Mass of air considered = 1 kg

Pressure, p_{1} = 7 bar = 7 × 10^{5} N/m^{2}

Temperature, T_{1} = 600 + 273 = 873 K

Pressure, p_{2} = 1 bar = 1 × 10^{5} N/m^{2}

Temperature, T_{2} = 250 + 273 = 523 K

Surrounding temperature, T_{0} = 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

s_{2} – s_{1} = c_{p} log_{e} \(\cfrac{T_2}{T_1}\) - R log_{e} \(\cfrac{p_2}{p_1}\)

Now, the change in availability is given by

Decrease in availability = 364.3 kJ/kg

(ii) The maximum work,

W_{max} = Change in availability = 364.3 kJ/kg. (Ans.)

(iii) From steady flow energy equation

Q + h_{1} = W + h_{2 }

W = (h_{1} – h_{2}) + Q

= c_{p} (T_{1} – T_{2}) + Q

= 1.005(873 – 523) + (– 9) = 342.75 kJ/kg

The irreversibility,

I = W_{max} – W

= 364.3 – 342.75 = 21.55 kJ/kg.