reducing S.C.at 0.48kV

[Evgueni]

Hello Gentlemen
It is my first time here.

There is short history about my problem. This project has radial distribution power system with the following voltage levels 115 kV , 13.8 kV , 4.16 kV , 0.48 kV.
After S.C. calculation I got S.C. as 30.7 kV r.m.s. on the bus 0.48 kV hich is connected with transformer 13.8/0.48 kV , 1600 kVA , Z%=6 , X/R=7.1 . Some loads connected with this transformer has very small value ,
as 1.5 kW. For this load the cable sizes 12 AWG or 10 AWG are reasonable .
To calculate cable current- thermal capability to withstand impact of the S.C. I used Std. 242-1986 p.328 formula (ICEA P-32-382-1969 [5] : Is.c.=CM/SQRT(Tf*Fac/(0.0297*LOG10((250+234)/(90+234))))/1000 ,
· For fault duration Tf I used 0.026 sec or 1.56 cycle
· Skin effect Fas was assumed as 1
· CM- was taken as 6529 Circular mils for 12 AWG
After calculation I got value as 2913 A or 3 kA for cable 12 AWG , so for
this particular case I have to reduce S.C. value from 30.7 kA to 3 kA (see
att.A)
1 question : Is it possible to reduce S.C. from 30.7 kA r.m.s. to 3 kA r.m.s. ?
2 question : If yes , could I make it by LV breakers with incorporated S.C. limiters or similar devises ?
3 question : If no , which alternative solution could I us ?

 

[mpparent]

Let me understand,...

You have about 31kA available at your main gear. What is the SC level at the load you are inquiring about?

Mike

 

[davidbeach]

A few feet of 12 AWG will significantly reduce your fault current. Through fault is what you would be concerned about and that is dependent on the fault current at the load end of the circuit, not the fault current at the source end of the circuit.

 

[BJC]

1. Yes, but why bother. I done projects where I would have been happy to get 30.7 kA.
42 KA is a standard rating for switchgear & panels.

You can reduce the 480 volt short circuit current by using more smaller 13.8/480 volt transformers but your going to spend more money on 13.8 switchgear than the cost of higher rated 480 volt breakers.

The location ( distance from the 13.8/480 xfmr) of your panels and mccs will do a lot to limit fault current.
Try simulating a 400 amp 480/277 volt panel 100, 150 and 200' away from your transformer. The fault current drops significantly with a cable run.

 

[cuky2000]

Evgueni,

I suggest not using the ICEA formula to determine the available SC at the end of the feeder. This formula is to determine the maximum short circuit withstands current of the cable.
For the short circuit reduction to 3 kA you have to add the enough cable length to have significant large impedance above 10 Ohms.
To reduce the SC available, current limiting protective device should be considered in addition to other impedance in the radial system.

 

[rcw retired EE]

Evgueni, My spreadsheet calculation says the impedance of 52 feet of #12 awg wire will knock 30.7 kA at 480 volts down to 3 kA. (82 feet of #10 awg does the same.) Any 480 V circuit with more than 50 feet of #12 wire (or 80 feet of #10) will have less than 3 kA for a fault at its load terminals. This means the wire will not melt for a bolted fault at the end of its run. (Assuming your 3 kA figure for withstand is correct.)

If the fault is not at the end of the run, the current will be higher and the wire will be damaged. But, if the fault is in the wire, it is damaged and will have to be replaced anyway.

Bottom line - small conductors are almost self protecting due to their high impedance.

 

[Evgueni]

Good morning Colleagues.
First I’d like to tell great thanks to everybody who responded to my question : mpparent , davidbeach , BJC , cuky2000 , rcwilson , and I’d like to provide some comments concerning previous discussion :
1. S.C. fault could occur in any place of the cable which is located in cable tray . The main goal of the cable limiter is to protect related cable from the arc flash , therefore prevent other components and cables against thermal effect.
2. The formula I used is for calculation of the maximum acceptable to S.C. withstand current for certain size of the cable and not available S.C. at the end of the cable.
Yesterday I got information from COPERBUSSMANN concerning the limiters. It was indicated that the limiters should be installed at the both ends of the cable . But this is not acceptable for the small application as 1.5 kW motors which has small connection box (the length of the smallest limiter is around 7.5”). Does anyone have an idea how to solve this problem?
An other item which I’d like to discuss is replacement of the fault cable. If S.C. occurs in the cable , and limiter opens ,therefor it’s means that the cable was under non- acceptable S.C. current and cable probably has physical , thermal damage of isolation . In this case the cable with limiters should be replaced by new ones. Why ? Because the cable limiter operates above certain S.C value .If this occurs it means that the cable subject to the physical and thermal damages and not good anymore. Am I right or not?
Several days ago I found in Internet one poor description about Current Limiting Circuit Breakers with combination of the dual-pivot contacts and single-pivot contacts. The work principle based on the high arc impedance during tripping action.(It could be SIEMENS product) Do you have some opinions about it?
Thanks in advance Evgueni.
Have a good weekend.

 

[davidbeach]

Cable limiters are intended to protect unfaulted cables in circuits where multiple cables are run in parallel my removing the connection to the faulted cable at both ends. If you have a single cable to a load, such as your small motor, there would be no need for a limiter at the load end. I would not use limiters for protection of single conductors per phase (not that I'd use them for many parallel installation either). The principal application of cable limiters is between transformers and the first downstream overcurrent device.

There are two conditions to be concerned about cable protection: through faults and cable faults.

For through faults, you need only consider the fault current at the load end of the cable and compare that with the withstand rating of the cable for the time it takes the source end overcurrent device to open. When you have multiple cables in parallel, through fault protection can be evaluated assuming that all cables equally carry the fault current. For nearly all real world circuits you will find that all cables are protected.

Cable faults will have a higher fault current, but will require replacement of the cable after the fault, so withstand is less of a concern. In runs where you have multiple cables in parallel, each in a separate conduit, looking at the withstand of N-1 cables vs. the load end fault current and clearing at the source end will allow evaluation as to whether or not the unfaulted cables will be damaged as they feed the fault from the load end of the paralleled set. Cable tray presents an additional complication in that a fault in one cable can spread to additional cables; limiters may speed up the clearing and limit the damage to other conductors. This additional protection will come at the cost of coordination with downstream overcurrent devices.

Any faulted cable will have to be replaced. Any adjacent cable showing any sign of insulation damage will have to be replaced. All other cables in the tray should be meggered and replaced if any sign of damage is shown.