Single Phase Inverters on 208 3 Phase

[wwhitney]

Desmos does not like sind, or I don't know how to use it.

As you already know since you've put Desmos into degree mode, you switch between degrees and radians via the bottom of the menu you get from the wrench icon on the upper right. Either way in Desmos you write sin(x). Since my typing doesn't have a little toggle switch, I'm going to write sin(x) for x in radians, and sind(x) for x in degrees. When you see sind(x), you just make sure Desmos is in degree mode and enter sin(x) into it.

But, I did take the point by point addition of A-B and B-C, determined the 60 degree phase shift between the two, used -30 to get the correct position, and added in the 294.449 peak of 208. Does that green line not represent the combination of A-B and B-C voltage waveforms?

No, you're again entering a formula for Green that is the wrong formula and would be assuming the answer anyway. Start with what you have correct (now I'm writing sin(x) instead of sind(x) since we're specifically talking about Desmos in degree mode):

Black: y=170sin(x)-170sin(x-120)
Red: y = 170sin(x-120) - 170 sin(x-240)

So to get Red-Black, enter the formula
Green y = (170sin(x-120) - 170sin(x-240)) - (170sin(x)-170sin(x-120))

If Desmos has a way to refer to the y values from a previous line, you could just enter
Green = Red - Black
for the appropriate syntax of Red and Black, but I don't know if it has that feature or what the syntax is.

Then the Green function is the actual difference of the Red and Black. But if you enter a new formula, not sure what color Desmos will use, but maybe it's Blue:

Blue y = 510sin(x-120)

Then you'll find that Blue and Green coincide. Which is a graphical proof of the statement in post #319 (with an extra factor of 170 everywhere).

Cheers, Wayne

 

[ggunn]

The 1000 foot view...

What is it that you are trying to show, exactly?

If you plot Van, Vbn, and Vcn together you will see three 60Hz waveforms 120 degrees apart. If you plot Vab, Vbc, and Vca you will likewise see three 60Hz waveforms 120 degrees apart, phase shifted from the first set. What else is there to see?

 

[ggunn]

The 1000 foot view...

What is it that you are trying to show, exactly?

If you plot Van, Vbn, and Vcn together you will see three 60Hz 120V waveforms 120 degrees apart. If you plot Vab, Vbc, and Vca you will see three 60Hz 208V waveforms 120 degrees apart, phase shifted from the first set. What else is there to see?

edited for clarity

 

[wwhitney]

If Desmos has a way to refer to the y values from a previous line, you could just enter
Green = Red - Black
for the appropriate syntax of Red and Black, but I don't know if it has that feature or what the syntax is.

It does actually, you enter on successive lines:

a(x) = foo
b(x) = bar
c(x) = a(x) - b(x)

And you can even name your function like v_a(x). To do that, type v, then _ to start a subscript, then your subscript (a in this case), then right arrow to leave the subscript, then the rest of it: (x) = blah.

Cheers, Wayne

 

[wwhitney]

Would the green line represent the combination of A-B and B-C as seen across B-N?

Here's the correct version of what you intend, with some nice symbolic names, just one of the voltages graphed as a dotted line, as well as 3 of the currents (the two inverter currents and their composite as regards current going into node b).

To set the magnitude of the currents, I arbitrarily made them 1/5 of the associated voltages. This corresponds to a 5 ohm load L-L, or a 208V inverter L-L that is putting out 208/5 = 41.6A (RMS).

Cheers,
Wayne

 

[wwhitney]

If you plot Van, Vbn, and Vcn together you will see three 60Hz 120V waveforms 120 degrees apart. If you plot Vab, Vbc, and Vca you will see three 60Hz 208V waveforms 120 degrees apart, phase shifted from the first set. What else is there to see?

To make a couple observations:

1) If you define Van, Vbn, and Vcn as you do, then noting that Vab = Van - Vbn, etc gives the results you say you will get.
2) And then if you take currents Iab and Ibc in phase with Vab and Vbc, the resulting current into node b is Ib = Ibc - Iab and is in phase with Vbn.

The last post shows (2) graphically; I think much earlier we saw (1) graphically, at least for the case of Vab.

Cheers, Wayne

 

[bellington]

As you already know since you've put Desmos into degree mode, you switch between degrees and radians via the bottom of the menu you get from the wrench icon on the upper right. Either way in Desmos you write sin(x). Since my typing doesn't have a little toggle switch, I'm going to write sin(x) for x in radians, and sind(x) for x in degrees. When you see sind(x), you just make sure Desmos is in degree mode and enter sin(x) into it.


No, you're again entering a formula for Green that is the wrong formula and would be assuming the answer anyway. Start with what you have correct (now I'm writing sin(x) instead of sind(x) since we're specifically talking about Desmos in degree mode):

Black: y=170sin(x)-170sin(x-120)
Red: y = 170sin(x-120) - 170 sin(x-240)

So to get Red-Black, enter the formula
Green y = (170sin(x-120) - 170sin(x-240)) - (170sin(x)-170sin(x-120))

If Desmos has a way to refer to the y values from a previous line, you could just enter
Green = Red - Black
for the appropriate syntax of Red and Black, but I don't know if it has that feature or what the syntax is.

Then the Green function is the actual difference of the Red and Black. But if you enter a new formula, not sure what color Desmos will use, but maybe it's Blue:

Blue y = 510sin(x-120)

Then you'll find that Blue and Green coincide. Which is a graphical proof of the statement in post #319 (with an extra factor of 170 everywhere).

Cheers, Wayne

Sorry, but my logic interferes with my graphing. I should be able to add the instantaneous values of A-B and B-C and end up with the same wave at 60 degrees between A-B and B-C. Why does it not?

 

[wwhitney]

Sorry, but my logic interferes with my graphing. I should be able to add the instantaneous values of A-B and B-C and end up with the same wave at 60 degrees between A-B and B-C. Why does it not?

For the questions we've been discussing, you don't want to perform the addition A-B + B-C, you want to perform the subtraction B-C - A-B.

But you are correct about A-B + B-C. Whenever A-B and B-C have the same magnitude, their sum will have phase shift equal to the average of the phase shifts of the summands. E.g. if they are 90 degrees apart, the sum will be in the middle, 45 degrees from either of the summands.

And for the further special case that A-B and B-C are 120 degrees apart, as in our application, their sum will have the same magnitude as each of the summands. Contrast this with the summands being 90 degrees, where the sum is sqrt(2) times the initial magnitude; 60 degrees apart, where the sum is sqrt(3) times the initial magnitude; or 0 degrees apart, where the sum is twice the initial magnitude (just like with normal numbers).

Cheers, Wayne

 

[bellington]

For the questions we've been discussing, you don't want to perform the addition A-B + B-C, you want to perform the subtraction B-C - A-B.

But you are correct about A-B + B-C. Whenever A-B and B-C have the same magnitude, their sum will have phase shift equal to the average of the phase shifts of the summands. E.g. if they are 90 degrees apart, the sum will be in the middle, 45 degrees from either of the summands.

And for the further special case that A-B and B-C are 120 degrees apart, as in our application, their sum will have the same magnitude as each of the summands. Contrast this with the summands being 90 degrees, where the sum is sqrt(2) times the initial magnitude; 60 degrees apart, where the sum is sqrt(3) times the initial magnitude; or 0 degrees apart, where the sum is twice the initial magnitude (just like with normal numbers).

Cheers, Wayne

With your help and additional instruction on the use of Desmos, I was able to get B-C minus A-B. But, I can't conceptualize how we can create a waveform 30 degrees before the two originals. If we only consider legs A-B and B-C with a load on each one where could we measure this particular waveform's 360 RMS voltage?

 

[wwhitney]

With your help and additional instruction on the use of Desmos, I was able to get B-C minus A-B.

Yes, you got it! See the left side of post #325 for a neater set of entries with more meaningful names to generate the graphs. The black, red, and green curves match between your version and my version, except I threw in a factor of 1/5.

But, I can't conceptualize how we can create a waveform 30 degrees before the two originals.

If you look at the graph, the green curve I(B) actually peaks 30 degrees after the red curve I(B-C) and 150 degrees after the black curve I(A-B).

If you take the negative of the black curve (since we are going to subtract black from red), then its peak shifts to 60 degrees after the red curve. Graph that if it isn't clear.

And now you are adding -Red to Black, aka -I(A-B) to I(B-C). As per your correct intuition from post #327, the sum will now peak half way in between the peaks of -Red and Black. Which is 30 degrees after the Red curve peaks.

If you find it helpful to look at all these graphs, that is a fine place to start learning these things. But in the long run the graphs are very cumbersome. So please do revisit Post #311 with the aim of ultimately doing such computations using vectors.

If we only consider legs A-B and B-C with a load on each one where could we measure this particular waveform's 360 RMS voltage?

You won't see it as a voltage on a 208Y/120V system.

If the two loads are supplied by a single set of 3 wires connected to A, B, and C, and each load is a 1 ohm resistance, so that for each load I = V = 120*sqrt(3) = 208A, then you will see the 360 RMS waveform as the combined current in the B wire. That's a lot of amps, which is why in my example I made the loads 5 ohms, so the currents are 1/5 as much. Also to keep the voltage values distinct from the current values.

Cheers, Wayne

 

[wwhitney]

And now you are adding -Red to Black, aka -I(A-B) to I(B-C). As per your correct intuition from post #327, the sum will now peak half way in between the peaks of -Red and Black. Which is 30 degrees after the Red curve peaks.

That should be -Black and Red, a mixup. -I(A-B) and I(B-C) are correct, though.

Cheers, Wayne

 

[Carultch]

With your help and additional instruction on the use of Desmos, I was able to get B-C minus A-B. But, I can't conceptualize how we can create a waveform 30 degrees before the two originals. If we only consider legs A-B and B-C with a load on each one where could we measure this particular waveform's 360 RMS voltage?

Keep it simple with unit amplitude and frequency. In other words, the period is 360 seconds, so 1 second is 1° worth of a cycle.

In degrees, the waveforms for phase A to neutral and phase B to neutral, are given by the following.
Va = sin(t)
Vb = sin(t - 120°)

We're interested in Va - Vb:
Va - Vb = sin(t) - sin(t - 120°)

I can't directly subtract sine waves that have different inputs, so I need trig functions that have the same input. Any linear combination of sine and cosine of the same input, is equivalent to a single trig function with a phase shift and an amplitude scaling. So we want to rephrase the phase shift, as a linear combination of sine and cosine.

Side note: this is what the vectors are representing. The axes of the "phase space" are sine amplitude and cosine amplitude. For a vector in this space, the x & y components tell us the linear combination of cosine & sine, while the magnitude and angle tell us the amplitude and phase.

The identity we want to use, is the following:
sin(x - y) = sin(x)*cos(y ) - cos(x)*sin(y )

Apply this for t and 120°:
sin(t - 120°) = sin(t)*cos(120°) - cos(t)*sin(120°)

Evaluate both trig functions of 120°:
cos(120°) = -1/2
sin(120°) = sqrt(3)/2

Thus:
Va - Vb = 3/2*sin(t) + cos(t)*sqrt(3)/2

Pull sqrt(3) out in front. That's just a constant, so call it k. Since 3 = sqrt(3)*sqrt(3), this leaves us with sqrt(3) in front of sin(t).
Va - Vb = k*(sqrt(3)/2*sin(t)+ cos(t)/2)

Why did I do this? Because now, I have k times a wave with unit amplitude. I can convert this back to the expression, k*sin(t + c).
sqrt(3)/2*sin(t)+ cos(t)/2 = sin(t + c)

Use the angle sum identity:
sin(t + c) = sin(t)*cos(c) + cos(t)*sin(c)

This means we want:
cos(c) = sqrt(3)/2 and sin(c) = 1/2
The corresponding c? c = 30°. Or 30° plus any integer multiple of 360° we want.

As you can see, this means:
sin(t) - sin(t - 120°) = k*sin(t + 30°)
This produces the phase lead of 30°, from subtracting a waveform with a phase lag of 120°.

You could see this as a phase lead of 30°, and you could also see it as a phase lag of 330°. It is mathematically the same. By convention, it's preferred to use the smallest number we can, for the phase shift.

 

[bellington]

Thanks to all of you. Great class! I absorbed a little. Now I know just enough of the math to get into more trouble. I'm still working on a graph that shows voltage and current through coil B as the result of AB and the result of BC in a manner that my head can contain.

Thanks again for your help and patience.

 

[bellington]

Keep it simple with unit amplitude and frequency. In other words, the period is 360 seconds, so 1 second is 1° worth of a cycle.

In degrees, the waveforms for phase A to neutral and phase B to neutral, are given by the following.
Va = sin(t)
Vb = sin(t - 120°)

We're interested in Va - Vb:
Va - Vb = sin(t) - sin(t - 120°)

I followed you fine to this point. After this point, I hear Charlie Brown's teacher. But, I appreciate the effort.

I can't directly subtract sine waves that have different inputs, so I need trig functions that have the same input. Any linear combination of sine and cosine of the same input, is equivalent to a single trig function with a phase shift and an amplitude scaling. So we want to rephrase the phase shift, as a linear combination of sine and cosine.

Side note: this is what the vectors are representing. The axes of the "phase space" are sine amplitude and cosine amplitude. For a vector in this space, the x & y components tell us the linear combination of cosine & sine, while the magnitude and angle tell us the amplitude and phase.

The identity we want to use, is the following:
sin(x - y) = sin(x)*cos(y ) - cos(x)*sin(y )

Apply this for t and 120°:
sin(t - 120°) = sin(t)*cos(120°) - cos(t)*sin(120°)

Evaluate both trig functions of 120°:
cos(120°) = -1/2
sin(120°) = sqrt(3)/2

Thus:
Va - Vb = 3/2*sin(t) + cos(t)*sqrt(3)/2

Pull sqrt(3) out in front. That's just a constant, so call it k. Since 3 = sqrt(3)*sqrt(3), this leaves us with sqrt(3) in front of sin(t).
Va - Vb = k*(sqrt(3)/2*sin(t)+ cos(t)/2)

Why did I do this? Because now, I have k times a wave with unit amplitude. I can convert this back to the expression, k*sin(t + c).
sqrt(3)/2*sin(t)+ cos(t)/2 = sin(t + c)

Use the angle sum identity:
sin(t + c) = sin(t)*cos(c) + cos(t)*sin(c)

This means we want:
cos(c) = sqrt(3)/2 and sin(c) = 1/2
The corresponding c? c = 30°. Or 30° plus any integer multiple of 360° we want.

As you can see, this means:
sin(t) - sin(t - 120°) = k*sin(t + 30°)
This produces the phase lead of 30°, from subtracting a waveform with a phase lag of 120°.

You could see this as a phase lead of 30°, and you could also see it as a phase lag of 330°. It is mathematically the same. By convention, it's preferred to use the smallest number we can, for the phase shift.

 

[bellington]

It is highly likely that one, or several of you, have made valiant efforts to explain this already, but why is the blue dashed line not the voltage across coil B (using a purely resistance load of 3 Ω) as the result of voltages AB and BC in this graph?

 

[Carultch]

It is highly likely that one, or several of you, have made valiant efforts to explain this already, but why is the blue dashed line not the voltage across coil B (using a purely resistance load of 3 Ω) as the result of voltages AB and BC in this graph?

Because the resistances are not in series, and you've used formulas that only apply for resistances in series.

Voltage across two points, is voltage at one point, minus voltage at the other point.
Sum of current entering a point, equals sum of current leaving a point.

This means, the interphase voltage is Va - Vb, or vice-versa. And the combined B-phase current due to loads AB and BC, is the vector sum of the currents of each of them.

I followed you fine to this point. After this point, I hear Charlie Brown's teacher. But, I appreciate the effort.

Here's what I'd like you to do to make sense of it.

In your Desmos file, set up:
V_A = sin(x)
V_B = sin(x - 120°)
V_AB = V_A - V_B

Now set up two sliders to define constants, which I'll call C and S. Give them both a range from -2 to +2, and a step of 0.01. Then let:
V_check = C*cos(x) + S*sin(x)

Adjust the two sliders until you get the graph of V_check to line up with the graph of V_B. Also try adjusting the sliders to get V_check to line up with V_AB. This is what I mean by a linear combination of sine and cosine.

Finding these two constants, is the key to using algebra to subtract V_A - V_B.

 

[bellington]

Because the resistances are not in series, and you've used formulas that only apply for resistances in series.

Voltage across two points, is voltage at one point, minus voltage at the other point.
Sum of current entering a point, equals sum of current leaving a point.

This is my problem. It appears to me that coil A and coil B are in series for AB voltage, and coil B and coil C are in series for BC voltage.

This means, the interphase voltage is Va - Vb, or vice-versa. And the combined B-phase current due to loads AB and BC, is the vector sum of the currents of each of them.

I was hoping I had accomplished currents by using A + B resistance across AB, then B + C currents across BC and adding the two B currents and voltages, and even power. Why does this not work?

Here's what I'd like you to do to make sense of it.

In your Desmos file, set up:
Va = sin(x)
Vb = sin(x - 120°)

This I can do.

Now set up two sliders to define constants, which I'll call C and S. Then let:
Vb_check = C*cos(x) + S*sin(x)

I don't know what I should do to set up two sliders called C and S, nor why I would slide constants.

Adjust the two sliders until you get the graph of Vb_check to line up with the graph of Vab.
This is what I mean by a linear combination of sine and cosine.

Finding these two constants, is the key to using algebra to subtract Va - Vb.

 

[Carultch]

This is my problem. It appears to me that coil A and coil B are in series for AB voltage, and coil B and coil C are in series for BC voltage.

I was hoping I had accomplished currents by using A + B resistance across AB, then B+C currents across BC and adding the two B currents and voltages, and even power. Why does this not work?

I will do that in Desmos without a clue what it means.

They may appear to be in series, but they aren't just two individual resistors in series. There are other sources & sinks at the point in between those resistors, so formulas for combining resistances in series won't directly work for this application. The idea that resistances simply add up, is only valid if your resistors are in series, with nothing else connected to the point between them.

Voltage is ACROSS two points. Current is THROUGH a point.

Voltage is additive if you have a sequence of points, and you are considering successive differences in voltage across each pair of points in that sequence. Given points X, Y, and Z, this means Vx - Vz = (Vx - Vy) + (Vy - Vz).

Voltage is subtractive when you start with voltages at two points, both relative to a common reference. Call that reference point 0. This means Vxy = Vx0 - Vy0.

Current is additive considering all paths into a point, and all paths out of a point. Sum of current in = sum of current out.

For all of the above
DC values add up directly
Instantaneous values add up directly
AC values have to add and subtract, either as vectors or waveform functions

 

[ggunn]

This is my problem. It appears to me that coil A and coil B are in series for AB voltage, and coil B and coil C are in series for BC voltage.

It depends on your point of reference. If the grounded center of the wye is your POR, Vab and Vbc are in parallel.

 

[bellington]

They may appear to be in series, but they aren't just two individual resistors in series. There are other sources & sinks at the point in between those resistors, so formulas for combining resistances in series won't directly work for this application. The idea that resistances simply add up, is only valid if your resistors are in series, with nothing else connected to the point between them.

If we open all breakers, except our inverters, then coils A-B are in series. What my graph should be showing is an inside-out look at the interactions of the two currents AB and BC and the impact across and through coil B.

Voltage is ACROSS two points. Current is THROUGH a point.

Yes. Sorry.

Voltage is additive if you have a sequence of points, and you are considering successive differences in voltage across each pair of points in that sequence. Given points X, Y, and Z, this means Vx - Vz = (Vx - Vy) + (Vy - Vz).

Voltage is subtractive when you start with voltages at two points, both relative to a common reference. Call that reference point 0. This means Vxy = Vx0 - Vy0.

Current is additive considering all paths into a point, and all paths out of a point. Sum of current in = sum of current out.

For all of the above
DC values add up directly
Instantaneous values add up directly

I'm missing why AC instantaneous values cannot also be added up directly. I do agree that it becomes magnitudes easier with vectors, or waveforms.

AC values have to add and subtract, either as vectors or waveform functions