NEC Table 220.88

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robh

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NEC Table 220.88 Optional Method-Permitted Load Calculations for Service and Feeder Conductors for New Restaurants


Hi all.
I am Manufacturing Plant Journeyman Electrician apprentice in Oregon. I am finishing up my schooling in a code class. We had a calculation to do last night in class and there was some confusion over this NEC table, and how to do the calculation.

The problem is as follows:

First Part: A new restaurant has a connected load of 324 kva, all electric equipment. What is the demand on the 3-phase 208/120v service?

According to the table this is my math:
10% (of amount over 200)+160
12.4+160=172
172/(208 x 1.732) = .47855 x 1000=478.55 Requires a 500 amp service.

Second Part: If the restaurant had NOT all electric equipment?

According to the table this is my math:
50% (of amount over 200)+200
62+200=262
262/(208 x 1.732) = .72727 x 1000=727.27 Requires an 800 amp service.

Is this right? If so why would a restaurant with the same load, but not all electric equipment require such a higher demand?

Thanks for the help.
Rob
 
Your math looks right. I would say the difference arises from the derating the total demand due to on off cycles of the equipment in the total electric installation. In the latter part, the assumption is probably that all of the electrical equipment could be on or starting at the same time.
Know What I Mean,
Vern
 
I agree with the assumption Vern is making regarding your specific question. This and other "inconsistencies" occur in Article 220 and the rest of the code for that matter. In the end, the overall demand load is far less than if no demand factors were permitted. There is no "one calculation fits all" approach.
 
robh said:
NEC Table 220.88 Optional Method-Permitted Load Calculations for Service and Feeder Conductors for New Restaurants

The problem is as follows:

First Part: A new restaurant has a connected load of 324 kva, all electric equipment. What is the demand on the 3-phase 208/120v service?

According to the table this is my math:
10% (of amount over 200)+160
12.4+160=172
172/(208 x 1.732) = .47855 x 1000=478.55 Requires a 500 amp service.

Second Part: If the restaurant had NOT all electric equipment?

According to the table this is my math:
50% (of amount over 200)+200
62+200=262
262/(208 x 1.732) = .72727 x 1000=727.27 Requires an 800 amp service.

Is this right? If so why would a restaurant with the same load, but not all electric equipment require such a higher demand?

Thanks for the help.
Rob
Click to expand...
Your caulations are correct. However, if the resturant was not all electric
the load would be much less and you would be using 50% of a much lower number. So assume the load is now only 150 kva. 50% over 200 kva = 0.
Load would be 150/.208x1.73 = 416 amps.
 
Thanks for the responses. I was thinking along the lines of Vern and bph. But Bob, I see your point if the load was lower on a not all eletric load, but what if its the same load? Do we kinda assume (for sake of argment?) that it is just a bigger resturant since the eletric load is that big? Thanks again for the awnsers.
Rob
 
Rob,
In the problem statement, the difference in the two scenarios was only that in case 1, the restuarant was all electric, and in case 2 it was NOT all electric. If you read more into it than that then you just missed the test question.
Know What I Mean,
Vern
 
It's good to pick apart the NEC and find flaws in the requirements. That is what inspires needed code change. In my opinion, Article 220 is a real mess and too conservative. While I hardly run into any "real world" application problems with Article 220, it sure is hard to use for the purposes of an exam question...
 
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