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Here is a simplified 3P electric motor below:

3P motor

This motor is simply combined of three coils facing toward to the rotator (in the center) 120 degree apart to each other.

The user wants his rotator to rotate (real work expected from the motor). Rotation force comes from the magnetic flux. And magnetic flux comes from the current flowing through the coils. So magnetic flux has nothing to do with voltage, only current is its concern. Am I right so far?

If so, look at the graph below:

graph

This is an illustration of the power factor. Current being out of phase to the voltage. Think that I am applying 3 phase AC voltage (120 degrees apart from each other) to ports 1-4, 3-6 and 2-5 in order. In that case, current flowing from the AC voltage source, directly flow through the coil as expected. But, I am going to pay for only real power? Isn't it right? So why do I want power factor to be high?

Secondly, take a look at the equivalent circuit of an inductor(or coil whatever) below. Even if I pay for the negative power, still though, why do I want power factor to be high?

inductance real power

To make power factor high, I have to increase the series resistance on the inductor(or coil whatever), that means, since I can not increase or decrease the voltage, I will get less current to flow on coils. Therefore, I will get less amount of magnetic flux. Therefore, I will get less performance from my motor.

This is maybe a really dumb question. I am aware of that, but please, can someone help me to understand it? I am trying to understand this for almost a month!

Alper91
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    If the motor isn't directly connected to the grid, but is supplied by a inverter, you will need to match the inverter to the apparent power. These are usually used in industrial context, so it won't apply to you directly, except if you are into some forms of DIY. Some consumer devices have an integrated motor-inverter system, but you usually won't buy the inverter separately. As far as I remember there could also be problems when you have a motor on a long supply cable, need to look it up though. – WalyKu Jul 25 '16 at 07:22

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But, I am going to pay for only real power? Isn't it right? So why do I want power factor to be high?

If you are a small user you will only pay for real power. Your poor power factor will make little difference to the electrical grid. It is not worthwhile for the power company to meter you for the reactive power (kVArh).

If you are a large user and your power factor is poor you will cause problems for the grid:

  • Your reactive load will cause higher currents to flow than would flow for the same (real) power in a resistive load. This will cause increased voltage drop along the supply line. If you are only metered for real power (kWh) then the power company has to bear the loss.
  • To solve this a dual meter is installed for large users. This measures both kWh and kVArh. Typically there is no surcharge if the power factor is kept >= 0.95 but penalties are applied if too much reactive power is used.

To make power factor high, I have to increase the series resistance on the inductor ...

No, we can correct for inductive power factor by adding capacitance across the supply.

enter image description here

Figure 1. Adding power-factor correction capacitors can partially (as in this example), fully or over-correct inductive power-factor. Source: PowerFactor.us.

In industrial applications a bank of power-factor capacitors is typically installed at the main distribution panel. A controller monitors the incoming voltage and current, calculates the power-factor and switches in contactors to connect the capacitors across the supply. Since most industrial premises are three-phase the capacitors are supplied in delta configuration inside a three-terminal metal case.

enter image description here

Figure 2. An industrial power-factor correction bank with controller (mounted on door), fuses, contactors (relays) and capacitors. Source: Direct Industry.

Ricardo
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Transistor
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    Is it fair to say that for domestic use the overall capacitive and inductive loads *average out* (over a local area) to a reasonably high power factor? As opposed to industrial use where there might be many multi-megawatt electric motors fed from one substation. – Andrew Morton Jul 24 '16 at 19:53
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    I don't know for sure but I suspect it's traditionally been quite high because the bulk of the load was incandescent lighting and heaters (water, cookers, etc.) which are all resistive. SMPS are now common in lighting, and most electronics (TV, computers, etc.) hence the push by the regulating authorities (CE, etc.) for good power factor in those too. Industrial loads would include much more inductance in big and many motors and fluorescent lamps. – Transistor Jul 24 '16 at 20:00
  • Finally! I got it. Thanks. So, power factor being low only hurts power delivery companies. But, please let me ask another question. I've read that electric motors have lower power factor when their load is low. So I buy an AC motor and run it with no load at home, does it mean I steal from power company? – Alper91 Jul 24 '16 at 20:19
  • @Alper91 please ask new questions by creating a new question after doing some research. Comments aren't for asking new questions. – user2943160 Jul 24 '16 at 20:25
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    @Alper91 It's not "stealing" if they're not even trying to charge you for it. – user253751 Jul 24 '16 at 21:15
  • "It is not worthwhile for the power company to meter you for the reactive power" - why? Just measure RMS current and RMS voltage separately, then multiply and bill that? – John Dvorak Jul 25 '16 at 04:08
  • @ JanDvorak: No, that wouldn't work. Both RMS readings will always be positive. e.g., 230 V rms x 10 A rms tells you 2300 kVA but nothing about power factor. See [Power Factor Measurement](http://etc.unitbv.ro/~olteanu/Tehnici%20de%20masurare%20in%20tc/Power%20Factor%20Measurement.pdf) around section 40.4. You'd have to meter everybody to find the culprits. With electronic metering coming in we'll probably see more of it. – Transistor Jul 25 '16 at 05:20
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To answer the title "Why do we want higher power factor in ac motors?": if you are a domestic user you pay for W. If you are an industrial consumer you pay for VA.

For example, if you are a water supply company, the largest cost of your operation is pumping the water. If you can increase the PF of your electrically-powered pumps then you can save a lot of money.

Related: Why do power companies never bother residential customers about power factors?

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Andrew Morton
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  • So lowering PF is actually good thing? But I've read on so many websites that high power factor means high efficiency. So is a good thing. What do you say about that? – Alper91 Jul 24 '16 at 19:43
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    Thank you for pointing out the abysmal error in my answer :) Corrected. – Andrew Morton Jul 24 '16 at 19:47
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Low power factor means that the load is drawing a higher current than is necessary to deliver a given amount of power. Since the excess current is not delivering power, the only cost connected with it is the cost of losses due to the current flowing in the transmission lines and other distribution equipment. There is also an indirect cost due to equipment being used to deliver current without power rather than delivering current with power. Most of the disadvantages are felt only by the power supplier. For small users, the suppliers just lump those costs into the total cost of doing business and don't have a metered charge for it. However, we all pay for power company inefficiencies indirectly in the price of the actual power that we purchase.

Induction motors are designed to operate at as high a power factor as possible without making the motor too large and expensive or inefficient. Users are expected to deal with the resulting power factor in whatever way is appropriate under their particular operating conditions. For small scale users, that generally means that no action is required. For large scale users, that may mean installing power factor compensation capacitors at each motor or in banks in their facility.