Minimum Fault Current 1

[111R]

I often see a "minimum fault" specified on some fault study software that seems to typically be around 30-60% of the maximum 3 phase fault current. What is this referring to?

Is there a common fault resistance value that is used for these calculations? Minimum fault current is confusing since a highly resistive fault due to a wire laying on a non-conductive surface could have negligible current flow.

 

[7anoter4]

In my opinion, it is referring to IEC 60909-0 Short-circuit currents in three-phase a.c. systems -Part O: Calculation of currents.
1 General 1.1 Scope
In general, two short-circuit currents, which differ in their magnitude, are to be calculated:
the maximum short-circuit current which determines the capacity or rating of electrical
equipment; and
the minimum short-circuit current which can be a basis, for example, for the selection of
fuses, for the setting of protective devices, and for checking the run-up of motors.
2.5 Minimum short-circuit currents
When calculating minimum short-circuit currents, it is necessary to introduce the following
conditions:
- voltage factor cmin, for the calculation of minimum short-circuit currents shall be applied according to table 1 ;
-choose the system configuration and the minimum contribution from power stations and
network feeders which lead to a minimum value of short-circuit current at the short-circuit
location;
- motors shall be neglected;
- resistances RL of lines (overhead lines and cables, line conductors, and neutral conductors)
shall be introduced at a higher temperature:
RL=[1+a*(Tc-20)]*RL20
where
RL20 is the resistance at a temperature of 20 oC;
Tc=is the conductor temperature in degrees Celsius at the end of the short-circuit duration;
a=is a factor equal to 0,004/K,

 

[jghrist]

In the USA, rural electric cooperatives have traditionally added a fault resistance of 30-40 ohms when calculating minimum fault currents on overhead distribution lines. See

https://www.rd.usda.gov/files/UEP_Bulletin_1724E-102.pdf,

page 6. The actual fault resistance can vary from zero up to very high values where the fault current is less than load current. My experience has been that using 30 ohms results in such low current values that it is difficult or impossible to provide overcurrent protection for minimum faults. Consider that with no line or source impedance, the fault current for a 7200 volt fault through a 30 ohm impedance is only 240A.

 

[RobertNC]

My experience with the use of minimum fault current values has been while performing sectionalizing studies of distribution feeders. The interrupting rating of a protective device is compared to the maximum (bolted) fault current that it would be subjected to. The pickup setting of the protective device is selected to be comfortably higher than expected load current but less than some calculated minimum fault current value.

At electric cooperatives, most engineers calculate minimum fault current by inserting a specific value of resistance into a fault (30 or 40 ohms for overhead line sections and 20 ohms for underground line sections). These resistance values were referenced in a Rural Electrification Authority (now Rural Utilities Service) document from the 1940s and most electric cooperative engineers still use them because they have no other guidance from RUS. For 12 kV distribution feeders, minimum fault currents calculated this way are typically in the range of about 100 to 200 Amps.

At utilities that do not fall under the authority of RUS, most engineers calculate minimum fault current as a percentage of the bolted fault value.

Distribution analysis software that I am familiar with (I've mostly used Milsoft's WindMil) will calculate minimum fault current using either of the above methods.

You are correct about a conductor laying on a non-conductive surface having negligible current flow. The reality is that real faults are usually either very low resistance, or they have significantly more than 40 ohms of impedance and will not likely be detected by overcurrent relays. I still calculate a minimum fault value (usually at either 25% or 50% of maximum fault current) because everyone else that I know does it too. I've been a distribution engineer for electric cooperatives and municipal electric utilities for the past 22 years.

 

[cranky108]

I believe the minimum fault impedance is used for motor start studies, where one is looking at the largest voltage drop during the starting of the motor.

 

[unclebob]

A plant may be feed by two transformers with the possibility of using both or only one of them to take all the load.

 

[Parchie]

@unclebob,
Quick thinking there!
Y'all,
As the word implies, minimum fault current refers to the minimum value of fault current; happens when the system is feed with the least number of sources of fault current connected to it.

 

[marks1080]

There can be a fair amount of 'art' to doing this as well. You may have a plant with two transformers as the example above, or you may be interested in the minimum fault current of a line that is connected via two stations with multiple companion lines and possible loop through via other stations, along with generation at some buses.

Now how do you determine minimum fault current? Do you switch off everything except the line in question? Well, no... you have to minimize sources in a way that makes sense. Look at the number of different contingencies that you are expected to ride through and use that criteria, plus a little common sense, to minimize your system and use that value as your minimum fault.

 

[piterpol]

¿What to do when the fault current is less than load current?, i.e. ¿how do you protect circuits under these conditions?

 

[davidbeach]

If it's a radial distribution feeder, the easiest thing to do is add a recloser out a ways where the load current is lower. For a very long time that was the only option, but now relays have load encroachment capabilities and you can set the relay for the fault and have it blocked when the current looks like load and trip when it looks like a fault.

 

[cranky108]

¿What to do when the fault current is less than load current?

Simple, add another feeder circuit to split the load.

 

[RobertNC]

If load current is higher than the calculated minimum fault current in a distribution sectionalizing study, I've typically just added the average load current to the minimum fault current to get an 'adjusted minimum fault current'.

This is more likely to be an issue when using an ohm value (like 30 ohms) for calculating the minimum fault current. If minimum fault current is calculated as a percentage of the bolted fault current value (like 25% or 50%), load currents approaching that value might justify upgrades to the distribution system anyway.

The maximum fault current for a protective device is calculated at the start of a line section, but the minimum fault current is calculated at the end of the line section. If minimum fault current is lower than load current on a particularly long section of line, I would follow davidbeach's advice and consider installing an additional recloser somewhere downline.

 

[juleselec]

From an overcurrent protection coordination standpoint, the minimum fault is the smallest fault current that represents a "real" short circuit for that system. Normally, something like 80%-90% of the minimum fault current is used to set a fast definite time or instantaneous element, i.e. you want the protection relay to trip as quickly as possible whenever there is a real fault. Anything lower than the minimum fault current is considered to be some other kind of fault and would be handled by other protection elements, e.g. inverse elements, thermal overload, stall protection, etc.

So the "minimum" part of the term refers to the overall system conditions that would lead to the lowest fault level (even a bolted short with zero fault impedance). This would include configurations and parameters that:

- Increase the upstream source impedance, e.g. n-1 contingency cases (single line, transformer, etc), minimum generation (in island systems), highest ambient temperature (i.e. higher resistances in upstream lines), etc
- Lead to the lowest voltages at the point of fault, e.g. transformer tap settings, shunt capacitor banks out of service / reactors in service, etc (in IEC 60909, this is assumed to be covered by the catch-all prefault voltage factor c)
- Have the least fault contributions from other sources, e.g. large induction motors out of service

Edit: some programs like ETAP also include manufacturer tolerances, for example transformer impedance tolerances of +/-10%. So you would use the negative tolerance for maximum fault and positive tolerance for minimum fault. The argument is that unless you have measured the impedance in the field, the manufacturer could be out by that much.

 

[waross]

It doesn't make that much difference with a high X:R ratio but Available Short Circuit Current Tests are conducted with the transformer at operating temperature. The resistive component of the current will be higher with a cold transformer. eg: A fault on energization of a cold transformer.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[315kVA]

Here is some experience from marine business: we calculate the maximum s.c. by taking into account all generators running in parallel and adding the contribution from the motors (except those fed through frequency converters). Minimum s.c. is calculated with only one generator connected (the smallest one if they have different ratings) and without motor load. This can be a case in harbour when you have mostly non-motor loads such as heating, lighting, galley and entertainment equipment and some motors such as ventilation fans which can be neglected to get a more conservative result.

 

[cranky108]

315, that may work for your application. But when your smallest sync. unit is 1 MW, and you are part of a 1.5 TW grid, that dos not work as expected.

If the intent is for a motor start study, then that should be stated. But have someone guess at what you are wanting is just asking for wrong information.