I need to conver 320g/kW.h into L/hr for a diesel generator 13kVa , at 50Hz with an output voltage of 230V.
If we assume output is actually 13 kVA and power factor is 1 (worst case), it is 13 kW x 320 g/kW·h = 4.16 kg/h As density of diesel fuel is around 0.85 kg/L, that is 4.9 L/h at max power. If at less than max output power, consumption will be somewhat less.
JohnS, in your last post you divided 4.16 by 0.85 to get 4.9. Shouldn't you have multiplied 4.16 by 0.85 to get 3.54............or am I mixed up?
4.16 kg/h x 1 L/0.85 kg = 4.9 L/h
I recommend always leaving the units attached to the numbers and working the units cancellation; it is always clear whether to multiply or divide. You can still make errors by using wrong conversion factors, but it is much easier to check all the inputs to the calculation.
Measure the engine's power output (kilowatts) and fuel flow (liters per hour). Divide.
It varies with the heat content of the fuel and the efficiency of the engine. For a particular engine and fuel, it varies considerably with torque and rpm (the plot is called an engine map, and shows brake specific fuel consumption contours vs torque and rpm).