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Thermo reversible heat engines physics question

  1. Jun 1, 2008 #1
    I am having trouble with this question...It is from "applied thermodynamics for engineering technologists" by Eastop (Q5.2 if anyone has it).....

    The Question:
    Two reversible heat engines operate in series between a source at 527°C and a sink at 17°C. If the engines have equal efficiencies and the first rejects 400kJ to the second, calculate:
    (i) the temp at which heat is supplied to the 2nd engine ALL DONE :-) =208.7°C
    (ii) The heat taken from the source
    (iii) The work done by each engine

    I have done part (i) and I know it's right but I have no idea what equations to use for the next 2 parts because heat and work are dependant on each other so I dont understand how I can find one without the other....

    Thankyou in advance
    Dave :-)
     
  2. jcsd
  3. Jun 1, 2008 #2
    [tex]\eta_{rev}=1-\frac{T_L}{T_H}[/itex]

    [tex]\eta_{actual}=\frac{W}{Q_{in}}[/itex]

    [tex]Q_{in}=W+Q_{out}[/itex] (1st law)

    [tex]\eta_{rev}=1-\frac{T_L}{T_H}=.319=\eta_{actual}=\frac{W}{Q_{in}}[/itex]

    [tex]\frac{Q_{1_{in}}}{400}-.319=1[/itex]

    Using the above formulas, I ended up with the work for each engine at 127.6kJ. Total heat from the source 527.6kJ. However, I took the mid resovoir temp to be 272 deg C, the average of the high and low. Can you tell me how you got 208.7?
     
  4. Jun 1, 2008 #3
    [tex]\eta_{rev}=1-\frac{T_L}{T_H}[/itex]

    [tex]\eta_{actual}=\frac{W}{Q_{in}}[/itex]

    [tex]Q_{in}=W+Q_{out}[/itex] (1st law)

    [tex]\eta_{rev}=1-\frac{T_L}{T_H}=.319=\eta_{actual}=\frac{W}{Q_{in}}[/itex]

    [tex]Q_{in}=W+Q_{out}----->Q_{in}-W=400kJ --->1-.319=\frac{400}{Q_{in}}[/itex]

    I MESSED my first response up. Using the above formulas, I ended up with the work for each engine at 187.4kJ. Total heat from the source 587.4kJ. However, I took the mid resovoir temp to be 272 deg C, the average of the high and low. Can you tell me how you got 208.7?
     
  5. Jun 1, 2008 #4
    The answer in the book is 208.7 Deg C....but I got it using:-

    [tex]\eta = 1-(T2/T1)[/tex]

    1 - (T2/T3) = 1 - (T3/T1)

    1 - (800/T3) = 1 - (T3/290) - Figures converted to Kelvin

    232000 = T3[tex]^{}2[/tex]

    T3 = 481.66K = 208.66 Deg C
     
  6. Jun 1, 2008 #5
    Ohh....derrr. My bad (I guess I've been paying more attention to the latex than the math). One more time.

    [tex]\eta_{rev}=1-\frac{T_L}{T_H}=1-\frac{481.7K}{800K}=.398=\eta_{actual}=\frac{W}{Q_{in}}[/itex]

    [tex]Q_{in}=W+Q_{out}----->Q_{in}-W=400kJ --->1-.398=\frac{400kJ}{Q_{in}}[/itex]

    [tex]Q_{Total Input}=664.5kJ[/itex]

    [tex]W_{1}=664.5kJ-400kJ=264.5kJ[/itex]

    [tex]W_{2}=400kJ*.398=159.2kJ[/itex]
     
    Last edited: Jun 1, 2008
  7. Jun 1, 2008 #6
    omg you're a champion!

    I totally didnt get the -W = 1 - 0.398 by using the T3 from part one
    Now it just all fits together...it's great

    Thankyou so much :-)
     
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