00:02
First, drawing pv diagram of pv diagram of diesel cycle.
00:11
So we are going to draw the pv diagram of the diesel cycle.
00:17
It will be very helpful in solving this problem.
00:21
So i am drawing here.
00:24
Just look at this.
00:25
This is the diagram.
00:26
Here it is p and here it is v.
00:28
It is 1, 2, 3, 4.
00:30
Going forward and solving this problem number a we can write as we know that t2 by t1 is equal to r to the power k minus 1 putting the given data so putting the given data in the question we have t2 is equal to t 1 r to power k minus 1 or 2 is equal to 300 into 16 to the power 1 .4 minus 1 or t 2 is equal to 300 into 16 to the power 1 .4 so t2 is coming as 909 .43 kelvin.
01:09
Now going forward we can write v2 by t2 is equal to v3 by t3.
01:17
So we can write t3 is equal to t2 into v3 by v2 is equal to t2 into rc.
01:28
So we can find it as t3 is equal to 909 .43 into 2.
01:34
That is 1818 .86 kelvin.
01:38
So t3 is coming as 1818 .18 .86 kelvin.
01:43
Now, going forward, solving problem number b.
01:46
So q -in is equal to cp t3 minus t2...