00:02
Hi, for this problem, we are told to draw the diesel cycle both in a ts diagram and in a pv diagram.
00:11
First of all, we are giving the compression ratio as 17.
00:16
The heat supplied as 1500 kilojoules per kilogram.
00:24
We are told that the inlet temperature is 35 degrees c and that the inlet pressure is one bar.
00:39
Okay, so in a pv diagram, this is what, this cycle looks like we'll first of all have isentropic compression from one to two have a constant pressure heat addition to three some isentropic expansion to four and then a constant volume heat rejection and for this place for this place the same thing on a t s diagram we'll have isentropic compression from one to two constant pressure heat addition which will have a higher slope to 3, isentropic expansion to 4, and a constant volume heat rejection.
01:53
Now, to calculate all the temperatures, we will take the processes individually.
01:59
From process 1 to 2, we have isentropic compression.
02:07
And it follows this relationship.
02:14
What it means, therefore, is that p2 will be p1 into v1 over v2 raised to paragana.
02:23
And v1 over v2 is simply the compression ratio therefore p2 will be p1 into the compression ratio to the power gamma and that will be 1 bar times 17 to the power 1 .4 and that will give us 57 .8 bar similarly using the equation of states for an ideal gas pv is equal to to rt.
03:03
Here, v here is in meter cube per kg.
03:08
Please know that this is specific volume.
03:11
You will see therefore that t2 over t1 will become v1 over v2 to the power gamma minus 1.
03:22
And that's the compression ratio to the power gamma minus 1.
03:27
Therefore, t2 will be t1 compression ratio to the power gamma minus 1.
03:32
And that will be 308 times.
03:34
Times 17 to the power 0 .4 and that will give you 956 .6 kelvin.
03:46
We move on to the next process.
03:56
It's an isobaric heat addition.
04:00
For that process, q in will be worth cp into t3 minus t2.
04:08
We were not giving cp.
04:11
We can say but cp over cv is equal to gamma.
04:15
Therefore cp will be gamma times cv if i make t3 the subject t3 will be what q in all over gamma time cv plus t2 and this will be 1 ,500 kilojoules per kg all over 1 .4 times 0 .718 kilojoules per kelvin kilo joles per k kgely, cancels kilojoules per kre, whatever i have here will be kelvin, plus 956 .6.
05:03
And this will give me 2455 .5 .1 kelvin.
05:15
P3 will be equal to p to y.
05:18
It's a constant pressure process.
05:27
Another thing we can see here is that for this constant pressure, process it should follow child's law therefore v3 over t3 should be equal to v2 over t2 and so we can say that v3 v2 is equal to t3 over t2 and this is usually what we call the cutoff ratio the cutoff ratio is v3 over v2 therefore my cutoff ratio will be what two five sorry sorry 2455 .1 all over 956 .6 .6 and that will be 2 .57.
06:25
Now let us move to the next process.
06:33
This is also an isentropic process and so we can write again that p3 over p4 will simply be what v4 over v3 to the power gamma and we can write this in this form v4 over v2 times v2 over v3 to the power gamma i haven't changed anything this is v4 is the same as v1 if you remember the last process is constant volume so this is the same thing as v1 over v2 times v2 over v3 to the power gamma you will see that this becomes the compression ratio this becomes one over the cutoff ratio to the power gamma this is what we call the expansion ratio r c over r p and our expansion ratio will be what 17 over 2 .57 our expansion ratio is 6 .62 and this ratio of pressures we can then say therefore because we are looking for p4 will be p3 all over that and that will give me 4 .1 also t4 will be t3 all over expansion ratio to the power gamma minus 1 and that will give me 252455 .1 all over 6 .62 to the power minus 1 gamma minus 1 rather and that will give me 1152 .7 kelvin now we have gotten all the temperatures and all the pressures the the second part is asking for the thermal efficiency.
08:52
And this is simply the work done over the heat supplied, which is the heat in minus heat out, all over heat in.
09:03
Let us find the heat out...