Question

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.

Answer 1

P₁=100kPa

T₁=25°C=298K

r=5

T₃=850°C=1123K

at T₁ and P₁,

h₁= $295.17+(\frac{298-295}{300-295})*(300.9-295.17)$

=298.608kJ/kg

(p_r)₁=$1.3068+(\frac{298-295}{300-295})*(1.386-1.3068)$

=1.35432

(p_r)₂=$(\frac{P_2}{P_1})*(p_r)_1$

=5*1.35432

=6.7716

At (p_r)₂ by interpolation,

h₂=$472.24+(\frac{6.7716-6.742}{7.268-6.742})*(482.49-472.24)$

=472.8186kJ/kg

At T₃,

h₃=$1184.28+(\frac{1123-1120}{1140-1120})*(1207.57-1184.28)$

=1187.77kJ/kg

(p_r)₃=$179.7+(\frac{1123-1120}{1140-1120})*(193.1-179.7)$

=181.71

(p_r)₄=$(\frac{P_4}{P_3})*(p_r)_3$

=$\frac{1}{5}*181.71$

=36.342

At (p_r)₄ by interpolation,

h₄=$756.44+(\frac{36.342-35.5}{37.35-35.5})*(767.29-756.44)$

=767.38kJ/kg

Back work ratio =$\frac{W_c}{W_t}=\frac{h_2-h_1}{h_3-h_4}$

=$\frac{472.8168-298.608}{1187.77-767.38}$

=0.4144

bwr=41.44%

Efficiency =$\frac{(1187.77-767.38)-(472.8168-298.608)}{1187.77-472.8168}$

=0.344

=34.4%