# Homework Help: 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
$$1 - \left(\frac{P_1}{P_2}\right)^\frac{\gamma - 1}{\gamma}$$
http://img50.imageshack.us/img50/7734/thermo1ym3.jpg [Broken]
http://g.imageshack.us/img50/thermo1ym3.jpg/1/ [Broken]

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.

Last edited by a moderator: May 3, 2017
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