Steam at 10 bar absolute pressure and 0.95 dry enters a super heater and leaves at the same pressure at 250°C. Determine the change in entropy per kg of steam. Take Cps = 2.25 kJ/kg K.
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Given that :
P =10bar
x = 0.95
tsup = 250°C
From Saturated steam table
tsat = 179.9
Now, entropy of steam at the entry of the superheater
s1 = sf1 + x1sfg1
= 2.1386 + 0.95 × 4.4478 = 6.3640 kJ/kg K
entropy of the steam at exit of superheater
s2 = sgf + Cps ln(Tsup/Tsat)
= 6.5864 + 2.25 ln (250+ 273/179.9 +273)
= 6.9102 kJ/kg K
Change in entropy = s2 – s1 = 6.9102 – 6.3640
= 0.5462 kJ/kg K