ME 18b
HW 4
Due 4 PM Wednesday April 28, 2010 Cheryl Geer’s Office 119 Thomas Number of classes attended: ________ Number of homework hours: ________ TAs Xiaobai Li Joules Gould 4.1Simple gas turbine power plant A gas turbine power plant has a compressor pressure ratio 20 in which air enters the compressor at 40°C 3 at a rate of 700 m /min. Air leaves the turbine at 650°C. Using variable specific heat for air and a compressor efficiency of 85% and a turbine efficiency of 88% determine: (a) the net power output, (b) the back work ratio, (c) the thermal efficiency and (d) which has the greatest impact on the back work ratio – the turbine or the compressor efficiency? Solution: State 1: T1 = 313K, h1 = 314 kJ/kg, Pr1 = 1.295 Isentropic Compression to State 2: = (20)1.295 = 25.9, then T2 = 448°C and h2 = 736 kJ/kg Isentropic turbine process T4 = 650°C, h4 = 959 kJ/kg, Pr4 = 66.84 = (20)66.84 = 1337, then T 3 = 1630°C and h3 = 2131 kJ/kg Mass flow rate m=
V, where
m = (1.113 kg/m3)(700 m3 /min)(1 min/60 sec) = 13.0 kg/s kg/s Compressor Work Wcomp,I = m(h2 – h1) = (13.0 kg/s)(736 – 314)kJ/kg = 5486 kW Wcomp,A = Wcomp,I /
comp
= 5486 kW/0.85 = 6454 kW
Turbine Work Wturb,I = m(h3 – h4) = (13.0 kg/s)(2131 – 959)kJ/kg = 15,240 kW Wturb,A = Wturb,I(
turb
)= 15240 kW(0.88) = 13,410 kW
Heat Input
1
Qin = m(h3 – h2) = (13.0 kg/s)(2131 – 736) kJ/kg = 18,140 kW (a) Net Power Out: Wnet = (13,410 – 6454) kW = 6950 kW (b) Back work ratio: rbw = Wcomp/Wturb = 6454/13,410 = 48.1% (c)
th
= Wnet/Qin = 6950/18140 = 38.3%
(d) Back work ratio for ideal cycle rbw,i = Wcomp,i /Wturb,i = 5486/15,240 = 36.0% ideal turbine, non ideal compressor: r bw,i = Wcomp,a /Wturb,i = 6454/15,240 = 42.3% non ideal turbine, ideal compressor: r bw,i = Wcomp,i /Wturb,a = 5486/13,410 = 40.9%
In this case the non-ideal compressor has the biggest impact on the back work ratio since it deviates from the ideal case more than the non-ideal turbine.
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4.2 Gas turbine with regenerator An automobile gas turbine is designed with a compressor pressure ratio of 8 and it also has a regenerator. Air enters the compressor at 17°C and 100 kPa. The cold air stream of the regenerator leaves it 10°C cooler than the hot air stream at the regenerator inlet. The turbine inlet temperature is 800°C. Use constant specific heat at room temperature and assume the compressor efficiency is 87% and the turbine efficiency is 93%. Find the rates of heat rejection and heat addition for the cycle when it produces 150 kW of net power. Solution: Isentropic compressor process = 290 K(8) T2,a = T1 + (T2,i – T1)/
comp
0.4/1.4
= 525 K
= 290 K + (525 – 290)/0.87 = 560 K
Isentropic turbine process = 1073 K(1/8) T4,a = T3 – (T3 – T4,i)(
turb)
0.4/1.4
= 592 K
= 1073 K - (1073 – 592)(0.93) = 626 K
Regenerator, cold stream exit temp T5 = T4,a – 10 K = 616 K Compressor work wc,a = Cp(T2,a – T1) = 1.004 kJ/kgK(560 – 290)K = 271 kJ/kg Turbine work wt,a = Cp(T3 – T4,a) = 1.004 kJ/kgK(1073 – 626)K = 449 kJ/kg Net Work wnet = wt,a + wc,a = (449 – 271) kJ/kg = 178 kJ/kg Mass Flow Rate m = Wnet /wnet = (150 kJ/s)/(178 kJ/kg) = 0.843 kg/s Heat Input Qin = m·Cp(T3 – T5) = (0.843 kg/s)(1.004 kJ/kgK(1073 – 616)K = 387 kW Heat Output Qout = Qin – Win = (387 – 150) kW = 237 kW
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4.3 Gas turbine with intercooling, reheating and regeneration Air enters a gas turbine with two stages of compression and two stages of expansion at 100 kPa and 22°C. The system uses a regenerator as well as reheating and intercooling as shown in the diagram below. The pressure ratio across each compressor is 4. 300 kJ/kg of heat is added to the air in each combustion chamber. The regenerator operates perfectly while increasing the temperature of the cold air by 130°C. Determine (a) The back work ratio of the cycle, (b) the system’s thermal efficiency assuming isentropic processes for the compressors and turbines (c) the regenerator effectiveness and (d) determine the thermal efficiency of the first stage and compare it to the overall cycle thermal efficiency. Use constant specific heat at room temperature.
Solution: Isentropic compressor process T4 =
= 295 K(4)
0.4/1.4
= 438 K
Compressor 2 has same inlet and outlet conditions as compressor 1 Regenerator process T5 = T4 + 130 K = (438 + 130) K = 568 K Heat Input Qin = Cp(T6 – T5), solve for T6: T6 = Qin /Cp + T5 = (300kJ/kg)/(1.004 kJ/kgK) + 568 K = 867 K Isentropic expansion process = 867 K(1/4)0.4/1.4 = 584 K Reheat Process Qrht = Cp(T8 – T7), solve for T8: T8 = Qrht /Cp + T7 = (300kJ/kg)/(1.004 kJ/kgK) + 584 K = 882 K Isentropic expansion process
4
= 882 K(1/4)
0.4/1.4
= 594 K
Compressor Work wcomp = 2Cp(T2 – T1) = 2(1.004 kJ/kgK)(438 – 295)K = 287 kJ/kg Turbine Work wturb = Cp[(T6 – T7) + (T8 – T9)] = (1.004 kJ/kgK)[(867 – 584) + (882 – 594)] = (285 + 290) kW = 575 kW Net Work for cycle wnet = wturb – wcomp = (575 – 287) kJ/kg = 287 kJ/kg Back work ratio for cycle rbw = wcomp /wturb = 287/575 = 50.1% Thermal efficiency for cycle = wnet /qin = 287/600 = 47.8%
th
Net work for first stage wnet = wturb – wcomp = (285 – 287/2) kJ/kg = 141 kJ/kg Thermal efficiency for first stage th,1
= wnet /qin = 141/300 = 46.9%
Regenerator Effectiveness R
= (T5 – T2)/(T9 – T2) = (568 – 438)/(594 – 438) = 83.6%
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