Power factor correction 1

[HamidEle]

The calculated capacitor to be needed to improve PF is 60kvar. But the closest standard size is 50Kvar. the next size is 75kvar. I am thniking of adding some heaters in the pump house to help improve the P.F. Is it workable solution? Is there any other alternative? We don't want to spend money on an automatic capacitor or a custom designed one, which is more expensive. I appreciate any inputs.

 

[davidbeach]

The heater will only improve your power factor by increasing the total load, certainly not an efficient solution. Pick a capacitor size and use it, you will either have more than the ideal correction or less than the ideal correction, such is life in the real world.

 

[itsmoked]

You don't really want a PF of one... How far will you miss PF=1 by if you use the 50Kvar?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[waross]

What is the monthly consumption of KWHrs and KVARHrs on your power bill? When doing a survey for a customer i request at least the last 12 month billing and prefer 24 months or more.
Are you in North America and are you penalized for a power factor of less than 0.9 or are you in a country or area where the penalty is less forgiving?
What is the nature of your load? Do you have a few large motors, lots of small motors or a mix of motors.
How many hours a day does your plant operate? How many days a week does your plant operate? Is there different loading on different shifts?
What type of plant do you have?
Do you have your own transformers?
Do you have any motors that typically run 24 hours per day?
These are all questions that I consider before applying the "Art" of power factor correction.
The science of power factor correction often costs a little more than the art of PF correction, but not as much as fully automatic correction.
After assembling the data I first ask;
How many KVARHrs per month do I need for 100% correction?
How many KVARHrs per month do I need for 90% correction?
How Many KVARs may I safely connect 24/7 and how many do I still need?
How many large motors can I correct to 100% with permanently connected capacitors and how many KVARs do I still need?
Are there any motors that run 24/7 and how much capacity may I permanently connect to them?
If there are a large number of smaller motors, I will consider correcting them in groups. A PLC may monitor the motors and connect a capacitor bank with a contactor whenever there are enough motors running to absorb the KVARs.
If the motors typically run together on a machine, I would not use a PLC but just use auxiliary contacts on the motor starters to pull in the capacitor contactor when all of the motors are running.
There are a few other tricks to the "art" of power factor correction.
The first rule of the 'Art" is do not do more than required.
If your penalty starts at 90% then correct to 90%.
If you incur a small penalty one or two months out of the year after correction, step back and look at the numbers. You may decide not to spend a few thousand dollars on added correction to save $30 or $50 a year in penalties.
If you care to provide some numbers and a description of your plant, we may make more specific suggestions.
respectfully

 

[wareagle]

Smoke posted "You don't really want a PF of one... How far will you miss PF=1 by if you use the 50Kvar? "

Question: Why not?

 

[waross]

Yes, in some parts of the world you will be penalized for any power factor other than 1.0 (100%), lagging or leading.
respectfully

 

[jraef]

Really? Wow. that seems draconian. Here in the US, the convention is to be as close as possible to .95pf, penalties for being at .99 would seem like a way of indirect taxation!

 

[wareagle]

The utility I deal with bills on KVA demand. So the closer to unity the better.

 

[dpc]

In the Northwest US, Bonneville Power's standard contract with local utilities includes a power factor penalty starting at 0.97, lagging OR leading (depending on time of day). There is also a 12 month ratchet so penalty is based on the worst power factor of the preceding 12 months.

Most local utilities pass this along to their customers.

 

[rmw]

Danged fine job waross. A star for you. I hope the OP gets back to you on this. The only thing that I didn't see you ask him is if there were any applications in his plant where he could utilize a synchronous motor(s) as part of the solution.

rmw

 

[Skogsgurra]

Hey!

I think that Smoked has a very valid point.

If difference between 50 and 60 kvar compensation is important, then the total load seems to be 1. very small, 2. very constant throughout day, week and year and 3. has a constant PF.

Most automatic (switched) compensation equipment have larger steps than 10 or 15 kvar. But it surely is possible to build equipment with smaller steps. Using one fixed compensation is a mistake. It is very difficult to imagine a load that is constant in both power and reactive power, as described above. Unless it is a process plant running an identical process for years.

Hamid: What is your power consumption (low/high) and power factor (also low/high) over the year?

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[HamidEle]

The power consumtion is 5kVA-15KVA and power factor is 0.90-0.99.The laods will be fed from 50KVA transformer, which is fed from 5MVA transformer. The 5MVA is almost unloaded.

The required Kvar is 55kVar-65kvar to achieve power factor of 0.90-0.995. If we use 50Kvar capacitor, the Power factor is below 0.90. If we use 75 Kvar, the power factor will be leading.

Skogsgurra, maybe use a fixed capacitors with small steps, like 10-15kvars.

 

[itsmoked]

Wow, someone cares about a power factor with a load of only 15kVA behind a 5MVA transformer.

Surprises me greatly.

Is this your 5MVA transformer?

Does the utility have a meter on the 15kVA side?

Are you doing this because of utility charges/penalties?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[HamidEle]

The utility has a meter on 25KV side, which is on the primary side of 5MVA transformer. 15kVA load is fed from 50KVA transformer. There would be a charge from Utility due to the low Power factor.

 

[davidbeach]

Deal with the power factor between the 5MVA transformer and the 50kVA. You also need to go back and check you numbers; there is no way that a 15kVA load could require 55-65kVAr to correct it back to a power factor between 0.9 and unity. Just can't happen.

 

[sslobodan]

Current power - P
your current power factor - k1
power factor you need - k2 (0.9)

Q=P*|{tg[arc(cosk1)] - tg[arc(cosk2)]}|

So recalculate like david said, because if you need 65Kvar for 15KV that means you have power factor around 0.2 now, and I don't know with what could you do that because it is almost pure inductive load. Must remind you that most of the engines have cosf around 0.8

 

[Skogsgurra]

Hamid,

I think that you should: A) either stop worrying or B) talk to someone that can explain this to you or C) dig out your text-books.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[HamidEle]

Remember that,when the 5MVA transformer is unloaded, the exciting current is 1.2%, which is almost purly inductive. That's why there is 65 kVars in the system, which will lead to lower Power factor of 0.20.

 

[davidbeach]

If you need to correct the excitation current power factor, you need to do that at the voltage level between the 5MVA transformer and the 50kVA transformer. Don't, DON'T, try to do that on the other side of the 50kA; you'll run into all sorts of voltage control issues.

 

[HamidEle]

I am just trying to correct the power factor between 5MVA and 50kVA transformer,which is 4160V level.