00:01
So in this problem, we're asked to repeat problem 160, assuming an isotropic pump efficiency or compressor efficiency of 70%.
00:11
Okay, so in the first case, we have a compressor, and the second case we had a pump, if remember right, because in the first case, we have saturated vapor of r134a.
00:21
In the second case, we have saturated liquid.
00:26
It's coming in at 10f, and leaving at 150 psi, we have one pound mass per second flowing.
00:38
And in this case, we are told, let's see here, in the ideal case, the entropy is the same because, again, it's an isotropic process in the real, in the ideal case.
00:58
So with a quality factor of one in a temperature, we can look up for this fluid, the pressure and the pressure, the entropy and the entropy in the inlet state.
01:12
And in the ideal case, then, that is also the exit entropy.
01:17
And so now we have in the ideal case the exit entropy and the pressure.
01:21
And so we can find the exit enthalpy and the exit temperature in the ideal case.
01:28
From that, we can figure out the work in the ideal case, and that is minus 15 .5 btu per power mass, so the specific work in the ideal case.
01:41
So this is 15 .5 ptu per power mass of work we need to put into the compressor.
01:48
Now, in the real case, we need to divide that by 0 .7 because of the efficiency loss.
01:55
And that winds up being minus 22 .1 pound mass, a btu per pound mass of work, we actually need to put into the compressor.
02:04
And again, going back to our energy equation, that has to be the inlet entropy minus the inlet entropy minus the exit entropy.
02:16
That allows us to find the exit entropy in the real case, which is higher than in the ideal case...