dsjr114
Member
- Location
- Warrington,Pa.,Bucks county
Can someone explain or show me the calculation to get a neutral calculated at 147,200 from a 196800va service
In reference to the Mike Holt NEC 2008 Commercial Calculation example under Section 11.13 video and as a general question...
Service load was 196,800VA @ 240V (820A). Breaker must be 1000A. Conductors are sized to the breaker with say 3 sets of 400 kcmil.
Neutral was calculated as 147,200VA @ 240V per 220.61 (613A) and adjusted by the demand factor of 70% for loads in excess of 200A yielding 489A. 489A divided by 3 raceways yield 163A, which requires a 2/0 neutral conductor per raceway.
However, 250.24(C)(2) says you can also size the neutral to the largest ungrounded service-entrance conductor per Table 250.66 but no smaller than 1/0 when paralleling. In this case, 400 kcmil phase conductors only requires a 1/0. (This method was used in the Section 11.14 example in the video and similarly here:
Which method do we use: 220.61 or 250.24(C)(2)? Method 1 (220.61) seems like it could vary based on load, which could change over the life of the building. Method 2 (250.24(C)(2)) seems to be undersized based on calculated neutral loads per Method 1.
Which method do we use: 220.61 or 250.24(C)(2)? Method 1 (220.61) seems like it could vary based on load, which could change
over the life of the building. Method 2 (250.24(C)(2)) seems to be undersized based on calculated neutral loads per Method 1.
In reference to the Mike Holt NEC 2008 Commercial Calculation example under Section 11.13 video and as a general question...
Service load was 196,800VA @ 240V (820A). Breaker must be 1000A. Conductors are sized to the breaker with say 3 sets of 400 kcmil.
Neutral was calculated as 147,200VA @ 240V per 220.61 (613A) and adjusted by the demand factor of 70% for loads in excess of 200A yielding 489A. 489A divided by 3 raceways yield 163A, which requires a 2/0 neutral conductor per raceway.
However, 250.24(C)(2) says you can also size the neutral to the largest ungrounded service-entrance conductor per Table 250.66 but no smaller than 1/0 when paralleling. In this case, 400 kcmil phase conductors only requires a 1/0. (This method was used in the Section 11.14 example in the video and similarly here:
Which method do we use: 220.61 or 250.24(C)(2)? Method 1 (220.61) seems like it could vary based on load, which could change over the life of the building. Method 2 (250.24(C)(2)) seems to be undersized based on calculated neutral loads per Method 1.
Which method do we use: 220.61 or 250.24(C)(2)? Method 1 (220.61) seems like it could vary based on load, which could change
over the life of the building. Method 2 (250.24(C)(2)) seems to be undersized based on calculated neutral loads per Method 1.