Assignment of Power Plant-I Designed by Sir Engr.

Masood Khan

SOLVED PROBLEMS OF CHAPTER # 9

TITLE: GAS TURBINE CYCLE

Gas Turbine Cycle Designed by Sir Engr. Masood Khan

PROBLEM # 9.1: T2 = 293 + (464.05 – 293)*1/0.8 = 506.81 K

A gas turbine has an overall pressure ratio of 5 & a Expansion process:
maximum cycle temperature of 550 C. The turbine drives T4S/T3 = (P4 / P3)γ-1/γ
the compressor and an electric generator, the mechanical T4S = T3*(P4 / P3)γ-1/γ = 823*(1/5) 1.333-1/1.333 = 550 K
efficiency of the drive being 97%. The ambient temperature From ŋIS,T:
is 20 C & the air enters the compressor at the rate of 15 ŋIS,C = (T3 – T4)/(T3 – T4S)
kg/s; the isentropic efficiencies of the compressor & the T4 = T3–(T3–T4S)*ŋis,c=823–(823–550)0.83=596.41 K
turbine are 80 & 83%. Neglecting changes in Kinetic Now work of expansion.
energy, the mass flow rate of fuel, & all pressure losses, Cpg(T3 – T4) = 1.15*(823 – 596.41) = 267.58 kJ/kg
calculates: Work of compression required = Cpa(T2 – T1)/ŋMS
(i) The power output; = 1.005*(506.81 – 293)/0.97 = 221.52 kJ/kg
(ii) The cycle efficiency; Generator work = (Expansion work – Compression
(iii) The work ratio. work)/drive efficiency
= (267.58 – 221.52)/0.97 = 47.48 kJ/kg
GIVEN DATA: Power output = (generator work)(mass flow rate)
Pressure ratio, P2 / P1 = 5 = 47.48*15 = 712.2 kW
Max temp, T3 = 550 C = 823 K Process in combustion chamber:
Mechanical efficiency of drive = 97 % Qs, heat supplied = Cpg(T3 – T2)
Atmospheric temp, T1 = 20 C = 293 K = 1.15(823 – 506.81) = 363.6185 kJ/kg
mo = 15 kg/s, ŋIS,C = 0.8 ŋIS,T = 0.83 Cycle efficiency = net work output / heat supplied
= 47.48 / 363.9185 = 0.1305
REQUIRED: Work ratio = net work output / gross work output
Power output. Cycle efficiency. Work ratio. = 47.48 / 267.58 = 0.177

DIAGRAM: PROBLEM # 9.2:

In a marine gas turbine unit a HP stage turbine drives the
SOLUTION: compressor, and an LP stage turbine drives the propeller
Compression process: through suitable gearing. The overall pressure ratio is 4/1,
T2S / T1 = (P2 / P1)γ-1/γ the mass flow rate is 60kg/s, the maximum temperature is
T2S = T1*(P2 / P1)γ-1/γ = 293*(5) 1.4-1/1.4 = 464.05 K 650 C, & the air intake conditions are 1.01 bar & 25 C. The
From Isentropic efficiency of compressor isentropic efficiencies of the compressor, HP turbine, & LP
Ŋis,c = (T2S – T1)/(T2 – T1) turbine, are 0.8, 0.83, & 0.85 respectively, & the
=> T2 = T1 + (T2S - T1)/ŋis,c mechanical efficiency of both shafts is 98%. Neglecting
Gas Turbine Cycle Designed by Sir Engr. Masood Khan

Kinetic energy changes, & the pressure loss in combustion, Cpg(T3 – T4) = 181.93/0.98 = 923 – T4 = 185.64/1.15
calculate: T4 = 923 – 185.64/1.15 = 761.57K
(i) The pressure b/w turbine stages; Now from ŋis,hpt
(ii) The cycle efficiency; Ŋis,hpt = (T3 – T4) / (T3 – T4S)
(iii) The shaft power. T4s= T3 –(T3 – T4)/ŋis,hpt= 923–(923–761.57)/0.83
T4s = 728.50 K
GIVEN DATA: Now, P3/P4 = (T3/T4S)γ-1/γ = (923/728.5) 1.333/0.333
Over all pressure ratio = 4/1 P3/P4 = 2.5786
Mass flow rate, mo= 60 kg/s Now since, P3 = P2 = 4.04 bar
Tmax = 650 C = 923 K P4 = P3 /2.5786 = 4.04 / 2.5786 = 1.5667 bar
Pa = P1 = 1.01 bar => P2 = 4.04 bar Pressure between turbine stages = P4 = 1.5667 bar
Ta = T1 = 25 C = 298 K Now, P5 = P1 = 1.01 bar
ŋis,c = 0.8 ŋis,hpt = 0.83 ŋiso,lpt = 0.85 P4/P5 = 1.5667/1.05 = 1.55
Shaft mechanical efficiency = 0.98 Now, T4/T5s = (P4/P5)γ-1/γ
T5s = 761.57 / (1.55) 1.333-1/1.333 = 682.59 K
REQUIRED: From Isentropic efficiency of LPT:
Pressure between turbine stages Ŋis,lpt = (T4 – T5) / (T4 – T5S)
Cycle efficiency. Shaft power. T5 = T4 – (T4 – T5S)*ŋis,lpt = 694.437 K
Work of expansion in LPT
DIAGRAM: Cpg(T4 – T5) = 1.15(761.57 – 694.437) = 77.20 kJ/kg
cycle = WNet/Qs= 77.20/Cpg(T3 – T2) = 0.15 = 15 %
SOLUTION: Shaft Power = mo* WLPT / Shaft Efficiency
Compression process: Shaft Power = 60*77.20/ 0.98 = 4726.53 kW
T2S/T1 = (P2 / P1)γ-1/γ
=> T2S = T1*(P2 / P1)γ-1/γ = 298*(4) 1.4-1/1.4 = 442.82 K PROBLEM # 9.3:
From Isentropic efficiency of compressor: For the unit of problem 9.2, calculate the cycle efficiency
Ŋis,c = (T2S – T1) / (T2 – T1) obtainable when a heat exchanger is fitted. Assume a
T2 = T1+(T2S – T1)/ŋis,c = 298+(442.82–298)/0.80 thermal ratio of 0.75.
T2 = 479.025K
Work of Compression DATA:
Cpa(T2 – T1) = 1.005*(479.025 – 298) = 181.93 kJ/kg From previous problem
This Work + Some Frictional Work = HPT Work Heat exchanger is fitted
Expansion Work of HPT = Compression Work/ŋm,shaft Thermal ratio = 0.75
Gas Turbine Cycle Designed by Sir Engr. Masood Khan

REQUIRED: Qs = 1.15(923 – 634) = 332.35 kJ/kg

Cycle efficiency ηThermal = Wnet / Heat Supplied = 86.28/332.35= 0.259

SOLUTION: PROBLEM # 9.4

Compression process: In a gas turbine generating set two stages of compression
T2S / T1 = (P2/P1)γ-1/γ = (4) 1.4-1/1.4 are used with an intercooler between stages. The HP
T2S = T1*1.49 = 298*1.49 = 442.82 turbine drives the HP compressor, & the LP turbine drives
From Isentropic efficiency of compressor: the LP compressor & the generator. The exhaust from the
Ŋis,c = (T2S – T1) / (T2 – T1) LP turbine passes through the heat exchanger, which
T2 = T1+(T2S – T1)/ŋis,c = 298+(442.82–298)/0.98 transfer heat to the air leaving the HP compressor. There is
T2 = 479K a reheat combustion chamber b/w turbine stages, which
Work of Compression = Cpa(T2 – T1) = 181.905 KJ/kg raises the gas temperature to 600 C, which is also the gas
Compression Work is supplied by the HPT temperature at entry to the HP turbine. The overall pressure
Expansion Work from HPT=181.905/Shaft efficiency ratio is 10/1, each compressor having the same pressure
Cpg(T4 – T5) = 181.905 / 0.98 ratio, & the air temperature at entry to the unit is 20 C. The
T5 = T4 – 181.905 / (0.98*Cpg) = 761.59K heat exchanger thermal ratio may be taken as 0.7, &
Now, ŋis,hpt = (T4 – T5) / (T4 – T5S) intercooling is complete between compressor stages.
T5s = T4 – (T4 – T5)/ŋis,hpt = 923–(923–761.59)/0.83 Assume isentropic efficiencies of 0.8 for both compressor
T5s = 728.5K stages, & 0.85 for both turbine stages & that 2% of the
Now P4/P5=(T4/T5S)γ-1/γ=(923/728.5) 1.333/0.333 = 2.5786 work of each turbine are used in overcoming friction.
P5 = P4/2.5786 = 4.04/2.5786 = 1.5666 bar Neglecting all losses in pressure, & assuming that velocity
Now P5/P6 = 1.5666/1.01 = 1.55 changes are negligibly small, calculates:
T5/T6S = (P5/P6)γ-1/γ = (1.55) 0.333/1.333 (i) The power output in kilowatts for a mass of 115
T6S = T5 / 1.550.333/1.333 = 682.61 K kg/s;
From ŋis,lpt = (T5 – T6)/(T5 – T6S) (ii) The overall cycle efficiency of the plant
T6 = T5 – (T5 – T6S)*ŋis,lpt = 686.559 K
Work of Expansion in LPT = Cpg(T5 – T6) DATA:
= 1.15(461.59 – 686.559) = 86.28 kJ/kg Tmax. = 600 C = 873 K T1 = 20 C = 293 K
Now Heat Exchanger Thermal Ratio is given by Over all Pressure Ratio = 10/1
Thermal Ratio =Temp Rise/max. Temp. diff. available Same pressure ratio through each compressor
0.75 = (T3 – T2) / (T6 – T2) Ŋis,C = 0.8, for each Compressor & Intercooling is
T3 = (T6–T2)*0.75+T2 = (686.559–479)*0.75 + 479) = 634 K complete
Heat supplied in Combustion Chamber = Cpg(T4 – T3) Ŋis,T = 0.85, for each Turbine
Gas Turbine Cycle Designed by Sir Engr. Masood Khan

2 % loss for each turbine power WLPT = Cpg(T8 – T9) =1.15(873–634.7) =274.045kJ/kg
Thermal Ratio of Heat Exchanger = 0.7 Generator Work
mo = 115 kg/s Reheat to Tmax. = Expansion Work–Compression Work–losses
= 274.045–143.21–2/100*(274.045) =125.3541 kJ/kg
REQUIRED: Shaft Power=125.3541*mo=125.3541*115=14146kW
Power Output, Overall Efficiency, Thermal Ratio = (T5 – T4)/(T9 – T4)
T5 = (T9–T4)*0.7+T4=(634.7–435.5)*0.7+435.5
DIAGRAM: T5 = 574.94 K
Qs = Cpg(T6 – T5)=1.15(873–574.94)=342.769kJ/kg
SOLUTION: Qs reheat =Cpg(T8–T7)=1.15(873–745.93) =146.13 kJ/kg
Compressor side: Total Heat Supplied=342.769 + 146.13 = 488.89kJ/kg
T2S/T1 = (P2/P1)γ-1/γ where P2/P1 = √10 = 3.16 Cycle Efficiency = Net work / Total Heat Supplied
T2S = T1*(P2/P1)γ-1/γ = 293*(3.16) 0.4/1.4 = 407 K ηCycle = 125.3541 / 488.89 = 0.256 = 25.6 %
Ŋis,C = (T2S – T1)/(T2 – T1)
T2 = T1+(T2s–T1)/ŋis,C = 293+(407–293)/0.8 = 435.5K PROBLEM # 9.5
Wc=Cpa(T2 – T1) = 1.005(435.5–293) = 143.21 kJ/kg A motor gas turbine unit has two centrifugal compressors
Same amount of work is required by HPC, which is provided by in series giving an overall pressure ratio of 6/1. The air
HPT leaving the HP compressors passes through a heat
Cpg(T6 – T7) – 2/100*Cpg(T6 – T7) = 143.21 exchanger before entering the combustion chamber. The
T6 – T7 – 2/100(T6 – T7) = 143.21/Cpg expansion is in two turbine stages, the first stage driving
T7 = T6 - 143.21/(Cpg*0.98) = 745.93 K the compressors & the second stage driving the car through
Ŋis,T = (T6 – T7)/(T6 – T7S) gearing. The gases leaving the LP turbine pass through the
T7S = T6 – (T6 – T7S)/ŋis,T heat exchanger before exhausting to atmosphere. The HP
T7S = 873 – (873 – 745.93)/0.85 = 723.5 K turbine inlet temperature is 800 C & the air inlet
Now P6/P7 =(T6/T7S)γ-1/γ = (873/723.5) 1.333/0.333 = 2.12 temperature to the unit is 15 C. The isentropic efficiency of
Now since P6 = P4 = 10.10 bar the compression is 0.8, & that of each turbine is 0.85. The
P7 = P6/2.12 = 10.10/2.12 = 4.76 bar mechanical efficiency of each shaft is 98%. The heat
P7 = P8 and P9 = P1= 1.01 bar exchanger thermal ratio may be assumed to be 0.65.
T8/T9S = (P8/P9)γ-1/γ Neglecting pressure losses & changes in Kinetic energy,
T9S = T8 / (P8/P9)γ-1/γ = 873 / (4.76/1.01) 0.333/1.333 = 592.68 K calculate:
Ŋis,T = (T8 – T9)/(T8 – T9S) (i) The overall cycle efficiency;
T9 = T8 – (T8 – T9S)*ŋis,T = 873 – (873–592.68)*0.85 (ii) Power developed.
T9 = 634.7K (iii) Specific Fuel Consumption.
Gas Turbine Cycle Designed by Sir Engr. Masood Khan

Calorific value of fuel = 42600 kJ/kg T6S = T5 – (T5 – T6)/ŋis,T = 1073–(1073–885.73)/0.85

And Combustion Efficiency = 0.97 T6S = 852.68 K
P6/P7 = (T6S/T5)γ-1/γ = (852.68/1073) 1.333/0.333 = 0.3985
DATA: P6 = 0.3985*P5 where P5 = 6.06 bar => P6 = 2.415 bar
Over all Pressure Ratio = 6/1 T6/T7S = (P6/P7)γ-1/γ
Tmax. = 800 C = 1073 K T1 = 15 C = 288 K T7S = T6 / (P6 / P7)γ-1/γ
Ŋis,C = 0.8 for Compressors Ŋis,T = 0.85 for Turbines T7s = 885.73 / (2.415/1.01) 0.333/1.333 = 712.4K
Ŋm,Shft = 0.98 Thermal Ratio = 0.65 Ŋis,T = (T6 – T7)/(T6 – T7S)
T7 =T6–(T6–T7S)*ŋis,T=885.73–(885.73–712.4)*0.85 =
REQUIRED: 738.39
Overall Cycle Efficiency. Power developed Thermal Ratio = (T4 – T3)/(T7 – T3)
Specific Fuel Consumption. T4 = T3 + (T7 – T3)* Thermal Ratio
Calorific value of fuel = 42600 kJ/kg T4 = 502.66 + (738.39 – 502.66)*0.65 = 655.88 K
And Combustion Efficiency = 0.97 Qs = Cpg(T5 – T4) = 1.15(1073–655.88) = 479.68 kJ/kg
WLPT =Cpg(T6–T7)=1.15(885.88–738.39)=169.44kJ/kg
DIAGRAM: Cycle Efficiency = 169.44/479.68 = 0.353 = 35.3 %
Shaft Power = 169.44*0.98*0.7 = 116.2
SOLUTION:
Compression process:
T3S/T1 = (P3/P1)γ-1/γ
T3S = T1*(P3/P1)γ-1/γ = 288*(6) 0.4/1.4 = 480.53 K
Assuming same pressure ratio for each compressor
P2/P1 = √6 = 2.45
T2S/T1 = (P2/P1) γ-1/γ
T2S = T1*(P2/P1) γ-1/γ = 288*(2.45) 0.4/1.4 = 372 K
From, ŋis,C = (T2S – T1)/(T2 – T1)
T3 = T2 + (T3S – T2)/ŋis,C = 502.66 K
Wc=Cpg (T2–T1)*2 = 1.005(393–288)*2 = 211.05kJ/kg
Expansion work of HPT = 211.05/ŋm, shaft
Cpg(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 Designed by S