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In a turbine the air expands from 7 bar, 600°C to 1 bar, 250°C. During expansion 9 kJ/kg of heat is lost to the surroundings which is at 1 bar, 15°C. Neglecting kinetic energy and potential energy changes, determine per kg of air : 

(i) The decrease in availability ; 

(ii) The maximum work ; 

(iii) The irreversibility. 

For air, take : cp = 1.005 kJ/kg K, h = cpT, where cp is constant.

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 Mass of air considered = 1 kg 

Pressure, p1 = 7 bar = 7 × 105 N/m2 

Temperature, T1 = 600 + 273 = 873 K 

Pressure, p2 = 1 bar = 1 × 105 N/m2 

Temperature, T2 = 250 + 273 = 523 K 

Surrounding temperature, T0 = 15 + 273 = 288 K 

Heat lost to the surroundings during expansion, 

Q = 9 kJ/kg.

(i) From the property relation, 

TdS = dH – Vdp

For 1 kg of air

s2 – s1 = cp loge \(\cfrac{T_2}{T_1}\) - R loge \(\cfrac{p_2}{p_1}\)

Now, the change in availability is given by

Decrease in availability = 364.3 kJ/kg

(ii) The maximum work, 

Wmax = Change in availability = 364.3 kJ/kg. (Ans.) 

(iii) From steady flow energy equation

Q + h1 = W + h

W = (h1 – h2) + Q 

= cp (T1 – T2) + Q 

= 1.005(873 – 523) + (– 9) = 342.75 kJ/kg 

The irreversibility, 

I = Wmax – W 

= 364.3 – 342.75 = 21.55 kJ/kg.

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