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Precept 5: Repowering of a cogeneration plant Mechanical Engineering - Energy Systems Authors: •Group xx: xxxxx xxxxx (xxxxxxxx), Fabio Santoro (xxxxxxxx) Lecturer:Prof. Stefano Consonni Teaching Assistants:Riccardo Cremona, Hamidreza Heydari and Nima Razmjoo Academic year:2025-2026Contents 1 Description2 2 Configuration A4 2.1 Performance indicators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 2.2 Condenser. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 3 Configuration B - solution53.1 HP HRSG section + HP turbine. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5 3.2 LP HRSG section + Attemperator. . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 3.3 LP turbine. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 3.4 Condenser. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 3.5 Extraction pump, LP feedwater pump and HRSG exit. . . . . . . . . . . . . . . . . .8 3.6 Numerical results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9 3.7 Performance indicators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12 References12 1 Energy Systems- Report Precept 5 - Group xx 1.Description The scope of these precept is to evaluate the feasibility for structural conversion of a power plant. The existing plant consists in a steam cycle electric power generation plant where the steam is provided by a boiler as represented in Figure1.Figure 1:Existing plant scheme In the new solution (Figure2), we manage to replace the boiler and its auxiliary devices with a gas cycle that produces electricity. The hot flue gases then enter the heat recovery steam generator (HRSG) to generate the steam that feeds the same existing steam plant. The steam plant operates in a bottoming configuration, where the main purpose is to produce electricity by means of two steam turbines operating at different pressure levels, while the residual heat is recovered for a heat user before re-entering the HRSG. In this plant, we use the same condenser as in the existing one. In the HRSG, there are several piping systems: economizers, evaporators, and superheaters for both pressure levels of the turbines. We need to compare the main performance indicators between the existing and the new configurations. We are also required to determine all the unknowns highlighted in system2. Extra data is provided in the following list:•The steam turbines, which don’t change between configurations A and B, are characterized by a coefficient function of the inlet propertiesφ=˙m [kgs ]√T[K] p[bar] and this coefficient remains constant between configurations A and B:φA=φ B; •The condenser has been designed for configuration A so, its dimension (surface area), the thermalproperty (heat transfer constant), the coolant mass flow rate and the inlet temperature of the coolant don’t change but other operative characteristic will be different. For the configuration A the inlet and outlet temperature of the cooling water respectively to 20◦ C and 30◦ C; •The steam turbines are real and they are affected by isoentropic efficienciesηs,H P= 84.34%and ηs,LP= 87.75%. The turbines are also affected by an organic efficiency (mechanical)η org= 99% and the electric generator efficiency isηel= 98.5%; •The inlet temperature of the HP turbine is538◦ C; •The extraction and feed-water pumps have a hydraulic efficiencyhhyd= 75%and a mechanical electrical efficiencyηem= 88%; •The auxiliary cycle consumption are˙ Waux= 2300kW (beyond the pumps); •The inlet temperature of the HP steam turbine is 538◦ C; 2 Energy Systems- Report Precept 5 - Group xx •The HRSG data are listed in the following table: Table 1:HRSG dataSymbolValueu.o.m.Description ∆T pp10◦ C Temperature difference at pinch point∆T S C7◦ C Subcooling temperature difference drum inlet∆T AP25◦ C Approach temperature differenceξ H RS G1%Heat losses of HRSG ∆pp  E C O15%Pressure drop in the economizers ∆pp  S H7%Pressure drop in the super-heater of the high pressure levelc P,F G1.1kJkg·K Flue gases specific heat capacityFigure 2:New plant scheme 3 Energy Systems- Report Precept 5 - Group xx 2.Configuration A 2.1.Performance indicators We will report in this section the computations and the results of the performance analysis of the existing configuration1. Net electric power generated is given by text: Pel,net,A= 64.8MW The thermal power cogenerated by the Heat User can be divided in two streams:Pth,cog,A= ˙m s,cog1(h in,H U−h out,H U) + ˙m s,cog2(h in,H U−h out,amb)≈83.71MW where the values of the enthalpies are extracted with the Excel water-vapor macro database [1]: ˙mscog1= 24.7kg/s˙m scog2= 8.1kg/sh in,H U=h(5bar,198◦ C)≈2851.60kJkg hout,H U=h(1.013bar,90◦ C)≈376.99kJkg h out,amb=h(1.013bar,15◦ C)≈63.08kJkg The power generated by the fuel can be extracted by the boiler efficiency definition: ηboiler=˙m steam∆h boiler z}|{ ˙ Qth,sc˙m f uel·LH V f uel |{z} ˙ Qf uel⇒ ˙ Qf uel=˙m steam∆h boilerη boiler≈241.42MW the values of the enthalpies are:˙msteam= 93.4kg/s∆h boiler=h out(90bar,538◦ C)−hout(110bar,243◦ C)≈2429.73kJ/kg The net electrical efficiency of the plantηel, the thermal efficiencyη thand total efficiencyη totare: ηel=P el,net˙ Qf uel≈26.84%η th=P th,cog˙ Qf uel≈34.67%η tot=P tot+P th,cog˙ Qf uel≈61.51% 2.2.Condenser As stated in the description section, the condenser device doesn’t change between the configuration A and B so, it is worth to determine all the missing design parameters to be used for the new plant. First, we determine the total heat transfer power discharged to the refrigerant: ˙ Qcond,A= ˙m cond·[h(0.08bar, x= 0.88)−h sat,L(0.08bar)]≈68497.54kW We assume for the specific heat capacity of water equal tocp,w= 4.186JkgK , we can determine the mass flow rate of the refrigerant, which would be the same value for the new configuration, as: ˙ Qcond,A= ˙m w,ref·c p,w(30−20) [K]⇒˙m w,ref=˙ Qcond,Ac p,w(30−20) [K]≈1636.35 kgs On the text it is specified that both the heat transfer area and coefficient don’t change, so we can conclude that also their product don’t change: AA=A B UA=U B ⇒(U A)A= (U A) B⇒U A=˙ Qcond,ALM T D A≈4285.93 kWK The condenser is a counter current heat exchanger design to allow that all the vapor content of the fluid exiting the low pressure turbine will be canceled to reach saturated liquid conditions, this makes the 4 Energy Systems- Report Precept 5 - Group xx heat exchanger indifferent to the heat capacity ratio. Both mass flow rate and specific heat capacity of the refrigerant don’t change, this implies that also the refrigerant heat capacity don’t change and, if we apply theε−N T Umethod, we can state that also the effectivenessεdoesn’t change: ˙mref ,A= ˙m ref ,B cp,ref ,A=c p,ref ,B ⇒Cref ,A=C ref ,B=C min (U A)A= (U A) B  ⇒N T U A=N T U B=N T U N T UA=N T U B=N T U⇒ε A=ε B=ε= 1−e−N T U ≈0.4651 3.Configuration B - solution In this section we illustrated the procedure to determine all the missing parameters. For the sake of simplicity we collected all the numerical results for the thermodynamic properties of the steam cycle in Table2and the gas cycle flue gases HRSG temperatures are in Table3. With the given data we can determine the remaining thermodynamic properties for the points(3),(4),(17)and(8). We know from the scheme that the last one is at saturated vapor conditions. In the low pressure steam drum we find(7)at saturated liquid conditions with the same pressure of(8)so, we know all the remaining properties of(7). Also the liquid stream(7*)is at the same pressure of the steam drum but at a lower temperature: T7∗=T 7−∆T S C All the other properties are given with the Excel macro. Let’s split the computations in sections: 3.1.HP HRSG section + HP turbine The first section to be analyzed is the section of the HRSG from(a)to(b), we comprehend also the high pressure turbine. Let’s refer to the T−˙ Qdiagram on the side, we can write several relations in function of the evaporation pressure pE V A,H P. Tb=T(p E V A,H P) + ∆T P P T13=T a−∆T S C= 556◦ C The temperature in(13)computed is too high and it reduced at 538◦ C to comply with the de- sign limit of the steam turbine. p11∗=p 11=p 12=p E V A,H P p13=p 12 1−∆pp S H,H P h13=h(p 13;T 13) T11=T 12=T(p E V A,H P)Figure 3:High pressure steam plant section h12=h sat,vap(p 12)h 11=h sat,liq(p 11)T 11∗=T 11−∆T S Ch 11∗=h(p 11∗;T 11∗) We can write the energy balance from(a)to(b)for the flue gases and from(11*)to(13)for the steam cycle: ˙mF G·(1−ξ H RS G)·c p,F G·(T a−T b) = ˙m 13(h 13−h 11∗)(a-b e.b.) 5T Q a b 11 11* 12 13 * Δ T AP Δ T PP Δ T SC SH HP EVA HP Energy Systems- Report Precept 5 - Group xx Now we define the relation of the coefficient function of the inlet properties of the high pressure turbines between configuration A and B: φH P,A=93.4 kgs √538 + 273.15K 90bar =const= ˙m 13·√T 13p 13=φ H P,B(phi HP eq.) All the previous expressions are dependent on thepE V A,H P. Here a possible optimization algorithm to evaluate the unknowns:Algorithm 1Optimization algorithm - HP HRSG section + HP turbine ▷Compute the steam turbine coefficient for configuration A:φ H P,A ▷Define the initial value forpE V A,H P(50bar) ▷Compute˙m13from (a-b e.b.)←p E V A,H P ▷Compute the turbine coefficientφH P,Bfrom (phi HP eq.)←˙m 13 ▷Compute theerror=|φH P,B−φ H P,A| ▷UpdatepE V A,H Ptill reach the error zeroThe remaining thermodynamic properties can be computed with the Excel macro and the mass flow rate with mass balance. We can also compute the thermodynamic properties at outlet of the high pressure turbine(14is)and(14)with the following expressions: h14is=h(p 14is;s 13)⇒h 14=h 13−η s,H P(h 13−h 14is) 3.2.LP HRSG section + Attemperator Now we focus on the low pressure side of the HRSG and the attemperator in order to compute the unknown variables. First let’s write all the expression with known values: Tc=T 7+ ∆T P PT 9=T b−∆T AP T7∗=T 7−∆T S Ch 7∗=h(p 7∗;T 7∗) h9=h(p 9;T 9)p 10=p 11∗1− ∆pp E C O h10is=h(p 10, s 7)h 10=h 7+h 10is−h 7η hyd We can write the mass balance of the attem- perator with˙m15′′ and˙m 10′ as unknowns: ˙m15′′ + ˙m 10′ = ˙m 17(attemp m.b.)Figure 4:Low pressure steam plant section We can also write the energy balance of the attemperator but we introduce a new unknownh15′′ ˙m15′′ h 15′′ + ˙m 10′ h 10′ = ˙m 17h 17(attemp e.b.) (15)is contact mixer (manifold), this means that the thermodynamic properties ofpare the same for each inlet or outlet stream. We have thath15′′ =h 15. Let’s write the energy balance of the manifold introducing a new unknown˙m9: ˙m14h 14+ ˙m 9h 9= ( ˙m 14+ ˙m 9)h 15(mani e.b.) Now we have four unknowns˙m9,h 15,˙m 15′′ and˙m 10′ with three equations, we can write the energy balance of the low pressure HRSG: ˙mF G·(1−ξ H RS G)·c p,F G·(T b−T c) = ˙m 9(h 9−h 7) + ˙m 10(h 11∗−h 10) + ( ˙m 10+ ˙m 10′ )(h 7−h 7∗) (b-c e.b.) 6T Q Δ T AP Δ T PP Δ T SC 7* * 7 8 8 9 10 11* b c ECO HP EVA LP SH LP Energy Systems- Report Precept 5 - Group xx To evaluate these unknowns we can write another optimization algorithm considering as ob jective the equivalence of theh17computed in (attemp e.b.) with the data provide:Algorithm 2Optimization algorithm - LP HRSG section + attemperator ▷Define the initial value for˙m 10′ (1.2kg/s) ▷Compute˙m9from (b-c e.b.)←˙m 10′ ▷Computeh15from (mani e.b.)←˙m 9 ▷Compute˙m15′′ from (attemp m.b.)←h 15 ▷Computeh17from (attemp e.b.)←˙m 10′ ,˙m 15′′ andh 15 ▷Compute theerror=|h17−hdata 17| ▷Update˙m10′ till reach the error zeroAll other thermodynamic properties can be evaluated with Excel macro. 3.3.LP turbine We found the thermodynamic properties of the manifold(15)so, with the mass balance we can compute the mass flow rate at the inlet of the low pressure turbine: ˙m14+ ˙m 9= ˙m 15′ + ˙m 15′′ ⇒˙m 15′ = ˙m 14+ ˙m 9−˙m 15′′ With the definition of isoenthalpic valve (h15′ =h 15), we can compute the inlet temperatureT 15′ in function of the unknown pressurep15′ : T15′ =T(p 15′ ;h 15)(Tin LP turb) As we done for the high pressure turbine, we can write the equivalence of the inlet coefficient also for the low pressure turbine: φLP,A=40 kgs √198 + 273.15K 4.9bar =const= ˙m 15′ ·√T 15′p 15′=φ LP,B(phi LP eq.) We can write an algorithm to find the pressure and temperatureAlgorithm 3Optimization algorithm - LP HRSG section + attemperator ▷Compute the steam turbine coefficient for configuration A:φ LP,A ▷Define the initial value forp15′ (5.2bar) ▷ComputeT15′ from (Tin LP turb)←p 15′ ▷ComputeφLP,Bfrom (phi LP eq.)←p 15′ ,T 15′ ▷Compute theerror=|φLP,B−φ LP,A| ▷Updatep15′ till reach the error zero3.4.Condenser Now we focus on the condenser that treats the outlet stream of the low pressure steam turbine. We need to remind all the assumptions made in Section2.2for the configuration A, we know the inlet temperature of the refrigerant (water) and given the effectiveness of the condenser, we can determine the new outlet temperature in function of the new unknown condensation temperature: ε=T w,out,B−T w,inT cond(p cond)−T w,in⇒T w,out,B=T w,in+ε(T cond(p cond)−T w,in)(Tw ref out eq.) 7 Energy Systems- Report Precept 5 - Group xx We can compute the heat power exchanged in the condenser with two distinct formulations: ˙ Qcond,B,1= ˙m w·c p,w·(T w,out,B−T w,in) ˙ Qcond,B,2= ˙m w·(h 16−h satL(p cond,B)(Qcond 1) (Qcond 2) We need to write the expressions of the thermodynamic properties for points(16is)and(16): h16,is=h(p cond,B;s 15′ )⇒h 16=h 15′ −η S T ,LP(h 15′ −h 16is)(h16 eq.) Given all these expressions we can write an optimization algorithmAlgorithm 4Optimization algorithm - Condenser ▷Define the initial value forp cond,B(0.1bar) ▷Computeh16from (h16 eq.)←p cond,B ▷ComputeTw,out,Bfrom (Tw ref out eq.)←p 15′ ▷Compute˙ Qcond,B,1from (Qcond 1)←T w,out,B ▷Compute˙ Qcond,B,2from (Qcond 2)←h 16,p cond,B ▷Compute theerror=|˙ Qcond,B,1−˙ Qcond,B,2| ▷Updatepcond,Btill reach the error zeroWith this algorithm we can evaluate also all the thermodynamic properties in(1)which are at satu- rated liquid conditions. 3.5.Extraction pump, LP feedwater pump and HRSG exit The extraction pump between(1)and(2)increases the condensed water pressure into the mixer(5) which is at atmospheric pressure so, we can set the outlet pressure of the pump asp2= 1.013bar. The thermodynamic properties for(2is)and(2)can be evaluated as: h2is=h(p 2, s 1)⇒h 2=h 1+h 2is−h 1η hyd We can write the energy balance of the mixer(5)to compute its thermodynamic properties: ˙m5h 5= ˙m 2h 2+ ˙m 3h 3+ ˙m 4h 4 The LP feedwater pump between(5)and(6)increases the pressure of the fluid in the mixer before entering the HRSG. Along the low pressure economizer the fluid pressure drops, so we can estimate the pressure at the feedwater pump outlet: p6=p 7∗1− ∆pp E C O so we can compute the remaining properties of(6is)and(6): h6is=h(p 6, s 5)⇒h 6=h 5+h 6is−h 5η hyd At last we can evaluate the flue gases temperature at the exit of the HRSG as: ˙mF G(1−ξ H RS G)c p,F G(T c−T d) = ˙m 6(h 7∗−h 6)⇒T d=T c−˙m 6(h 7∗−h 6)˙m F G(1−ξ H RS G)c p,F G 8 Energy Systems- Report Precept 5 - Group xx 3.6.Numerical results In Table2we find all thermodynamic properties and mass flow rate of the steam cycle, we clarify here the colored legend:•text: data given in the Figure2; •text: data required with a mark"?"in Figure2; •text: data computed for completeness and counter check of the results. In Table3we find the numerical results for the HRSG temperatures. In Figure5we plotted theT−˙ Q diagram of the HRSG. In Figure6we represented the steam cycleT−sdiagram, each mass stream is represented with different lines. Table 2:Steam Cycle thermodynamic and mass flow rate resultsipThsvxstatus˙m [bar][ ◦ ][kJ/kg][kJ/(kgK)][m 3 /kg][kg/s] 1 0.0938 44.554 186.57 0.6328 0.00101 0 SL 35.54 2is 1.0130 44.554 186.65 0.6328 0.00101 - sub-cool 35.54 2 1.0130 44.572 186.67 0.6330 0.00101 - sub-cool 35.54 31.013090.00 376.99 1.1926 0.00104 - sub-cool24.70 41.013015.00 63.08 0.2245 0.00100 - sub-cool8.10 5 1.0130 57.52 240.81 0.7999 0.00102 - sub-cool 68.34 6is 6.3529 57.53 241.36 0.7999 0.00102 - sub-cool 68.34 6 6.3529 57.58 241.55 0.8005 0.00102 - sub-cool 68.34 7* 5.4000 147.76 622.64 1.8190 0.00109 - sub-cool 68.34 7 5.4000 154.76 652.83 1.8901 0.00110 0 SL - 85.4000 154.76 2751.52 6.7947 0.34857 1 SV 9.82 95.1000259.54 2980.60 7.3005 0.47404 - steam9.82 10is 69.4609 155.49 659.85 1.8901 0.00109 - sub-cool 56.98 10 69.4609 156.05 662.18 1.8958 0.00109 - sub-cool 56.98 10’ 69.4609 156.05 662.18 1.8958 0.00109 - sub-cool1.54 11* 59.0417 267.54 1172.42 2.9517 0.98510 - sub-cool 56.98 11 59.0417 274.54 1208.32 3.0178 0.00132 0 SL - 1259.0417 274.54 2785.60 5.8977 0.03302 1 SV 56.98 1354.9088538.00 3518.01 7.0416 0.06578 - steam56.98 14is 5.1000 197.93 2850.92 7.0416 0.41441 - steam 56.98 145.1000247.41 2955.39 7.2526 0.46248 - steam56.98 15 5.1000 249.19 2959.09 7.2597 0.46418 - steam - 15’4.5803248.32 2959.09 7.3084 0.51692 - steam35.54 15” 5.1000 249.19 2959.09 7.2597 0.46418 - steam31.26 16is 0.0938 44.554 2307.46 7.3084 13.80 0.8855 L-V 35.54 160.093844.554 2387.28 7.5597 14.321890.9188 L-V35.54 175.0000198.00 2851.60 7.0520 0.42302 - steam32.80 Table 3:HRSG temperatures iabcd T i[◦ C]581.00 284.54 164.76106.99 9 Energy Systems- Report Precept 5 - Group xx Figure 5:HRSGT− ˙ Qdiagram10 Energy Systems- Report Precept 5 - Group xx Figure 6:Steam CycleT−sdiagram 11Γ∆ΓΘΓΛΓΞΓΠΓΣΓΥΓΦΓΨ ΓΩΓαΓβΓγΓδΓϵΓζ ΓηΓθΓιΓκΓλΓµΓιΓνΓξ ΓπΓρ ΓΣΓ∆ ΓΘΓ∆Γ∆ ΓΘΓΣΓ∆ ΓΛΓ∆Γ∆ ΓΛΓΣΓ∆ ΓΞΓ∆Γ∆ ΓΞΓΣΓ∆ ΓΠΓ∆Γ∆ ΓΠΓΣΓ∆ ΓΣΓ∆Γ∆ ΓΣΓΣΓ∆ ΓσΓτΓυ ΓϵΓτΓγΓϕΓβΓχΓγΓτ Γσ ΓθΓ∆ΓψΓρ ΓΘΓΛΓΣΓΥΓΞΓΠΓΦΓΘΓΦ ΓΘΓ∆ΓλΓΘΓ∆Γω ΓΘΓΘΓΘ ΓΘΓΘΓΘΓΛΓΘΓΞΓΘΓΥΓΨΓΘΓΦΓ∆ΓΘ ΓΘΓΦ ΓΘΓΦΓ∆ΓΘ ΓΘΓΠ ΓΘΓΣΓλΓΘΓΣΓε ΓϑΓΘΓΣΓωΓηΓϕΓβΓχΓγΓϕΓβΓτΓϖΓϱΓχΓγΓςΓτ ΓΣΓ∆ ΓΥΓ∆ ΓΦΓΘΓ∆ ΓΦ ΓΦΓ∆ ΓΘΓ∆Γ∆ ΓΘΓΘΓΘΓ∆ ΓΘΓΘΓ∆ ΓΘΓΛΓ∆ ΓΘΓΞΓ∆ ΓΘΓΠΓ∆ ΓΘΓΣ ΓΦΓ∆ ΓΨΓ∆ ΓϑΓ∆ ΓΘΓΣ ΓΘΓΣΓ∆ ΓΘΓΣΓΘΓ∆ ΓΘΓΥΓ∆ ΓΘΓ∆ ΓΛΓ∆ ΓΣ ΓΘΓΣΓ∆ ΓΘΓΦ ΓΘΓ∆ΓΘΓ∆ ΓΘΓΦ ΓΘΓΦΓ∆ ΓΘΓΦΓ∆ΓΘΓ∆ ΓΘΓΦΓ∆ΓΛ ΓΘΓΦΓ∆ΓΛΓ∆ ΓΞΓ∆ ΓΣ ΓΘΓΦΓ∆ΓΛΓ∆ ΓΠΓ∆ ΓΣ Energy Systems- Report Precept 5 - Group xx 3.7.Performance indicators The thermal power of the steam turbines and the pumps: ˙ WS T ,th= ˙m 13(h 13−h 14) |{z} HP ST+ ˙m 15′ (h 15′ −h 16) |{z} LP ST≈52.38MWth ˙ Wpump,th= ˙m 1(h 1−h 2) |{z} extraction pump+ ˙m 5(h 6−h 5) |{z} LP feedwater pump+ ( ˙m 10+ ˙m 10′ )·(h 10−h 7) |{z} HP feedwater pump≈0.601MWth The electric power of the same devices are:˙ WS T ,el=˙ WS T ,th·η org·η el≈51.08MWel ˙ Wpump,el=˙ Wpump,th·η em≈0.683MWel The total electric power of the new configuration is:˙ Wel,net,B=P el,net,GT+˙ WS T ,el−˙ Wpump,el−˙ Waux≈190.10MWel The total electric power just for the steam cycle is:˙ Wel,net,B,sc=˙ WS T ,el−˙ Wpump,el−˙ Waux≈48.10MWel The total thermal power entering the steam cycle is:˙ Qth,in,sc= ˙m F G·(1−ξ H RS G)·c p,F G·(T a−T d)≈213.70MWth We can compute the electrical efficiency of the steam cycleηsc,el: ηsc,el=˙ Wel,net,B,sc˙ Qth,in,sc≈22.51% We can estimate the input thermal power of the fuel in the gas turbine as:˙ Qf uel,GT=P el,net,GTη GT≈411.59MW The input thermal power of the fuel corresponds to the input thermal power of the whole system in configuration B. With this and the power produced by the gas turbine, we can compute the total output thermal power released by the gas turbine as: ˙ Qth,out,GT=˙ Qf uel,GT−P el,net,GT≈269.59MW The thermal power of the fuel for the configuration B is almost double the one for configuration A and, since the heat power for the Heat User doesn’t change in value (we can verify by the power balance), the thermal efficiency associated to the whole plant is approximately half the original configuration. We can also compute the electric and total efficiency. ηth,B=P th,cog˙ Qf uel,GT≈20.34%(= 58.66%·η th,A) ηel,B=˙ Wel,net,B˙ Qf uel,GT≈46.19%(= 1.72·η el,A)η tot= 66.52% The electric efficiency is almost doubled while, the overall efficiency is slightly higher than the case A. There is an alternative formulation for the electric efficiency through the combine cycle efficiencyηcc by first computing the recovery cycle ratior: r≜˙ Qth,in,sc˙ Qth,out,GT≈0.793η cc,el=η GT ,el+r·η sc,el·(1−η GT ,el)≈46.19% 12 Energy Systems- Report Precept 5 - Group xx References [1]Keith Woodbury, Bob Taylor, Joseph Chappell, Kenny Mahan, Jesse Huguet, and Troy Dent. Ther-modynamics.https://excelinme.net/thermodynamics/, 2025. Excel in Mechanical Engineering Research Team. 13