Answer to Question #128834 in Mechanics | Relativity for Ariyo Emmanuel

Question

A fluid is heated reversibly at a constant pressure of 1.05 bar until it has a specific volume of 0.1m3/kg. It is then compressed reversibly according to a law pv= Constant to pressure of 4.2bar, then allowed to expand reversibly according to a law pv1.7= Constant, and is finally heated at constant volume back to the initial conditions. The work done in the constant pressure process is 1515Nm, and the mass of fluid present is 0.2 kg. Calculate the network of the cycle and sketch the cycle on a p-v diagram.

Answer 1

Calculation of the net work done on fluid in the cycle.

Sketch.

a)

$w_{12}=P(V_1-V_2)=515J$

$=0.515kJ$

$V_1=\frac{0.515}{105}+0.1\times0.2$ =0.02490m^3

b)

$P_2V_2=P_3V_3$

$V_3=(\frac{105}{420})0.1.0.2=0.005m^3$

$W_{23}=P_3V_3In\frac{V_2}{V_3}=420\times0.005In\frac{0.02}{0.005}=2.911kj$

c)

$P_4=P_3(\frac{V_3}{V_4})^{1.3}=4.20(\frac{0.005}{0.0249})^{1.3}=0.52 bar$

$W_{34}=\frac{P_3V_3-P_4V_4}{1-1.3}=\frac{420\times0.005-52\times0.0249}{-0.3}=-2.684kJ$

d)

$W_{41}=0$

Therefore the net work done on the fluin in cycle is :

$W_{net}=W{12}+W_{23}+W{34}+W{41}$

$=0.515+2.911-2.684+0=0.742kJ$

$=742J$