00:01
We need to find the efficiency of the brayton cycle.
00:03
So we know that heat transfer occurs only along the isobaric processes.
00:24
So in this case, the isobaric processes would be from b to c and from a to d, or rather from d to a.
00:30
And we know that the heat transfer doesn't occur from a to b or from c to d because again, these are adiabatic expansions.
00:39
So given this, we can say that, that q sub h is going to be equal to, rather the heat input would be equal to q sub b, c, which would be equal to n, the number of moles times the specific heat, rather the molar heat capacity at a constant pressure times the change in temperature.
01:01
So t sub c minus the temperature at b.
01:05
The heat exhaust is going to be equal to the heat transfer associated with process d .a.
01:12
And this is going to to be equal to the number of moles times the molar heat capacity at a constant pressure, times the difference between the temperature at d minus the temperature at a.
01:24
And so finally, we also know that the efficiency is going to be equal to 1 minus q sub l divided by q sub h.
01:33
So at this point, we can say that e is going to be equal to 1 minus n sub c p times t sub d, minus t sub a divided by n sub c p t sub c minus t sub b these are going to cancel out and we finally have this expression now we're going to use the ideal gas law in order to substitute for these and say that e is now going to be equal to one minus p sub d v sub d minus p sub a v sub a divided by p sub c v sub c minus p sub b b b this is going to equal furthermore one minus p sub a times v sub d minus v sub a divided by p sub b v v sub b v v c minus v sub b this is we can cancel out we other we can factor out a p a pressure sub a and a pressure sub b because these are again i so barric processes, which means that these have a constant pressure.
02:48
Now, we are going to say that process a, b is, of course, adiabatic as it's labeled.
03:00
And we can say that p sub a, v .a to the gamma power would be equal to p sub b, v sub b to the gamma power.
03:08
So we can say that v.
03:09
A is going to equal v.
03:11
Sub b times p sub b, divided by p sub a, to the one over gamma power.
03:19
The same exact thing for process cd.
03:25
Process cd is also adiabatic...