Mass of air, m = 8 kg
Temperature, T1 = 650 K
Pressure, p1 = 5.5 bar
Atmospheric pressure, p0 = 1 bar
Atmospheric temperature, T0 = 300 K
For air : cv = 0.718 kJ/kg K ; cp = 1.005 kJ/kg K.
(i) Change in available energy (for bringing the system to dead state),
= m[(u1 – u0) – T0∆s]
Using the ideal gas equation,
∴ Change in available energy
= m[(u1 – u0) – T0∆s] = m[cv(T1 – T0) – T0∆s]
= 8[0.718(650 – 300) – 300 × 0.288] = 1319.2 kJ
Loss of availability per unit mass during the process
= p0 (v0 – v1) per unit mass
Total loss of availability = p0(V0 – V1)
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. cp (T1 – T0)
= 8 × 1.005 (650 – 300) = 2814 kJ
Change in entropy during cooling
∆s = mcp loge \(\left(\cfrac{T_1}{T_0}\right)\)
= 8 × 1.005 × loge \(\left(\cfrac{650}{300}\right)\) = 6.216 kJ/K
Unavailable energy = T0 ∆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.