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LM6000 Gas Turbine Outputs/Inputs

  1. Jun 5, 2012 #1
    I am doing research and was curioius as to how I would obtain useful numbers for efficiencies from the following information, concerning a CHP aero derivative turbine,

    Power KW = 43822
    Heat Rate LHV KJ/kWH = 8980

    From this, can I say that if I ran the machine for an hour I would use 43,882KWH and receive 109311.54KWH of heat out put [ from, (8980kj/kwh x 43822kwh)*1000/(3.6MJ)) = 109311.54KWH ]

    However, when I use the table for running at full load, the electrical efficiency of the machine is 39%. I considered that this may be total efficiency but 39% for a CHP seems way too small, should be closer to 60-70 right?

    Which means that I put in 43822KWH/.39 = 112364.1KWH but got out 109311+43882=153193KWH.

    As I am sure that this is not an over-unity machine, can anyone show me where I am going wrong?
     
  2. jcsd
  3. Jun 5, 2012 #2

    russ_watters

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    Staff: Mentor

    It's a generator: kJ/kWh is how much heat you need to input for every kWh you get in output. You've got the efficiency equation upside-down!
     
  4. Jun 6, 2012 #3
    Hm, why aren't both units in energy then, one's in power. I realized I wasn't de-rating the heat output so I think that's where I was going wrong. I guess what I'm trying to find out is the following,

    "If I was a unit of fuel going into the generator how many units of energy would I make, how much heat would I make and how much waste would I make."

    I apologize for my inexperience, I am very unfamiliar with CHP machines.
     
  5. Jun 7, 2012 #4

    brewnog

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    A kWh is an energy unit, not power.

    kJ/kWh is the normal way to express fuel consumption for an engine fuelled by gas. Convert to kJ/kWs (divide by 3600), take the reciprocal, and that is your efficiency.

    For overall efficiency, you also need to consider the amount of heat you are recovering through your CHP system.
     
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