Reverse of carnot cycle

  1. In the textbook, it reads: the reverse of carnot cycle is just a refrigerator or heat pump. In this sense, except for the direction of carnot cycle is reverse, could I say the way to calculate the efficiency of the carnot refrigerator is the same? That is, could be still use

    [tex]\eta = 1 - \frac{T_{cold}}{T_{hot}}[/tex]

    to calculate the efficiency of Carnot refrigerator?

    Thanks
     
  2. jcsd
  3. russ_watters

    Staff: Mentor

    Chegg
    No: since the cycle is being viewed from the opposite direction (basically), the efficiency is the inverse.
     
  4. You are right. I deduce the efficiency for the Carnot refrigerator similar to the steps for deducing the efficiency for Carnot engine. I've got something like this

    [size=+2]
    [tex]
    \eta = \frac{T_h}{T_h-T_c}
    [/tex]
    [/size]

    But with this expression, if [tex]T_c[/tex] is just a bit below [tex]T_h[/tex] but still be positive, the efficiency will be large than one? How's that possible? How to explanation this? What's the physical significance of this result?
     
  5. russ_watters

    Staff: Mentor

    Well, since these devices don't do mechanical work, the output is moved heat. In fact, for both heat pumps and air conditioners, the amount of heat moved is greater than the input power.
     
  6. re Russ Watters:
    An air conditioner IS a heat pump -- just turn it around in the window!
    iiuc, modern combination A/Cs-heat pumps (Mitsubishi's/Fujitsu's, etc) don't mechanically switch air flows, they re-route the refrigerant through different coils, or some such thing.

    To the OP, when you run a heat engine in reverse (ie, A/Cs, heat pumps), you no longer speak of efficiency, but of COP -- coefficient of performance, or whatever else engineers like to call it. And as you observed, it is always greater than 1, just as efficiency is always less than 1.

    There is no violation of thermo here, just a statement that if heat can drive a motor, a motor can pump heat. This pumped heat, plus the inevitable decay of the motor's work into heat, gives the appearance of more than you put in, but it really is not.
     
    Last edited: Dec 24, 2008
  7. russ_watters

    Staff: Mentor

    Different way of saying the same thing, but yes, you are correct. It might lead to a better explanation, for people confused about the COP concept....

    If a window air conditioner moves 1 kW of heat energy from the cold coil (the evaporator coil) to the hot coil (condenser coil) while consuming 0.5kW of electrical energy, it dissipates a total of 1.5 kW at the hot coil. That's an efficiency of .5/1.5=.333. Flip it around in the window and you get 1.5 kW of heating for an electrical input of 0.5 kW, or a COP of 1.5/.5=3.0.
     
    Last edited: Dec 24, 2008
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