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 so for 1 hour = 61.02kWh Engine output Syn speed of generator = and the engine uses Therefore So if I use the 42.5MJ/L and multiply but 24 L i get 1020MJ/L Overall efficiency when supplying the load. Have I managed to do this correctly ?...
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.