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Two stage turbocharger heat pump.

  1. Dec 3, 2011 #1
    So, I've been thinking a lot on this lately. I want to use two turbochargers, one with just the compressor - and one with both compressor and turbine.

    Doing calculations at 10 deg C. Doing the math for Norway because commercially avatible evap heat exchangers doesn't really work well in our climate.

    This is how it is going to work.
    A 2,5kW electric motor (90% efficient, controller losses included) drives the first compressor (comp1), delivering ~2 bar (a little over 2:1 pressure ratio) at 80% efficiency. Temperature within the air reaches 430 deg C. (((283*2)/0.8)-273)

    Between the first compressor and the second, there will be a need for a heat exchanger. Calculating 80% efficiency, new temperature out is 85 deg C. The reason for this is that alu (responsible for most of the parts in the compressor) melts at 660 degC, something that would easily be achieved.

    Pressure drop after intercooler? No, p will remain constant - only T changes.

    The second compressor(comp2) (80% efficient) further compresses the mixture to 3-3.5 bar. This is driven by its own turbine. Air temperature out from compressor is 500 degC. ((((273+85)*1.75)/0.8)-273)

    This turbine regenerates (65% efficient) the energy lost from the first compression. After this the air exits. Very low temperature might be a concern here - dont want things to freeze up lol :P (getting to this at the end).

    This is the complete loop:
    Free air 10C -> compressor1 -> 430C, 2Atm -> Heat ex1 -> 85C, 2Atm -> compressor 2 -> 500C, 3.5Atm -> Heat ex2 -> 100C, 3.5Atm -> Turbine -> 133K, -140C (!), 1Atm

    I want to build this some time in the next year. I've been looking at popular turbos and have more than one concern. Here is some.

    1. Driving the first compressor at ~90k rpm. Bigger spinners require less rpm, but more power. Also - side loads if I want to use a gear ratio might lead to upgrades in the bearings.

    2. How to calculate total efficiency? 400% achievable? If I have mass flow, this should be ok - rigth? I'm a bit lost in energy in gas at temps - been a while since i worked with physics xD

    Just throwing it out here for constructive critisicm :)
     
  2. jcsd
  3. Dec 3, 2011 #2
    R= 8.3145 J/mol K.

    So 22,4 l air weighs in at around 29gram. The turbos I'm looking at puts out from ~20 lb/min to 50lb/min. Quite how to read what flow the turbine needs isn't obvious, I'll look into that later..

    For 20 lb/min:
    ~9000 grams=312 mol gas.
    Temperature difference: 10C to-140C = 150K
    Energy out per min: 390kJ = 6,5kW:
    Eout/Energy in (2,5kW - totally not sure of this)= 260%

    For 50lb/min:
    Energy out: 16,25kW
    Eout/Ein=650%

    The question is really how much energy in to spin the turbo...
     
  4. Dec 3, 2011 #3
    The temperatures must be in degree Kelvin.

    Regards
     
  5. Dec 3, 2011 #4
    They are, as you quoted. (look at the calculations)
     
  6. Dec 3, 2011 #5
    Yeah. I agree. I was referring to the representations. If its Kelvin it should be 'degree K'. I got confused as you later refer to 85 degree C in your original post, which is a wide difference to the values of 430 and 500.
     
  7. Dec 3, 2011 #6
    I assume you're not confused anymore, but the intercooler brings the temp down from 430 to 85 degC. The challenge lies within extracting energy from the air, using and reusing as much as possible as input energy. Is 600% just a load of bull - or is there some possibillity this might work?
     
  8. Dec 5, 2011 #7
    So, I'm going to find the energy needed to compress the gas.

    I'm picturing a huuge cylinder able to keep 9000g of gas, witch gives ~200000 litres of gas.
    A cylinder 20meter in diameter. Areal (Pi*100dm*100dm). Heigth=6,34dm.

    This heigth is reduced to 3,17dm as the pressure is increased to 2 bar. What will the stored energy in a spring be?

    The force on the "piston" will be pressure times areal. The areal is 314m2. The pressure is ,relatively to 100Pa, 100Pa. Total force is therefore 31400N.

    If [tex]E_{p} = \int^{x}_{0} F \; dx, Where F =31400N over 3.17dm, this gives 9950 joules.

    And that seems somewhat little. That's a measly 165 watts of contious work, something must be wrong?

    The gas volume from start to end will be reduced by 50%, how do I go about to make my calculations work?
     
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