1. Limited time only! Sign up for a free 30min personal 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!

Thermo Efficiency question

  1. Jul 1, 2011 #1
    Thermo Efficiency question!!

    1. The problem statement, all variables and given/known data
    Figure P22.53 represents n mol of an ideal monatomic
    gas being taken through a cycle that consists of two
    isothermal processes at temperatures 3Ti and Ti and two
    constant-volume processes. For each cycle, determine, in terms of n, R, and Ti , (a) the net energy transferred
    by heat to the gas and (b) the efficiency of an engine
    operating in this cycle.

    2. Relevant equations



    3. The attempt at a solution
    Now, I've solved part a), and get W = 2nRTln2 (difference between top and bottom portions), can't get part b). I know that efficiency = energy sought/ energy cost = W/Q_h, but they don't really specify in the question where the energy intake is, or energy cost. The answer is 0.273.... I've found that for each cycle:
    the sides both equal dU = dQ = 3nRT, the top Q = 3nRTln2, bottom Q = nRTln2. Seems like the 'energy cost' based on e = .273 and W = 2nRTln2 should be 3+3ln2, no idea how that comes about though... perhaps somehow the 'cost' is over the top and side??

    Ari
     

    Attached Files:

  2. jcsd
  3. Jul 1, 2011 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Re: Thermo Efficiency question!!

    Determine the direction of the heat flow in each part of the graph. To do that you must be careful about the sign. Heat flow into the gas is positive. Heat flow out is negative. Total heat flow in is Qh. Total heat flow out is Qc. What is W in terms of Qh and Qc? From that you should be able to determine efficiency.

    AM
     
  4. Jul 2, 2011 #3
    Re: Thermo Efficiency question!!

    Alright. Makes sense. W = Q_h - Q_l, and Q_h is all the positive contributions to the work, while Q_l is all the negative ones. When I do my calculations, I get the first two legs coming out positive, and the second two legs coming out negative, thus Q_h = 3nRT + 3nRTln2. Thanks Andrew.

    Ari
     
  5. Jul 2, 2011 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Re: Thermo Efficiency question!!

    Generally correct. Strictly speaking Qh, being heat flow, consists of all the positive contributions to the heat flow, not the work. (This heat flow enables greater work to be done in the expansion phase).

    You can determine the direction of heat flow qualitatively from the First Law. A constant volume increase in pressure requires heat flow into the gas (dQ = dU > 0). An isothermal expansion requires heat flow in (dQ = dW > 0).

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Thermo Efficiency question
  1. Thermo Question (Replies: 6)

  2. Thermo question (Replies: 1)

  3. Thermo Questions (Replies: 1)

Loading...