Generator Not Producing FLC (Full Rated Current)

Thanks for the answer.

How could i draw rated current i.e. 69amps? I tried overloading Genset and it didnt help.

Connect it to a 200 AMP load.

You could put an inductance in series with the load bank, calculated to give PF 0.8 or lower.

Note that the PF 0.8 is a rather arbitrary figure. In practice if the PF of the load is > 0.8, maximum current is lower, if it's < 0.8 max current is constant at 69 amp, power is lower.

Thank you very much Codemaster.

"Note that the PF 0.8 is a rather arbitrary figure. In practice if the PF of the load is > 0.8, maximum current is lower, if it's < 0.8 max current is constant at 69 amp, power is lower." that would support my argument.

Apply some loads whose PF is 0.8, such as motors. Then you can get more amps out without bumping up against the kW limit.

What is the thinking behind carrying out this test?

• Is the generator manufacturer's published rating inadequate evidence for having selected the right one?
• Is it to prove the circuit overload protective devices work correctly without using the manufacturers' evidence?
• Some other reason?

Hello PWSlack,

Customer wants to see the Genset generating 69amps per phase while connecting to the full load.

As the Genset data plate says stand by current rating = 69amps.

Customer is questioning if Manufacturer gives 69amps rating why can't he draw 69amps from Genset.

OK. That question has been answered above. The next step is to educate the Customer with the answer and point out that the only useful load that the Customer can have is the non-reactive part, which is lower than the amps stated owing to the power factor on the equipment and cabling connected.

Do keep in mind that in such a position the equipment is operating right on the edge of the settings of its circuit protective device(s). If a margin is required to sustain the load during minor peturbations and excursions above that setting then a larger genset is in order, as suggested above. Which is probably why the Customer wants a test - a suspicion that it just ain't big enough.

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Engines are rated by shaft power kW, deducting the generator losses gives the maximum rated electrical power kW of a genset.

Generators are rated by volts and amps. For the stator (usually output) winding the current heats winding resistance and sets the limit at which insulation gets too hot. For the rotor, the current needed (and thus heating) is determined by the power factor - an 0.8 lagging power factor load needs substantially (~x 2) more rotor current than a unity power factor (resistive) load - the load current is 25% greater at 0.8 power factor.

The standard power factor for generators is 0.8 lagging. This corresponds to the power factor of the typical industrial load.

You should also know that the "headline" brochure kW rating is often not "continuous", but the maximum for 1 hour, with a life penalty if used more than occasionally.

You have said it that you are using an unity power factor load bank, there is such a thing in the market called inductive load bank, where you can test the genset at 40 kW unity p.f. and add a 30 kVAR inductive load to get a 50kVA load. The problem is the consultant ( or whoever he is) do not know electrical formula because he is only looking at the alternator nameplate details.

Hi Strava,

Let me explain you why you are not able to load the DG set to the Full Capacity on Unity PF.

Your Generator is Rated 50 KVA @ 0.8 PF, which means it can be loaded to 40 KW load and the Engine has been selected to deliver this 40 KW only.

Now if we calculate the KVA for 40 KW at Unity PF it Would be 40 KVA and hence the Full load Current remains at 55.6 Amp.

If you try to load the Genset above this 40 KW the Engine will be overloaded and speed will droop, DG will Trip.

It just ain't big enough, is it? Someone has screwed up, perhaps?

How embarrassing.

what is the terminal voltage of the Generator at the time of loading?