Fault Currents and Time Current Curvex
Fault Currents and Time Current Curvex
(OP)
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
Thank you very much






RE: Fault Currents and Time Current Curvex
A. - I don't think it makes a difference. Since the fault is balanced, there would be no current from neutral to ground so the presence of ground wire doesnt affect the circuit.
RE: Fault Currents and Time Current Curvex
Is a 3 phase bolted fault, 3 phases shorted or 3 phases & ground shorted?
<<Reference
1. ANSI/IEEE Standard C37.010-1979 "American National Standard Application Guide for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current Basis."
Reference 1 par 5.1.2 "Types and Severity of System Short Circuits." A three-phase power system is subjected to the following types of faults:
1) Three-phase ungrounded fault (worst) (Three line conductors are connected together with zero impedance)
2) Three-phase grounded fault (less worst) (Three line conductors are connected to ground. The ground is assumed grounded and conductive with some impedance to the source; therefore, this fault is usually less severe than 1).)
3) Phase-to-phase ungrounded fault
4) Phase-to-phase grounded fault
5) Phase-to-ground fault (least severe); however, it may produce a higher fault current than any three-phase fault.>>
On a TCC, is the scale for 3 phase circuits, the line to line or line to ground voltage?
<<Please, clarify this since TCC appears to stand for Time-Current Curve, normally, and voltage would be a parameter not shown on any scale. It would indicate to which voltage base are TCCs or coordination charts referred to. Usually, the voltage is phase-to-phase; but it may be phase-to-neutral without any major problem, if properly identified.>>
RE: Fault Currents and Time Current Curvex
"3 phases shorted or 3 phases & ground shorted".
As jbartos mentioned, these are the appropriate fault types to study if you want a worst-case scenario. The other faults give lower currents.
RE: Fault Currents and Time Current Curvex
To jbartos: The voltage I''m referring to is what I've seen on a TCC where on the current scale it says " Current x 10 at 2400V" In the application I'm looking at this is line to ground but what if I need to look at a three pase fault on the same TCC?
RE: Fault Currents and Time Current Curvex
Take a simple example with a 13.8-2.4 kV transformer. Assume a 65A fuse protecting the transformer HV, with a 1600A power circuit breaker on the LV side.
To coordinate these two devices, select the base study voltage as 480V (arbitrary, on the assumption that you will want to coordinate other downstream breakers at this voltage). You will need to draw the fuse curve on the 480V base, so that 100A on the fuse curve would be equivalent to (100x13800/480) = 2875A. This is where you would line up the 100A line of the fuse characteristic curve, if you were doing this manually (using a somewhat antique fixture known as a light table or light box) - in today's world you will probably be using one of the available software programs, which make life so much easier.
The current multiplier is selected so that the curves end up around the middle of the log-log graph sheet. For our example, it would be appropriate to select a multiplier of 100.
After drawing the fuse curve you would then add the other curves as required to verify & document the coordination. Your graph sheet would be labelled "Current x 100 @ 480V"
You will normally draw separate curves for phase and ground faults, but the principles are the same.
RE: Fault Currents and Time Current Curvex
You didn't mention a transformer so I'm assuming you're working all at one voltage level.
Whether the load (motor) is actually connected delta or wye (and hence current flowing to ground or not) is pretty much irrelevant from a protection standpoint since motor protection curves will be expressed in line current whether the motor is delta or wye.
All of the above is in response to your statement "the application I'm looking at is line to ground...". Can you clarify that?
RE: Fault Currents and Time Current Curvex
I repeat there is no difference between: "3 phases shorted or 3 phases & ground shorted".:
<< It is a common practice to refute statements, theorems, lemmas, etc. rather than to repeat own statement without any substantiation or proof.
Three-phase-to-phase fault has the following relationship for its line (or phase) currents:
Ia' + Ib' + Ic' = 0 .................................................. Eq.1
This relationship is not so obvious for three-phase-to-phase-to-ground
(Ia + Ig/3) + (Ib + Ig/3) + (Ic + Ig/3) = Ig > 0 .......... Eq.2
Ia + Ib + Ic = 0 ...................................................... Eq.3
where Ib is the ground current that can be fairly large.
Since (based on the same supplied power/energy)
Ia'=(Ia + Ig/3) > 0 => Ia = Ia' - Ig/3 > 0 or Ia' > Ia
Ib'=(Ib + Ig/3) > 0 => Ib = Ib' - Ig/3 > 0 or Ib' > Ib
>>
Ic'=(Ic + Ig/3) > 0 => Ic = Ic' - Ig/3 > 0 or Ic' > Ic
Therefore, the fault caused by Ia', Ib', and Ic' is actually energetically bigger than the fault caused by Ia, Ib, and Ic fault currents.
RE: Fault Currents and Time Current Curvex
I made two assumptions: #1 - the voltages, loads, and impedances in the pre-fault system are balanced; and #2 - there is no fault resistance.
I believe these are both commonly-made simplifying assumptions. Perhaps in overhead lines there is more attention to fault resistance, but the assumption of zero fault resistance is always conservative [unless I believe your post above which indicates that ungrounded/infinite ground resistance is the worst case... but I don't believe it... more on that at the end of this message]. Also, given the fact that the poster is a self-proclaimed newby, I don't think that any overly-complicated analysis assumptions are warranted.
IF you can accept my two assumptions, THEN the ungrounded three-phase fault gives zero voltage at the fault point where the three phases are shorted (Due to symmetry). Therefore, adding a ground wire to this fault point would not change the currents or voltages, and the grounded/ungrounded 3-phase symmetrical faults give identical results.
I have to say I'm confused by you discussion of Ia, Ib, Ic (grounded) and Ia' Ib' Ic' (ungrounded). For your discussion to be relevant, let's assume a voltage imbalance.....no problem with that. You predict that Ia'>Ia, Ib'>Ib, and Ic'>Ic... i.e. all three fault currents are higher in the ungrounded 3-phase fault than in the grounded 3-phase fault? Adding a parallel path (through ground) decreases fault current? Intuitively, that just doesn't seem right. Seems more likely that subtracting the vector quantity Ig/3 will increase some of the three currents and decrease others. Maybe I have misinterpreted the meaning of your symbols.
Once again... sorry to be a little bit argumentative. Just trying to defend my little slice of cyberspace
respectfully
electricpete
RE: Fault Currents and Time Current Curvex
Again, thats to all for some very interesting discussions. I'm on a learning curve and very much appreciate the information.
RE: Fault Currents and Time Current Curvex
There is no residual current, hence Ig in your equation is zero. Ia=Ia', Ib=Ib', Ic=Ic' and there is no difference between the three phase and three phase plus ground fault current.
Newbie_Bill, as I said above, you probably need to draw separate curves for the phase and ground faults, although I would draw them both at 4160V.
RE: Fault Currents and Time Current Curvex
1. To fully understand the Ig path one may need some background in common-mode currents and differential-mode currents and their paths.
2. There are different aspects of faults:
2.1 The maximum magnitude of the fault current. Please, notice that single line to ground fault may have the higher fault current magnitude than the any three-phase-to-phase fault.
2.2 The fault judged energetically are in context with fault damages due amount of energy developed by the fault.
3. Ultimately, protective devices, as hardware, are developed by their manufacturers who make practical, power distribution as well as market oriented research, assumptions, development and production. Sometimes, the textbook or EPRI research kinds of assumptions are not good enough. Somehow, the more mature engineering, design and practice are needed to perfect the hardware and satisfy customers/clients requirements.
4. Under assumptions that the power source parameters and transmission parameters stay the same, and Ig is different from zero, the Ia', Ib' and Ic' > Ia, Ib, and Ic, respectively.
4. One must exercise utmost caution when dealing with three-phase-to-phase faults, three-phase-to-phase-to-ground faults, line-to-line faults, line-to-line-to-ground faults, and line-to-ground faults, especially, when line-to-line voltage base is used and line-to-neutral voltage base is used. Also, there are relationships between line-to-line faults and three-phase-to-phase faults, etc. that have not been invoked yet.
RE: Fault Currents and Time Current Curvex
Points 1, 3: The method of symmetrical components has been in general use by generations of electrical engineers over the course of many decades for calculating balanced and unbalanced system fault conditions, and this method states that a for a balanced 3-phase fault, whether or not the fault point involves ground, there is no zero sequence current flowing and hence no residual current. All fault current calculation methods in use today are based on the application of the symmetrical component method.
Point 2: I agree fully that a line to ground fault can result in a higher current than the 3-phase fault. However, this would be as a result of the system grounding configuration that could result in a low zero sequence impedance and has nothing to do with the point under discussion.
Point 4: The basic assumption that Ig is different from zero is not correct for the balanced 3-phase fault condition
RE: Fault Currents and Time Current Curvex
1. Please, notice that 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."
does not use a word "balanced" or "balanced fault". These words frequently appear in textbooks to elucidate the symmetrical components, simplify their mathematics, disregard Voltage or Current versus Time effects, shorten a paper to be published, to fit normal "six" or so page limit in some society Journals, etc. In practice, there is mostly unbalanced load, somewhat unbalanced power supply voltage source, Voltage or Current versus Time effects, etc.
2. If anyone has any suggestions or opinions that the balanced loads options shall be included in standards and those ungrounded three-phase-to-phase fault be the same (energetically) as the three-phase-to-phase grounded faults, one may wish to contact:
http://www.ansi.org
or
http://www.ieee.org
with ones suggestions. Who knows, may be the next revision of those standards will reflect them?
3. Incidentally, Referenced ANSI/IEEE Standard C37.010-1979
states "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. The corresponding value for grounded three-phase fault is 58 % when Xo=X1 and up to 75 % on an effectively grounded system. See IEEE Std 32-1972, Requirements, Terminology, and Test Procedures for Neutral Grounding Devices."
RE: Fault Currents and Time Current Curvex
On the subject of the 2400v label on a time-current curve for 4160 system... I have no suggestions as to what they're trying to convey. Anyone else?
RE: Fault Currents and Time Current Curvex
Is a 3 phase bolted fault, 3 phases shorted or 3 phases & ground shorted?
Original answer, by electricpete:
I don't think it makes a difference. Since the fault is balanced, there would be no current from neutral to ground so the presence of ground wire doesnt affect the circuit.
This answer is correct in terms of the magnitude of fault current. jbartos post of May 22 states that the grounded 3-phase fault has a higher magnitude current than the ungrounded 3-phase case - THIS IS NOT SO. The standard quoted relates to the interrupting duty of circuit breakers, not to the magnitude of fault current.
I maintain that the current magnitude is the same for both the grounded and ungrounded 3-phase fault. The term "bolted fault" refers to a metallically connected fault (such as safety grounds being left on a transmission line), rather than a flashover through air, which involves arc resistance and consequently lowers the fault current.
RE: Fault Currents and Time Current Curvex
1. There is no need to be burden with symmetrical component approach and tedious mathematics that many textbooks include. Incidentally, they often use line to neutral positive sequence model for the balanced three-phase fault, not further specified, if they happen to treat the three-phase faults.
2. Simply, imagine a squirrel-cage motor with winding ends brought to the terminal box so that it can be connected into y and then switched to delta connection.
2a. Consider y-connection first with its neutral point grounded (that can be done). The motor is connected to the small protective device that will see it as a three-phase-to-phase to ground fault (it may be balanced).
Power can be arbitrarily set for this illustrative comparison of faults.
Py=3 x Vln x Ia = 3 x Z x I**2 = 3 x (Vln**2)/Z
e.g. on 208V/120V 3phase 60Hz system, there may be
Py= 3 x (120**2 V)/200ohms = 216 VA
2b. Consider delta connection, ungrounded, that will represent the three-phase-to-phase ungrounded fault (balanced).
Pdelta= 3**0.5 x Vac x Ia = 3**0.5 x (Vab**2)/Z
for the same Z and 208V/120V 3phase 60Hz system
Pdelta=3**0.5 x (208**2)/200ohms = 375 VA
and 375/216 = 3**0.5 as it should be (see textbooks) for this examples.
For larger and larger HP rated motor the power will be increasing, more and more representing the discussed faults with the delta connection consuming more energy, and more current with the 3**0.5 multiplier higher than the Py. This result suggests what the ANSI/IEEE standard clearly states that "...three-phase ungrounded fault imposes the most severe duty on a circuit breaker..." which anyone is welcome to challenge.
3. Answer to: Newbie_Bill (Visitor) May 22, 2001
To jbartos: The voltage I''m referring to is what I've seen on a TCC where on the current scale it says " Current x 10 at 2400V" In the application I'm looking at this is line to ground but what if I need to look at a three phase fault on the same TCC?
<<If you happen to have motors connected into y-connection, then the current is line current and voltage is line-to-neutral, Van, and you tend to protect it by a proper protective devices at the appropriate voltage, which you state so on the chart. If you happen to have delta connected load, transformer, then the delta connection will need line-to-line voltage Vab, or 3**0.5 x Vln and the current will be related to this voltage and aligned appropriately with the load. Normally, in three-phase three-wire system, the voltages are line-to-line and currents are line currents, Ia, Ib, and Ic. See Reference: ANSI/IEEE Std 242-1986 "An American National Standard IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems," (Buff Book), on page 544 Figure 240 "Selection of Main Overcurrent Relay Curve.">>
RE: Fault Currents and Time Current Curvex
Do the impedances in your delta ungrounded three-phase fault represent fault (arc) impedances, or source impedance?
IF we ASSUME zero fault (arc) impedance, THEN the only remaining impedance is the source impedance feeding the fault (transformers, cable etc). If I short these together at the point of the fault (no fault resistance), I end up with an ungrounded-wye configuration, not a delta configuration. (agreed?)
Am I correct in concluding that you have assumed a non-zero fault (arc) impedance to give your delta model of the ungrounded 3-phase fault?
RE: Fault Currents and Time Current Curvex
jbartos - that is an interesting example and helps me start to understand where you're coming from.
Do the impedances in your delta ungrounded three-phase fault represent fault (arc) impedances, or source impedance?
<<Yes, in this very approximate example to grasp the gist of the problem. If one happens to be lucky enough and is awarded by some research contract/grant, e.g. from EPRI or similar entity (about $500,000 or more), one may add very impressive mathematics and proofs.>>
IF we ASSUME zero fault (arc) impedance, THEN the only remaining impedance is the source impedance feeding the fault (transformers, cable etc). If I short these together at the point of the fault (no fault resistance), I end up with an ungrounded-wye configuration, not a delta configuration. (agreed?)
<<Not quite, do not forget that you may end up with 3**0.5 x infinity which is different from infinity without the square root of 3 multiplier in the abstract world admitting the infinity reasoning. In the real world, there will always be very small impedance Z forming delta-connection for the three-phase-to-phase ungrounded fault and very same small impedance Z forming Y-connection for the three-phase-to-phase grounded fault. This appears to be a clue resolving the fallacy of these fault differences, which anyone may fall for.>>
Am I correct in concluding that you have assumed a non-zero fault (arc) impedance to give your delta model of the ungrounded 3-phase fault?
<<Yes, in the real world. Not, in the abstract world.>>
<<Comment: There have been many books, textbooks, and papers, I was reading, and I cannot seem to find any suitable detailed treatment of those two faults. If you happen to know any, please, post them.>>
RE: Fault Currents and Time Current Curvex
In looking at your "simplified" squirrel cage motor example, the two cases that you propound are not equivalent. If you use the example of the wye connected motor with neutral grounded for the analog of a 3-phase grounded fault, then you must use the same model, with the neutral ungrounded, for the 3-phase ungrounded fault. Introducing the delta connected model is not valid, as this represents a totally different condition. In any case, I have a basic problem with your statement that "you may end up with 3**0.5 x infinity which is different from infinity without the square root of 3 multiplier in the abstract". When translated into real world terms, infinity is still infinity (actually, I think that you really meant "zero" impedance here) and any statement to the contrary is splitting hairs.
However, I digress. The central point here is that when you extend your analysis using the same model with different connections for the two cases, you will see that there is no difference between the current flowing whether your motor neutral is grounded or ungrounded. This comes back to the original question.
I think that you are trying to reinvent the wheel here. The method of symmetrical components is not a "burden", nor a mathematical abstraction - it is used daily by practicing engineers involved in real world power systems. I would urge any engineer new to the field to get a solid understanding of this topic, in order to grasp the basics of balanced and unbalanced system operations.
RE: Fault Currents and Time Current Curvex
RE: Fault Currents and Time Current Curvex
RE: Fault Currents and Time Current Curvex
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.
RE: Fault Currents and Time Current Curvex
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.
RE: Fault Currents and Time Current Curvex
- 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
RE: Fault Currents and Time Current Curvex
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.
RE: Fault Currents and Time Current Curvex
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
RE: Fault Currents and Time Current Curvex
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.
RE: Fault Currents and Time Current Curvex
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.\\\
RE: Fault Currents and Time Current Curvex
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.
RE: Fault Currents and Time Current Curvex