208V 2 Pole Load

[wwhitney]

A load of 10kw on A and B phases has 10kw load on each phase not 5 kw on each. That is why we get 48 amps on A and B phases. Now we add a load to B and C phase and we still have 48 on Phase A but how do we only have 83 amps on B phase. Why not 96. How do we get 83 amps.

Say you have a 120/240V system A - N - B, and you have 48A from a load on A-N. Now you add a 48A load on B-N. The current on N cancels to 0A instead of adding to 96A. Instead of double the current, you get no current.

A similar thing happens in 3 phase when you have a load A-B and a load B-C and look at the current on B. But in 3 phase the cancellation is a smaller effect, as A and C are only 60 degrees apart (from the point of view of B). Now instead of double, you get sqrt(3) times the current.

Cheers, Wayne

 

[Carultch]

If you have two 10kVA loads, each 208 volts single phase, and each is connected to phases A and B on a panel, those currents would just 'add'? So the current on each phase would be 96 amps (48 for each load)? I'm assuming so since the currents for both loads would be coming in at the same phase / time.

Yes. If you have two loads, both with unity power factor across the same pair of phases, the current does simply add. The vectors that are adding are in the same direction, and therefore to find the new magnitude of them, you can simply add magnitudes of the originals. Simply adding magnitudes of vector quantities is only valid when they both have the same direction. Since phased quantities of current and voltage are also modeled as vectors, these quantities simply add, only when they are in phase with each other.

 

[synchro]

A load of 10kw on A and B phases has 10kw load on each phase not 5 kw on each. That is why we get 48 amps on A and B phases. Now we add a load to B and C phase and we still have 48 on Phase A but how do we only have 83 amps on B phase. Why not 96. How do we get 83 amps.

The current on B phase is the sum of the currents from the load across A-B and the load across B-C. These currents are not in phase with each other but instead are 60° apart, because the 208V across the A-B and B-C loads are shifted 60° from each other.
If they were in-phase (i.e., at 0° between them) then their currents would add directly and you'd get the 96 amps you mentioned.
Because they are shifted 60 degrees, the portion of each load current that's in-phase with the other (and also in-phase with the B-N voltage) is cos(60°/2) x 48A ≈ 0.866 x 48A ≈ 41.6A. And so when they add together the resultant B phase current is 2 x 41.6A ≈ 83A.

Where did the "missing" current go? It's not really missing. There is a portion (i.e., "component") of each of the A-B and B-C load currents that is not in phase with each other, or with the B-N voltage, but instead is 90° apart from it (one is +90° and the other one -90°). Notice that this makes these two current components 180° from each other, and so this current flows through the A-B load into the B-C load and vice-versa without any contribution to the current on phase B. This is the same reason that no current will appear on the neutral of a split-phase 120V/240V system when the L-N currents on the two phases are balanced, because of their 180 degree relationship.

How much current flows between the A-B and B-C loads and does not go through phase B? That is somewhat academic, but it would be sin(60°/2) x 48A = 0.5 x 48A = 24A.
And so the bottom line is that the A-B and B-C load currents each contribute 41.6A to the B-phase current, which will then be 41.6A + 41.6A ≈ 83A.
24A of these load currents will flow from one load to the other and therefore does not appear on the B-phase. As mentioned, 41.6A of the 48A will be in-phase with the B-N voltage and 24A will be 90° from it. This is consistent with Mr. Pythagorus"s Theorem: √ ( 41.6² + 24²) ≈ 48.

It might be helpful to think of this like two people pushing on a pole. They can't occupy the same point in space, and so the force they can apply will not be completely is the same direction. For example, say they are each pushing at a 30° angle to the pole. Then cos(30°) ≈ 0.866 or 86.6% of their force gets applied along the direction of the pole. The rest of the force each person applies is effectively one person pushing against the other.

 

[Carultch]

Say you have a 120/240V system A - N - B, and you have 48A from a load on A-N. Now you add a 48A load on B-N. The current on N cancels to 0A instead of adding to 96A. Instead of double the current, you get no current.

A similar thing happens in 3 phase when you have a load A-B and a load B-C and look at the current on B. But in 3 phase the cancellation is a smaller effect, as A and C are only 60 degrees apart (from the point of view of B). Now instead of double, you get sqrt(3) times the current.

Cheers, Wayne

To add a visual to this, see below:


This is drawn to scale, such that 1 inch = 10 Amps. The three brown lines indicate the WYE pattern of each phase. The dashed lines indicate the delta pattern between phases. The black lines indicate the current vectors. All given load currents are assumed to be ideal loads with no phase shifts or distortions, and all voltage sources are assumed to be ideal AC sources as well, as perfect sine waves that are 120 degrees apart in phase.

The maximum possible angle between a vector along the line between A and B (for current on a load across A and B), and a vector along the line between B and C, is 120 degrees. Pick up the two lines, and translate them to connect them end-to-end, so they span across the axis of the B-phase as much as possible. We reverse these vectors as needed to achieve the maximum possible combination on the B-phase, because the current will eventually flow in both directions, and we are ultimately interested in maximum magnitudes. What is the total length of the line formed by the remaining side of the triangle, when we make it as long as possible when projected onto the B-phase? 8.3138 inches. This corresponds to a current on the B-phase of 83.1A, which comes from 48A*sqrt(3).

Now when we look at it from the perspective of the other two phases. The maximum possible combination of Iab and Ibc as vectors that could be drawn in either direction, when projected onto A, is when they form sides of the equilateral triangle, and the vector sum is the third side. This vector sum has a magnitude of 48A. When measuring current on A, from just these two phase-to-phase loads alone, we will get 48A, since only the magnitude ends up getting measured. But, as I have learned in this thread, when there are phase-to-neutral loads on A as well (Ia0), they don't add up conventionally with this value. It needs to add up as vectors with Ia0 to get Ia. The same thing also happens with the C-phase current.

 

[Dennis Alwon]

Let's see if I have this right. If I take a meter and read the amps on phase B with no other loads except the 10kw load across A and B then I will read 48 amps. If I add another 10kw load from B to C I would still read 48 amps on the wire going to the breaker on the B phase but If I put my meter across the B phase feeder then I will read 83 amps. I this correct?

 

[winnie]

Let's see if I have this right. If I take a meter and read the amps on phase B with no other loads except the 10kw load across A and B then I will read 48 amps. If I add another 10kw load from B to C I would still read 48 amps on the wire going to the breaker on the B phase but If I put my meter across the B phase feeder then I will read 83 amps. I this correct?

Exactly.

I think post 21 is the clearest statement of why this is without going into the math. The AB load is not in phase with the BC load, so you get a smidge of cancellation in the total current flow.

-Jon

 

[synchro]

Let's see if I have this right. If I take a meter and read the amps on phase B with no other loads except the 10kw load across A and B then I will read 48 amps. If I add another 10kw load from B to C I would still read 48 amps on the wire going to the breaker on the B phase but If I put my meter across the B phase feeder then I will read 83 amps. I this correct?

Correct. The current drawn by each L-L load depends only on the voltage supplied to it by its 2-pole breaker, which to keep it simple, we assume will be a constant voltage independent of the various load currents. So you will still read 48A going to the load on A-B when you put a load on B-C. The vector summation of the 48A currents from the loads on A-B and those on B-C occurs on the B-phase busbar, and results in the 83A you will measure on the B-phase feeder.

 

[Dennis Alwon]

Thank you all. It just seems so odd to me that I keep refusing to see it. I think I finally got it... now if I can remember it with this old brain... lol

 

[GoldDigger]

That's true, but doesn't change the validity of my comments, as far as I can see.

The magnitude of the vector sum of I_ba and I_ca is correctly given by the sqrt expression, but the direction of the vector sum is parallel to I_a0 only if I_ba = I_ca or I_a0 = 0. When non-parallel, the magnitude of the final vector sum will be less than the expression you listed.

Cheers, Wayne

It seems to me that the definition of parallel is problematic when one of the vectors is the zero vector. The "simple" definition of parallel in terms of having the same direction does not apply to the zero vector, which does not have a unique direction.
One formal definition that two vectors are parallel if and only if one vector is a scalar mutiple of the other could go either way, depending on whether the scalar is allowed to be zero.
The unqualified formal definition that two vectors are parallel if and only if their dot product is equal to the product of their magnitudes, combined with the unqualified formal definition that two vectors are perpendicular if and only if their dot product is zero gives the troubling result that the zero vector is both parallel to and perpendicular to every other vector.

 

[Grouch1980]

To add a visual to this, see below:
View attachment 2558735

The 208 volt line segments are 60 degrees apart from each other correct? So, when you look at their sine waves, with all 3 interposed together on a graph, one 208 volt sine wave is at 0 degrees, the second one is at 60 degrees, the third one at 120 degrees, and then at 180 degrees we're back at the first 208 volt sine wave?

 

[Carultch]

The 208 volt line segments are 60 degrees apart from each other correct? So, when you look at their sine waves, with all 3 interposed together on a graph, one 208 volt sine wave is at 0 degrees, the second one is at 60 degrees, the third one at 120 degrees, and then at 180 degrees we're back at the first 208 volt sine wave?

They will look like they are 60 degrees apart if you plot Va-Vb and Va-Vc on the same plot. Notice that the issue is that Va is the positive probe both times, in this example.

If you give each of the phases a chance at being connected to the positive probe and each of the phases a chance at being connected to the negative probe, your voltage differences will be Vab=(Va-Vb), Vbc=(Vb - Vc), and Vca=(Vc-Va). These three waveforms will be 120 degrees apart from each other, and each is 30 degrees from a phase-to-neutral waveform. Since 120 degrees is the supplement of 60 degrees, this is why they look like they are 60 degrees apart in the delta formation.

Here's a plot, given unit amplitude of the phase-to-neutral waveforms and a 12 second period, to keep the numbers simple. The solid black/red/blue waveforms are phase-to-neutral, and the dashed brown/purple/navy (blended colors) are phase-to-phase waveforms. (Va-Vb) is 30 degrees ahead of Va; (Vb-Vc) is 30 degrees ahead of Vb, and (Vc-Va) is 30 degrees ahead of Vc. Given a 12 second period, 1 second = 30 degrees.

 

[Grouch1980]

Since 120 degrees is the supplement of 60 degrees, this is why they look like they are 60 degrees apart in the delta formation.

I understand supplementary angles, that they have to add up to 180 degrees (ok, I googled it)... how does that correlate though to the 60 degrees shown in the delta formation?

 

[Grouch1980]

I understand supplementary angles, that they have to add up to 180 degrees (ok, I googled it)... how does that correlate though to the 60 degrees shown in the delta formation?

I think I have it? If you extend the Iab, Ibc, and Ica lines further out, there you get 120 degrees between them, on the other side of the 60 degree angles.

 

[Carultch]

I think I have it? If you extend the Iab, Ibc, and Ica lines further out, there you get 120 degrees between them, on the other side of the 60 degree angles.

Indeed, that's the idea. It's the angles between the vectors, rather than the angles between the line segments. They are related to each other, but with the angles between the vectors, which direction the vector is pointing also makes a difference.

 

[Grouch1980]

Indeed, that's the idea. It's the angles between the vectors, rather than the angles between the line segments. They are related to each other, but with the angles between the vectors, which direction the vector is pointing also makes a difference.

Thanks!

 

[Eclectic engineer]

This discussion has been very educational and has changed how I think about 3-Phase Panels. I would like to share a couple examples here are get you guys' feedback. Given below are couple screenshots of 6 circuits connected to a 3-Phase panel.
I would like to know the total amp draw on each phase using the principles discussed above. My initial thinking was to directly add all the amp draws from each circuit. But that is definitely not the right way to do it. Hence for Example 1, I am thinking if the total load on each phase is equal to 80 * Cos (30) = 69.2A. Is this right?
Example 2 is unbalanced and I am not sure how to formulate the phase load.


Example 1 - Equally balanced phases


Example 2 - Unbalanced phases

 

[wwhitney]

If you can assume that all the circuits have the same power factor, then for each example first combine all the A-B loads, all the B-C loads, and all the A-C loads.

Now to get the current on line A, you need to add the contributions from A-B and from A-C. Since we assumed the same power factor, those currents will be 60 degrees out of phase. You need to do vector addition.

I guess the most direct way to get the magnitude of the vector sum is the law of cosines for vector addition: for vectors X and Y with angle theta between them, | X + Y | = sqrt(|X|² + |Y|² + 2 |X| |Y| cos theta). In our case, theta is 60 degrees, and cos 60 degrees = 1/2, so the last term just becomes |X| |Y|. Note then that when |X| = |Y|, you just get |X+Y| = sqrt(3) * |X|, which is a familiar factor.

So for example, in your second case, the A-B loads total 60A and the A-C loads total 20A. That means the line current on A will be sqrt(60*60 + 20*20 + 20*60).

Cheers, Wayne

 

[Eclectic engineer]

So using that formula, in Example 1, all phases would be loaded at 69.28A as |X| = |Y|.
In example 2,
Phase A = 72.11A
Phase B = 87.17A
Phase C = 52.91A
Can you confirm if you get the same values?

If I were to wrap a amp meter around each of the three phases, I would read the values calculated above. Is that right?

 

[wwhitney]

Yes, you punched the computations into your calculator correctly. And yes, we are calculating the line currents, which an ammeter could measure.

Note that the above only works when all the loads are line-to-line. When you mix in line-neutral loads, the math gets more complicated, as now you may have to sum 3 different vectors. The line-neutral current will be at 30 degrees to each of the line-line currents involving that line.

Cheers, Wayne

 

[wwhitney]

When you mix in line-neutral loads, the math gets more complicated, as now you may have to sum 3 different vectors. The line-neutral current will be at 30 degrees to each of the line-line currents involving that line.

In which case if you assume unity power factor (all currents are in phase with their voltages), the computation isn't too bad. Say the A-N current is X, the A-B current is Y, and the A-C current is Z. Then we could write those as vectors as:

X = (|X|,0)
Y = (cos 30 |Y|, sin 30 |Y|)
Z = (cos 30 |Z|, - sin 30 |Z|)

cos 30 = sqrt(3)/2, and sin 30 = 1/2, so that means X+Y+Z = (|X| + sqrt(3)/2 (|Y|+|Z|) , (|Y| - |Z|) / 2). And to get |X+Y+Z| (the line current an ammeter would show on A), you just take the square root of the sum of the squares of the two components of X+Y+Z.

Cheers, Wayne