Short Circuit Calculation

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Markdegracia0370

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Hi guys;

Can you help with this, in how to calculate the short circuit calculation of this single line diagram to properly size the protections.

Please see attached file
 

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Is the 4160 system utility or customer owned? You can start with just figuring out what you get at the secondary of the transformer assuming infinte primary fault current. This would be full load current divided by the impedance, or in your case about 31.4K. Then you can use this calculator to find the available fault current at the end of your conductors:

https://www.mikeholt.com/freestuff-shortcircuitcalculations.php

Generally I would go with those values, but if it really would save you a bunch of money and hassle if the fault current was lower, you could take into account the impedance of the primary, but it can be hard to get values for this from the utility. Dont forget about your series ratings if applicable.
 
electrofelon said:
Is the 4160 system utility or customer owned? You can start with just figuring out what you get at the secondary of the transformer assuming infinte primary fault current. This would be full load current divided by the impedance, or in your case about 31.4K. Then you can use this calculator to find the available fault current at the end of your conductors:

https://www.mikeholt.com/freestuff-shortcircuitcalculations.php

Generally I would go with those values, but if it really would save you a bunch of money and hassle if the fault current was lower, you could take into account the impedance of the primary, but it can be hard to get values for this from the utility. Dont forget about your series ratings if applicable.
Click to expand...

Hi sir;

The 4160 is our system utility.
 
Since you are using this to size the equipment, the infinite bus method should be used as noted above for at the secondary teminals of the transformer. Then this current is used to determine the fault currents downstream.

I would think this is design work that would have to stamped by a PE and that PE should know how to do this.

Also, if you are also going to do an arc flash study, you will need the available fault current from the utility. You will need the available fault current anyhow to label the service equipment in accordance with NEC 2017 Art. 110.16(B), assuming you are in the USA
 
Markdegracia0370 said:
Hi guys;

Can you help with this, in how to calculate the short circuit calculation of this single line diagram to properly size the protections.

Please see attached file
Click to expand...

that is involved and should be done by someone familiar with that work, imo using a program, although this could be done by hand
why are they asking you to do it if you don't know how?
is this actual work or an educational exercise?

get utility sc data, mva, x/r, kA, etc.

1 label all Z on the drawing, xfmrs, conductors
2 determine any motor contributions
3 calculate sc I at end of main cb feeder (min) and at the CB lugs (max) (basically add all series Z, divide into system V)
3 calculate sc I at the end of each feeder (min) and at the CB lugs (max)

why is the main that feeds the bus the same size cond as all the sub-feeders?
 
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As other's have said, it isn't possible to determine with the information provided. The determination of the available fault current is usually done with software by a professional engineer. The process involves reducing the impedance network to a Thevenin equivalent impedance measured from the point of interest. All conductor sizes/lengths, motor horsepower, would need to be known/estimated.

The transformer secondary conductors are incorrect. You'll need 6 sets of 350, (assuming conductors in conduit without derating) or some other combination providing required ampacity. To properly calculate the available fault current at the switchboard you will need to know the impedance of the conductors, which requires that the correct size, length and # of sets be determined.

It's not likely that a distribution center (Switchboard?) at this ampacity will be available with bus that is braced for anything less than 65 kaic at 480V, same goes for the 3200 amp insulated case breaker. You will have to be careful if you are using distribution panels in the switchboard, I would recommend matching the breakers AIC rating to the switchboard bus rating. This allows for a fault duty of 2.78% equivalent impedance, which the switchboard would likely never see, even if the utility replaces the transformer with a lower impedance version in the future.
 
Markdegracia0370 said:
Hi guys;

Can you help with this, in how to calculate the short circuit calculation of this single line diagram to properly size the protections.

Please see attached file
Click to expand...

Something is odd about this SLD. You have cable going between the LV SWGR main and the LV SWGR bus. This would be directly connected together unless mounted separately; but why, when most manufacturers provide it internal. Also, why so many 400A breakers? A typical LV SWGR lineup has 4 breakers per vertical section, and the cable would directly go out to the motor control, and not need an intermediate 400A breaker.

Do you also need to set protection on the VCB ahead of the transformer?

Curious as to what is this facility?

Regardless, you need the SC info at the HV side of transformer to proceed.
 
kingpb said:
Something is odd about this SLD. You have cable going between the LV SWGR main and the LV SWGR bus. This would be directly connected together unless mounted separately; but why, when most manufacturers provide it internal. Also, why so many 400A breakers? A typical LV SWGR lineup has 4 breakers per vertical section, and the cable would directly go out to the motor control, and not need an intermediate 400A breaker.

Do you also need to set protection on the VCB ahead of the transformer?

Curious as to what is this facility?

Regardless, you need the SC info at the HV side of transformer to proceed.
Click to expand...

This is the substation set-up we design at underground (70meters below the ground). From VCB to the Distribution Panel (8 units of 400A CB) was installed at Underground and the recloser/Cutout was installed at the surface.
Its very hard to calculate the SC due to continuous advancement of the underground operation and of course our cable will also consistently extending.
 
I assume you have the fuse curve for the fuses ahead of the VCB. If so, then you can use that to see how much fault current is going to flow before those fuses trip. Also, by plotting the device and transformer curves you will be able to see how you can coordinate such that the settings will trip the devices as soon, and as quickly as possible. The risk is that for some reason you have such a low amount of fault current flowing the relay will not pick-up and trip on instantaneous and you could have some time delay before the short time pick-up sees it.

If you have enough current on the HV side to supply the inrush to the transformer, then it would be reasonably safe to estimate that the fault current will be enough greater than that and the relay on the VCB or the fuse will detect it and trip. But in the end, a short circuit analysis still needs to be completed. Another issue that could arise is that the fault current could be low enough that if you used a maintenance switch on the ACB, the switch setting may not be low enough and you could falsely think you are being protected.
 
Markdegracia0370 said:
This is the substation set-up we design at underground (70meters below the ground). From VCB to the Distribution Panel (8 units of 400A CB) was installed at Underground and the recloser/Cutout was installed at the surface.
Its very hard to calculate the SC due to continuous advancement of the underground operation and of course our cable will also consistently extending.
Click to expand...

google 'msha short circuit program'
free basic program for ug mining
https://arlweb.msha.gov/TECHSUPP/ACC/shortcircuit/shortcuit.htm
 
For IEC World, IEC 60076-5 indicates, for up to 24 kV systems, 500 MVA as a maximum short-circuit power. However the tabulation starts from 7.2 kV only. A 40 kA utility circuit breaker it is usually provided.
The minimum short-circuit current -if it will be for 4.16 kV as reserve protection for secondary, it has to be for the transformer impedance and the supply cable length up to another circuit breaker. In my opinion, you may increase the transformer impedance with 7.5% tolerance and a 300 ft. low voltage cable[ 100 m].
However, NEC Art.450 Table 450.3(A) recommends Maximum Rating or Setting of Overcurrent Protection for Transformers Over 1000 Volts (as a Percentage of Transformer-Rated Current)
600% of transformer rating in order to overcome the inrush current when connected.
The low voltage setting 125% the transformer rated current.
 
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