Question

0.2 m³ of air at 4 bar and 130°C is contained in a system. A reversible adiabatic expansion takes place till the pressure falls to 1.02 bar. The gas is then heated at constant pressure till enthalpy increases by 72.5 kJ. Calculate : 

(i) The work done ; 

(ii) The index of expansion, if the above processes are replaced by a single reversible polytropic process giving the same work between the same initial and final states. 

Take c_p = 1 kJ/kg K, c_v = 0.714 kJ/kg K.

Answer 1

Initial volume, V₁ = 0.2 m³ 

Initial pressure, p₁ = 4 bar = 4 × 10⁵ N/m² 

Initial temperature, T₁ = 130 + 273 = 403 K 

Final pressure after adiabatic expansion, 

p₂ = 1.02 bar = 1.02 × 10⁵ N/m² 

Increase in enthalpy during constant pressure process 

= 72.5 kJ. (i) 

Work done :

Process 1-2 : Reversibe adiabatic process :

where, R = (cp – cv) = (1 – 0.714) kJ/kg K 

= 0.286 kJ/kg K = 286 J/kg K or 286 Nm/kg K

m = \(\cfrac{4\times10^5\times0.2}{286\times403}\) = 0.694 kg.

Process 2-3. Constant pressure :

Work done by the path 1-2-3 is given by 

W_1–2–3 = W_1–2 + W_2–3

Hence, total work done = 85454 Nm or J.

(ii) Index of expansion, n : 

If the work done by the polytropic process is the same,