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Monthly Column by Eddie Rodriguez                      

Making Table 220.55 Easy

Previous Column's by Eddie:
Lets Talk AC Wiring
Puck Lights

Before we can get into some of the cool tips about table 220.55 {not only used to calculate service and feeder loads but conductor  sizes also} Lets get the notes down pat. After teaching hundreds of students how to use this table. I have found it easiest to turn the notes of table 220.55 to steps. They now become nine steps one built off another rather then a bunch of small print notes. OK step one is the foundation of all the steps witch stems from note 4. Read the article and enjoy it don’t worry about notes at the end of the article I will set them all up for you. The first step deals with one range regardless of the size as long as its not over 12 kva or smaller then 3 kva (column “C”) then the code see it as a 8 kva range. The 8 kva comes from the table, column “C”.  Remember this 8 kva and were it came from cause it keeps coming up throw the steps. So one single range at 8000 va / 240 volts = 33 amps  there’s  your load and step one. This calculation is the exact reason why when wear doing a rough we know we can safely run a number 8 wire and be code compliant. I am not saying you cannot go larger with the feeder. Now step two is a single range over 12 kva. This step can be found in note one. For ranges 12 kva through 27 kva. In ower case we will use a 14kva range. From the 14 kva we will subtract 12 kva with a remainder of 2 kva left. The 12 kva comes from the notes {a given number}{14 kva- 12 kva = 2 kva}. Next we increase the remainder by 5% for each kva, this will include any major fraction (2 kva x 5% = 10%). From column ‘C” 1 range we get 8000 va added to the remainder {8000 x 110% = 8800va}. Most of the rest of steps are similar to step two , you will find a slight different twist. Which we will see now in step three. Step three is more then one  range unequal size over 12 kva. This step taken from note 1. So say we have 3 ranges 15.6 kva each. Here we take the range size subtract 12 kva (given number) {15.6 kva – 12 kva = 3.6 kva}. You see the basic step again. Multiplying the remainder by the 5%{4 x 5% = 20%}. Now going to table 220.55 column “C” down to 3 ranges and we come up with 14 kva added to the remainder {14 kva x 120% = 16.8 kva}. Which gives use a calculated load of 16.8 kva.Moveing along we go to step four, more then one range not over 12 kva.This step is an easy one, being The ranges are not over 12 kva you would just need the table 220.55.For example if we had 5 ranges- 9 kva each range. Not over 12 kva we go to column “C”. Going down to 5 ranges we get a 20 kva total calculated load {5 units} 9 kva = 20 kva. Lets try more the one range unequal ratings over 12 kva. Well call this step five and its from note two. Here we have 3/9 kva ranges and 3/14 kva ranges. Any ranges under12 kva is now treated as a 12 kva range you can find this in the note 1. This step requires us to average the ranges, so the 3/9 kva ranges are now 3/12 kva ranges added to the three 14 kva ranges we will end up with 78 kva.  

          3/9 kva {min 12 kva} 3 ranges x 12 kva range = 36 kva

          3/14 kva                  3 ranges x 14 kva range = 42 kva

                                                                           78 kva

  Now we need to get the average of the 6 ranges


                                78 kva /6 = 13 kva


and  now back to the basic step. Subtracting the 12 kva from 13 kva {13 kva – 12 kva = 1kva}. Don’t forget the 12 kva is a given from the notes. Off to the table 220.55,column “C” 6 ranges we get 21 kva. Here we add the 5% to the remainder 1 kva (21 x 5% = 22.05 kva). So we are left with a total calculated load of 22.05 kva for all 6 ranges and completion of step five.  Steps six and seven are a breeze to figure out. Logically well do six first, hang in there were on the home stretch. More then one range, equal ratings over 1 ¾ kva less then 3 ½ kva note 3.A good example would be 10/3 kva ranges. First thing you must do is multiply the size of ranges by the quantity of ranges 3 kva  x 10 units = 30 kva. The size of these small ranges sends us to the table 220.55 column “A”. Column “A” and “B” are in percentages unlike column “C” witch is in kva ratings. Following down column “A” to 10 units and we get a multiplier of  .49%. Multiple the total kva ratings of all the ranges times the percent from column “A” 30 kva x .49% = 14.70 kva. 14.70 kva is our total calculated load and another step under our belts. OK step seven here we have more the one range equal ratings not over 3 ½ kva less then 8 ½ kva.  This step is in note three as it was in step six the only difference is the slightly larger ranges. We will work with eight 6  kva ranges. Multiply the range kva times the amount of ranges (6 kva x 8 units =48 kva). The total is multiplied by the percent from column “B” of table 220.55, 8 ranges gives you .36% {48 kva x .36% 17.28 kva}. A total load of 17.28 kva. Step eight is pretty easy, it deals with a counter mount cooking unit or one oven. Found in note 4 also, you simply you use the name plate.   The last step is actually two part example. One counter mounted cooking unit and up to two ovens.  Just different  variables of a broken  up range. All can be found in note 4. Lets call first example 9A this will be a combination of one counter mount cooking unit (cook top) and one oven. Lets go with a 6 kva cook top and 3 kva oven. You simply add the two and treat them as one unit 6 kva = 3kva = 9 kva. Take it to column “C” (not over 12 kva no % added) one range 8000 va. Now 9B a 6 kva cook top and two 4 kva ovens. Add up all units (6 kva + 4 kva+ 4 kva = 14 kva) and treat as one. Here its over 12 kva so we must subtract 12 kva from 14 kva (14 kva – 12 kva = 2 kva) then add the 5% for each remainder (8 kva x 110% = 8.8 kva). The calculated load for all units would be 8800 va.And finally we finishing the last step. 

                                        Just  a couple of FYI’s I have run into taking various test. One question was what is the max size range you can put on a 40amp breaker while Appling table 220.55. We now have some experience with this table so rather then trying to figure this all out, we now can see the answer is a 16kva range. Small thing like this answer can save time on a exam ( 16 kva – 12 kva = 4 kva)(8000 va x 120% = 9600 va / 240 amps = 40 amps). Keep in mind this answer is baring all other codes. Another question I have gotten on exams is calculating a range load with out the use of the table in front of me. Unexpected this could be a nightmare. But lets take a look at the way this table is designed and you will see this is not so hard to answer. The first column is simple ,its numbered 1-25 for the most part. This is the amount of units we are dealing with. Then numbered in groups 26-30.Any other amount you wont be questioned on. The last column is what we want to know. From all the work we have already done we alamaticly know one range is at 8000 va. The next 4 numbers are increased by 3 from the last number (8,11,14,17,20). From hear the numbers simple increase by one all the way down to 25 ranges. Keep this format of the table and you will be pretty safe.

A small example is; what is the damned factor for 15 ranges. We know we start with 8000va for one range. The next 4 are 11,14,17,20, increasing by 3.This covers us to 5 ranges, were dealing with 15.Ten more ranges increased by 1 will give us a demand of 30 kva. Well that covers table 220.55 pretty complete.  Hope you enjoyed the article




 1. One Range { note 4 }      Less Then 12 kva




           {Table 220.55 column  C }  ®   8000Va /  240 V = 33 amps



2. One Range { note 1 }       More Then 12 kva


                                                            14 kva Range


                                                                                  note 1


                        {Range size }  ®    14 kva Range  - 12 kva = 2 kva


                                                                    2 kva X 5% = 10%



       {Table  1 Range  Col.C } ®      8000 kva X 110% = 8800kva



3. More the one range, Equal ratings {note 1}   Over 12 kva 


          Example ;  3 / 15.6 kva Ranges

                                                                                  note 1


                                         (Range size) ® 15.6 kva – 12kva = 3.6 kva


                                                                         4 X 5% = 20%


                                                                    14kva X 120% =16.8 kva


                                                              3 Ranges

                                                              column C


 4. More then one range , Equal ratings { Column C }  Not Over 12 kva


                                                              {5 units } 9 kva = 20 kva

­                      ­

                                                                               ranges    column C




5. More then one range, Unequal Ratings {note 2 } Over 12 kva


               Example ;    3/ 9 kva Ranges

                                  3/ 14 kva Ranges

                                                                    note 2


                                                3/9 kva  {min 12 kva } 3 ranges X 12 kva {range} = 36 kva

                                                 3/ 14 kva                      3 ranges X 14 kva {range} = 42 kva


           AVERAGE                       78 kva / 6 = 13 kva


                                                         13 kva –12 kva = 1 kva


                                                          21 kva X 105% = 22.05

                                                              ­         {5%}

                                                         column C


6. More then one range, Equal Ratings {note 3 } Over 1 3/4 kva   ®   less then 3 ½ kva 


                 Example : 10 3 kva ranges


                                                        3 kva ranges X 10 units = 30 kva


                                                               30 kva X 0.49% = 14.70 kva


                                                                          Column A


7. More then one range , Equal Ratings {note 3 } 3 ½ kva  ®  8 ¾  kva 


                  Example ; 8  6 kva ranges


                                                           6 kva ranges X 8 units = 48 kva


                                                             48 kva X 0.36% = 17.28 kva


                                                                     Column B



8. One Counter – Mounted Cooking  Unit   Or  One Oven


                                                             NAMEPLATE RATEING     



9 One Counter – Mounted Cooking  Unit  And  Up to Two ovens


A.    1 cook top & 1 oven 

                   Example ;   6 kva  cook top

                                       3 kva  oven


                                                                     6 kva + 3 kva  = 9 kva

                                                            { not over 12 kva no % added }


                                                                              Column C


                                                                     One  range  8000 kva




                   Example;   1/   6 kva cook top

                                     2/   4 kva ovens


                                                            6 kva  + 4 kva + 4 kva = 14 kva


                                 { total ranges }  ®   14 kva – 12 kva = 2 kva


                                                                    8 kva X 110% = 8.8 kva


                                                                  Column C



Monthly Column by Eddie Rodriguez      


For a simple one unit ac condenser all calculations are done for you. The nameplate contains all you need to know. First the conductor, size, the name plate will have the term “MAX CIRCUIT AMPACITY” on it. This term tells you what size wire you need. You will also need the assistance of table 310.16.The 75
degree column would be used due to the terminal temperature rating{NEC 110.14}.
Now the over current device. Again the nameplate does all your work. The term “MAX SIZE OVERCURENT DEVISE” tells you what size fuse or breaker you will need. You can always check section 240.6. This section list all breakers and fuses made. Hear you would pick the device closet to the nameplate with out going over. Keep in mind that if the nameplate specifies fuses you must use them. This is not a code but a manufactures requirement or you will void any guarantees. Ok yes you can tie in the NEC 110.3 (B) which states you must follow the instructions of equipment installed. So so far anything discussed can be found in Article 440 of the NEC.Keep in mind not every code section is listed in this article .This is to keep it simple, I don’t think you need a bunch of numbers to teach proper installations. So if for some reason the over current device trips on start up you can up the breaker size accordance of section 440.22. Personally I haven’t had a case yet for this need, but hay somebody must have for it to be in the code. Very often forgotten is work clearance when installing AC disconnect. Don’t forget 110.26 Spaces About Electrical Equipment. This section requires 3’ in front of the disconnect must remain clear. This is for at least 30” horizontal of the disconnect from grade to a min of 6’6”. So this means above and below disconnect. Next months article you don’t want to miss, “Lets simplify the range table 22.55”.

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Monthly Column by Eddie Rodriguez   

Puck Lights When Can I use Them?

Countless times I have walk up to a roughed out kitchen counter and I see wire set up for low voltage under counter lighting. I have seen every type of wire tried.  Zip cord, uf stranded out door cable, and even thermostat/door bell wire. None of these wires are approved in the walls for this type of installation. Another installation that’s not code compliant is 6 or so puk lights wired with Romex. The splices have no boxes and the lights no connectors. This is actually a class three wiring type installation. Class three wiring requires boxes, wire nuts, and connectors. Basically most of the wiring we do is class three wiring, we will get into this more shortly. So class 3 is not the solution to a code compliant puck light installation. Many electrians have come to me and said so how do I install this lights? My customer must have them. Well really there not desighined for this type installation. They were actually designed for bookshelves with all wiring in the shelves and cord and plug connected. Till the 2005 NEC code came out, and introduced some changes and new sections, I had no answer for them. Article 411 and 725 now offer some reprieve for these lights. The start to the answer begins with the additions to section 411.4 in the 2005 codebook. Section 411.4 deals with locations not permitted for lighting systems operating at 30 volts or less if not installed to the requirements to (A) or (B).(A) where concealed or extended through a building wall.(B) with in 10’of pools, spas, fountains or similar locations.(A) is ower concern for the time being.(A)(1) allows any chapter 3 wiring method.Basicaly standard wiring methods like romex or uf cable. This wiring method requires proper connectors, boxes and proper connectors for entering wire into light fixture. If your luckily enough to find connectors for these lights it all comes out looking gordy and ugly under the counters. Remember all splices must remain accessible. So 411.4(A)(1) is not the answer to ower problems. But 411.4 (A)(2) is the new entree that will help us now. Hear we are required to supply the lights by a class 2 power source and in accordance with section 725.52.. A small summery might help here. Ok so if we follow 411.4 and 725.52 we can install these puck lights with lets say romex cable. Splices under the counter, with no boxes or connectors .All this and staying with in the bounders of the code. Great just what we are looking for lets go on. After looking all thought Article 725. As long as we stay with in class 2 power source. We can install these puck lights with any Chapter 3 Wiring method or class 2 wire. This installation would allow open splices under the counter. What we need to keep in mind is that we must not exceed the limitations of article 725 for a class 2 power source. In the back of the codebook are two tables 11(A) and 11(B). These tables explaining the limitations of class 1,2,and 3 power sources. Class 2 Limits us to 250va max. Most puck lights are about 50 watts. This leaves us with only allowing 4 lights on a circuit.

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