Energy Efficiency

Discussion in 'Homework Help' started by Kayne, Jun 14, 2010.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    I have been given this in my revew question

    A diesel plant uses 100mL of Fuel in 25 seconds, when driving a 10 pole, 3 phase generator delivering a load at 50Hz, 415v ,100A, 0.8pf lagging. Assume the generator efficiency is 0.85 and the fuel energy is 42.5MJ/L. (3.6MJ = 1kWh)

    Calculate the following

    Generator output

    Power = 1.73\times Volts \times I \times Pf
    1.73\times 415 \times 100 \times 0.85=61.02kW

    so for 1 hour = 61.02kWh

    Engine output

    Syn speed of generator = (120\times f)\div P
    (120\times50)\div 10 = 600rpm \rightarrow 360000rphr

    and the engine uses  100mL = 25sec
    Therefore

    400mL = 1min \rightarrow 24L = 1 Hr

    So if I use the 42.5MJ/L and multiply but 24 L i get 1020MJ/L

    1020 \div 3.6 = 283.33 kwh

    Overall efficiency when supplying the load.

    61.02 \div 283.33 \times 100 = 21.5%



    Have I managed to do this correctly ?...
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Firstly you've put the wrong value for the load pf - given as 0.8 while you've used 0.85 which is the quoted gen efficiency.

    You've also gone about solving this in a rather convoluted manner.

    Take the fuel for instance -

    Consumption is 100 mL in 25 sec or 4 mL per second.

    This equates to (4/1000)*42.5E6 Watts [Joules/sec] = 170kW

    Which is the maximum achievable power (at 100% efficiency) from the fuel at the indicated rate of consumption.

    This is required to operate a load of

    P_{load}=\sqrt{3}*V*I*(pf)=\sqrt{3}*415*100*0.8=57.5kW

    I would think the overall efficiency would then just be

    \eta=\frac{P_{load}}{P_{fuel}}*100%

    You would not include the generator efficiency in the overall efficiency calculation. You would however need this to find the engine output power - which you could do by working back from the generator output power.
     
    Last edited: Jun 14, 2010
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