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.

 

[crshears]

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]

 

[Bloozntooz7868]

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.

 

[crshears]

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

CR

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

 

[rhatcher]

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.

 

[rhatcher]

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.

 

[crshears]

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]

 

[waross]

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

 

[Bloozntooz7868]

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?

 

[rhatcher]

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.