00:01
So here, we have the quick diagram.
00:03
We have a turbine, steady, single flows for both the inlet and the exit.
00:09
Steady single flows, then adiabatic and reversible.
00:22
So we can first say that the specific work for the turbine equaling then h -7 -3 minus h -7 -4, s -4 -4 equaling s -3 for the turbine, we know that we can use the property.
00:37
Relation equation 832 taking a look at table 8 .5 for k if s sub 4 equaling s sub 3 within the turbine we can say that then the temperature sub 4 equaling the temperature sub 3 multiplied by the pressure at 4 divided by the pressure at 3 raised to the k minus 1 over k power this is going to be equal to 930 rather 920 23 .2 multiplied by 100 divided by 170 raised to the .286 power and this is giving us then 793 .2 kelvin.
01:23
We can say that then the specific work for the turbine equaling h sub 3 minus h sub 4.
01:29
This is going to be equal to the specific heat capacity multiplied by t sub 3 minus t sub 4.
01:40
This would be equaling then to 1 .004 multiplied by 9223 .2 minus 793 .2.
01:54
And this is giving us then the specific work being 130 .5 kilojoules per kilogram.
02:09
The rate of work then, the time rate of work for the turbine would then be equaling to the mass transfer rate multiplied by the specific work.
02:18
And once we take the product, this is 13 .05 kilowatts, given that the mass transfer rate is 0 .1 kilograms per second.
02:32
So given this, we can then take a look at the compressor...