Answer to Question #284308 in Mechanical Engineering for Talha

Question

Unit mass of certain fluid is contained in a

cylinder 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 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 0.05m

Answer 1

$V_1=0.05m^3$

$V_2=2V_1=2*0.05=0.1m^3$

$P_2=P_1(\frac{V_1}{V_2})^n$ , Where n=2

$P_2=20(\frac{1}{2})^2=5bar$

$P_3=P_2=5bar$

$V_3=V_1=0.05m^3$

$W=W_{1-2}+W_{2-3}+W_{3-1}$

$=\frac{P_1V_1-P_2V_2}{n-1}+P_2(V_3-V_2)+0$

$=\frac{20*10^2*0.05-5*10^2*0.1}{2-1}+5*10^2(0.05-0.1)$

=25kJ