Question

A fluid at a pressure of 3 bar and specific volume of 0.18 m³/kg_m contained in a piston cylinder arrangement expands reversibly to a pressure of 0.6 bar according to relation p = c/v² where c is constant. Compute the work done in cal/kg_m by fluid on the piston.

Answer 1

pV² = C

p₁ = 3 bar = 3 x 10⁵ N/m²

V₁ = 0.18 m³/kg

W = area under pV graph

W = $\smallint$ pdV = $\smallint$ C/V² dV = -C(1/V₂ - 1/V₁)

W = C(1/V₁ - 1/V₂)

C = pV² = p₁V₁² = 3 x 0.18² = 0.0972 bar (m³/kg²)

V₂ = (C/p₂)^1/2 = (0.0972/0.6)^1/2 = 0.402 m³/kg

W = 0.0972 x (1/0.18 - 1/0.402) = 29840 Nm/kg = 7127 cal/kg