1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Carnot efficiency

  1. Mar 17, 2006 #1
    Suppose you had the choice of either raising the high-temperature reservoir a certain number of kelvins, or lowering the low-temperature reservoir the same number of kelvins. Which would you choose (assuming you wanted to increase the efficiency, of course)?

    After doing experimenting with some random numbers, I thought the answer to be: lower the low-temperature reservoir.

    For example,
    1 - 100/300 = 66.7%
    1 - 95/300 = 68.3%
    1 - 100/305 = 67.2%
    both of the last two are greater than the first, but the second has the greatest efficiency.
    However, my book says to raise the high-temperature reservoir. Is my book wrong?
     
  2. jcsd
  3. Mar 17, 2006 #2

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    Could you tell us what the question asks, exactly? I'm thinking there's something missing. Carnot efficiency approaches 100% as the cold resevoir approaches zero kelvins; for a constant temperature difference, the greatest efficiency will be found for the coldest cold resevoir. This is what you have noticed. But it would take more energy to remove 5 K from the cold than it would to add 5 K to the hot.
     
  4. Mar 17, 2006 #3
    The exact question reads:
    "The Carnot efficiency shows that the greater the temperature difference of the reservoirs of a heat engine, the greater the efficiency. Suppose that you had the choice of either raising the high-temperature reservoir a certain number of kelvins, or lowering the low-temperature reservoir the same number of kelvins. Which would you choose (assuming you wanted to increase the efficiency, of course)?"
     
  5. Mar 17, 2006 #4
    The carnot heat engine efficiency is given to us as:

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

    What can you conclude?

    In other words, what happens as the denominator, [itex] T_H [/itex] approaches infinity? similarly, what happens as the numerator, [itex] T_L [/itex] approaches zero?

    Edit: I think you and Chi are right and your book is wrong.
     
    Last edited: Mar 17, 2006
  6. Mar 17, 2006 #5

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    I've been over this several ways, but if you look at the other version of the carnot efficiency formula, it seals it:
    e=(Thot-Tcold)/ Thot

    this says that if the difference in temperature is the same, the greater that T-hot is, the lesser the efficiency is.
     
  7. Mar 17, 2006 #6
    As I am sure you are already aware, your formula is the same thing Chi Meson. The formula[tex] \eta_{th, rev} = 1 - \frac{T_L}{T_H} [/tex] is derived from the reduction of your formula. (Just simplify your fraction)
     
    Last edited: Mar 17, 2006
  8. Mar 18, 2006 #7

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    Yeah. It's just that looking at the other version (of the same thing) it was more obvious to me that the book is incorrect. You can simply say that efficiency is inversely proportional to T hot (if delta T is constant).
     
    Last edited: Mar 18, 2006
  9. Mar 19, 2006 #8
    I thought my book was wrong.
    Thanks, all.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Carnot efficiency
Loading...