I have a generator set that is oversized, it is capable of produce 180 kVA but I just demand 35 kW, it works with propane gas as fuel for primary engine.

My question is, which are the disadvantages of this over sizing, does the gen set consume less fuel in a low load scenario than in a 80% capacity scenario? In the case of diesel generators when are low current loads the genset present wet stack and unburned fuel accumulation in the cylinders but as this genset works with propane gas which burn in a "cleaner" manner I think it could not has this problem but maybe others.

Can I say the analogy: when there is no load the engine is working part time and the other times is rotating freely without resistance or less resistance to rotate (inertia)? I mean if the current that the load let pass to the generator increase, the magnetic field force that oppose rotation will also increase, but this increase in force is proportional to the consumption in a linear manner or more in a logarithmic one?

enter image description here

Supposing the x axis is rpm and y axis fuel consumption.

  • 1
    \$\begingroup\$ Generators will usually run at a constant speed. The fuel consumption per kW could increase or decrease as load increases, you'll have to check your manual. \$\endgroup\$
    – vir
    Oct 12, 2021 at 20:19
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    \$\begingroup\$ Tip: use 2 x <Enter> for paragraph breakes. (Fixed.) You forgot to label the red and blue curves on your graph. \$\endgroup\$
    – Transistor
    Oct 12, 2021 at 20:19

2 Answers 2


Wet stacking, where unburned fuel condenses in the exhaust system, does not occur with gaseous fuels because they are gaseous and therefore do not condense at standard atmospheric conditions.

You are correct that under low loads, the engine is turning with less resistance (from the alternator) than it is at higher loads. Power is the product of torque and angular velocity (P = τω), and the engine controls work to maintain angular velocity as a constant. This means that under normal operating conditions torque is proportional to power. Fuel efficiency varies over the range of torque output by the engine.

For fuel efficiency at various loads, you should check the manufacturer's literature that came with your generator. It should include some information about fuel consumption based on load. It may not have data all the way down to 24% loaded as you are apparently running yours, but you can get an idea.

For the sake of demonstration, I retrieved the literature for a Caterpillar DG150-2 which is similar in rating to yours (150 kW).

LP Gas Usage on 150 kW generator

You can see that the efficiency drops by about 16% as load decreases to half. It will drop even more as load reduces.

Also worth noting: The power (in kW) must be produced by the engine. Yours is probably capable of about 144 kW (assuming a typical power factor of 0.8). In determining the loading on the engine, you need to compare the power.

The apparent power rating (in kVA) relates to the current-carrying capacity of the alternator's windings. (You didn't state your generator's voltage, so I can't say what its current rating is.)

For example, if your load had a (ridiculously low) power factor of 0.14, you would require 250 kVA for your 35 kW load (35 / 0.14 = 250). Even though the engine would easily handle the power, the alternator windings could not handle the current and the output circuit breaker would probably trip. As long as your 35 kW load has a power factor of more than 0.194, your 180 kVA alternator can handle it (35 / 180 ≅ 0.194)

  • \$\begingroup\$ I would expect max efficiency to happen at max BMEP / torque. \$\endgroup\$ Oct 12, 2021 at 20:57
  • \$\begingroup\$ Can I increase this power factor with a capacitor Bank?or wont it affect the power factor of the genset \$\endgroup\$
    – TTT2
    Oct 13, 2021 at 5:33
  • \$\begingroup\$ @hacktastical Since it's a synchronous generator (i.e. ω is maintained constant) and P=τω, maximum torque and power occur at the same time: maximum loading. \$\endgroup\$
    – Theodore
    Oct 13, 2021 at 12:49
  • \$\begingroup\$ @TFM Since this generator can handle 180 kVA, you really only need to worry about power factor correction if the power factor of your load really is 0.194 or lower, which is not likely. (35/180≅0.194) \$\endgroup\$
    – Theodore
    Oct 13, 2021 at 12:52

The generator is very efficient at converting mechanical energy into electrical energy. 95% would not be unreasonable. The engine that drives the generator is very inefficient at converting chemical energy into mechanical energy. Most of the chemical energy is lost as waste heat. An efficiency of 33% might be a reasonable estimate.

With that in mind, the efficiency of your overall system, engine and generator will be predominantly determined by the (in)efficiency of your engine. How does it change with speed and load? That all depends upon what engine it is. The manufacturer may know. Someone who tests the engine may know. But without some kind of data, we here on the internet cannot know.


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