208 3PH question

Yes, but !! Be careful in caculating VA from current measured with a clamp-on type of ammeter, though. In as much as E and I are out of phase, any leg current measured with clamp-ons will not be accurate. Full load current will be equal to 3 x that of current in each leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

Thanks.

In the formula above, Amps would be the circuit current. Full load current (as in FLA) is the same as the current on one leg, if we're talking about the same thing. The only way you can triple the amps in this case is if you multiply by 120, which makes sense with 208 wye since it could be supplying three 120 V circuits with a neutral. That math, however, doesn't work with other three phase voltages.

The formula as given is for the full load rated current of a 3 phase motor, either delta or wye; makes no difference. If one were to actually mesasure the KVA in each leg it would be 1/3 the KVA as calculated for full load. The math works with all 3-phase.

Bob Swinney

In the formula above, Amps would be the circuit current. Full load current (as in FLA) is the same as the current on one leg, if we're talking about the same thing. The only way you can triple the amps in this case is if you multiply by 120, which makes sense with 208 wye since it could be supplying three 120 V circuits with a neutral. That math, however, doesn't work with other three phase voltages.

KVA is not current, and the formula is not for current. Full load current is not equal to 3x that of current in each leg.

Hmmnnn - - what part of KVA did you find beneath your level of comprehension.? Actually the concept is fairly simple for most people.

KVA is not current, and the formula is not for current. Full load current is not equal to 3x that of current in each leg.

It's pretty funny how you keep pressing a point when it's obvious you don't have a clue. Your statement that full load current is equal to 3x that of current in each leg is dead wrong. Stick to your original statement instead of saying something else. How is full load current equal to 3x the current in one leg?

What's that Lassie? You say that ATP* fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Sun, 28 Dec 2008 21:47:26 -0500:

I just remembered that I will need to know the power factor to figure out what the watts used is.

Any way to do that without any fancy equipment? It probably changes as the load changes.

I'm trying to figure out what it costs to run one of the shops air compressors. They are 25hp screw type. The motor runs constant, and a valve shunts the output when pressure is reached. I measured about

10A unloaded and 15A loaded on one leg with a clamp on ammeter. I thought there would be more difference. Must be the power factor.

Apples and oranges. Watts = power = work.

VA is volts x amps but ELI the ICE man comes into effect.

IOW the voltage and current waveforms are not in phase causing power to be lower than just volts x amps. The amps that have to be carried down the utility are real to the utility so in an industrial setting VA is what one is billed at for demand.

In a resistive circuit V * A * 1.732 or so would be the right number for watts or VA.

Wes

A quick search on rotary air compressors turned up a 25 HP with an 18.5 KW motor, which would be over 51 full load amps per phase before efficiencies are even considered if my calcs are right. 15 amps would be pretty lightly loaded. Take a look at this chart from WEG:

They give efficiencies and power factors at different loads. I don't get involved with large motors like this very often, maybe Bruce can comment.

Wes, does any meter with current transformers measure VA as opposed to a smaller service with an inline wattmeter, which should measure actual watts?

One may look at 3 phases as originating in 3 separate generators. In fact, some early 3-phase was distributed over 6 wires to illustrate the 3 generator concept. Of course the voltages and currents in each phase (leg) were identical. The legs (phases) are 120 degrees apart. The formula for each leg taken seperately is KVA = (Volts x Amps) / 1000. The three legs combined at

120 degrees have KVA = (volts x amps x 1.73) / 1000.

Wes, does any meter with current transformers measure VA as opposed to a smaller service with an inline wattmeter, which should measure actual watts?

Home meters have often not measured true watts. The expensive ones do but they get close and don't play to much with phase angles.

Mart>> snipped-for-privacy@privacy.net (dan) wrote:

A watt-hour-meter measures true watts.

Some utilities charge power factor penalties.

I'd welcome a cite to same; with the tariff as well.

The Thomson moving coil watt hour meter, the one with the spinning disk, measures honest-to-gosh watthours. It's the gold standard of the industry.

Only recently have solid-state meters supplemented them, usually where time-of-day billing and/or remote metering is wanted. They can also read VARS I suppose; it's just a bit more firmware code. I expect they will someday replace all Thomson meters.

Outside of those, I have never seen a residential installation that even measured VARS, much less billed for them. Some do log a peak load reading. [They had a needle that was pushed up and had friction against falling back to zero...]

In some industrial installations, you can be charged a "power factor penalty" but I've never seen any recording metering of same; merely the utility tests it every so often. I suppose such is possible on the largest [steel mill, auto plant] consumers that buy at the 132KV and above level.

Some confusion may come from the well known fact that current and voltage are out of phase in an inductive circuit. As Wes pointed out ELI is the ICE man, a great memory aid which has voltage leading current in L; current leading voltage in C. Power (heat) developed in an inductor is the vectorial product of current and voltage. Power, however, is a sort of absolute quantity. Power does not add vectorially. If 3 equal quantities comprise the whole, then each is equal to 1/3 of the whole. If voltage is constant in 3 equal "packages" of power, then total power must be 3 x the current in each package.

Bob Swinney

It's pretty funny how you keep pressing a point when it's obvious you don't have a clue. Your statement that full load current is equal to 3x that of current in each leg is dead wrong. Stick to your original statement instead of saying something else. How is full load current equal to 3x the current in one leg?

I'm quite aware of the math involved. I'm not confused. Your statement that full load current is equal to 3x that of current in each leg is still wrong, no matter how you try to rationalize it, because we don't express current in a three phase circuit that way. If a three phase motor has a full load current of 30 amps, that means each leg carries 30 amps, not 10. It's a convention.

Your other assertion:

"Yes, but !! Be careful in caculating VA from current measured with a

is also wrong, because VA is not the same as watts. VA is apparent power and it IS measured using ammeters and voltmeters.

Apologies to Dan who started this long and annoying thread with an honest question.

The simple answer is No; it is not practical to measure current in the individual phases. This pulsating current cannot easily be measured. I was wrong to try to show that the current was 1/3 of rated load current. It can be seen, though, that Power in each phase must be equal to 1/3 total Power if the phases are balanced. Simple logic supports this.

Instantaneous current is another matter. In a 3-Phase motor the phases are separated by 60 electrical degrees. At any point in time instantaneous current will vary in accordance with which phase is being considered and with the point of time in the cycle. At say, time

0 of the sine wave, Phase 1 current is at zero crossing (0 degrees), thus its instantaneous current would be I total x sin 0. At this same time Phase 2 is at 60 degrees and its instantaneous current would be I total I x sin 60. Also at this same time Phase 3 is at 120 degrees and its instantaneous current would be I total x sin 120. Current of the 3 phases adds and the resultant could be stated: I instantaneous at time zero = I total x (sin 0 + sin 60 + sin 120). I instantaneous at any time = I total x [sin theta + sin (theta + 60) + sin (theta + 120)] These effects can be viewed by displaying 3 phases with a common time base on an oscilloscope.

Bob Swinney

I'd welcome a cite to same; with the tariff as well.

The Thomson moving coil watt hour meter, the one with the spinning disk, measures honest-to-gosh watthours. It's the gold standard of the industry.

Only recently have solid-state meters supplemented them, usually where time-of-day billing and/or remote metering is wanted. They can also read VARS I suppose; it's just a bit more firmware code. I expect they will someday replace all Thomson meters.

Outside of those, I have never seen a residential installation that even measured VARS, much less billed for them. Some do log a peak load reading. [They had a needle that was pushed up and had friction against falling back to zero...]

In some industrial installations, you can be charged a "power factor penalty" but I've never seen any recording metering of same; merely the utility tests it every so often. I suppose such is possible on the largest [steel mill, auto plant] consumers that buy at the 132KV and above level.

That will be a big surprise to electricians. It can and should be measured, using an analog meter or a digital meter that calculates the root mean square. AC voltage is also cyclic, does that mean we shouldn't measure it?

I was wrong to try to show that the current was 1/3 of

Yes, which was never in dispute.

None of which is relevant to someone attempting to estimate the power use of a compressor.