# Answer to Question #154612 in Mechanical Engineering for Fredison styles

Question #154612

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

1
2021-01-18T01:12:27-0500 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

W1-2 = 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):

W2-3 = 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):

W3-4 = 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

W4-1 = (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 = W1-2 + W2-3 + W3-4 + W4-1 = 15.9073 kJ + 0.625 kJ + 0 kJ - 10.7833 kJ = 5.749 kJ

Wnet = 5.749 kJ

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