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Why are transformer ratings always given in VA and not in Watt?

Because they seldom power purely resistive loads. Reactive power is expressed in VA. This is to prevent over loading a transformer that drives say, rack of relays and solenoids which draw current that isn't necessarily dissipated as heat. The transformer has to supply it none the less and _it's_ resistive losses don't care about the phase angle. Actually, most AC loads have some reactance and should be expressed in VA. Heaters and light bulbs are about the only exception as nearly pure resistance.

Regards
cww

cww,

I have a silly question for you - it's really a moot point. Do you find that it's common practice (in your industries) to express reactive power in VA instead of VAR? Technically VA=Watt, right? But since everybody uses "W" as a unit, VA can be understood to be reactive. I've only been involved in a few power (generation) projects, but they all used VARs (Volt-Amp-Reactive) when describing reactive power. Anyone else have experience with this?

----
Nathan Boeger
Inductive Automation
"Design Simplicity Cures Engineered Complexity"

VA is total "apparent power" supplied. VAR, or more correctly, VAr (little case "r") is the reactive component of the apparent power; it is the imaginary power (power that does no net "work") supplied by the current which is 90degrees out of phase with the voltage. Watts is the real power delivered by the current that is exactly in phase with the voltage. As the phase angle between the current and voltate waveforms go from 0 (exactly in phase) to 90degrees (or PI/2 in radians) out of phase, then the Watts goes from maxmium to zero, while the VAr goes from 0 to maximum.

VA (apparent power), Watts, and VAr are related by pythagoras theorem:
VA^2 = W^2 + VAr^2.

Hope this helps.

That clears the notation up perfectly - thanks.

----
Nathan Boeger
http://www.inductiveautomation.com
"Design Simplicity Cures Engineered Complexity"

Responding to Anonymous' Nov 28, 9:14 pm comment.... I'm impressed:

You're one of few posters that has correctly applied the rules of capitalization, i.e., names of people!

Furthermore, you correctly use the lower case "r' to denote reactive power!

Then why did you not apply the capitalization rule to Pythagoras?

(Just messin' around)

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[tal-2@webtv.net] (Cepsicon@aol.com)

Except that the names of units named after people or derived from units named after people should not be capitalised when spelt out in full... The
abbreviations are capitalised.

So 50 watts = 50 W; 500 volts = 0.5 kV, 60 hertz = 60 Hz.

And before someone gets excited about nitpicking, just think of the difference a 50 mw generator and a 50 MW one!

Bruce

Thank you, that may be more informatiom than I need.

I have a Transformer 250KVA & wanted to know the Max current, or the real watts. I believe it's 1:2 ratio. With 120Vac applied to H1.3 & H2.4 windings in parallel. Secondaries X1, X2, X3, X4 in series.
110 Vac output + or -5 vac. I have not loaded it up as yet.

If you have time to maybe send me the math formula or example.

In the project, I am feeding this transformer with a Auto Trans. of 4.2 Amp.

Thank You from Larry
AgonyAt48 @ aol. com

(VA != W) except where (PF == 1).

VA = the vector sum of VA(reactive component) and VA(resistive component).

I'm going to sound like Phil Corso saying this, but there was a typographical
error in my previous post. It should have read:

W = the vector sum of VA(reactive component) and VA(resistive component).

Sorry, Michael, but the relationship between Apparent Power, Active Power, and Reactive Power is:

VA(apparent) = vector sum of W (active)+ VAr (reactive)!

Note: VAr could be inductive or capacitive!

Phil C.

In reply to Phil Corso - You are correct. I should have let my first version stand, as I replaced the wrong VA with W. The resistive component is the in-phase component, which is the same thing as the true power (watts).

Yes, it's fairly common and understood that VA could be reactive and W are resistive. I think VAR implies a known reactive load. Transformers especially simply use VA ratings, since they have no idea what you are going to do with them. I think power people have tended towards VAR since they started worrying about power factors and non linear loads. Actually, in power dist. these days there are very few loads that can be accurately described with either Watts or VA/VAR, loads are complex and non-linear and can have extremely high peak to average ratios with solid state ballasts and VFDs. We have large loads that have perhaps a 20 degree conduction angle at most.

Regards

cww

The voltage rating of a transformer has to do with the insulation strength - the current rating with the amount of copper present. These two are
independent. The VA rating of a transformer is a short-hand way of combining the two, also of use since there will usually be at least 2 windings involved. You can't operate a transformer winding rated at 400 V, 1600 kVA by applying 200 V and expect it to handle 8 kA.

Bruce

I hope this will help you:

Watt = Volt x Current x cos(phi)
VAR = Volt x Current x sin(phi)
**note :(where phi = phase angle between Voltage and Current)
and.
VA is vector sum of Watt and VAR,
->VA^2 = Watt^2 + VAR^2

thanks
FX Bambang

Responding to krish's Nov 22, 2:42pm query:

Generators, transformers and transmission-lines are in a class of electrical equipment normally rated in kVA rather than kW for the following reasons:

a) Losses and heating are usually determined by Voltage and Amperes, independent of power factor.

b) Their cost, as well as volume and weight, are generally proportional to their kVA rating.

Although not part of the original query, one can ask, "Why aren't motors in the same class?" They are, at least in the USA! Rated output Hp is
approximately equal to input kVA. However, motors built to metric-standards are rated in output kW!

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[tal-2@webtv.net] (Cepsicon@aol.com)

Transformers are rated in VA because manufacturer does not know the power factor of the load which you are going to connect. In case of motors manufacturer knows exactly the power factor of it at full load, that is why motors are rated in KW. So customer should not exceed the VA rating of the transformer For example: there is a 100KVA transformer and you can connect 80KW (maximum)@ 0.8pf motor to it.

Come back with any questions

Because the power is nothing but a multiplication of voltage & ampere and the major losses which occurs in the transformer are core loss and copper loss. Now, core loss is occurs due to voltage & the copper loss is due to current. Hence, the transformer rating is always given in VA not in Watt.

AC power (watts) is only equivalent to VA when voltage and current are in phase (power factor is 1.0). Since transformer losses are proportional to VA and not power, transformer ratings are given in VA. To give a practical example, if you had a very large capacitive or inductive load, you could over-heat the transformer while still consuming little or no power at the load.

Related question, I am computing power loading to my UPS and one of the equipment given only the secondary power rating. I wonder, can I assume that secondary power is about equal to primary power to the transformer? Or what is the formula or relationship of primary to secondary, given
Secondary 24V; 2A
Primary 120V

Hi guys,

Sorry this is a little off-topic but I thought that I could get the answer here.

For example, let's say that you hook up a 12v halogen light bulb up that uses up 50VA(W). Does this light bulb use up the same amount of electricty as a 220V light bulb at 50W would (in terms of the electricity bill that you receive)? My friend thinks so but I don't because I remember learning that transformers change the voltage from 220 to 12, and thus sending the wattage used in the opposite direction, which would equal 12*50/220=2 watts. Am I wrong, and is this the right way to calculate this? I would also appreciate the actual formula and a link to a site of this formula, thanks.

Sorry Anonymous, but your friend is right... VA remains the same. Your calculation is flawed, that is, you have erroneously multiplied the 12-V supply by the 220-V current. The 12-V current should have been 50/12=4.25 A! Thus, 12x4.25 = 50VA.

Regards, Phil Corso, PE (Cepsicon@aol.com)

50 watts is 50 watts. That is the amount of power that is being consumed. A transformer changes voltage and current, not power (other than you lose a small amount of power due to losses in the transformer). So:

50VA at 12V draws 50VA/12V = 4.17 amps (at 12 volts)

50VA at 220V draws 50VA/220V = 0.23 amps (at 220 volts)

Note though that a lot of household compact fluorescent bulbs (I am not speaking of halogen bulbs here) will show two "power" ratings. One is the rating for how much power it actually consumes, and the other is the power consumption of an equivalent incandescent bulb. A compact fluorescent bulb that is approximately equivalent in visible light output to a 60W incandescent bulb will actually consume 13 or 14 watts. The larger number (60W in this example) has nothing to do with actual power consumption though. It is simply present as a guide to help you to select the correct bulb size when replacing an existing incandescent bulb.

Why is it that people say transformers are rated in VA because they deal with voltage and ampere? All electrical devices deal with voltage and ampere (I thought everyone knew that).

Also why do some people say that VA is independent of the power factor? VA, the apparent power (modulus of the complex power), has a direct relationship with the (real) power and power factor (p.f).

Apparent power = S = ã3*V*I
Power factor = p.f = cosĮ

Real power = P = ã3*V*I*cosƒÆ

Replacing ã3*V*I in the Real power equation with S and cosƒÆ with p.f.

Real power = P = S*p.f
There's the relationship between the two. I'm not sure how you can talk about apparent power without power factor.

P.S. There's nothing with using VAR instead of VAr when referring to reactive power.

Because it is taken from this role p=VIcosQ