How about a more realistic 80 C? G = ________ - ( _______ ) ( __________ ) = ____________ J E = -( _______ J) / ( ____ ) ( ___________ J/V ) = ________ V By correcting for air ( rather than using 100 % pure O2) and allowing for humidified air.

V ~ 1.10 V at 80C and 1 atm pressure

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To calculate maximum possible fuel cell voltage (at open circuit) G = -nFE0 where F = Faradays constant = 96485 coulombs (C)

1 C = 1 J/V E0 = -G/nF where n = 2 electrons E0 = -(237224 J) / (2) (96485 J/V ) = 1.23 V But. PEM fuel cells do not operate at 25 C. How about a more realistic 80 C? G = -285840 - ( 353 ) (-163.14) = -228252 J E = -(228252 J) / (2 ) (96485 J/V ) = 1.18 V By correcting for air ( rather than using 100 % pure O2) and allowing for humidified air.

V ~ 1.10 V at 80C and 1 atm pressure

Lets assume we have a PEM fuel cell with 100 cm2 electrode area operating at cell voltage Vcell =0.7 V and a cell current icell =0.6 amp/cm2 ( i.e. 60 amps) P total = P electrical + P thermal P thermal = P total - P electrical

= ( V ideal x icell) (Vcell x icell) = (Videal Vcell) x icell

= ( ____ - _____) x ( ______ )

P thermal = __________ W ( _______ %) P total = __________ W P electrical = __________ W ( _______ %)

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Lets assume we have a PEM fuel cell with 100 cm2 electrode area operating at cell voltage Vcell =0.7 V and a cell current icell =0.6 amp/cm2 ( i.e. 60 amps) P total = P electrical + P thermal P thermal = P total - P electrical

= ( V ideal x icell) (Vcell x icell) = (Videal Vcell) * icell

= (1.23- 0.7)V * ( 60 A )

P total = 73.8 W P thermal = 31.8 W (43.1 %) P electrical = 42 W ( 56.9 %)