Question

A steam power plant operates on the simple ideal rankine cycle. the steam enters the turbine at 4 MPa and 500 C and leaves it at 50 kPa and 150 C. the water leaves the condenser as a saturated liquid and is subsequently displaced to the boiler by means of a pump at a temperature of 85 C, which is the isentrophic efficiency of the turbine?

Answer 1

Given data:

Steam enters turbine at 4 MPa and 500° C     and leaves at 50 kPa   and 150° C.

From steam table

Temp	Pressure	h   (KJ/kg)	s (KJ/kg K)
500° C	4 MPa	3445.2	7.09

and

Temp	Pressure	hf	hfg	hg	sf	sfg	sg
150° C	50 kPa			2780.1			7.94
	50 kPa	340.47	2305.4	2645.9	1.091	6.5029	7.5939

Hence In real condition;

Net work produced by turbine per kg of steam = 3445.2 - 2780.1 =665.1 KJ/kg

For Isentropic condition , exit condition of steam will be such that

S_inlet = S_exit

Dryness fraction must be = 0.9225    For Isentropic condition in turbine.

Hence at this condition, Enthalpy of steam = 340.47 + 0.9225*2305.4 = 2467.228

Hence in Isentropic condition, work produced by turbine = 3445.2 - 2467.228 = 977.972 KJ/kg

Therefore

Hence isentropic efficiency of turbine = 68 %