Wattage calculations using only two phases 208

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aeroguy

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I am embaresed even to have to ask this question, however I apparently dont know and all peers seem to have a different opinion. The more I think about it, the more I become unsure.

I am trying to measure total wattage of a strip heater. The strip heater is attached to two phases of 208V. Amperage drawn on both legs is appox. 7A.

Should the formula used be;

1) I (A phase) + I (B phase) X V

2) I (any phase) X V X 1.73

3) I (any phase) X V

4) I (any phase) X V X PF X 1.73 (and if so how do you calculate the PF on a heater?)

Please help.
 
In a purely resistive load the power factor is 1. So in your scenario Watts is simply Volts across the load X Amps through the load (Case 3). And while the current in both legs is approximately 7A, they better be identical.
 
The fact that the heater is connected between two phases of a 120/208 volt system is not relevant, all that matters is the voltage accross the heater (208 volts) and the current through the heater (7 amps)

If accuracy is required it might be best to measure the voltage accross the heater, but in practice assuming nominal voltage should be fine.
 
aeroguy said:
OK. And so the fact that amperage draw on both legs is 7A is irrelevant?
Click to expand...
That's merely a matter of physics. How could it be anything else without a third circuit conductor?

The current is the same throughout any simple, single-pathway series circuit.
 
aeroguy said:
OK. And so the fact that amperage draw on both legs is 7A is irrelevant?
Click to expand...
This is a common misconception that comes about from people thinking that a "hot" wire is a "supply" and the "neutral" is a "return". By extension, a phase-to-phase connection is using two "supply" wires, so half the current must come from each wire. This is not how electrical systems work.
 
aeroguy said:
OK. And so the fact that amperage draw on both legs is 7A is irrelevant?
Click to expand...
For this circuit, yes. You essentially want to know what the load is doing.

The current through the load is what you want, and it is the same on the incoming leg as it is on the outgoing leg. Either one will tell you what goes through the load.

The leg paths, material, etc, along with the current, might be important for other stuff, but not for what you want to know in this case.
 
Ahh. Ye Olde Oregon Fudge Factor.

The transformer as an _aggregate unit_ is supplying 1456VA.

But if you consider the individual coils which make up the transformer, you get a different answer.

Consider that this load is the only load on the transformer. You have two secondary coils, each supplying 120V and each carrying 7V. Each coil is supplying 840VA, for a total of 1680VA.

The reason for this is that the voltage applied to the heater is not in phase with the individual coil voltages. The voltage applied to the heater is the vector sum of the two coil voltages, and is 30 degrees out of phase with each of them. Since the current through the heater is in phase with the voltage applied to the heater, the current through the entire circuit (_both_ coils and the heater) is 30 degrees out of phase with the individual coil voltages.

Imagine that instead you were to connect this transformer to a pair of 7A 120V loads (connected line-neutral). You would have more power delivered to the load, with the same transformer heating.

Of course, most such transformers serve many loads, with relatively balanced three phase loading. Once the 3 phase load is balanced, the difference between load VA and transformer 'internal' VA goes away.

-Jon
 
090729-1459 EST

aeroguy:

When you have any new problem you should try to analyze the basic system. Just plugging values into some equation can get all sorts of incorrect answers.

Velocity (speed) is equal to distance divided by time. What distance and what time? Consider three points A, B, and C. These form a right triangle of AB = 1 mile, BC = 1 mile, and AC = 1.414 miles. I drive from A to B at 1 MPH and B to C at 1 MPH. But if I calculate my speed from A to C to get to C at the same time as going from A to B to C, then it is 0.707 MPH. This is because I went 1.414 miles in 2 hours, instead of 2 miles in 2 hours.

The answer to your original question has been provided several times. Fundamentally you needed to look at the problem as one resistor of resistance of R with an applied voltage of V and then I = V/R. Further noting that nothing else was connected to either supply line.

.
 
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