Simple 3 stage engine

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1. Jul 27, 2017

Toby_phys

A possible ideal-gas cycle operates as follows:

1. From an initial state ($p_1$, $V_1$) the gas is cooled at constant pressure to ($p_1$, $V_2$); Lets call the start and end temperature $T_1$ and $T_2$

2.The gas is heated at constant volume to ($p_2$, $V_2$);Lets call the start and end temperature $T_2$ and $T_3$

3.The gas expands adiabatically back to ($p_1$, $V_1$). Lets call the start and end temperature $T_3$ and $T_1$

Assuming constant heat capacities, show that the thermal efficiency η is

$$\eta=1-\gamma\frac{V_2/V_1 -1}{p_2/p_1-1}$$

Efficiencey is defined as: $$\eta=\frac{W}{Q_h}$$ the work done over the heat entered. The heat enters at stage 2 (and some leaves at stage 1 but that doesn't matter). So I need to find the heat entered at stage 2 and the work done.

**Stage 1:**

From the ideal gas equation we get:
$$p_1V_1=nRT_1, \ \ \ \ p_2V_2=nRT_2 \implies \frac{T_2}{T_1}=\frac{V_2}{V_1}$$

The work done is just force times distance which is pressure times change in volume:

$$\Delta W=-p_1\Delta V=-p_1(V_2-V_1)$$

**Stage 2:**

It doesn't change in volume and so no work is done. However heat is put into the system, increasing the pressure. We need to find this heat.

$\Delta U= Q_h$

For an ideal gas we have:
$$\Delta U= C_v\Delta T=C_v(T_3-T_2)$$

Where $C_v$ is heat capacity at constant volume.

**Stage 3:**

Stage 3 is adiabatic so $\Delta U=\Delta W=C_v(T_1-T_3)$

We also have, using the ideal gas law:
$$T_3=\frac{p_2V_2}{p_1V_1}T_1$$

Let us sub this into the efficiency:

$$\eta=\frac{C_v(T_1-T_3)-p_1(V_2-V_1)}{C_v(T_3-T_2)}$$

If we get $T_3$ and $T_2$ in terms of $T_1$ and sub these in we get:

$$\eta=-1-\frac{p_1(V_2-V_1)}{C_vT_1(\frac{p_2V_2}{p_1V_1}-\frac{V_2}{V_1})}$$
And with the ideal gas law, with $n=1$ for simplicity we get $T_1=\frac{p_1V_1}{R}$

$$\implies\eta=-1-\frac{R(V_2/V_1-1)}{C_vT_1(\frac{p_2V_2}{p_1V_1}-\frac{V_2}{V_1})}$$

$R=C_p-C_v$ and $\gamma=C_p/C_v$

$$\implies\eta=-1-\frac{(\gamma-1)(V_2/V_1-1)}{C_vT_1(\frac{p_2}{p_1}-1)\frac{V_2}{V_1}}$$

I have no real clue really if this is right or wrong.

2. Jul 27, 2017

kuruman

How did you get this for an adiabatic process?

3. Jul 27, 2017

Toby_phys

$dq=0$ for adiabatic process I think

4. Jul 27, 2017

TSny

This can't be correct since it yields $\eta > 1$. Note $p_2 > p_1$ and $V_2 < V_1$.

OK. But $W$ represents the work done by the gas. Later in your work, it looks like you calculate the work done on the gas. This can lead to sign errors.

Looks like maybe you made an error in getting the -1 part of the second equation above. Note that the denominator of the first equation does not cancel the first term of the numerator of the first equation to yield -1.

I think the algebra will be simpler if you use the fact that the work done by the gas during the cycle equals the total heat added during the cycle. (WHY?)

The heat added consists of heat added at constant volume and heat added at constant pressure. So you can write the work as the sum of two terms with the first term involving $C_V$ and the second term involving $C_P$.

You know that the heat , $Q_h$, is for the constant volume process and therefore involves $C_V$. So, when you set up the efficiency, you will find that the ratio of $C_P$ to $C_V$ will pop up in the algebra and that will give you the $\gamma$ factor you need.

Last edited: Jul 27, 2017