# The air, in a steady flow, enters the system at a pressure of 8 bar and 180°C

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The air, in a steady flow, enters the system at a pressure of 8 bar and 180°C with a velocity of 80 m/s and leaves at 1.4 bar and 20°C with a velocity of 40 m/s. The temperature of the surroundings is 20°C and pressure is 1 bar. Determine :

(i) Reversible work and actual work assuming the process to be adiabatic ;

(ii) Irreversibility and effectiveness of the system on the basis of 1 kg of air flow.

Take for air : cp = 1.005 kJ/kg K ; R = 0.287 kJ/kg K.

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Initial pressure of air, p1 = 8 bar

Initial temperature of air, T1 = 180 + 273 = 453 K

Final pressure of air, p2 = 1.4 bar

Final temperature, T2 = T0 = 20 + 273 = 293 K

Surroundings’ pressure, p0 = 1 bar

Mass of air = 1 kg

Initial velocity of air, C1 = 80 m/s

Final velocity of air, C2 = 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,

Wrev = 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 = Wrev – Wactual

= 181.49 – 163.2 = 18.29 kJ/kg.

Effectiveness, 