Fred,
Definately is 'Homework'.
I am not about cheating, I am all about learning as you hopefully know by now.
So where I have got to:
I have crudly sketched a PV and a TS diagram and have calauated as follows.
Known values:
R= 0.287kJ/kgK
Change in Entropy (in Expansion which should also be the same for Contraction)
Delta s = 0.25 kJ/kg
Net Work W = 100 kJ/kg
Aboslute Temperature at point 1 on a PV diagram i.e. before isothermal expansion = 750 K
Pressure at point 1 = 800 kN/m^2
My first step was to find the lower temperature and amount of heat rejected during isothemal compression. I note that the area of the square created in a TS graph will equal the net work. i.e. the product of change in entropy and change in temperature will be the net work. Therefore dividing the net work by the change in entropy should give the change in Temperature i.e.
Delta T = W/Delta s = 100 kJ/kg / 0.25 kJ/kgK = 400 K.
Ergo T2 i.e. lower Temperature = T1 - Delta T = 750K - 400K = 350K
Next I note that the heat rejected by the cycle in isothermal compression is equal to the product of the Temperature (T4) after compression the change in entropy. i.e.
Qc = T4*Delta s = 350K*0.25kJ/kgK = 87.5 kj/kg
To verify this I looked at the heat taken in during isothermal expansion as Qh = T1*Delta s = 750K*0.25kJ/kgK = 187.5kJ.
This holds true with the first law i.e. Sum of Q = Sum of W noting heat in is positive and heat out is negative
187.5 + (-87.5) = 100
Then the thermal Efficiency being T1-T2/T1 = 750-350/750 = 0.53 (reocurring).
Then I move onto the final part which I am not sure about.
I note that the isothermal work done in expansion is an identity with the heat taken in.
The heat taken in Qh = mRT1In(r) Noting this problem is in specific terms I have negated m as mass of air is not given (as a punt although the units come out as desired following this approach).
Ergo:
187.5 kJ/kg = R*T1*In(r)
= 0.287*750*In(r)
187.5 =215.25*In(r)
0.871=In(r)
0.871= log (base e) (V2/V1)
e^0.871 = V2/V1
e^0.871 = 2.389
Then I have taken that for an isothermal process P1*V1/T1 = P2*V2/T2. T will cancel as Temerature is constant in an isothermal process leaving P1*V1 = P2*V2
This can then be rearanged to give P1/P2 = V2/V1
Ergo: P1/P2 = 2.389
As I have P1 (800 kN/m^2) I can derive P2 as 334.9 kN/m^2.
Or at least this is my thinking.
A
