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]

Q.Is a 3 phase bolted fault, 3 phases shorted or 3 phases & ground shorted?

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.

 

[jbartos]

Suggestions:
Is a 3 phase bolted fault, 3 phases shorted or 3 phases & ground shorted?
<<Reference
1. ANSI/IEEE Standard C37.010-1979 &quot;American National Standard Application Guide for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current Basis.&quot;
Reference 1 par 5.1.2 &quot;Types and Severity of System Short Circuits.&quot; 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.>>

 

[electricpete]

I repeat there is no difference between:
&quot;3 phases shorted or 3 phases & ground shorted&quot;.

As jbartos mentioned, these are the appropriate fault types to study if you want a worst-case scenario. The other faults give lower currents.

 

Thank You all for your responses!
To jbartos: The voltage I''m referring to is what I've seen on a TCC where on the current scale it says &quot; Current x 10 at 2400V&quot; 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?

 

[peterb]

When you are dealing with the coordination of a set of devices operating at different voltages (say on HV and LV sides of a transformer), you need to have a common basis. The time current curves are drawn for the voltage level of most of the devices in question, with other device curves converted to the study voltage level.
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 &quot;Current x 100 @ 480V&quot;

You will normally draw separate curves for phase and ground faults, but the principles are the same.

 

[electricpete]

I think that you are safe to assume in your case that line current is the proper current to look at during 1- three phase fault; 2 - overcurrent relay time-current curves; 3 - motor protection curves.

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 &quot;the application I'm looking at is line to ground...&quot;. Can you clarify that?

 

[jbartos]

Suggestion to: electricpete (Electrical) May 21, 2001
I repeat there is no difference between: &quot;3 phases shorted or 3 phases & ground shorted&quot;.:
<< 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.

 

[electricpete]

jbartos - I always appreciate your thoughtful input. I apologize for being curt in my previous post. It's always good to explore the differences. Although I know you are well aware of all the subtleties of fault current analysis. It is the nature of the medium that we end up quarelling over minor differences. Never the less I feel complelled to defend my post.

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 ;-) I Look forward to more discussions on this or other posts.

respectfully
electricpete

 

electricpete - What I mean by line to ground is that I'm looking at a TCC for a 4160Y distribution circuit and it's voltage scale is 2400 V. I'm therefore a little confused on plotting a 3 phase fault current on that scale and then putting the recloser curve on that 2400 V scale for a 4160 (three pase) fault.

Again, thats to all for some very interesting discussions. I'm on a learning curve and very much appreciate the information.

 

[peterb]

jbartos, your equations don't conform with standard symmetrical component analysis. Your equation (2) assumes that there is ground (zero sequence) current flowing, which is not the case for a balanced three phase fault. The fact of the matter is that, for a balanced 3 phase fault, there is no unbalanced current (negative or zero sequence) flowing in the fault.
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.

 

[jbartos]

Suggestions:
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.

 

[peterb]

Responding to jbartos post of May 24, I am at a loss to know the basis for stating that the current in the 3-phase to ground fault is different from that in a 3-phase fault.

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

 

[jbartos]

Suggestions:
1. Please, notice that the referenced ANSI/IEEE Standard C37.010-1979 &quot;American National Standard Application Guide for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current Basis.&quot;
does not use a word &quot;balanced&quot; or &quot;balanced fault&quot;. 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 &quot;six&quot; 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 &quot;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.&quot;

 

[electricpete]

my apologies for giving an incorrect answer on the subject of grounded vs ungrounded 3-phase faults. jbartos has given the definitive answer, since it comes from the standard.

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?

 

[peterb]

Original question:
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 &quot;bolted fault&quot; 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.

 

[jbartos]

Suggestion to the previous posting Peterb May 29, 2001:
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 &quot;...three-phase ungrounded fault imposes the most severe duty on a circuit breaker...&quot; 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 &quot; Current x 10 at 2400V&quot; 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 &quot;An American National Standard IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems,&quot; (Buff Book), on page 544 Figure 240 &quot;Selection of Main Overcurrent Relay Curve.&quot;>>

 

[electricpete]

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?

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?

 

[jbartos]

Suggestions/comments on: electricpete (Electrical) May 30, 2001
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.>>

 

[peterb]

Sorry, jbartos, I still think that there is a basic fallacy in your argument and the example that you have provided.

In looking at your &quot;simplified&quot; 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 &quot;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&quot;. When translated into real world terms, infinity is still infinity (actually, I think that you really meant &quot;zero&quot; 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 &quot;burden&quot;, 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.