Kirchoff's law

[bcorbin]

Something else to consider:

The 'waves' that we are talking about are _not_ actual waves in the wire.
-Jon

Oh, but they are waves. Longitudinal waves. Just like sound waves.

 

[rattus]

Let's not totally scare the OP away just yet.

"Eighty-two . . . eighty-two . . . eighty-two; two hundred forty six total." ~ Rainman

Let's not confuse him either.

 

[rattus]

Square one, a direct answer:

Square one, a direct answer:

I am trying to gain a better understanding of phase angles, like why, when you have 60A per phase, that the sum of a 3 phase 3 wire system is not 180A. Also the word potential and what it means has me confused. While I'm not looking for simple answers, somthing to tecnical could cause my head to explode .

The phase angle is a measure of the time delay of a sinusoidal wave relative to a predetermined reference point. We express the phase angle in degrees where a full cyle (16.7msec @ 60Hz) is 360 degrees.

Then the waves in Larry's diagram could represent the phase/line currents in a balanced wye,

Ia = 60Arms @ 0 (0 msec)
Ib = 60Arms @ -120 (5.6 msec)
Ic = 60Arms @ -240 (11.1 msec)

Kirchoff's current law says that the sum of all currents (In included) into or out of a node is zero. Then,

In = -(Ia + Ib +Ic)

A quick look at Larry's diagram shows that this sum is always zero, therefore,

In = 0

Now let Larry's diagram represent the L-L voltages,

Vab = 208Vrms @ 0
Vbc = 208Vrms @ -120
Vca = 208Vrms @ -240

Kirchoff's voltage law says that the sum of the voltages around a closed loop is zero. This can be demonstrated by scaling the various voltage waves at any point in time.

 

[charlie b]

If you grow more corn than you can eat, you can store some of the extra in a barn. Then when you are not growing any more, you can take what is in the barn and eat that. But you cannot store ?current? in the same way. Every electron that moves away from the source along one wire will have to make its way back to the source via another wire (or via some other path). There is no place along the circuit that can store the ?extra current,? until there is a need for it or until there is another place to store it.

A three-phase WYE system will have four wires that can carry current. Any current that leaves the source on Phase A will have to return to the source via Phase B, or Phase C, or the neutral wire. If the loads are balanced, then all the current will flow in Phases A, B, and C, and no current will flow in the neutral.

For a specific example, suppose at a given moment the current in Phase A is at its maximum value. What is going on in Phases B and C at that same moment is that each has a value of current that is exactly equal to one half the peak value, and the current in each is in the opposite direction as the current in Phase A. Thus, you are adding ?peak current? (Phase A) to ?negative one half peak current? (Phase B) and again to ?negative one half peak current? (Phase C). You can see this if you look at Larry?s sketch of three phase currents as shown in post #10. You can look at any point in time, as shown in that sketch, and see that if you add the values of the three phases, you will get zero.

 

[spsnyder]

Verify Charlies observation mathmatically as a proof. Lets say a wave peaks at 90-degrees (a sin-wave w/ eqn, y=sin(x)). The sin (90) = 1. Now calculate the instaneous value of the other two waves at that time (90-degrees)

Sin(90+120) = Sin (210) = -.5
Sin(90+240) = Sin (330) = -.5

Note that the addition of 120-degrees is due to the fact the the waves are identical only separated in time by 120-degrees. Try it with any value other than 90-degrees and you will see it always adds to 0.

 

[LarryFine]

But you cannot store “current” in the same way. Every electron that moves away from the source along one wire will have to make its way back to the source via another wire (or via some other path). There is no place along the circuit that can store the “extra current,” until there is a need for it or until there is another place to store it.

Well, in a manner of speaking, you can store electricity in a system for a short duration: a quarter of a cycle. This is what reactive current does, especially with capacitance.

Now, I don't mean to include an actual charged capacitor in this sense, because to be fair, Charlie said storing current, not voltage. Also, there's electrostatic charge like a Van de Graff genny.

 

[winnie]

[....]This is what reactive current does, especially with capacitance.

[...]Also, there's electrostatic charge like a Van de Graff genny.

Arguably, there are both examples of how Kirchoff's laws are _approximations_.

Kirchoff's Voltage Law says that if you go around a loop, the total voltages sum to zero. But this is _not_ true if the loop encloses a changing magnetic field.

Kirchoff's Current Law says that the sum of the current flowing into and out of a node must be zero. Again, this is not strictly true, because charge can pile up at a given node (say the dome of a Van de Graff generator)...but only if there is a changing electric field.

Note that this does become a semantics argument; for example the voltage induced in a loop by a changing magnetic field can be accounted for (lumped) into the circuit in a way that makes sense and lets you apply the KVL. See http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm, I believe lecture #16 for the argument about KVL that I am poorly parroting.

-Jon

 

[K2500]

For a specific example, suppose at a given moment the current in Phase A is at its maximum value. What is going on in Phases B and C at that same moment is that each has a value of current that is exactly equal to one half the peak value, and the current in each is in the opposite direction as the current in Phase A. Thus, you are adding “peak current” (Phase A) to “negative one half peak current” (Phase B) and again to “negative one half peak current” (Phase C). You can see this if you look at Larry’s sketch of three phase currents as shown in post #10. You can look at any point in time, as shown in that sketch, and see that if you add the values of the three phases, you will get zero.

So, If this system was unbalanced.
Ia=61 @ 0
Ib=46 @ -120
Ic=62 @ -240

Then, I would take Ia+(-1/2Ib)+(- 1/2Ic)=In
61-23-31=7

This gives me In=7, or 7A neutral current.

If I just shifted values I would say 11A.

Is there a big flaw in my math?

 

[rattus]

Arguably, there are both examples of how Kirchoff's laws are _approximations_.

Kirchoff's Voltage Law says that if you go around a loop, the total voltages sum to zero. But this is _not_ true if the loop encloses a changing magnetic field.

Kirchoff's Current Law says that the sum of the current flowing into and out of a node must be zero. Again, this is not strictly true, because charge can pile up at a given node (say the dome of a Van de Graff generator)...but only if there is a changing electric field.

Note that this does become a semantics argument; for example the voltage induced in a loop by a changing magnetic field can be accounted for (lumped) into the circuit in a way that makes sense and lets you apply the KVL. See http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm, I believe lecture #16 for the argument about KVL that I am poorly parroting.

-Jon

Winnie, I must disagree. Kirchoff's laws hold in any case. You must of course draw the circuit completely which means you must insert emfs induced by magnetic fields. Furthermore, if a node stores charge, it must be modeled as a capacitor, then it is no longer a node, it is a circuit element.

 

[Smart $]

So, If this system was unbalanced.
Ia=61 @ 0
Ib=46 @ -120
Ic=62 @ -240

Then, I would take Ia+(-1/2Ib)+(- 1/2Ic)=In
61-23-31=7

This gives me In=7, or 7A neutral current.

If I just shifted values I would say 11A.

Is there a big flaw in my math?

Yep!

The values you have given are, or at least assumed to be root mean square (RMS) values. AC current having a sinusoidal waveform when plotted has an RMS value of the peak-to-peak range divided by the square root of 2. We typically use RMS values when discussing AC to liken it to DC effectiveness (and use DC formulas). When plotted, RMS has the effect of chopping off sinusoidal-waveform peaks and putting the area removed into the plains. You end up with two lines parallel to the Time axis (the left-to-right going one) and, for a given current, the upper line would be at positive one-half the RMS value and the lower line would be at negative one-half the RMS value, and the effective current is the relative value between lines, which is the RMS value. The area between the lines is equal to the area between the sinusoidal waveform and any steady state voltage reference (the 0V reference is generally used as such) within the peak-to-peak range.

Thus the error in your math is you are not using instantaneous values.

The correct formula for neutral RMS current is:

In = √(Ia?+ Ib?+Ic?–Ia?Ib–Ib?Ic–Ia?Ic)​

Determining the phase angle of the neutral current is a bit more complex algebraicly.

 

[rattus]

A simpler explanation:

A simpler explanation:

Yep!

The values you have given are, or at least assumed to be root mean square (RMS) values. AC current having a sinusoidal waveform when plotted has an RMS value of the peak-to-peak range divided by the square root of 2. We typically use RMS values when discussing AC to liken it to DC effectiveness (and use DC formulas). When plotted, RMS has the effect of chopping off sinusoidal-waveform peaks and putting the area removed into the plains. You end up with two lines parallel to the Time axis (the left-to-right going one) and, for a given current, the upper line would be at positive one-half the RMS value and the lower line would be at negative one-half the RMS value, and the effective current is the relative value between lines, which is the RMS value. The area between the lines is equal to the area between the sinusoidal waveform and any steady state voltage reference (the 0V reference is generally used as such) within the peak-to-peak range.

Thus the error in your math is you are not using instantaneous values.

The correct formula for neutral RMS current is:

In = √(Ia?+ Ib?+Ic??Ia?Ib?Ib?Ic?Ia?Ic)​

Determining the phase angle of the neutral current is a bit more complex algebraicly.

The RMS value of a sinusoidal voltage or current is simply the peak value (not peak to peak) multiplied by 0.707 or divided by sqrt(2). The RMS vallue is also known as the "effective" value because that value of a DC voltage or current would have the same heating "effect". One cannot see this "equivalent" DC voltage or current on a scope, although it is sometimes plotted on paper.

For example, the common 120Vrms household voltage has peak values of +/- 170V. However, the RMS value is always positive.

I agree with Smart that Charlie's example was based on instantaneous values and also I agree with his formula for In in RMS values.

 

[charlie b]

So, If this system was unbalanced.

Ia=61 @ 0
Ib=46 @ -120
Ic=62 @ -240
-
-
-
-
(etc.)

Is there a big flaw in my math?

Yes. There is no reason to presume that an unbalanced set of loads would have their phase angles at 0, 120, and 240. They won't be, and thus the rest of your steps are invalid. The formula posted by Smart $$ is valid.

 

[rattus]

Yes. There is no reason to presume that an unbalanced set of loads would have their phase angles at 0, 120, and 240. They won't be, and thus the rest of your steps are invalid. The formula posted by Smart $$ is valid.

First off Charlie, it is entirely possible that the load currents are separated by exactly 120 degrees, say in the case of incandescent lights.

Secondly, are you sure that the algebraic formula applies for any set of angles? I think it does not.

It is obvious that three equal currents separated by 120 degrees add to zero.

Now consider three equal currents at 0, 90, and -90 degrees. It is immediately obvious that they do not add to zero. However, the algebraic forumula evaluation yields zero.

Enlighten me if I am wrong.

 

[charlie b]

The algebraic formula does not take angles into account. It uses the RMS values of currents in the three phases, to obtain the RMS value of current in the neutral. So a set of currents at 0, +90, and -90 degrees will not sum to zero, using that formula.

I have a copy of a proof of the validity of that formula. Actually, it is not the entire proof. It is a set of instructions on how to construct the proof. It tells you what to multiply, and which Trig Identities to apply, but it does not show you the math. I received it as a Word file in an email from another member of this Forum, who had said he would send it to anyone who asked for it.

It has been a while since I performed any detailed analysis of unsymmetrical loading, and I fear I don?t have the time now to review my old textbooks. What I recall, and it may well be wrong, is that the mechanism by which you obtain an imbalanced loading condition is going to drag the angles between the three phase currents away from the balanced 0, 120, 240 pattern. That does not depend on the power factor of the loads either.

 

[K2500]

I was just going for hypothetical, I realize that I'm in over my head, but was just trying to get a feel for the math. I have'nt learned a thing yet without getting my hands dirty.

By the way, my new answer is 15.52Arms...

 

[Smart $]

The RMS value of a sinusoidal voltage or current is simply the peak value (not peak to peak) multiplied by 0.707 or divided by sqrt(2).

Umm, that woud be zero-crossing-to-peak value. But nonetheless, you are correct in the traditional sense. Yes, I erred in my description, but the reason I put it in those terms is because years ago (more than I care to admit :wink: ) I worked with DC offset voltages (or DC with a ripple) and had to adjust my thinking to match. So to correct my earlier description, the RMS value of a sinusoidal voltage or current is:

V~rms = |ΔV ? 2√2|, and
I~rms = |ΔI ? 2√2|, respectively.​

However, your description will suffice for most.

 

[Smart $]

First off Charlie, it is entirely possible that the load currents are separated by exactly 120 degrees, say in the case of incandescent lights.

Not Charlie, but yes, it is possible... but in such a case, the neutral current has to be zero. If there is ever any balancing ["linear"] current on the neutral, no more than one of the connected loads are operating at the base phase angle. The easiest explanation is that any current on a neutral conductor has to have a phase angle. If that current balances three conducting lines of a mwbc, it can only be the inverse phase angle of one line, which would also mean the other two lines' loads are equal to each other but not that of the in-phase the load. In fact, the in-phase load must have a current greater that that required to balance the loads on the other two lines, for if it were less, the neutral current's phase angle would be the same as rather than the inverse of line.

For example, we have two equal and purely resistive mwbc loads, the neutral current would be of the same angle as the third line... such as Load1 is 10A on Line A @ 0? and Load2 is 10A on Line B @ 120?, the neutral current is 10A @ 240?, which is the angle of Line C. Now if we add a load current on Line C and ramp it up, the neutral current decreases, while the angle stays the same, until the load reaches 10A where neutral current will be 0A. After passing 10A, the neutral current's angle reverses to 60? (the inverse of 240?) to balance the excess of Line C's load. If we increase or decrease the load on either Line A or Line B, the neutral current's angle shifts away or towards that line, respectively, to counter the imbalance.

Once the above is understood, basic "phaseology" applies... where the phase angle of voltages to either side of a load is other than 180?, the phase angle of the load itself is never in-phase. Line-to-line loads utilize this very concept!

Secondly, are you sure that the algebraic formula applies for any set of angles? I think it does not.

It is obvious that three equal currents separated by 120 degrees add to zero.

Now consider three equal currents at 0, 90, and -90 degrees. It is immediately obvious that they do not add to zero. However, the algebraic forumula evaluation yields zero.

Enlighten me if I am wrong.

Umm, the formula is for a 120? 3? 4W voltage system. I believe it to be impossible to have three equal currents at those angles on 4W mwbc of such a system.

 

[rattus]

Umm, that woud be zero-crossing-to-peak value. But nonetheless, you are correct in the traditional sense. Yes, I erred in my description, but the reason I put it in those terms is because years ago (more than I care to admit :wink: ) I worked with DC offset voltages (or DC with a ripple) and had to adjust my thinking to match. So to correct my earlier description, the RMS value of a sinusoidal voltage or current is:

V~rms = |ΔV ? 2√2|, and
I~rms = |ΔI ? 2√2|, respectively.​

However, your description will suffice for most.

Of course it will suffice because it is the classic formula for the RMS value of a sinusoidal voltage or current and that is what we are discussing. We should not confuse the issue by using vague notation like delta V, or by using peak to peak values. We should use the simplest notation possible, that is:

Vrms = Vpk x 0.707
Irms = Ipk x 0.707

 

[Smart $]

Of course it will suffice because it is the classic formula for the RMS value of a sinusoidal voltage or current and that is what we are discussing. We should not confuse the issue by using vague notation like delta V, or by using peak to peak values. We should use the simplest notation possible, that is:

Vrms = Vpk x 0.707
Irms = Ipk x 0.707

You go right ahead and do that... once again, if you like.

 

[rattus]

Huh?

Huh?

From Smart:
Quote from Smart:

"Not Charlie, but yes, it is possible... but in such a case, the neutral current has to be zero. If there is ever any balancing ["linear"] current on the neutral, no more than one of the connected loads are operating at the base phase angle. The easiest explanation is that any current on a neutral conductor has to have a phase angle. If that current balances three conducting lines of a mwbc, it can only be the inverse phase angle of one line, which would also mean the other two lines' loads are equal to each other but not that of the in-phase the load. In fact, the in-phase load must have a current greater that that required to balance the loads on the other two lines, for if it were less, the neutral current's phase angle would be the same as rather than the inverse of line."

Smart, you need to explain this one a little better. I cannot see why In has to be zero for the case of 120 degree separation between load currents. If I am missing something, please educate me.