Power draw on 208V single phase ckt served by 208/120V t

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dinos

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I have a 2-pole circuit breaker (in a panel fed from a 208/120V 3PH/4W transformer) serving a load which does not use a neutral (i.e. it is only fed with an A phase conductor and a B phase conductor only) and I measure 10A on each pole of this circuit breaker. What is the VA consumed by the load?

One part of me says its:
2*(120V*10A)=2400VA.

The other part of me says its:
208V*10A=2080VA.

My gut feeling is that if I were to connect a power analyzer to the panel main (with voltage leads on A,B,C,N and CT's on A,B,C), with only this one circuit in operation, it would report 2400VA as the power draw.

Can someone please offer their opinion on which is correct and why?

Thank You.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Consumed by the load, 2080VA, but the xfrmr thinks it is delivering 120x10x2, or 2400VA. If you are concerned about the xfrmr load, I would go with 2400VA. Whole nuther thread on this subject.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Originally posted by rattus:
Consumed by the load, 2080VA, but the xfrmr thinks it is delivering 120x10x2, or 2400VA. If you are concerned about the xfrmr load, I would go with 2400VA. Whole nuther thread on this subject.
Click to expand...
Yes, I'm interested. :) Why is this the case?
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Peter, the way the xfrmr sees it, it is delivering 10A @ 120V from two legs of the wye. That is 1200VA per leg for a total of 2400VA. The line to line voltage though is only 208V as a result of the vector algebra, so that 10A produces only 2080VA in the load itself.

Now, let the load be resistive. We have 2080KW dissipated in the load. But what about that 2400VA? Well, the currents in this case are displaced 30 degrees from the phase voltages. Then real power is

120 x 2 x cos(30)= 2080W.

The 1.154 comes from the inverse of cos(30). It is a matter of the phase angles involved.

I see no value in making up Byzantine formulas to solve dog simple problems as in the Oregon test. Sounds like a trick question.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Ok, I think I get it. This is the difference between true power and apparent power?

Right, wrong, not even close?
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Peter, my example demonstrates the effect of power factor on real power. PF is the cosine of the difference in phase angle between voltage and current. In this case it is cos(30) which is 0.866.

Apparent power delivered by each winding is simply,

Pa = Vphase x Iphase. (1200VA)

Real power delivered by each winding is,

Preal = Vphase x Iphase x PF. (1040W)

Now load current is in phase with the line to line voltage, so PF = 1 in the load itself.

Now connect this same load, 20.8 Ohms, (ignoring temperature effects), across 240V from a 120/240 service.

The current is now simply,

I = 240/20.8 = 11.54A and

Pa = 11.54A x 240V = 2769VA, and

Preal = 11.54A x 240V x cos(0) = 2769W (from the xfrmr and to the load)

Because now all currents and voltages are in phase.

The second example is simpler because it is a single phase scenario.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Hey Rattus,

How can the first example of a load on line A and line B of a 120/208 3 ph 4w be anything but a single phase load without having the other line C connected.

I see how the vector sum gives 208, but wouldn't an ocilloscope show one resultant sine wave with 208V rms amplitude.

In the 120/240 single phase example or in the centertapped winding of the 120/240 3ph 4w,also single phase, there wouldn't be any displacement of current from voltage, so no vector sum required. One single sine wave 240V rms amplitude.

Any comment?

Jim
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Originally posted by ccjersey:
Hey Rattus,

How can the first example of a load on line A and line B of a 120/208 3 ph 4w be anything but a single phase load without having the other line C connected.

Reply: It is a single phase load, but the xfrmr is 3-phase wye, and we are using two of the phases, say Va @ 0 and Vc @ -120. Then Vac is at 30 degrees.

I see how the vector sum gives 208, but wouldn't an ocilloscope show one resultant sine wave with 208V rms amplitude.

Reply: Yes, if you used a dual trace scope and subracted Vc from Va, that is, Vac = Va - Vc.

In the 120/240 single phase example or in the centertapped winding of the 120/240 3ph 4w,also single phase, there wouldn't be any displacement of current from voltage, so no vector sum required. One single sine wave 240V rms amplitude.

Reply: The two legs of the 120/240 service are 180 degrees apart which means the two windings in series sum up to 240V. No phase angles to consider unless you have a reactive load.

Any comment?

Jim
Click to expand...
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Originally posted by rattus:
Consumed by the load, 2080VA, but the xfrmr thinks it is delivering 120x10x2, or 2400VA. If you are concerned about the xfrmr load, I would go with 2400VA. Whole nuther thread on this subject.
Click to expand...
The 10A load in each transformer "phase" is not "in-phase" with either source. VA in the transformer will equal VA in the load.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Engy, how do you reconcile the fact that 208x10 is not equal to 2x120x10?

I claim it is because the load sees 208V while the transformer windings see a total of 240V while delivering 10A.

The tranformer rating is of utmost importance. You cannot afford to overload it. Would you treat it any differently if it were driving loads to neutral which provided the same current and phase? The xfrmr does not know the difference! Its iron and copper don't know the difference. It still gets hot.

Also, phase is not a factor in VA ratings.


I think we have stumbled onto one of life's great mysteries here.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

First, lets go back to dino's original question. If the metering is properly installed, it would still indicate a load of 2080VA.

The "physical" impact to the transformer is equivalent to 2400VA; 1200VA delivered from each of two windings.

Begin with assumption that the transformer output is actually rated in Amps rather than VA. Further assume that it was "fully-loaded" on all three windings at 10A. The total output would be 3600VA; each winding contributing 1200VA.

This is where RMS come in. The energy (heat) of a 10A output over indefinite time at steady state on each winding is equivalent to 1200VA at 120Volts. This is not a powerfactor issue but simply a current limit on the windings.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Bob,

I would have to say that 2400VA would be indicated if the meters measured phase voltages and phase currents.

In practice it seems that KVA ratings are really current ratings since we usually operate at a fixed voltage. Your opinion?
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Originally posted by rattus:
Bob,

I would have to say that 2400VA would be indicated if the meters measured phase voltages and phase currents.

In practice it seems that KVA ratings are really current ratings since we usually operate at a fixed voltage. Your opinion?
Click to expand...
As, I said if the metering were set up properly; i.e., phase to phase voltage reading. As stated, line to neutral would (improperly) indicate 2400VA.

Yes, most folks forget that kVA ratings generally assume a relatively fixed voltage. In reality, transformers are current limited.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

If this load was a 3 phase load connected from phase to phase as dino said, there would be no discussion. However the phase current in a delta connection would be Ia = 10 x 1.73 = 17.3 amps.
The power would be 17.3 x 208 x 1.73 = 6225 watts.
What we actually have is 1/3 of the 3 phase load
and the power is 1/3 of 6252 watts = 2075 watts.
A slight error from the 208 x 10 = 2080 watts.

Rattus posted
Now, let the load be resistive. We have 2080KW dissipated in the load. But what about that 2400VA? Well, the currents in this case are displaced 30 degrees from the phase voltages.
Click to expand...
There is nothing dino's post that indicates the power factor of the load. Therefore there is not a phase angle difference in the voltage and the current. The cos 0 degrees = 1.

Rattus Post
Apparent power delivered by each winding is simply, Pa = Vphase x Iphase. (1200VA)
Real power delivered by each winding is,
Preal = Vphase x Iphase x PF. (1040W)
Click to expand...
You are confusing delta and wye connections.
In a wye connection the power is Vphase x Iphase x Number of phases x PF. In a delta hookup the power is Iline x Vline x PF x number of phase to phase connections.

Rattus Post
I claim it is because the load sees 208V while the transformer windings see a total of 240V while delivering 10A.
Click to expand...
How can the transformer voltage be different than the voltage at the load?

[ March 03, 2005, 05:36 PM: Message edited by: bob ]
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Originally posted by bob:
There is nothing dino's post that indicates the power factor of the load. Therefore there is not a phase angle difference in the voltage and the current. The cos 0 degrees = 1

Reply: Bob, in my post I set the condition of a resistive load. That sets the phase angles, and the phase angle of the load current is that of the line to line voltage, which leads Va by 30 degrees. The PF in the two legs of the wye is cos(30) or 0.866. The PF in the load is cos(0) or 1.0.

You are confusing delta and wye connections.
In a wye connection the power is Vphase x Iphase x Number of phases x PF.

Reply: Bob, that is what I did to compute the power delivered by Va and Vc. That is 1040W per phase which adds up to 2080W in the load.

How can the transformer voltage be different from the voltage at the load?

Reply: Not sure what you mean. The phase voltages of a wye are smaller than the line to line voltages. This is a 120/208 wye.[/QB]
Click to expand...
The point is that the individual transformers are delivering more apparent power than real power because there is a phase difference between Vphase and Iphase.
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Rattus Post
Reply: Bob, in my post I set the condition of a resistive load. That sets the phase angles, and the phase angle of the load current is that of the line to line voltage, which leads Va by 30 degrees. The PF in the two legs of the wye is cos(30) or 0.866. The PF in the load is cos(0) or 1.0.
Click to expand...
You are incorrect. The fact that there is a 120 degrees seperation does not mean that the voltage and current is out of phase by 30 degrees. In a resistive load the current and voltage are in phase. If the load has inductors or capactiors the voltage and current will be out of phase.

Reply: Bob, that is what I did to compute the power delivered by Va and Vc. That is 1040W per phase which adds up to 2080W in the load.
Click to expand...
Thats not what you said. You said
Peter, the way the xfrmr sees it, it is delivering 10A @ 120V from two legs of the wye. That is 1200VA per leg for a total of 2400VA.
Click to expand...
This is a phase to phase connection and not phase to neutral.

Reply: How can the transformer voltage be different from the voltage at the load?

Reply: Not sure what you mean. The phase voltages of a wye are smaller than the line to line voltages. This is a 120/208 wye.[/QB]
Click to expand...
I am replying to your statement
Engy, how do you reconcile the fact that 208x10 is not equal to 2x120x10?
I claim it is because the load sees 208V while the transformer windings see a total of 240V while delivering 10A
Click to expand...
The voltage at the load and transformer is 208 volts less any VD. Its not 240 volts.

[ March 03, 2005, 08:20 PM: Message edited by: bob ]
 
Re: Power draw on 208V single phase ckt served by 208/120V t

Bob and Engy, let me start over.

Assumptions:

Wye phase voltages:

Van = 120 @ 0
Vbn = 120 @ 120, unloaded
Vcn = 120 @ 240

Line to line voltages:

Vac = 208 @ 30, 20.8 Ohm load
Vba = 208 @ 150
Vcb = 208 @ 270

Wye phase currents:

Ia = 208 @ 30/20.8 Ohms = 10A @ 30
Ib = 0
Ic = -10A @ 30 = 10A @ 210, return path
In = 0

We are dealing with one current and three phase angles. Only one voltage can be in phase with the current.

I am saying that the phase currents and phase voltages in this wye are separated by 30 degrees. That is to say, the PF in the xfrmr windings is 87%. PF in the load is 100%

Draw out the vector diagrams, and you will see.

[ March 03, 2005, 09:52 PM: Message edited by: rattus ]
 
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