Fault Currents and Time Current Curvex 2

Is a 3 phase bolted fault, 3 phases shorted or 3 phases & ground shorted? On a TCC, is the scale for 3 phase circuits, the line to line or line to ground voltage? I'm new to all this, any good resources to get further info on?

Thank you very much

 

[electricpete]

To me, jbartos' example illustrates one point - IF we assume fault resistance, THEN the grounded and unbalanced 3-phase faults will be different (even if balanced). The reason being that we would be forced to connect the fault resistance phase-phase for ungrounded fault, but phase-ground for grounded fault. I agree with peterb that in the fault studies I have seen (within power plant distribution), zero fault impedance has been assumed.

 

[peterb]

Think about it, electricpete. The wye connected motor analogy is valid for both the grounded and ungrounded case. In fact, replace the motor windings with resistors if you like, to simulate fault resistance, and the result is the same - THERE IS NO DIFFERENCE IN THE FAULT CURRENT whether the neutral (or fault point) is grounded or not. Why would you want to connect the fault resistance phase-phase? It needs to be connected phase-neutral for the analogy to hold.

 

[electricpete]

peterb - could be true what you're saying. But it doesn't seem physically likely to connect all 3 phases to some fourth point which is NOT ground (again assuming there is an arc/fault resistance... irrelevant otherwise). I'll grant you it's also pretty remote to connect all three phases to each other through a fault resistance, although perhaps a credible scenario with overhead transmission lines.

I never intended to disagree with anything you said, peterb. I was just trying to acknowledge jbartos' point that there are apparently more varied fault calculation scenario's discussed in standards than those in calculations I have seen where fault resitance is assumed zero and voltage is assumed balanced.

 

[jbartos]

Suggestions: References:
1. Stevenson, Jr. W. D., "Elements of Power System Analysis," Third Edition, McGraw-Hill Book Co., 1975
2. Wadhwa C. L., "Electrical Power Systems," Second Edition, John Wiley & Sons, New York, 1991
3. IEEE Std 399-1990 "IEEE Recommended Practice for Industrial and Commercial Power Systems Analysis"
4. Gross C. A., "Power System Analysis," John Wiley & Sons, New York, 1979
5. Bergen A. R., "Power Systems Analysis," Prentice-Hall, Inc., Englewood Cliffs, NJ, 1986
6. Gungor B. R., Power Systems," Harcourt Brace Jovanovich, Publishers, New York, 1988

1) Reference 1 covers three-phase faults on pages 282-289 without use of symmetrical components. Chapter 13 "Unsymmetrical Faults" covers Line-to-Line Fault, Double-Line-to-Ground Fault, and Line-to-Ground Fault. It appears that Prof. and Dr. Stevenson treated symmetrical components what they are worth.
2) Reference 2 covers three-phase-to-phase-to-ground fault under symmetrical components; however, no sign of the ungrounded three-phase-to-phase treatment. This is a fairly meety book on Electrical Power System otherwise, authored by Professor of Electrical Engineering Delhi College of Engineering in Delhi, India, and Dean of Faculty of Technology at University of Delhi, Delhi, India.
3) Reference 3 par 3.2.9 "The Symmetrical Component Analysis," pages 52-56 does not include three-phase faults.
4) Reference 4 only addresses in Section 8-2 "The Balanced Three Phase Fault" that is a general balanced three-phase-to-phase-to-ground fault treated by symmetrical components. Dr. Gross biography includes Professor of Electrical Engineering at Auburn University in Alabama. Dr. Gross received the Outstanding Teacher Award for three consecutive years.
5) Reference 5 includes the symmetrical three-phase-to-phase-to-ground short circuit in Example 13.10 on page 449. Else, it includes Single-Line-to-Ground fault, Double-Line-to-Ground fault, and Line-to-Line Fault treated by symmetrical components. Dr. Bergen is a Professor in Department of Electrical Engineering and Computer Science at University of California, Berkeley, CA.
6) Reference 6 Chapter 10 covers "Conventional Fault Studies" including Three-phase-to-phase (balanced) fault. Paragraph 10.2 states that a three-phase fault is an unbalanced condition. Rather, it is a condition wherein all three lines of a system are shorted (a-b-c) or grounded (a-b-c-g) at a point. However, this is where the fallacy is, namely, the approach to those points is different. The a-b-c approaches to that point over the delta connection where there are phase-to-phase ungrounded voltages, while the a-b-c-g approaches to the point over the y-connection and the voltages are always grounded. Obviously, both approaches may be balanced. One needs a good understanding of the infinitesimal calculus (limits, limiting processes) and some background, how to treat the infinity (oo), since the equations for the power (and current, when voltage is kept constant and different from zero) included Z impedance in the denominator. Once the Z is approaching to zero the power is approaching to 3**0.5 x oo for delta (perhaps motor, could be heater, load bank, etc.) and to the oo for the y-connection. Whenever, the Z is different from zero, e.g. 0.01 or 0.0001 or etc., the power (current) for the delta connection will be 3**0.5 x oo, and the oo for the y-connection.
The symmetrical components. Luckily, one may notice a puzzling remark in Reference 6 on a page 207, which states "Note that the zero-sequence line voltage is always zero, even though zero-sequence phase voltages may exist. For this reason, it is not possible to construct a complete set of symmetrical components of phase voltages even when the unbalanced system of line voltages is know." This may be interpreted as a weakness of the symmetrical component approach, and a potential answer to the missing treatment or the three-phase-to-phase-to-ground faults in Reference 1 through 5 and a dubious treatment of this fault in the Reference 6 about a-b-c and a-b-c-g, which seems to be adhered to in some of the above postings without any proofs or references. Professor and Dr. Gungor is with Department of Electrical Engineering, University of South Alabama, Mobile, AL.
The referenced ANSI/IEEE Standard C37.010-1979 "American National Standard Application Guide for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current Basis" is in agreement with my "motor delta and y-winding concept of proof" by stating that "In general, the three-phase ungrounded fault imposes the most severe duty on a circuit breaker, since the first phase to interrupt has a normal-frequency recovery voltage of approximately 87 % of system phase-to-phase voltage.

 

[peterb]

Final words from me on this subject, I promise.
- I maintain that the analogy, to be valid, must use the same wye connected load model for both cases
- A wye connected motor is an example of a circuit that is normally connected with the common point ungrounded; think also about a wye-delta started motor. In any case, the occurrence of the 3-phase ungrounded fault itself connnects the three phase leads together to a common point
- Taking the example under discussion, I would like to see the difference calculated for the fault currents approached from the point of view of decreasing the fault impedances to zero (ie a bolted fault, which was the original question). This would preferably be done with a realistic value of source impedance
- The assumption of zero fault resistance leads to the worst-case value of fault current and is used for sizing switchgear and setting relays; source voltage is assumed balanced because it normally is balanced

 

[jbartos]

Suggestion:
A simpler approach to the ungrounded three-phase-to-phase fault and grounded three-phase-to-phase fault difference in magnitude, to avoid any difficult mathematics that could be viewed as a burden.
1. Consider a three-phase load (motor, heater or load bank) in y-connection that has its neutral grounded for simplicity, at first. Then, the line current in phase a, Iay, can be
Iay = Valn/Z = 120V/2ohms = 60Amps, or
Iay = Vab/Z = 208V/(sqrt3 x 2ohms) = 60Amps
where Valn is line to neutral (which is purposely considered grounded for this type of fault), Z is impedance between the line "a" and neutral "n", which may simulate the grounded three-phase-to-phase balanced fault if Vab, Vbc, and Vca ares used. Now, supposing that Z is reduced to 1ohm due to the more severe grounded three-phase-to-phase fault. Then
Iay = Valn/Z = 120V/1ohms = 120Amps, or
Iay = Vab/Z = 208V/(sqrt3 x 1ohms) = 120Amps
Next, supposing the Z is reduced to Z = 0ohm. Then
Iay = Valn/Z = 120V/0ohms = oo (infinity)Amps, or
Iay = Vab/Z = 208V/(sqrt3 x 0ohms) = oo (infinity)Amps
2. Next, consider a three-phase load (motor, heater or load bank) in delta-connection with the same Z, which has no neutral accessible and no ground. Then, the current in each Z, Iz, can be
Iz = Vab/Z = sqrt3 x Valn/Z = 208V/2ohms = sqrt3 x 60Amps = 104Amps
where Vab is line a to line b voltage, Z is impedance between the line a and line b, which may simulate the ungrounded three-phase-to-phase balanced fault if Vab is used. Now, supposing that Z is reduced to 1ohm due to the more severe ungrounded three-phase-to-phase fault. Then
Iz = Vab/Z = sqrt3 x Valn/Z = 208V/1ohms = 208Amps, or
Next, supposing the Z is reduced to Z = 0ohm. Then
Iz = Vab/Z = sqrt3 x Valn/Z = 208V/0ohms = sqrt3 x oo (infinity)Amps = oo'Amps
Iz is related to Iad (delta connection line current) by Iz = Iad / sqrt3. This means
Iad = sqrt3 x Iz = sqrt3 x Vab/Z = sqrt3 x sqrt3 x Valn/Z, or
Iad = sqrt3 x 208V/2ohms = sqrt3 x 104Amps = 180Amps, or
Iad = sqrt3 x 208V/1ohms = sqrt3 x 208Amps = 360Amps, or
Iad = sqrt3 x 208V/0ohms = sqrt3 x oo'Amps = oo''Amps
3. Finally, a comparison of the ungrounded three-phase-to-phase fault current (as demonstrated over delta connection) to the grounded three-phase-to-phase current (as demonstrated over y-connection) reveals:
Iad/Iay = (sqrt3 x Vab/Z)/(Vab/Z)=sqrt3, or
The line current Iad for the ungrounded three-phase-to-phase (balanced) fault current (as demonstrated over the delta connection) is sqrt3 times bigger than the line current Iay for the grounded three-phase-to-phase (balanced) fault current (as demonstrated over the y-connection) for the same terminal voltage Vab=Vbc=Vca=208V across y-connection and delta-connection.
4. The power requirement or duty for a circuit breaker or switchgear is:
4a. The grounded three-phase-to-phase fault
Py = 3 x Valn x Iay = 3 x (Vab/sqrt3) x Iay = sqrt3 x Vab x Iay, in VAs
4b. The ungrounded three-phase-to-phase fault
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x sqrt3 x Iay = 3 x Vab x Iay, in VAs
4c. A comparison of power requirements for the ungrounded three-phase-to-phase fault to the grounded three-phase-to-phase fault reveals:
Pd/Py = 3 x Vab x Iay/( sqrt3 x Vab x Iay) = sqrt3, or
The ungrounded three-phase-to-phase fault has sqrt3 times bigger power requirement for its circuit breaker or switchgear.
5. Safety related aspects:
Any circuit breaker or switchgear rated for the grounded three-phase-to-phase fault in the ungrounded three-phase three-wire power supply system (or similarly for higher number of phases power supply system) with ungrounded three-phase-to-phase faults is incorrectly rated and poses a potential hazard to its surroundings including personnel. Notice, that the medium voltage higher power rated circuit breakers/switchgear may explode and injure personnel. Therefore, there is a legitimate safety concern and the strict adherence to the cited industry standard is necessary to ensure the overall safety.

 

[jbartos]

The following are:
1) Additions: delta - y transfiguration
2) Clarifications: bringing ungrounded three-phase-to-phase fault to direct comparison to grounded three-phase-to-phase fault
3) Corrections (I beg your pardon): The ungrounded three-phase-to-phase fault is 3 times more powerful than the grounded three-phase-to-phase fault
to my previous posting:
A simpler approach to the ungrounded three-phase-to-phase fault and grounded three-phase-to-phase fault difference in magnitude, to avoid any difficult mathematics that could be viewed as a burden.

1. Consider a three-phase load (motor, heater or load bank) in y-connection that has its neutral grounded for simplicity, at first. Then, the line current in phase a, Iay, can be
Iay = Valn/Z = 120V/2ohms = 60Amps, or
Iay = Vab/(sqrt3 x Z) = 208V/(sqrt3 x 2ohms) = 60Amps
where Valn is line to neutral (which is purposely considered grounded for this type of fault), Z is impedance between the line "a" and neutral "n", which may simulate the grounded three-phase-to-phase balanced fault if Vab, Vbc, and Vca ares used. Now, supposing that Z is reduced to 1ohm due to the more severe grounded three-phase-to-phase fault. Then
Iay = Valn/Z = 120V/1ohms = 120Amps, or
Iay = Vab/(sqrt3 x Z) = 208V/(sqrt3 x 1ohms) = 120Amps
Next, supposing the Z is reduced to Z = 0ohm. Then
Iay = Valn/Z = 120V/0ohms = oo (infinity)Amps, or
Iay = Vab/(sqrt3 x Z) = 208V/(sqrt3 x 0ohms) = oo (infinity)Amps
2. Next, consider a three-phase load (motor, heater or load bank) in delta-connection with the same Z, which has no neutral accessible and no ground. Then, the current in each Z, Iz, can be
Iz = Vab/Z = sqrt3 x Valn/Z = 208V/2ohms = sqrt3 x 60Amps = 104Amps
where Vab is line a to line b voltage, Z is impedance between the line a and line b, which may simulate the ungrounded three-phase-to-phase balanced fault if Vab is used. Now, supposing that Z is reduced to 1ohm due to the more severe ungrounded three-phase-to-phase fault. Then
Iz = Vab/Z = sqrt3 x Valn/Z = 208V/1ohms = 208Amps, or
Next, supposing the Z is reduced to Z = 0ohm. Then
Iz = Vab/Z = sqrt3 x Valn/Z = 208V/0ohms = sqrt3 x oo (infinity)Amps = oo'Amps
Iz is related to Iad (delta connection line current) by Iz = Iad / sqrt3. This means
Iad = sqrt3 x Iz = sqrt3 x Vab/Z = sqrt3 x sqrt3 x Valn/Z, or
Iad = sqrt3 x 208V/2ohms = sqrt3 x 104Amps = 180Amps, or
Iad = sqrt3 x 208V/1ohms = sqrt3 x 208Amps = 360Amps, or
Iad = sqrt3 x 208V/0ohms = sqrt3 x oo'Amps = oo''Amps
3. Finally, a comparison of the ungrounded three-phase-to-phase fault current (as demonstrated over delta connection) to the grounded three-phase-to-phase current (as demonstrated over y-connection) reveals:
Iad/Iay = (sqrt3 x Vab/Z)/(Vab/(sqrt3 x Z))=3, or
The line current Iad for the ungrounded three-phase-to-phase (balanced) fault current (as demonstrated over the delta connection) is 3 times bigger than the line current Iay for the grounded three-phase-to-phase (balanced) fault current (as demonstrated over the y-connection) for the same terminal voltage Vab=Vbc=Vca=208V across y-connection and delta-connection.
4. The power requirement or duty for a circuit breaker or switchgear is:
4a. The grounded three-phase-to-phase fault
Py = 3 x Valn x Iay = 3 x (Vab/sqrt3) x Iay = sqrt3 x Vab x Iay, in VAs
4b. The ungrounded three-phase-to-phase fault
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x 3 x Iay = sqrt3 x 3 x Vab x Iay, in VAs
4c. A comparison of power requirements for the ungrounded three-phase-to-phase fault to the grounded three-phase-to-phase fault reveals:
Pd/Py = sqrt3 x 3 x Vab x Iay/( sqrt3 x Vab x Iay) = 3, or
The ungrounded three-phase-to-phase fault has 3 times bigger power requirement for its circuit breaker or switchgear.
5. Safety related aspects:
Any circuit breaker or switchgear rated for the grounded three-phase-to-phase fault in the ungrounded three-phase three-wire power supply system (or similarly for higher number of phases power supply system) with ungrounded three-phase-to-phase faults is incorrectly rated and poses a potential hazard to its surroundings including personnel. Notice, that the medium voltage higher power rated circuit breakers/switchgear may explode and injure personnel. Therefore, there is a legitimate safety concern and the strict adherence to the cited industry standard is necessary to ensure the overall safety.

Suggestion: Please notice, that the ungrounded three-phase-to-phase fault aligned with delta connection (e.g. motor, heater, load bank, etc.) can be transfigured (changed) to y-connection by:
Zay = (Zad x Zbd)/(Zad + Zbd + Zcd) = (Z x Z)/(Z + Z + Z) = Z/3, in ohms
Zby = (Zbd x Zcd)/(Zad + Zbd + Zcd) = (Z x Z)/(Z + Z + Z) = Z/3, in ohms
Zcy = (Zcd x Zad)/(Zad + Zbd + Zcd) = (Z x Z)/(Z + Z + Z) = Z/3, in ohms
Then, the ungrounded three-phase-to-phase fault power for delta-connection transfigured to the ungrounded y-connection is:
Delta connection power is:
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x sqrt3 x Iz = 3 x (Vab**2) / Z, in VAs
Transfigured to y-connection:
Pytransfigured= (Vab**2) / (Z/3) = 3 x [(Vab**2)] / Z = 9 x [(Valn**2)] / Z, in VAs
The y-connection power is:
Py = 3 x Valn x Iay = 3 x Valn x Valn/Z = 3 x (Vab**2)/(3 x Z) = (Vab**2)/Z, in VAs
Therefore, again by comparison of the ungrounded three-phase-to-phase power Pd to the grounded three-phase-to-phase power Py:
Pd/Py = Pytransfigured/Py = [3 x (Vab**2) / Z]/ [(Vab**2)/Z] = 3

 

[rhatcher]

jbartos,
I am still trying to understand the basics presented in your previous post and would like some help with the mathematics and theory you presented on 3 phase power. Specifically, I am having trouble following the two mathematical proofs you provided in the last post showing that the ratio Pd/Py=3. This is important to me as I have always thought that Pd/Py=sqrt3 and had a whole set of (imaginary?) equations and relationships in my mind supporting this (no wonder people give me wierd looks sometimes...I hope you can clear this up for me as I am now somewhat embarrassed!). Anyway, I want to understand this but it may be that you provided too much information for me to grasp at one time as I was merely confused when I read and tried to understand your 1st proof but became utterly confounded when confronted with your 2nd proof.

Cut and pasted from your from your previous post:

(1st Proof)

4a. The grounded three-phase-to-phase fault
Py = 3 x Valn x Iay = 3 x (Vab/sqrt3) x Iay = sqrt3 x Vab x Iay, in VAs
4b. The ungrounded three-phase-to-phase fault
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x 3 x Iay = sqrt3 x 3 x Vab x Iay, in VAs

(2nd Proof)

Delta connection power is:
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x sqrt3 x Iz = 3 x (Vab**2) / Z, in VAs
Transfigured to y-connection:
Pytransfigured= (Vab**2) / (Z/3) = 3 x [(Vab**2)] / Z = 9 x [(Valn**2)] / Z, in VAs
The y-connection power is:
Py = 3 x Valn x Iay = 3 x Valn x Valn/Z = 3 x (Vab**2)/(3 x Z) = (Vab**2)/Z, in VAs

That is pretty complicated for me so I tried to look at it this way as I thought it would clear it up for me.

1st Proof 2nd Proof
Pd = sqrt3 x 3 x Vab x Iay = 3 x (Vab**2) / Z
Py = sqrt3 x Vab x Iay = (Vab**2)/Z

I am confused because in each proof it is shown that Pd/Py=3, but when I try to compare one proof to the next, for example Py in Proof 1 (call it Py1) compared to Py in Proof 2 (Py2), the equivalency is beyond my grasp. Of course, I sometimes do not know when to quit so I tried substitution between the two proofs to see if that would clear it up for me. Wow...I knew I was out of my league when I discovered that:

Pd2/Py1= sqrt3 but Pd1/Py2=sqrt3*3

So, I am admitting your superior understanding of the topic as this makes no sense to me at all. This being the case it goes without saying that I did not understand:

Pytransfigured= (Vab**2) / (Z/3) = 3 x [(Vab**2)] / Z = 9 x [(Valn**2)] / Z, in VAs

Please be kind enough to share your knowledge of this with me so that I may have a chance at understanding the rest of your explanation of 3 phase fault currents.

Thank you in advance.

 

[jbartos]

Suggestions/Answers to: rhatcher (Electrical) Jun 8, 2001
marked by /// \\I am having trouble following the two mathematical proofs you provided in the last post showing that the ratio Pd/Py=3. This is important to me as I have always thought that Pd/Py=sqrt3 and had a whole set of (imaginary?) equations and relationships in my mind supporting this. Anyway, I want to understand this but it may be that you provided too much information for me to grasp at one time as I was merely confused when I read and tried to understand your 1st proof but became utterly confounded when confronted with your 2nd proof.
Cut and pasted from your from your previous post:
(1st Proof)
4a. The grounded three-phase-to-phase fault
Py = 3 x Valn x Iay = 3 x (Vab/sqrt3) x Iay = sqrt3 x Vab x Iay, in VAs
4b. The ungrounded three-phase-to-phase fault
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x 3 x Iay = sqrt3 x 3 x Vab x Iay, in VAs
(2nd Proof)
Delta connection power is:
Pd = sqrt3 x Vab x Iad = sqrt3 x Vab x sqrt3 x Iz = 3 x (Vab**2) / Z, in VAs
///Notice that the Iad is line current for the delta connection and Iz is current in the delta leg Z. They are related by Iad = sqrt3 x Iz, leading to:
sqrt3 x Vab x sqrt3 x Iz = 3 x Vab x Iz
Then, the Iz = Vab/Z, which when substituted yields
3 x Vab x Vab/Z = 3 x (Vab**2) / Z \\
Transfigured to y-connection:
Pytransfigured= (Vab**2) / (Z/3) = 3 x [(Vab**2)] / Z = 9 x [(Valn**2)] / Z, in VAs
///Vab = sqrt3 x Valn has been used in the above equation.\\
The y-connection power is:
Py = 3 x Valn x Iay = 3 x Valn x Valn/Z = 3 x (Vab**2)/(3 x Z) = (Vab**2)/Z, in VAs
///Notice that Iay = Valn/Z and Valn = Vab/sqrt3 or Valn**2 = (Vab**2)/3 \\
That is pretty complicated for me so I tried to look at it this way as I thought it would clear it up for me.

1st Proof 2nd Proof
Pd = sqrt3 x 3 x Vab x Iay = 3 x (Vab**2) / Z
Py = sqrt3 x Vab x Iay = (Vab**2)/Z

I am confused because in each proof it is shown that Pd/Py=3, but when I try to compare one proof to the next, for example Py in Proof 1 (call it Py1) compared to Py in Proof 2 (Py2), the equivalency is beyond my grasp. Of course, I sometimes do not know when to quit so I tried substitution between the two proofs to see if that would clear it up for me. Wow...I knew I was out of my league when I discovered that:

Pd2/Py1= sqrt3
/// Pd2/Py1 = [ 3 x (Vab**2) / Z]/[sqrt3 x Vab x Iay] = sqrt3 x Vab / (Z x Iay) =
= sqrt3 x Vab / Valn = sqrt3 x Vab / (Vab / sqrt3) = 3 , i.e. not sqrt3 \\
but Pd1/Py2=sqrt3*3
///Pd1/Py2 = [sqrt3 x 3 x Vab x Iay]/[(Vab**2)/Z] = sqrt3 x 3 x Z x Iay/Vab =
= sqrt3 x 3 x Valn / Vab = 3 x (sqrt3 x Valn)/Vab = 3 x Vab/Vab = 3, i.e. not sqrt3 x 3 \\\
So, I am admitting your superior understanding of the topic as this makes no sense to me at all. This being the case it goes without saying that I did not understand:

Pytransfigured= (Vab**2) / (Z/3) = 3 x [(Vab**2)] / Z = 9 x [(Valn**2)] / Z, in VAs
///Notice that
(Vab**2) / (Z/3) = 3 x (Vab**2) / Z for Zay = Zby = Zcy = Z/3 transfigured delta impedances into y-connection.
Vab = Valn x sqrt3 or (Vab**2) = 3 x (Valn**2)\\
Please be kind enough to share your knowledge of this with me so that I may have a chance at understanding the rest of your explanation of 3 phase fault currents.
Thank you in advance.
///You are welcome.
Please, notice that the motor coil, for example, has Pycoil = Valn x Iay for y-connection and Pdcoil = Vab x Iz, then
Pdcoil/Pycoil = Vab x Iz / (Valn x Iay) = (Vab x Vab/Zay)/(Valn x Valn/Z) = (Valn x Valn / (Z/3))/(Valn x Valn / Z) = 3
However, there is one sqrt3 legitimate, namely
Iz = Iad/sqrt3 = 3 x Iay/sqrt3 = sqrt3 x Iay
which means that the delta leg current is related to the y-connection line (identical to leg) current by Iz/Iay=sqrt3
which is causing a big problem in understanding in comparison to
Iad/Iay = 3, which represents delta-connection line current over y-connection line current.\\\

 

[rhatcher]

I am at a loss. I would like to work through this but right now I need a drink (actually, six or more sounds better). It has simply been one of those weeks for me. I hope you guys don't ban me for this....I will come back in the morning and in the light of day may see the error of my ways. In the meantime though (just in case the morning sheds no new light), will someone give me a reality check by either bailing me out or confirming that I am lost?....

Please note that being lost when you know that you are lost is not as bad as thinking that you might be lost but not knowing for sure......Of course, if you are lost and don't know it then all is well...maybe that is what they mean by the saying "ignorance is bliss".....that sounds appealing right now so as I said before, I am going now to find some "bliss" at the local drinking establishment and will be back in the morning to face reality.

 

[jbartos]

Cheer up! It is not a very easy stuff considering what you can find in many books or literature on this topic. I got me too.