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Refrigeration: Coefficient of Performance (Carnot & Lorenz)

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Cooling of a food from 293K (20 ⁰C), freezing it at 272K (-1⁰C), and further cooling to 253K (-20 ⁰C); using cooling water rising in temperature from 293K (20 ⁰C) to 303K (30 ⁰C) to accept heat. The specific heat of the unfrozen food is taken as 3.2 kJ/kg K, that of the frozen food as 1.7 kJ/kg K, and the enthalpy of the freezing as 240 kJ/kg. Calculate the Lorenz coefficient of performance and compare it with Carnot coefficient of performance.

    2. Relevant equations

    3. The attempt at a solution

    Lorenz COP:
    For cooling water side, the average absolute temperature is simply the logarithmic mean absolute temperature:

    T2,avg = (303-293)/ln(303-293) = 298 K

    For the food, the entropy change can be calculated as follows:

    s1'- s1'' = (1.7 ln (272/253)+240/272+3.2 ln (293/272))=1.243 kJ/kgK

    The heat per unit mass is:

    q1=1.7×(272-253)+240+3.2×(293-272)=339.5 kJ/kg

    The average absolute temperature of acceptance of heat:

    T1,avg=339.5/1.243 = 273.1 K

    The Lorenz COP is thus:
    COP= (273.1)/(298-273.1)=10.97

    For the Reversed Carnot cycle, it has to operate between the highest and lowest temperatures to satisfy the requirement for heat transfer. This leads to:

    COP= (253)/(303-253)=5.06

    Are the calculations correct? I tried to confirm the answer with my teacher, who told me that the Carnot COP is incorrect. He says the correct answer is:

    Carnot COP = Tlow/(Thigh - Tlow)
    COP= (293)/(293-293) = ∞

    He told me that in Carnot refrigeration, the temperature of the food and the cooling water remain constant (unchanging) and thus we should use the initial temperatures for both the food (293K before cooling) and the water (293K before receiving any heat). Is he correct? I couldn't seem to really understand the reason behind his answer. Could someone please kindly verify the answers?

    Thank you for the help.
  2. jcsd
  3. Feb 23, 2016 #2
    If I recall, the Carnot system measures separately; the amount of thermal energy removed from one medium and the amount of thermal energy added to another medium. It is not so much a measure of how much heat has been added to or taken away from the medium you are most interested in, food temperature in your example, but rather it is a measure of the efficiency of the system.

    In most systems; there will be additional heat exchange which is the result of the mechanical heat generated by the system, for example friction, which does not directly imply that all the heat was removed from the source medium. This would have to be accounted for in the equation, if measuring at the output side of the system.

    In a Carnot system, which implies it is a reversible engine, we can presume that this extra heat loss, friction for example, will be present on the same side whether we are using the engine to heat or cool either the medium we are most interested in. As such, we should be able then to measure the amount of increase in temperature of one side which should exactly equal the amount of decrease in temperature at the other side.

    Using your numbers above then, COP= (253)/(303-253)=5.06, would be inaccurate because you are trying to measure both transactions in a single step and ignoring the system in the middle, where the Carnot measure is more interested in the efficiency of the system than either of the temperature of the cooler frozen food, which dropped 40 degrees in temperature or the warmer water, which only increased 10 degrees in temperature.
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