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Heat pumps and efficiency

  1. Apr 13, 2012 #1
    In my thermodynamics class we're talking about heat pumps (i.e. refrigerators and such). When talking about a typical heat engine (carnot or not) it makes sense to me to talk about the efficiency of the engine [itex]η=1-\frac{Q_{l}}{Q_{h}}[/itex], because it clearly represents how much bang you get for you buck (to be colloquial). But when you discuss heat pumps, it's possible to have an efficiency greater than 1, and I have a difficult time understanding this. It's clear from the definition in the case of a refrigerator for example, because you have [itex]η=\frac{Q_{l}}{W_{in}}[/itex]. In Blundell & Blundell's Concepts in Thermal Physics p.133, they reason this is because efficiency is essentially "what you want to achieve" divided by "what you have to do." Can someone clarify this idea of efficiency in the context of heat pumps or explain why it makes sense to talk about an efficiency [itex]\geq 1[/itex]?
  2. jcsd
  3. Apr 13, 2012 #2
  4. Apr 13, 2012 #3


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    Imagine you are pumping water from a low reservoir to a high reservoir. The potential energy added to the water is mgh and the energy required of the pump is measured in joules. Divide one by the other and you get efficiency. Eout/Ein = efficiency.

    But what if you didn't care about the mechanical energy in moving the water. What if what you cared about was how many fish you transported during this process. Then you do Fout/Ein = Fish Transportation Coefficient of Performance (FTCOP). Would you agree that depending on the units, this number could be tiny or huge, but it doesn't have anything to do with thermodynamic efficiency and a number greater than 1 doesn't imply a violation of conservation of energy?

    With some devices it just so happens that you don't care about the mechanical efficiency of the pump. A heat pump is one of them. So is a cooling tower for that matter. What you care about in a heat pump or cooling tower is the amount of heat energy transferred into or out of the working fluid. That's why it isn't a "water pump" any more, it is a "heat pump". And my example above would be a "fish pump".

    Now a fish pump isn't a thermodynamic cycle in the sense that a carnot heat engine is. But a heat pump or chiller is and that's why it is confusing to talk about units of energy in and out and get an answer greater than 1. But consider the following typical performance of different parts of a chiller plant tied to an chilled water air conditioner:

    Condenser water pump (through cooling tower): 0.1 kW/Ton
    Cooling tower fan (optional): 0.1 kW/Ton
    Chilled water pump (through system): 0.2 kW/Ton
    Chiller: 0.7 kW/Ton
    Air handler fan: 0.2 kW/Ton

    Taken together, this system has a coefficient of performance of 1.3 kW/Ton. In the real world, we use kW/Ton because it is more useful than a non-dimensional COP. But that's a COP of 2.7 for the entire system. Only one component of that system is a "heat pump" in the sense your book is teaching you -- a vapor compression refrigeration cycle. But every component involves moving heat around using a pump, compressor or fan, at a COP of much better than 1.0.
  5. Apr 13, 2012 #4


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    Not sure whether I'm making the same point as Russ...

    I depends whether you choose to consider heat as merely another form of energy or as something a bit different.
    When we say that a power station is only 40% efficient, it may nevertheless be close to its theoretical limit of, say, 50% efficiency, as prescribed by the laws of thermodynamics. So is it really 40% efficient or 80% efficient?

    With the first interpretation, it is entirely consistent to consider a heat pump as having greater then 100% efficiency. Indeed, if you use the electricity from the power station to drive a heat pump to generate the same temperature difference that the power station employed, the system as a whole would have a theoretical efficiency of 100%.
  6. Apr 14, 2012 #5


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    'Efficiency' refers to (Work Out)/ (Work or Energy In). A heat pump doesn't actually produce 'work out' so 'efficiency' is not really the right term to use. So there's nothing to worry about and no violation. If you tried to use the heat that your heat pump pushed out in some sort of maximally efficient heat engine, then you would be limited by thermodynamic efficiency in the actual work you could achieve with it. You would be back to the conventional 'efficiency' idea kicking in.
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