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%.