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5 kg of air at 550 K and 4 bar is enclosed in a closed system. 

(i) Determine the availability of the system if the surrounding pressure and temperature are 1 bar and 290 K respectively. 

(ii) If the air is cooled at constant pressure to the atmospheric temperature, determine the availability and effectiveness.

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Mass of air, m = 5 kg 

Temperature, T1 = 550 K 

Pressure, p1 = 4 bar = 4 × 105 N/m2 

Temperature, T2 = T0 = 290 K 

Pressure, p2 = p0 = 1 bar = 1 × 105 N/m2

(i) Availability of the system : 

Availability of the system is

Availability of the system

= m [cv (T1 – T0) – T0∆s] 

= 5[0.718 (550 – 290) – 290 × 0.246] = 576.7 kJ.

(ii) Heat transferred during cooling 

Q = m × cp × (T1 – T0

= 5 × 1.005 × (550 – 290) 

= 1306.5 kJ ...... heat lost by the system

Change of entropy during cooling 

Unavailable portion of this energy

= T0 . (∆S) = 290 × 3.216 = 932.64 kJ 

∴ Available energy = 1306.5 – 932.64 = 373.86 kJ.

Effectiveness, ∈ = \(\cfrac{Available \,energy}{Availability \,of \,the \,system}\) 

\(\cfrac{373.86}{576.7}\)

= 0.648 or 64.8%.

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