g/kwh to gal/h

Originally Posted by JohnS

Lets avoid issues of the "power factor" of the load, and assume power factor is 1, so 180 KVA = 180 kW.

213 g/kWh x 180 kW = 38340 g/h or 38.340 kg/h
A liter of diesel fuel is about 0.85 kg, so 45.1 L/h or 11.9 US gallons per hour at full load. If generator is not putting out full load, consumption will be less, but not linearly with load (it still consumes fuel just to spin, even if no load is connected.)

Thanks a lot John..You made it really simple.

I applied the same logic to calculate the consumption for a Air Conditioning Unit (115 Tons) that had "Specific fuel consumption at maximum torque : 200 g/KWh.". Here the output of the engine was expressed in BHP. I simply converted BHP to KW (Mulitplied BHP by 0.7457 to get KW) and then got the calculation done as you mentioned

Thanks all the way from India

Dear A+,

I had to do the conversion this moring for myself. Back to basics.
I used info available at various conversion sites and please feel free to check my figures.

1 gram = .0001 Kg
1 Kg = 2.2046 lbs
7.2 lbs = 1 gallon of diesel

1 gram or .001 Kg x 2.2046 lbs/Kg = .0022046 lbs.
.0022046 lbs x 1 gal /7.2 lbs = .000306 gallon

So 1 gram = .000306 gallon

Your consumption is given in g/kWh and you want gal/kWh so the only thing that changes is from

So let's take a consumption of 211 g/kWh x .000306 gallon = .064566 gallon and your genset is rated at 180 kW = 11.62 gallons per hour


Hopes this helps

Originally Posted by A+ Ano

I am trying to compare fuel consumption ratings on Diesel electric generators
The american ones show gals per hour of consumption at 50% load 75% load & 100% load which is what I am familiar with.
All the other generator sets are rated by g/kwh which I think is grams per Kilowatt hour?
What I'm having trouble with is a gram is a weight not a volume? Do I multiply the total kilowatt hours which in this case is 60kw x the grams per hour(in this case 230) 60x230=13,800 grams
Do I have to figure out what a gal of diesel fuel wieghs? and the convert the grams to weight?

Any help would be great.
Thanks A+

Rated Power kva 20
Max power kva 22.5
PHASE three
Frequency (HZ) 50HZ
Rated voltage (V) 220～240/110～120；400/230V
Rated Current (A) 34.7A
Bore*Stroke 100x115
Displacement(L) 3.61
12h Power kw(ps) 26.5
Fuel Consumpion g/kw* h 258.4
LUBE CONSUMPTION g/kw h 2.04
Compression ratio 18.0:1

Hi, please help me I need to know what is the fuel consumption in litres for the above specs

[JohnS]

Originally Posted by Unregistered

Rated Power kva 20
Max power kva 22.5
PHASE three
Frequency (HZ) 50HZ
Rated voltage (V) 220～240/110～120；400/230V
Rated Current (A) 34.7A
Bore*Stroke 100x115
Displacement(L) 3.61
12h Power kw(ps) 26.5
Fuel Consumpion g/kw* h 258.4
LUBE CONSUMPTION g/kw h 2.04
Compression ratio 18.0:1

Hi, please help me I need to know what is the fuel consumption in litres for the above specs

Lets assume power factor = 1, so rated power is 20 kW. At 20 kW, fuel consumptuion is
20 kW x 258.4 g/kWh = 5168 g/h or 5.168 kg/h
Diesel varies, but 0.84 kg/l is a good typical figure for #2 diesel, so
5.168 kg/h x 1 L/0.84 kg = 6.15 L/h

I need to conver 320g/kW.h into L/hr for a diesel generator 13kVa , at 50Hz with an output voltage of 230V.

[JohnS]

Originally Posted by Unregistered

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?

[JohnS]

Originally Posted by Unregistered

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.

Right you are. Thanks for the clarification.

If a 500 kW diesel generator has a fuel consumption of 208 g/kWh at full load. How much it will be at 50% and 25% load? Thanks