Answer to Question #134600 in Mechanical Engineering for Afrasyab

Question #134600

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
2020-09-23T14:55:10-0400


Given information

Initial pressure, "p_1=100kPa"

Initial temperature, "T_1=25^0 C"

Compression ratio, r =5

Turbine inlet temperature, "T_3=850^0 C"

At temperature "T_1=25^0 C=298 K" and the pressure "p_1=100 kPa"

"h_1=295.17+{(298-295)\/(300-295)}*(300.9-295.17)=298.608 kJ\/kg"

"(p_r)_1=1.3068+{(298-295)\/(300-295)}*(1.386-1.3068)=1.35432"

"(p_r)_2=(p_2\/p_1)((p_r)_1= 5 *1.35432=6.7716"

At "(p_r)_2=6.7716"

By interpolating

"h_2=472.24+{(6.7716-6.742)\/(7.268-6.742)}*(482.49-472.24)=472.8168 kJ\/kg"

At temperature "T_3=850^0 C=1123 K"

"h_3=1184.28+{(1123-1120)\/(1140-1120)}*(1207.57-1184.28)=1187.77 kJ\/kg"

"(p_r)_3=179.7+{(1123-1120)\/(1140-1120)}*(193.1-179.7)=181.71"

"(p_r)_4=(p_4\/p_3)((p_r)_3= 1\/5 *181.71=36.342"

At "(p_r)_4" =36.342

By interpolation

"h_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    

"\u03b7={(1187.77-767.38)-(472.8168-298.608)}\/1187.77-472.8168"

"\u03b7=0.344"

η=34.4 %


The back work ratio

"bwr = W_c\/W_t=(h_2-h_1)\/(h_3-h_4)"

"bwr =(472.8168-298.608)\/(1187.77-767.38)"

"bwr =0.4144"

bwr = 41.44 %



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