Resultant currents from Parallel Generators with different PF 7

[Bloozntooz7868]

I am trying to make sense of currents I am seeing on 2 parallel incomers to a switchboard. We have a 480V Emergency Board with a main incomer from our grid (made up of 9 Turbine Generators), and a Diesel Generator (DG) incomer with a rated power of 1310kW. The intention is that if normal power is lost the diesel Gen supplies the board to power essential supplies.
Operations perform a weekly test on the DG and run it in parallel with the main for a 4 hour load run. When in parallel the system is configured to run the DG at rated power (1310kW) with a PF of 0.99 (of course, when in stand-alone operation, it will supply the demand of the load). It does this automatically with a 2.5 minute loading ramp (soft load), and a 1.5 minute off-loading ramp. The system works without any flaws.

Observing the currents on both the incomer Cutler-Hammer circuit breaker (which have inbuilt Digitrip protection relays that show phase current) during the loading and unloading sequences, I see values that I am having some trouble understanding:

1) With the main breaker only closed (grid supplying the board), I observe phase current of 1330A on the relay. This more or less equals the summation of the outgoing load breakers (which also have Digitrip units). No problem there.

2) With the DG running in parallel I observe phase currents of 1600A on DG incomer (more than the load requirement), but yet 710A on Main incomer. It is this that I am trying to understand and I believe it is to do with the PF 0.99 that the DG is forced to run at, while in parallel with a system that has its own dynamic PF based on the wider load.

Some round figure calculations:

3) The phase current (via the main breaker) of the Emergency Board is 1330A. Neglecting any imbalance, and assuming a PF of 0.9 (we do not have the ability to check P, Q, S, or PF on the board) this equates to 980KW and 483KVAr actual load.

4) With the DG running in parallel, the DG incomer shows 1600A phase current at PF 0.99, which equates to 1317KW (which tallies with the rated output of 1310KW) and 188KVAr.

5) Again, with the DG running in parallel, the main incomer shows 710A which equates to 531KW and 260KVAr (this is my problem area)

6) The resultant of the load demand (from point 3) and the ‘supplied’ DG values (from point 4) is: 980KW-1317KW = -337KW and 483KVar-188KVar = +295KVAr

I would assume that the -337KW would be the real power export back to the grid, and the +295KVAr would be the required reactive power import from the grid (due to the DG running at 0.99PF and not fully supporting excitation requirements)?

Q1) So how come I am seeing 710A phase current on the main incomer breaker? This equates to (again using system 0.9PF) 531KW and 260KVAr (the Digitrip relays on the breakers do not recognise direction so I am not able to determine if this current value is import or export).

Q2) How can I determine resultant currents when I have an export of real power but an import of reactive power?

The only figures I see that are close are the 260KVAr main incomer reactive power calculation (from point 5) that is near to the calculated requirement of 295KVAr (from point 6) that the loads on the board need.

I may be looking at this all wrong so all help is appreciated, and thanks in advance for your time responding.

 

[davidbeach]

Assume that you have a load pf of 0.668 and try your numbers again.

 

[squeeky]

Yes David, hit the nail on the head. Assumptions are 99% of failure. Lighting loads, essential services, are notorious for having bad power factors. If you have modern electronic starters and LED lighting add some harmonics 5th and 7th to mess things around.

I would look at the capability diagram of the generator (Alternator). To have a machine made to run at 0.99 pf is not impossible but probably special built. Normally 0.8 pf is the normal area to design at. Running at 0.99 will give you max kilowatts but you will be generating at an under excitation to the duty point and at the point of importing vars if any inductive load switches out. I would say the Diesel Generator is being used to correct the power factor. If you work out the standard rating of the 1310kW to the alternator size you get 1323kVA and 1638kVA using 0.99 and 0.8 pf. It is more likely to be a standard size of 1650kVA unit. This would have about an 1850kW engine. A look at the engine rating may help.

Point 5. Your kVA cannot be smaller than your kW. Assuming you have them the wrong way around, the pf is 0.48. This is nearer the figure David gave you.

Have another look and come back.

Just a warning that most utilities that permit parallel connections without some sort of co-generation agreement would have some pretty stringent rules about connecting to the grid and then supplying to the grid. Your protection should disconnect you automatically for loss of mains. You don’t want to resynchronize onto an auto re-closer that is doing 3 or 4 switching operations. When you export you should trip your mains on reverse power. You are using energy and supplying it for free to the grid.

 

[Bloozntooz7868]

David / Squeeky,

Thank you for your replies.

I appear however to have not made it clear on the actual way our DG operates. When it runs in parallel with the main incomer, the AVR (type: Magnamax DVR2000C) operates in PF control mode and forces the DG to output at 0.99PF (as opposed to there being harmonics etc, that make it operate that way as suggested). It is deliberately set to run at near unity PF when in parallel, via a contact input to it. When it runs in island mode (after loss of the min incomer), the contact opens an then it operates as a stand alone unit supplying the load. For the complete picture, there is a Woodward DSLC Load Controller and Synchronizer which performs the overall management.
Another point is that we are not connected to a utility grid. When I talked about 'grid' I am referring to our internal network of 9 Turbine Generators.

The result is that in parallel operation we have the DG outputting 1600A @ 480V 0.99PF (1317kW / 188KVAr) which is more real power than the loads on the board require (980KW), but less Reactive power than the loads need (483KVAr), and it is doing all this in parallel with our 'grid' that is operating at the system PF (typically say 0.85-0.9). I am assuming that the excess real power would be exported via the main incomer breaker back to our 'grid', but that there would be reactive power imported via the main breaker to support the load????

This begs the question of currents. If I am assuming this P & Q flow is correct, then how do you make calculations on currents when there is a flow of KW in one direction through a breaker, along with a counter flow of KVAr??? My whole goal is to make sense of the currents I am seeing on the circuit breakers.

I hope I have explained this a bit clearer than the first time.

Thanks for your time.

 

[davidbeach]

No, I understand what you're saying about the mode of operation. It's just that your numbers don't support the assumption of a pf of 0.85-0.9.

Make a spreadsheet with:
[ul]
[li]Assumed pf[/li]
[li]calculated load kW[/li]
[li]calculated load kVAr[/li]
[li]DG kW[/li]
[li]DG kVAr[/li]
[li]resultant kW[/li]
[li]resultant kVAr[/li]
[li]resultant current[/li]
[/ul]
then vary your assumed pf.

I rounded things slightly differently than you and have the DG at 1316/192.

Assuming a pf of 0.9 results in:
[ul]
[li]Assumed pf = 0.9[/li]
[li]calculated load kW = 995[/li]
[li]calculated load kVAr = 482[/li]
[li]DG kW = 1316[/li]
[li]DG kVAr = 192[/li]
[li]resultant kW = -322[/li]
[li]resultant kVAr = 290[/li]
[li]resultant current = 360[/li]
[/ul]

Reducing the pf increases the resultant pf and eventually a bit of trial and error produces:
[ul]
[li]Assumed pf = 0.6684[/li]
[li]calculated load kW = 739[/li]
[li]calculated load kVAr = 822[/li]
[li]DG kW = 1316[/li]
[li]DG kVAr = 192[/li]
[li]resultant kW = -577[/li]
[li]resultant kVAr = 630[/li]
[li]resultant current = 710[/li]
[/ul]

But there are also a bunch of unsupported assumptions in there also; your DG may not actually be running with a pf of 0.99, your voltage may not be precisely 480 under both conditions; your current reading may be in RMS rather than just the 60Hz fundamental and harmonics will complicate the calculations, etc.

 

[rhatcher]

Bloozntooz7868,

Your assumption about the power factor of the grid breaker is incorrect. The power factor of the grid breaker is the power factor of the power flow through the breaker, not the power factor of the grid or the emergency switchboard to which it is connected.

In this case, the grid breaker, as seen from the emergency switchboard, is consuming KW and supplying KVAR. The power factor is, therefore, leading. A rough calculation (I do not have a proper calculator handy) indicates a power flow of about 590 KVA with a leading power factor of 0.9.

 

[davidbeach]

If the load power factor is in fact leading, the load pf would be about 0.853. Somehow, though, I think it more likely to be a bad lagging power factor than a leading power factor, but there is not enough information available to determine leading or lagging.

 

[Bloozntooz7868]

rhatcher thank you,

I was thinking it was along the lines of a PF problem but I was missing the 'how' to it all. The PF through the breaker makes perfect sense - you have cleared that up.

-----

David (thank you again),

I'm getting there, but I am having a little trouble understanding where your calculated resultant current came from (which would be the answer to my whole question) for your example at PF 0.6684

Going back to your previous example with PF 0.9:
resultant kW = -322 (this = -430A)
resultant kVAr = 290 (this = 800A)
Resultant current = 800+(-430) = 370A (same as your 360A in reality)

but using PF 0.6684 I get:
resultant kW = -577 (this = -1038A)
resultant kVAr = 630 (this = 1018A)
Resultant current = 1018+(-1038) = 20A (not 710A like what you put)

I feel a bit awkward about asking this as I'm sure it's very simple, but am I correct in calculating the resultant current with just addition like above, or is there something else to it?

 

[rhatcher]

DavidBeach,

Nice (LPS) job with your calculations! I agree 100% with your results!

(I started my post of 10:23 before your post of 10:11 showed up. Otherwise, I would have said this the first time.)

 

[davidbeach]

You can't add the real and reactive currents; they're at right angles to each other. Calculate kVA and the find the current. Or, use the currents but calculate the square root of the sum of the squares.

 

[squeeky]

Are you generating at 480v and stepping up via transformers to an MV bus? Are all the generating units close to one another? I agree that at your point of connection all the power factors will be the same (similar)but will differ at the point of generation. Why are you in pf mode? Are you using the alternator as a synchronous capacitor to correct your power factor? If not, why are you setting it at 0.99pf. Do you have access to your capability diagram? This will show you where is is designed to run. It sounds like your DG is very small compared to the installed capcity of your Turbine driven loads so it will not be able to make any difference to the voltage. I'm scratching here and learning as well.

 

[Bloozntooz7868]

David,

Thank you for your patience here, and can I ask that you stretch it a bit more... I think I'm having a case of "can't see the wood for the trees".

Could you please spell out for me your calculations on how you arrived at a resultant current of 710A when using 0.6684 PF. Nothing I'm doing is coming anywhere near that.

-----

squeeky,

We generate at 12.47kV and step down to the Plant requirements of 2.4kV and 480V. The four DG's we have are Standby Gen's (not for PF correction) and are small in relation to the 90MW TG capacity (as for the close part - well, they are all contained on the site). They supply the essential boards to bring the Plant up from black, and are exercised weekly and set to operate at 0.99PF - as for the philosophy to this, I have not been able to determine, because it was all done before I started here, and the grown-ups involved in this choice have moved on - it is a kind of 'this is what we have' scenario, but I have motivated an engineer to look into it for me. I have access the the capability diagram, but it is not something that is directly impacting the situation I'm questioning.

 

[davidbeach]

Oh, I don't know, maybe an error in my spreadsheet.

How about a pf of 0.826?

If you multiply the resultant kVA by 0.831 instead of dividing by 0.831 then you get 710A at .6684, but if you actually do the math right, it appears to be at 0.826 where the current works out to be 710A. Sorry about that.

To finish correcting the record, the leading power factor would be 0.9525.

 

[Bloozntooz7868]

David,

Thank you very much - my figures are now working out and, as near as doesn't matter, matching yours (I used 0.8275 for the load PF). A purple Star for you.

Thanks for you help and patience - much appreciated, and thanks to the other responders for their input as well.

 

[rhatcher]

Bloozntooz7868 (and DavidBeach),

I noticed the error in DavidBeach's calculations last night when I set out to explain to you how to calculate the resultant current from the kW and kVA. I had previously checked his results and declared that I agreed 100%. However, the one thing that I did not check was the current of the grid. That calculation did not prove to be correct. That result proves everything else wrong.

So, I have prepared a speadsheet that does calculate the answer to your question. The good news is that the power factor of your switchboard is much better than 0.668. There is no bad news except that I misled you for about 24 hours by agreeing to the previous result.

The speadsheet takes the inputs of load amps, DG amps, DG power factor, grid amps, and bus voltage. You vary the load power factor until the output indicates the grid current that you are measuring. In the case that you described, the load power factor is 0.873. The results follow and the spreadsheet is attached.

Load kVA: 1105.74
Load kW: -965.31
Load kVAR: -539.29
Load PF: 0.87
Load Amps: 1330.00

DG kVA: 1330.22
DG kW: 1316.91
DG kVAR: 187.65
DG PF: 0.99
DG Amps: 0.99

Grid kVA: 497.27
Grid kW: -351.60
Grid kVAR: 351.64
Grid PF: 0.71
Grid Amps: 709.89

I hope you find this helpful.

 

[rhatcher]

I forgot the spreadsheet.

 

[rhatcher]

Sorry. The attachment should be here this time. I will add that there was a cut and paste error in my previous post. The DG amps in my previous post should read 1600.

 

[Bloozntooz7868]

Thank you rhatcher - that is what I had as well. It has been a good learning exercise for me as I was initially very unsure how to approach combining supplies with differing Power Factors, and making sense of the results.

Like I say, thanks to everyone for their inputs - I have already passed this knowledge onto my peers, so many more people are benefiting from this than just me! A great example of what Eng-Tips is all about.

Ciao,
David.

 

[crshears]

I was initially very unsure how to approach combining supplies with differing Power Factors, and making sense of the results.

Are we starting around the mulberry bush yet one more time?

Perhaps I'm being persnickety, but according to my training and experience the 'power factor' of a system is determined by the loads it supplies - - or more precisely stated the vectorially summed power factor of all of the devices connected to that system and the devices comprising that system. [Among 'system devices' I would include cables and high-voltage circuits, since these can supply significant amounts of lagging reactive power.]

By this reasoning, there is no such thing as "combining supplies with differing Power Factors," since the power factor of each synchronous generator is determined by its field excitation, and induction generators absorb lagging VARs according to their design curves and the local sytem conditions. Indeed it can be very misleading to even speak of "combining supplies with different power factors," as if the power factor of a given machine is somehow inherent to the machine itself and not subject to control by means of excitation adjustment - - but it IS in fact adjustable, except in the case of the induction generator mentioned.

Awaiting sharpening if required...



CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[Bloozntooz7868]

Are we starting around the mulberry bush yet one more time?

- what on earth are you talking about? Since when have I ever had interaction with you about such things?

the 'power factor' of a system is determined by the loads it supplies

- is that not what we have been striving to work out?

no such thing as "combining supplies with differing Power Factors,"

- well I was simply making the analogy of two generation systems running in parallel with one of them forced to operate at a fixed PF.

What’s the point in entering a thread in such an aggressive way and then not actually make your point clear? If you think that everything that has been discussed before is nonsense then please educate us all and put your mathematical solution down. Don’t just fly in like a seagull, ‘dump’ on everyone with words, then fly away. Your own profile states “Strong ops exp; weak in plant design” – I would suggest using some of your self-confessed ‘strong’ side and providing positive input to the thread with an alternative solution as your goal. I’m all ears for learning.