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 m^{3}/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

Expert's answer

Solution:

Given Data:

p_{1} = 6.15 bar; p_{2} = 2.15 bar; v_{2} = 0.22 m^{3}/kg p_{4} = 0.82 bar; W_{(p=c)} = 0.625 kJ; m = 0.32 kg

Process 1-2 (Isothermal):

p_{1}v_{1} = p_{2}v_{2}

v_{1} = (p_{2}v_{2})/p_{1} = (2.15 bar x 0.22 m^{3}/kg x 0.32 kg) / 6.15 bar = 0.0246 m^{3}

W_{1-2} = p_{1}v_{1}ln(v_{2}/v_{1}) = 6.15 bar x 0.0246 m^{3} x ln (0.22 m^{3}/kg x 0.32 kg ÷ 0.0246 m^{3}) = 15.9073 kJ

Process 2-3 (Constant Pressure):

W_{2-3} = p_{2}(v_{3} - v_{2}) = 0.625 kJ

v_{3} = 0.625 kJ / 215 kJ/m^{3} + (0.22 m^{3}/kg x 0.32 kg) = 0.0733 m^{3}

Process 3-4 (Constant Volume):

W_{3-4} = 0

Process 4-1 (Polytropic):

p_{4}/p_{1} = (v_{1}/v_{4})^{n} =

0.82 bar / 6.15 bar = (0.0246 m^{3} / 0.0733 m^{3})^{n}

ln (0.1333) = n* ln (0.3356)

n = 1.8456

W_{4-1} = (p_{4}v_{4} - p_{1}v_{1}) / (n - 1) = (82 kJ/m^{3} x 0.0733 m^{3} - 615 kJ/m^{3} x 0.0246 m^{3}) / (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

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