Initial pressure of air, p_{1} = 8 bar

Initial temperature of air, T_{1} = 180 + 273 = 453 K

Final pressure of air, p_{2} = 1.4 bar

Final temperature, T_{2} = T_{0} = 20 + 273 = 293 K

Surroundings’ pressure, p_{0} = 1 bar

Mass of air = 1 kg

Initial velocity of air, C_{1} = 80 m/s

Final velocity of air, C_{2} = 40 m/s.

**(i) Reversible work and actual work : **

Availability of air at the inlet

Availability of air at the inlet

= 1.005 (453 – 293) – 293 (– 0.159) + \(\cfrac{80^2}{2\times10^3}\)

= 160.8 + 46.58 + 3.2 = 210.58 kJ

Availability at the exit

∴ Availability at the exit

= – 293 (– 0.09656) + \(\cfrac{40^2}{2\times10^3}\) = 29.09 kJ/kg

Reversible/theoretical work which must be available,

W_{rev} = 210.58 – 29.09 = 181.49 kJ.

Actual work developed can be calculated by using the energy equation for adiabatic steady flow process as follows :

**(ii) Irreversibility and effectiveness : **

Irreversibility, I = W_{rev} – W_{actual}

= 181.49 – 163.2 = 18.29 kJ/kg.

Effectiveness,