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Steam expands adiabatically in a turbine from 20 bar, 400°C to 4 bar, 250°C. Calculate :

(i) The isentropic efficiency of the process ; 

(ii) The loss of availability of the system assuming an atmospheric temperature of 20°C ; 

(iii) The effectiveness of the process ; The changes in K.E. and P.E. may be neglected.

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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,

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