CHAPTER 3 MODELING OF SOLID OXIDE FUEL CELL GAS TURBINE HYBRID SYSTEM. Solid oxide fuel cells consists of a solid electrolyte (zirconia), which is a

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1 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] 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

2 85 electrolyte, and flow channels for air delivery & collection and fuel. Figure 3.1 Schematic of a Solid Oxide Fuel Cell[8] H or a hydrocarbon like methane is supplied on the anode and air or O on the cathode side of the fuel cell. H and CO (if H is not pure) diffuse through the porous anode to the three phase boundary formed by the electrolyte, the gaseous H and anode Similarly, O 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 H 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 : O 4e O At the anode : H O 4HO 4e (3.1) (3.)

3 86 Water gas shift at anode : CO HO CO H (3.3) If CO is present in the H stream, the CO reacts with H O via a water gas shift reaction that produces H and CO. 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, we get, G -G 1 = nrt ln(p/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)

4 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, m n PP M N G Go RT ln a b PP A B (3.10) where G 0 is the standard Gibbs free energy change for the reaction, G mg ng ag bg o o o o o M N A B (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 (3.1) 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 = -W e (3.14) i.e. the change in Gibbs free energy of the reaction is equal to the maximum electrochemical work, W e, 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

5 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, W e = n e FE (3.15) where F is the total charge of 1 mole of electrons, known as Faraday s constant. Therefore G = -n e FE (3.16) Applying equation 16 to equation 10, we get what is known as Nernst equation, m n RT PP M N EEo ln n a b e F P A P B (3.17) where E is related to G by Equation 15. For the reaction occurring in an SOFC, 1 H O HO (3.18) the reversible potential can be written as, E RT P HO Eo ln 1/ F PH P O (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

6 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.. Figure 3. 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

7 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

8 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. 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.. 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

9 9 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, th, cell ne FE LHV (3.0) If fuel input energy is considered, the overall conversion efficiency of the fuel cell system is given by W / m LHV overall e f * (3.1) Where, mf is the mass of fuel consumed A term, called ideal efficiency, is defined for a fuel cell as : ideal G/ H (3.) 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 88K. The second law efficiency or exergetic efficiency for the electrochemical process is given by the following expression: actual work W e II ideal reversible work G (3.3) 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 (3.4)

10 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. 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

11 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 MW 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

12 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

13 96 combined cycle, a computer code has been developed through C, which consists of several control loops (presented in Appendix A.) 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 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.. 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

14 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: 98K Turbine efficiency(η gas turbine) : 0.84 Pressure : KPa 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. [5]:.5 Combustor : 5% Properties of fuels SOFC LHV of methane, Haseli et al. [65] : kj/kg Fuel utilization factor( U ): 85% LHV of natural gas, Kotas et al. [149]: f kj/kg SOFC stack temp(tstack) : 173K HHV of coal gas, Kotas et al. [149] : kj/kg Current density : 0.3A/cm LHV of ethanol, Kotas et al. [149] : 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 cm Specific chemical exergy of natural gas, Kotas et al. [149]: 497 kj/kg Air utilization factor(u a): 5% Specific chemical exergy of coal gas, Kotas et al. [149]: kj/kg Turbine inlet temp: (TIT) : Specific chemical exergy of ethanol, > C Kotas et al. [149] : kj/kg

15 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 Thermodynamic analysis of the GT-SOFC system The analysis of individual components, their chemical reactions, exergy and energy calculations are discussed below Solid Oxide Fuel Cell (SOFC) The following chemical reactions that took place generally in SOFC during power generation [148] Anode : H O H O e (3.5) CO O CO e (3.6) Cathode : O 4e O (3.7) In the current analysis, it is assumed that fuel reacts with HO and releases H and CO. CO again reacts with HO in shift produces H. 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 H and CO. during shift reaction the CO is converted into CO Reforming : CH 4 HO CO 3H (3.8) Shifting: CO HO CO H (3.9)

16 99. If Ethanol is the fuel based on Douvarttzides et al. [63], One mole of ethane releases six moles of H and two moles of CO. C H 5OH H O CO 6 3 H (3.30) Steam reforming reaction: 3. If coal gas is used as Fuel, the following chemical reactions take place in the coal gasifier C H O CO H (3.31) 3C H O CO CH (3.3) 4 Overall reaction is given by C HO CO H (3.33) The hydrogen is generated through the above chemical reaction and it is supplied to the fuel cell. C H5OH H O CO 4 H (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 CH 4 O CO HO (3.35) 1 O H O (3.36) H 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 E E 0 RT P ln 8F P CH CO P O 4 P H O (3.37)

17 100 Based on the Uechi et al. [148], ideal voltage values for an intermediate temperature of SOFC operating at C and C are 0.99V and 0.91V respectively. The DC power produced by the SOFC is given by Haseli et al. [65] Pele, DC V j (3.38) c A c Where, V c E Vloss and Vloss Vactivation Vohmic Vconcentrat ion (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 P Vp ( mv) 59ln (3.40) P1 The effect of temp is given by V ( mv) 0.008( T T ) j (3.41) T 1 Rate of heat production is, Q,, gen Fc Pele DC 1 10,kW (3.4) Vc 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, P 7 ma, FC V ele, DC, kg/sec (3.43) c

18 101 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 T Ex, thermal m Cp T T0 T0 ln T0 Thermal exergy, (3.44) Mechanical exergy, P x, m R T0 ln (3.45) P 0 E mechanical xi Chemical exergy, E x, chemical m R T0 xi ln (3.46) yi 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] 0 p 4 p 0 a 3 a 0 reaction I T S S S S S (3.47) SOFC Irreversibility of the chemical reaction explained by Srinivas et al. [151], is given by S m LHV T 0 fc 1 (3.48) rxn f 0 G Where H (3.49) Energy balance The energy balance is made based on the Haseli et al. [5] and given below

19 ffc f ffc f ele, DC 4 4 m h m U LHV m 1 U hf, in P m h 0 (3.50) Exergy balance Exergy balance is made based on the Haseli et al. [65] m m U m P E, D 0 33 m ffc fm ft ffc ch, f f 4 4 ele, DC x SOFC ex, FC m m U m m ch, f (3.51) Pele, DC (3.5) ffc fm ft ffc f 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 : U m h Q m h Q 0 m (3.53) 3 f ffc 4 combustion 5 5 loss Q combustion Q loss m U m LHV ffc 1 f ffc (3.54) m U m LHV ffc 1 f ffc 1 (3.55) combustion The exergy balance of combustion chamber based on the exergy values is discussed below Uf ch, f m ffc fm ft fch m5 ExDcc m4 1 (3.56) 4 m ffc 4 5 The irreversibility in combustion chamber S S CC p 5 p 0 S S S a 4 a 0 rxn 0 I (3.57)

20 103 Where T S m LHV 1 S rxn 0 reac tan ts fcc f (3.58) m55 m44 Exergy efficiency, ex, CC 100 (3.59) m U m Compressor ffc 1 f ch, f Irreversibility in the compressor, T S I compressor fcc phf chf (3.60) 0 S1 Exergy efficiency, m (3.61) compressor m h h Recuperator Irreversibility in the recuperator, I m recuperato r m (3.6) Exergy efficiency, m 3, 100 (3.63) ex recuperator m Gas Turbine Rate of exergy loss in the gas turbine, I gas turbine 5 T0 S5 S6 m (3.64) Exergy efficiency, W gas turbine gas turbine 100 (3.65) m Performance of the plant Net power developed by SOFC stack, P FC, AC inverter Pele, DC (3.66) Net power developed by the gas turbine, P gen P (3.67) gen gas turbine Total net power developed by the system, P net PFC, AC P gen (3.68) The total heat supplied to the system, Q total m ffc U f LHV CH 4 Q combustion (3.69) The total thermal efficiency of the cycle,

21 104 Pnet th, cycle 100 (3.70) Q tot The exergy efficiency of the cycle, Pnet ex, cycle 100 (3.71) m f fm ft f, ch 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 = kn/m (or) kpa P Let P 1 = pressure ratio in the compressor = 4 (Assuming that the pressure ration in fuel cell is same as the Pressure Ratio) Therefore P = x 9 = kpa The pressure of air at the outlet of the recuperator by considering the pressure losses in the heat exchanger P3 = P * Pressure drop in the Recuperator = KPa = kpa Pressure of working fluid at the outlet of the SOFC is also calculated by considering the pressure drop in the SOFC

22 105 P4 = P3 * Pressure drop in the SOFC = x = 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 = = 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 = / = kpa Pressure of working fluid at the outlet of the Recuperator = P7 = P1 = 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 = m = m3 = 3.57 X 10-7 x x P V ele, Dc c kg/s = kg/s Mass flow rate of fuel entering the SOFC

23 106 mf, fc = ma, fc x 1 AFR = x 1 = kg/s. 65 Mass flow rate of fluid entering the SOFC = m4 = ma, fc + mf,,fc = kg/s. Mass flow rate of fuel entering the combustion chamber = mf, cc = 0.3 x mf, fc = 0.3 x = kg/s. Mass flow rate of fluid leaving the combustion chamber = m5 = m4 + mf, cc = = kg/s. Therefore, m5 = m6 = m7 = kg/s. Temperatures :- Temperature of air at the inlet of the compressor = T1 = 98 K (ambient temperature is assumed) Temperature of air at the exit of the compressor is calculated by considering the isentropic efficiency T 1 rp 1 T *1 ac 1 = K The amount of heat generated in the fuel cell is estimated by the equation provided by [17 ] Qgen, FC = P ele Dc, 1.5 x kw 1000 Vc The temperature of working fluid leaving the SOFC = T4 T Q T gen, FC 4 0 cpg, fc Cpg, fc = kj/kg K Therefore, T4 = K

24 107 Temperature of gases entering the gas turbine T5 is assumed as 150K, which is due of the variable parameter i.e. turbine inlet temperature (TIT) T5 = 150K Temperature of gases leaving the gas turbine by considering the isentropic efficiency of the turbine and is given as T P 5 T 1gt 1 P g g Let g = 1.33 for the gases leaving the gas turbine T6 = K Temperature of gases leaving the recuperator,t7 = 670K Therefore, Temperature of air leaving the recuperator T T m C T T g, cc pg m afc C pafc T3 = K Estimation of Exergy values of each state of the cycle Thermal exergy of air entering the air compressor[149] T 1 Eth m * 1 a cpa T1 T0 T0 ln T0 Let T1 = T0 & P1 = P0 Therefore, E th = 0 1 Exergy Evaluations :- Physical exergy is defined as the sum of the thermal exergy and mechanical exergy.

25 108 Thermal exergy of a fluid at a temperature T and the reference temperate T0 is given th T mcp T T0 T0ln T0 Mechanical exergy of a fluid at a temperature P and the reference pressure P0 is given m P mrt0 ln P0 yi Chemical exergy = m mrt0 yi ln xi 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 :- T = K, P = kpa m = kg/s. T0 = 98.0K, P1= kpa th * ln 98 = kW m, *0.87*98*ln = kW

26 109 ch, = 0 State 3 :- T3 = K P3 = kpa m3 = m = m1 = Kg/s. R = 0.87 kj/kgk th,3 Cpa = 1.03 kj/kgk *1.03* ln 98 = kW m, *0.87*98*ln = kW ch,3 = 0 (No change in chemical composition of air) State 4 :- T4 = K mf,fc = kg/s. P4 = kpa Cpg = kj/kgk Rg = 0.96 kj/kgk m4 = m= + mf,fc = kg/s. th, *1.168* ln 98 = kW m, *0.96*98*ln = kW

27 110 State 5 :- T5 = 150 K Rg = 0.96 m5 = m4 + mf,cc Cpg = kj/kgk P5 = kpa mf,cc = kg/s = Kg/s th, *1.545* ln 98 = kW m, *0.96*98*ln = kW State 6 :- T6 = K P6 = kpa Cpg = kj/kg K m5 = m6 = kg/s R = 0.96 kj/kg K th, *1.545* ln 98 = kW m, *0.96*98*ln State 7 :- = kW T7 = 670 K Cpg = kj/kgk

28 111 P1 = P7 = kpa m7 = m6 = m5 = kg/s R = 0.96 kj/kg K th, *1.545* ln 98 = 75.45kW m,7 = 0 Physical exergy of fuel :- Fuel Cell : P8 = 450 kpa Rg = Ru / M = kj/kg K mf,fc = m8 = kg/s. m, * *98*ln = kW Combustion Chamber : P9 = 430 kpa Rg = mf,cc = m9 = kg/s. m, * *98*ln = 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, & 3

29 11 State 4 : - ch,1,,3 = 0 (working fluid is air only) In the fuel cell, air composition changes [17] x = , 0 N x =0.059, 0 O x =0.003, 0 CO x = HO y N = , y O =0.151, y CO =0.066, y HO= yi ch mrt0 yi ln xi ln ln ch, * ln ln = x 0.96 x 98 [ ( ) ] = x 0.96 x 98 x = kW State 5:- At this state gas composition changes to x N = , O x = 0.151, x CO = 0.066, x HO= y N = , y O = , y CO = , y HO= ln ln = 0.96* ln ln = x 0.96 x 98 [

30 113 = x 0.96 x 98 ( ) = kW ch,6,7 = kW Exergy of Fuel :- Specific exergy of fuel [65] = f = KJ/Kg Total Exergy of fuel supplied to the Fuel Cell f = mf,fc x f = x f,8 = kW Total exergy of fuel supplied to combustion chamber = mf,cc x f = x f,9 = kW Total exergy of fuel supplied to the cycle = total = ch,4 + ch,5 +f,8 +f,9 + m,8 + m,9 + = = kW Exergy Destruction[150] in each component of the system :- 1. Compressor :- T P Ic ma * T0 * Cpa *ln R ln T1 P * *ln 0.87 ln

31 * I c = kW. Recuperator :- IRecup = Exergy of product gases Exergy of air T mg, cc * Cpg T6 T7 T0 ln T Exergy of product gases = *1.545* ln 670 = = kW T magc * Cpa T3 T T0 ln T Exergy of air = *1.03* ln = = kW IRecup = kW 3. Fuel Cell :- T P T P I m * T * C ln R ln R ln E D SOFC g, fc 0 pgfc g a x rxn, fc T0 P0 T0 P0 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 Lower calorific value ExDrxn,fc = x x ( ) = kW

32 115 ISOFC * ln 0.96ln * KW ln 0.87 ln = x 98 [ ] = x 98 x ( ) ISOFC = kW 4. Combustion Chamber :- T P T P I m * T 98 C ln R ln C ln R *ln E D cc gcc 0 pgcc g pa, cc a x ran, cc T0 P0 T0 P0 The irreversibility in combustion reaction[151] is given by ExDrxn,cc = T0(S)rxn = mf,fc(lhv)f(-1) = x x ( ) = kW Icc * ln 0.96 ln * kw ln 0.87 ln = x 98 [ ] = Icc = kW

33 Gas Turbine :- T5 P5 I gt mg, cc * T0 Cpg ln Rg *ln T6 P6 = * *ln 0.96ln = [ ] Igt = kW Exergy efficiency of System Components: 1) Compressor:- ex, c m 1 x x1 m h h 1 1 x100 = t m m C T T 1 pa, fc % ) Recuperator:- ex, recup m3 m e e x3 x 6 x6 x7 100 = % x 3) Solid Oxide Fuel Cell:- P de, Dc ex, SOFC Pde, Dc ExD SOFC 100 = %

34 117 4) Combustion Chamber:- ex, cc m f, fc * 1 U f * f M f, cc m a f EP[4] E D x cc = % ) Gas Turbine:- ex, gt [5] [5] [6] [6] t m t m = % Performance of the Cycle:- The total hat supplied to the system Qtotal = (mf,fc + mf,cc) x LHV = ( ) = kW Net Power developed by the fuel cell PFC,AC = inverter x Pele, Dc = 0.89 x = 000kW 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(T-T1)] = [ x ( )] -[ x 1.03 x ( )

35 118 = Pgt = kW Pgen = 0.95 x Pgt = kW Total net power developed by the system Pnet = PFC,AC + Pgen = = kW Total thermal efficiency of the system (I law efficiency) P th Q net tot x 100 = % x Total exergy (II Law) efficiency of the cycle Pnet ex 57.57% total