Generators and Diesel Consumption

Great link, Guest! I'll get quite a lot of use out of that chart.

Good chart,

easy to convert to "real" figures (litres/hr)

Chas

In order to generate the stated electrical power (Pe) of 320A at 407V which is about 180 kVA (calculated as Pe= I*V* sqrt3 (assuming 3 phase)), the diesel engine needs to transfer an amount of mechanical power (Pm) to the generator (overcoming torque and turning it at a certain rotational speed). There are various losses in the energy transfer/generation process. Lets assume it is about 10%. Therefore Pm = Pe*1.1 thus Pm = 180*1.1=198. Say 200kw... As your engine is capable of much higher output it is clearly only running at a part load less than 30 % (200kw/650kw). This would normally cause it to be less fuel efficient than the best quoted fuel efficiency figures which are normally determined at full load. Specific fuel consumption figures for the size and type of diesel implied typically range between 200g/kWh and 250g/kWh. Some of the very best (such as certain MTU engines) quote figures as low as 190g/kWh. The fuel consumption would therefore be the product of the engine power output (Pm) and the Specific Fuel Consumption. If we accept the engine power output as 200 kW and assume a good average of 230g/kWh it gives a fuel consumption of 46 000 g/h. Since the specific gravity of diesel is about 0.86 it means that 46 000 g/hour would equal 53 l/hour (46 kg/0.86 kg/l). Hope this helps a bit more than just figures. Regards PJT

Genset's rated current is 917 A. But current derived is averagely 360A (320-400 A). Isn't this rated current is the maximum current possible. then load will become 360/917=40%. Is it rite or kW conversion is necessary to determine load.

Hi Tabbymulla I think you are quite right. The rated current is the maximum current possible on a continuous basis. Calculating it in that way is a good way of establishing what % load the engine is running at. I assume this is in order to use the tables suggested by Guest. Either way you still need to get an indication either of the maximum power output in kW (to use the tables suggested) or of the kW output at the operating conditions (to use Specific Fuel Consumption (SFC)). One must keep in mind that there are losses in the process of transferring mechanical power to electrical power and therefore the engine is producing more power (and using more fuel) than what would seems to be suggested by the electrical output of the generator. Regards PJT (Just a last comment. My previous calcs were based on 320 A and not 360 A. At 360 A the fuel consumption seems to be heading for 75 l/h).

My sincerest apologies. I realised that I made a mistake in multiplying by 0.8 when calculating Pe. I therefore repost the affected part with the corrected values. In order to generate the stated electrical power (Pe) of 320A at 407V which is about 226 kVA (calculated as Pe= I*V* sqrt3 (assuming 3 phase)), the diesel engine needs to transfer an amount of mechanical power (Pm) to the generator (overcoming torque and turning it at a certain rotational speed). There are various losses in the energy transfer/generation process. Lets assume it is about 10%. Therefore Pm = Pe*1.1 thus Pm = 226*1.1=248.6 kW As your engine is capable of much higher output it is running at a part load less than 40 % (248.6kw/711 kw)... ... If we accept the engine power output as 250 kW and assume a good average specific fuel consumption of 230g/kWh it gives an average fuel consumption of 57500 g/h. Since the specific gravity of diesel is about 0.86 it means that 57500 g/hour would approach 67 l/hour (57.5 kg/0.86 kg/l). I hope there are no more gremlins! Regards