Efficiency of heat engine question

[JoeyBob]

Homework Statement

see attached

Relevant Equations

PV=nRT

So efficiency is W/Qin.

W= 0 for isochoric processes and for the isobaric, P(change in V). So W=Pi(Vi-Vf)+Pf(Vi-Vf)

Qin is negative Qs.This would happen at step 2 and 3. For the isobaric, Q=ncv(change in T) and for isochoric, Q=ncp(change in T).

Now if I put everything in the equation I get (Pi(Vi-Vf)+Pf(Vi-Vf))/(ncv(change in T)+ncp(change in T))

Now my problem obviously is that I don't know what a single one of these variables are. All I know are the changes in temp from other steps. For instance, the change in T from d to a is 17.4K. But this doesn't exactly help me calculate anything since i don't have mol, volume, or anything at all really.

 

[haruspex]

I get (Pi(Vi-Vf)+Pf(Vi-Vf))/(ncv(change in T)+ncp(change in T))

Check your signs. What would you expect if Pi=Pf?

You need to involve the given temperature differences, and you have not used your relevant equation.

 

[JoeyBob]

Check your signs. What would you expect if Pi=Pf?

You need to involve the given temperature differences, and you have not used your relevant equation.

I think the bottom signs need a negative? Just looking at the graph I know that Pi can't be Pf (they have different height).

The problem with PV=nRT is that I don't know n. Or P. Or V. But let's say I substitute change in T into the equation everywhere.

Pi(nR(change T)/P) + Pf(nR(change T)/P) / -ncv(change T) - ncp(change T)

I know I didn't label everything precisely, but this gives the general idea.

I guess stuff can cancel out? But my problem now is about the (change T). I get (T change) at 4 and 1 in the diagram. But there are two spots in my fraction where change in T is from 2 and 3.

Arent I missing these values?

 

[haruspex]

Just looking at the graph I know that Pi can't be Pf (they have different height).

It's a sanity check. What should you get if Pi were equal to Pf? The equation should still be valid.

I don't know n

You know it is constant.
Write out the PV/T expressions for the four states and set them equal.

 

[JoeyBob]

It's a sanity check. What should you get if Pi were equal to Pf? The equation should still be valid.

You know it is constant.
Write out the PV/T expressions for the four states and set them equal.

What do you mean by set them equal?

I could try to simplify my equation above to get (R(change T1 + change T2))/(-cv(change T1) - cp (change T2)) but i don't think its right.

 

[vela]

I think you can't come up with a numerical answer without at least one of the temperatures.

 

[haruspex]

What do you mean by set them equal?

R and n are constant, so PV/T should be the same at all four points, no? From which ...

I think you can't come up with a numerical answer without at least one of the temperatures.

... it is possible to find all the temperatures, no?

 

[vela]

it is possible to find all the temperatures, no?

No, I don't think so. Were you able to?

 

[Chestermiller]

You do not need to know anyone of the temperatures to get the efficiency.

 

[vela]

I think you can't come up with a numerical answer without at least one of the temperatures.

Never mind. I didn't see you were given $T_a = T_c$.

 

[Chestermiller]

From the given data, you know the 4 temperature differences so you know the 4 heats. From the 4 heats, you know the net work. You do not need to use the ideal gas law to solve this problem.

 

[JoeyBob]

From the given data, you know the 4 temperature differences so you know the 4 heats. From the 4 heats, you know the net work. You do not need to use the ideal gas law to solve this problem.

Okay so I think I've gotten somewhere but I am not sure how to use heat to get work.

What I did for each temperature changes was use the equation Q=nc(T change). the c is either (3/2)R or (5/2) R depending on if its isochoric or isobaric. Now I have all the Qs, some positive, some negative. The negative ones would be what work is divided by, right? Sure, n remains, but I am hoping it will be canceled out.

Now I know work = 0 for ischoric systems, so I just need to find the work for the isobaric systems. But work = P(change in volume) for isobaric systems, values I don't have. Now if I think about the units for work (Joules) I can see that its the same unit used for heat.

But I am not sure how to directly relate the two (the Q that I've calculated and the work)

Ive now tried W=nR(T change) for isobaric but when I do W/Q, where Q was the negative heats I got. my answer is too large.

 

[Chestermiller]

What is the change in internal energy over the entire cycle?

 

[JoeyBob]

What is the change in internal energy over the entire cycle?

It would be the sum of the work minus the sum of the heat. Idk if its because I am calculating it wrong or not, but I get 0. But this seems to be wrong because if the internal energy didnt change, doesn't that mean the efficiency was 100%?

 

[JoeyBob]

What is the change in internal energy over the entire cycle?

Okay I finally figured out how to do but would really appreciate if you could explain why it works and why my earlier method was wrong (I hate this chapter so much).

As you said, I calculated Q using nc(T change). Did this for each step. Some Qs were negative, some were positive. Now I then calculated work using W=nR(T change) for the isobaric processes. Isochoric work is 0.

Method that was wrong:

I summed up the NEGATIVE Qs I got and made them positive. I summed up the work. I divided the work by these Qs. Answer is too big.

Method that got right answer:

I summed up the POSITIVE Qs. I summed up the work. I divided the work by these Qs. Answer was right.

Why is it the Qs that are positive?? I thought it was the ones that were less than 0 that I use for efficiency calculations?