Calculating Diesel Consumption in DG While Running in Full-Load

i am using about mathematical calculation

those calculation should be related with Power, speed of the DG and other datas

pls give mathematical calculations to derive the values

with regards,

prakash

Your answer can not be answered with the supplied information. efficiency and fuel flow rates vary with manufacturer, use, and with individual generators.

To estimate your fuel usage, empty the tank, then fill with a measured amount of fuel. Time how long it takes for the generator to run out of fuel with whatever load you want on it. Take the amount of fuel and divide by the time and you should be able to get gal/hr or liters/hr.

In very general terms, a diesel will burn about 250ml/kW/hr, but that figure varies with the engine model, load and speed. To get an accurate figure at a particular load, you need to have the manufacturer's figures.

In general terms, fuel burn KW will be around 2.5 to 3.3 KW per KW generated.

In more specific terms, consult the engine manufacturer's BSFC (Brake Specific Fuel Consumption) chart for the BSFC at various RPMs, and use this number to determine fuel burn at full KW output. BSFC varies from engine to engine, and varies by RPM.

The numbers can be converted to whatever usits of consumption you prefer.

BSFC varies from engine to engine, and varies by RPM...

and load, with BSFC generally being lowest at the engine's torque peak load and rpm.

Incidentally, the 250ml/kW/hr figure I quoted is a typical BSFC for a small diesel, with the kW representing the mechanical output or the engine. If such an engine is powering a generator, then the electrical output is likely to be 85% of the mechanical input to the generator (depending upon generator efficiency). This would yield an overall efficiency of about 27% -- a bit lower than the figures you quoted, but I am thinking in terms of very small (portable) diesel generators. Some huge diesels are much more efficient.

I do not wish to appear pedantic, Ken........but units are l/kWh..........not l/kW/h, I mention this only because, if someone wanted to carry out any calcs, incorrect units can cause a problem.

Funny you should mention it. Actually, what I'd intended to write was g/kW/h, not ml/kW/hr. Mathematically, g/kW/h is the same as g/kWh (just as 32 feet per second squared is the same as 32 feet per second per second.)

I searched around to see if I could find any support for my logic, and found only this, where BSFC is expressed as .39lb/hp/hr (instead of .39lb/hp.hr). I prefer the .39lb/hp/hr (or the rough equivalent 240g/kW/hr) but only, I suppose, from experience flying. There, you see gauges reading in pounds per hour or gallons per hour. So then it seems natural to think of lbs/hr/hp I suppose.

With some struggle, I found this textbook, which suggests that the x/y/z logic is preferred. I suppose if you are thinking "how many units of fuel does this engine use per hour when it is producing 10 hp" then the x/y/z/ logic makes sense.

But in any case, I think you're right -- most manufacturers use g/kWh rather than g/kW/h.

An after thought (i.e., rambling even more): I suppose it has to do with thinking in terms of power or in terms of energy. If you are thinking "how much fuel flow will produce 40 kW of power", then the x/y/z logic makes sense. If you are thinking in terms of energy: "how much fuel will produce 40 kWh of energy" then the x/(y*z) logic makes sense. Either way, the calculation comes out OK, if you keep your wits about you: 2/2/2 = 1/2 2/(2*2) = 1/2