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Welcome to this module in the Durbin's Learning system. We are going to introduce you today to the calculations required by the NEC 2011 to determine feeder and branch circuit loads for cooking equipment in a dwelling.

Upon completion of this module you will be able to :

1. Determine the demand on a dwellings service of the cooking appliances

2. Determine the demand for large cooking appliances rated 12.1kW to 27kW

3. Determine the demand for medium cooking appliances rated 8.75kW through 12 kW

4. Determine the demand for small cooking appliances rated less than 8.75kW

5. Determine the demand on the branch circuit feeder for all of the above type cooking appliances

6. Determine the demand for cooking appliances used for instructional purposes.

You will also be very familiar and comfortable with using Table 220-55 of the NEC.

Note that this module was written according to the 2011 edition of the NEC.

For the purposes of this module all household cooking appliances are assumed to be single phase, 240V appliances on single-phase services.

This is some of the terminology you will encounter in this module.

Demand

Demand Factor

Cooking Appliances

This module has been written in accordance with the 2005 National Electric Code. The primary difference between the 2002 and 2005 Code as applies to cooking calculations is the re-designation of the Table 220.55 All column references in this module are to the 2005 Code.

You should have a 2005 Code book, a pencil, a calculator and paper to take notes on.

**Article 220.55 Electric Ranges and Other Cooking Appliances- Dwelling Unit(s**)- The demand load for household electric ranges, wall-mounted ovens, counter-mounted cooking units, and other household cooking appliances individually rated in excess of 1 3/4 kW shall be permitted to be computed in accordance with Table 220.55.

You should have a 2005 Code book, a pencil, a calculator and paper to take notes on.

The above is the official way of writing the following:

Cooking equipment that has a nameplate rating that is greater, but not including, 1.75kW has a very small chance of operating at full nameplate rating for any length of time....SO... we do not have to account for the total nameplate rating. We can calculate it at an amount that is less. How much less? We will learn that over the next few pages. The method for determining how much less will come from Table 220.55 and it's notes.

Cooking equipment that has a nameplate rating that is 1.75kW or less does not use a lot of power. They are very small and for the most part have only two settings, off and on. A good example of this type of appliance is a built in food warmer.

These appliances use their full nameplate every time they are turned on. We have to calculate these appliances at their full nameplate rating.

Remember: You must never use these appliances in conjunction with Table 220.55. These appliances' demand will be added to the cooking equipment demand, at 100%, after all other calculations have been completed.

Step 1. Find the "number of appliances" in the first column.

Step 2. Look under column C and cross index.

This one gets a little trickier.

You will notice the table does not have a number. Rather is states " 15kW + 1kW for each range"

Mathematically you would determine the demand this way:

15,000 + ( number of ranges X 1,000)= demand

In this case it would be:

15,000 + ( 35 X 1000)= 50,000

Your answer should be 50kW

This is the standard cooking appliance or perhaps it would be better to say the basis of Table 220.55. Most older style ranges fell into this category, particularly before the self cleaning oven. The demand factor system was developed based on these ranges. Later accommodations were made for smaller cooktops and wall mounted ovens and for large rated, self-cleaning ovens.

Determine demand from Table 220.55, Column C.

**Example:** What is the demand load for two ranges rated at 10kW?.

Step 1. Find the "number of appliances" in the first column of Table 220.55. In this case "2".

Step 2. Run your finger across to Column C. It should read 11.

This means that the demand load for both ranges is 11kW.

This is the total demand load. They are not 11,000 each but 11,000 for the combined load.

If the ranges had been one 9kW and one 10kW, the process would have been the same and the answer would have been the same. This method can only be used with ranges rated 8.76kW through 12kW.

Determine demand from Table 220.55, Column C.

Step 1. Find the "number of appliances" in the first column of Table 220.55. In this case "2".

Step 2. Run your finger across to Column C. It should read 11.

This means that the demand load for both ranges is 11kW.

This is the total demand load. They are not 11,000 each but 11,000 for the combined load.

If the ranges had been one 9kW and one 10kW, the process would have been the same and the answer would have been the same. This method can only be used with ranges rated 8.76kW through 12kW.

Step 1. Find the "number of appliances" in the first column.

Step 2. Look under column C and cross index.

Your answer should be 25kW.

Step 1. Find the "number of appliances" in the first column.

Step 2. Look under column C and cross index.

This one gets a little trickier.

You will notice the table does not have a

number. Rather is states " 25kW + 3/4kW for each range"

Mathematically you would determine the demand this way:

25,000 + ( number of ranges X 750)= demand

In this case it would be:

25,000+ ( 65X 750)= 73,750

Your answer should be 73.75kW

Table 220.19, Note 3: Over 1-3/4 kW through 8-3/4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all ranges rated more than 1-3/4 kW but not more than 8-3/4 kW and multiply the sum by the demand factors specified in Column A or B for the given number of appliances.

Step 1: Total the nameplate ratings of all cooking appliances.

Step 2: Multiply sum by the demand factor of Column A or B.

Column A = over 1-3/4 kW but less than 3-1/2 kW

Step 1: Add nameplate, 2 units x 3 kW = 6 kW

Step 2: Multiply sum by demand factor

Table 220-55, Column A = 2 units = 75% demand factor

(2 units x 3 kW per unit) x .75 demand factor = 4.5 kW

Net computed load (demand load) = 4.5 kW

Table 220-55, Column A= 49% demand factor for ten units

10 units x 2 kW each x .49 demand factor = 9.8 kW

Demand load = 9.8 kW

Example:

Column B demand factor for 4 units = 50%

4 units x 6 kW per unit x .5 DF = 12 kW net computed demand

Any demand load calculated using Columns A or B can also be calculated using Column C.

Example: What is the demand load for 8 ranges rated 8 kW?

8 kW can be Column C (not over 12 kW rating)

8 kW can be Column B (3-1/2 kW to 8-3/4 kW rating)

Column C = 23 kW demand load for 8 units

Column B = demand factor for 8 units = 36%

8 units x 8 kW per unit x.36 DF = 23.04 kW demand load

In this case, the Column C answer is the smallest and should be used as the demand load.

Example: 15 ranges rated 7.8 kW each. What is the

minimum demand load?

Column C = 30 kW demand load for 15 units

Column B = demand factor for 15 units = 32%

15 units x 7.8 kW per unit x .32 DF = 37.44 kW demand load

The Column C answer of 30 kW is the minimum

demand load.

Note: You must always use the smallest load when taking the Journeyman's test.

Note 1 is used to calculate the demand load for individual ranges over 12 kW through 27 kW when all the nameplate ratings are the same. Column C is the only column used in this category. A major fraction is .5 or larger.

Example: Find the demand load for 10 ranges rated 15 kW each.

Step 1: Use Column C to find the demand load for 10 units.

Column C,10 units = 25 kW

Step 2: Subtract 12 kW from nameplate rating.

15 kW nameplate rating -12 kW = 3 kW in excess of 12 kW

Step 3: Increase the Column C demand load by 5% for each 1 kW

or major fraction that the nameplate rating exceeds 12 kW.

Note: The Column C demand load is 25 kW (Step 1). The

nameplate rating (15 kW) exceeded 12 kW by 3 kW (Step 2).

25 kW increased by 15% is the same as

25 kW x 1.15 = 28.75 kW

Net computed demand load = 28.75 kW

Example: What is the demand load for a 17.5 kW range?

Step 1: Column C = 8 kW for one unit

Step 2: 17.5 kW exceeds 12 kW by 5.5 kW. Since .5 is a major

fraction, 5.5 kW is the same as 6 kW.

Step 3: 6 kW x 5% = 30% increase of the Column C answer.

8 kW increased by 30% is the same as 8 kW x 1.3 = 10.4 kW

Net computed demand load = 10.4 kW

Here is a trick to help you remember the process.

Use the example above.

Step one. Write down the number of ranges you are calculating.

In this case: 1

1

Step two. Draw a line approximately 2 inches long.

1 ----------------------

Step three. Write down the value found in Column C for the number of ranges.

1---------------------- 8,000

Step four. Write down the rounded off rating of the range under the line.

1---------------------- 8,000

18

Step five. Write the number 12 under the range rating. This number will always be 12.

1-------------------- 8,000

18

12

--------

Step six. Subtract 12 from the first number and write down your answer.

1 ----------------------- 8,000

18

12

--------

6

Step seven. Multiply your answer by 5. It will represent the 5 %. Be sure to not multiply by 5%.

1 ------------------- 8,000

18

12

_____

6 x 5= 30

Step eight. Now multiply the number next to the line by 1. _____. The blank will be the answer to your multiplication in step seven.

1 ---------------- 8,000

18 x1.30

12 10,400

______

6 x 5= 30

Answer= 10.4 kW

Note two calculations are very similar to Note 1 calculations. This Note is used for those situation when multiple ranges need to be calculated and the nameplate ratings of the ranges fall within 8.76 kW and 27 kW.

The Note describes this as "Over 8 3/4 kW through 27 kW of unequal ratings." The key words of this sentence are over and unequal.

Over: This means the rating of 8 3/4 kW can not be included in this note.

Unequal: This means that there must be more than one range and that at lest one range must have a rating that is different from the rest of the ranges.

The next part of the note instructs us to average the value of the ranges. Then the note instructs us to change the rating of any range rated less than 12 kW to 12 kW before we total the values of the ranges.

This was a wording change in the 1999 edition of the NEC. The previous wording caused confusion about the use of the table. This wording clarifies the intent of the CMP.

According to this wording we simply add the values of our nameplates and then divide the sum by the number of ranges. All we need to remember is that for any range that has a nameplate value of less than 12 kW, we must consider that range to have a value of 12 kW for the purposes of the average.

The last step of the calculation is a direct quote from Note 1. A helpful reminder for Note 2 calculations is to think of it as a Note 1 calculation.

Start off the same as a Note 1 calculation. Write down the number of ranges and then draw a line and write down the value of Column C. The next step would be to write the rounded off nameplate rating. If there are multiple ranges with different nameplate ratings you would not be able to determine the value to write down. The average value seems to be the logical answer." This reminds us to look at Note 2. Get an average. The rest of the calculation is identical to a note 1 calculation.

An apartment complex has 5 apartments. The apartments are equipped with different ranges. The nameplate ratings of the ranges are: 8.97 kW, 9.65 kW, 13.6 kW, 21.2 kW and 23.6 kW. What is the demand of the cooking equipment on the main service entrance?

Step one. Write down the number of ranges and the value found in Column C.

5----------------20,000

Step two. Write down the rounded nameplate value of the ranges. We will need to get an average value.

8.97 becomes 12

9.65 becomes 12

13.6

21.2

+23.6

82.4

Now divide the sum by the number of ranges.

82.4/5=16.48

Round this number off to 16 and write it under the line.

5--------------20,000

16 x1.20

12 24,000

4x5=20

The answer is: 24 kVA

Note three is used for small ranges. In the 1970s architects moved away from the traditional all-in-one range. They moved to the counter mounted cook-top and wall mounted oven. Typically homes are designed with a single cook-top and two wall mounted ovens. This is still common today, however we now see a mix of ranges and cook tops.

The problem with the counter mounted cook top and wall mounted ovens was with their rating size. They typically have ratings less than 8 kW. The entire purpose of Article 220.55 was to reduce the amount of load we used to size our conductors. This was done as a practical, cost saving measure. The typical cooking appliance is not going to place it's full load on the circuit for an extended length of time. Therefore why spend the extra money to size the conductor for the full load unnecessarily. Table 220.55 was developed and worked very well for determining the size of conductor needed. Until the small cooking appliances showed up on the scene.

Each of the appliances, the cook top and the wall mounted ovens are considered a separate piece of cooking equipment. Let's take a look at a typical situation to demonstrate why Table 220.55 had to be modified to accommodate these small units.

Let's suppose we have a home that is equipped with a 3.2 kW cooktop and a 5.1 kW wall mounted oven and a 4.9 kW wall mounted oven.

In applying these appliances to Table 220.55 we can simply use Column C. They are all rated less than 12 kW. According to Column C the demand that is to be applied would be 14,000VA.

Now let's add the nameplate ratings of the appliances together.

3.2

5.1

+ 4.9

13.2 kW

Notice that the actual nameplate ratings added together is less than the adjusted demand factor. Table 220.55 took us in the opposite direction it was intended. Thus the NEC code making panel added Columns A and B and Note 3 to the table.

The first step will always be to add the nameplate ratings of the appliances together.

Here is the format that can be used when doing note three calculations. This format will help you keep the steps in order and stay organized.

We will use the following example.

A single family dwelling is equipped with a 3.2kW cook-top and two 4.6kW wall mounted ovens. What is the total cooking equipment demand on the main service entrance.

Step One: On a piece of paper write down

A B C

Step Two: Under each column write down the name plate ratings of each appliance in the appropriate column. Table 220.55 Column A has a range of (Less than 3.5kW) and Column B has a range of (3.5kW to 8.75kW).

A B C

3.2 4.6

4.6

Step Three: Write down the total number of appliances there are in each column. Write this next to the letter for the column. This is an obvious number when doing single family calculations. It is important to establish the habit however, because multi-family dwelling calculations can be very confusing on this point.

A 1 B 2 C

3.2 4.6

4.6

Step Four: Add the totals in each column.

A 1 B 2 C

3.2 4.6

+ 0 +4.6

3.2 9.2

Step Five: Note three instructs us to apply the demand factor found in Table 220.55 to each column. Find the demand factor for each column. Use the amount of appliances you wrote down next to the column letter to determine the demand factor.

For example. Find Column A on Table 220.55. Next to A on our sheet we have written 1. Cross index 1 and Column A on Table 220.55. You will get a result of 80. This means 80%. If we do the same for Column B and 2 we get a result of 65 or 65%. Now multiply the sums in each column by these numbers.

A 1 B 2 C

3.2 4.6

+ 0 +4.6

3.2 9.2

x.80 x .65

2.56 5.98

Step Six: Now simply add the results. 2.56 + 5.98= 8.54kW

Step Seven: Because Note 3 is optional, we need to compare our results from Columns A and B to the result from Column C. The correct answer will be the lower of the two totals.

A 1 B 2 C

3.2 4.6 14 kW

+ 0 +4.6

3.2 9.2

x.80 x.65

2.56 + 5.98 = 8.54kW

In this example we would use the 8.54kW.

On the next page you will practice some Note 3 calculations. Then we will take a look at applying Note 3 to multi-family dwellings.

Multi-family Note 3 calculations are done the exact same way as single family dwellings. The format we taught you is applicable to both calculations. You will need to be sure to differentiate between the amount of apartments and the amount of appliances. This is the most confusing factor of the calculation.

Let's do one together.

What is the total cooking equipment demand on one meter bank of an eighty apartment complex? The apartments are divided evenly between four meter banks. Each apartment is equipped with a 3.41 kW cooktop and two 8.73 kW wall mounted ovens. The service is 230 volts, single-phase.

Note: In this problem we are only looking for the demand on a single meter bank. The complex has eighty apartments: however only twenty apartments will be on a single meter bank.

Step One: Write A B C on a piece of paper.

A B C

Step Two: Divide the appliances up according to their respective columns and write them down under the appropriate column and total the kW for each column.

A 20 B 40 C

3.41 8.73

x20 +8.73

68.2 17.46

x 20

349.2

Step Three: Write the number of appliances in each column next to the letter for that column. See above. Notice we wrote the number of appliances. It is very easy to fall into the trap of writing down the number of apartments.

Step Four: Apply the demand factor from each column to the total kW in each column.

A 20 B 40 C

3.41 8.73

x 20 +8.73

68.2 17.46

x .35 x 20

23.87 349.2

x .22

76.824

Step Five: Add the totals from Column A and Column B.

23.87 + 76.824 = 100.694

Step Six: Compare the total to the total found in Column C for 60 appliances.

A 20 B 40 C

3.41 8.73 25,000

x 20 +8.73 + ( 750 x 60)

68.2 17.46 70,000

x .35 x 20

23.87 349.2

x .22

76.824

23.87 + 76.824= 100.694

In this example the value found in Column C is less than the volume found in Columns A and B. We are allowed to use either method according to Note 3. Therefore we will use the smaller of the two values to determine the minimum demand on the main service entrance.

Look over the example and compare it to the NEC and Table 220.55. If you have any questions concerning Table 220.55 and Notes 1, 2 and 3, you should stop the module at this point. Use the form below to ask questions. We will respond to your questions within one business day. Then you can continue to the next section prepared

All of the calculations you have learned up to this point have determined the demand of the cooking equipment on the main service of a dwelling unit. This helps us to determine the service size and wire size needed to feed the service.

A circuit breaker (OCPD) will need to be installed in the panel to provide overcurrent protection for the wire that feeds the range or cooking equipment in the house. This circuit breaker will need to be sized according to the NEC. The wire from the panel to the cooking equipment will also have to be sized according to the NEC.

Several code references must be consulted to determine the size of the OCPD and the wire. First we must determine the load or demand on the branch circuit. Branch circuit demand requirements are found in Part 1 of Article 220.

Specifically, 220.14(B) contains the requirements for "Electric Dryers and Household Cooking Appliances."

"Load computations shall be permitted as specified in 220.54 for electric dryers and in 220.55 for electric ranges and other cooking appliances."

This statement simply means we have a choice. We can either size the OCPD and wire using the full nameplate value or we can apply all the things we have learned up to now and "squeeze" the nameplate using Table 220.55. This will allow us to make a safe installation at a lower cost which can be passed onto our customers. Everything in 220.55 that applies to service entrances also applies to branch circuits.

There is one twist. Note number 4 to Table 220.55

Note 4 to Table 220.55 is used exclusively when determining loads for a branch circuit.

You should make a note in your Code books that this note can never be applied to service demand sizing calculations.

The note applies to three very specific situations.

"It shall be permissible to compute the branch-circuit load for one range in accordance with Table 220.55."

This statement is simply a confirmation of 220.14(B).

"The branch-circuit load for one wall mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance."

This is a specific requirement. A single range can be squeezed with the table. A single cook top or wall mounted oven can not. It is important we always remember this distinction.

"The branch-circuit load or a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be computed by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range."

Two points:

1. Notice the mandatory language. You do not have to use Table 220.55 and squeeze a branch circuit load, but if you do....you have to do it this way. No options.

2. This note is taking into consideration that a combination of a cooktop and one or two wall mounted ovens, all on the same circuit, is functionally the same as a single range. Therefore the NEC is not only allowing you to treat it as a single range, they are requiring you to treat it as a single range.

This mandatory language is a good example of the precise language of the NEC. Why is it mandatory to treat the combination as a single unit? The preceding sentence informed us that a single cook top or wall mounted oven could not be squeezed. Any squeeze allowed would have to be done by doing some sort of combination. The CMP is simply defining the only acceptable combination.

Impressive is the precision of the language of the NEC.

Now let's take a look at some other references which will apply

Article 210.55 contains the requirements for sizing branch circuits. By sizing we mean determining the minimum amount of ampacity the branch circuit must be able to carry.

The requirement is found in 210.19(A)(1)

The branch-circuits conductors must have an ampacity that is equal to at least the maximum load served. They must also be capable of carrying 125% of the maximum load of any continuous duty loads.

The NEC defines a continuous duty load as:

"A load where the maximum current is expected to continue for 3 hours or more."

Does a piece of cooking equipment fall in to the category of continuous duty loads? Will it operate for a period of 3 hours or more at maximum current? Perhaps when cooking a pot roast or a large turkey the oven may operate for more than 3 hours but not at maximum current. The maximum current rating is based upon the oven operating at maximum and all the burners operating at high. The oven will turn on and off as it reaches temperature and falls below temperature. The burners will also fluctuate in current as they reach their desired temperature. The end result is there is no time that the cooking equipment can be expected to operate at maximum current for three hours straight. Therefore the conductor for the branch circuit must be rated to carry 100% of the maximum load served.

Remember this "maximum load served" an be determined by either the nameplate or by using table 220.55 and "squeezing" the load. The choice is yours." Also keep in mind, that when a test asks for the minimum it is expected.

Article 210.19 (A)(3) contains specific requirements for branch circuits for household cooking equipment.

"Branch-circuit conductors supplying household ranges, wall-mounted ovens, counter-mounted cooking units, and other household cooking appliances shall have an ampacity not less than the rating of the branch circuit and not less than the maximum load to be served."

This statement is a clarification that a cooking load is not considered to be a continuous duty load. The circuit must be sized according to 100% of the maximum load. Keep in mind that 220.14(B) allows us to adjust this maximum expected load using 220.55.

"For ranges of 8 kW or more rating, the minimum branch-circuit rating shall be 40 amperes."

This is a specific requirement for ranges rated over 8 kW. This is the nameplate rating, not the adjusted rating. It does not apply to counter-mounted appliances or wall-mounted ovens. This requirement supercedes all calculations. The minimum branch circuit can be 40 amperes regardless of the adjustment allowed in 220.55.

There are two exceptions to this rule.

"Exception No. 1: Tap conductors supplying electric ranges, wall-mounted electric ovens, and counter-mounted electric cooking units from a 50-ampere branch circuit shall have an ampacity of not less than 20 amperes and shall be sufficient for the load to be served. The taps shall not be longer than necessary for servicing the appliance."

This exception is specific to those situations that we discussed in the note 4 portion of this module. The typical method of wiring would be to run a single cable from the panel to the kitchen area and terminate it in a J-box. Then run individual cables from the J-box to the cook top and wall mounted ovens. The NEC allows for these conductors to be of a smaller size according to the tap conductor rules found in Article 240.21. This exception gives some specific requirements for these tap conductors.

1. They must be as short as possible and still allow for servicing. In other words they must be kept to a minimum but the CMP wanted to ensure that this length requirement did preclude the ability to remove the equipment for servicing.

2. If the branch circuit is a 50 amp circuit then the conductors must have an ampacity that is at least 20 amperes. All other branch circuits must follow the tap conductor rules found in article 240.21.

Exception No. 2 pertains only to neutral conductor sizing. This will be covered in a future module when we cover neutral load calculations.