Can the Maximum Efficiency of a Thermal Cycle Be 2/5?

[Clara Chung]

Homework Statement

Attached below

Relevant Equations

Attached below

Attempt:
a) Leg 4 and Leg 1 W=P_0(V_1-V_0)+P_1(V_0-V_1)
b) Leg 4: Q= C_v*ΔT=C_v*ΔP*V_0/nR = 3/2*(V_0*P_0-V_0*P_1)
Leg 1: Q=C_p*ΔT=C_p*P_0*ΔV/nR = 5/2*(P_0*V_1-P_0*V_0)
Q=3/2*(V_0*P_0-V_0*P_1)+ 5/2*(P_0*V_1-P_0*V_0)
c) W/Q = [P_0(V_1-V_0)+P_1(V_0-V_1)] / [3/2*(V_0*P_0-V_0*P_1)+ 5/2*(P_0*V_1-P_0*V_0)]
No idea how to get 2/5

 

[Chestermiller]

Please try to factor your expressions and also to express as much as possible in terms of $\Delta p$ and $\Delta V$.

 

[Clara Chung]

Please try to factor your expressions and also to express as much as possible in terms of $\Delta p$ and $\Delta V$.

Ok..
I got W/Q = ΔpΔV / ((3/2)*Δp*V₀ + (5/2)*ΔV*P₀). I don't know how to continue because I can't see any constraint on the variables...

 

[Chestermiller]

Ok..
I got W/Q = ΔpΔV / ((3/2)*Δp*V₀ + (5/2)*ΔV*P₀). I don't know how to continue because I can't see any constraint on the variables...

Much better. Now, if I write $p_0=p_1+\Delta p$, I obtain:
$$\frac{W}{Q}=\frac{\Delta p\Delta V}{[\frac{5}{2}\Delta p\Delta V+\frac{5}{2}p_1\Delta V+\frac{3}{2}V_0\Delta p]}$$
Does this give you any ideas?

 

[Clara Chung]

Much better. Now, if I write $p_0=p_1+\Delta p$, I obtain:
$$\frac{W}{Q}=\frac{\Delta p\Delta V}{[\frac{5}{2}\Delta p\Delta V+\frac{5}{2}p_1\Delta V+\frac{3}{2}V_0\Delta p]}$$
Does this give you any ideas?

I try to divide the numerator and denominator by \Delta p\Delta V, so 1/ (5/2 + 5/2 p-1/ \Delta p + 3/2 * V_0/\Delta V) , because \Delta p and \Delta V are positive, the maximum value 2/5 occurs when p_1 and V_0 = 0?

 

[Chestermiller]

I try to divide the numerator and denominator by \Delta p\Delta V, so 1/ (5/2 + 5/2 p-1/ \Delta p + 3/2 * V_0/\Delta V) , because \Delta p and \Delta V are positive, the maximum value 2/5 occurs when p_1 and V_0 = 0?

Yes. They can both approach zero. See your diagram.