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why neglect load current in fault studies? 2

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15thaugust

Electrical
Jan 25, 2015
3
Why load current is usually neglected in fault studies? Fault current is usually 5-6 times the load current. How we can neglect load current which is as high as 20%?

regards

Sunil
 
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Several ways of looking at the same thing:
The fault current is the maximum current the transformer is capable of.
The only impedance considered for for fault current is the transformer impedance.
Fault current is based on bolted fault conditions, or zero impedance. Putting a load impedance in parallel with a zero impedance bolted fault still results in zero impedance.
Under bolted fault conditions, the load voltage is zero and with zero voltage, the load current is zero.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I agree with @waross. I remember my school instructor talking about motor load current (can't remember the actual terminology) - where if there are enough motors in the system and a fault occurs, there may be enough current contribution from the motor that it plays a factor in the actual fault study... but overall (and so far in my career) I've seen load current neglected in fault studies...



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The Wye Wye Transformer Connection video:
 
The current contributed to a fault by motors is not load current, although it is a multiple of the motor ful load current. This is regenerated power and is in the opposite direction to load current. The motor contribution acts as a current source feeding the fault in parallel with the transformer current.
The motors act as induction generators for a brief time. The actual time depends on the inertia of the motors and, for overhauling loads, may be continuous.
In any event, the motor contribution is greatest at the moment of the fault and contributes to the fault current and to the maximum magnetic forces on the conductors carrying the combined fault current.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Number one, fault current is more likely 10 times or more load current, unless you are in a very weak area.
Number two, you can't simply add load current to fault current because load current is at an angle near zero, and fault current is closer to 80 to 90 degrees. When you do the vector math, you will find the added current is not very much.
Also, the load does decrease during a fault because of the drop in positive sequence voltage.

So if you want to include the load current in a fault, you need to first determine what the faulted condition load current is.
In the case of large motor load, the inductive feedback might exceed the normal load (rare on transmission, which is why I don't consider it).
Also, you need to consiter the contribution from grounded wye capacitor banks in your fault calculations.
 
Thank you cranky108. If we consider load current as 1 pu resistive and fault current as 5 pu inductive and add them vectorially, the result would be sqrt(26) which is very close to 5. So even if load current is 20% we can ignore it. Beautiful explanation.

Thanks
 
What load current. When the transformer is delivering full fault current the load voltage will be zero. Zero Volts implies zero Amps. Zero at any phase angle is still zero! It is safe to ignore the zero load current.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Bill, that's a great answer in the radial distribution feeder world. In networked transmission systems things get more complicated.

Load in the faulted phase may go to zero, but the unfaulted phases still wants to do its own thing. The direction on load current vs. fault current can either desensitize or increase the sensitivity of the protection, particularly when dealing with distance protection.

It was a huge wake up call for me when I transitioned from the radial world to the networked world. Somethings I thought I knew turned out to not apply at all any more.
 
Thank you David.
I took the question to be in relation to fault studies where the load current could be ignored. I am going to do some browsing to try to get a better feeling for the implications of load current on fault protection in the networked world.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I don't always neglect load current in radial systems once I experienced a miss-coordination caused by it. Instantaneous pickup on a feeder was set the same as the definite time pickup of the main. Neglecting load current, all was well since the delay provided a sufficient coordinating interval. The real event was fault that was below the pickup of the feeder instantaneous, but the fault current combined with load current on the healthy feeders to put the main above pickup.
 
Thanks for sharing that stevenal.
I see my error.
The maximum fault current is limited by the transformer impedance and the load impedance is effectively shorted out.
However in the event of a far out fault, where the conductor impedance combines with the transformer impedance to further limit the fault current, there may be upstream issues.
My comments are valid for the maximum fault current, but there may be issues with fault currents that are less than the maximum.
The current on the feeder under consideration will not be more than the transformer can supply, but the less than maximum fault current may combine with loads on healhty feeders and result in coordination issues.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Simply the fault current is the upset limit of a large load current with a greater angle the system can provide on the faulted phase(s), whether it started with 10% or 100% of full load current on the faulted phase is out of the equation.
 
If you are that concerned, you can always do the two step approach: Run the load flow to see what the normal flows are. Then do the fault study and superimpose the two results.

Our protection folks typically use Aspen to run their fault studies. I found you can add load to those models if you like but most protection folks do their studies without load.

Like Cranky says, I'm used to fault currents being much higher than the load current, but if you situation has a 5 to 1 ratio of fault to load current, then superposition will work.
 
There is a methodology that does include the pre-load in fault calculations, especially mapping network currents. And the motor internal rotor voltages would be less than 1.0 pu if the motor is loaded.
 
It seems to be implied so far that by fault current is meant three-phase faults. However, with earthfaults, especially with high impedance earthed systems, load current can make the difference as to whether a directional relay operates correctly. Also, with distance relaying, load current can affect phase selection (particularly a problem with the older type single measurement element relays)as well. With remote zero sequence infeeds one often has to deal with load current superimposed on zero sequence current which requires prudent consideration when employing distance relays.
 
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