A cylinder contains 1 kg of a certain fluid at an initial pressure of 20 bar. The fluid is allowed to expand reversibly behind a piston according to a law pV² = constant until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston regains its original

Question

A cylinder contains 1 kg of a certain fluid at an initial pressure of 20 bar. The fluid is allowed to expand reversibly behind a piston according to a law [latex]p V^2=[/latex] constant until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston regains its original position ; heat is then supplied reversibly with the piston firmly locked in position until the pressure rises to the original value of 20 bar. Calculate the net work done by the fluid, for an initial volume of 0.05 m³.

Accepted Answer

Refer Fig. 2.39.

Mass of fluid, m = 1 kg

[latex]p_1=20[/latex] bar [latex]=20 imes 10^5[/latex] N/m²

[latex]V_1=0.05[/latex] m³

Considering the process 1-2

[latex]p_1 V_1^2=p_2 V_2^2[/latex]

∴ [latex]p_2=p_1\left(\frac{V_1}{V_2} ight)^2=20\left(\frac{V_1}{2 V_1} ight)^2[/latex] [latex]\left[\because \quad V_2=2 V_1( ext { given }) ight][/latex]

[latex]=\frac{20}{4}=5[/latex] bar

Work done by the fluid from 1 to 2 = Area 12 ML1 = [latex]\int_1^2 p d V[/latex]

i.e., [latex]W_{1-2}=\int_{v_1}^{v_2} \frac{C}{V^2} d V ext {, where } C=p_1 V_1^2=20 imes 0.05^2[/latex] bar [latex]m ^6[/latex]

[latex] herefore \quad W_{1-2}=10^5 imes 20 imes 0.0025\left[-\frac{1}{V} ight]_{0.05}^{01}[/latex]

[latex]=10^5 imes 20 imes 0.0025\left(\frac{1}{0.05}-\frac{1}{0.1} ight)=50000[/latex] Nm

Work done on fluid from 2 to 3

= Area 32ML3 [latex]=p_2\left(V_2-V_3 ight)=10^5 imes 5 imes(0.1-0.05)=25000[/latex] Nm

Work done during the process 3-1

= 0, because piston is locked in position (i.e., Volume remains constant)

∴ Net work done by the fluid

= Enclosed area 1231 = 50000 – 25000

= 25000 Nm.

Question 2.22: A cylinder contains 1 kg of a certain fluid at an initial pr...

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