0:00
Hi everybody.
00:01
So for this one, we're repeating problem 1250.
00:06
Okay, so number of 50 for an isotropic compressor efficiency of 82.
00:12
Okay.
00:13
So for this, we need to find the r, which is going to be 0 .75.
00:19
The numbers come from the charts and the problem 50, okay? so 0 .75, 0 .583 plus 0 .7 .5 plus 0 .5.
00:30
2 .5 .0 .2765, which equals 0 .45785, kilojoules per kilogram times k.
00:44
And now we have your cp mix, okay.
00:49
And that is going to be 0 .75 times 2 .25 plus 0 .25 plus 0 .25 times 1 .25 times 1 .75 .7.
01:01
Equals 2 .132 kilojoules per kilogram kelvin.
01:09
Okay, so now we have your cd mix, which is going to be a cp minus your r.
01:19
So we got 2 .132 minus 0 .457 -85, which is going to equal to 1 .674 .6, 4 .7, which is going to equal to 1 .674.
01:32
415 kilojoules per kilogram times kelvin and now we need to find the ratio of specific heat which is going to be k equals cpcb and that's ratio of specific heat which is going to be 2 .132 divided by 1 .67415 which is 1 .2734.
01:57
Okay and now we can plug it into the equation for it, which is pressure 2, pressure 1, and now we have k minus 1 over k...