Initial pressure of steam, p1 = 20 bar
Initial temperature of steam, t1 = 400°C
Final pressure of steam, p2 = 4 bar
Final temperature of steam, t2 = 250°C
Atmospheric temperature, = 20°C (= 293 K).
Initial state 1 : 20 bar, 400°C ; From steam tables,
h1 = 3247.6 kJ/kg ; s1 = 7.127 kJ/kg K
Final state 2 : 4 bar 250°C ; From steam tables,
h2′ = 2964.2 kJ/kg, s2′ = 7.379 kJ/kg K
The process is shown as 1 to 2′ in
s1 = s2 = 7.127 kJ/kg K
Hence, interpolating,
(i) Isentropic efficiency :
(ii) Loss of availability :
Loss of availability
= b1 – b2
= h1 – h2′ + T0 (s2′ – s1′)
= 3247.6 – 2964.2 + 293 (7.379 – 7.127)
= 283.4 + 73.83 = 357.23 kJ/kg.
(iii) Effectiveness :
Effectiveness,