Volt Ampes vs. Watts

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Wireit

Member
I am having difficulty in getting the volt amps for a copier machine. The name plate shows a 9 amp load at 120 volts. I believe one would take the amps times volts (120X9=1080) and get the watts. How does one get the volt amps for this copier?
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
It would be safe to use the Volts x amps = VA becasue the affect of pf will be low.

If you wanted to be conservative, use a pf of 0.85

Therefore: (120 x 9)/.85 = 1,270 VA
 

rattus

Senior Member
Wireit said:
I am having difficulty in getting the volt amps for a copier machine. The name plate shows a 9 amp load at 120 volts. I believe one would take the amps times volts (120X9=1080) and get the watts. How does one get the volt amps for this copier?

The apparent power is,

Pa = 120V x 9A = 1080 VA

The real power could be somewhat less,

Pr = 120V x 9A x PF Watts
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Simple thing to remember: since PF is always (equal to or) less than 1, so watts will be (equal to or) lower than VA.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
The issue here, and the reason we have two difference answers posted, is the nature of the current measurement. When you say that the current draw is 9 amps, you are describing the current that is associated with apparent power (VA), not the current associated with real power (watts). So the answer given by Rattus is correct. There is no need to account for power factor, if you are given current and voltage, and are looking for VA.
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
charlie b said:
The issue here, and the reason we have two difference answers posted, is the nature of the current measurement. When you say that the current draw is 9 amps, you are describing the current that is associated with apparent power (VA), not the current associated with real power (watts). So the answer given by Rattus is correct. There is no need to account for power factor, if you are given current and voltage, and are looking for VA.

And how is the answer I provided not correct?
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
kingpb said:
And how is the answer I provided not correct?
Because it took the value of VA, and divided it by some other number, and claimed to get VA as the answer.
kingpb said:
Therefore: (120 x 9)/.85 = 1,270 VA
The VA is the 120 x 9. The value of ?9 amps? is the current in the wire, the current that takes into account both real and reactive loads. We don?t generally speak of ?real current, reactive current, and apparent current.? But if we did, the 9 amps would be the ?apparent current.?
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Charlie, if you don't mind, I'd like to throw in one of my 'plain English' explanations. (I hope it's accurate!)


Volts is volts and amps is amps. The circuit supplies 120v, and the load current is 9 amps. The circuit doesn't care that, because of the reactive load, they don't peak at the same instant. The circuit voltage is still 120, and the conductors must still carry 9 amps; that's volt-amps.

A watt-hour meter, which is basically an electric motor whose speed depends on voltage and current, will spin faster when the voltage and current peak simultaneously than it will if the two are offset in time, even with the same voltage and current. It responds to watts.

Think of it like this: volt-amps aren't higher than watts because of poor power factor, watts are lower than volt-amps because of it. A watt-meter reads more and more inaccurately as power factor declines. Amperage is higher than a watt-meter indicates woth a poor PF.

Conductors have to be sized to carry both the usable current and the reactive current to deliver the former to the load. Power companies don't like low-PF customers because they have to build their systems to be capable of supplying both the usable and 'wasted' current.

A motor 'converts' watts, and not VA, into horse-power. If a motor has a poor PF, it causes more current to flow in its supply conductors than it is capable of turning into torque, so the conductors must be upsized for a given HP rating, and the POCO's charge more per measured watt/hr.

PF-improving measures don't eliminate reactive current, they merely localize the current to the conductors between the load and the PF-correcting device, and alleviate the rest of the supply system of the burden. That can end up costing less than the conductors would.

In up-sized-neutral MWB computer-circuits, the neutral 'shouldn't' be carrying any more current than the ungrounded circuit conductors, yet even with balanced loads, the neutral heats up more than the others do. That's current that is not being consumed by the loads.

If there's a low PF, the metered wattage would read to be a bit less than volts x amps, because the voltage and current don't peak at the same time, and the motor, i.e., the W/Hr meter, spins slower.

The point is that, when given the load's amp rating, the PF is not important. If it's given in VA, Kva, etc., then the PF matters for sizing the conductors, the service, etc. We have to pay to supply current we don't get any benefit from.

Gee, I hope this makes as much sense to you as I think it did to me. I'm open to criticism and corrections, as if I could stop you guys! :)
 
boy do I feel stupid!

boy do I feel stupid!

LarryFine said:
Charlie, if you don't mind, I'd like to throw in one of my 'plain English' explanations. (I hope it's accurate!)


Gee, I hope this makes as much sense to you as I think it did to me. I'm open to criticism and corrections, as if I could stop you guys! :)

Well, I call my self an "electrician", have been doing this for 15 years, and am planning to start my own "side" business with a friend soon, but reading all these posts from you older guys who actually LEARNED this stuff makes my head spin.

I REALLY need to get some theory and code classes under my belt. But in the meantime, thanks a bunch for all the great info you all post here. Couldn't ask for better "teachers"!:D
 

sreeja

Member
When the alternating current supply is rectified to obtain the direct
current needed by most electronics, the current waveform may be far from
sinusoidal, for example "square-wave"-like or a series of impulses.
Here the power must be determined by calculating the product of the
instantaneous voltage and current, and integrating the product over a full
cycle. Watts will generally not be equal to volts times amps.
 

chaterpilar

Senior Member
Location
Saudi Arabia
My two cents.....the answer is 120 X 9=1080 Voltamperes (Va) or 1.080Kva

Assuming, the supplier of the copier ( some rhyming..) had told you that the load is 1.080 kva then your system pf becomes relevant..and current drawn will vary depending upon the pf.

Larry,.. ^5 to your answer it was definitely plain English..
 

cadpoint

Senior Member
Location
Durham, NC
LarryFine said:
Charlie, if you don't mind, I'd like to throw in one of my 'plain English' explanations. (I hope it's accurate!)

...

If there's a low PF, the metered wattage would read to be a bit less than volts x amps, because the voltage and current don't peak at the same time, and the motor, i.e., the W/Hr meter, spins slower.

The point is that, when given the load's amp rating, the PF is not important. If it's given in VA, Kva, etc., then the PF matters for sizing the conductors, the service, etc. We have to pay to supply current we don't get any benefit from.

Gee, I hope this makes as much sense to you as I think it did to me. I'm open to criticism and corrections, as if I could stop you guys! :)

I think you did A fine explaination of what you said, I just want to argue with You and Charlie... Only cause I recall PF as something else...(just a little) and based on Google!, shoot!
http://books.google.com/books?id=8Wdcx-DwUrEC&pg=PA158&lpg=PA158&dq=pf+in+motor+rating&source=web&ots=DevOvLoVS_&sig=z4ooilOS7W55WXk6nL8yTEPopYk#PPA158,M1

this is based on PF in respects to motor rating or "pf in motor rating"
 

crossman

Senior Member
Location
Southeast Texas
LarryFine said:
Gee, I hope this makes as much sense to you as I think it did to me. I'm open to criticism and corrections, as if I could stop you guys! :)

I gotta take you up on this!

I have participated on this forum sporadically since 2005. I have learned that it is okay to disagree and even to argue to a certain extent if everyone keeps an open mind and is able to accept logic when things go badly. I know in the past, I have made some errors, and was glad when enough info was presented to make me realize my errors. If I am in error on this post, I want to hear what is right.

Larry, thanks for making the points you did, however I have 2 questions followed by my thoughts.

LarryFine said:
In up-sized-neutral MWB computer-circuits, the neutral 'shouldn't' be carrying any more current than the ungrounded circuit conductors, yet even with balanced loads, the neutral heats up more than the others do. That's current that is not being consumed by the loads.

I am thinking the "triplen" current is the culprit and the major factor in these cases. And that triplen current is absolutely being used by the loads. The reason it exists is because of the nature of the "switching power supplies" in the loads. These switching power supplies conduct only at specific points on the incoming sine wave voltage. And the problem is, with 208Y/120, when "A" phase loads are conducting, "B" phase loads and "C" phase loads have stopped conducting, so the current in the "A" phase loads has no choice but to go back to the source on the neutral. Then the same thing happens with "B" phase loads, and so on. So my conjecture is that the triplen harmonic is being used by the loads.

Some extra thoughts: The current on the neutral ends up being closer to a square wave with a frequency of 180 Hertz rather than a nice clean sine wave. And, with the square wave, the effective, or RMS current is greater than a sine wave current. The RMS current of a sine wave is:

Irms = Ipeak x .707

but the RMS current of the "near square wave" triplen current is probably closer to:

Irms = Ipeak x .9

It is absoultely possible for the RMS current in the neutral to be larger than the RMS current in any phase.

Question 2:

cadpoint said:
The point is that, when given the load's amp rating, the PF is not important. If it's given in VA, Kva, etc., then the PF matters for sizing the conductors, the service, etc. We have to pay to supply current we don't get any benefit from.

Isn't having the load's amp rating essentially the same thing as having the KVA since amps is converted to voltamps simply by multiplying by the voltage (and 1.73 if 3-phase)? Did you mean to say KW where you said VA and Kva above? If I am given the KVA of a load, I do not need to use power factor to compute the amps for the sizing of wires. The KVA takes into account the maximum current already. The KVA will never be lower than the KW, and in most cases, the KVA will be higher, so we better be sizing to the KVA of a load.

Now, if we were given the KW of a load that contained significant reactance, then we would have to take pf into account, but I don't know of any equipment containing reactance that would give the KW instead of KVA.

If the load is mostly resistive, the nameplate may indicate KW, and we still do not take into account the pf, say like a resistive-element water heater.

I do not know of any loads we would encounter where the manufacturer would put info on the nameplate where we would have to take the power factor into account to calculate the feeder or branch circuit load.
 

cadpoint

Senior Member
Location
Durham, NC
For The Record!

For The Record!

The PF is what the Windings with give you with applied Volts and Amps applied, be it 85% effient or 80% of the applied service.
 

jojo

Member
Location
Philippines
Occupation
Electrical Engineer
"I am having difficulty in getting the volt amps for a copier machine. The name plate shows a 9 amp load at 120 volts. I believe one would take the amps times volts (120X9=1080) and get the watts. How does one get the volt amps for this copier?"

since this is single-phase:
Real power(watts) = (line voltage)(line current)(power factor)
Apparent power (volt-amperes) = (line voltage)(line current)

They're right, real power value is always less than the apparent power value because of the power factor.
 
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