# Cyclic heat engine

Isothermal? Whoa! Are you looking at the same picture I'm looking at?

The graph is Pressure versus Volume.

It's a horizontal line on the graph for 1-2 and 3-4 so it IS isobaric... *rubs eyes* Yeah, thats what it looks like.

Yes, it is Isobaric and it is isothermal (in this case). Go read your book about how a carnot cycle works, what does it tell you for process 1-2?

Now whats confusing me is that we are looking at a cyclic process and asked to apply it to a Carnot efficiency.

Now, if I understand correctly from my book, isobaric does not mean isothermal. I know that isotherms are always curved on a P versus T graph so if I proceed to draw isotherms....

http://xs73.xs.to/pics/06121/aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa.png [Broken]

So basically, I don't understand how you achieve isothermal in any case...

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I know what a Carnot engine is. Two adiabatic processes plus two isothermal.
Well, the question says it's isobaric and isometric. But then it says, "What would be the efficiency of a Carnot engine operating with the same high-temperature and low-temperature reservoirs?" Does this mean that we are to assume that the graph is the Carnot engine? or that the low and high temperature reservoirs of the graph are that of the Carnot engine?
If it is the former, then how would I determine which processes are the adiabatic and which are the isothermal?

I thought your graph has 1-2 labeled 650K. That would make all values of T from 1-2 = 650k.

LOL! Well, I thought that too... But then it would have to have been curved... I know you CAN calculate at any point the T because that is definitely a given on any Pressure versus Volume graph... So tell me, did I explain that part correctly now that we've cleared up the confusion?

I got it! BOOOO

I thought your graph said 1-2 was a constant temperature line!

<Pulls my hair out>

None of it is constant temperature. You have to find the gas constant using:

$$PV=mRT$$

Then do as you guys tried by going around the entire process.

You will find that $$T_H = 835.714$$

and $$T_L = 390$$

use that in carnot and you will get:

$$\eta_{th,rev} = 1- \frac{T_L}{T_H} = 1 - .4666 = .5333$$

That was an AWFUL problem in my opinion.

cyrusabdollahi said:
I got it! BOOOO
Oh darn it :-(

Now this is the part where you explain how you did it with your ever-knowing knowledge and mind.

Oh wait.. Now I forget something... T(h) would be the hottest temperature and T(c) would be the coldest?

I kept thinking initial and final Explain it to me again

No, I just stole what you did. (with some refining) :rofl:

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Wow, thanks a lot!
I kept taking the inital temperature as the hot too!

Tell your teacher that problem is unclear.