00:01
The heat rejected net cycle work and efficiency as well as the us power, the cycle, the engine, you need to support the current equation.
00:19
Okay? so we have this led to the temperature entropy, kilo -quered, per kilogram, parcaravan, that's your entropy.
00:48
Right, so it's going to give us something that looks like the cycle, so we're going to have something that looks like, yeah.
01:21
So let's see something like this.
01:24
Okay, so this point one, point two.
01:32
I want to twist, eccentric compression.
01:35
2 -2 -3 is constant pressure heat addition.
01:40
3 to 4 is eccentric expression.
01:43
And then 4 -2 -1 is called your heat rejection.
01:53
So we can have the so we can rotate the in -bar.
02:06
We have a big meter, we have a critical number, all right? we have a computer, we use two big meter.
02:18
We have something around with, this is one, two, two, three, four.
02:38
All right, so we have that.
02:40
So to solve the problem, we apply the government principles for some.
02:45
Some problems in the problems in diesel cycle.
02:55
So if we consider process 2 to 3, process 2 to 3, we want to see how can get the temperature, the maximum temperature.
03:04
So process 2 to 3, we have a report here, q, it's called a q, the heat supply, 2, 3, the energy supply, okay, so let's use your cp.
03:36
Let's hit at constant pressure.
03:39
The temperature at point three, that at point two.
03:45
Now, for the temperature at point two, we are going to compact what we are given.
03:50
We are given 1000, 1001 and 1 ,000 and 1 ,000 and 1d001 and heat at degree farides.
04:08
So we can just convert it to kelvin.
04:12
We're going to let's convert it to degrees celsius by applying that is in degrees celsius is equal to use this particular pharoid to what we have in farro height which is a temperature in farro height and us 32 all right multiply it by five okay divide by nine so we multiply everything by 5 and divide by 9 so if we do that and plug anything in here we'll get this uh this one uh the same thing as uh before to 21 sorry we give you for this one for g2 2 2 we give it all 70 so 5997 okay so plug in everything that is there 597 7 0 .8 degrees celsius so if you add 273 to 8 plus 273 to 173 to be compared to caravine which is 177 .8 so we can bring it here now i will take rcpy we take rcp, so the heat a constant pressure to be 1 .15, okay, kilo per kilogram per kelvin, all right? so if we plug in what we have yet, so let's take this on to 1 .10, 15 kilojou per kilogram per kelvin, and this is, we don't know this one yet, you're welcome for the maximum temperature.
06:51
This is 870, we just put forth, just mark it's t2.
06:59
1870, 80, 18 kelvin.
07:04
Alright, so let's change our q, heat supply q2 to 3 was given for us as 1 and 88 btu.
07:14
So let's come back to 1 ,88 btu.
07:18
We'll do some compression.
07:24
188 btu.
07:28
So we can just do our conversion.
07:29
We give you since one btu will give you 10 is given in the question one btu as british tamil 5 .0 6 .0.
07:44
Yeah for 188 btu is equal to 188 okay financial and 5 5 .10 .06 okay so we should give you so we plug in every so you're going to have it to be 198, 351, 1 .2 .2.
08:27
0 .2 .2.
08:29
So we can just take it to, so we can plug it in here, but we provide it to kilojo.
08:37
So we are assuming that everything is in kilojou per kilogram.
08:42
So this one will now give us 190 tickets, divide by 1 ,001, 2, 3, 190 ,000 ,000.
08:50
Point there, 35, so plug it here, 198, 1 ,3, 35.
08:59
So if you do this, you can just make t3, there the service of the formula, let's block this one out, just for us, if you do this on then, t3, making the service of the formula, you'll have it to be 1000, 143.
09:26
1 2 .2 8 kelvin.
09:27
So this is a .28 kelvin.
09:36
This is the maximum temperature you are talking about, okay? as a maximum temperature.
09:45
And you can decide to convert it to, you can try to convert it to the far -hite.
09:56
If you want to convert it to far -hite, you can just do that and get it in the paroite.
10:07
She's saying that you assume okay so we can leave it like that can leave it like that all right let's continue like that so you can do the conversion later if you want to so be the big parts want to look at so if we look at also with the same process process two to process two 3 rights we know that at that point is a constant project process yet we are going to have v3 apply the gas law v2 over t3 is equal to v3 um 3 so we can just make a v3 the the formula to make v3, the 3 decides on the formula, this is equal to v2, t3 over t2.
12:15
So we let's look for the volume of v2.
12:18
You know v1 already, that v1 is equal to 6 .5 for big meter.
12:27
So v1 is equal to 6.
12:37
Point five two point five four cubic feet all right so you can convert this one to you can convert into volume by applying one cubic feet because of 0 .083 cubic meter so if you do that in cubic meter this will give you 0 .25 0 .1 -85 okay cubic liter then you and also look for what would be v2.
13:37
So v2 now will not give us, we know that compression ratio that is given to us is a 15.
13:43
And compression ratio arrow is equal to b1 over b2.
13:52
So if we make b2 the circle of the formula, b2 would be equal to b1 over arrow.
14:03
And arrow is our compression ratio, which is actually equal to.
14:07
15 okay even in the question 15 so if you do this now this is also a we want you know v1 already is 0 .1 5 cubic meter all over 15 so if you do if you do the division i have to be 0 point uh v3 sorry v2 is equal to 0 .101 2 0 .0 .012 the big meter.
15:06
Alright.
15:06
So we haven't gotten that.
15:08
We know v2 now.
15:11
So we can plug in all the details we have here now.
15:14
Plug it in there.
15:16
We're going to have v3.
15:19
And using this our equation to be 0 .012.
15:27
That's our t3 in kelvin, which we've got to be 1 ,000.
15:33
Of 3 .28 1 ,000, 143, 43, 1 ,000 for 3 .28, 3 .2 8.
16:12
Let's block this on out.
16:14
So if you divide this on now by, please this one is not part of that's, okay? that's a different thing we got earlier on.
16:22
So i'm blocking it out.
16:23
Alright? so if you divide this on now, it's going to give you by t2 now, which is a, it around 70 is 70 .8 okay to give you a degree equal to 2 .0140 .0 .0 .0 .0 .0.
16:53
0 .014 cubic meter.
17:12
So that's the goal for point three.
17:16
All right.
17:17
Now haven't gotten that, we can continue now to get a, we're actually trying to get in p.
17:24
So we proceed to this way.
17:27
We're going to have our, we find what we call the cutoff ratio, i'll also see, it's actually our v3 over v2.
17:41
So we know v3 already, there will no v2 also in going to at 1 .2.
17:46
So if you do that division, just plug in, you arrive at 1 .17.
17:51
It's a cutoff ratio.
17:54
Then we can come here now and apply for the expansion process the expansion process which is process process process work now process process uh three to four it's expansion process so we are going to use this equation of this relation t4 temperature point four about temperature at point three is equal to the cut -up ratio i just talked about about the pressure ratio pressure ratio and the ratio specific heats one is a power -head so if we love things into this one this will give us t4 just so just make t4 the solvit of the formula you have the strategy equal to t3 k -3 into a cutoff ratio with pressure ratio and pressure ratio into ratio specific heats plus one so, this is 1 ,040, this is 143 .28, 28.
19:44
And into cut up ratio 1 .17...
