Question #170739

An ideal Brayton-cycle gas turbine with regeneration but no reheat or intercoolingoperates at full load with an inlet temperature to the compressor of 60oF and an inlet temperature to the turbine of 1540oF. The cycle is also to be operated at part load at a pressure ratio of 3.0 with a turbine inlet temperature of 1100o F. Compare the thermal efficiency and network output per lbm flow for the two operating points if the working fluid (air) is treated as a perfect gas and the maximum possible regeneration is obtained.


1
Expert's answer
2021-03-11T06:56:28-0500


Full Load

1.T1=520R,P1=1atm1. T_1=520R, P_1=1atm

121 \to 2 isentropic compressor

T2=T1(P2P1)k1k=540R(6)1.411.4=867.6RT_2=T_1(\frac{P_2}{P_1})^{\frac{k-1}{k}}=540 R(6)^{\frac{1.4-1}{1.4}}=867.6R

454 \to 5 isentropic turbine

T5=T4(P5P4)k1k=2000R(16)1.411.4=1199RT_5=T_4(\frac{P_5}{P_4})^{\frac{k-1}{k}}=2000 R(\frac{1}{6})^{\frac{1.4-1}{1.4}}=1199R

ηt=WtWcQin\eta_t=\frac{W_t-W_c}{Q_{in}}

Wtm=(h4h5)=cp(T4T5)=0.24(20001199)=192Btu/lbm\frac{W_t}{m}=(h_4-h_5)=c_p(T_4-T_5)=0.24(2000-1199)=192 Btu/lbm


Wcm=(h2h1)=cp(T2T1)=0.24(867.6520)=83.4Btu/lbm\frac{W_c}{m}=(h_2-h_1)=c_p(T_2-T_1)=0.24(867.6-520)=83.4Btu/lbm


Qinm=(h4h3)=cp(T4T3)=Wtm=192Btu/lbm\frac{Q_{in}}{m}=(h_4-h_3)=c_p(T_4-T_3)=\frac{W_t}{m}=192 Btu/lbm

ηt=19283.4192=56.2\eta_t=\frac{192-83.4}{192}=56.2 %


Wnetm=WtWcm=109Btu/lbm\frac{W_{net}}{m}=\frac{W_t-W_c}{m}=109 Btu/lbm


Part Load

121 \to 2 isentropic compressor

T2=T1(P2P1)k1k=711.7RT_2=T_1(\frac{P_2}{P_1})^{\frac{k-1}{k}}=711.7R

454 \to 5 isentropic turbine

T5=T4(P5P4)k1k=1139.8RT_5=T_4(\frac{P_5}{P_4})^{\frac{k-1}{k}}=1139.8R

Wtm=cp(T4T5)=0.24(15601140)=101Btu/lbm\frac{W_t}{m}=c_p(T_4-T_5)=0.24(1560-1140)=101 Btu/lbm


Wcm=cp(T2T1)=0.24(711.7520)=46Btu/lbm\frac{W_c}{m}=c_p(T_2-T_1)=0.24(711.7-520)=46Btu/lbm


Qinm=Wtm=101Btu/lbm\frac{Q_{in}}{m}=\frac{W_t}{m}=101 Btu/lbm

ηt=10146101=54.4\eta_t=\frac{101-46}{101}=54.4 %


Wnetm=10146=55Btu/lbm\frac{W_{net}}{m}=101-46=55 Btu/lbm



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