Efficiency is work/heat supplied...
work=area of the square and heat can be found out by using Cp and Cv respectively BC,AD and AB,CD respectively...
answer is (b)2/21
Akash Anand I guess you are wrong
Upvote·0· Reply ·2013-03-02 03:41:41
Efficiency is work/heat supplied...
work=area of the square and heat can be found out by using Cp and Cv respectively BC,AD and AB,CD respectively...
answer is (b)2/21
Answer is A ..and if any one got this try to provide the solution as well.
Oh...got it...
Work done=Area of square
= 2
Heat =MCΔT
For AB and CD---use Cv
and for BC and DA use Cp
It will come as 21/2. [Cv=fR/2 and Cp=(f+2)/2*R. Where f=degree of freedom which in this case=3]
Since efficiency=W/Q
=4/21
efficiency = work done / total heat supplied to the system
work done from graph is 2 p0vo clearly
heat is supplied in process AB and BC
during AB heat supplied ={(3pov0-p0vo)/R}*3R/2
during BC heat supplied ={(6povo-3povo/R)}*5R/2
so total heat supplied is 21pov0/2
therefore answer is (a) 4/21