Mass of air considered = 1 kg
Pressure, p1 = 7 bar = 7 × 105 N/m2
Temperature, T1 = 600 + 273 = 873 K
Pressure, p2 = 1 bar = 1 × 105 N/m2
Temperature, T2 = 250 + 273 = 523 K
Surrounding temperature, T0 = 15 + 273 = 288 K
Heat lost to the surroundings during expansion,
Q = 9 kJ/kg.
(i) From the property relation,
TdS = dH – Vdp
For 1 kg of air
s2 – s1 = cp loge \(\cfrac{T_2}{T_1}\) - R loge \(\cfrac{p_2}{p_1}\)
Now, the change in availability is given by
Decrease in availability = 364.3 kJ/kg
(ii) The maximum work,
Wmax = Change in availability = 364.3 kJ/kg. (Ans.)
(iii) From steady flow energy equation
Q + h1 = W + h2
W = (h1 – h2) + Q
= cp (T1 – T2) + Q
= 1.005(873 – 523) + (– 9) = 342.75 kJ/kg
The irreversibility,
I = Wmax – W
= 364.3 – 342.75 = 21.55 kJ/kg.