Initial pressure of steam, p_{1} = 20 bar

Initial temperature of steam, t_{1} = 400°C

Final pressure of steam, p_{2} = 4 bar

Final temperature of steam, t_{2} = 250°C

Atmospheric temperature, = 20°C (= 293 K).

Initial state 1 : 20 bar, 400°C ; From steam tables,

h_{1} = 3247.6 kJ/kg ; s_{1} = 7.127 kJ/kg K

Final state 2 : 4 bar 250°C ; From steam tables,

h_{2}′ = 2964.2 kJ/kg, s_{2}′ = 7.379 kJ/kg K

The process is shown as 1 to 2′ in

s_{1} = s_{2} = 7.127 kJ/kg K

Hence, interpolating,

**(i) Isentropic efficiency :**

**(ii) Loss of availability : **

Loss of availability

= b_{1} – b_{2}

= h_{1} – h_{2}′ + T_{0} (s_{2′} – s_{1}′)

= 3247.6 – 2964.2 + 293 (7.379 – 7.127)

= 283.4 + 73.83 = 357.23 kJ/kg.

**(iii) Effectiveness : **

Effectiveness,