# Homework Help: Efficiency of an imaginary ideal-gas engine 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
$$\eta = 1 - \gamma \frac{\frac{V_{1}}{V_{2}} - 1}{\frac{P_{3}}{P_{2}} - 1}$$
http://img134.imageshack.us/img134/4534/thermo3eo8.jpg [Broken]
http://g.imageshack.us/img134/thermo3eo8.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)
$$\eta = \frac{|W|}{|q_{in}|}$$
$$TV^{\gamma - 1} =$$ constant
$$PV^{\gamma} =$$ constant

3. The attempt at a solution
First of all, $$\eta = \frac{|W|}{|q_{in}|}$$

1->2
We have $$dP = 0$$ and $$P = constant$$
$$W = \int^{V_{2}}_{V_{1}} dV$$
$$W = -P_{2} (V_{2} - V_{1})$$
Since this is an ideal gas, we know:
$$dq = C_{P} dT$$
Therefore,
$$q = C_{P} (T_{2} - T{1})$$

2->3
We have $$dV = 0$$ and $$dw = 0$$ since the process is isochoric.
Therefore,
$$dE = dq$$
Since this is an ideal gas $$dE = C_{V} dT$$
Therefore,
$$dq = C_{V} dT$$
$$q = C_{V} (T_{3} - T{2})$$
By hypothesis,
$$w = 0$$

3->1
This is an adiabatic process, so $$dq = 0$$
Therefore,
$$dE = dw$$
Since this is an ideal gas $$dE = C_{V} dT$$
Therefore,
$$W = \int^{T_{1}}_{T_{3}} C_{V} dT$$
$$W = C_{V} (T_{1} - T_{3})$$
Also, by hypothesis,
$$q = 0$$

Plugging all this into the equation for $$\eta$$, we get:
$$\eta = \frac{P_{2} (V_{2} - V_{1} ) - C_{V} (T_{1} - T_{3})}{C_{P} (T_{2} - T_{1}) + C_{V} (T_{3} - T_{2})}$$

Now, I don't know how to manipulate this to get it in the form that the problem asked for.

Any help would be greatly appreciated. Thanks.

Last edited by a moderator: May 3, 2017
2. Sep 14, 2008

### Andrew Mason

Your solution does not provide a solution for T1.

The key here is the adiabatic condition:

$$PV^\gamma = K$$

Try applying the adiabatic condition to find the work done from 3 to 1.

AM

3. Mar 11, 2009

### belalegosi

Sorry to revive a dead thread but I still can't figure it out.

4. Dec 4, 2009

### FrankymyBoy

Sorry to do likewise but I'm in the same boat.

5. Dec 4, 2009

### Redbelly98

Staff Emeritus
You'll need to show how far you got with solving the problem before you can receive help. That's our forum policy, and the reason why belalegosi got no help last March.

6. Dec 4, 2009

### Andrew Mason

In looking at this again, there is a much easier approach than the one taken by the original poster. You do not need to use the adiabatic condition at all. You don't have to calculate the work done either. Just the heat flows. Use the expression for efficiency in terms of Qin and Qout only.

Qin and Qout can be determined as the OP has done. That will give you an expression for efficiency involving T1, T2 and T3. Then use the ideal gas equation to work out expressions for the temperatures in terms of P and V.

AM

7. Dec 4, 2009

### ehild

You need only the absorbed heat in the denominator, Cv(T3-T2).

PV =RT and R = Cp-Cv.

ehild

8. Dec 6, 2009

### Andrew Mason

As ehild has pointed out, the denominator is Qin which does not include $Q_{1-2}$. The latter represents the heat flow out of the system.

To find the efficiency, use $$\eta = W/Q_{in} = \frac{Q_{in} - Q_{out}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}$$

In this case $Q_{in}$ is the heat flow from 2-3 and $Q_{out}$ is the heat flow from 1 to 2. Using the ideal gas law to put T1/T2 in terms of V1/V2 and T3/T2 in terms of P3/P2.

AM

Last edited: Dec 6, 2009