Question #238847

a)        Air enters the compressor of a gas turbine at 100 kPa and 25°C. For a pressure ratio of 5 and a maximum temperature of 850°C determine the back work ratio and the thermal efficiency using the Brayton cycle.



1
Expert's answer
2021-09-21T02:06:42-0400



Given information;

Initial pressure, p1=100kPap_1=100kPa

Initial temperature, T1=25°CT_1=25° C

Compression ratio, r =5

Turbine inlet temperature, T3=850°CT_3=850°C

At temperature T1=25°C=298KT_1=25°C=298 K

and the pressure p1=100kPap_1=100 kPa


h1=295.17+(298295)/(300295)(300.9295.17)=298.608kJ/kgh_1=295.17+{(298-295)/(300-295)}*(300.9-295.17)=298.608 kJ/kg


(pr)1=1.3068+(298295)/(300295)(1.3861.3068)=1.35432(p_r)_1=1.3068+{(298-295)/(300-295)}*(1.386-1.3068)=1.35432


(pr)2=(p2/p1)((pr)1=5×1.35432=6.7716(p_r)_2=(p_2/p_1)((p_r)_1= 5 ×1.35432=6.7716


At (pr)2=6.7716(p_r)_2=6.7716

By interpolating

h2=472.24+(6.77166.742)/(7.2686.742)(482.49472.24)=472.8168kJ/kgh_2=472.24+{(6.7716-6.742)/(7.268-6.742)}*(482.49-472.24)=472.8168 kJ/kg


At temperature T3=850°C=1123KT_3=850° C=1123 K

h3=1184.28+(11231120)/(11401120)×(1207.571184.28)=1187.77kJ/kgh_3=1184.28+{(1123-1120)/(1140-1120)}×(1207.57-1184.28)=1187.77 kJ/kg


(pr)3=179.7+(11231120)/(11401120)×(193.1179.7)=181.71(p_r)_3=179.7+{(1123-1120)/(1140-1120)}×(193.1-179.7)=181.71


(pr)4=(p4/p3)((pr)3=1/5×181.71=36.342(p_r)_4=(p_4/p_3)((p_r)_3= 1/5 ×181.71=36.342



At (pr)4(p_r)_4 = 36.342

By interpolation

h4=756.44+(36.34235.5)/(37.3535.5)×(767.29756.44)=767.38kJ/kgh_4=756.44+{(36.342-35.5)/(37.35-35.5)}×(767.29-756.44)=767.38 kJ/kg


We know the thermal efficiency of the cycle

η=(1187.77767.38)(472.8168298.608)1187.77472.8168η=\dfrac{(1187.77-767.38)-(472.8168-298.608)}{1187.77-472.8168}


η=0.344η=34.4%η=0.344\\ η=34.4 \%


The back work ratio

bwr=Wc/Wt=(h2h1)/(h3h4)bwr = W_c/W_t=(h_2-h_1)/(h_3-h_4)


bwr=(472.8168298.608)(1187.77767.38)bwr =\dfrac{(472.8168-298.608)}{(1187.77-767.38)}


bwr=0.4144bwr=41.44%bwr =0.4144\\ bwr = 41.44 \%

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