NEC Table 220.88

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deee

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I am not sure how to use Table 220.88 and cant find the answer I am looking for. I have seen Mike Holt do a calculation as referred to the note at the bottom of the table.

Question: What is the Calculated load for an all electric restaurant that has a total connected load of 325 kva at 120/208 3phase paralleled in 2 raceways. Size the Conductors?

First 200 kva at 80% and the remainder from 201 to 325 at 50%.
200 x .80 = 160
125 x .10 = 12.5
160 + 12.5 = 172.5

172.5/(208x1.732)=x
172.5/(360.3)=.4787
1000x.4787= 479 amps

479/2 (Raceways)=239 amps
Conductors per Table 310.15(B)(16) 75 degree C is 250 kcmil
The Grounded conductor per 250.24(C)(2) not smaller than 1/0
The Grounding Electrode Conductor per Table 250.66 (equivalent of the parallel conductors is 500 kcmil) is 1/0

Total Connected All Electric Restaurant
Load (kva) Calculated Loads (kva)
My question is in Table 220.88 where it states " 201-325 10% (amount over 200) + 160" where does the (amount over 200) + 160 come into play?
 
You got lucky on the first calculation. If the load is 325 then you use that row not 0-200 then 201-325

201-325kva says take 10% of (125) , the amount over 200 and add 160. You got the right answer but did not calculate it accordingly. The next time it may not work. I didn't check the rest yet
 
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NEC

Dennis
I don't really understand the table. The way I learned it, is the way I originally posted it. So any information would be appreciated.

If I read you right, you are saying to do 125 kva at 50% and add 160 to get 172.5 kva
 
deee said:
I am not sure how to use Table 220.88 and cant find the answer I am looking for. I have seen Mike Holt do a calculation as referred to the note at the bottom of the table.

Question: What is the Calculated load for an all electric restaurant that has a total connected load of 325 kva at 120/208 3phase paralleled in 2 raceways. Size the Conductors?

First 200 kva at 80% and the remainder from 201 to 325 at 50%.
200 x .80 = 160
125 x .10 = 12.5
160 + 12.5 = 172.5

172.5/(208x1.732)=x
172.5/(360.3)=.4787
1000x.4787= 479 amps

479/2 (Raceways)=239 amps
Conductors per Table 310.15(B)(16) 75 degree C is 250 kcmil
The Grounded conductor per 250.24(C)(2) not smaller than 1/0
The Grounding Electrode Conductor per Table 250.66 (equivalent of the parallel conductors is 500 kcmil) is 1/0

Total Connected All Electric Restaurant
Load (kva) Calculated Loads (kva)
My question is in Table 220.88 where it states " 201-325 10% (amount over 200) + 160" where does the (amount over 200) + 160 come into play?
Click to expand...

Assuming you are using Table 220.88 in lieu of Part III.

Step 1 (0.10 x (325 - 200)) + 160.0 =
Step 2 12.5kVA + 160.00 kVA = 172.5 kVA

Service Size : 172.5 x 1000
__________ = 479.1666 (479 Amps)
208 V x 1.732
(or 360)

Now you got the right kVA in your method but it may not hold TRUE in other examples. As an FYI, there is a GREAT example in the 2014 NEC Handbook that you may want to take a gander at.
 
deee said:
Dennis
I don't really understand the table. The way I learned it, is the way I originally posted it. So any information would be appreciated.

If I read you right, you are saying to do 125 kva at 50% and add 160 to get 172.5 kva
Click to expand...
Yes but I think you meant to write 125 kva at 10% not 50%
 
deee said:
My question is in Table 220.88 where it states " 201-325 10% (amount over 200) + 160" where does the (amount over 200) + 160 come into play?
Click to expand...

It is the amount of your calculated load that is OVER 200. In your example, the amount over 200 is 125 ( difference between 200 and 325 ), then you apply the 10% to that figure, then add the 160 to it. This is your demand load.
 
NEC

NEC

Dennis
I have done the calculations both ways and used different values and obtained the same numbers. I just did not understand this table. The note at the bottom of the table. says in short " To calculate the total connected load, multiply the total connected load by this single demand factor" assuming that the 10% is the single demand factor that is being referred to.
 
deee said:
Dennis
I have done the calculations both ways and used different values and obtained the same numbers. I just did not understand this table. The note at the bottom of the table. says in short " To calculate the total connected load, multiply the total connected load by this single demand factor" assuming that the 10% is the single demand factor that is being referred to.
Click to expand...


Yes it works the same for that example but try using it both ways for 500 kva and see what happens
 
If the kva is 0-200 then you use that first row-- row 1
If the kva is 201-325 then you only use the second row-- row 2
if the kva is 326-800 then you only use Row 3
and If the load is over 800 then use only Row 4
 
deee said:
Dennis
I have done the calculations both ways and used different values and obtained the same numbers. I just did not understand this table. The note at the bottom of the table. says in short " To calculate the total connected load, multiply the total connected load by this single demand factor" assuming that the 10% is the single demand factor that is being referred to.
Click to expand...
Well, I can only say that this is how the handbook does it, How Charles Miller does it.....and how we do it. Here is an extract from a column published in Electrical Contractor Magazine on the subject.

If the total connected load is more than 200 kVA, an additional math step is required. For example, what is the optional method service load calculation for a new restaurant with a total connected load of 394 kVA? This new restaurant will be all electric. Start by looking in the ?Total Connected Load (kVA)? column, and find the row that includes 394. Since 394 is between 326 and 800 kVA, it is in the third row. Follow this row across to the center column because this restaurant will be all electric. Next, calculate the amount over 325. The amount over 325 is 69 kVA (394 ? 325 = 69). Now multiply the amount over 325 by 50 percent (69 50% = 34.5). Finally, add 34.5 to 172.5 kVA (34.5 + 172.5 = 207). The optional method service load calculation for this new restaurant is 207 kVA (see Figure 3)."

http://www.ecmag.com/section/codes-standards/branch-circuit-feeder-and-service-calculations-part-lxi

 
In your example you take 10% of the amount over 200kva and you get 125. So 10% of 125= 12.5kva
now add 160 to it.... 160+12.5= 172.5 kva

Try the total load at 300kva instead of 325kva

Your way
.8 X 100= 80kva
160+ 80= 240kva. You see that is higher than the 325 kva load which gave us 172.5kva

My way
100 X .10= 10kva
10kva + 160= 170kva

It doesn't work both ways
 
NEC

NEC

Dennis
If I do this 500 kva calculation using the Mike Holt method I get the same answer of 260 kva as I did using the method you laid out. I like yours better because it is much easier and faster.

Mike Holt Method (500 kva connected load)

First 200 at 80% 200x.80=160
201 to 325 at 10% 125x.10=12.5
326 to 500 at 50% 175x.50=87.5

160+12.5+87.5=260 kva

Your Method (Dennis)

50% (amount over 325) 500-325=175
175x.50=87.5
87.5=172.5=260
 
In your example you take 10% of the amount over 200kva and you get 125. So 10% of 125= 12.5kva
now add 160 to it.... 160+12.5= 172.5 kva

Dennis Alwon said:
Try the total load at 300kva instead of 325kva

Your way
.8 X 100= 80kva
160+ 80= 240kva. You see that is higher than the 325 kva load which gave us 172.5kva

My way
100 X .10= 10kva
10kva + 160= 170kva

It doesn't work both ways
Click to expand...



Im not sure how to box your response in, but I see what you are saying. Very good info. I will use your way from now on. Thanks a bunch
 
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deee said:
Dennis
If I do this 500 kva calculation using the Mike Holt method I get the same answer of 260 kva as I did using the method you laid out. I like yours better because it is much easier and faster.

Mike Holt Method (500 kva connected load)

First 200 at 80% 200x.80=160
201 to 325 at 10% 125x.10=12.5
326 to 500 at 50% 175x.50=87.5

160+12.5+87.5=260 kva

Your Method (Dennis)

50% (amount over 325) 500-325=175
175x.50=87.5
87.5=172.5=260
Click to expand...


So using mike's method what happened to the adding 160-- I have no idea what you did but it isn't right
 
It works out the same if you only apply the % factor to the amount within each range then sum them up. The single demand factor is two-fold, essentially giving you the factor for the amount within the range and summing with the the pre-calculated value for the amount within the lower ranges.
 
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