84

CHAPTER 3

MODELING OF SOLID OXIDE FUEL CELL – GAS

TURBINE HYBRID SYSTEM

“A model is neither true nor false – it is more or less useful”

- (Stafford Beer, 1985)

3.1 Solid Oxide Fuel Cell

Solid oxide fuel cells consists of a solid electrolyte (zirconia), which is a

ceramic and a good conductor of oxygen ions. This property of zirconia

was first discovered by Nernst in late 1890’s. Though the technology

has evolved in these hundred years and production methods have

improved, zirconia is still considered to be the best electrolyte for solid

oxide fuel cells. Zirconia starts conducting oxygen ions when its

temperature is above 700°C and thus solid oxide fuel cells are best

suited for co-generation. Waste heat from the fuel cell can be utilized

in a bottoming cycle and power generation efficiencies of more that

60% are achievable. Also, within the SOFC operating temperature

range, emissions of NOx are likely to be very small resulting in a

cleaner environment [19].

3.1.1 Thermodynamics and Electrode kinetics of SOFCs

A solid oxide fuel cell is an electro chemical reactor which converts

hydrogen and oxygen into electricity. Figure 3.1 shows the schematic

diagram of taking place in a solid oxide fuel cell. It basically consists

of two porous electrodes (anode and cathode) separated by a ceramic

85

electrolyte, and flow channels for air delivery & collection and fuel.

Figure 3.1 Schematic of a Solid Oxide Fuel Cell[8]

H2 or a hydrocarbon like methane is supplied on the anode and

air or O2 on the cathode side of the fuel cell. H2 and CO (if H2 is not

pure) diffuse through the porous anode to the three phase boundary

formed by the electrolyte, the gaseous H2 and anode Similarly, O2

diffuses through cathode to three phase boundary of the cathode side

where it accepts electrons from the cathode and gives oxygen ions.

These oxygen ions travel through the porous electrolyte and react with

H2 to produce electrons and water at the anode and thus voltage is

generated between two electrodes. The two electrodes can be

connected via an external circuit and an electrical current can be

generated [19].

The general reactions in Fuel Cell are:

At the cathode : 2 4 2O e O (3.1)

At the anode : 2 22 2 4 4H O H O e (3.2)

86

Water gas shift at anode : 2 2 2CO H O CO H (3.3)

If CO is present in the H2 stream, the CO reacts with H2O via a water

gas shift reaction that produces H2 and CO2.

Before we begin to look at how the electromotive force (EMF) and thus

power is produced in a fuel cell, it is necessary to understand some

basic thermodynamic concepts [19]. Consider the following

thermodynamic relation for a reversible process when there is no shaft

work extracted and the system is restricted to do only expansion work:

dG = VdP – SdT (3.4)

if the process is isothermal, the above equation reduces to:

dG = VdP (3.5)

using the ideal gas relation, PV=nRT in Equation 4 where n is the

number of moles of the gas, we get

dG = nRTdP/P (3.6)

Integrating the above Equation from state 1 to state 2, we get,

G2-G1 = nRT ln(P2/P1) (3.7)

If the state 1 is replaced with some standard reference state, with

Gibbs free energy G° and standard pressure P°, the Gibbs free energy

per unit mole at any state ‘i’ is given by,

gi = g° + RTln(Pi/P°) (3.8)

Consider that the following chemical reaction takes place at constant

temperature and pressure,

aA + bB mM + nN (3.9)

87

where a, b, m, and n are the stoichiometric coefficients of the

reactants A and B and the products

M and N, respectively. Now, Equation 8 takes the following form,

lnm n

M No a b

A B

P PG G RT

P P

(3.10)

where G0 is the standard Gibbs free energy change for the reaction,

o o o o

o M N A BG mg ng ag bg (3.11)

gi°, are the standard Gibbs free energies of the constituents.

Equation 10 gives the Gibbs free energy change for the reaction. To

find that relation, consider the following thermodynamic identity for a

reversible process, (dQ = TdS)

dG = -W + PdV + VdP – SdT (3.12)

At constant temperature and pressure, the above equation can be

written as,

dG = - W + PdV (3.13)

Since it is a non-expansion work, equation 13 takes the form,

dG = -We (3.14)

i.e. the change in Gibbs free energy of the reaction is equal to the

maximum electrochemical work, We , that can be extracted when

reactants A and B react to give products M and N under constant

temperature and pressure conditions through a reversible reaction.

The Electro Motive Force produced due to half-cell reactions

drives the electrons to move from anode to cathode. If ne mole of

88

electrons move from anode to cathode per unit time and the Electro

Motive Force of the cell is E, the power produced is simply EMF

multiplied by the current,

We = ne FE (3.15)

where F is the total charge of 1 mole of electrons, known as Faraday’s

constant.

Therefore G = -ne FE (3.16)

Applying equation 16 to equation 10, we get what is known as Nernst

equation,

ln

m n

M No a b

e A B

P PRTE E

n F P P

(3.17)

where E° is related to G° by Equation 15.

For the reaction occurring in an SOFC,

2 2 2

1

2H O H O

(3.18)

the reversible potential can be written as,

2

2 2

1/2ln

2

H O

o

H O

PRTE E

F P P

(3.19)

This maximum theoretical voltage, E, is also known as “Open Circuit

Voltage” and can be measured when there is no current in the circuit.

Also, it can be observed, that to get the maximum Open Circuit

Voltage, a high concentration of reactants is required [19].

Equation 19 gives the maximum Open Circuit Voltage but this

is not the operating voltage of the fuel cell. The operating voltage is

89

always less than the OCV due to the losses associated with the

current production. There are three major types of voltage losses as

shown in the Figure 3.2.

Figure 3.2 Current-Voltage Characteristics of a Fuel Cell Operating at

1073K [19]

Activation loss is associated with the energy intensive activities of

the forming and breaking of chemical bonds at the electrodes. At the

cathode, the oxygen enters a reaction site and draws electrons from

the catalyst to form oxygen ions. The produced ions form bonds with

the catalyst surface while electrons remain near the catalyst until

another oxygen molecule starts to react with the catalyst, thus

breaking the bond with the ion. The energy input to break the bond

with the ion finds whether the electron will bond again with the

catalyst, or will remain with the ion. The same procedure occurs at

90

the anode also. The incoming hydrogen is broken up into it’s

components by the catalyst where it draws oxygen ions to form water

and electrons are released on the anode. The amount of energy needed

for these activities of breaking and forming of chemical bonds comes

from the fuel, and thus reduces the overall energy the cell can

produce. If the reaction rate increases (high current density), the fuel

flow rate must also increase, which increases the kinetics and thus

lowers the energy required to break bonds. Therefore when the

current requirement is low, the overall cell polarization is dominated

by the activation losses. Other factors, which lowers the activation

polarization, are increasing temperature, active area of the electrode,

and activity of electrodes by the use of suitable catalyst.

Ohmic loss is caused by the electrical resistance the charge has to

overcome when traveling across the different materials or interfaces of

the cell. The resistances of the electrodes, current collectors and the

electrolyte are all factors which add to the energy loss. Resistance is

added by the electrodes because of the contact resistance through the

electrode material itself, with the current collectors and with the

electrolyte. The electrolyte can add to the ohmic polarization through

the resistance to ionic flow [19].

Concentration loss is also known as diffusion polarization. It results

from restrictions to the transport of gases to the chemical reaction

sites. This usually occurs at high current densities because the rate at

which the fuel (hydrogen) is consumed at reaction sites is higher than

91

the rate of diffusion. The scarcity of hydrogen at the chemical reaction

sites effectively reduces the electrode activity leading to a

corresponding loss in output voltage. This polarization is also affected

by the physical restriction of the transfer of a large atom, oxygen, to

the chemical reaction sites on the cathode side of the fuel cell.

Concentration polarization can be reduced by increasing the fuel

concentration and gas pressure, using high surface area electrodes, or

using thinner electrodes which shortens the path of the gas to the

reaction sites [19].

The combination of all the three polarizations affects the overall

operating voltage. Each polarization dominates at a different current

density range. Figure 3.2 shows that when there is no current in the

circuit, the Open Circuit Voltage is reduced by the activation

polarization. As the current increases, the activation polarization

continues to decrease the operating voltage but the rate of reduction

decreases in a parabolic manner. For moderate current densities, the

ohmic polarization dominates and the polarization curve remains

more or less a straight line as shown in the figure 3.2. There is an

inflection point observed at a certain value of the current density and

afterwards the concentration polarization dominates.

As mentioned earlier, the efficiency of the fuel cell is not

restricted by the Carnot limit. Because of the isothermal nature, most

of the energy released in the chemical reaction is converted to

electrical energy, instead of being consumed to raise the products

92

temperature. Hence, the electrochemical processes in the cell offer

high generation efficiencies. The first law efficiency of a SOFC based

upon the lower heating value, is written as,

,e

th cell

n FE

LHV

(3.20)

If fuel input energy is considered, the overall conversion efficiency of

the fuel cell system is given by

*/overall e fW m LHV (3.21)

Where, mf is the mass of fuel consumed

A term, called ideal efficiency, is defined for a fuel cell as :

/ideal G H (3.22)

Which is simply the ratio of available Gibb’s free energy to the total

enthalpy of reaction. For a hydrogen-oxygen cell, operating at

standard condition, the value of this ratio is about 83%. This value

shows the enormous potential of a fuel cell. To achieve a matching

efficiency a Carnot engine would be required to exchange heat with a

source of about 1773K, while rejecting heat to sink at 288K.

The second law efficiency or exergetic efficiency for the

electrochemical process is given by the following expression:

II

Wactual work ideal reversible work

e

G

(3.23)

Considering the fuel cell as steady flow device, the second law

efficiency can also be conveniently expressed as

II

exergy out + work output

exergy in

(3.24)

93

where, the exergy values of the inlet and outlet streams are the values

obtained by adding the physical as well as chemical exergy terms of

the respective streams (by neglecting exergy associated with KE, PE

and other kind of energy).

3.2 Modeling of Solid Oxide Fuel Cell-Gas Turbine Combined

Cycle Power Plant For Different Fuels

A solid oxide fuel cell (SOFC) is an electrochemical conversion device

that produces electricity directly from oxidizing a fuel. Fuel cells are

characterized by their electrolyte material; the SOFC has a solid oxide

or ceramic, electrolyte. Advantages of this class of fuel cells include

high efficiency, long-term stability, fuel flexibility, low emissions, and

relatively of low cost. The largest disadvantage is the high operating

temperature which results in longer start-up times and mechanical

and chemical compatibility issues.

Solid oxide fuel cells are a class of fuel cell characterized by the

use of a solid oxide material as the electrolyte. In contrast to proton

exchange membrane fuel cells (PEMFCs), which conduct positive

hydrogen ions (protons) through a polymer electrolyte from the anode

to the cathode, the SOFC uses a solid oxide electrolyte to conduct

negative oxygen ions from the cathode to the anode. The

electrochemical oxidation of the oxygen ions with hydrogen or carbon

monoxide thus occurs on the anode side. They operate at a very high

temperature, typically between 500 and 1,000°C. At these

temperatures, SOFCs do not require an expensive platinum catalyst

94

material, as is currently necessary for lower temperature fuel cells

such as PEMFCs, and are not vulnerable to carbon monoxide catalyst

poisoning. However, vulnerability to sulfur poisoning has been widely

observed and the sulfur must be removed before entering the cell with

the use of adsorbent beds or other means. Solid oxide fuel cells have a

wide variety of applications from use as auxiliary power units in

vehicles to stationary power generation with outputs from 100 W to 2

MW.

3.2.1 Thermodynamic configuration of GT-SOFC Fuel Cell

Schematic diagram of the GT-SOFC based combined cycle for

power generation system considered in the present investigation and

the corresponding T-S diagram is shown in Figs. 3.3(a) & 3.3(b). It

consists of a compressor, a gas turbine, a combustion chamber, SOFC

and a recuperator. The air is pressurized in the compressor and

preheated in the recuperator is supplied into the cathode of the fuel

cell. The outlet air from the cathode is used to burn the residual

hydrogen, carbon oxide and fuel in the anode outlet gas. The products

of chemical reaction are very lean and hence additional amount of fuel

is injected into the combustion chamber in order to stabilize the

combustion. The extra fuel is supplied for increasing the turbine inlet

temperature. The flue gas from combustion chamber is expanded in

the turbine for power generation and the exhaust gases used to

preheat the compressor outlet air in the recuperator for analyzing the

95

Fig. 3.3(a) Schematic diagram of the GT-SOFC combined cycle power

generation system

Fig.3.3(b) T-S diagram of GT-SOFC Combined cycle Power generation

system

96

combined cycle, a computer code has been developed through “C”,

which consists of several control loops (presented in Appendix A.2) to

estimate the fluid thermodynamic properties and exergy values at

various conditions. The thermodynamic properties of gases are

evaluated using relations provided by Turns et al [147] and details are

presented in Appendix A.1. The effects of various parameters like

compressor pressure ratio, turbine inlet temperature and ambient

temperature are studied. Standard cycle analysis conditions and input

parameters for the simulation are shown in Table 6.1

3.2.2 Assumptions of used for the analysis

The following are the assumptions made based on the Uechi et al.

[148] are made in the model development

1. All gases behave like ideal gases.

2. Gas leakages are negligible.

3. Chemical reactions proceed to equilibrium states.

4. Internal distributions of temperatures, gas composition, and

pressure in each system component are uniform.

5. The system and component performance are calculated only for

steady state conditions.

6. The temperatures at the outlets of anode, cathodes and reformer

are equal to the cell temperature

7. In the combustor, the residual species from the anode and the

injected fuel are burnt

97

Table 3.1 Standard cycle analysis conditions and input

parameters for the simulation of GT-SOFC combined cycle

power generation system.

Ambient conditions Gas Turbine Cycle

Temperature: 298K Turbine efficiency(η gas turbine) : 0.84

Pressure : 101.325KPa Compressor efficiency(η compressor): 0.81

Pressure losses [65] Recuperator efficiency(η recuperator): 0.8

Recuperator air side: 4% AC Generator efficiency(η Generator): 0.95

Recuprator gas side: 4% Combustor efficiency(η combustorr): 0.98

Fuel cell stack : 4% Steam to carbon ratio (SCR) Haseli et al.

[52]: 2.5

Combustor : 5% Properties of fuels

SOFC LHV of methane, Haseli et al. [65] : 50050

kJ/kg

Fuel utilization factor( fU ): 85% LHV of natural gas, Kotas et al. [149]:

47141 kJ/kg

SOFC stack temp(Tstack) : 1273K HHV of coal gas, Kotas et al. [149] :

38380.65 kJ/kg

Current density : 0.3A/cm2 LHV of ethanol, Kotas et al. [149] :

26947.86 kJ/kg

DC – AC Inverter Efficiency(η

inverter) : 0.89

Specific chemical exergy of methane,

Haseli et al. [65]:51840 kJ/kg

Cell area(Ac): 834 cm2 Specific chemical exergy of natural gas,

Kotas et al. [149]: 49272 kJ/kg

Air utilization factor(U a): 25% Specific chemical exergy of coal gas,

Kotas et al. [149]: 38764.45 kJ/kg

Turbine inlet temp: (TIT) :

>10000C

Specific chemical exergy of ethanol,

Kotas et al. [149] : 29746.01 kJ/kg

98

8. Heat loss is negligibly small.

9. In the Fuel cell, all reactants generate their ideal no. of electrons

and no fuel or oxidant crosses the electrolyte.

10. Fuel is provided at the required system pressure.

3.2.3 Thermodynamic analysis of the GT-SOFC system

The analysis of individual components, their chemical reactions,

exergy and energy calculations are discussed below

3.2.3.1 Solid Oxide Fuel Cell (SOFC)

The following chemical reactions that took place generally in SOFC

during power generation [148]

eOHOHAnode 2: 2

2

2 (3.25)

eCOOCO 22

2 (3.26)

2

2 24: OeOCathode (3.27)

In the current analysis, it is assumed that fuel reacts with H2O and

releases H2 and CO. CO again reacts with H2O in shift produces H2.

The heat required for reformer is supplied by the SOFC. The chemical

reactions for different fuels are given below:

1. If methane is the fuel, it is first transformed to synthetic gas by

steam and produces H2 and CO. during shift reaction the CO is

converted into CO2

Reforming 224 3: HCOOHCH (3.28)

Shifting: 222 HCOOHCO (3.29)

99

2. If Ethanol is the fuel based on Douvarttzides et al. [63], One mole

of ethane releases six moles of H2 and two moles of CO.

2252 623 HCOOHOHHC (3.30)

Steam reforming reaction:

3. If coal gas is used as Fuel, the following chemical reactions take

place in the coal gasifier

2 2 22 2C H O CO H (3.31)

2 43 2 2C H O CO CH (3.32)

Overall reaction is given by

2 2C H O CO H (3.33)

The hydrogen is generated through the above chemical reaction and it

is supplied to the fuel cell.

2252 42 HCOOHOHHC (3.34)

Reforming is endothermic reaction and shifting is exothermic reaction.

The net reaction in the reformer is endothermic reaction.

The net cell reaction for methane is written as

OHCOOCH 2224 22 (3.35)

OHOH 2222

1 (3.36)

The energy interactions of the cell require the evaluation of both the

current and voltage. The reversible cell voltage, E, is defined by Haseli

et al. [65] by considering Nernst equation is

OHPP

OPP

F

RTEE

CO

CH

2

2

2

2

0

2

4ln8

(3.37)

100

Based on the Uechi et al. [148], ideal voltage values for an

intermediate temperature of SOFC operating at 8000C and 11000C are

0.99V and 0.91V respectively.

The DC power produced by the SOFC is given by Haseli et al. [65]

cc AjVDCPele , (3.38)

Where, lossc VEV and ionconcentratohmicactivationloss VVVV (3.39)

The actual cell voltage ‘Vc’ depends upon the operating parameters like

the current density (j), operating pressure and temperature etc. Fuel

cell hand book includes empirical formulae that correlate the

performance of an SOFC to these parameters.

The effect of pressure based on the Williams [19] is given by

2

1

( ) 59lnp

PV mv

P (3.40)

The effect of temp is given by

2 1( ) 0.008( )TV mv T T j (3.41)

Rate of heat production is,

6

,, 10125.1

c

DCeleFcgenV

PQ ,kW (3.42)

The oxygen required for the chemical reaction is normally supplied

from air. The air flow is usually well above the stoichiometric amount,

normally twice higher. Where λ is the stoichiometric ratio

Mass flow rate of air usage,

c

DCele

FCaV

Pm

,7

, 1057.3 , kg/Sec (3.43)

101

The exergy of fuel is being considered on the sum of thermal,

mechanical and chemical exergies. The detailed equations are

provided by Kotas [149] and gives below

Thermal exergy, , 0 0

0

lnx thermal p

TE m C T T T

T

(3.44)

Mechanical exergy,

0

0, lnP

PTRmE mechanicalx (3.45)

Chemical exergy,

i

iichemicalx

y

xxTRmE ln0, (3.46)

The thermal exergy depends on the temperature of the fuel cell

and the mechanical exergy depends on the compression pressure

ratio. The entropy values at different points are required for the

evaluation of irreversibility.

The irreversibility in the solid oxide fuel cell is estimated based on the

equations available in the Cengel and Boles [150]

003040 reactionSa

Sa

Sp

Sp

STSOFC

I (3.47)

Irreversibility of the chemical reaction explained by Srinivas et al.

[151], is given by

10 ffcrxn LHVmST (3.48)

Where 0357.150050

51840

H

G (3.49)

Energy balance

The energy balance is made based on the Haseli et al. [52] and given

below

102

3 3 , 4 41 , 0ffc f ffc f ele DCm h m U LHV m U hf in P m h (3.50)

Exergy balance

Exergy balance is made based on the Haseli et al. [65]

0,,44,33 SOFCxDCeleffchffcftfmffc DEPmUmmm

(3.51)

3344,

,

,

mmUmm

P

fchfffcftfmffc

DCele

FCex (3.52)

3.2.3.2 Combustion Chamber

The products from the SOFC are further heated in the combustion

chamber by supplying adequate quantity of fuel in order to raise the

temperature. The unburnt fuel in the SOFC is also burnt in the

combustion chamber.

The Energy balance of Combustion Chamber is given below :

05543 losscombustionffcf QhmQhmUm (3.53)

LHVffc

mf

Uffc

mcombustion

Q

1 (3.54)

LHVcombustionffc

mf

Uffc

mloss

Q

11

(3.55)

The exergy balance of combustion chamber based on the exergy

values is discussed below

ccxCHftfmffcfchffc DEmfmUfmm 554,44 1 (3.56)

The irreversibility in combustion chamber

00405

rxnSa

Sa

Sp

Sp

SCC

I (3.57)

103

Where 1tan0 fLHV

fccmtsreacSTrxnS (3.58)

Exergy efficiency,

1001 ,

4455

,

chfphffccfchfffc

CCexmUm

mm (3.59)

3.2.3.3 Compressor

Irreversibility in the compressor, 120 SSTI compressor (3.60)

Exergy efficiency,

100121

121

hhm

mcompressor

(3.61)

3.2.3.4. Recuperator

Irreversibility in the recuperator,

132766 mmI rrecuperato (3.62)

Exergy efficiency,

100766

232,

m

mrrecuperatoex (3.63)

3.2.3.5. Gas Turbine

Rate of exergy loss in the gas turbine,

6505 SSTmI turbinegas (3.64)

Exergy efficiency,

100655

m

W turbinegas

turbinegas (3.65)

3.2.3.6 Performance of the plant

Net power developed by SOFC stack, DCeleinverterACFC PP ,, (3.66)

Net power developed by the gas turbine, turbinegasgengen PP (3.67)

Total net power developed by the system, genACFCnet PPP , (3.68)

The total heat supplied to the system,

combustionCHfffctotal QLHVUmQ 4

(3.69)

The total thermal efficiency of the cycle,

104

100, tot

netcycleth

Q

P (3.70)

The exergy efficiency of the cycle,

100

,

,

chfftfmf

netcycleex

m

P (3.71)

3.2.3.7 Model Calculations :-

Calculation of the Temperature, Pressure and Mass flow rate fluid at

each state points of the cycle.

Pressures :-

Pressure of air at the inlet to the compressor is assumed as ambient

temperature

P1 = 101.325 kN/m2 (or) kPa

Let 2

1

P

P = pressure ratio in the compressor = 4

(Assuming that the pressure ration in fuel cell is same as the Pressure

Ratio)

Therefore P2 = 101.325 x 9 = 405.3 kPa

The pressure of air at the outlet of the recuperator by considering the

pressure losses in the heat exchanger

P3 = P2 * Pressure drop in the Recuperator

1100

= 405.3 KPa 4

1100

= 389.088 kPa

Pressure of working fluid at the outlet of the SOFC is also calculated

by considering the pressure drop in the SOFC

105

P4 = P3 * Pressure drop in the SOFC

1100

= 389.088 x 5

1100

= 369.634 kPa

Pressure of working fluid at the outlet of the Combustion Chamber is

also calculated by considering the pressure drop in the Combustion

Chamber.

P5 = P4 * Pressure drop in the CC

1100

= 369.634 5

1100

= 351.152 kPa

Pressure of working fluid at the outlet of the recuperator is also

evaluated by considering the pressure drop in the recuperator

P6 = P1 / Pressure drop in the Recuperator

1100

= 101.325 / 4

1100

= 105.547 kPa

Pressure of working fluid at the outlet of the Recuperator

= P7 = P1 = 101.325 kPa

Let the Pressure of fuel being supplied to SOFC and combustion

chamber = P8 = P9 = 390 kPa

Mass Flow Rates :-

Mass flow rate of air through SOFC [19] is given by

ma, fc = m1 = m2 = m3 = 3.57 X 10-7 x x ,ele Dc

c

P

V

kg/s = 4.50258 kg/s

Mass flow rate of fuel entering the SOFC

106

mf, fc = ma, fc x 1

AFR= 4.50258 x

1

65= 0.06927 kg/s.

Mass flow rate of fluid entering the SOFC

= m4 = ma, fc + mf, ,fc = 4.57185 kg/s.

Mass flow rate of fuel entering the combustion chamber

= mf, cc = 0.3 x mf, fc = 0.3 x 0.06927 = 0.020781 kg/s.

Mass flow rate of fluid leaving the combustion chamber

= m5 = m4 + mf, cc = 4.57185 + 0.020781 = 4.592631 kg/s.

Therefore, m5 = m6 = m7 = 4.592631 kg/s.

Temperatures :-

Temperature of air at the inlet of the compressor = T1 = 298 K

(ambient temperature is assumed)

Temperature of air at the exit of the compressor is calculated by

considering the isentropic efficiency

1

2 1

1* 1

p

ac

rT T

=476.797K

The amount of heat generated in the fuel cell is estimated by the

equation provided by [17 ]

Qgen, FC = , 1.251 1694.14294

1000

ele DcPx kW

Vc

The temperature of working fluid leaving the SOFC = T4

gen, FC

4 0

,

Q

pg fc

T Tc

Cpg, fc = 1.2168 kJ/kg K

Therefore, T4 = 1094.2937K

107

Temperature of gases entering the gas turbine T5 is assumed as

1250K, which is due of the variable parameter i.e. turbine inlet

temperature (TIT)

T5 = 1250K

Temperature of gases leaving the gas turbine by considering the

isentropic efficiency of the turbine and is given as

1

56 5

6

1 1

g

g

gt

PT T

P

Let g = 1.33 for the gases leaving the gas turbine

T6 = 979.21661K

Temperature of gases leaving the recuperator,T7 = 670K

Therefore, Temperature of air leaving the recuperator

,

3 2 6 7

g cc pg

afc pafc

m CT T T T

m C

T3 = 836.82568K

Estimation of Exergy values of each state of the cycle

Thermal exergy of air entering the air compressor[149]

1

11 0 0

0

* lnth a pa

TE m c T T T

T

Let T1 = T0 & P1 = P0

Therefore,1thE = 0

Exergy Evaluations :-

Physical exergy is defined as the sum of the thermal exergy and

mechanical exergy.

108

Thermal exergy of a fluid at a temperature T and the reference

temperate T0 is given

0 0

0

lnth cp

Tm T T T

T

Mechanical exergy of a fluid at a temperature P and the reference

pressure P0 is given

0

0

lnm

PmRT

P

Chemical exergy = 0 ln i

m i

i

ymRT y

x

The exergy of the working fluid can be determined at each state of the

system using the above equations.

State 1 :-

T1 = T0, P=P0

Therefore, th,1 = 0, m1=0, ch1=0

State 2 :-

T2 = 476.797K, P2 = 405.3 kPa

m2 = 4.50258 kg/s.

T0 = 298.0K, P1=101.325 kPa

2

476.7974.50258*1.23 476.797 298 298ln

298th

= 178.43324kW

,2

405.34.50258*0.287*298*ln

101.325m

= 533.8448kW

109

ch,2 = 0

State 3 :-

T3 = 836.82508 K m3 = m2 = m1 = 4.50258 Kg/s.

P3 = 389.088 kPa R = 0.287 kJ/kgK

Cpa = 1.023 kJ/kgK

,3

836.825684.50258*1.023* 836.82568 298 298ln

298th

= 991.012kW

,3

389.0884.50258*0.287*298*ln

101.325m

= 518.1248kW

ch,3 = 0 (No change in chemical composition of air)

State 4 :-

T4 = 1094.2937 K mf,fc = 0.06927 kg/s.

P4 = 369.634 kPa

Cpg = 1.2168 kJ/kgK

Rg = 0.2926 kJ/kgK

m4 = m= + mf,fc = 4.57185 kg/s.

,4

1094.29374.57185*1.2168* 1094.2937 298 298ln

298th

= 2273.4087kW

,4

369.6344.57135*0.2926*298*ln

101.325m

= 515.9139kW

110

State 5 :-

T5 = 1250 K Cpg = 1.2545 kJ/kgK

Rg = 0.2926 P5 = 351.152 kPa

m5 = m4 + mf,cc mf,cc = 0.020781 kg/s

= 4.592631 Kg/s

,5

12504.592631*1.2545* 1250 298 298ln

298th

= 3023.185510kW

,5

351.1524.592631*0.2926*298*ln

101.325m

= 446.6875kW

State 6 :-

T6 = 979.21661 K

P6 = 105.547 kPa

Cpg = 1.2545 kJ/kg K

m5 = m6 = 4.592631 kg/s

R = 0.2926 kJ/kg K

,6

979.216614.592631*1.2545* 979.2166 298 298ln

298th

= 1882.256kW

,6

105.5474.592631*0.2926*298*ln

101.325m

= 16.34778kW

State 7 :-

T7 = 670 K Cpg = 1.2545 kJ/kgK

111

P1 = P7 = 101.325 kPa m7 = m6 = m5 = 4.592631 kg/s

R = 0.2926 kJ/kg K

,7

6704.592631*1.2545* 670 298 298ln

298th

= 752.245kW

m,7 = 0

Physical exergy of fuel :-

Fuel Cell : P8 = 450 kPa

Rg = Ru / M = 0.518354 kJ/kg K

mf,fc = m8 = 0.06927 kg/s.

,8

4500.06927*0.518354*298*ln

101.325m

= 15.953kW

Combustion Chamber :

P9 = 430 kPa

Rg = 0.518354

mf,cc = m9 = 0.020781 kg/s.

,9

4300.020781*0.518354*298*ln

101.325m

= 4.639kW

Thermal exergies of fuel supplied to Fuel Cell and Combustion

Chamber are assumed to be Zero since the temperature difference is

negligible.

Chemical exergies of fuel and air :-

At the states 1, 2 & 3

112

ch,1,2,3 = 0 (working fluid is air only)

State 4 : -

In the fuel cell, air composition changes [17]

2

0

Nx = 0.7748, 2

0

Ox =0.2059, 2

0

COx =0.003, 2

0

H Ox =0.019

2Ny = 0.76887,

2Oy =0.15122, 2COy =0.0266,

2H Oy =0.05327

0 ln i

ch

i

ymRT yi

x

,4

0.76887 0.151220.76887 ln 0.15122ln

0.7748 0.20594.57185* 0.2926 298

0.0266 0.053270.266ln 0.05327 ln

0.0003 0.019

ch

= 4.57185 x 0.2926 x 298 [-0.0059 + (-0.046675) +

0.1192979 + 0.0549178]

= 4.57185 x 0.2926 x 298 x 0.1216407

= 48.491kW

State 5:-

At this state gas composition changes to

2Nx = 0.76887,

2Ox = 0.15122, 2COx = 0.0266,

2H Ox = 0.05327

2Ny = 0.76325,

2Oy = 0.13547, 2COy = 0.03753,

2H Oy = 0.06752

=

0.76325 0.135470.76325ln 0.13547ln

0.76887 0.151220.2926*298

0.03753 0.067520.03753ln 0.06752ln

0.0266 0.05327

= 4.592631 x 0.2926 x 298 [-0.005599 – 0.014899 +

0.0129189 + 0.0160057

113

= 4.592631 x 0.2926 x 298 (0.0084266)

= 3.3744kW

ch,6,7 = 3.3744kW

Exergy of Fuel :-

Specific exergy of fuel [65] = f = 51840 KJ/Kg

Total Exergy of fuel supplied to the Fuel Cell

f = mf,fc x f

= 0.06927 x 51840

f,8 = 3590.9568kW

Total exergy of fuel supplied to combustion chamber

= mf,cc x f

= 0.020781 x 51840

f,9 = 1077.28704kW

Total exergy of fuel supplied to the cycle

= total = ch,4 + ch,5 +f,8 +f,9 + m,8 + m,9 +

= 48.491 + 3.3744 + 3590.9568 + 1077.287

+ 15.953 + 4.639

= 4740.701kW

Exergy Destruction[150] in each component of the system :-

1. Compressor :-

2 2

0

1 1

* * *ln lnc a pa

T PI m T C R

T P

467.797 405.3

4.50258*298 1.023*ln 0.287 ln298 101.325

114

4.50258*298 0.480807 0.397866

Ic = 111.2876kW

2. Recuperator :-

IRecup = Exergy of product gases – Exergy of air

Exergy of product gases = 6

, 6 7 0

7

* lng cc pg

Tm C T T T

T

= 979.21661

4.592631*1.2545* 979.21661 670 298ln670

= 1130.0.116kW

Exergy of air = 3

3 2 0

2

* lnagc pa

Tm C T T T

T

= 836.82508

4.50258*1.023* 836.82568 476.797 298ln476.797

= 886.2039kW

IRecup = 886.2039kW

3. Fuel Cell :-

3 34 4

, 0 ,

0 0 0 0

* * ln ln lnSOFC g fc pgfc g a x rxn fc

T PT PI m T C R R E D

T P T P

Exergy loss in chemical reaction in the fuel cell is given by [151 ]

ExDrxn,fc = T0(S)rxn = mf,fc(LHV)f(-1)

Specific exergy of fuel 51840

1.035764Lower calorific value 50050

ExDrxn,fc = 0.06927 x 50050 x (1.035764-1)

= 123.9924kW

115

1094.2937 369.6341.2168*ln 0.2926ln

298 101.3254.57185* 298 123.9924

836.82568 389.0881.099ln 0.287ln

298 101.325

SOFCI KW

= 4.57185 x 298 [1.582778 – 0.378677

– 1.13474 + 0.38615]+123.99

= 4.57185 x 298 x (0.455511) + 123.9924

ISOFC = 744.5858kW

4. Combustion Chamber :-

5 5 4 4

0 , ,

0 0 0 0

* 298 ln ln ln *lncc gcc pgcc g pa cc a x ran cc

T P T PI m T C R C R E D

T P T P

The irreversibility in combustion reaction[151] is given by

ExDrxn,cc = T0(S)rxn = mf,fc(LHV)f(-1)

= 0.020781 x 50050 x (1.035764-1)

= 37.197744kW

1250 351.1521.2545* ln 0.2926ln

298 101.3254.592631* 298 37.197744

1094.2937 369.6341.182ln 0.287 ln

298 101.325

ccI kW

= 4.592631 x 298 [1.798708 – 0.363668 – 1.5375

+0.371429] + 37.197744

= 368.112059 + 37.197744

Icc = 405.3098kW

116

5. Gas Turbine :-

5 5

, 0

6 6

* ln *lngt g cc pg g

T PI m T C R

T P

= 1250 351.152

4.592631*298 1.2545*ln 0.2926ln979.21661 105.547

= -1368.604038 [0.306281 – 0.351724]

Igt = 62.193473kW

Exergy efficiency of System Components:

1) Compressor:-

2 11

,

1 2 1

100x x

ex c

mx

m h h

=

2 2

1 , 2 1

712.27886.48%

823.568

t m

pa fcm C T T

2) Recuperator:-

3 2

6

3

,

6 7

100x x

ex recup

x x

m

m e e

= 886.20227

100 78.42%1130.0166

x

3) Solid Oxide Fuel Cell:-

,

,

,

100de Dc

ex SOFC

de Dc x SOFC

P

P E D

= 2247.19092

100 75.11%2991.77588

117

4) Combustion Chamber:-

, ,

,

* 1 *

[4]

af fc f f f cc m f

ex cc

x cc

m U M

EP E D

= 1215.28528

100 74.9907%1620.58105

5) Gas Turbine:- 10 11

,[5] [5] [6] [6]

ex gt

t m t m

= 1560.1068

100 96.166%1622.2998

Performance of the Cycle:-

The total hat supplied to the system

Qtotal = (mf,fc + mf,cc) x LHV

= (0.06927 + 0.020781) 50050

= 4507.05255kW

Net Power developed by the fuel cell

PFC,AC = inverter x Pele, Dc

= 0.89 x 2247.19092

= 2000kW

Net power developed by the Gas Turbine

Pgen = gen x P gas turbine

Power developed by the gas turbine

Pgt = WT - WC

= [mg, cc x Cpg, cc X (T5-T6)] – [ma, fc x Cpa, fc X(T2-T1)]

= [4.592631 x 1.2545 (1250-979.21661)]

-[4.50258 x 1.023 x (476.797 - 298)

118

= 1560.106476 - 823.56389

Pgt = 736.542586kW

Pgen = 0.95 x Pgt = 699.71546kW

Total net power developed by the system

Pnet = PFC,AC + Pgen

= 2000 + 699.71546

= 2699.71546kW

Total thermal efficiency of the system (I law efficiency)

100netth

tot

Px

Q

= 2699.71546

100 59.899%4507.05255

x

Total exergy (II Law) efficiency of the cycle

57.57%netex

total

P