Efficiency of an engine performing the Joule ideal-gas cycle.

1. Sep 13, 2008

Jacobpm64

1. The problem statement, all variables and given/known data
The figure represents a simplified PV diagram of the Joule ideal-gas cycle. All processes are quasi-static, and $$C_P$$ is constant. Prove that the thermal efficiency of an engine performing this cycle is
$$1 - \left(\frac{P_1}{P_2}\right)^\frac{\gamma - 1}{\gamma}$$

2. Relevant equations
$$PV = RT$$
$$\gamma = \frac{C_P}{C_V}$$
$$dE = dq + dw$$ (should have strokes through the d's on dq and dw, but I don't know how to latex inexact differentials)
$$\eta = 1 - \frac{|q_{out}|}{|q_{in}|}$$
$$TV^{\gamma - 1} =$$ constant
$$PV^{\gamma} =$$ constant

3. The attempt at a solution
First of all, $$\eta = 1 - \frac{|q_{out}|}{|q_{in}|}$$
$$q_{in}$$ is only path 2->3 and $$q_{out}$$ is only path 4->1.

2->3
Since we have an ideal gas,
$$dq = C_{P}dT$$
$$q_{in} = \int^{T_3}_{T_2} C_{P}dT$$
However,
$$dT = \left(\frac{\partial T}{\partial P}\right)_{V} dP + \left(\frac{\partial T}{\partial V}\right)_{P} dV$$
since dP = 0 in the 2->3 process, we have:
$$dT = \left(\frac{\partial T}{\partial V}\right)_{P} dV$$
Now, using the ideal-gas equation of state and solving for T:
$$T = \frac{PV}{R}$$
Differentiating:
$$\left(\frac{\partial T}{\partial V}\right)_{P} = \frac{P}{R}$$
Now, substituting into the above expression, we get:
$$q_{in} = \frac{C_{P} P_{2}}{R} \int^{V_{3}}_{V_{2}}dV$$
So,
$$|q_{in}| = \frac{C_{P} P_{2}}{R} (V_{3} - V_{2})$$

Now, considering process 4->1

4->1
We have the same process as above, but with different pressures and volumes. Therefore,
$$q_{out} = \frac{C_{P} P_{1}}{R} \int^{V_{1}}_{V_{4}} dV$$

Since q_{out} is negative, we switch signs,
$$q_{out} = \frac{C_{P} P_{1}}{R} (V_{4} - V_{1})$$

Plugging into the efficiency formula $$\frac{C_{P} P_{1}}{R}$$ cancels, and we get:
$$\eta = 1 - \frac{P_{1} (V_{4} - V_{1})}{P_{2} (V_{3} - V_{2})}$$

Now, I'm pretty sure I have to use the identities:
$$P_{1} V^{\gamma}_{4} = P_{2} V^{\gamma}_{3}$$
$$P_{1} V^{\gamma}_{1} = P_{2} V^{\gamma}_{2}$$

I have tried dividing these two equations so that all P's cancel.
I have also tried subtracting the equations.

I can't, for the life of me, get my efficiency in the form that the problem asks me to put it in.

Any help would be greatly appreciated. Thanks.

2. Sep 14, 2008

alphysicist

Hi Jacobpm64,

To cut down on the algebra, try writing the adiabatic relations as

$$P_{1}^{1/\gamma}\ V_{4} = P_{2}^{1/\gamma}\ V_{3}$$
$$P_{1}^{1/\gamma}\ V_{1} = P_{2}^{1/\gamma}\ V_{2}$$

and use them to eliminate $V_4$ and $V_1$. Do you get the answer?

3. Sep 14, 2008

Jacobpm64

I sure did get the answer!

You don't know how much you're helping me.

This course is really hard to me, and I appreciate the help that I'm getting here.

I have one question though.

How did you come up with that adiabatic relation? (I didn't know that it was one).

4. Sep 14, 2008

alphysicist

It's the same adiabatic relation as you had in your post. Just raise both sides to the $$1/\gamma$$ power to move the exponent onto the P instead of the V:

\begin{align} P_{1} V^{\gamma}_{1} &= P_{2} V^{\gamma}_{2} \nonumber\\ \left( P_{1} V^{\gamma}_{1}\right)^{1/\gamma} &= \left(P_{2} V^{\gamma}_{2} \right)^{1/\gamma}\nonumber\\ P_{1}^{1/\gamma}\ V_{1} &= P_{2}^{1/\gamma}\ V_{2} \nonumber \end{align}

Since in your problem you were trying to cancel the V variables in your numerator and denominator, I think it gives easier algebra to use the adiabatic relation in a form that has V to the first power