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    Breaker size in main panel to support sub panel?

    Asked Nov 8, 2007, 01:57 PM 7 Answers
    Hello,

    I am wiring a sub-panel to the outside of my home (50 feet from main 200 amp box). I ran 3 wire - 6 gauge. What is the largest 2 pole breaker I can put in the main panel to supply the sub-panel.

    TYPE NM-B 6-3 WITH GROUND 600 VOLTS CIRTEX-I EZ-STRIP (R) PLUS 17:11 2/5/07 SB/RC

    Thank you in advance!
    - Mark

    Last edited by Smooth Fx; Nov 8, 2007 at 07:16 PM.
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    KISS's Avatar
    KISS Posts: 12,505, Reputation: 838
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    #2

    Nov 8, 2007, 02:11 PM
    Report back with kind of cable and conduit size if used.

    While your at it run an extra wire. You need ground, L1, L2 and Neutral. The neutral bus must be isolated in the sub-panel.
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    Smooth Fx's Avatar
    Smooth Fx Posts: 4, Reputation: 1
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    #3

    Nov 8, 2007, 07:00 PM
    The cable comes with ground. This is what it says on the outer wrap:
    TYPE NM-B 6-3 WITH GROUND 600 VOLTS CIRTEX-I EZ-STRIP (R) PLUS 17:11 2/5/07 SB/RC

    It is run w/o any conduit, other than the heavy black wrap it comes in.

    Thank you again!
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    tkrussell's Avatar
    tkrussell Posts: 9,638, Reputation: 722
    Senior Electrical & Lighting Expert
     
    #4

    Nov 9, 2007, 02:17 AM


    The NM-B #6-3 With GROUND already has a ground. A separate ground wire from panel to panel is not needed nor allowed.

    This cable may have a maximum of 60 amp breaker.

    Keep the neutral isolated from the equipment ground at the subpanel, by using the insulated neutral bar for the white, and a separate equipment grounding bar bolted to the panel metal box for all green and bare grounds.
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    Smooth Fx's Avatar
    Smooth Fx Posts: 4, Reputation: 1
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    #5

    Nov 9, 2007, 06:57 AM
    Thank you! I was under the impression, the black and the red could handle 50 amps a piece, so I was going to use a larger breaker than the 50 amp recommended at my local home supply store; but I had no idea how big was safe. Thank you, thank you!

    Is there a calculation formula to determine this? I hate to throw up a breaker question every time I go down this road.

    Nice to get such a fast response too!
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    KISS's Avatar
    KISS Posts: 12,505, Reputation: 838
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    #6

    Nov 9, 2007, 07:16 AM
    Tk:
    Check my logic. You have to use 120 V for the service because of the neutral. K is ~ 12.9 for copper and CMA for #6 is 26251. For a 3% drop at 50' (100' total), the load is 74 amps, but you need to derate it for continuous use (divide / 1.25), so therefore the max continuous load is 59.2. Since you allowed to size up, the next breaker size is 60 A.

    The problem is backwards. Normally you have the load. Derate if continuous (more than 3 hrs). Go to the ampacity table. Pick a wire size that is greater. Check the voltage drop. Size conductor up (lower AWG). Check again. Size to next breaker size.

    Include issues about cable type and number of conductors in conduit.
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    tkrussell's Avatar
    tkrussell Posts: 9,638, Reputation: 722
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    #7

    Nov 9, 2007, 09:26 AM


    I am not following you, such as the statement 'For a 3% drop at 50' (100' total), the load is 74 amps, ".

    Seems your making it harder than necessary.

    Question is what can be max size breaker on #6 NM cable. #6 NM cable max rating is 60 Deg C column at 55 amps, allowed to upsize breaker rating to next standard size, which is 60.

    The max continuous load on a 60 amp circuit is 48 amps, (80% of 60). I am not considereing this load at continuous, so 60 amps is the max.

    If the load is at 60 amps at 120 volts for 50 feet, the Vd will be 2.9 %, just under the recommended 3% maximum Vd for a feeder.

    Using the K factor is not the most accurate method of calculating voltage drop.

    Use: 2 x Length x Amps x Ohms per 1000 foot of chosen wire / 1000 = Voltage drop
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    KISS's Avatar
    KISS Posts: 12,505, Reputation: 838
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    #8

    Nov 9, 2007, 11:15 AM
    What I did was to use the Voltage Drop Calculator here http://www.electrician2.com/calculat...cpd_ver_1.html
    with Cu, 50 feet, 120 1ph, #6, 26244 for cir mils for #6, and k=12.9 for Cu
    Solved it iteratively for a 3% drop and got 74 A. I divided that by (1/0.8) and got 59.2.Then upsized to 60A

    In esscence 74 Amps in #6 copper at 50 feet away continuous will cause a 3% drop.
    Apply the derating and up-size, so, that the wire and the protection element are correct and get 60 amps for the overload protection.

    A 60 amp breaker cannot handle 60 Amps continuous by design and you have to derate the breaker to get the maximum continuous load of 48 Amps.

    Both ways to me make sense. I never think like someone else does. The iterative calculator tool made it easy. Oh well!

    For reference, the FUNDAMENTAL formulas are R=pL/A or Resistance = resitivity * length divided by cross-secional area and ohms law with modifications that make the numbers convient.

    Thanks
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