# Complex Power with power factor

1. Nov 4, 2013

### ken15ken15

By definition,Ia=(S/3V)*
I want to know if there is power factor, say cosϕ=0.8
Should Ia=(S/(3V*0.8)) or just Ia =(S/3V)*?
Because in the reference book, it sometimes consider the power factor but sometimes don't, I am quite confused with it.

2. Nov 4, 2013

### psparky

Ia=?
S=?
3V=?

Your question needs some theory infused to make sense.

The power triangle consists of three things. Watts, Vars and KVA

In the case of motors, Watts is the actual power delivered to shaft. (horizontal part of power triangle) KVA is the amount of power that it takes to get this power to the shaft (hypotenuse of the triangle) and the complex or reactive or imaginary power is the (vertical part of the triangle.)

YOu can see that KVA will always be the biggest number because it is the hypetunuse. If you put an ampmeter on a three phase running motor, you will take that amperage number, multiply by your voltage (480 for example) then multiply by the square root of 3 (1.73) to get your KVA.

Lets say you measured 52 amps on your motor....a 40 HP three phase 480 volt motor.

If you do the math: 52*480*1.73=43.18 KVA
If you divide 43.18 by 745 (745 watt per HP) you will get 57.9 HP!
But this motor only delivers 40 HP at the shaft!

So you can say it takes 57.9 HP of electrical power to produce 40 HP of real power to a shaft!

40 HP X 745 = 29.8 KWatts.

29.8 KWatts is the horizontal piece of triangle.
43.18 KVA is the hypotenuse of the triangle.

Right there you can get the power factor.....29.8/43.18 which is a PF of .7!

Or you can divide 40HP by 57.9 HP which is a PF of .7 as well.....same, same.

The amps of the power factor can be figured the same way, if know any of the two amperages, you can find the third. If you know the KVA or the Watts, you can figure the other one by the power factor. All related.

Vars is the reactive power ( or imaginary power....or complex power).
Vars is simply the horizontal piece of the triangle. You can use pythagream theory or simple trigonometry to figure Vars.

Although Vars are considered "imaginary", if you put 100 amps of imaginary amps thru a 12 guage wire, you will instantly smoke that wire. Amps are amps.

Capacitors improve power factor by reducing inductive power (from motors) Inductive power is a 90degree vector pointing straight down, Capacitive power is a 90 degree vector pointing straight up. By adjusting capacatince, you can "tune" that inductive power to near zero....and a power factor of near 1. Factories never quite go to PF of 1 because it induces harmonics to the system, but that's another story.

Capactors will reduce KVARS, reduce KVA and reduce power consumption. This is power factor correction.

It should also be noted that Heaters and lights are mostly purely resistive. Watts equals KVA and there is essentially a power factor of 1 in the case of purely resistive circuits.

Any questions?

Last edited: Nov 4, 2013
3. Nov 4, 2013

### the_emi_guy

Let's be careful not to imply that the power factor is introducing loss. Your example presumes we have a lossless system (1watt of electrical power becomes 1watt of mechanical power to the shaft).

Power factor does not introduce loss. In your example the electrical source needs to be able to cough up 57.9HP of *instantaneous* electrical power, while providing an average of 40HP which is delivered to the shaft.

Think of applying a voltage source across an ideal capacitor. We can measure AC volts and AC amps, multiply them and get a non-zero VA value. Power factor is 0, real power transfer is 0, losses are 0. The source is not sourcing any net power to the capacitor averaged over a cycle, but it is still being asked to supply instantaneous voltage and current.

4. Nov 5, 2013

The explanations of psparky and the the-emi-guy remarks are corrects, of course. I should add a remark about losses. [I am sure for psparky and the the-emi-guy nothing here it is new!]
For single-phase [or two-phase] induction motor the product I*V [where I it is the current the motor draws from supply source, and V it is the voltage measured at motor terminals] is not the actual power required from the supply system but an "apparent" power. [In D.C. system it is the actual, indeed].
In A.C. system the actual power- which the motor requires from the supplier is of course I*V*cos(I,V) [cos(I,V)=power factor p.f.]. By-the-way, in three phases system the apparent power is S=sqrt(3)*I*VL-L [VL-L= line-to-line voltage]- or S=3*I*VL-N [VL-N=line-to-neutral].
But still is not the actual power delivered by the motor at its shaft, since there are a lot of losses as in the stator winding, the magnetic circuit, the rotor winding [or squirrel cage], the mechanical losses-ventilation, bearings’ friction. The efficiency is the ratio of the “clean” delivered power and the required power["active"] from the electricity supply source.
A word about harmonics also. Usually formulae are referring to the first harmonics [fundamental].
However-and in our era more and more-due to use of inverter controlled driving the current and voltage reaches the motor terminals is full of harmonics which may reduce the power factor and rise the losses. There was an attempt to introduce another value VADistortiont[abandoned]. Briefly, there is S-apparent power [unit VA],P-active power [unit watt], Q -reactive power [unit VAR]
See:
http://www.lsczar.info/doc/31 What is Wrong.pdf
See Total Harmonic Distortion Voltage[THDv] also.
In complex numbers S=V*I* where I* is conjugate current .See [for instance]:
http://www.electrical4u.com/complex-power-active-reactive-and-apparent-power/

5. Nov 5, 2013

### psparky

Correction:

Vars is the vertical piece of the triangle.