Power Factor correction of a Diesel GenSet

[Mark_Djorn]

Hi All,

During my exercises on paper, we always assume the source from grid has Tension and Currentnin phase.
Today, for the first time, I was learning about Diesel Genset and noticed its plate saying 2700 MVA / 2000 MW. So basically, somehow a source has its own Power Factor (PF).

1. Can anyone explain me how a source can have its PF besided the load it is attached to?
2. Can I correct my source (GenSet) PF so that I can maximise its utilisation and bring that to feed a 2300 MW load for example? In my mindset, if I improve the source PF to 1, I could actually power a load which could be 2300 MW or higher?

The only info I can find for point number 2 is that 2000 MW refers.to the Diesel Engine and 2700 MVA refers to the Generator. Yet thenGenSet PF is 0.74 which is the ratio 2000/2700.

I don't understand how we can refer to electrical and mechanical power and then state that PF is 0.74 is the actual ratio of thise two powers. Sorry guys, I got a bit lost.
Thank you for your help. If you have/find videos on this subject, feel free to post the link. It helps me a lot to understand. An explanation would be highly appreciated too.
Mark

 

[ElectroChem]

Hi All,

During my exercises on paper, we always assume the source from grid has Tension and Currentnin phase.
Today, for the first time, I was learning about Diesel Genset and noticed its plate saying 2700 MVA / 2000 MW. So basically, somehow a source has its own Power Factor (PF).

1. Can anyone explain me how a source can have its PF besided the load it is attached to?
2. Can I correct my source (GenSet) PF so that I can maximise its utilisation and bring that to feed a 2300 MW load for example? In my mindset, if I improve the source PF to 1, I could actually power a load which could be 2300 MW or higher?

The only info I can find for point number 2 is that 2000 MW refers.to the Diesel Engine and 2700 MVA refers to the Generator. Yet thenGenSet PF is 0.74 which is the ratio 2000/2700.

I don't understand how we can refer to electrical and mechanical power and then state that PF is 0.74 is the actual ratio of thise two powers. Sorry guys, I got a bit lost.
Thank you for your help. If you have/find videos on this subject, feel free to post the link. It helps me a lot to understand. An explanation would be highly appreciated too.
Mark

It's not that this genset has a PF of its own, more that the nameplate is describing the nature of the load it can support.

This generating unit can drive a load of up to 2000MW (which btw is a MASSIVE diesel motor, like run-a-city-power-station-size, so I'm thinking 2700KVA/2MW is more likely). If you try and load it past this point, the prime mover will be overloaded, you will get frequency and voltage drops as it slows down.

The 2700MVA listed does correspond to a PF of 0.74, in this case it is saying that driving a highly out-of-phase load imposes extra burdens on the generator, so if you have a load drawing 2700MVA at 0.4pf then it could only be about 1000MW of true power. You can't push past the rated MW value, this reflects the capacity at unity pf.

To your last point, the 2700/2000 is not the ratio of mechanical to electrical power. Generators are rated by their electrical output, there will be generator/motor losses added to that to properly size the diesel motor. Reactive loads on the system are not true power, so they don't impose a simple extra load on the motor, it's more complicated.

 

[Mark_Djorn]

It's not that this genset has a PF of its own, more that the nameplate is describing the nature of the load it can support.

This generating unit can drive a load of up to 2000MW (which btw is a MASSIVE diesel motor, like run-a-city-power-station-size, so I'm thinking 2700KVA/2MW is more likely). If you try and load it past this point, the prime mover will be overloaded, you will get frequency and voltage drops as it slows down.

The 2700MVA listed does correspond to a PF of 0.74, in this case it is saying that driving a highly out-of-phase load imposes extra burdens on the generator, so if you have a load drawing 2700MVA at 0.4pf then it could only be about 1000MW of true power. You can't push past the rated MW value, this reflects the capacity at unity pf.

To your last point, the 2700/2000 is not the ratio of mechanical to electrical power. Generators are rated by their electrical output, there will be generator/motor losses added to that to properly size the diesel motor. Reactive loads on the system are not true power, so they don't impose a simple extra load on the motor, it's

Thank you for your quick and clear explanation.
Apologies and yes... it is actually 2.7 MVA and 2 MW. Great from you to spot that mistake.

Ok, from my understanding, the GenSet is rated at 2.7 MVA; whereas 2 MW is the rated power as maximum allowance to avoid an overload of the equipment, if the load's PF was equal or lower that 0.74 (2000/2700).

Question: my reading says that the load is not actually inductive, in fact it has a resistive nature and a capacitive component (because mainly electronics with capacitive component).

Now my next question is:

3.a
Since the load is resistive/capacitive, can I assume that I will not encounter that issue represented by the 0.74 PF?

3.b
Or perhaps, if the load is capacitive, I will have to consider a negative PF of -0.74 where the ratio remains 2000/2700?

4. Given the two options above, can the PF be corrected to make full use of the 2.7 MVA and therefore feed a resistive/capacitive load of 2.3 MW?

In my simplistic way to see things, if my load is resistive/capacitive (and let's consider the worst case scenario where the PF is -0.74), I could correct that by adding an inductive load (in series or parallel? Not sure) exactly as I would add a capacitive load in parallel to an inductive motor. Is that feasible?
Thank you again

 

[ElectroChem]

Thank you for your quick and clear explanation.
Apologies and yes... it is actually 2.7 MVA and 2 MW. Great from you to spot that mistake.

Ok, from my understanding, the GenSet is rated at 2.7 MVA; whereas 2 MW is the rated power as maximum allowance to avoid an overload of the equipment, if the load's PF was equal or lower that 0.74 (2000/2700).

Question: my reading says that the load is not actually inductive, in fact it has a resistive nature and a capacitive component (because mainly electronics with capacitive component).

Now my next question is:

3.a
Since the load is resistive/capacitive, can I assume that I will not encounter that issue represented by the 0.74 PF?

3.b
Or perhaps, if the load is capacitive, I will have to consider a negative PF of -0.74 where the ratio remains 2000/2700?

4. Given the two options above, can the PF be corrected to make full use of the 2.7 MVA and therefore feed a resistive/capacitive load of 2.3 MW?

In my simplistic way to see things, if my load is resistive/capacitive (and let's consider the worst case scenario where the PF is -0.74), I could correct that by adding an inductive load (in series or parallel? Not sure) exactly as I would add a capacitive load in parallel to an inductive motor. Is that feasible?
Thank you again

3 a/b - Generators hate leading (capacitive) power factors. The reactive power pushes voltage back into the windings, causing issues with regulation. You would absolutely have to take that into account.
That said, it's quite uncommon to have heavily leading loads outside of some very esoteric electronics, VSDs and power supplies usually try and modulate their pf close to unity.

4. A 2700KVA/2000KW genset is rated to 2000KW of real power at unity power factor, or at pf<1 as long as KVA does not exceed 2700KVA. There are no clever tricks to extract the 'full' 2700KVA, if you have a 2.3MVA resistive load then you need a genset sized to suit.

PFC loads are always added in parallel with what they're correcting. Not sure how much node analysis you've studied, but a perfectly balanced PFC (system load brought to unity) is operating in parallel resonance, with all the reactive power cycling from the PFC to load and back again so none is supplied/consumed by the source. Establishing a series resonant circuit on a 2700KVA system would be... spectacular... briefly.

 

[Lucien Nunes]

@ElectroChem gives some great explanations there. The key is that the real and apparent power ratings are dictated by separate considerations and both apply at once.

The generator's rated apparent power in MVA is limited by the alternator. The maximum allowable current will be the MVA divided by the rated voltage. If this is exceeded, either the windings will overheat or the regulation will fail, even if the MW real power is within spec. This might happen if the load pf is very low.

The generator's rated real power in MW is limited by the diesel motor. If you exceed the MW limit it might overheat, exceed emissions limits or fail to maintain speed, even if the MVA apparent power is within spec. This might happen if the pf is high.

The manufacturer will fit a diesel that can provide as much mechanical power as is needed to drive a typical load of typical power factor up to the MVA rating of the alternator. If you want a unit that will generate its rated MVA at unity pf, i.e. MW = MVA, you have to pay extra to get a bigger diesel!

 

[Mark_Djorn]

@ElectroChem gives some great explanations there. The key is that the real and apparent power ratings are dictated by separate considerations and both apply at once.

The generator's rated apparent power in MVA is limited by the alternator. The maximum allowable current will be the MVA divided by the rated voltage. If this is exceeded, either the windings will overheat or the regulation will fail, even if the MW real power is within spec. This might happen if the load pf is very low.

The generator's rated real power in MW is limited by the diesel motor. If you exceed the MW limit it might overheat, exceed emissions limits or fail to maintain speed, even if the MVA apparent power is within spec. This might happen if the pf is high.

The manufacturer will fit a diesel that can provide as much mechanical power as is needed to drive a typical load of typical power factor up to the MVA rating of the alternator. If you want a unit that will generate its rated MVA at unity pf, i.e. MW = MVA, you have to pay extra to get a bigger diesel!

Thank you both @ElectroChem and @Lucien Nunes . Your explanations were thorough, yet simple for me to understand.

An extra details I should have mentioned: the equipment in question is not a standard GenSet but a DRUPS. Perhaps this may change how the equipment would behave with capacitive load. However, I understand what you mean when saying that electronic components may have their own way to modulate PF.

Just to recap:
1. I have no way to improve my (seamingly) capacitive load PF close to unity in such conditions and pretend to be able to supply more than what the Diesel Engine is rated for.
2. 2000 kW is what exactly I can expect to supply (and no more, in general) unless if I upsize my Engine.

I would like to ask questions on power factor correction but it will be more appropriate for me to open a new topic. I would like to truly thank you for your passionate approach to this topicm you have both been very helpful.