Calculation of the net work done on fluid in the cycle.
Sketch.
a)
w12=P(V1−V2)=515J
=0.515kJ
V1=1050.515+0.1×0.2 =0.02490m^3
b)
P2V2=P3V3
V3=(420105)0.1.0.2=0.005m3
W23=P3V3InV3V2=420×0.005In0.0050.02=2.911kj
c)
P4=P3(V4V3)1.3=4.20(0.02490.005)1.3=0.52bar
W34=1−1.3P3V3−P4V4=−0.3420×0.005−52×0.0249=−2.684kJ
d)
W41=0
Therefore the net work done on the fluin in cycle is :
Wnet=W12+W23+W34+W41
=0.515+2.911−2.684+0=0.742kJ
=742J
Comments
Thanks for the answer please continue helping me