Is this formula correct?

[widgitmaster]

Is this formula correct, electricity is not one of my strengths!

Back in 2000 I wrote a few hundred web pages for my computer club, and I had found this info online, assuming it was accurate! But have received an email informing me it is flawed! Please help!
http://bugclub.org/beginners/math/WattsVoltsAmps.html


Converting KW to AMPS:

Multiply KW by 1000/voltage and then by power factor
Example:
30KW multiplied by 1000v = 30,000
30,000 divided by 480 = 62.5 x .90 = 56.25amps
(for 3 phase divide by 1.73)



Eric

[Al_The_Man]

It is basically flawed because it does not differentiate between AC and DC or assumes a resistive load for some of the formula.
for DC watts=Ia x V, and will apply on AC if the load is resistive.
This is why the power rating for an AC inductive (or capacitive) device is stated in VA.
Because lets say with an inductive device on AC the Voltage and the Current may not peak, co-incidentally.
For example, say a perfect inductor where the current will lag the voltage by 90° and the voltage is 100v and the current 1 amp.
The apparent power is 1x100=100w but the actual power is 0.
The voltage is at peak when the current is passing through zero.
The power factor is the ratio between the actual power and the apparent power.
You need to know the power factor to calculate the actual power.
Al.

[widgitmaster]

I think I might have asked the wrong question, as I need a correct formula for the web page! As I stated earlier, I know little and made an attempt to write some web pages for the Computer Club, to bring them into the 21st century!

So if possible, could someone please supply me with a functional formula or a set for both AC & DC?

Eric

[pastera]

The use of power factor in your equation would compensate for the reactive components in the load.
The problem is when there is a large amount of harmonic content in the current (voltage and current wave shape differ greatly). The equation given would be inaccurate under this condition.

If this is for basic wire sizing, then the equations given should be accurate enough for most needs when the power factor is used. Do you really need to be accurate to a couple of percent for wire sizing when you should have safety factors of 1.5-2.5?

Aaron

[widgitmaster]

Ok, so this is still confusing, as AL states it is flawed, and ARON says is acceptable!

Here is the statement from the email I received:

I just wanted to point out an error I noticed in the conversion of KW to Amps (item number 6). The procedure states to multiply the KW*1000/voltage by the power factor, but it should actually be divided by power factor:

PF = True Power/Apparent Power = Watts/VA, thus Amps = Watts/(V*PF)

So, in the example, the current should be: 30KW *1000/(480*.9) = 69.4 amps

Thanks, again, for providing such a handy website. I also enjoyed the beginners section on computer hardware.

This emailer states the flaw is in the formulas math!

I'm now thinking the emailer is flawed!
Eric

[Al_The_Man]

I assumed you were refering to the page link, not the formulae per-se.
Unless you have a specific reason I wouldn't be too concerned with power factor for hobbiest or computer applications.
But the power in any phase, assuming, a balanced load =
the line voltage times the line current divided by the square root of 3.
For the three phases, the total power is three times the power in
any phase.
3 X the line voltage X the line current divided by the square
root of 3.
3 divided by the square root of 3 simplifies to just the
square root of 3.
Multiplying that, by the power factor converts volt-amps to watts assuming the power factor is other than one.
Once you get into unbalanced 3 phase loads its a whole new complicated ball game.
Al.

[kreutz]

Power factor(PF) = Real Power / Apparent Power

That is , PF= Power / V * I

If you want to find the Current then:

I = Power / (V * PF)

Power factor takes care of the AC /DC dilemma because on DC PF=1 (voltage and current are in phase)

[Al_The_Man]

BTW Power factor is a decimal number between 0 and 1, (one being ideal).
Al.

[ViperTX]

Well Eric.....after a bottle of wine....your first posting seems correct for a 90 percent power factor...