00:01
We'll be looking at refrigeration, heat pump and air conditioning system.
00:06
For this particular problem, we have this ideal vapor compression, refrigeration system.
00:16
One to determine the compression cycle, one to determine the compressor power imputes.
00:24
Also, the rate at which heat is transferred to the working flu that goes through the condenser on the equation of performance.
00:35
Now we are two that superheated people enter the compressor, okay? right, at 70khal.
00:46
So let's say something like this, right? so let's say at this pressure, the pressure year is a 70kla with a 0 .7 par.
01:05
All right okay so we'll call here we can just let us complete we'll have this okay so we call year point one we call here point one prime here point two prime here point two year point three point four all right and the compressor we have the condenser the trucking taking place here this is here operator right so we'll write this on where so we have this point one alright, so let's say these are temperature axis with celsius, and the two that this is a 120 kiosklae the condenser so we're seeing him as a 12.
02:02
So the pressure there is actually 12 .4 bar, okay, when you convert it to bar, why we're given that the temperature, okay, for the evaporator here people entered at minus 18 okay enter the the compresor at minus 18 degrees celsius okay so that is minus 18 that's at the evaporator minus 18 degrees celsius okay so have this now we call these are entropy in kilo joe gram kelvin, right? so we can just, we're given this, the revision capacity to be equal to 8 tons.
03:22
So we can just, if we carry out our interpolation, considering the evaporator, and to get the values for the entropy, the entropy so we are going to have our sf we'll give us uh...
03:51
Then we're going to give us 0 .0294 okay okay that is entropy and saturated of the liquid line 0 .02 2 .0 .094 kilo kilo per kilogram per kilogram per kelvin okay so we'll have that then we hope also look at s .r .1 which is also equal to sg, the people line along the entropy the project along the self -term vapor line.
04:41
You're going to give you equal to 0 .994, 84 kilojou per kilogram per kelvin.
04:54
Right? so i can just look for the difference in entropy.
05:00
During the saturdaysipoland liquid line.
05:03
We're going to have this to be 4 to 0 .919.
05:09
9 .19 kilo per kilogram per kilogram for kelvin.
05:14
So, if you follow your interpolation, you're going to have also, let me bring this here, your h at f also, we're going to give you 6 .926.
05:37
65 kilo per kilogram, kilo per kilogram, right? and they're going to have this again, which at point one, you're going to give you, of course, is simply not at least, g to 226 .5 or 6 kilo -g per kilogram.
06:13
Hfg, it will give you 219 .6 0 .5 kilogram.
06:27
Alright? so, to further our interpolation, and we're going to have that pressure of 0 .7 bar, okay? 0 .7 bar, okay.
06:52
We're going to have, sorry, h1 ,000.
07:02
Sorry, s1 prime, you're going to give us 1 .0 .03469 kilo -jou per kilogram.
07:16
And h1 prime, you're going to give us 24 -242 .25 kilo -ch kilo per kilogram.
07:31
Sorry, the entropy that we have here, s -1 price.
07:35
1 .03 -469 kilo per kilogram...