Amperage calculation

If I have a resistance heating load of 50 KW at 480V and I need to find the amp draw would I need to use 1.73 in my calculation if only two legs of the 480V 3 PH system are being used? Also Power Factor would not play a role seeing as this is a resistive load, correct?

Thanks
 

david luchini

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With a 480V, single phase load, you would not use 1.732 in your amperage calculation.

And you are correct, power factor would not play a part in a resistive load.
 

GoldDigger

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If only two legs are being used, then the line current in those two legs will be exactly equal to the current in the load.
But the power factor of the line current will not be 1.
If you have three such two line loads connected in a balanced way, the current in each line will be 1.73 times a single load current because of cancelation, and then the PF of the line current will be 1.
From the POCO point of view they cannot tell the difference between the unbalanced line to line resistive load and a combination of a partly capacitive line to neutral load on one leg and the correspondingly partly inductive wye load on the other.

If you are looking at a pure delta supply, all of my "neutral" references involve a hypothetical neutral point. :)
 
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charlie b

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I don't understand your explanation, Golddigger. Power factor has nothing to do with how much load is on any one phase-to-neutral load, or any single phase line-to-line load, or on any balanced or imbalanced three phase loading. If you have nothing other than a single phase 480 volt load on a three phase 480 volt system, and if that one load is purely resistive, the power factor, as seen by either that load or the system as a whole, will be 1.0.
 
If only two legs are being used, then the line current in those two legs will be exactly equal to the current in the load.
But the power factor of the line current will not be 1.
If you have three such two line loads connected in a balanced way, the current in each line will be 1.73 times a single load current because of cancelation, and then the PF of the line current will be 1.
From the POCO point of view they cannot tell the difference between the unbalanced line to line resistive load and a combination of a partly capacitive line to neutral load on one leg and the correspondingly partly inductive wye load on the other.

If you are looking at a pure delta supply, all of my "neutral" references involve a hypothetical neutral point. :)
So if I used all three phases and balanced the two line loads between them I would have to use 1.73 in my calculation, correct?
 

GoldDigger

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Charlie, consider a three phase wye system.
Now connect one line to line resistive load to it.
Look at the phase angle of the line current relative to the line to neutral voltage.
You get non-zero angle, hence PF is not 1.
If the supply is wye, then it will have to deliver a VA which is greater than the wattage of the load.
And if that supply is a generator or inverter, it will feel that VA in the firm of IR losses. But the prime mover will not have to deliver greater power.

Tapatalk!
 
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david luchini

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Charlie, consider a three phase wye system.
Now connect one line to line resistive load to it.
Look at the phase angle if the line current relative to the line to neutral voltage.
You get non-zero angle, hence PF is not 1.
Power factor is the ratio of true power to apparent power. For a purely resistive load, the power factor is 1.
 

charlie b

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Look at the phase angle of the line current relative to the line to neutral voltage. You get non-zero angle, hence PF is not 1.
I don't agree that the phase angle between the line current and the line to neutral voltage will be non-zero. There is no inductive or capacitive load to cause the current being driven by that voltage to be out of phase with that voltage.

 

GoldDigger

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But since the other end of the load is at a different phase voltage, the resulting current will not be in phase with the line voltage.
One vector current cannot be in phase with two different phase voltages at the same time.
Another view to take:
If you multiply each line voltage times the line current , without considering power factor and add the two line VA figures you will get sqrt3/2times the actual load power.

Please do not dismiss this by saying that you cannot add the line power values that way because that is exactly what you would do for wye loads, and by the principles of superposition and equivalence the source cannot tell the difference.

Tapatalk!
 
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charlie b

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I will take it one step further. Start with no load at all on the three phase wye system. There are no line currents, and each of the line-to-neutral voltages will be out of phase with each other by exactly 120 degrees. Now connect the single phase, 480 volt, purely resistive load under discussion, and put it on phases A and B. You will have a line current in phase A, and it will be in phase with the phase A to neutral voltage. You will have a line current in phase B, and it will be in phase with the phase B to neutral voltage. But the three phase-to-neutral voltages will no longer be separated from each other by 120 degrees. That is the result of the unbalanced loading. I no longer recall how to do the matrix algebra needed to work out the symmetrical components that would tell me the new phase angles between the voltages. But SKM could do that job for me.
 

GoldDigger

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I will take it one step further. Start with no load at all on the three phase wye system. There are no line currents, and each of the line-to-neutral voltages will be out of phase with each other by exactly 120 degrees. Now connect the single phase, 480 volt, purely resistive load under discussion, and put it on phases A and B. You will have a line current in phase A, and it will be in phase with the phase A to neutral voltage. You will have a line current in phase B, and it will be in phase with the phase B to neutral voltage. But the three phase-to-neutral voltages will no longer be separated from each other by 120 degrees. That is the result of the unbalanced loading. I no longer recall how to do the matrix algebra needed to work out the symmetrical components that would tell me the new phase angles between the voltages. But SKM could do that job for me.
Now you have stumbled off the deep end Charlie.
Drawing current from one or more legs, even if unbalanced, is not going to change the line voltages other than by the small amount of the IR drop.
Unbalance in this case is seen only in current, not in voltage.
And, just for grins, if the vector voltage changed that much the line to line voltage would have to change too, along with the load current.


Tapatalk!
 

david luchini

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Another view to take:
If you multiply each line voltage times the line current , without considering power factor and add the two line VA figures you will get sqrt3/2times the actual load power.

Please do not dismiss this by saying that you cannot add the line power values that way because that is exactly what you would do for wye loads, and by the principles of superposition and equivalence the source cannot tell the difference.
I think you mean you will get 2/sqrt3 times the actual load power.

And yes, you cannot add line power values that way...there are NO line connect loads in the example.

Consider a load that is 10A, 208V, single phase, or 2080VA. If you wanted the same load divided evenly between to L-N voltages, you would have (2) 1040VA loads, and the line current on each line would be 8.67A. As you can see the current for the same amount of load being line-n connect vs. l-l connected is not the same. You are taking the approach that the current drives the load, when it is the load that drives the current.
 

GoldDigger

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Thanks for the correction.
What I am saying is that for a wye power source, there is absolutely no way to tell the difference between a line to line resistive load and the equivalent pair of complex impedance line to neutral loads, so the two ways of calculating the power must give the same results.
And a wattmeter would agree.

You have a greater line current than you would with the divided (and different resistance) line to neutral loads, so you must have a different power factor.

You cannot add the VA on a per line basis, but you MUST be able to add the power on a per line basis.

Tapatalk!
 
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david luchini

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Thanks for the correction.
What I am saying is that for a wye power source, there is absolutely no way to tell the difference between a line to line resistive load and the equivalent pair of complex impedance line to neutral loads, so the two ways of calculating the power must give the same results.
And a wattmeter would agree.

You have a greater line current than you would with the divided (and different resistance) line to neutral loads, so you must have a different power factor.

You cannot add the VA on a per line basis, but you MUST be able to add the power on a per line basis.
A wattmeter wouldn't care what the power factor is. If you had a 2000 watt resistive load vs. a 2000 watt load with an 80% power factor, wouldn't you want the wattmeter to read 2000 watts in both cases?

You are trying to force L-N voltage into a circuit that has no L-N connected loads. If you had a wye system, it has 3 L-N voltages and 3 L-L voltages. IF a single resistive load is connected L-L, there is NO L-N currents, only L-L currents. The L-N voltage isn't relevant to the circuit. You have L-L current and L-L voltage with the same phase angle. The power factor is 1.

Apparent power S is the sum of the Real power P plus j times the Reactive power Q: S = P + jQ. The magnitude of S = sqrt(P^2 + Q^2). And power factor is the ratio of real power to apparent power: pf = P/S.

With the single resistive load as above, we know P=2000 w and Q=0 var, so S = sqrt(2000^2 + 0^2) = 2000.

So the power factor of the circuit = 2000/2000 = 1.
 

Smart $

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... You are trying to force L-N voltage into a circuit that has no L-N connected loads. If you had a wye system, it has 3 L-N voltages and 3 L-L voltages. IF a single resistive load is connected L-L, there is NO L-N currents, only L-L currents. The L-N voltage isn't relevant to the circuit. You have L-L current and L-L voltage with the same phase angle. The power factor is 1.

Apparent power S is the sum of the Real power P plus j times the Reactive power Q: S = P + jQ. The magnitude of S = sqrt(P^2 + Q^2). And power factor is the ratio of real power to apparent power: pf = P/S.

With the single resistive load as above, we know P=2000 w and Q=0 var, so S = sqrt(2000^2 + 0^2) = 2000.

So the power factor of the circuit = 2000/2000 = 1.
Forget the the load "end" for a minute. What are the apparent and real power values of each secondary winding?
 

GoldDigger

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Forget the the load "end" for a minute. What are the apparent and real power values of each secondary winding?
<Engage sarcasm>
Obviously when you have a wye source and a balanced delta load there cannot be any current in the generator windings since there is no current in the neutral.
And an unbalanced line to line load cannot possibly work. <Disengage sarcasm>

Tapatalk!
 

winnie

Senior Member
Now connect the single phase, 480 volt, purely resistive load under discussion, and put it on phases A and B. You will have a line current in phase A, and it will be in phase with the phase A to neutral voltage. You will have a line current in phase B, and it will be in phase with the phase B to neutral voltage.
I have to disagree on this point. The current through a resistive load will be in phase with the voltage applied to that load, and thus in phase with the A to B voltage. For a single circuit, the current must be the same throughout, and _not_ in phase with either of the phase to neutral voltages.

A single phase _resistive_ load will have unity power factor as compared to the line-line voltage, but a 0.86 power factor when referenced to the line-neutral voltages.

-Jon
 

winnie

Senior Member
With the single resistive load as above, we know P=2000 w and Q=0 var, so S = sqrt(2000^2 + 0^2) = 2000.

So the power factor of the circuit = 2000/2000 = 1.
The circuit in question does not consist of a single resistor. At a minimum it consists of a resistor and _two_ transformer secondary coils.

The power factor in the resistor is 1; I agree with your calculation above for this _portion_ of the circuit.

The power factor in each of the transformer secondary coils is less than 1.

The unbalanced loading on the three phase source causes this power factor to be present _on the source_.

-Jon
 
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