KW, KVA, KVAR relationship

[Chitina Jake]

I provide O&M for a small remote utility in Alaska. we are an isolated grid with less than 100 customers, running john deere 4045 gensets, averaging 50KW. I have noticed that we have a PF of leading .7 and I wanted to correct it. one of my objectives is to "sell" the correction to my board of directors, and I wanted to try to include a fuel savings as well as equipment longevity. I have been researching the relationship between KW, KVA, KVAR and PF and I had a question relating KVA and KW:

If I were to correct my PF to .9, would that equate to a fuel savings because the engine doesn't have to produce the torque to create the extra VAR's? or is the KVA/KVAR to KW relationship purely theoretical to describe the difference in phase angle between the voltage and amps?

If I reduce my VAR's does the engine produce my KW more efficiently, so it consumes less fuel? we average 3GPH and at $3.50/gallon of fuel, I am looking to reduce costs as much as possible to keep electrical prices low

 

[LionelHutz]

The prime mover load produces the kW. Changing the VARs produced doesn't affect the engine load.

Correcting the power factor could cause a reduction in kW due to less cable or transformer losses with the lower current, but it is a secondary effect. A bigger kW change might be possible by changing the system voltage.

 

[crshears]

Welcome to the forums, JC.

Original Poster [OP]: I have noticed that we have a PF of leading .7

My first questions are, "Measured where? And how have you determined PF is leading?"

Since you just joined and your profile doesn't tell us much about you, my response may be a bit pedantic, but I've learned to live with that.

As a retired power system / grid operator with 40 years of experience, I can state with confidence that in the utility industry, the convention is that inductive reactive power is drawn by motor and transformer loads; this means that reactive power demand and supply is always spoken of in terms of the supply of, or demand for, lagging reactive power.

I have always insisted therefore that one should not speak of a system's power factor, as PF is always equipment-specific; one should instead say that a system or part of it is either mainly inductive or capacitive.

I do not wish to offend, but I will say again, as I have written elsewhere, power factor is always associated with current flow at a particular point. So when you state that you have a leading power factor, I would expect to see that when two of your generators are supplying the load, with each unit being supplied with equal amounts of excitation current, if at that time both units' excitation systems are placed on manual and the excitation current of one is increased slightly, the stator current of that machine should decrease, since that unit will now be absorbing less kvars and therefore slide down the left slope of its V capability curve. The companion unit will correspondingly have its stator current increase, since it will now be absorbing more of the system's lagging reactive power, sliding further up the left leg of that unit's capability curve.

If, on the other hand, upon excitation increase that unit's stator current increases while its companion unit's stator current decreases, the power factor of both generators is in fact 0.7 lagging, not leading.

A way to confirm that your system is actually capacitive is that if there ever is a complete power system interruption and you are re-starting from scratch, once a generator is at operating speed and voltage, the machine's terminal voltage will rise when load is first applied. If instead the voltage drops and excitation must be increased to return voltage to its correct value, the system is inductive, not capacitive.

As to correcting power factor, the operator must always ask, "Why?" All that excessive current due to reactive power flows does is produce unnecessary heat in equipment windings, be they generator or transformer. As long as the currents flowing do not exceed the continuous capability curve of the machines or equipment in question, there is generally little justification for seeking to improve power factor.

Things can of course get much more complicated, but at the basics level, this should work for you.

 

[waross]

The low PF is caused by your customers.
To deliver 50 kW at 0.7 PF, your lines and transformers must carry the current 50/0.7 or 71 KVA.
Put another way, to carry 50 kW at 0.7 PF, your lines and transformers must be 43% oversized.
With the current at 143% of what it would be at unity PF, your line losses and transformer losses (I²R) will be 1.43^2 or 200%.
You cannot correct this at the source.
Options;
1. Institute PF penalties on your customers.
2. If your revenue supports your operating expenses, just live with it.
3. Identify your customers with the poorest PF and work with them to affect corrections.

The dollars and cents: Compare your total billed KWHrs with your total delivered KWHrs. How much does the difference cost you per month.
The actual cost of the poor PF may not be enough in real dollars to worry about.
The wasted KWHrs, use your cost of production, rather than the billing price.
A good rule of thumb is 13 KWHrs per gallon of diesel.
KWHr losses divided by 13 equals gallons of fuel approximately.
Come back with a description of your loads for further advice.

I don't usually mention my bona fides, but I was for a time the system engineer of a small system with about 5000 services.
After I had left the area, the Utility paid my plane fare and expenses to return from Canada to Central America to solve a billing issue.
I discovered a wiring mistake in a metering circuit that resulted in a large seafood processing plant being overcharged about $100,000 over the course of a year.
Yes, I used to do this stuff for a small diesel plant.

Has the ice gone out yet?
Many years ago, during my "yondering years", I canoed the Big Salmon River and spent some time bar tending at the Carmacks Hotel.
Not too far from you. grin

 

[Chitina Jake]

We do have a Voltage spike when we start from a dead bus. we only run 1 generator (117KW, unless the inrush load requires 2, then the secondary unit shuts down after a few minutes. Our power factor is measured both at the generator breakers, measured through our Woodward easy Gen 3000, and at the main bus through our totalizer. Approximately 2 miles of our transmission lines run underground inducing a capacitive load and, in the summer when our inductive loads are highest (Air conditioners, chest freezers, and so on), our PF improves to .78. I have Ignition software for monitoring and trending.

Each generator does have different voltage swings and react differently to sudden loads. 2 units are due for engine overhauls and all 3 generators are due for refurbishing. I have attached pictures of the Easy Gen info, generator data plate and switchgear.

Back to my original question,

How could I calculate an increase in efficiency? Could I correct to a .9 PF if corrected at the Transformer?

Here's what I have so far:

Given this snapshot @ 2.5 gallons/hour and $3.50/gallon, corrected to a 0.8 PF:

((58-(41.9/.8))*.8)/41.9)*2.5*24*365*3.5= $8,311 in fuel savings.

I know my math is off somewhere due to my lack of understanding in the relationships between KW KVA and KVAR, which was the purpose of my original post, and I do apologize for the confusion. I did a different calculation which I believe was correct but lost the sheet (I think my kids drew on it and threw it away). Please help correct my math. If additional info is needed, I probably have it in my trending software.

 

[Chitina Jake]

The low PF is caused by your customers.
To deliver 50 kW at 0.7 PF, your lines and transformers must carry the current 50/0.7 or 71 KVA.
Put another way, to carry 50 kW at 0.7 PF, your lines and transformers must be 43% oversized.
With the current at 143% of what it would be at unity PF, your line losses and transformer losses (I²R) will be 1.43^2 or 200%.
You cannot correct this at the source.
Options;
1. Institute PF penalties on your customers.
2. If your revenue supports your operating expenses, just live with it.
3. Identify your customers with the poorest PF and work with them to affect corrections.

The dollars and cents: Compare your total billed KWHrs with your total delivered KWHrs. How much does the difference cost you per month.
The actual cost of the poor PF may not be enough in real dollars to worry about.
The wasted KWHrs, use your cost of production, rather than the billing price.
A good rule of thumb is 13 KWHrs per gallon of diesel.
KWHr losses divided by 13 equals gallons of fuel approximately.
Come back with a description of your loads for further advice.

I don't usually mention my bona fides, but I was for a time the system engineer of a small system with about 5000 services.
After I had left the area, the Utility paid my plane fare and expenses to return from Canada to Central America to solve a billing issue.
I discovered a wiring mistake in a metering circuit that resulted in a large seafood processing plant being overcharged about $100,000 over the course of a year.
Yes, I used to do this stuff for a small diesel plant.

Has the ice gone out yet?
Many years ago, during my "yondering years", I canoed the Big Salmon River and spent some time bar tending at the Carmacks Hotel.
Not too far from you. grin

We have about 2 miles of underground, which I have been told is causing our capacitive load. in the summer, the inductive loads increase our PF to .78

Our current revenue barely breaks even, but I have overhauls I need to do, so I have been trying to "trim the fat". I have only been with the company for 2 years and have been trying to improve the efficiency, but that requires some purchases that we currently are struggling to afford, especially since many of our grants were revoked. we are primarily residential, with only a couple of commercial companies that operate in the summer and the DOT which operates year-round. so far, our operation isn't interested in metering/monitoring KVAR. I am trying to find any improvements that can affect a cost savings in 5 years.

The Ice is still here but melting quickly. The roads have melted, and we expect a toasty summer, maybe even in 90's

 

[crshears]

Just to help you double check your math, this page https://www.allaboutcircuits.com/te...ent/chpt-11/true-reactive-and-apparent-power/ gives a diagrammatic tutorial on real, apparent and reactive power.

Like Bill [waross] said, power factor is determined by the load, not the generators, and if you have cables producing excessive lagging VARs the only way to cancel these out is with appropriately placed reactors to absorb some of them.

If a generator is running solo, you won't have any reactive load sharing capabilities in play, and the generator power factor will be what it is. Again, re-balancing inductances and capacitances will have little if any effect on fuel economy; all you can do is reduce the unnecessary flow of reactive current so as to have the maximum real power transfer capability available.

 

[Chitina Jake]

Just to help you double check your math, this page https://www.allaboutcircuits.com/te...ent/chpt-11/true-reactive-and-apparent-power/ gives a diagrammatic tutorial on real, apparent and reactive power.

Like Bill [waross] said, power factor is determined by the load, not the generators, and if you have cables producing excessive lagging VARs the only way to cancel these out is with appropriately placed reactors to absorb some of them.

If a generator is running solo, you won't have any reactive load sharing capabilities in play, and the generator power factor will be what it is. Again, re-balancing inductances and capacitances will have little if any effect on fuel economy; all you can do is reduce the unnecessary flow of reactive current so as to have the maximum real power transfer capability available.

visited the website and I do have one question,
If I reduce the KVA that should equate to a reduction in amps, correct? if there is an amp reduction, shouldn't I be able to calculate an efficiency increase? the website has a lot to digest, which is why I am turning to this forum, so someone smarter can explain why an increase in PF has a negligible effect on fuel economy, preferably with numbers and how they interrelate. So far, I have only heard that it will have a very small impact on efficiency, but very little in the way of evidence I can digest. I have found lots of equations and theory's, but I am having difficulty determining where, when and if I should apply them. I have had a smattering of Trig 20 years ago so I know I can understand the results, but I just need some guidance

 

[waross]

What is your total delivered KWHr per week or per month. This is the total of all customer KWHrs as measured by the customer meters.
What is your total produced KWHr per the same period. This is the total KWHrs as shown on your generator or plant meters.
Subtract. The difference is mostly line losses.
Improving the PF will reduce these losses.
This is your main saving.
What type of meters are you using for your customers?
If you are using electronic meters, they may already be measuring customer KVARHrs.
This will help you to determine which customers are the worst offenders for low PF.

 

[waross]

Are you familiar with Pythagoras' Theorem?

W² + VAR² = VA²

kW² + KVAR² = KVA²

kWHr² + KVARHr² = KVAHr²

Power Factor = kW/KVA

41.9 kW, squared = 1755.61
40.1 KVAR squared = 1608.01
Sum = 3363.62 Root = 58 KVA
41.9 kW / 58 KVA = 0.722

 

[wcaseyharman]

Here’s how I think I would go about calculating losses and seeing if I could get some savings in energy:

Let’s think about the generator only (put a reactor off your switchgear) for simplicity.

This is how I would go about it:

Losses are I^R.
If you get the R off the generator datasheet:
Current is 72A. (From your picture)
Losses are 72^2*Rgen*3=current losses

(the 3 comes from resistance per phase)

Current at 1.0 PF (kva=kw, from the picture looks like you’re generating 41kW)
41kw/(480*sqrt(3))
~50A
50^2*Rgen*3= new losses

Btus/kwh=3412
Btus in a gallon of diesel: 138000

Savings per hour: ((Old losses - new losses)/1000)*3.50*3413/138000.

Assuming .02 ohms per phase (off a 250kw gen datasheet online)
Current losses - 311W
New losses - 150W
Savings- 161 W

I get about $0.01 per hour if I did my math right.

Edit: whoops, I didn’t account for engine thermal efficiency. Assuming the engine is roughly 25% efficient, that BTU/kW becomes 4*3412 =13,648.

So doing my math again that comes to about $0.05 per hour for the generator efficiency improvement, which is about $450 per year. There are a few assumptions in this math that would need to be tweaked for your application.

If you corrected to .9 your current would be:

41/.9=45kVA
45kVA/(480V*sqrt3) = 54A



And the same math as above to calculate losses.

If you move your power factor correction out toward the cables you can add the resistance of the additional equipment that sees a current reduction and calculate the losses in the same way.

-unrelated to your question-

Out of curiosity, have you compared where you are operating to the gen capability curve? Smaller gens often have really poor leading PF capability, you could be overheating the stator end iron without knowing it - the rating of .8 PF is almost always the lagging point, the corresponding leading PF could be much higher. I think my company did that carrying 10miles of 34.5kV cables many years ago on a 2.5MW gen and burned it up.

 

[edison123]

"If I reduce the KVA that should equate to a reduction in amps, correct?"

Correct.

"if there is an amp reduction, shouldn't I be able to calculate an efficiency increase?"

The reduced current means less I^2R loss in the generator, cables, transformer, which is indeed a power saving. You need to know the resistance values of the generator, cable and transformer (if any) to calculate the power saving.

If your long cables are causing a leading pf, then you need inductive var compensation at the farthest end to bring the pf close to unity. Since your pf is changing from season to season, this inductive var compensation should be automatically switchable on/off. Find out the cost of this inductive var compensation and offset it against your calculated power savings. You will most likely find it that it is not economical.

Bill - why are you shouting?

 

[waross]

Bill - why are you shouting?

Frustration.
Until the kW supplied by the generator is compared with the kW delivered to the consumers, we don't know if we are dealing with pennies, dollars or big dollars.
Actually I was trying to emphasize to the Original Poster the importance of finding out just how much dollar loss we are dealing with and whether it is worth correcting.
I encountered an extreme situation where a consultant proposed about $1000 to correct a power factor that was incurring a penalty of less than $10 per month.
Time to stop playing with percentages and get some real dollar numbers.

 

[crshears]

Hi JC,

The theory that follows holds up in real life.

The power triangle shows [1] real power, where voltage and current are in phase; the product of volts x amps equals watts, 746 of which make one horsepower, which involves doing actual work, which requires energy expenditure, in your case the burning of Diesel fuel, and [2] reactive power, where the voltage and current are 90° out of phase with each other. This means that when current is present there is no voltage, and when voltage is present there is no current, meaning no watts in either case, meaning no theoretical work, meaning no theoretical energy expenditure and therefore, again theoretically, no fuel burnt.

We know that in real practice the current in inductive devices and capacitors will never be exactly 90° out of phase, due to resistance; nevertheless, although there may be small amounts of fuel saved by getting generator power factor closer to unity, the savings are not typically significant, particularly when a cost/benefit analysis is performed.

 

[LionelHutz]

How could I calculate an increase in efficiency?

The prime mover (engine) produces kW. Changing the kVAR only does not change the kW load on the prime mover hence doesn't change the amount of fuel used. You can't change just the kVAR in practice, but assuming you could it doesn't affect the fuel used.

The kW is used by customers and by resistive losses in the system leading to the customer. The first can't be reduced unless the customer reduces their usage.

The total current is a result of the kW and kVAR draw by each customer and each component in the system. If the customer usage doesn't change then you can only reduce the kVAR part of the current in the system. Hence, you can only reduce the kVAR part of the resistive losses in your system.

The point where you correct for the kVAR is the point where reducing the resistive losses stops. For example, if you correct at your generator then only the resistive losses in the generator get reduced. If you correct at the customer then the resistive losses of the system all the way to that correction point might be reduced.

Customers typically cause a lagging power factor. You say there is a leading power factor due to the underground cable. That cable is probably correcting any customer lagging power factor. This means you're probably not able to do anything past the generator side of that underground cable. This leaves you possibly only able to practically reduce the losses in the components generator to the underground cable connection.

If you did want to correct the whole system, then you are likely correcting lagging power factor at the customers which would cause a further leading power factor on your generator side due to the underground cable which then means you're further correcting the cables. So, you'd be correcting the customers with capacitors and the underground cables with inductors.

Reducing 71A to maybe 50A at best appears that maybe you just won't be able to save much attempting to bring the power factor closer to unity.