• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Cyclic heat engine

  • Thread starter endeavor
  • Start date
66
0
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.
 
2,903
13
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?
 
66
0
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...
 
Last edited by a moderator:
176
0
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?
 
2,903
13
I thought your graph has 1-2 labeled 650K. That would make all values of T from 1-2 = 650k.
 
66
0
LOL! Well, I thought that too... But then it would have to have been curved... o_O 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?
 
2,903
13
I got it! BOOOO
 
2,903
13
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:

[tex] PV=mRT [/tex]

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

You will find that [tex] T_H = 835.714 [/tex]

and [tex] T_L = 390 [/tex]

use that in carnot and you will get:

[tex] \eta_{th,rev} = 1- \frac{T_L}{T_H} = 1 - .4666 = .5333 [/tex]

That was an AWFUL problem in my opinion.
 
66
0
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.
 
66
0
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 o_O Explain it to me again
 
2,903
13
No, I just stole what you did. (with some refining) :rofl:
 
Last edited:
176
0
Wow, thanks a lot!
I kept taking the inital temperature as the hot too!
 
2,903
13
Tell your teacher that problem is unclear.
 

Related Threads for: Cyclic heat engine

Replies
18
Views
2K
Replies
1
Views
188
  • Posted
Replies
2
Views
4K
  • Posted
Replies
5
Views
2K
  • Posted
Replies
3
Views
6K
  • Posted
Replies
12
Views
2K
  • Posted
Replies
7
Views
2K
  • Posted
Replies
6
Views
6K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top