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    Calculate current limiting effect of conductor length..

    1. An industrial control panel is labeled with an SCCR of 5kA
    2. The feeder supplying it utilizes a J class 25 amp fuse with a let-through current of 5.5kA (UL 508A table SB4.2)
    3. I want to prove that a 50 foot length of #10 cable supplying power to this 5 kA rated panel will effectively limit the available fault current to a value below 5kA.

    Can it be as simple as taking the resistance of 50 feet of #10 at 0.05 ohms times 2 (out and back) for a total of 0.10 ohms, using I=E/R and coming up with 480 V / 0.1 Ohms = 4800 Amps?

    #2
    Not sure of the engineering viability of doing so, but I used a "short circuit program" that Bussmann has on line and came up with a value of 400 amps.

    You can download the program here:
    http://www.cooperbussmann.com/software/index.aspx
    Last edited by augie47; 12-28-10, 02:52 PM.
    At my age, I'm accustomed to restaurants asking me to pay in advance, but now my bank has started sending me their calendar one month at a time.

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      #3
      Read this thread
      http://forums.mikeholt.com/showthrea...nt+calculation

      You need a little more info but go from step 4 on and get your answer.
      Last edited by zog; 12-28-10, 03:03 PM.

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        #4
        Rich,
        That is basically it. I ran the calcullation on my computer and got a shade over the 5000. I would do as Augie and Sog said. Use the online calculators. You can include more information and get even a lower fault current. This installation is called a series rating where you have a fuse and a breaker in series. The combination is tested by the mfg to verify that it works correctly.

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          #5
          Thanks

          Thanks guys, that was just what I was looking for. I had the right concept but it needed refining.

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            #6
            Originally posted by richwaskowitz View Post
            1. An industrial control panel is labeled with an SCCR of 5kA
            2. The feeder supplying it utilizes a J class 25 amp fuse with a let-through current of 5.5kA (UL 508A table SB4.2)
            3. I want to prove that a 50 foot length of #10 cable supplying power to this 5 kA rated panel will effectively limit the available fault current to a value below 5kA.

            Can it be as simple as taking the resistance of 50 feet of #10 at 0.05 ohms times 2 (out and back) for a total of 0.10 ohms, using I=E/R and coming up with 480 V / 0.1 Ohms = 4800 Amps?

            I would think that you need to take the source impedance into consideration. A 1500kVA would supply a higher fault current then a 500kVA transformer for example.

            Maybe I am overlooking something but I dont believe you can just divide the bus voltage by the impedance without knowing the upstream system capacity and impedances?

            Comment


              #7
              Transformer Secondary Fault Currents

              Kind of on topic. The following site has table that list fault current on the secondary of various size transformers.

              IT list transformer size, cable size and the fault for cables of 10-150 in ten ft intervals.


              http://www.alabamapower.com/architec...208%202003.pdf

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                #8
                I don't think a transformer or system ahead of it comes into play since the assumption is that there is enough fault current ahead of the fuse, that the maximum 5.5KA is going to be available. That would be the conservative approach. In this case the fuse maximum let-through is acting the same as a transformer would do.

                The cable will also act as a choke and have some limiting of the initial 5.5KA current. I modeled it in ETAP and get 2.7kA at the panel, and by hand I got 3.2kA. The hand calc is using the MVA method and it is on the conservative side, so the results make sense to me.
                "Just because you're paranoid, doesn't mean they're not out to get you"

                Comment


                  #9
                  Originally posted by kingpb View Post
                  I don't think a transformer or system ahead of it comes into play since the assumption is that there is enough fault current ahead of the fuse, that the maximum 5.5KA is going to be available. That would be the conservative approach. In this case the fuse maximum let-through is acting the same as a transformer would do.

                  The cable will also act as a choke and have some limiting of the initial 5.5KA current. I modeled it in ETAP and get 2.7kA at the panel, and by hand I got 3.2kA. The hand calc is using the MVA method and it is on the conservative side, so the results make sense to me.
                  That's close to what I got with a spreadsheet that does a calculation based on the point to point method described in the Bussman SPD handbook. I got a little over 2.5KA.

                  The previous answer of 400 amps seems low, and the other answer of over 5KA seems high.

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