Question

In a gas turbine power plant, the compressor work is 500 kJ/kg, the heat supplied is 1 MJ/kg and the turbine work is 800 kJ/kg.The thermal efficiency of plant is:
a) 60%
b) 30%
c) 38%
d) 50%

Answer 1

The Thermal efficiency of plant is 30 %  

Explanation:

Given as :

For a gas turbine plant

The compressor work = $W_c$  = 500  KJ/Kg

The turbine work         = $W_t$  = 800 KJ/Kg

The heat supplied       = Q   = 1 MJ/Kg  = 1000 KJ/Kg

Let Thermal efficiency of plant = $\eta$  

According to question

Thermal efficiency of plant = $\dfrac{Turbin work - compressor work}{heat supplied}$

Or,                                     $\eta$ = $\dfrac{W_t - W_c}{Q}$

Or,                                      $\eta$ = $\dfrac{800 KJ/Kg - 500 KJ/Kg}{1000KJ/Kg}$

Or,                                      $\eta$ = $\dfrac{300 KJ/Kg}{1000KJ/Kg}$

∴                                         $\eta$ = $\dfrac{30}{100}$

i.e                                       $\eta$ = 30%

So, Thermal efficiency of plant = $\eta$ = 30%

Hence, The Thermal efficiency of plant is 30 %  . Answer