9 Single Phase Ranges on Three Phase Feeder

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jbwhite

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This is a study question for the masters test that a friend asked about.

A school has 9 10kw ranges for a cooking class. The ranges are 240v single phase, the feeder is 240v three phase. What is the amperage for the ranges on the feeder.

I went to table 220.55 and looked up 9 ranges in columb C and got 24kw.

Next I divided by 240 volts and got 100 amps. Then I divided by 1.73 and got 57.80 amps.

I ignored 220.86 because I did not see that it applied to the standard method.

The book answer was 88 amps.

What did I do wrong?
 
Now I think I got it..

220.55 second part is :

Where two or more single-phase ranges are supplied by
a 3-phase, 4-wire feeder or service, the total load shall be
calculated on the basis of twice the maximum number connected
between any two phases.

So as rabbitgun Says 3 is the max number connected between two phases.
2 X 3 = 6
Table 220.55 says for six ranges.
21 KW
21000/240 = 87.5
Round Up: 88
 
I'd fail you.

I think the answer of 21KW whilst (presumably) being correct by the book, is wrong: because this is a school, all nine ranges are likely to be used for identical purposes at the same time. This is different to the normal diversity approach where a mix of workload is assumed.
 
dbuckley said:
I'd fail you.

I think the answer of 21KW whilst (presumably) being correct by the book, is wrong: because this is a school, all nine ranges are likely to be used for identical purposes at the same time. This is different to the normal diversity approach where a mix of workload is assumed.
Click to expand...
So, this is becoming interesting. If you go to article 220.56 Kitchen Equipment- other than dwelling units... use table 220.56. Table 220.56 says 6 and over 65%. So we take 9 ranges at 10kw= 90000 and multiply that by .65 and we get 58,500. Divided by 240 volts and voila we get 244 amps. For 3 phase we divide again by 1.73 and we get 141 amps. A far cry from 88 which I think was calculated wrong to begin with since there was no division by 1.73 for 3 phase.
 
dbuckley said:
I'd fail you.

I think the answer of 21KW whilst (presumably) being correct by the book, is wrong: because this is a school, all nine ranges are likely to be used for identical purposes at the same time. This is different to the normal diversity approach where a mix of workload is assumed.
Click to expand...
I was looking at table 220.55 not five which says to use that table for: "This table also applies to household cooking appliances rated over 1 3⁄4 kW and used in instructional programs."
 
I still don't see how JBwhite's equation works if it is 3 phase but it appears to get the right answer. Any help out there.
 
Dennis, the ranges are single phase, so the loads will likely be laid out:
1.AB
2.CA
3.BC
4.AB
5.CA
6.BC
7.AB
8.CA
9.BC

So we have (6) ranges pulling from A, (6) pulling from B, and (6) pulling from C. That's how he's using (6) ranges.

Also, as far as diverity is concerned, your max output will be reached with (4) burners on high and the bake setting (both upper and lower coil) on 500. Still an unlikely scenerio. Each range has its own diversity as it cycles to maintain the desired heat level.
 
Jps1006 said:
Dennis, the ranges are single phase, so the loads will likely be laid out:
1.AB
2.CA
3.BC
4.AB
5.CA
6.BC
7.AB
8.CA
9.BC

So we have (6) ranges pulling from A, (6) pulling from B, and (6) pulling from C. That's how he's using (6) ranges.

Also, as far as diverity is concerned, your max output will be reached with (4) burners on high and the bake setting (both upper and lower coil) on 500. Still an unlikely scenerio. Each range has its own diversity as it cycles to maintain the desired heat level.
Click to expand...
I get the part opf the phases thus 6 ranges, but it is fed with a 3 phase feeder so we ignore that because they are single phase load and divide by 240 instead of 240*1.73???
 
Dennis Alwon said:
I still don't see how JBwhite's equation works if it is 3 phase but it appears to get the right answer. Any help out there.
Click to expand...

I could come up with the answer by fixing the voltage to 208.

2x3=6...21kw/2=10500kw plus the 21kw=31500/360=87.5 amps.. this was using the example D5 in annex D.
Rick
 
quote from page 722 NEC
===========================
Minimum Size Feeders Required from Service Equipment to Meter Bank (For 20 Dwelling Units ? 10 with Ranges)
For 208Y/120-V, 3-phase, 4-wire system,
Ranges: Maximum number between any two phase legs = 4 2 ? 4 = 8.
Table 220.55 demand = 23,000 VA
Per phase demand = 23,000 VA ? 2 = 11,500 VA

Equivalent 3-phase load = 34,500 VA

Net Calculated Load (total):
40,590 VA + 34,500 VA = 75,090 VA
75,090 VA ? (208 V)(1.732) = 208.4 A
============================

It looks like whoever wrote the practice book that I am looking at made an error. The example above shows converting my original answer to an equivalant three phase load.

I wonder which way my friend should answer on the test.

It is all about passing, isn't it :D
 
I am sorry for being a pest on this one but I still feel like something is not right. First of we keep getting different answers and I have a question about the following statement anyway.

Article 220.55 second part is :
Where two or more single-phase ranges are supplied by
a 3-phase, 4-wire feeder or service, the total load shall be
calculated on the basis of twice the maximum number connected
between any two phases.

Article 220.55 refers to Dwelling Units-- Now I know Table 220.55 Note 5 says that we can use this table on household cooking appliances rated over 1 3/4 kw and used in instructional program, however the converse is not necessarily true. Based on this article we figured 6 ranges based on twice the maximum number connected between any two phases, however this is not a dwelling unit. This is a school so why can we work backwards and assume that article 220.55 would apply. Do you see what I am getting at??? If this was not a instructional program with single phase ranges on a 3 phase feeder would you still calculate the ranges based on the second paragraph of article 220.55???? I would love some input.
 
Dennis Alwon said:
If this was not a instructional program with single phase ranges on a 3 phase feeder would you still calculate the ranges based on the second paragraph of article 220.55???? I would love some input.
Click to expand...

If this was an other than dwelling or instructional, and you used table 220.56 with (9)10kw ranges, the demand factor would be @ 65%=58,500kw. If you used the 2nd paragraph of 220.55 method (2x3=6 ranges) it would defeat the 65% demand factor, because it would put you at 21kw.
Rick
 
RUWired said:
If this was an other than dwelling or instructional, and you used table 220.56 with (9)10kw ranges, the demand factor would be @ 65%=58,500kw. If you used the 2nd paragraph of 220.55 method (2x3=6 ranges) it would defeat the 65% demand factor, because it would put you at 21kw.
Rick
Click to expand...
I'm sorry my question was a rhetorical question to make a point of what I was saying above that.
 
Article 220.55 says to use the table and how including the second paragraph.

The note to the table says that this includeds instructional facilities, or what ever the wording is.

so the second paragraph does apply. if you use 208v for the three phase the calc from anex D works and the answer is right.

I think the test writer expected us to take the leap of faith from 240 volt cooking appliances to on a three phase 208 wye feeder.

poor test writing if you ask me, but the only explaination that I can think of.
 
jbwhite said:
Article 220.55 says to use the table and how including the second paragraph.

The note to the table says that this includeds instructional facilities, or what ever the wording is.
Click to expand...
I think you are missing my point and perhaps that is my fault. Your first sentence says article 220.55 states..... Why would you start at article 220.55 if that is for dwelling units. I get the fact that Table 220.55 Note 5 talks of instructional programs but that only relates to Table 220.55-- you went over to article 220.55 second paragraph which although it is 3 phase still relates to dwelling units. I am not sure you can do this.
 
jbwhite said:
ok then how should i be doing it?
Click to expand...
I am not saying you are wrong I am just confused by it. It doesn't make sense. I guess what I would have done was multiplied the 9 10kw ranges and got 90000 then multiplied by Table 220.56--65%. and get 58,500 and then divide by 360 and you get 162.5. I have no idea if that is correct or not.
Edited -- forget it --- this is definitely not correct
 
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