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Carnot cycle work & efficiency

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    The oxygen contained in a thermally insulated container will be cooled down to its boiling point (-183 C), and then condensed. Thereby one uses a reverse Carnot process operating between the oxygen gas (instantaneous) temperature and the room temperature (17 C). What is the minimum amount of work that must be supplied from outside to 30.0 kg liquid oxygen to be produced if you start with oxygen at room temperature. The Enthalpy of vaporization for oxygen is ##214\cdot 10^3 J/kg## and the heat capacity can be set to ##117 J/(kg \cdot K)## in the entire range.
    (The question was translated so if there's any confusion just ask!)

    2. Relevant equations
    Carnot cycle efficiency (is this true both for an heat engine and a heat pump?)
    ##\eta = \frac{T_1-T_2}{T_1}##

    3. The attempt at a solution
    ##m = 30kg##
    Enthalpy of vaporization:
    ##H = 214\cdot 10^3 J/Kg##
    heat capacity
    ##c = 917J/(kg\cdot K)##
    ##T_1 = 290K##
    ##T_2 = 90K##

    Not really sure how to even start but lets give it a try:
    The energy needed to vaporize the gas should be
    ##E_1 = m \cdot H = 6.42 MJ##
    But the efficiency for a carnot cycle operating between ##T_1## and ##T_2## would be
    ##\eta = 0.69##
    so the work needed would be
    ##E_1/\eta = 9.3MJ##
    The work needed to cool the gas is harder since the temperature is changing. The total energy should be
    ##E_2 = mc(T_2-T_1) = -5.5MJ##
    And then maybe I could use something along the lines of
    ##dW_2 = \frac{mcdT}{\eta}##
    and then
    ##W_2 = mT_1c\int \frac{dT}{T_1-T} = - mT_1cln(T_1-T_2)##
    but really I'm clueless at this point.
     
    Last edited: Mar 29, 2015
  2. jcsd
  3. Mar 29, 2015 #2

    Andrew Mason

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    The Carnot cycle in reverse is a heat pump or refrigerator. The Coefficient of Performance (COP) of the reverse Carnot cycle is:

    COP = ##\frac{Q_c}{W} = \frac{Q_c}{Q_h - Q_c} = \frac{1}{\frac{Q_h}{Q_c} - 1} = \frac{1}{\frac{T_h}{T_c} - 1}##

    So ##dW = \frac{dQ_c}{COP}= \frac{mc dT}{COP}##

    AM
     
    Last edited: Mar 29, 2015
  4. Mar 29, 2015 #3
    The problem is the chapter this is on doesn't take up the COP at all (nor do any other chaptors in this book) so I'm not sure that's what I'm supposed to use.
    I also seem to be getting a divergent integral?
    ##\frac{1}{\frac{T_h}{T_c}-1} = \frac{T_c}{T_h-T_c}##
    ##dW = mc \int_{T_h}^{T_c} \frac{T}{T_h-T}dT = [-T_hln(T_h-T) -T ]_{T_h}^{T_c} = lim_{T\to T_h} T_h ln(T_h-T) =-\infty##
     
  5. Mar 29, 2015 #4

    Andrew Mason

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    Sorry. I confused you with the error in my post (since corrected). It should be:

    ##W = \int_{T_h}^{T_c}dW = mc \int_{T_h}^{T_c} (\frac{T_h}{T} - 1)dT ##

    AM
     
  6. Mar 29, 2015 #5
    Cheers! I seem to be getting the right answer now!
    I guess even if the book doesn't mention it it's just like thermal efficiency but for an heat pump instead so maybe the author left that as part of the exercise.
     
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