Power Factor 1

[astrolab2000]

In our plant, we operate a turbine-generator in over-excitation such that we generate a capacitive 8 MVARS.

Also, in parallel, we purchase from our utility 7.4 MWATTS. We have a power factor of 0.8 on this incoming line. Inductive load. So, Apparent power is 9.25MVA. Consequently, reactive power for this incoming line is 5.55 MVARS.

What is the TOTAL REACTIVE power of our system ?
8MVARS + 5.55MVARS = 13.55 MVARS ? or ( 8 MVARS - 5.55 MVARS ) = 2.45 MVARS capacitive

Thanks

Dan

 

[tinfoil]

If by TOTAL, you mean NET:

From the utility' point of view (POV), you are furnishing total net recative power of 2.45MVAR while consuming 7.4MW of real power, so you have a leading power factor of ~0.95, usually denoted as -0.95

 

[jghrist]

If your system is an inductive load to the utility and the utility furnishes 5.55 Mvar (we'll neglect any transformer reactive loss), and you also operate the generation over-excited and generate 8 Mvar, then your total reactive load is 13.55 Mvar.

 

[astrolab2000]

Hello,

Thanks a lot for your replies. However, I still have the same problem since some think that here the total MVARS will be 13.55, and some others think that total MVARS will be 2.55.

What I would believe is the following :

Assuming that S = P -jQ for capacitive and S = P + jQ on the inductive side, what I get, is the following.

From our first source of power, we have
S = 16.56 MVA
P = 14.5 MWATTS
Q = 8MVARS (capacitive since our generators runs over-excited)
Power factor = 0.875

From our second source of power (the incoming line from our utility).

We purchase P = 7.4MWATTS
Power factor = 0.8
S = 9.25 MVA
Q = 5.55 MVARS
Load of our mill is inductive

So, having
S1 = 14.5 - J8
S2 = 7.4 + J5.55

I would believe that

S TOTAL = 21.9MWATTS - J2.45

Consequentely, S TOTAL would be equal to 22.03 MVA
Overall Power factor = 0.994

Am I right to think that, or are some of my assumptions wrong ?

Please help

Thank in advance

Dan

 

[Marmite]

Assuming your generator runs synchronised to the grid, at the utility meter from your figures you are importing 5.55MVAr inductive, but this is despite generating 8MVAr of reactive power on site. This means your overall site requirement is 13.55MVAr inductive. Normally the utility meter is the most important point to measure the power factor/inductive demand because the utility will penalise you severely in the tariff for poor power factor.
Regards
Marmite

 

[jghrist]

The confusion is the sign of the reactive power at the generator. If the generator is over-excited, it is supplying reactive power to the system. Measuring power generated (into the system) would give P = +14.5 MW, Q = +8 Mvar. You are not generating a capacitive 8 Mvar. If you had an 8 Mvar capacitor bank, it would supply Q = +8 Mvar into the system, the same as your generator is doing. The term "generating a capacitive 8 MVARS" is confusing the situation.

With 13.55 Mvar, the plant power factor is 85% lagging, which is reasonable. With -2.45 Mvar, the plant power factor is 99.3% leading, which is not reasonable unless you have a bunch of capacitors in the plant.

 

[LionelHutz]

Your plant is at 13.55MVAR reactive and the generator is supplying 8MVAR of this which makes the incoming hydro system see the remaining 13.55MVAR - 8MVAR = 5.55MVAR of this reactive plant load. That would be the answer.

Adding capacitors to your system will remove some of the inductive load that the hydro system sees. You have to add the plant capacitor size to the inductive VARs of the hydro metering to get your plant VARs without the capacitor. In other words - take away plant capacitors and hydro meter reads a worse inductive power factor.

If your plant was at 2.45MVAR capacitive and the generator was supplying 8MVAR of inductive power then the utility would see -2.45MVAR + 8MVAR = 5.55MVAR inductive load. The big burning question though would be why run the generator at 8MVAR when running at 2.45MVAR would be perfect. Anyways, you said the generator is supplying capacitive VARs which immediately makes this wrong.

 

[astrolab2000]

Thanks to all of you, for your great help. you are so kind for having helped me. I have just discovered this web site some days ago, and really it is fantastic.

To be honest with you, I may miss some experience in the heavy industry, so I have been confused by some readings where it was written that kind of things :

SOURCE : WIKIPEDIA
"It is also possible to effect power factor correction with an unloaded synchronous motor connected across the supply. The power factor of the motor is varied by adjusting the field excitation and can be made to behave like a capacitor when over excited."

SOURCE : ABB
"Shunt capacitors, lightly loaded transmission lines and over-excited synchronous machines are capacitive loads and act as sources for reactive power resulting in a leading power factor"

Another source
"At very low power levels the synchronous motor can be made to look capactivie and can be used as a continually adjustable power factor corrector."


Then, with those readings, I mad a misinterpration and I thought that the MVARS produced by an over-excited generator was CAPACITIVE. But, this is not a good way to see the problem.

In fact, I understand now that in overexcitation, the generator will DELIVER VARS instead of consuming (in under-exitation).
So I must then add this VARS to the VARS delivered by the utility, and finally you are right, 13.55 MVARS would be the total MVARS delivered in the case I submitted to you.

Thank a million to all !

Best Regards,

Dan

 

[slavag]

Hi Dan.
Please pay attention on ABB and Wikpedia source:
synchronous motor ,not generator.
It's very popular in plants with big syn motors, used them as sync compensator and correct power factor.
More from this, I know one place with 3 big syn motors, them all times work in OE and utillities pay to this plant for export of MVAR's ( it's some week line).
Regards and Good Luck.
Slava

 

[slavag]

Sorry, weak line end.

 

[rmw]

I don't think it is because the line is weak. If it were weak, it would fall. It is because sending vars long distances over transmission lines heats the lines and increases the I^2R losses but if the vars can be created in close proximity to where they are needed for whatever it is out there that needs vars (motors need vars for example) they can save the transmission losses and run their generators way far away at better power factors and produce more real power.

The above said by a ME, now all you electrical guys jump in there and straighten me out.

rmw

 

[jghrist]

rmw,
Not all weakness is mechanical. A line could be considered electrically weak if it has large watt or var losses.

 

[slavag]

Rmw, you are right.
Syn compensators and capasitor banks used for "generated" MVAR's in long transmission line for:
1. Save losses ( as you said)
2. Increase voltage in "middle" of line.
or near to power station for save MVAR's generation in the power station
or as said jghrist : line electrically weak ( this is my case)
Regards.
Slava