00:01
So the work required for an aviabatic process can be written as w comp is equal to c .p.
00:10
Times t2 minus t1.
00:12
So we're going to substitute the known values.
00:15
That's 1 ,0005.
00:20
Excuse me.
00:23
Times 690 kelvin's minus 280 kelvin's.
00:27
And you're left with 400 ,012 .50 joules per kilogram.
00:38
And so we know that w comp is 412 .05 kilojoules per kilogram.
00:49
So for the thermal efficiency of the brayton cycle is then given by n -thermal is equal to w -net over q -n.
01:11
And so that is, if we're going to substitute into the thermal efficiency formula, we get w -turbine minus w -comp.
01:20
Over cpt3 minus t2.
01:28
And we're going to rearrange to solve for w turbine.
01:33
So we get w turbine is equal to in thermal times cp times t3 minus t2 plus w comp.
01:46
And plug in what you know.
01:48
0 .424 times 1 ,0005 times 1600 minus 690.
01:55
Plus 412 .05, and you get that equaling 386 .90 kilojoules per kilogram.
02:08
And then for the power output of the cycle is the turbine work multiplied by the mass flow rate.
02:14
So p equals m times w turbine, which is 0 .2222 times 386 .9 .9 .2 .2 .2 times 386 .9.
02:29
And you get 85 .98 kilowatts.
02:36
So the power output of the brayton cycle is 85 .98 kilowatts.
02:41
So for the heating rate, we have qn is equal to cpt3 minus t2.
02:55
And you plug in what you know, 1 ,05 times 1 ,600 minus 690...