Question

Calculate the fraction of the vapor and the fraction of the liquid when the pressure of saturated liquid water is changed from 1400 kPa to 50 kPa in a throttling process

A steam plant is operating under Rankine cycle. If the heat added to the boiler is 4255 kJ/kg and the heat rejected in the condenser is -2180 kJ/kg, calculate the thermal efficiency of the cycle.

Answer 1

1. Enthalpy of saturated water at 1400 kPa= 830.13 kJ/kg

Enthalpy of saturated water at 50 kPa = 340.48 kJ/kg

Enthalpy of saturated steam at 50 kPa = 2645.21 kJ/kg

Since throttling is an iso-enthalpic process, hence we can equate the enthalpies before and after the process.

If x is the vapor fraction at 50 kPa, then:

830.13 = 2645.21x + 340.48(1-x)

Solving for x, we get -->

x = 0.2125 (fraction of vapor)

1-x = 0.7875 (fraction of liquid)

2. Work obtained = heat added - heat rejected = 4255 + 2180 = 2075 kJ/kg.

Thermal efficiency = work obtained/heat added = 2075/4255 = 0.488​​ OR 48.8%