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Resultant currents from Parallel Generators with different PF
7

Resultant currents from Parallel Generators with different PF

Resultant currents from Parallel Generators with different PF

(OP)
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.

RE: Resultant currents from Parallel Generators with different PF

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

RE: Resultant currents from Parallel Generators with different PF

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.

RE: Resultant currents from Parallel Generators with different PF

(OP)
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.

RE: Resultant currents from Parallel Generators with different PF

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:
  • Assumed pf
  • calculated load kW
  • calculated load kVAr
  • DG kW
  • DG kVAr
  • resultant kW
  • resultant kVAr
  • resultant current
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:
  • Assumed pf = 0.9
  • calculated load kW = 995
  • calculated load kVAr = 482
  • DG kW = 1316
  • DG kVAr = 192
  • resultant kW = -322
  • resultant kVAr = 290
  • resultant current = 360
Reducing the pf increases the resultant pf and eventually a bit of trial and error produces:
  • Assumed pf = 0.6684
  • calculated load kW = 739
  • calculated load kVAr = 822
  • DG kW = 1316
  • DG kVAr = 192
  • resultant kW = -577
  • resultant kVAr = 630
  • resultant current = 710
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.

RE: Resultant currents from Parallel Generators with different PF

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.

RE: Resultant currents from Parallel Generators with different PF

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.

RE: Resultant currents from Parallel Generators with different PF

(OP)
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?

RE: Resultant currents from Parallel Generators with different PF

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.)



RE: Resultant currents from Parallel Generators with different PF

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.

RE: Resultant currents from Parallel Generators with different PF

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.

RE: Resultant currents from Parallel Generators with different PF

(OP)
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.

RE: Resultant currents from Parallel Generators with different PF

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

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.

RE: Resultant currents from Parallel Generators with different PF

(OP)
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.

RE: Resultant currents from Parallel Generators with different PF

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.

RE: Resultant currents from Parallel Generators with different PF

I forgot the spreadsheet.

RE: Resultant currents from Parallel Generators with different PF

(OP)
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.

RE: Resultant currents from Parallel Generators with different PF

Quote (Bloozntooz7868)

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]

RE: Resultant currents from Parallel Generators with different PF

(OP)

Quote (crshears)

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?

Quote (crshears)

the 'power factor' of a system is determined by the loads it supplies
- is that not what we have been striving to work out?

Quote (crshears)

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.

RE: Resultant currents from Parallel Generators with different PF

Hello Bloozntooz7868,

Thank you for getting out your file; I stated clearly that I was awaiting sharpening if required, and I wasn't kidding.

I thought I made my point abundantly clear; but I've been wrong before, and I'm sure I will be again. And I wasn't going for aggressive at all...was I acting aggressively?

Moving on:

With respect, I submit that you are assuming facts not in evidence.

To begin with, I by no means thought that "everything that has been discussed before is nonsense" as you asserted; on the contrary I found the thread to be exemplary in the courtesy exhibited by all participants in checking and re-checking their spreadsheets and calculations to provide much valuable information and education to us all, including me. Indeed the thread was of such quality that I felt no need to chime in, since I couldn't have said any of what was said any better.

My issue was not with any of the mathematics in the thread since, as you have taken pains to point out, that is not my strong suit. Rather, I took issue with the wording in your last submission to the thread, since it seemed to suggest that after having participated in such an excellent circle of discussion you appeared to me to be straying into tangential thinking.

Someone once said, "Ideas have consequences." I agree; therefore I consider it of high importance to express oneself clearly and unambiguously.

If what you wrote wasn't what you meant...

As for positive contributions, I make them when I can.

Thanks for listening.

CR

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

RE: Resultant currents from Parallel Generators with different PF

(OP)
And I apologise for maybe being a little sensitive.

The analogy I made was just that - analogy. I see that it could be construed as me still not understanding that ways and the means of what is happening despite such valuable input, but I can assure you that is not the case.

Thank you for your input as well.

All the best.

RE: Resultant currents from Parallel Generators with different PF

Hello David, no apology needed; "sharpening" is not synonymous with sniping. smile

CR

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

RE: Resultant currents from Parallel Generators with different PF

crshears,

Bloozntooz7868's response on 22 Jul 14 10:17 made perfect sense based on the question that he posed and the answers that were presented to that question.

It would seem that you either misunderstood the question (reading comprehension) or you misunderstood the answer (engineering math comprehension).

In either case, I agree with bloozntooz7868's assesment that your response was aggressive, inappropriate, and completely lacking in contributatory substance.

RE: Resultant currents from Parallel Generators with different PF

Bloozntooz7868,

There is no need to apologize. You apparently understand, at least to some degree, the situation at hand. If you have further questions about the calculations then one of us in the forum will be glad to help. Like many things in engineering, it's all about the math.

RE: Resultant currents from Parallel Generators with different PF

Quote (rhatcher)

Like many things in engineering, it's all about the math.

...er...ahem...

...as politely as I can, I must disagree; though it's MOSTLY about the math, it's not ALL about the math.

I attempted to convey in my response to David that I was not taking issue with the engineering math involved, since I do not consider that my strong suit; and if it isn't considered arrant hubris to say so, I thought I got that message across quite clearly [textual composition].

In my view David's subsequent response showed that he grasped this [reading comprehension].

I also thought I showed my concern [questions of "tone" notwithstanding] that, based on the wording chosen, there might still be some miscomprehension occurring, concern that David also seems to have recognized and acknowledged...

A fellow operating engineer I held in high esteem once said, "It's all in how you explain it." I personally think he did a disservice to the math, but he did get his point across [conversational excellence].

CR

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

RE: Resultant currents from Parallel Generators with different PF

2
Many years ago, we tried to improve the power factor of one of two generators running in parallel.
By tweaking both the governor control and the voltage setting we eventually achieved 100% PF. Then we looked at the instruments on the other machine.
One machine was producing most of the kW and no VARs.
The other machine was producing few kW and all the VARs. The PF was extremely low. We never did that again.
Stepping back from the detailed calculations, look at the power triangles.
A normal power triangle shows the real power as the base line, the reactive power as the altitude and the apparent power as the hypotenuse. The real current, the reactive current and the apparent current are proportional to the length of the various sides of the triangle.
For example if your plant is running at 90% PF, and the base line is scaled as 100% then the the altitude representing the reactive power (or current) will be about 48% and the hypotenuse or apparent power or current will be 111%
(Draw a right triangle with a base of 100 and sides of 48 and 111.)
Now if the DG is replacing 80% of the real load and none of the reactive load we will have the following triangles:
DG, A straight line of 80% and zero altitude.
Turbine generators, The original base line of 100% becomes 20% The altitude remains 48% and the hypotenuse becomes 52%.
(Draw a right triangle with a base of 20 and sides of 48 and 52.)
Your Ammeters will be reading 80% from the DG and 52% from the TGs.
The ammeters will read apparent current or the hypotenuse value.
If you draw a power triangle representing the plant conditions and then subtract the DG contribution from the base of the triangle the resulting triangle should represent your readings.
I would suggest running at a PF closer to the plant PF, but not lower than 80%.
Note, if the DG produces more real power than the plant is using you may motor the turbines. Not a good idea.
"We have always done it this way:, may be paraphrased as;
"We have always done it wrong and we will continue to do it wrong!"

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Resultant currents from Parallel Generators with different PF

(OP)
Waross,

Quote (waross)

If you draw a power triangle representing the plant conditions and then subtract the DG contribution from the base of the triangle the resulting triangle should represent your readings.

I like the alternative way of looking at this very much - thank you very much indeed.

I have attached a hand-written pdf file depicting this. Looking at it this way makes me think how simple the original question was to answer (and why didn't I think of doing that...) - just goes to show, a picture is worth a thousand words.

What would you call the new angle created by the resultant current - leading?

RE: Resultant currents from Parallel Generators with different PF

Nice explanation Waross. LPS for you.

Bloozntooz7868, if you are following this then you are on the way to fully understanding the question at hand.

I want to add to Waross's excellent explanation by expanding the view. Most of the times that you see the power triangle presented it is the classic coordinate system with the origin at the bottom left, the X axis heading horizontilly to the right, and the Y axis heading vertically straight up. Picture an 'L' shape where the horizontal leg is the same length as the vertical leg. The Y axis represents kVARS and the X axis represents kW.

This is only part of the picture. The fact is that this is only one quadrant, or one quarter, of the complete power picture.

The entire picture, all four quadrants, looks like this: "+". Now, considering that the vertical axis is kVar and the horizontal axis is kW, we need to add '+' and '-' symbols to depict consumption or production of each type of power. For the vertical axis, kVAR, the '-' is at the top and the '+' is at the bottom. For the horizontal axis, kW, the '+' is at the left and the '-' is at the right. Finally, consider that 0 degrees is the right horizontal axis and the degrees rotate counterclockwise so that 90 degrees is straight up.

So, the top right quadrant, with '-' kVAR and '-' kW, is considered the 'motoring' region. The bottom left quadrant, with '+' kVAR and '+' kW, is considered the 'generating' region. The other two quadrants do not fit standard applications and are not defined by names. However, operation in those quadrants is possible and does happen. Your situation is a prime examplle. In the bottom right quadrant you are consuming kW and producing kVAR. In the top left quadrant you are producing kW and consuming kVAR.

The purpose of this depiction of electrical power is the vectorial summation of what is going on. Consider a simple situation where one motor is powered by one source. The motor will appear in the top right quadrant with the power triangle hypotenuse, the kVA vector, pointed to the top right. The generator will appear in the bottom left quadrant with the power triangle hypotenuse, the kVA vector, pointed to the bottom left.

In this case, the motor kVA vector and generator kVA vector will be equal in amplitude and exacty 180 degrees apart. The sum of power consumed and power produced will thus be zero. Clearly, if you were to draw the kW and kVAR legs of each triangle, they will also be 180 degrees opposite in angle and equal in amplitide, thus summing to zero. Power produced equals power consumed, as it should.

In your case, it is a little more complicated. You have switchboard loads in the top right motoring quadrant. You also have a DG essentially straddling the left two quadrants, pointing almost straight left, but with a power factor of 0.99 leaning slightly into the bottom left generating quadrant. Finally, you have the grid tie in the lower right quadrant, one of the 'unnamed' quadrants' ('+' kVAR / '-' kW').

When you add the three vectors together the sum will be zero. Of course, this all requires complex math with imaginary numbers. Vector math is in polar coordinates such as (100<45), (kVA amplitude and power angle), or rectangular coordinates such as (70.71 + j70.71), (kW + kVAR).

With regard to whether a power factor is leading or lagging, devices operating in the lower two quadrants, '+' kVAR, are considered to have a leading power factor. The top quadrants, '-' kVAR, are considered to be lagging.




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