# Efficiency of a Heat Engine

Redbelly98
Staff Emeritus
Homework Helper
Looking good.

Just a side note: those p1 v1f terms could just as easily be p3 v3f, since that is a constant quantity along path 31. This might come in handy later, when you have to manipulate expressions to get things to look like the given answer from post #1.

What's left now are Q12 and Q23.

Q23 (the heat out) isn't really necessary for calculating efficiency, since that is really Qc here and doesn't enter into the efficiency calculation.

So we just need Q12 (the heat in, or energy cost). There are a few ways one might do this:

1. Use Q = ∫ T dS along path 12
2. Use Q = ΔU + W along path 12
3. Use ΔU = 0 = (Q12+Q23+Q31) - (W12+W23+W31) around the entire cycle

The key is to pick which one is the easiest way to arrive at the answer.

I'm going to be away from the computer for several hours, until at least 6 pm (Eastern USA time zone). See how far you can get, then I can check back later.

Once you have Q12, you'll need to calculate the final answer:

Efficiency = (W12+W23+W31) / Q12

and that will involve playing with the algebra to get it in the form given in the problem statement,

$$e = 1 - \frac{1}{f}\left(\frac{1 - \frac{p_3}{p_1}}{1 - \frac{v_1}{v_3}}\right)$$

where "f" is really γ, and I have fixed a typo (v1/v3 instead of v3/v1).

Good luck!

Okay, so I have chosen method 2,

Q = U + W along path 12

so

W = (V3 - V1)P1

$$\Delta U = C_V \Delta T$$ (P const.)

Since Ideal gas, use ideal gas law,

$$T = \frac{PV}{nR}$$

Thus:

$$T_1 = \frac{p_1v_1}{nR}$$

$$T_2 = \frac{p_1v_3}{nR}$$

So:

$$\Delta T = T_f - T_i = T_2 - T_1 = \frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}$$

and thus:

$$\Delta U = C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR})$$

thus:

$$Q_{12} = (C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1$$

Now efficiency,

$$e = \frac{W_{12} + W_{23} + W{31}}{Q_{12}}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1 + 0}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

I am assuming there is a way to simplify this...

I will keep looking, but I am sure I can get the 1 at the neginning from the last part of the fraction, and the 1/gamma isclose, as I have 1/Cv, and gamma is just CP/Cv,

But how can I get rid of the nR's?

Redbelly98
Staff Emeritus
Homework Helper
Good, that answer is a correct expression. And yes, it must be simplified to get the asked-for expression.

It's all algebra from this point onward.

• Anything you know that relates Cv, Cp, gamma, and nR to one another
• p1 v1f = p3 v3f (because path 31 is adiabatic)

Like I said, it's just playing around with the algebra at this point.

Good luck!

Okay, so:

$$\gamma = f = \frac{C_p}{C_v}$$

$$C_p - C_v = nR$$

So:

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(C_V (\frac{p_1v_3}{C_p - C_v} - \frac{p_1v_1}{C_p - C_v}) + (V_3 - V_1)P_1}$$

Now the equation has a 1/\gamma, so would this be on the right path?

$${C_v} \frac{C_p}{f}$$

$$e = \frac{\left[\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)}\right] - \left[\frac{p_1v_1^f}{v_3^{f - 1}(-f + 1)}\right] + (v_3 - v_1)p_1}{(\frac{C_p}{f} (\frac{p_1v_3}{C_p - C_v} - \frac{p_1v_1}{C_p - C_v}) + (V_3 - V_1)P_1}$$

and since p1v1 ^f = v3p3^f

$$e = \frac{\frac{p_1v_1^f}{v_1^{f - 1}(-f + 1)} - \frac{p_3v_3^f}{v_3^{f - 1}(-f + 1)} + (v_3 - v_1)p_1}{(\frac{C_p}{f} (\frac{p_1v_3}{C_p - C_v} - \frac{p_1v_1}{C_p - C_v}) + (V_3 - V_1)P_1}$$

Does this look okay?

Andrew Mason
Homework Helper
Those are the right limits, but in the wrong order. To see why, note that path 31 starts at "3" and ends at "1". So the limits are

$$\left[ \text{...} \right] ^{v1}_{v3}$$

EDIT: including figure from 1st post:
There is a simple way to do this. Apply the first law. There is no change in U in one cycle. So the work is simply Qh-Qc = Q12-Q23. You just have to work out the temperatures to determine what those heats are.

AM

Okay so:

$$Work = Q_1_2 - Q_3_1$$

$$Q_{12) = (C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1$$

Okay so Q23

Q = U + W along path 23

W = 0

$$\Delta U = C_P \Delta T$$ (V const)

using Ideal gas law, T = PV/nR

$$\Delta T = \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}$$

$$\Delta U = C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}$$

$$Q_{23} = C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}$$

So:

$$e = \frac{Work}{Q_{12}}$$

$$Work = \left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)$$

$$e = \frac{\left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$

So does this look okay now?

Andrew Mason
Homework Helper
...
So:

$$e = \frac{Work}{Q_{12}}$$

$$Work = \left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)$$

$$e = \frac{\left(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1\right) - \left(C_P \frac{P_3V_3}{nR} - \frac{P_1V_3}{nR}\right)}{(C_V (\frac{p_1v_3}{nR} - \frac{p_1v_1}{nR}) + (V_3 - V_1)P_1}$$
Try:

$Q_{12} = n\gamma C_v(T_2-T_1)$ and $Q_{23} = nC_v(T_3-T_2)$

$$\eta = W/Q_h = \frac{Q_{12} - Q_{23}}{Q_{12}} = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

Work out T1 and T3 using:

$$TV^{\gamma-1} = const.$$

ie.

$$T_3 = T_1\left(\frac{V_1}{V_3}\right)^{\gamma-1} = \frac{P_1V_1}{nR}\left(\frac{V_1}{V_3}\right)^{\gamma-1}$$

$$T_2 = P_1V_3/nR$$

AM

I see. So:

$$Q = n C_P \Delta T$$

From the expression for gamma,

$$C_p = f C_V$$ (I am using f because gamma doesn't show well on latex)

so

$$Q = n f C_V \Delta T$$

$$Q_{12} = n f C_V (T_2 - T_1)$$

$$Q_{23} = n C_V (T_3 - T_2)$$

$$e = W/Q_h = \frac{Q_{12} - Q_{23}}{Q_{12}}$$

$$\frac{Q_{12} - Q_{23}}{Q_{12}} = \frac{nf C_V (T_2 - T_1) - n C_V (T_3 - T_2)}{n f C_V (T_2 - T_1)}$$

Cancels to:

$$\frac{Q_{12} - Q_{23}}{Q_{12}} = \frac{f(T_2 - T_1) - (T_3 - T_2)}{f(T_2 - T_1)}$$

$$TV^f = c$$

$$T_{1}V_{1}^{f -1}= T_{3}V_{3}^{f-1}$$

$$T_{1}= T_{3}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}$$

Use ideal gas law:

T = PV/nR

$$\T_1= \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}$$

$$T_2 = \frac{P_1V_3}{nR}$$

$$T_3 = \frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}}$$

so:

$$\frac{(f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}) - (\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3})}{f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}})}$$

Okay so far?

Andrew Mason
Homework Helper
I
$$\frac{(f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}) - (\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3})}{f(\frac{P_1V_3} - \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}})}$$

Okay so far?
Notice that:

$$\eta = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

reduces to:

$$\eta = 1 - \frac{1}{\gamma}\frac{(T_3 - T_2)}{(T_2-T_1)}$$

So you just have to show that:

$$\frac{(T_3 - T_2)}{(T_2-T_1)} = \frac{(1-\frac{P_3}{P_1})}{(1-\frac{V_3}{V_1})}$$

AM

Notice that:

$$\eta = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

reduces to:

$$\eta = 1 - \frac{1}{\gamma}\frac{(T_3 - T_2)}{(T_2-T_1)}$$
I see that now.

$$\eta = \frac{(\gamma(T_2-T_1) - (T_3 - T_2)}{\gamma(T_2-T_1)}$$

is the same as:

$$\eta = \frac{(\gamma(T_2-T_1)}{\gamma(T_2-T_1)} - \frac{(T_3 - T_2)}{\gamma(T_2-T_1)}$$

Which cancels down to:

$$\eta = 1 - \frac{1}{\gamma}\frac{(T_3 - T_2)}{(T_2-T_1)}$$

Okay so:

$$\frac{(T_3 - T_2)}{(T_2-T_1)}$$

$$T_1= \frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}}$$

$$T_2 = \frac{P_1V_3}{nR}$$

$$T_3 = \frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}}$$

So:

$$\frac{(\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3}{nR})}{(\frac{P_1V_3}{nR} -\frac{P_3V_3}{nR}\frac{V_{3}^{f-1}}{V_{1}^{f -1}})}$$

Factorise out:

$$\frac{(\frac{P_1V_1}{nR}\frac{V_{1}^{f-1}}{V_{3}^{f -1}} - \frac{P_1V_3}{nR})}{\frac{V_3}{nR}(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$

and:

$$\frac{\frac{p_1}{nR}(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{\frac{V_3}{nR}(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$

Does this look okay so far?

I am also assuming that the nR can cancel, so:

$$P_1(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{V_3(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$

Does this look right?

Redbelly98
Staff Emeritus
Homework Helper
To be honest I have lost track. However, it would be desirable to simplify the term
v1 v1γ-1
The idea is, once you have an expression for the efficiency, for you to manipulate and simplify it algebraically and get

$$e = 1 - \frac{1}{\gamma}\left(\frac{1 - \frac{p_3}{p_1}}{1 - \frac{v_1}{v_3}}\right)$$​

which, as I had mentioned earlier, has a typo corrected from what was given in Post #1 (v1/v3 instead of v3/v1).

Oops, my Latex went a bit wrong. it should have been:

$$\frac{P_1(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{V_3(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})}$$