NEC Table 220.86

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erickench

erickench

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Hi, it's me again with another one of my dumb questions. How do you calculate a load for a school using NEC Table 220.86? Anybody have an answer? I would appreciate it. The table doesn't make sense. What is meant by the first three VA/ft-squared? I mean I could see applying a demand factor to an initial amount of VA followed by a different demand factor for any VA beyond that. But this table is using a unit amount per square foot.:confused:
 
Okay, I'm going to try to answer my own question like I did with the other thread. NEC Table 220.12 states that the unit load for a school is 3VA/square-foot. Lets assume that no additional load is added to this. The demand factor would be %100 as indicated in NEC table 220.86 because the total load divided by the area would be 3VA/square-foot. Now if we were to add additional load and then divide by the total area in square-foot we would come up with a number higher than 3VA/square-foot. The demand factor would be %75 up to 20/VA-square-foot. And if we keep adding additional load and then divide by total area we would excede 20/VA-square-foot and the demand factor would then be %25. Correct me if I'm wrong.:mad:
 
You're wrong

You're wrong

220.12 is for minimum lighting branch circuit requirements only. 220.86 is for sizing your service load. To use 220.86 the school must be heated electrically or cooled electrically. You must take the total connected load (including loads from 220.12 if your actual branch circuit loads are smaller). It is then very simple to take the total connected load and multiply the appropriate factors.

Ex.

Total square feet 1000
Total connected load 25000 VA
Total of 25 VA/sqft

3W/sqft *1000sqft = 3000W
17W/sqft * 1000sqft = 17000W*.75 = 12750W
5W/Sqft * 1000sqft = 5000W*.25 = 1250W

Total Feeder or service demand = 17000 VA

Hope this helps.
 
Well I have to say it was pretty confusing. I could'nt understand what this "plus" nonsense was all about. It's written into the table first 3 VA/ft-squared plus 3 to 20 VA/ft-squared. It would take some pretty tricky figuring for this type of problem.
But what I don't understand is where do you get this 17W/sqft? Do you just subtract 3VA/ft-squared from
20VA/ft-squared? And why use 5W/sqft? Why not just apply the 25% demand factor to the remainder?
 
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The answer is yes on the 17W/sq ft, it is the total amount subject to the 75% factor (since we have already taken the 3W/sqft @ 100%). In my example the remainder is 5W/sq ft so yes, just take whatever is above 20W/sq ft and multiply by 25%. There is actually a pretty good explanation in the handbook also.
 
Okay, here are three examples. Tell me if these are appropriate.


1500 sqft 20,000VA

3x1500=4500VA multiply by 100% Demand Factor and you get 4500VA

10.3x1500=15500VA multiply by 75% Demand Factor and you get 11625VA

Total=16125VA

Note: (20,000-4500) divide by 1500 is equal to 10.3



1500 sqft 30,000VA

3x1500=4500VA multiply by 100% Demand Factor and you get 4500VA

17x1500=25,500VA multiply by 75% Demand Factor and you get 19125VA

Total=23625VA

Note: (30,000-4500) divide by 1500 is equal to 17



1500 sqft 40,000VA

3x1500=4500VA multiply by 100% Demand Factor and you get 4500VA

17x1500=25,500VA multiply by 75% Demand Factor and you get 19125VA

6.67x1500VA=10,000VA multiply by 25% Demand Factor and you get 2500VA

Total=26125VA

Note: (40,000-25,500-4500) divide by 1500 is equal to 6.67

:D
 
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