In a Brayton cycle, the air enters the compressor at 300 K and maximum temperature of cycle is 1200 K. What will be the thermal efficiency of cycle for maximum power output ?

Concept:

A gas turbine work on Brayton cycle, Thermal Efficiency of Brayton cycle is given by 

 \({\eta _{th}} = 1 - \frac{1}{{{{\left( {{r_p}} \right)}^{\frac{{\gamma - 1}}{\gamma }}}}}\)where rp = pressure ratio, \(r_p=\frac{P_2}{P_1}\)

For Maximum work : \({r_p} = {\left( {\frac{{{T_3}}}{{{T_1}}}} \right)^{\frac{\gamma }{{2\left( {\gamma - 1} \right)}}}}\)

Where T3 is the maximum temperature in the cycle which is at the inlet of the turbine,

and T1 is the minimum temperature in the cycle which is at the inlet of the compressor now if we put the maximum compression ratio in the thermal efficiency formula then 

\({\eta _{th}} = 1 - \sqrt {\frac{{{T_1}}}{{{T_3}}}} = 1 - \sqrt {\frac{{{T_{min}}}}{{{T_{max}}}}} \)

Calculation:

Given, temperature limits, T3 = 1200 K, and T1 = 300 K

So \({\eta _{th}} = 1 - \sqrt {\frac{{{T_1}}}{{{T_3}}}} = 1 - \sqrt {\frac{{{T_{min}}}}{{{T_{max}}}}} \)

\({\eta _{th}} = 1 - \sqrt {\frac{{300}}{{1200}}} = 0.5\;or\;50\% \)