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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,

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