Answer:
a) W =400 kJ
b) W = 0 kJ
c) W =-160.944 KJ
Explanation:
Given
Process 1 ---> 2
The relation of the process P = constant
Pressure of point (1) P1 = 10 bar = P2
Volume of point (1) V1 = 1 m^3
Volume of point (2) V2 =4 m^3
The relation of the process V = constant
Process 2 ---> 3
The relation of the process V = constant
V3 = V2
Pressure of point (3) P3 = 10 bar
Volume of point (3) V3 = 4 m^3
Process 3 ---> 1
The relation of the process PV = constant
Required
Sketch the processes on the PV coordinates
The work for each process in kJ
Solution
The work is defined by
W=[tex]\int\limits^a_b {x} \, dx[/tex]
a=V2
b=V1
x=P
dx=dV
Process 1 ---> 2
P3 = P4 = 5 bar
W=[tex]\int\limits^a_b {x} \, dx[/tex]
a=V3
b=V2
x=4
dx=dV
putting the value of a, b, x, dx in above integral
W=400 kJ
Process 2 ---> 3
V = constant Then there is no change in the volume,hence W = 0 kJ
Process 3 ---> 1
By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1
W=[tex]\int\limits^a_b {x} \, dx[/tex]
a=V1
b=V3
x=1V^-1
dx=dV
putting the value of a, b, x, dx in above integral
W=| ln V | limit a and b
= -160.944 KJ
