00:02
Hi, for this problem, we are giving properties of dry air and we are told that for a brating cycle, we should first of all draw the cycle, find the turbine exit temperature, a network, and the thermal efficiency.
00:20
Now we were giving the pressure ratio as it, giving the minimum pressure or the p -in as 310kcal, the maximum temperature which is usually t3 as 1160 kelvin we're also giving the isentropic efficiency of the compressor as 0 .8 and the isentropic efficiency of the turbine at 0 .82.
01:04
Sorry this was 0 .75.
01:12
Now to draw this cycle on a ts diagram for a gas turbine cycle we know that we usually start with isentropic compression from 1 to 2 s but because the turbine is not 100 % efficient the real compression work or the work spent compressing the gas is more than the isentropic or the ideal work then we have this constant pressure heat addition from 2 to 3 and then again from 3 to 4 s if this was purely isentropic this would have a 4 s what as usual, the work done by the turbine is less than the ideal work.
02:07
So this would be actual 4.
02:10
And then we go back to where we started with the heat rejection at constant pressure.
02:21
So this would be the actual line, not the other one.
02:25
And let me just change that also.
02:29
This would be the actual line.
02:32
This should be straight.
02:43
And another way of drawing this.
02:45
This is my compressor usually mounted on the same shaft as the turbine and of course air comes in here we live from here so this is point one and point two we have a combustion chamber here and leaves the combustion chamber this is point three i usually fuel is added here and then it goes into the turbine this is the compressor this is the turbine now the air leaves the turbine and this becomes points four and then we usually have heat rejection in between and i'll call this a cooler now we don't want to use um assume constant cp and all of that we want to check use the temperature variation of cpcv and what that means is that we will have to use the tables for ascentropic properties of air and how do we do that that's what i want to show here for process 1 to 2 that process is is centropic compression so we can use the table now we're told that t1 is 310k now we go to this table and that was why i brought in the table here the first column is temperature here we have the entropy the static enthalpy here we have the relative pressure which is very very important so we will come here and at t1 equal to 310k we can see that this is what the enthalpi is and this is what the relative pressure is those are the two important variables for me and i will just come here and write at t1 equal to 310k relative pressure is 1 .5546 static enthalpy is 310 .2 .246 .540 .2 .24 per kilogram now but this relative pressure we found here obeys this ratio between p2 and p1 so now we have pr1 we don't have pr2 but we know p2 over p1 and what is this p2 over p1 it's simply the pressure ratio which we're given at the initial as one of the variables describing this gas turbine plant.
05:59
The pressure ratio is simply the ratio of p to 1, p1, the ratio across which isentropic compression occurs, and this is equal to 8.
06:09
We can see therefore that pr2 is going to be 8 times pr1, and that will give us that pr2 is 8 times 1 .5546, and that should give me 12 .4368.
06:37
Now using pr2 we will go back to the table and read what will then be t2 but in this instance in this instance that will be t2 isentropic if this was purely isentropic we'll use it from tables let me write it here from tables we'll find t2s and would find h2s let me try and do that okay so what i did was i brought in another part of the table i couldn't get the whole table at once and here i want us to see something we don't exactly have pr equal to 12 points 4368 sorry that's not the portion i want never mind let me rewrite what i just cleaned that is 12 .66 okay but we have this this region here so whatever we are looking for we are looking for pr2 equal to 12 .4368 that means it is somewhere between 11 .86 and 12 .66.
08:49
So we have to do a linear interpolation between these t2s values between 550 and 560 and between this values also for the static entelope 555 .7 and 565 .17 and that's what i will do here.
09:09
Now i will just say from tables when pr2 is equal to 11 .86, we have that t2s is what 55k, h2s is 555 .7574.
09:41
Likewise.
09:41
When pr2 is 12 .66 t2s is 560k and h2s is 565 .17, ply jones per p .j.
10:02
And so doing a linear interpolation, how do we do that? simply the t2s i need minus the lower value of t2s, which is 550, all over the higher value, 560 minus the lower value will be equal to 12 .4368 minus the lower value there was 11 .86 all over the higher value was 12 .66 minus 11 .86.
10:43
You will see therefore when i solve this that t2s would be equal to.
10:53
You will see therefore that t2s will be equal to.
10:57
Let me see 557.
11:01
K how we do the same thing for the enthalpy h2 minus the lower value there was 555 .74 divided by the higher value 565 .17 minus 555 .75 .7.
11:26
That's still going to be equal to 12 .3 .12 .4368 minus 11 .86686.
11:39
All over 12 .66 minus 11 .86.
11:43
And you'll find that h2, this is h2s rather not h2, because it's all assuming isentropic conditions.
11:52
And this would be 562 .5 .54 kilojoules per cage.
12:01
Now i'll try to manage the remaining space i have on my white spot.
12:09
Now that we found this, we can then call on the isentropic efficiency, of the compressor and it has this formula 82s minus h1 all over h2 minus h1 and we'll find that the actual h2 therefore will simply be h1 plus one over the isentropic efficiency of the compressor into h2s minus h1 and you'll find therefore that h2 will be equal to when i just put in the variables i have.
12:48
We found h1...