00:01
In this problem, let's see here, we got another, not an idea, a reheat ranking cycle, but not ideal.
00:10
And so in this case, we have a high pressure side up here at 10 megapascals.
00:19
Low pressure side were given this time of 10 kilpascals.
00:25
We're actually told what the power output is, and that's 80 megawatts.
00:29
The temperature at, let's see here, i think i wrote that wrong again.
00:38
T3 and t5, not t4.
00:43
So up here, we are at 500 degrees celsius.
00:48
The pressure here is a 1 megapa scale.
00:53
We're given that.
00:55
The efficiency of the turbine is now 80%, and that of the pump is 95%.
01:00
So at 3, we can get the entropy and the entropy at 3, because we know the pressure and temperature.
01:11
So the entropy at 4 prime, straight down here, is the same as entropy at 3.
01:17
So we can get the isotropic enthalpy at 4.
01:23
And from that, and from these two things, and the turbine efficiency, we can have find the enthalpy for the actual entropy.
01:36
Now at 5, we have pressure and temperature, so we can get enthalpy and entropy.
01:43
And then we know the isotropic case, the entropy at 6 is the same as entropy of 5.
01:49
So that would give us a quality factor in the isotropic case of 94 .9%.
01:56
And then the enthalpy in the is in the isentropic case would be 2 ,461 kilojoules per kilogram.
02:05
Now we can then use this and this and our efficiency to get the enthop the actual entropy at six and that is 2 ,665 kilojoules per kilogram.
02:19
Now it turns out and i drew this wrong and that is actually a saturated vapor...