# Simple efficiency question

1. Apr 18, 2009

### diegzumillo

Hello again, all! I have another basic thermodynamics question :P This one came from a 2008 admission test. Seems simple enough, but as usual, my answer doesnt match =P

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

A monoatomic ideal gas performs, reversibly, the cycle shown in the diagram attached in this post. The values of $$P_0$$ and $$V_0$$ are, respectively, $$1\times 10^5 Pa$$ and $$100cm^3$$. The area of the interior of the cycle is 15J. What is the efficiency of this cycle?

2. Relevant equations

$$e=\frac{W}{Q_h}$$
e is the efficiency, W is work done by the system and $$Q_h$$ is the energy absorbed by the system (heat).
This is probably the only equation relevant here..

3. The attempt at a solution

The problem seemed straigh forward: The area enclosed by the cycle is the work done. The area under the curve $$ab$$ is the $$Q_h$$, right? If so, calculating this is trivial and results in an efficiency of, approximately, $$0,43$$.
However, this is not right. The correct answer should be $$0,136$$.

#### Attached Files:

• ###### problem_efficiency_engine.jpg
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2. Apr 19, 2009

### diegzumillo

No ideas? :P

3. Apr 19, 2009

### Redbelly98

Staff Emeritus
Ah, no. That area would be the work done from a to b.

Q is T dS, and W is P dV.

The easiest way to get Q, for any single path of this cycle, is to use

ΔU = Q - W

since ΔU and W are fairly straightforward to calculate.

Last edited: Apr 19, 2009
4. Apr 19, 2009

### abhikesbhat

Redbelly you are smart!!

5. Apr 19, 2009

### diegzumillo

Thanks Redbelly :) I got carried away by the relation $$W=Q_h-Q_c$$ and assumed something wrong :P

Can I classify this system as cyclic? If so, isn't the internal energy variation supposed to be zero?

6. Apr 20, 2009

### Redbelly98

Staff Emeritus
Gosh, thank you.

Yes, the entire process is cyclic, and ΔU is zero for the entire cycle.

But ... to calculate Qh, you'll need to consider each individual subpath, and whether heat is flowing into or out of the system for that path. Heat in contributes to Qh, while heat out does not.

Since ΔU is not necessarily zero for each subpath, it needs to be considered.

7. Apr 24, 2009

### diegzumillo

Hello again! I let go of this problems for a few days. But today I took another look at it, and I think I have a solution :)

We'll use this equation for heat
$$Q=C\Delta T$$
where this C depends on the path.
We can start by calculating the heat for the path BC and CA, we'll call them $$Q_{bc}$$ and $$Q_{ca}$$, respectively. We cannot calculate directly $$Q_{ab}$$ because we only know $$c_p=3/2 R$$, for constant pressure, and $$c_v=5/2 R$$, for constant volume. So we have
$$Q_{bc}=\frac{c_v}{R}(P_c V_c - P_b V_b)=-45J$$
$$Q_{ca}=\frac{c_p}{R}(P_a V_a - P_c V_c)=-50J$$
(I'm calling $$P_i$$ and $$V_i$$ for better understanding)

We can see that both of those values represent heat leaving the system. Now we'll use the internal energy relations to find $$Q_{ab}$$. The internal energy relations for each path are:
$$\Delta U_{bc}=Q_{bc}-W_{bc}$$ (Yes, $$W_{bc}$$ is zero)
$$\Delta U_{ca}=Q_{ca}-W_{ca}$$
Since we know that
$$\Delta U=\Delta U_{ab}+\Delta U_{bc}+\Delta U_{ca}=0$$
and
$$\Delta U_{ab}=A_{ab}-W_{ab}$$
We have
$$Q_{ab}=W_{ab}-(Q_{bc}-W_{bc}+Q_{ca}-W_{ca})$$
$$Q_{ab}=W-(Q_{bc}+Q_{ca})=110J$$
Wich is our heat transfered into the system! ($$Q_h$$)

In possession of these values, we can now calculate the efficiency of this cycle:
$$e=\frac{W}{Q_h}=13,6%$$

=D

(Btw, for some strange reason, I can't visualize correctly the formulas. So there may be some errors. I'll correct them as soon as I'm able to read what I wrote :P)

8. Apr 24, 2009

### Redbelly98

Staff Emeritus
Latex equations have been having problems for several days now.

Much of what you wrote can be done without Latex. You can get the Greek letter Δ here:
https://www.physicsforums.com/blog.php?b=347 [Broken]

Also:

[noparse]a[/noparse] for subscript a
[noparse]2[/noparse] for superscript 2

Last edited by a moderator: May 4, 2017
9. Apr 24, 2009

### diegzumillo

Yep.. just saw the anouncement. What a bummer!

Since there is no estimate for the return, I'll post the solution again! (But I'll keep that one there... it took me a while to do! :P)

10. Apr 24, 2009

### diegzumillo

The solution again. Simplified notation version :D

We'll use this equation for heat
Q=CΔT
where this C depends on the path.
We can start by calculating the heat for the path BC and CA, we'll call them Qbc and Qca, respectively. We cannot calculate directly Qab because we only know cp=3/2 R, for constant pressure, and cv=5/2 R constant volume. So we have
Qbc=cv/R (PcVc-PbVb)=-45J

Qca=cp/R (PaVa-PcVc)=-50J

(I'm calling Pi and Vi for better understanding before using the values given by P0 and V0)

We can see that both of those values represent heat leaving the system. Now we'll use the internal energy relations to find Qab. The internal energy relations for each path are:
ΔUbc=Qbc-Wbc (Yes, Wbc is zero)
ΔUca=Qca-Wca

Since we know that
ΔU=ΔUab+ΔUbc+ΔUca=0
and
ΔUab=Qab-Wab
We have
Qab=Wab-(Qbc-Wbc+Qca-Wca)
Qab=W - (Qbc+Qca) = 110J
Wich is our heat transfered into the system! (Qh)

In possession of these values, we can now calculate the efficiency of this cycle:
e=W/Qh=13,6%

=D

Thanks again Redbelly! And btw, that's a really helpful list of symbols! :) Especially for this dark times without latex.. lol

11. Apr 24, 2009

### Redbelly98

Staff Emeritus
It looks like the values of cv and cp have been swapped here, but you did use the correct values in the calculation. Good job, you nailed this one!

You're welcome. "Dark times", LOL