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 ?...

 

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.