Gas Turbine Cycle-Thermodynamics-Problem Solution, Exercises of Thermodynamics

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SOLVED PROBLEMS OF CHAPTER # 9

TITLE: GAS TURBINE CYCLE

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, P 2 / P 1 = 5 Max temp, T 3 = 550 C = 823 K Mechanical efficiency of drive = 97 % Atmospheric temp, T 1 = 20 C = 293 K mo^ = 15 kg/s, ŋIS,C = 0.8 ŋIS,T = 0.

REQUIRED: Power output. Cycle efficiency. Work ratio.

DIAGRAM:

SOLUTION: Compression process: T2S / T 1 = (P 2 / P 1 )γ-1/γ T2S = T 1 (P 2 / P 1 )γ-1/γ^ = 293(5) 1.4-1/1.4^ = 464.05 K From Isentropic efficiency of compressor Ŋis,c = (T2S – T 1 )/(T 2 – T 1 ) => T 2 = T 1 + (T2S - T 1 )/ŋis,c

T 2 = 293 + (464.05 – 293)*1/0.8 = 506.81 K

Expansion process: T4S/T 3 = (P 4 / P 3 )γ-1/γ T4S = T 3 (P 4 / P 3 )γ-1/γ^ = 823(1/5) 1.333-1/1.333^ = 550 K From ŋIS,T: ŋIS,C = (T 3 – T 4 )/(T 3 – T4S) T 4 = T 3 – (T 3 – T4S)ŋis,c=823–(823–550)0.83=596.41 K Now work of expansion. Cpg(T 3 – T 4 ) = 1.15(823 – 596.41) = 267.58 kJ/kg Work of compression required = Cpa(T 2 – T 1 )/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) = 47.4815 = 712.2 kW Process in combustion chamber: Qs, heat supplied = Cpg(T 3 – T 2 ) = 1.15(823 – 506.81) = 363.6185 kJ/kg Cycle efficiency = net work output / heat supplied = 47.48 / 363.9185 = 0. Work ratio = net work output / gross work output = 47.48 / 267.58 = 0.

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

REQUIRED:

Cycle efficiency

SOLUTION: Compression process: T2S / T 1 = (P 2 /P 1 )γ-1/γ^ = (4) 1.4-1/1. T2S = T 1 1.49 = 2981.49 = 442. From Isentropic efficiency of compressor: Ŋis,c = (T2S – T 1 ) / (T 2 – T 1 ) T 2 = T 1 +(T2S – T 1 )/ŋis,c = 298+(442.82–298)/0. T 2 = 479K Work of Compression = Cpa(T 2 – T 1 ) = 181.905 KJ/kg Compression Work is supplied by the HPT Expansion Work from HPT=181.905/Shaft efficiency Cpg(T 4 – T 5 ) = 181.905 / 0. T 5 = T 4 – 181.905 / (0.98Cpg) = 761.59K Now, ŋis,hpt = (T 4 – T 5 ) / (T 4 – T5S) T5s = T 4 – (T 4 – T 5 )/ŋis,hpt = 923–(923–761.59)/0. T5s = 728.5K Now P 4 /P 5 =(T 4 /T5S)γ-1/γ=(923/728.5) 1.333/0.333^ = 2. P 5 = P 4 /2.5786 = 4.04/2.5786 = 1.5666 bar Now P 5 /P 6 = 1.5666/1.01 = 1. T 5 /T6S = (P 5 /P 6 )γ-1/γ^ = (1.55) 0.333/1. T6S = T 5 / 1.550.333/1.333^ = 682.61 K From ŋis,lpt = (T 5 – T 6 )/(T 5 – T6S) T 6 = T 5 – (T 5 – T6S)ŋis,lpt = 686.559 K Work of Expansion in LPT = Cpg(T 5 – T 6 ) = 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 = (T 3 – T 2 ) / (T 6 – T 2 ) T 3 = (T 6 – T 2 )0.75+T 2 = (686.559–479)0.75 + 479) = 634 K Heat supplied in Combustion Chamber = Cpg(T 4 – T 3 )

Qs = 1.15(923 – 634) = 332.35 kJ/kg ηThermal = Wnet / Heat Supplied = 86.28/332.35= 0.

PROBLEM # 9.

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

DATA: Tmax. = 600 C = 873 K T 1 = 20 C = 293 K Over all Pressure Ratio = 10/ Same pressure ratio through each compressor Ŋis,C = 0.8, for each Compressor & Intercooling is complete Ŋis,T = 0.85, for each Turbine

2 % loss for each turbine power Thermal Ratio of Heat Exchanger = 0. mo^ = 115 kg/s Reheat to Tmax.

REQUIRED: Power Output, Overall Efficiency,

DIAGRAM:

SOLUTION: Compressor side: T2S/T 1 = (P 2 /P 1 )γ-1/γ^ where P 2 /P 1 = √10 = 3. T2S = T 1 (P 2 /P 1 )γ-1/γ^ = 293(3.16) 0.4/1.4^ = 407 K Ŋis,C = (T2S – T 1 )/(T 2 – T1) T 2 = T 1 +(T2s–T 1 )/ŋis,C = 293+(407–293)/0.8 = 435.5K Wc=Cpa(T 2 – T 1 ) = 1.005(435.5–293) = 143.21 kJ/kg Same amount of work is required by HPC, which is provided by HPT Cpg(T 6 – T 7 ) – 2/100Cpg(T 6 – T 7 ) = 143. T 6 – T 7 – 2/100(T 6 – T 7 ) = 143.21/Cpg T 7 = T 6 - 143.21/(Cpg0.98) = 745.93 K Ŋis,T = (T 6 – T 7 )/(T 6 – T7S) T7S = T 6 – (T 6 – T7S)/ŋis,T T7S = 873 – (873 – 745.93)/0.85 = 723.5 K Now P 6 /P 7 =(T 6 /T7S)γ-1/γ^ = (873/723.5) 1.333/0.333^ = 2. Now since P 6 = P 4 = 10.10 bar P 7 = P 6 /2.12 = 10.10/2.12 = 4.76 bar P 7 = P 8 and P 9 = P 1 = 1.01 bar T 8 /T9S = (P 8 /P 9 )γ-1/γ T9S = T 8 / (P 8 /P 9 )γ-1/γ^ = 873 / (4.76/1.01) 0.333/1.333^ = 592.68 K Ŋis,T = (T 8 – T 9 )/(T 8 – T9S) T 9 = T 8 – (T 8 – T9S)ŋis,T = 873 – (873–592.68)0. T 9 = 634.7K

WLPT = Cpg(T 8 – T 9 ) =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 Shaft Power=125.3541mo=125.3541115=14146kW Thermal Ratio = (T 5 – T 4 )/(T 9 – T 4 ) T 5 = (T 9 – T 4 )0.7+T 4 =(634.7–435.5)*0.7+435. T 5 = 574.94 K Qs = Cpg(T 6 – T 5 )=1.15(873–574.94)=342.769kJ/kg Qs reheat =Cpg(T 8 – T 7 )=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.

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.