Energy Efficiency

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

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    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

    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
    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


    I would think the overall efficiency would then just be


    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