Answer to Question #154612 in Mechanical Engineering for Fredison styles

Question

A fluid at 6.15 bar is expanded reversibly according to law PV = constant to a pressure of 2.15 bar until it has a specific volume of 0.22 m³/kg. It is then cooled reversibly at a constant pressure, then is cooled at constant volume until the pressure is 0.82 bar; and is then allowed to compress reversibly according to a law PVn = constant back to the initial conditions. The work done in the constant pressure is 0.625 kJ, and the mass of fluid present is 0.32 kg. Calculate the value of n in the fourth process, the net work of the cycle and sketch the cycle on a P-V diagram

Answer 1

Solution:

Given Data: 

p₁ = 6.15 bar; p₂ = 2.15 bar; v₂ = 0.22 m³/kg p₄ = 0.82 bar; W_(p=c) = 0.625 kJ; m = 0.32 kg  

Process 1-2 (Isothermal):

p₁v₁ = p₂v₂

v₁ = (p₂v₂)/p₁ = (2.15 bar x 0.22 m³/kg x 0.32 kg) / 6.15 bar = 0.0246 m³

W_1-2 = p₁v₁ln(v₂/v₁) = 6.15 bar x 0.0246 m³ x ln (0.22 m³/kg x 0.32 kg ÷ 0.0246 m³) = 15.9073 kJ

Process 2-3 (Constant Pressure): 

W_2-3 = p₂(v₃ - v₂) = 0.625 kJ

v₃ = 0.625 kJ / 215 kJ/m³ + (0.22 m³/kg x 0.32 kg) = 0.0733 m³

Process 3-4 (Constant Volume): 

W_3-4 = 0

Process 4-1 (Polytropic):

p₄/p₁ = (v₁/v₄)ⁿ =

0.82 bar / 6.15 bar = (0.0246 m³ / 0.0733 m³)ⁿ

ln (0.1333) = n* ln (0.3356)

n = 1.8456

W_4-1 = (p₄v₄ - p₁v₁) / (n - 1) = (82 kJ/m³ x 0.0733 m³ - 615 kJ/m³ x 0.0246 m³) / (1.8456 - 1) = -10.7833 kJ

Wnet = W_1-2 + W_2-3 + W_3-4 + W_4-1 = 15.9073 kJ + 0.625 kJ + 0 kJ - 10.7833 kJ = 5.749 kJ

W_net = 5.749 kJ