Fuel Consumption
Hey all :P
Output power - 60MW
Efficiency - 42%
Fuel density 0.85
Calorific Value 45MJkg
How would I work out the fuel consumption lph?
Thank youu.
Fuella
Output power - 60MW
Efficiency - 42%
Fuel density 0.85
Calorific Value 45MJkg
How would I work out the fuel consumption lph?
Thank youu.
Fuella
Efficiency = (Power output / Power input) x 100 % = 42%
Rearranging the formula, Power input needed to produce 60MW output = 60x106/0.42 = 142.86MW
Power = Joules / Second
So 142.86x106 Joules are burned through each second to produce 60MW.
The calorific value of the fuel is given as is 45MJ per kg, then 142.86MJ will burn fuel at 142.86/45 = 3.175 kg/second.
Or 3.175 x 60 x 60 = 11,429 kg per hour.
1 litre of water @ STP = 1 kg
Fuel relative density is given as 0.85 x 1 kg = 0.85 kg / litre
Volume of fuel used = mass / relative density = 11429 / 0.85 = 13445 litres / hour.
Rearranging the formula, Power input needed to produce 60MW output = 60x106/0.42 = 142.86MW
Power = Joules / Second
So 142.86x106 Joules are burned through each second to produce 60MW.
The calorific value of the fuel is given as is 45MJ per kg, then 142.86MJ will burn fuel at 142.86/45 = 3.175 kg/second.
Or 3.175 x 60 x 60 = 11,429 kg per hour.
1 litre of water @ STP = 1 kg
Fuel relative density is given as 0.85 x 1 kg = 0.85 kg / litre
Volume of fuel used = mass / relative density = 11429 / 0.85 = 13445 litres / hour.
(edited 10 years ago)
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