marina outlets

[Cmc]

What is the service conductor load for the following 20 outlets at a marina? 17-30 amp recpt. 3-50 amp recpt. 555.12 T please show me the math

 

[Little Bill]

What is the service conductor load for the following 20 outlets at a marina? 17-30 amp recpt. 3-50 amp recpt. 555.12 T please show me the math

The (17) 30A would fall under the 15-30 column which is 70%
The (3) 50A would fall under the 1-4 column which is 100%

You can do the math.

 

[Cmc]

marina outlets

Sir i done that math and the answer come up wrong :huh:

 

[MAC702]

Since you've already done it, show us what you did and maybe we can spot something.

 

[Cmc]

marina outlets

Since you've already done it, show us what you did and maybe we can spot something.

I just figured it. You take the 17x30amp=510A then take the 3x50 amp=150 add them together = 660A x 70% demand factor=462 amps that is the correct answer:happyno:

 

[MAC702]

With Little Bill's info:

17 * 30 A * 0.7 = 510 A * 0.7 = 357 A

Then add the 3 * 50 A * 1.0 = 150 A

357 A + 150 A = 507 A

 

[buzzbar]

The (17) 30A would fall under the 15-30 column which is 70%
The (3) 50A would fall under the 1-4 column which is 100%

You can do the math.

If this is for the total service load, with the (20) shore power receptacles being there as a load, you would figure ALL of them, times 70%, since the demand applies to the TOTAL number of shore power receptacles. The code says nothing about separating the 30A shore power from the 50A shore power. Also, if you have a 30A and a 50A shore power for one slip, you only have to count the higher of the two (50A), not both.


So, having said that, you would do the following:
(17) x 3600 = 61,200
(3) x 12,000 = 36,000
61,200 + 36,000 = 97,200
97,200 * .7 = 68,040

Single phase 120/240 would be 68,040/240= 283.5 amps
Three phase 120/208 would be 68,040/360 = 189 amps