Mass of air, m = 8 kg

Temperature, T_{1} = 650 K

Pressure, p_{1} = 5.5 bar

Atmospheric pressure, p_{0} = 1 bar

Atmospheric temperature, T_{0} = 300 K

For air : c_{v} = 0.718 kJ/kg K ; c_{p} = 1.005 kJ/kg K.

(i) Change in available energy (for bringing the system to dead state),

= m[(u_{1} – u_{0}) – T_{0}∆s]

Using the ideal gas equation,

∴ Change in available energy

= m[(u_{1} – u_{0}) – T_{0}∆s] = m[c_{v}(T_{1} – T_{0}) – T_{0}∆s]

= 8[0.718(650 – 300) – 300 × 0.288] = 1319.2 kJ

Loss of availability per unit mass during the process

= p_{0} (v_{0} – v_{1}) per unit mass

Total loss of availability = p_{0}(V_{0} – V_{1})

Loss of availability = \(\cfrac{1\times10^5}{10^3}\) (6.891 – 2.713) = 417.8 kJ

(ii) Heat transferred during cooling (constant pressure) process

= m. c_{p} (T_{1} – T_{0})

= 8 × 1.005 (650 – 300) = 2814 kJ

Change in entropy during cooling

∆s = mc_{p} log_{e} \(\left(\cfrac{T_1}{T_0}\right)\)

= 8 × 1.005 × log_{e }\(\left(\cfrac{650}{300}\right)\) = 6.216 kJ/K

Unavailable energy = T_{0} ∆S

= 300 × 6.216 = 1864.8 kJ

Available energy = 2814 – 1864.8 = 949.2 kJ.

Effectiveness, ∈ = \(\cfrac{Available \,energy}{Change \,in \,available \,energy}\)

= \(\cfrac{949.2}{1319.2}\) = 0.719.