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minimum number of cables required to safely ground transformer

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    minimum number of cables required to safely ground transformer

    4/0 grounded cable is rated at 43kA for 15cycles. What is the minimum number of cables per phase required to safely ground a 2500kVA 5.75%Z transformer on the 480V side?

    A-1
    B-2
    C-3
    D-4

    Looking for help on this. XFMR fault current is 52,297 on the 480V side assuming 3phase? my guess is 2 cables per phase.

    480*1.732 = 831.36
    2500000 / 831.36 = 3007.12
    3007.12 / 0.0575 = 52297

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    Wanderer20001us is offline Junior Member Pro Subscriber
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    Wouldn't you need to know some more information about the system (i.e. what is the upstream protective device and clearing time, etc.)?

    Quote Originally Posted by samair99 View Post
    4/0 grounded cable is rated at 43kA for 15cycles. What is the minimum number of cables per phase required to safely ground a 2500kVA 5.75%Z transformer on the 480V side?

    A-1
    B-2
    C-3
    D-4

    Looking for help on this. XFMR fault current is 52,297 on the 480V side assuming 3phase? my guess is 2 cables per phase.

    480*1.732 = 831.36
    2500000 / 831.36 = 3007.12
    3007.12 / 0.0575 = 52297

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    Yes.

    Quote Originally Posted by Wanderer20001us View Post
    Wouldn't you need to know some more information about the system (i.e. what is the upstream protective device and clearing time, etc.)?
    Your calculations are correct, I came up with the same answer. For this specific question I do not think that clearing time will affect the answer. The clearing time will change the PPE requirement if you are doing an arc flash analysis.

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  7. EEBSMSPE96 is offline Junior Member Pro Subscriber
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    The correct answer is B-2.

    Quote Originally Posted by samair99 View Post
    4/0 grounded cable is rated at 43kA for 15cycles. What is the minimum number of cables per phase required to safely ground a 2500kVA 5.75%Z transformer on the 480V side?

    A-1
    B-2
    C-3
    D-4

    Looking for help on this. XFMR fault current is 52,297 on the 480V side assuming 3phase? my guess is 2 cables per phase.

    480*1.732 = 831.36
    2500000 / 831.36 = 3007.12
    3007.12 / 0.0575 = 52297
    The correct answer is B - 2 cables per phase. The 52,297Amps of available fault current you calculated is correct. The equation to calculate the maximum available fault current is I[availablesc] = (KVA x 100)/(1.732 x KV[lowside] x %Z) = (2500 x 100) / (1.732 x 0.480 x 5.75) = 52,298Amps. The rating of 43kA for 15 cycles for 4/0 personal protective ground cable is right out of Table 1 of ASTM Standard F855-09 for Protective Ground Cable ratings (See Grade 5). It is very important to note that Table 1 applies only to electrical systems with X/R ratios less than or equal to 1.8. Table 2 applies to systems with X/R ratios greater than 1.8. Since they gave you the 43kA rating in this problem, that's what you had to use to solve this problem. In the real world, where X/R ratios are rarely less than or equal to 1.8, I personally use the ratings that are given in Table 2 since the X/R ratio is greater than 1.8 in most electrical systems.

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