# Efficiency of a Heat Engine

1. Feb 20, 2009

### TFM

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

A hypothetical engine, with an ideal gas as the working substance, operates on the cycle shown in Figure 1. Show that the efficiency of this engine is

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

Where $$\gamma = \frac{c_p}{c_v}$$

2. Relevant equations

Efficiency = Benefit/Cost

Benefit = Work out

Cost = Heat in

PV = nRt

$$\Delta U = Q_{hot} - Q_{cold} - Work$$

Work = pdv

3. The attempt at a solution

The Graph is attached, but basically it has a Adiabatic curve, which at the bottom goes up vertically, then left horizontally, back to make a cycle.

So far I have:

E = Work/Cost

Work = pdv

Cost = Q

$$w = pdv$$

Since p is not constant, use ideal law,

$$w = \frac{nRt}{v}dv$$

thus

$$w = \frac{nRt}{v}dv$$

$$w = nRt [ln(v)]^{v_1}_{v_2}$$

I also have:

$$e = 1 - \frac{Q_{cold}}{Q_{hot}}$$,

from

$$\Delta U = Q_{hot} - Q_{cold} - Work$$

Which is cylci and this delta u = 0.

Many thanks,

TFM

#### Attached Files:

• ###### Heat Engine Graph.jpg
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2. Feb 20, 2009

### Redbelly98

Staff Emeritus
That's a good start. By calculating "Work out" and "Heat in", you'll be able to figure out the efficiency.

There's a problem with
Note that T is not constant along path 31, so it must be treated as a function of V in doing the integral. Better yet forget T, keep things in terms of P. How are P and V related for an adiabatic process? (Hint: it involves γ)

3. Feb 21, 2009

### TFM

Well, that would be:

$$PV^{\gamma} = Constant$$

Would you rearrange for p, and insert into:

work = pdv

4. Feb 21, 2009

### Redbelly98

Staff Emeritus
Yes. And another hint:
"the constant" = P1 V1γ or P3 V3γ (Choose either one)

5. Feb 21, 2009

### TFM

Well, wouldn't it be:

$$p_1v_1 = p_3v_3$$

?

6. Feb 21, 2009

### Redbelly98

Staff Emeritus
No. That would be true for an isothermal process, since
PV = nRT = constant only if T=constant​

But this is an adiabatic process. It's not isothermal, so T changes.

7. Feb 21, 2009

### TFM

I see, so I should rearrange like so,

$$pv^{\gamma} = C$$

$$p = \frac{C}{v^{\gamma}}$$

insert c

$$p = \frac{C}{v^{\gamma}}$$

insert into work:

$$Work = \frac{C}{v^{\gamma}}dv$$

Integrate:

$$Work = C\int v^{-\gamma}}dv$$

$$Work = C\left[ \frac{\gamma + 1}{v^{\gamma + 1}} \right]^{limits}$$

So does this look right?

8. Feb 21, 2009

### Redbelly98

Staff Emeritus
Looks good up to here.

Mmm, no, try that integral again.

9. Feb 21, 2009

### TFM

Okay, the integral of x^a is (x^a+1) / (a + 1)

so if we say a = -gamma, x= v

$$\left[ \frac{v^{-\gamma} + 1}}{-\gamma + 1} \right]$$

$$C\left[ \frac{1}{(-\gamma + 1)v^{-\gamma + 1}} \right]$$

$$C = p_1V_1^{\gamma}$$

$$p_1V_1\left[ \frac{1}{(-\gamma + 1)v^{-\gamma + 1}} \right]$$

Does this look okay now?

10. Feb 21, 2009

### Redbelly98

Staff Emeritus
Uh, the algebra looks rather messy here. Can you double-check it and clean things up?

11. Feb 21, 2009

### TFM

Hmm, can't see why they thing split in two. So:

{I'll use f for gamma, because gamma doesn't seem clear in latex}

$$C\int{v^{-f}}dv$$

$$C\left[\frac{v^{-f + 1}}{-f + 1}\right]$$

$$C\left[\frac{1}{v^{f + 1}(-f + 1)}\right]$$

$$C = pv^f$$

$$\left[\frac{pv^f}{v^{f + 1}(-f + 1)}\right]^{limit}_{limit}$$

Does this look okay?

12. Feb 21, 2009

### Redbelly98

Staff Emeritus
Two things:

1. in the denominator, the exponent should be f-1, not f+1

2. in the numerator, let's call those p1 and v1^f, since they are constants.

Otherwise it looks good.

13. Feb 21, 2009

### TFM

okay so:

$$\left[\frac{p_1v_1^f}{v^{f - 1}(f - 1)}\right]^{limit}_{limit}$$

Better?

14. Feb 21, 2009

### womfalcs3

The contents inside the parentheses were correct before. You just had to change the exponent.

15. Feb 21, 2009

### TFM

So:

$$\left[\frac{p_1v_1^f}{v^{f - 1}(f + 1)}\right]^{limit}_{limit}$$

Is this better?

16. Feb 21, 2009

### Redbelly98

Staff Emeritus
Compare that with post #11, and my comments in post #12. Does it look like you made only the changes I said, or did you change something else too? (Hint: I wouldn't ask that question if what you wrote were correct.)

As a side note, this problem involves a lot of algebraic manipulation. And there is more of that to come. You will have to be very careful with the algebra if you are going to get this one.

I'm willing to help with reasoning things out, etc., but when symbols mysteriously change, or +/- signs keep appearing/disappearing for no reason, it gets very tedious for anybody assisting you.

Please, please try to be careful with the algebra. Double check things yourself, please don't keep relying on us to catch these numerous little errors.

17. Feb 21, 2009

### TFM

Okay from P11:

$$\left[\frac{pv^f}{v^{f + 1}(-f + 1)}\right]^{limit}_{limit}$$

So,

1. in the denominator, the exponent should be f-1, not f+1

$$\left[\frac{pv^f}{v^{f - 1}(-f + 1)}\right]^{limit}_{limit}$$

2. in the numerator, let's call those p1 and v1^f, since they are constants.

$$\left[\frac{p_1v_1^f}{v^{f - 1}(-f + 1)}\right]^{limit}_{limit}$$

Better?

18. Feb 22, 2009

### Redbelly98

Staff Emeritus
Yes, good.

19. Feb 22, 2009

### TFM

Okay so:

$$Work = \left[\frac{p_1v_1^f}{v^{f - 1}(-f + 1)}\right]^{limit}_{limit}$$

So I need to put this back into:

$$\Delta U = Q_{hot} - Q_{cold} - Work$$

$$\Delta U = Q_{hot} - Q_{cold} - \left[\frac{p_1v_1^f}{v^{f - 1}(-f + 1)}\right]^{limit}_{limit}$$

Now for the cycle $$\Delta U = 0$$

$$0 = Q_{hot} - Q_{cold} - \left[\frac{p_1v_1^f}{v^{f - 1}(-f + 1)}\right]^{limit}_{limit}$$

Thus

$$\left[\frac{p_1v_1^f}{v^{f - 1}(-f + 1)}\right]^{limit}_{limit} = Q_{hot} - Q_{cold}$$

Is this okay?

20. Feb 22, 2009

### Redbelly98

Staff Emeritus
Let's come back to that expression later. In the interest of organizing our thoughts, I suggest making a table showing W and Q for each of the three paths:

Work:
W31 = ?
W12 = ?
W23 = ?

Heat:
Q31 = ?
Q12 = ?
Q23 = ?

Once you fill in that table it should be easier to figure out the total Work, and the Heat Input, and from there figure out the efficiency.

We'll come back to the W31 expression we've been working on for the last 7-or-so posts, but first:
On which paths, if any, is W or Q equal to 0? Fill in those parts of the chart first, and we'll go from there.