00:01
So here we have an air turbine.
00:05
We have an a, this is adiabatic, reversible process.
00:13
We can then say that the specific work equaling the inlet enthalpy minus the exit enthalpy.
00:23
We know that the exit entropy equals the inlet entropy for part a.
00:28
We can use, again, table a7.
00:33
And we can say that then the inlet for the inlet state, the inlet entropy is equaling 1 ,277 .8 kilojoules per kilogram.
00:47
And then we have the inlet entropy equaling 8 .34596.
01:00
And this would be kilojoules per kilogram per kelvin.
01:07
We can say that then the entropy at exit, equaling the entropy at the inlet plus the ideal gas constant times the natural log of the exit pressure divided by the inlet pressure we can actually solve equaling then 8 .3 4596 plus 0 .287 multiplied by the natural log of 100 divided by 800 and so we we find that then this is equaling 7 .7492...