The two strength factors that relate to all three competitive forces are
1 answer:
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Answer:
a) W =400 kJ
b) W = 0 kJ
c) W =-160.944 KJ
Explanation:
<u>Given </u>
<u><em>Process 1 ---> 2 </em></u>
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
<u>Process 2 ---> 3 </u>
The relation of the process V = constant
V3 = V2
Pressure of point (3) P3 = 10 bar
Volume of point (3) V3 = 4 m^3
<u>Process 3 ---> 1 </u>
The relation of the process PV = constant
<u>Required </u>
Sketch the processes on the PV coordinates
The work for each process in kJ
<u>Solution </u>
The work is defined by
W=
<em>a=V2</em>
<em>b=V1</em>
<em>x=P</em>
<em>dx=dV</em>
<u>Process 1 ---> 2 </u>
P3 = P4 = 5 bar
W=
<em>a=V3</em>
<em>b=V2</em>
<em>x=4</em>
<em>dx=dV</em>
putting the value of a, b, x, dx in above integral
W=400 kJ
<u>Process 2 ---> 3 </u>
V = constant Then there is no change in the volume,hence W = 0 kJ
<u>Process 3 ---> 1 </u>
By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1
W=
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
