Mcconkey-chapter 9 Solution
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Assignment of Power Plant-I
Designed by Sir Engr. Masood Khan
SOLVED PROBLEMS OF CHAPTER # 9
TITLE:
Gas Turbine Cycle
Designed by Sir Engr. Masood Khan
PROBLEM # 9.1: A gas turbine has an overall pressure ratio of 5 & a maximum cycle temperature of 550 C. The turbine drives the compressor and an electric generator, the mechanical efficiency of the drive being 97%. The ambient temperature is 20 C & the air enters the compressor at the rate of 15 kg/s; the isentropic efficiencies of the compressor & the turbine are 80 & 83%. Neglecting changes in Kinetic energy, the mass flow rate of fuel, & all pressure losses, calculates: (i) The power output; (ii) The cycle efficiency; (iii) The work ratio. GIVEN DATA: Pressure ratio, P2 / P1 = 5 Max temp, T 3 = 550 C = 823 K Mechanical efficiency of drive = 97 % Atmospheric temp, T 1 = 20 C = 293 K o ŋIS,C = 0.8 ŋIS,T = 0.83 m = 15 kg/s, REQUIRED: Power output. Cycle efficiency.
GAS TURBI TURBI NE CYCLE
Work ratio.
T2 = 293 + (464.05 – 293)*1/0.8 = 506.81 K Expansion process: γ-1/γ T4S/T3 = (P4 / P3) T4S = T3*(P4 / P3)γ-1/γ = 823*(1/5) 1.333-1/1.333 = 550 K From ŋIS,T: ŋIS,C = (T3 – T4)/(T3 – T4S) T4 = T3 – (T3 – T4S)*ŋis,c=823 – (823 – 550)0.83=596.41 K Now work of expansion. C pg(T3 – T4) = 1.15*(823 – 596.41) = 267.58 kJ/kg Work of compression required = Cpa(T 2 – T1)/ŋMS = 1.005*(506.81 – 293)/0.97 = 221.52 kJ/kg Generator work = (Expansion work – Compression work)/drive efficiency = (267.58 – 221.52)/0.97 = 47.48 kJ/kg Power output = (generator work)(mass flow rate) = 47.48*15 = 712.2 kW Process in combustion chamber: Qs, heat supplied = Cpg(T 3 – T2) = 1.15(823 – 506.81) = 363.6185 kJ/kg Cycle efficiency = net work output / heat supplied = 47.48 / 363.9185 = 0.1305 Work ratio = net work output / gross work output = 47.48 / 267.58 = 0.177
Gas Turbine Cycle
Designed by Sir Engr. Masood Khan
PROBLEM # 9.1: A gas turbine has an overall pressure ratio of 5 & a maximum cycle temperature of 550 C. The turbine drives the compressor and an electric generator, the mechanical efficiency of the drive being 97%. The ambient temperature is 20 C & the air enters the compressor at the rate of 15 kg/s; the isentropic efficiencies of the compressor & the turbine are 80 & 83%. Neglecting changes in Kinetic energy, the mass flow rate of fuel, & all pressure losses, calculates: (i) The power output; (ii) The cycle efficiency; (iii) The work ratio. GIVEN DATA: Pressure ratio, P2 / P1 = 5 Max temp, T 3 = 550 C = 823 K Mechanical efficiency of drive = 97 % Atmospheric temp, T 1 = 20 C = 293 K o ŋIS,C = 0.8 ŋIS,T = 0.83 m = 15 kg/s, REQUIRED: Power output. Cycle efficiency.
Work ratio.
DIAGRAM: SOLUTION: Compression process: T2S / T1 = (P2 / P1)γ-1/γ γ-1/γ 1.4-1/1.4 T2S = T1*(P2 / P1) = 293*(5) = 464.05 K From Isentropic efficiency of compressor Ŋis,c = (T2S – T1)/(T2 – T1) => T2 = T1 + (T2S - T1)/ŋis,c
Gas Turbine Cycle
Kinetic energy changes, & the pressure loss in combustion, calculate: (i) The pressure b/w turbine stages; (ii) The cycle efficiency; (iii) The shaft power. GIVEN DATA: Over all pressure ratio = 4/1 Mass flow rate, mo= 60 kg/s Tmax = 650 C = 923 K Pa = P1 = 1.01 bar => P 2 = 4.04 bar Ta = T1 = 25 C = 298 K ŋis,c = 0.8 ŋis,hpt = 0.83 ŋiso,lpt = 0.85 Shaft mechanical efficiency = 0.98 REQUIRED: Pressure between turbine stages Cycle efficiency. Shaft power. DIAGRAM: SOLUTION: Compression process: T2S/T1 = (P2 / P1)γ-1/γ
T2 = 293 + (464.05 – 293)*1/0.8 = 506.81 K Expansion process: γ-1/γ T4S/T3 = (P4 / P3) T4S = T3*(P4 / P3)γ-1/γ = 823*(1/5) 1.333-1/1.333 = 550 K From ŋIS,T: ŋIS,C = (T3 – T4)/(T3 – T4S) T4 = T3 – (T3 – T4S)*ŋis,c=823 – (823 – 550)0.83=596.41 K Now work of expansion. C pg(T3 – T4) = 1.15*(823 – 596.41) = 267.58 kJ/kg Work of compression required = Cpa(T 2 – T1)/ŋMS = 1.005*(506.81 – 293)/0.97 = 221.52 kJ/kg Generator work = (Expansion work – Compression work)/drive efficiency = (267.58 – 221.52)/0.97 = 47.48 kJ/kg Power output = (generator work)(mass flow rate) = 47.48*15 = 712.2 kW Process in combustion chamber: Qs, heat supplied = Cpg(T 3 – T2) = 1.15(823 – 506.81) = 363.6185 kJ/kg Cycle efficiency = net work output / heat supplied = 47.48 / 363.9185 = 0.1305 Work ratio = net work output / gross work output = 47.48 / 267.58 = 0.177
PROBLEM # 9.2: In a marine gas turbine unit a HP stage turbine drives the compressor, and an LP stage turbine drives the propeller through suitable gearing . The overall pressure ratio is 4/1, the mass flow rate is 60kg/s, the maximum temperature is 650 C, & the air intake conditions are 1.01 bar & 25 C. The isentropic efficiencies of the compressor, HP turbine, & LP turbine, are 0.8, 0.83, & 0.85 respectively, & the mechanical efficiency of both shafts is 98%. Neglecting
Designed by Sir Engr. Masood Khan
C pg(T3 – T4) = 181.93/0.98 = 923 – T4 = 185.64/1.15 T4 = 923 – 185.64/1.15 = 761.57K
Now from ŋis,hpt Ŋis,hpt = (T3 – T4) / (T3 – T4S) T4s= T3 – (T3 – T4)/ŋis,hpt= 923 – (923 – 761.57)/0.83
T4s = 728.50 K γ-1/γ 1.333/0.333 Now, P3/P4 = (T3/T4S) = (923/728.5) P3/P4 = 2.5786 Now since, P3 = P2 = 4.04 bar P4 = P3 /2.5786 = 4.04 / 2.5786 = 1.5667 bar Pressure between turbine stages = P 4 = 1.5667 bar Now, P5 = P1 = 1.01 bar P4/P5 = 1.5667/1.05 = 1.55 Now, T4/T5s = (P4/P5)γ-1/γ 1.333-1/1.333 T5s = 761.57 / (1.55) = 682.59 K From Isentropic efficiency of LPT: Ŋis,lpt = (T4 – T5) / (T4 – T5S) T5 = T4 – (T4 – T5S)*ŋis,lpt = 694.437 K Work of expansion in LPT C pg(T4 – T5) = 1.15(761.57 – 694.437) = 77.20 kJ/kg cycle = W Net/Qs= 77.20/Cpg(T 3 – T2) = 0.15 = 15 % o Shaft Power = m * WLPT / Shaft Efficiency Shaft Power = 60*77.20/ 0.98 = 4726.53 kW
Gas Turbine Cycle
Kinetic energy changes, & the pressure loss in combustion, calculate: (i) The pressure b/w turbine stages; (ii) The cycle efficiency; (iii) The shaft power. GIVEN DATA: Over all pressure ratio = 4/1 Mass flow rate, mo= 60 kg/s Tmax = 650 C = 923 K Pa = P1 = 1.01 bar => P 2 = 4.04 bar Ta = T1 = 25 C = 298 K ŋis,c = 0.8 ŋis,hpt = 0.83 ŋiso,lpt = 0.85 Shaft mechanical efficiency = 0.98 REQUIRED: Pressure between turbine stages Cycle efficiency. Shaft power. DIAGRAM: SOLUTION: Compression process: T2S/T1 = (P2 / P1)γ-1/γ γ-1/γ 1.4-1/1.4 => T2S = T1*(P2 / P1) = 298*(4) = 442.82 K From Isentropic efficiency of compressor: Ŋis,c = (T2S – T1) / (T2 – T1) T2 = T1+(T2S – T1)/ŋis,c = 298+(442.82 – 298)/0.80 T2 = 479.025K Work of Compression C pa(T2 – T1) = 1.005*(479.025 – 298) = 181.93 kJ/kg This Work + Some Frictional Work = HPT Work
Expansion Work of HPT = Compression Work/ŋ m,shaft
Gas Turbine Cycle
Designed by Sir Engr. Masood Khan
C pg(T3 – T4) = 181.93/0.98 = 923 – T4 = 185.64/1.15 T4 = 923 – 185.64/1.15 = 761.57K
Now from ŋis,hpt Ŋis,hpt = (T3 – T4) / (T3 – T4S) T4s= T3 – (T3 – T4)/ŋis,hpt= 923 – (923 – 761.57)/0.83
T4s = 728.50 K γ-1/γ 1.333/0.333 Now, P3/P4 = (T3/T4S) = (923/728.5) P3/P4 = 2.5786 Now since, P3 = P2 = 4.04 bar P4 = P3 /2.5786 = 4.04 / 2.5786 = 1.5667 bar Pressure between turbine stages = P 4 = 1.5667 bar Now, P5 = P1 = 1.01 bar P4/P5 = 1.5667/1.05 = 1.55 Now, T4/T5s = (P4/P5)γ-1/γ 1.333-1/1.333 T5s = 761.57 / (1.55) = 682.59 K From Isentropic efficiency of LPT: Ŋis,lpt = (T4 – T5) / (T4 – T5S) T5 = T4 – (T4 – T5S)*ŋis,lpt = 694.437 K Work of expansion in LPT C pg(T4 – T5) = 1.15(761.57 – 694.437) = 77.20 kJ/kg cycle = W Net/Qs= 77.20/Cpg(T 3 – T2) = 0.15 = 15 % o Shaft Power = m * WLPT / Shaft Efficiency Shaft Power = 60*77.20/ 0.98 = 4726.53 kW
PROBLEM # 9.3: For the unit of problem 9.2, calculate the cycle efficiency obtainable when a heat exchanger is fitted. Assume a thermal ratio of 0.75. DATA: From previous problem Heat exchanger is fitted Thermal ratio = 0.75
Designed by Sir Engr. Masood Khan
REQUIRED: Cycle efficiency
Qs = 1.15(923 – 634) = 332.35 kJ/kg ηThermal = Wnet / Heat Supplied = 86.28/332.35= 0.259
SOLUTION: Compression process: γ-1/γ 1.4-1/1.4 T2S / T1 = (P2/P1) = (4) T2S = T1*1.49 = 298*1.49 = 442.82 From Isentropic efficiency of compressor: Ŋis,c = (T2S – T1) / (T2 – T1) T2 = T1+(T2S – T1)/ŋis,c = 298+(442.82 – 298)/0.98 T2 = 479K Work of Compression = C pa(T2 – T1) = 181.905 KJ/kg Compression Work is supplied by the HPT Expansion Work from HPT=181.905/Shaft efficiency C pg(T4 – T5) = 181.905 / 0.98 T5 = T4 – 181.905 / (0.98*Cpg) = 761.59K Now, ŋis,hpt = (T4 – T5) / (T4 – T5S) T5s = T4 – (T4 – T5)/ŋis,hpt = 923 – (923 – 761.59)/0.83 T5s = 728.5K γ-1/γ 1.333/0.333 Now P4/P5=(T4/T5S) =(923/728.5) = 2.5786 P5 = P4/2.5786 = 4.04/2.5786 = 1.5666 bar Now P5/P6 = 1.5666/1.01 = 1.55 γ-1/γ 0.333/1.333 T5/T6S = (P5/P6) = (1.55) 0.333/1.333 T6S = T5 / 1.55 = 682.61 K
PROBLEM # 9.4 In a gas turbine generating set two stages of compression are used with an intercooler between stages. The HP turbine drives the HP compressor, & the LP turbine drives the LP compressor & the generator. The exhaust from the LP turbine passes through the heat exchanger, which transfer heat to the air leaving the HP compressor. There is a reheat combustion chamber b/w turbine stages, which raises the gas temperature to 600 C, which is also the gas temperature at entry to the HP turbine. The overall pressure ratio is 10/1, each compressor having the same pressure ratio, & the air temperature at entry to the unit is 20 C. The heat exchanger thermal ratio may be taken as 0.7, & intercooling is complete between compressor stages. Assume isentropic efficiencies of 0.8 for both compressor stages, & 0.85 for both turbine stages & that 2% of the work of each turbine are used in overcoming friction. Neglecting all losses in pressure, & assuming that velocity changes are negligibly small, calculates: (i) The power output in kilowatts for a mass of 115 kg/s;
Gas Turbine Cycle
Designed by Sir Engr. Masood Khan
REQUIRED: Cycle efficiency
Qs = 1.15(923 – 634) = 332.35 kJ/kg ηThermal = Wnet / Heat Supplied = 86.28/332.35= 0.259
SOLUTION: Compression process: γ-1/γ 1.4-1/1.4 T2S / T1 = (P2/P1) = (4) T2S = T1*1.49 = 298*1.49 = 442.82 From Isentropic efficiency of compressor: Ŋis,c = (T2S – T1) / (T2 – T1) T2 = T1+(T2S – T1)/ŋis,c = 298+(442.82 – 298)/0.98 T2 = 479K Work of Compression = C pa(T2 – T1) = 181.905 KJ/kg Compression Work is supplied by the HPT Expansion Work from HPT=181.905/Shaft efficiency C pg(T4 – T5) = 181.905 / 0.98 T5 = T4 – 181.905 / (0.98*Cpg) = 761.59K Now, ŋis,hpt = (T4 – T5) / (T4 – T5S) T5s = T4 – (T4 – T5)/ŋis,hpt = 923 – (923 – 761.59)/0.83 T5s = 728.5K γ-1/γ 1.333/0.333 Now P4/P5=(T4/T5S) =(923/728.5) = 2.5786 P5 = P4/2.5786 = 4.04/2.5786 = 1.5666 bar Now P5/P6 = 1.5666/1.01 = 1.55 γ-1/γ 0.333/1.333 T5/T6S = (P5/P6) = (1.55) 0.333/1.333 T6S = T5 / 1.55 = 682.61 K From ŋis,lpt = (T5 – T6)/(T5 – T6S) T6 = T5 – (T5 – T6S)*ŋis,lpt = 686.559 K Work of Expansion in LPT = C pg(T5 – T6) = 1.15(461.59 – 686.559) = 86.28 kJ/kg Now Heat Exchanger Thermal Ratio is given by Thermal Ratio =Temp Rise/max. Temp. diff. available 0.75 = (T3 – T2) / (T6 – T2) T3 = (T6 – T2)*0.75+T 2 = (686.559 – 479)*0.75 + 479) = 634 K Heat supplied in Combustion Chamber = C pg(T4 – T3)
PROBLEM # 9.4
Gas Turbine Cycle
DATA: Tmax. = 600 C = 873 K T1 = 20 C = 293 K Over all Pressure Ratio = 10/1 Same pressure ratio through each compressor Ŋis,C = 0.8, for each Compressor & Intercooling is complete Ŋis,T = 0.85, for each Turbine
Designed by Sir Engr. Masood Khan
2 % loss for each turbine power Thermal Ratio of Heat Exchanger = 0.7 o m = 115 kg/s Reheat to Tmax. REQUIRED: Power Output,
In a gas turbine generating set two stages of compression are used with an intercooler between stages. The HP turbine drives the HP compressor, & the LP turbine drives the LP compressor & the generator. The exhaust from the LP turbine passes through the heat exchanger, which transfer heat to the air leaving the HP compressor. There is a reheat combustion chamber b/w turbine stages, which raises the gas temperature to 600 C, which is also the gas temperature at entry to the HP turbine. The overall pressure ratio is 10/1, each compressor having the same pressure ratio, & the air temperature at entry to the unit is 20 C. The heat exchanger thermal ratio may be taken as 0.7, & intercooling is complete between compressor stages. Assume isentropic efficiencies of 0.8 for both compressor stages, & 0.85 for both turbine stages & that 2% of the work of each turbine are used in overcoming friction. Neglecting all losses in pressure, & assuming that velocity changes are negligibly small, calculates: (i) The power output in kilowatts for a mass of 115 kg/s; (ii) The overall cycle efficiency of the plant
Overall Efficiency,
DIAGRAM: SOLUTION: Compressor side: γ-1/γ T2S/T1 = (P2/P1) where P2/P1 = √10 = 3.16 γ-1/γ 0.4/1.4 T2S = T1*(P2/P1) = 293*(3.16) = 407 K Ŋis,C = (T2S – T1)/(T2 – T1) T2 = T1+(T2s – T1)/ŋis,C = 293+(407 – 293)/0.8 = 435.5K Wc=C pa(T2 – T1) = 1.005(435.5 – 293) = 143.21 kJ/kg Same amount of work is required by HPC, which is provided by HPT C pg(T6 – T7) – 2/100*Cpg(T6 – T7) = 143.21 T6 – T7 – 2/100(T6 – T7) = 143.21/Cpg T7 = T6 - 143.21/(Cpg*0.98) = 745.93 K Ŋis,T = (T6 – T7)/(T6 – T7S) T7S = T6 – (T6 – T7S)/ŋis,T T7S = 873 (873 745.93)/0.85 = 723.5 K
WLPT = C pg(T8 – T9) =1.15(873 – 634.7) =274.045kJ/kg Generator Work = Expansion Work – Compression Work – losses = 274.045 – 143.21 – 2/100*(274.045) =125.3541 kJ/kg o Shaft Power=125.3541*m =125.3541*115=14146kW Thermal Ratio = (T5 – T4)/(T9 – T4) T5 = (T9 – T4)*0.7+T4=(634.7 – 435.5)*0.7+435.5 T5 = 574.94 K Qs = C pg(T6 – T5)=1.15(873 – 574.94)=342.769kJ/kg Qs reheat =C pg(T8 – T7)=1.15(873 – 745.93) =146.13 kJ/kg Total Heat Supplied=342.769 + 146.13 = 488.89kJ/kg Cycle Efficiency = Net work / Total Heat Supplied ηCycle = 125.3541 / 488.89 = 0.256 = 25.6 %
PROBLEM # 9.5 A motor gas turbine unit has two centrifugal compressors in series giving an overall pressure ratio of 6/1. The air leaving the HP compressors passes through a heat exchanger before entering the combustion chamber. The expansion is in two turbine stages, the first stage driving the compressors & the second stage driving the car through gearing. The gases leaving the LP turbine pass through the heat exchanger before exhausting to atmosphere. The HP turbine inlet temperature is 800 C & the air inlet
Gas Turbine Cycle
Designed by Sir Engr. Masood Khan
2 % loss for each turbine power Thermal Ratio of Heat Exchanger = 0.7 o m = 115 kg/s Reheat to Tmax. REQUIRED: Power Output,
Overall Efficiency,
DIAGRAM: SOLUTION: Compressor side: γ-1/γ T2S/T1 = (P2/P1) where P2/P1 = √10 = 3.16 γ-1/γ 0.4/1.4 T2S = T1*(P2/P1) = 293*(3.16) = 407 K Ŋis,C = (T2S – T1)/(T2 – T1) T2 = T1+(T2s – T1)/ŋis,C = 293+(407 – 293)/0.8 = 435.5K Wc=C pa(T2 – T1) = 1.005(435.5 – 293) = 143.21 kJ/kg Same amount of work is required by HPC, which is provided by HPT C pg(T6 – T7) – 2/100*Cpg(T6 – T7) = 143.21 T6 – T7 – 2/100(T6 – T7) = 143.21/Cpg T7 = T6 - 143.21/(Cpg*0.98) = 745.93 K Ŋis,T = (T6 – T7)/(T6 – T7S) T7S = T6 – (T6 – T7S)/ŋis,T T7S = 873 – (873 – 745.93)/0.85 = 723.5 K γ-1/γ 1.333/0.333 Now P6/P7 =(T6/T7S) = (873/723.5) = 2.12 Now since P6 = P4 = 10.10 bar P7 = P6/2.12 = 10.10/2.12 = 4.76 bar P7 = P8 and P9 = P1= 1.01 bar T8/T9S = (P8/P9)γ-1/γ γ-1/γ 0.333/1.333 T9S = T8 / (P8/P9) = 873 / (4.76/1.01) = 592.68 K Ŋis,T = (T8 – T9)/(T8 – T9S) T9 = T8 – (T8 – T9S)*ŋis,T = 873 – (873 – 592.68)*0.85 T9 = 634.7K
Gas Turbine Cycle
Calorific value of fuel = 42600 kJ/kg And Combustion Efficiency = 0.97 DATA: Over all Pressure Ratio = 6/1 Tmax. = 800 C = 1073 K T1 = 15 C = 288 K Ŋis,C = 0.8 for Compressors Ŋis,T = 0.85 for Turbines Ŋm,Shft = 0.98 Thermal Ratio = 0.65 REQUIRED: Overall Cycle Efficiency. Power developed Specific Fuel Consumption. Calorific value of fuel = 42600 kJ/kg And Combustion Efficiency = 0.97 DIAGRAM: SOLUTION: Compression process: γ-1/γ T3S/T1 = (P3/P1) γ-1/γ 0.4/1.4 T3S = T1*(P3/P1) = 288*(6) = 480.53 K Assuming same pressure ratio for each compressor P2/P1 = √6 = 2.45 T2S/T1 = (P2/P1) γ-1/γ
WLPT = C pg(T8 – T9) =1.15(873 – 634.7) =274.045kJ/kg Generator Work = Expansion Work – Compression Work – losses = 274.045 – 143.21 – 2/100*(274.045) =125.3541 kJ/kg o Shaft Power=125.3541*m =125.3541*115=14146kW Thermal Ratio = (T5 – T4)/(T9 – T4) T5 = (T9 – T4)*0.7+T4=(634.7 – 435.5)*0.7+435.5 T5 = 574.94 K Qs = C pg(T6 – T5)=1.15(873 – 574.94)=342.769kJ/kg Qs reheat =C pg(T8 – T7)=1.15(873 – 745.93) =146.13 kJ/kg Total Heat Supplied=342.769 + 146.13 = 488.89kJ/kg Cycle Efficiency = Net work / Total Heat Supplied ηCycle = 125.3541 / 488.89 = 0.256 = 25.6 %
PROBLEM # 9.5 A motor gas turbine unit has two centrifugal compressors in series giving an overall pressure ratio of 6/1. The air leaving the HP compressors passes through a heat exchanger before entering the combustion chamber. The expansion is in two turbine stages, the first stage driving the compressors & the second stage driving the car through gearing. The gases leaving the LP turbine pass through the heat exchanger before exhausting to atmosphere. The HP turbine inlet temperature is 800 C & the air inlet temperature to the unit is 15 C. The isentropic efficiency of the compression is 0.8, & that of each turbine is 0.85. The mechanical efficiency of each shaft is 98%. The heat exchanger thermal ratio may be assumed to be 0.65. Neglecting pressure losses & changes in Kinetic energy, calculate: (i) The overall cycle efficiency; (ii) Power developed. (iii) Specific Fuel Consumption.
Designed by Sir Engr. Masood Khan
T6S = T5 – (T5 – T6)/ŋis,T = 1073 – (1073 – 885.73)/0.85 T6S = 852.68 K γ-1/γ 1.333/0.333 P6/P7 = (T6S/T5) = (852.68/1073) = 0.3985 P6 = 0.3985*P5 where P 5 = 6.06 bar => P 6 = 2.415 bar γ-1/γ T6/T7S = (P6/P7) γ-1/γ T7S = T6 / (P6 / P7) 0.333/1.333 T7s = 885.73 / (2.415/1.01) = 712.4K Ŋis,T = (T6 – T7)/(T6 – T7S) T7 =T6 – (T6 – T7S)*ŋis,T=885.73 – (885.73 – 712.4)*0.85 = 738.39 Thermal Ratio = (T4 – T3)/(T7 – T3) T4 = T3 + (T7 – T3)* Thermal Ratio T4 = 502.66 + (738.39 – 502.66)*0.65 = 655.88 K Qs = C pg(T5 – T4) = 1.15(1073 – 655.88) = 479.68 kJ/kg WLPT =C pg(T6 – T7)=1.15(885.88 – 738.39)=169.44kJ/kg Cycle Efficiency = 169.44/479.68 = 0.353 = 35.3 % Shaft Power = 169.44*0.98*0.7 = 116.2
Gas Turbine Cycle
Calorific value of fuel = 42600 kJ/kg And Combustion Efficiency = 0.97 DATA: Over all Pressure Ratio = 6/1 Tmax. = 800 C = 1073 K T1 = 15 C = 288 K Ŋis,C = 0.8 for Compressors Ŋis,T = 0.85 for Turbines Ŋm,Shft = 0.98 Thermal Ratio = 0.65 REQUIRED: Overall Cycle Efficiency. Power developed Specific Fuel Consumption. Calorific value of fuel = 42600 kJ/kg And Combustion Efficiency = 0.97 DIAGRAM:
Designed by Sir Engr. Masood Khan
T6S = T5 – (T5 – T6)/ŋis,T = 1073 – (1073 – 885.73)/0.85 T6S = 852.68 K γ-1/γ 1.333/0.333 P6/P7 = (T6S/T5) = (852.68/1073) = 0.3985 P6 = 0.3985*P5 where P 5 = 6.06 bar => P 6 = 2.415 bar γ-1/γ T6/T7S = (P6/P7) γ-1/γ T7S = T6 / (P6 / P7) 0.333/1.333 T7s = 885.73 / (2.415/1.01) = 712.4K Ŋis,T = (T6 – T7)/(T6 – T7S) T7 =T6 – (T6 – T7S)*ŋis,T=885.73 – (885.73 – 712.4)*0.85 = 738.39 Thermal Ratio = (T4 – T3)/(T7 – T3) T4 = T3 + (T7 – T3)* Thermal Ratio T4 = 502.66 + (738.39 – 502.66)*0.65 = 655.88 K Qs = C pg(T5 – T4) = 1.15(1073 – 655.88) = 479.68 kJ/kg WLPT =C pg(T6 – T7)=1.15(885.88 – 738.39)=169.44kJ/kg Cycle Efficiency = 169.44/479.68 = 0.353 = 35.3 % Shaft Power = 169.44*0.98*0.7 = 116.2
SOLUTION: Compression process: γ-1/γ T3S/T1 = (P3/P1) γ-1/γ 0.4/1.4 T3S = T1*(P3/P1) = 288*(6) = 480.53 K Assuming same pressure ratio for each compressor P2/P1 = √6 = 2.45 T2S/T1 = (P2/P1) γ-1/γ γ-1/γ 0.4/1.4 T2S = T1*(P2/P1) = 288*(2.45) = 372 K From, ŋis,C = (T2S – T1)/(T2 – T1) T3 = T2 + (T3S – T2)/ŋis,C = 502.66 K Wc=C pg (T2 – T1)*2 = 1.005(393 – 288)*2 = 211.05kJ/kg Expansion work of HPT = 211.05/ ŋm, shaft C pg(T5 – T6) = 211.05/0.98 T6 = T5 – 211.05/(0.98*1.15) = 1073 – 211.05/(0.98*1.15) = 885.73 K ŋis,T = (T5 – T6)/(T5 – T6S)
Gas Turbine Cycle
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Gas Turbine Cycle
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