Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics and conservation of energy

  1. Dec 15, 2014 #1
    I have a question about conservation of energy in my system

    A heat pump allows you to move a quantity of heat with a smaller energy input. It is most efficient for small temperature differences.

    A heat engine allows you to convert a temperature difference into mechanical work. It is most efficient for large temperature differences.

    Since their efficiencies are inversely proportional, if you use a heat pump to create a temperature gradient you won't be able to get as much energy back out with a heat engine.

    Now imagine we have 500 kJ of high temperature thermal energy from an industrial process. The heat was 'created' through regular heating, let's assume at 100% efficiency for simplicity. The industrial heat drives a heat engine and the temperature difference is such that we get a 50% efficiency, i.e. 250 kJ of mechanical power out.

    The heat from the industrial process is added via a heat exchanger.

    Let's say we preheat the heat exchanger via heat pump, but by only a small amount and therefore at high efficiency e.g. COP=4. e.g. we have added 100 kJ heat at a cost to us of 25 kJ .I'm guesstimating the numbers for the purpose of the explanation, but we now add 400 kJ from the industrial process so that there is 500 kJ in the heat exchanger and get the same power output of 250 kJ, of which 20% is attributed to the preheating.

    In the latest scenario instead of getting 250 kW mechanical power out at a cost of 500 kW, we achieve the same 250 kW at a cost of 425 kW. We have therefore increased efficiency. What I am worried about is that when I look at the preheat on its own, it feels like we have violated the conservation of energy. Can someone please provide explain either where the concept breaks the laws of thermodynamics or if not, where does the energy come from?

    Don't get too stuck up on why we are doing this, i've left out the application because of IP owned by someone else.
     
  2. jcsd
  3. Dec 15, 2014 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You want to "fire" a boiler with a heat pump, and you are calling the "exhaust" heat from the heat pump a "free" energy source. It does NOT work that way. The heat pump will move heat from low T to high T, but if you do not furnish it a low T energy source, it cannot move anything.
     
  4. Dec 15, 2014 #3
    Please read again, you've not understood it. I didn't mention free energy once. I am trying to figure out if I can increase efficiency by preheating the heat exchanger with a pump because I can add 100 kW of heat at a cost of 25 kW.

    I'll then be adding heat from a higher temperature industrial process, it's not waste heat but it can be thought of on a similar basis, though I can't elaborate on why I am doing this.

    Since the heat engine is running between ambient and a high temperature (much higher than that of the preheat), I get 50% efficiency. By first preheating with a heat pump rather than doing all the heating with the heat from the industrial process, it would appear that I increase efficiency above 50% (not sure where you got the idea that I was trying to get it above 100%, I am trying to reduce losses).

    Obviously the basis for the operation of the heat engine is that it takes energy out of the temp gradient between the hot heat exchanger and ambient, so the heat exchanger cools down.

    I'm asking because I may have missed something, as something about it feels odd to me, but you've not explained why it wont work.
     
  5. Dec 15, 2014 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No. You have to provide the other 75 kW the heat pump is pumping; your 100 kW preheat costs you to 25 kW to run the heat pump plus the 75 kW it pumps.
     
  6. Dec 15, 2014 #5

    The 100 kW is thermal energy moved from cold to hot. it costs 25 kW to transfer the energy because it is increasing the thermal gradient. If what you said was true, heat pumps would not exist as you may as well use a convection heater.

    Normally if you then run a heat engine you'd get back less than the 25 kW, but because I am adding extra heat from the industrial process the heat engine efficiency increases. The conversion of heat from the industrial process to mechanical work is 50% efficient. The preheat is meant to increase efficiency slightly.
     
  7. Dec 15, 2014 #6

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You meant "75 kW" rather than 100. What happens to your cold reservoir as you pump from it at 75 kW?
     
  8. Dec 15, 2014 #7
    The heatpump compresses gas, which heats it. the heat is exchanged with the heat exchanger. Gas is then expanded, it cools to below ambient. The surrounding air heats it. The air can be considered an infinite source for practical purposes.
     
  9. Dec 15, 2014 #8

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Let's back up and start again. This looked to me (not to anyone else that I'm aware of) on first (and second, and third reading like a Rube Goldberg end run).
    Then the outlet T of the unspecified process is elevated by however much results from the preheat.
    The heat that has not been removed from that original process stream is the unaccounted heat in your heat balance/conservation of energy, which on "nth" reading is what bothered you enough to request another view of the idea. I was answering an entirely different question from what you asked --- my bad.
     
  10. Dec 15, 2014 #9

    russ_watters

    User Avatar

    Staff: Mentor

    The "industrial process" is waste heat, right? So instead of getting 500 kJ of free energy you are getting 400 kJ of free energy and paying for the other 100 kJ. That's not better, it is worse.

    Also, and maybe this is more germane to the question, you are not adding the heat energies at the same temperature, so you can't take credit for the same efficiency of the output from them. You've kept the efficiencies separate on the input, so keep them separate at the output: what happens to the efficiency of your use of the "industrial process" energy if you reject the heat form it at a higher temperature?
     
    Last edited: Dec 15, 2014
  11. Dec 15, 2014 #10

    Ok thanks, I can see how you misunderstood, I asked the question because I was concerned there might be a Rube Goldberg paradox and was trying to see if someone else saw one.
     
  12. Dec 15, 2014 #11

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Did I figure out correctly what your question was?
     
  13. Dec 15, 2014 #12

    But its not waste heat, I said you could treat it as if were but I can't say exactly what the system is used for and therefore using 400 kJ instead of 500 kJ of the 'waste heat' is a saving.


    Could you elaborate on your second point. In my design the preheat is added at a lower temperature than that of the final temperature and so both the heat pump and heat engine will be efficient. Unfortunately as the heat engine runs and the heat exchanger cools, efficiency will fall from 50%, but this will happen by the same amount regardless of whether i preheat since the target temperatures are still the same.
     
  14. Dec 15, 2014 #13
    I think so but by this stage I'm getting confused
     
  15. Dec 15, 2014 #14

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think what Russ was pointing out is the following:
    If the "waste" heat from the heat exchanger would have been dumped into a cooling tower/whatever, literally wasted, you pick up whatever you get from your heat engine for "free." If you throw in another 25 kW to drive a heat pump that in some ways degrades performance of the heat exchanger, you still get the heat engine output, but you've invested another 25 kW to power it.
     
  16. Dec 15, 2014 #15

    russ_watters

    User Avatar

    Staff: Mentor

    The output is the same total, but the outputs from each process are different. You are assuming each is 50% efficient in providing you output, but they aren't the same. If you keep them separate on the input you must keep them separate on the output and calculate your efficiencies for them separately. And you haven't calculated the effect of rejecting heat from the high temp process at a higher temperature -- that's where your extra loss is that you haven't accounted for.
     
  17. Dec 15, 2014 #16

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    I suggest that the OP should back off a bit and consider what he is trying to accomplish.

    If you are trying to analyze a complex system, you should assume that energy is conserved, and use that principle to aid in your analysis. Anywhere where energy conservation appears to be violated means that the analysis must be in error. That applies to the system as a whole, and to subsets of the system if you draw imaginary boundaries around them.

    If you are trying to prove that energy conservation really works, then you should reduce your experiments or analysis to the smallest simplest system you can imagine. Indeed, a single particle's interactions. Then you can use higher level of abstractions to simplify larger systems. If energy is conserved in every possible elementary particle comprising a system, then it is conserved in the composite system.

    But it makes no sense to dream up a system so complex that I confuse myself, then to post a non-mathematical non-graphical verbal description of it to a PF forum, and ask others to point out my error.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Thermodynamics and conservation of energy
Loading...