Generators given in kVA but loads given in kW

If I have a generator at 13kVA 0.8 PF. It is capable of producing ~10.4kW.

I have a load of 10 kW but at a 0.7 PF and therefore consuming ~14.3 kVA.

What is the relationship here?

I thought this was simple, but nowhere I can find really addresses this.

Generators and transformers are rated in both kVA and kW. The voltage tends to be fixed by the speed or excitation. The current is limited by the I2R Joule heating in the windings. The windings therefore impose the kVA limit. If the generator is not called upon to produce much power, it can reduce the power it calls for from the engine. It's not uncommon for a generator to have a higher kVA rating than kW rating, so that it can drive loads with a poor power factor.

Your load is more extreme (0.7 PF at 10 kW) than the generator's windings are sized to drive, it draws more current at 10 kW dissipation, ie uses more kVA, than the generator is rated to deliver, even though the generator's engine could deliver the kW.

That said, when a generator supplies too much current, its windings heat up faster, and if allowed to continue heating, to a higher temperature than they are rated to withstand. For the small overload you are talking about here, unless the generator is monitored by electronics (a fuse would not have that fine discrimination), it's likely that it would be able to run your load, for a short time at least, while the windings were still initially cool. Continued running at that higher load would result in excess temperatures, and greatly reduced generator lifetime, and maybe even prompt failure from overheating.

What is the relationship here?

Your load requires more kVA than the generator is rated for and therefore, you have a problem to solve. Both kVA and kW ratings for the load must be within the capabilities of the generator.

This does indeed look simple:

At the same rated voltage, the load would draw more current than the generator is specified to provide.

Depending on the type of load and generator, you could try to operate the generator at reduced voltage - with a load where current is proportional to voltage, a reduction by 9 % would keep the generator in spec; you'd need to check if you're OK with load power reduced by 17 %.

Another option is trying to improve power factor.
(You'd get an idea why a provider would limit/charge PF.)

You can try to measure power factor or RMS current - if the load is specified to be no worse than 0.7, but actually is no worse than 0.8, or load current is no more than generator rated current, you should be OK.

• (Is consuming VAs imprecisely phrased?) Commented Aug 22, 2023 at 5:50

Well, notice that apparent power is given by:

$$\text{S}\space\left[\text{VA}\right]=\text{V}_\text{RMS}\space\left[\text{V}\right]\cdot\text{I}_\text{RMS}\space\left[\text{A}\right]\tag1$$

And real power is given by:

$$\text{P}\space\left[\text{W}\right]=\underbrace{\text{V}_\text{RMS}\cdot\text{I}_\text{RMS}}_{=\space\text{S}}\cdot\cos\left(\varphi\right)\tag2$$

Where $$\\cos\left(\varphi\right)\$$ is the power factor.

So, we get:

$$\text{P}=13\cdot1000\cdot\frac{4}{5}=10400\space\text{W}\tag2$$ $$10000=\text{S}\cdot\frac{7}{10}\space\Longleftrightarrow\space\text{S}=\frac{100000}{7}\approx14285.714\space\text{VA}\tag3$$