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Cyclic heat engine

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http://img20.imageshack.us/img20/2964/physics16ri.th.png [Broken]
The working substance of a cyclic heat engine is 0.75kg of an ideal gas. The cycle consists of two isobaric processes and two isometric processes as shown in Fig. 12.21 (image above). What would be the efficiency of a Carnot engine operating with the same high-temperature and low-temperature reservoirs?

I don't know how to solve this problem. At first, I thought to simply use the ideal gas law as in,
PV/T = P2V2/T2
and to use that to go through each process, from 1 to 2, 2 to 3, and 3 to 4. I ended up with a temperature of 390K at point 4. The Carnot efficiency would then be 40% (1 - 390/650). However, that is wrong and that solution did not even use the mass of the ideal gas.

I then thought about using the mass to find the number of moles, PV=nRT. I found n = 0.80996 mol. But I don't know where to go from here...
 
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2,903
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What does the area of that graph tell you?
 
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The area shows the amount of work done.
So W = Qin - Qout = 500 J
but now what?
 
2,903
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What did your book get? I got 40% doing it another way, but maybe I am wrong. Is this the same book that said to raise the temperature of the heat source :uhh: ?
 
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The book says 53%, which would mean that the low temperature reservior is 305.5K. However, I do not know how to arrive at this answer.
 
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I'm gonna take a stab at this... Although I never liked Thermodynamics that much X_X


Isobaric means that pressure is constant and delta P = 0

Isobaric meaning:
[tex]W = P \triangle V[/tex]
[tex]\triangle U = \frac{3/2}n R \triangle T[/tex]

Isovolumetric meaning:
[tex]\triangle W = 0[/tex]
[tex]\triangle U = \frac{3/2}n R \triangle T[/tex]

NOTE: Pressure is in kPA so convert to Pa so that we can work with the units.

First thing we do is calculate the Temperature at all the points using PV=nRT.
[tex]T = PV/nR[/tex]

Then apply the formula for the Carnot efficiency:

[tex]e = 1 - T_c/T_h[/tex]
 
2,903
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I'm sitting here scratching my head, because I did it your way and got the same answer still, 40%. Are you SURE thats the right solution?
 
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Da-Force, I don't understand how to solve it your way. What answer did you get?

cyrusabdollahi, maybe my book is wrong. (It wouldn't be the first or second time)
 
2,903
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Did they give you a worked out solution or just a number?

Also, is this for a thermo-class or physics class, so I can judge if im making this problem overly complicated.
 
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Just a number.
This is for a basic physics class.
 
2,903
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I really dont know what to tell you, I think the answer is 40%.

What a poorly written problem. They say Ideal gas but tell you nothing about the type of gas?

Da-Force said:
First thing we do is calculate the Temperature at all the points using PV=nRT.
You cannot do that, Pv=RT is not true at all points along the curve. From 1-2 and 3-4 you have heat transfer. If you use Pv=RT along those process curves you would get a changing temperature, which you do not have.

Also, I do not see how this:

[tex]\triangle U = \frac{3/2}n R \triangle T[/tex]

Is necessary to solve the problem.
 
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cyrusabdollahi said:
I really dont know what to tell you, I think the answer is 40%.

What a poorly written problem. They say Ideal gas but tell you nothing about the type of gas?



You cannot do that, Pv=RT is not true at all points along the curve. From 1-2 and 3-4 you have heat transfer. If you use Pv=RT along those process curves you would get a changing temperature, which you do not have.

Also, I do not see how this:

[tex]\triangle U = \frac{3/2}n R \triangle T[/tex]

Is necessary to solve the problem.
Let me explain that reasoning.

We are given a graph of P versus V.

We know n, R and we can get P and V from the graph.

So even if there is heat transfer, we can STILL calculate temperature at all points of a P versus V graph. The delta U was to explain how isovolumetic and isobaric relate. Although it is not needed in the problem, I just include as helpful info :-)
 
2,903
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So even if there is heat transfer, we can STILL calculate temperature at all points of a P versus V graph.
No, you cannot. Pv=RT will not work if you have heat flow into a system. Just try it yourself for 1-2,and see what happens to Pv=RT. R is a constant, P is a constant. V is changing, but you were TOLD T is a constnat. Your equation is now nonsense. Exactly because of heat flow.

You can use Pv=RT for 4-1 and 2-3 because in those steps there is no heat flow.
 
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*tries the problem* I got 40% efficiency when I did it at first.. Then I tried it the way I explained and I seem to get the 53% so I am not sure who's right anymore o_O

I found a problem like this in my 1000 page physics book and they did something very similar to mine although it was a different scenario.
 
2,903
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What did you do to get the value?
 
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Da-Force, can you explain your method of finding the answer? I don't understand what you said you did.
 
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Da-Force said:
*tries the problem* I got 40% efficiency when I did it at first.. Then I tried it the way I explained and I seem to get the 53% so I am not sure who's right anymore o_O

I found a problem like this in my 1000 page physics book and they did something very similar to mine although it was a different scenario.
Ack! Edit o_O

I got 40%... I messed up reversing one of the numbers around... o_O

But, I examined it was taking the efficiency as the Work(out)/Work(in) and I got 60% and so I'm looking at this here and I agree the book is wrong.
 
2,903
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:rofl: you got me all excited for nothing! I will beat you!
 
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I'm wondering why you asked for the area enclosed by the curve.. If I remember my physics..

W = Q(h) - Q(c)

e = 1 - Q(c)/Q(h) = W/Q(h)

So... This would be efficiency for a regular engine.. *coughs*

Carnot engine is temperature.
So, now I hate this problem... Unless book is wrong, I like the 40% answer...
 
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well, I guess my book is wrong once again...
thanks for the effort though!
 
2,903
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Because carnot means that [tex] \frac{Q1}{Q2} = \frac {T1}{T2} [/tex]
 
66
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Where entropy is conserved.. I'll give ya your props :-p

I just did it with Q1/Q2 and now I think I've done it wrong cause I got 60%.
 
176
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hmm... about this:
cyrusabdollahi said:
No, you cannot. Pv=RT will not work if you have heat flow into a system. Just try it yourself for 1-2,and see what happens to Pv=RT. R is a constant, P is a constant. V is changing, but you were TOLD T is a constnat. Your equation is now nonsense. Exactly because of heat flow.

You can use Pv=RT for 4-1 and 2-3 because in those steps there is no heat flow.
I don't think it says T is constant, but anyways I tried using PV=T for each step, 1-2, 2-3, 3-4, and I got T2=835.7K, T3=501.4K, and T4 = 390K, and then using T4, T1 = 650K
Then, e = 1 - T4/T1 = 0.40
 
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Thank you endeavor.. o_O

Someone actually understands what I was trying to say with my big post LOL

I've been going through all the possible variants for T and I'm not getting anything close to 53% other than the 60% mentioned above.
 
2,903
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T is a constant. I dont think you guys understand a carnot cycle. 1-2 and 3-4 are reversible isothermal compression and expansion processes. T is a constant in these steps.

Phases 2-3 and 4-1 are Reversible adiabatic expansion/compression processes and the temperature varies from TL to T-H.
 

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