Capacitor ratings and installation

Technical information on the installation of capacitors. Rule one: Call an electrician. November 29, 2000

Can anyone tell me how to determine the size of capacitors needed for a 100 hp 480 volt motor and how to wire it to correct power factor and lower demand charges?

Forum Responses
You should not do it by yourself if you are not an electrician. An electrician would first measure the actual power factor you are having when the motor is running. 100HP=75 kW (I suppose it is 3phase) and in case you now have p.f. lower than 0.7 you could connect ca 3x10 - 20 kVAr. The connection depends on your system connection (again the electrician should know--in delta or star and with corresponding voltage). And connect it over a switch after the motor is running, switch it off before the motor switched off. The more overdimensioned motor for the driven fan or pump, the lower (worse) p.f. you will get.

Capacitors will not improve your demand factor--there is a lot of bad info about this. Demand is measured over 15 or 30 minutes, so a small change for a few seconds will not change the demand. Just like turning motors on all at once or staggered--the demand is the same. There are only a few cases where capacitors are worth the money--seldom in wood products operations.

Gene Wengert, forum technical advisor

I do not know what is meant by demand factor, but once you dimension the capacitors correctly, you should reduce your bill since the current incoming from the power source to the connection point of the capacitors will be reduced.

This has 2 benefits:
1. Reduced current, hence reduced meter counting
2. Reduced loss in the incoming cable that now carries less current--for the unnecessary heat (energy) of the cable you pay for.

Contact your electric company and see what they have to say about capacitors in your application. I have yet to find a forest products company that can show a financial benefit in installing them. Their benefit is in reducing the power factor, but most utilities do not have a power factor charge (or it is very small). In any case, demand and overall consumption of power will stay the same.

Gene Wengert, forum technical advisor

We do pay $400-$850/month for power factor and $850/month for demand. The electric company reccomended that we get capacitors to reduce the power factor and demand. We contacted two electrical engineers and they both said that it would reduce the power factor and demand. The electric company even offered to finance it. They said that the savings will pay for itself in less than a year.

The electric company and engineers are also looking at putting in a waste burning power plant with us.

Your ultility is unusual with such a large power factor. Indeed, capacitors will reduce this charge; sounds like a winner to me; are you at the end of the line perhaps?

Gene Wengert, forum technical advisor

No. We are not at the end of the line but we have a copper end doging mill with two 100hp motors and we cut about 4 logs a minute with three to four passes per log. We are jaming the blades pretty heavy and this is post oak that we are cutting, which is pretty hard. I believe this is what has caused our power factor jump. We did not use it last month and the powerfactor charge dropped by $400. The engineer showed up today and we have a meeting in the morning with the electric company's rep. I guess in about two months we will know if it works or I spent some more money for nothing.

It is advisable that the capacitors are connected as much as possible near to the center of the "bad" load which causes the bad power factor. Then, the "bad" features of the load current will be compensated by the "good" current of the capacitors, so the incoming current from the source to this nod is cured/reduced, so no unnecessary cable heat, hence savings and no unnecessary voltage drop at the load.

A little side note here--residential power meters only read what is called "REAL" power, i.e. what you would have with a power factor of 0.0. If the voltage sine wave is perfectly in sync with the current sine wave, frequency wise, then you have 0.0 power factor. If the voltage wave leads the current wave by say, 10 degrees, then you have 17% power factor (cos(10)=0.17). If the current wave lagged by 90 degrees you would have a 100% power factor. Induction, which is present in all motors, causes the power factor to rise (become > 0). Capacitance in the system causes the power factor to decrease.

The reason utility companies don't like shops that have large motors is that they can cause their system power factor to increase, preventing residential meters from reading correctly. For residential customers, the larger the power factor is on the incoming line, the less your bill will be. Utility companies will provide large banks of capacitors around industrial complexes to correct the system power factor, or they will request the offending customer to do so, or they will charge more. Industrial power meters will correctly read power usage where large power factors are present.

By the way, the opposite of REAL power is called "IMAGINARY" power in engineering jargon. So with a power factor of 17%, you have 83% Real power and 17% Imaginery power. I didn't invent the term; that's just what they taught me in school. Also, AC motors generate more power as the power factor decreases. You don't want a power factor much over 20%. Ideally less than 10%.

I have to oppose you. The power factor, or the cosinus, can (almost) never be near 0 but 1 or less than 1. What you speak about is, in fact, sinus. But power factor is cosinus--ratio of the "real" power to the "imaginary" power. Sinus is ratio of the "reactive" power to the "imaginary" power--can be 0 in the best (non-inductive) case. Root of Cosinus=Root of (1-sinus^2).

The phase displacement angle (called "fee"--Greek letter) is the angle between the voltage and the lagged current. In the best case when no inductive load is present but pure resistance only, as e.g. lighting bulb or a heater, the current is in line with the voltage - no lagging--0 degree--the cosinus = 1 (sinus=0--no reactive power). The "real" power (kW) is equal to the "imaginary" power (kVA), because in the formula of real power output (the power we can make use of) of the consumer (1-phase):

P=Voltage(U) x Current (I) x cosinus

the cosinus is 1, hence does not change the result.

The same is valid for the 3-phase consumers, just multiplied by root of 3.

Once an inductive portion is present as it is the case at all motors, the current is lagging with an increasing angle behind the voltage, resulting in a lower (but worse) cosinus, coming under 1 to 0.9 and further down, so decreasing the real power output of the motor (in comparison with the imaginary power input). The motor pulls more current--refer to the formula above for:

I = P/U/cosinus.

The cosinus at rated power is to be found printed on the motor label and it is usually around 0.85 - 0.70. If the motor is running with a lower load then rated, or just without any load, the cosinus will be worse, hence lower. But also when the motor is overloaded then the cosinus goes down. When the rotor is locked, then it comes near zero (momentary case at the starting).

The reactive (inductive) power (kVAr) needed for magnetizing the motor can be compensated by another reactive power of an opposite vector direction which is done by a power capacitor (kVAr). The balance of this equation results in a higher (better) cosinus near to the 1 (usually not much better than 0.9). The supply utilities do not want to have it better than 1 since it would bring problems with stability of the system. Better than 1 means not over 1 but again under 1 with a minus prefix (e.g. -0.9) what means a capacitive cosinus. (By the way, in school I used to hear jokes that the Bolsheviks' artillery during their Great Revolution on Lenin's advice could fire with angles of cosinus larger than 1.)

So, the "imaginary" power of the feeding system is always increased by the bad (when not compensated) reactive power. This brings a need of a higher rating of feeding cables and transformer. That's why you can read the rating at transformer in kVA, which figure has to be always higher than the sum of all loads in kW, hence diminished by the resulting cosinus (power factor). A transformer 250 kVA will be fully loaded when feeding 200 kW of motors with power factor=0.8 (near the transformer). 1HP=0.75 kW.

The comments below were added after this Forum discussion was archived as a Knowledge Base article (add your comment).

Comment from contributor C:
I wish to correct an earlier post. Reactive loads do not cause any meters to read incorrectly. Residential meters only read *real* power - regardless of the power factor of the load. The reason utilities have power factor charges is that they have to provide that *reactive* power in the first place. To measure it requires different metering schemes which are not cost effective to put into every service.

*Reactive* power is very much a real quantity. It flows along the wire the same as the *real* power. The resistance of the utilities wire causes a voltage drop that increases as the current draw increases. The utility has to provide each customer all of the power they demand, within a specified voltage range. Therefore, this voltage drop has to be remedied. The wire also has a thermal limit on how much current can travel through it. The end result is that they charge the businesses that use the most *reactive* current (3 phase customers) a surcharge to offset the cost of supplying this current, and they try to create the *reactive* current as close to the load as possible (with capacitor banks) to limit the current flow over the entire line.

Comment from contributor A:

I am an electrician with a utility and we deal with power factor correction all the time. I run into problems with low power factors within the plant (.08 to .17) when using power monitors for loading purposes. The range seems small until the range of horsepower changes that the power factor causes is considered. For example, I have a 25 hp motor running a grizzley. The trip hp with a power factor of .08 and .17 is the differnce between a quick trip (.2 sec) and destroyed equipment. The higher power factor will trip the motor long before the lower power factor. .08 pf seems low, but it is confirmed by instrumentation. 2.53 kw or so equals 25 hp at .08 pf. My problem is that the power factor varies so much within the plant, that it is difficult to set the trip relays to avoid nuisance trips and still protect the equipment.