Diesel generator -diesel consumption

Too little information. Look in the user manual.

If all you need is a ball park you could chain the component efficiencies.

kva*(Kwh/kva)*Gen eff*Diesel eff*gallon/BTU=gallons/hr

kva*(48/60)*.90*.40*(3412 BTU/1kwh)*(1 gallon/130,000 BTU)=

I get 1.46 gallons /hr for a 62.5kva genset.

Cross checking by estimating hp gives powering a 67 hp engine for one hour. This compares to the smallest 4 cylinder production engines which MIGHT give an hour of power for a gallon and a half. However, don't bet on it. 40% efficiency for a desiel engine is best case. If you can not get a direct read on the projected fuel consumption you could refine the estimate by looking at the gen/alternator set speed and the engine power factor at that speed.

1800 rpm gens are too slow for best engine power and 3600 or higher engine rpms give a smaller engine and better efficiency but potentially shorter service life.

Hope this helps.

Sincerely,

Mr. Gee

GA Mr. Gee for a good back of the envelope estimate. These are usually good enough to get that ROM estimate. I might go lower on the engine efficiency perhaps 30-35% but given the information in the original question, this is all one can expect.

I just looked at a 62.5 KVA gen set and the consumption at full load was 4.3 GPH. Just for your information.

The simplest and easiest way is to check the manufacturer's specification sheet. It will be shown there.

Double check all these posts against 0.335 pounds of diesel per horsepower per hour. You can convert this to commie units! To get hp, take the kVA x .8 if your gen set name plate shows 0.8 PF rating, then x 1.341 to get commie units then / .95 (average generator eff).

Consumption of fuel should refer to the output. The other way what we do is to caculate Electrical Units (KWH)/Liters or gallons. we take measurements of Electircal Energy in KWH and diesle consumed for one hour & compare.

I agree. The unstated assumptions in my rough cut answer were full output and near perfect efficiency. Very questionable assumptions. If the user manual data provided by the commoner are correct and I believe they are then that particular units net thermodynamic efficiency is really pretty low.

That is the type of thermodynamic inefficiency that a diesel engine can show when run slowly.

Even so, only hitting 35% of best case is pretty bad. I wonder if I may have made a mistake in my estimation method. If its there I didn't find it. But, it is very hard for me to accept that this system is that bad. That result would give an engine efficiency of only about 15%. I know that running slow can cut the engine efficiency a lot, but cutting it more than half seems unrealistic. If someone has further thoughts on this subject I would be interested to listen.

Thanks in advance,

Mr. Gee

we have 500 kva & 370 kva generator we need the diesel consumption for both generators. how to calculate ?