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# minimum number of cables required to safely ground transformer

1. Junior Member
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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

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

Originally Posted by samair99
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.

Originally Posted by Wanderer20001us
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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## The correct answer is B-2.

Originally Posted by samair99
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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