# In a turbine the air expands from 7 bar, 600°C to 1 bar, 250°C.

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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. 