# Work and heat transferred in ideal engine

1. Nov 29, 2007

### El Hombre Invisible

[SOLVED] Work and heat transferred in ideal engine

1. The problem statement, all variables and given/known data

System is ideal engine containing one mole of ideal gas.

System in initial state P1, V1, T1.
System undergoes free expansion along adiabat to P2, V2, T2.
System undergoes isothermal compression to P3, V3, T3.
System is heated along isochore back to P1, V1, T1.

Give the work and heat for each path in terms of T1, T2, and $$C_{P}, C_{V} and \gamma$$. Show that the efficiency of the engine is:

$$\eta = 1 - \frac{T_{2} ln(T_{1}/T_{2})}{T_{1} - T_{2}}$$

2. Relevant equations

From the above we see that V3 = V1 and T3 = T2.
Convention used is dU = dW + dQ

$$PV = nRT$$
$$W = -\int P dV$$
$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$
$$C_{P} - C_{V} = nR$$

3. The attempt at a solution

$$Q = 0$$
$$W = C_{V}(T_{1} - T_{2})$$

For the third, isochoral path:

$$W = 0$$ since the volume is constant
$$Q = C_{V}(T_{1} - T_{2})$$

For the second, isothermal path, well: internal energy is constant, therefore:

$$W = -Q$$
$$W = -\int _{V2} ^{V1} \frac{nRT_{2}}{V} dV = (C_{P} - C_{V})T_{2} [ln(V_{2}) - ln(V_{1})] = (C_{P} - C_{V})T_{2} ln\left(\frac{V_{2}}{V_{1}}\right)$$

This is where this gets awkward and I'd appreciate someone checking what I've done. Because I have W in terms of V but I want it in terms of T, I used:

$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$
and took the $$\gamma - 1$$ root, giving:

$$T_{1}^{\frac{1}{\gamma - 1}}V_{1} = T_{2}^{\frac{1}{\gamma - 1}}V_{2}$$

Substituting into by expression for W:

$$W = (C_{P} - C_{V})T_{2} ln\left(\left(\frac{T_{1}}{T_{2}}\right)^\frac{1}{\gamma - 1}\right) = (C_{P} - C_{V})T_{2} \frac{ln(T_{1}/T_{2})}{\gamma - 1}$$

This seems an extreme solution. It also seems wrong since evaluating $$\eta = \frac{W}{Q}$$ does not give the desired equality. Any obvious errors?

Cheers,

El Hombre

2. Nov 30, 2007

### siddharth

Won't the work done in the free expansion from (P1, V1, T1) to (P2, V2, T2) be 0 by definition? So, while the pressure and the volumes change, the temperature won't.

3. Dec 2, 2007

### El Hombre Invisible

Apologies, that should have been 'adiabatic expansion' not 'free expansion'. Turned out my solution was correct: I had stupidly used the net heat instead of the heat in when evaluating the efficiency though, so I thought, as I had expected, my method was wrong. All's well that ends well.

Thanks anyway...

El Hombre