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8 kg of air at 650 K and 5.5 bar pressure is enclosed in a closed system. If the atmosphere temperature and pressure are 300 K and 1 bar respectively, determine : 

(i) The availability if the system goes through the ideal work producing process. 

(ii) The availability and effectiveness if the air is cooled at constant pressure to atmospheric temperature without bringing it to complete dead state. Take cv = 0.718 kJ/kg K ; cp= 1.005 kJ/kg K.

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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 × log\(\left(\cfrac{650}{300}\right)\) = 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.

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