Power factor correction from generation stand point 1

[sparky1976]

Dear Forum,

I just got question from the operation people on How the power factor correction in the system can calculate in saving of fuel ?
I could not give them direct answer, my answer was, the power factor correction could reduce the amps flow in equipment (generator, trf's,transmission line)and because of that less heat generated and make the life expetancy longer.

Could any one enlighten me the relation/equation from the primemover (the mechanical energy) to electrical energy generated from generator ?

I still confused because the mechanical energy use watt and reactive energy used var (vars that we want to reduce from power factor correction),can any one suggest a book on this matter ?

What are the advatages and disadvantages of capasiotr bank to the system (its a Island with average 100 MW load and average Pf 0.86) we want the pf at 0.90.

Regards
Pitat

 

[Alex68]

Hi yamin
Consider a 1ph system.
apparent power is A=V*I=(P^2+Q^2)^1/2

Reducing the Q flow, you reduce the flowing current
I= P/(V*cosPhi) where cosPhi is the power factor.

If you increase the cosPhi with constant P and V, then the I decreases.

The Joule losses are L=R*I*I, that is losses depend on the I squared!
A deltaI=-10% means a deltaL=-20%
in this way you are saving primary energy value equal to deltaL

hope this helps you

 

[electricuwe]

With a better power factor you reduce losses in the lines, the transformers and generator windings. How much you really can save depends on you system:

You will have significant savings in a rural power system with long lines, small transformers and generators.

In a system where the load is directly connected to a large synchronous generator it might be the most economic way to supply the reactive power by the generator.

 

[jbartos]

Suggestion: The X MW rated generator will need X + loss rated prime mover. If the power factor reduces loss that is one way to save energy and the production cost. Another way is to deliver to customer electrical energy with as high power factor as possible. This means that the voltage drop across the transmission line will be smaller and the customer will have higher input voltage to its loads. Once the loads have higher voltage, they will draw more current and the kWhour meter will record more kWhours, which in turn brings better income to the electricity producer.

 

[cuky2000]

Dear Yamin:

Improving the power factor will reduce reactive power since the short interconnected line and transformer are mainly reactive power components.

In reference to your question “How the power factor correction in the system can calculate in saving of fuel?” I believe that there is not significant fuel saving improving the power factor from the generator side because the fuel consumption is proportional to the mechanical output on the prime mover that equalize to the active power delivered by the generator plus the losses.

The generator active power will be the sum of the active load producing energy plus the heat generated in the resistance of the short interconnected line (inland operation) and transformer (if any). Assuming constant load, the heat difference for two-power factors scenario are small compare with the total power delivered.

With constant load, the Improvement the power facto from 86 & to 90% will have tendency to increase the supply voltage, but not significant fuel saving in the prime mover. Witch is approximately constant for practical purposes.

 

[Bung]

Further to jbartos post, nobody can deliver energy at a high power factor - the power factor is entirely load dependent, ie the load sets what the pf will be.
Bung
Life is non-linear...

 

[sparky1976]

All,
Thanks for the suggestions, but nobody mention about the highs and lows of capacitor banks. I know that It can cause volatge surge but is there any lows of capaciotr banks ?

Thanks
Pitat

 

[infomaniac]

Yamin, capacitors will reduce the overall current as long as they are not oversized to cause a leading power factor (at which point the current will begin to increase again). You do have to be careful on the amount of capacitance you add to a generator. In general, you do not want to add more capacitive impedance than the synchronous reactance of the generator (x''). There is a book called Power Capacitor Handbook by T. Longland, TW Hunt, A Brecknell that covers this issue in depth.

 

[electricuwe]

Capacitors may suffer from harmonic voltages. If your system contains a significant amount of nonlinear load, such as converters, you should only use capacitors with series reactors.

See

http://www.elma.olsztyn.pl/english/index2.htm

for choosing a MV capacitor bank.

 

[sparky1976]

Infomaniac thx for the book info,
Electric cuwe we have 4 converters for 4 8 MW sync motors.

For infomation we donot sell electricity to public, we use for our own production.

For cuky2000, please correct me if my assumption is wrong.
Here's the assumption:
I have 100 MW load with pf 0.85 ,the loads are motors so its inductive.
my apparent power 100MW / 0.85 = 117,6 MVA
if I can get the pf to unity so my real power = apparent power 100 MVA. so the difference in apparent power is 17 MVA.
I assume the prime mover to produce this 17 MVA difference also need fuel.
So the price for 1 MVA is 100 USD (diesel fuel).
The saving calculation go like this:
17MVA X 24 hrs X 365 days X 100 USD = 14892000 USD/year.

Again please correct me if the my rough estimation is wrong.

thanks
Pitat

 

[bigamp]

Yamin,

Your assumption is not correct.

The diesel fuel goes into producing MW load (active power) not reactive power.

Apparent power (MVA) is the square root of the sum of the real power squared and the reactive power squared. If you put in capacitance you will reduce the reactive power, but not the active power hence the MVA apparent power will be reduced but the fuel usage will not.

If you improve the power factor of your load, you will find that the generator excitation does not need to be so high to maintain voltage. Since the generator excitation power ultimately comes from the prime mover, there will be a tiny reduction in fuel usage, due to less generator excitation being required, if you improve your load power factor.

Electricuwe, great website reference for MV power factor correction.

Regards