Range Calculator by Gerald Newton electrician2.com
NEC 220.55 Feeder and Service Range, Stove Top, and Oven Calculator for Dwellings
and cooking appliances used in instructional programs (see examples at bottom of page)

 

Section 1

Each unit load required to be
Greater than 1 3/4 KW and less than or equal to 12 KW
(See NEC Table 220.55 and Note 3)

 

Final Output for Section 1  KW

Inputs

Subdomain Analysis

    Each unit load required to be
greater than 1 3/4 KW and less than or equal to 12 KW
  greater than 1 3/4 KW to less than
3 1/2 KW

Col A
3 1/2 KW to 8 3/4 KW


Col B
not over 12 KW



Col C
over 12 KW through 27 KW


Note 1

over 8 3/4 KW to 27 KW


Note 2

over 1 3/4 through 8 3/4 KW


Note 3
   

A

Select KW

B

Select fraction
KW

C
Select Number
of units with KW value

D

rangea

E

subtotal

F

rangeb

G

subtotal

H

rangec

I

subtotal

J

ranged

K

subtotal

L

rangee

M

subtotal

N

rangef

O

subtotal

P

1    
2    
3    
4    
5    
6    
7    
8    
9    
10    
 

Unit Output Summary

Total Connected Load KW for each subdomain

Col A Col B Col C     Note 2 Note 3

Total Number of Units for each subdomain

Units Units Units     Units Units

At least one value is in subdomain

Y? Y? Y?     Y? Y?

Demand Load in KW

col a demand col b demand col c demand        

Calculated Demand Load in KW 

            col a , b, or col c

 

    sum col a and b                

 

Section 2

Each Range load required to be
Over 12 KW to 27 KW and of the same rating
(See NEC Table 220.55 Note 1)

 

Final Output for Section 2  KW

Inputs

 
   
Enter over 12 KW to 27 KW
(All ranges of the same rating)

Summary for over 12 KW to 27 KW ranges of the same ratings

    Select KW Select fraction
KW
Select Number
of units with KW value
            over 12 KW through 27 KW      
                       

Total Connected Load KW

                     

Number of Units

                     

At least one value in domain

                     

Total Demand in KW

            Total        
 

Section 3

Each Range load required to be
Over 8 3/4 to 27 KW unequal ratings
(See NEC Table 220.55 Note 2)

 

Final Output for Section 3 KW

Inputs

 

Enter over 8 3/4 KW to 27 KW
(Ranges of the unequal rating)

Summary over 8 3/4 to 27 KW unequal ratings

    Select KW Select fraction
KW
Select Number
of units with KW value
                  over 8 3/4 KW to 27 KW    
27                    
28                    
29                    
30                    
31                    
32                    
33                    
34                    
35                    
36                    

Summary for over 8 3/4 KW to 27 KW of unequal ratings

Total Connected Load KW

                     

Number of Units

                     

At least one unit in Domain

                     

Total Demand in KW

                     

Examples

Example 1
(from 2005 NEC Handbook Annex D Example D6)

Ranges all of the same rating
24 ranges each rated 16 KW.
Maximum demand for 24 ranges at 12-KW ratings is 39 KW.  16 KW exceeds 12 KW by 4.

5 per cent x 4 = 20 per cent
39 KW x 20 per cent = 7.8 KW
39 + 7.8 = 46.8 KW

Total demand is 46.8 KW
(Verified using Section 2 of Calculator)

 

Example 2
(from 2005 NEC Handbook Annex D Example D6)

Ranges of unequal rating
5  ranges at 12 KW
2 ranges at 12 KW
20 ranges at 13.5 KW
3 ranges at 18 KW

5 x 12 = 60 KW
2 x 12 = 24 KW
20 x 13.5 = 270 KW
3 x 18 = 54 KW

total           30 ranges   408 KW

408 / 30 = 13.6 KW (average per Note 2 to Table 220.55
From Column C Table 220.55 demand for 30 ranges of 12 KW rating is 15 KW + 30 ranges = 45 KW.
13.6 exceeds 12 KW by 1.6.  Round off to whole number or 2.
5 per cent x 2 = 10 per cent ( 5 per cent increase for each KW over 12 KW)
45 KW x 10 per cent = 4.5 KW

Total Demand is 45 KW + 4.5 KW = 49.5 KW
(Verified using Section 3 of Calculator)

Example 3
(from Author)

Ranges of unequal rating
5  ranges at 12 KW
2 ranges at 12 KW
20 ranges at 13.5 KW
40 ranges at 18 KW

5 x 12 =      60 KW
2 x 12 =      24 KW
20 x 13.5 = 270 KW
40 x 18 =    720 KW

total           67 ranges     1074 KW

1074 / 67 = 16 KW (average per Note 2 to Table 220.55
From Column C Table 220.55 demand for 67 ranges of 12 KW rating is 25 KW +( 67 x 0.75 ) = 75.25 KW.
16 exceeds 12 KW by 4. 
5 per cent x 4 = 20 per cent ( 5 per cent increase for each KW over 12 KW)
75.25 KW x 20 per cent = 15 KW

Total Demand is 75.25 KW + 15 KW = 90.3  KW
(Verified using Section 3 of Calculator)

Example 4
(from Harman Guide to the National Electrical Code)

Ranges all of the same rating
5 ranges each rated 16 KW.
Maximum demand for 5 ranges at 12-KW ratings is 20 KW.  16 KW exceeds 12 KW by 4.

5 per cent x 4 = 20 per cent
20 KW x 20 per cent = 4 KW
20 + 4 = 2 KW

Total demand is 24 KW
(Verified using Section 2 of Calculator)

Example 5
(from Harman Guide to the National Electrical Code)

Ranges of unequal rating
1 range at 16 KW
1 range at 17 KW
1 range at 10 KW
 

1 x 16 =       16 KW
1 x 17 =      17 KW
1 x 10 =      10KW

total             3 ranges        43 KW

43 / 3 = 14.3 KW (average per Note 2 to Table 220.55)
From Column C Table 220.55 demand for 3 ranges of 12 KW rating is 14 KW.
14.3 exceeds 12 KW by 3.  (round up)
5 per cent x 3 = 15 per cent ( 5 per cent increase for each KW over 12 KW)
14 KW x 15 per cent = 2.1 KW

Total Demand is 14 KW + 2.1 KW = 16.1  KW
(Verified using Section 3 of Calculator)

Example 6
(from Harman Guide to the National Electrical Code)

Ranges of equal rating using Column B and C of Table 220.55.
50  ranges at 6 KW
 

From column B demand for 50 each  6 KW ranges is 20 per cent of total load.
50 x 6 =300 KW;  20 per cent of 300 is 60 KW.


From Column C Table 220.55 demand for 50 ranges of 12 KW rating is 25 KW + 0.75 x number of ranges.
0.75 x 50 = 40; 
25 KW + 40 = 65 KW
Total Demand is 65  KW
(Verified using Section 1 of Calculator)

By Note 3 to Table 220.55 either Column C or Column B may be used. 
Minimum Demand is 60 KW.
(Verified using Section 1 of Calculator)

Example 7
(from Harman Guide to the National Electrical Code)

Ranges of equal rating using Column A and C of Table 220.55.
10 counter top ranges at 3 KW
 

From column A demand for 10 each  3 KW ranges is 49 per cent of total load.
10 x 3 =30 KW;  49 per cent of 30 is 14.7 KW.


From Column C Table 220.55 demand for 10 ranges of 3 KW rating is 25 KW .

Total Demand is 25  KW
 

By Note 3 to Table 220.55 either Column C or Column A may be used. 
Minimum Demand is 14.7 KW.
(Verified using Section 1 of Calculator)

Example 8
(from Harman Guide to the National Electrical Code)

Ranges of equal rating using Column C of Table 220.55.
50 counter top ranges at 12 KW.


From Column C Table 220.55 demand for 50 ranges of 12 KW rating is 25 KW + 0.75 x number of ranges.

 0.75 x 50 = 37.5
25 KW + 37.5 = 62.5 KW

Total Demand is 62.5  KW
(Verified using Section 1 and Section 3 of Calculator)

Example 9
(from Author)

Ranges of unequal rating in Columns A, B, and C.
2  ranges at 2 KW
1 range at 4 KW
1 range at 12 KW
 

2 x 2 =       4 KW
1 x 4 =       4 KW
1 x 12 =    12 KW

total             4 ranges     20 KW


From Column C Table 220.55 demand for 4 ranges of 12 KW rating is 17 KW.
Ranges are in Table 220.55 Columns A, B, and C but are not all 8 3/4 KW or less therefore Note 3
cannot be used.

Total Demand is 17 KW
(Verified using Section 1 of Calculator)

 

Example 10
(from Harman Guide to the National Electrical Code)

Ranges supplied by  3-phase 208Y/120 volt circuit.
Section 220.55 provides that the total load shall be based on the basis of twice the maximum
number of ranges connected to any two phases.

Given 30 each 12 KW ranges.

number of ranges /3 = 10 ranges/phase

2 x number of ranges/phase = 20

Using Section 1 the total demand load for 20 each 12 KW ranges is 35 KW.

The demand load for each two phases is 35/2 = 17.5 KW.

The total demand load is 3 x 17.5 = 52.5 KW

The line current is:

I = 52,500 / (1.732 x 208) = 52,500 / 360.3 = 145.7 amperes

 

Example 11
(from Author)

Ranges of unequal rating.
6  ranges at 3 KW
10 ovens at 5 KW
14 ranges at 6 KW
45 ranges at 12 KW
25 ranges at 15 KW
1 range at 20 KW

All these values do fit into either Section 1, 2, or 3.  We therefore have to decide on an alternate method.
One method is to use 8.8 KW for any value below 8 3/4 KW giving the following:

30 ranges at 8.8 KW
45 ranges at 12 KW
25 ranges at 15 KW
1 range at 20 KW

Total Demand is 105.8 KW
(Use Section 3 of Calculator)

A second method is to divide the problem into two categories.
One for all the values 12 KW and lower and the other for values greater than 12 KW then add the two results.

6  ranges at 3 KW
10 ovens at 5 KW
14 ranges at 6 KW
45 ranges at 12 KW

Using Section 1 the demand load is:  81.25 KW

25 ranges at 15 KW
1 range at 20 KW

Using Section 3 the demand load is:  47.15 KW

The sum of 81.25 and 47.15 is 128.4 KW

Example 12
(from Author)
This example illustrates a possible solution.  For installations such as this a legally licensed
and bonded professional electrical engineer should be consulted.

Ranges of unequal rating supplied by  3-phase 208Y/120 volt circuit.
6  ranges at 3 KW
10 ovens at 5 KW
14 ranges at 6 KW
45 ranges at 12 KW
25 ranges at 15 KW
1 3-phase range at 20 KW

This problem does not fit nicely within the rules of Section 220.55 or Table 220.55 and accompanying Notes.

Divide ranges to balance load and find how many are on each two phases.

For the 3-phase range divide load by three and apply to each phase.

Phases  AB                    AC                BC
           2 - 3 KW          2 - 3 KW      2 - 3 KW
           4 - 5 KW          3 - 5 KW      3 - 5 KW
           5 - 6 KW          5 - 6 KW      4 - 6 KW
          15 - 12 KW      15 - 12 KW  15 - 12 KW
           8 - 15 KW         8 - 15 KW     9 - 15 KW
             1 - 7 KW          1 - 7 KW     1 - 6 KW
_________________________________________
       35 - 366 KW        34 - 363 KW   34 - 359 KW

Select phases with maximum number of ranges which is AB.

2 x number of ranges/phase = 2 x 35 = 70

Double the number of ranges for each category and use 8.8 KW for any value below 8 3/4 KW.

Using 24 - 8.8 KW, 30 - 12 KW, and  16 - 15 KW use section 3 to find demand.

Demand is 81.38 KW using Section 3.

81.38 KW /2 = 40.69 KW

The total demand load is 3 x 40.7 = 122.1 KW

The line current is:

I = 122,100 / (1.732 x 208) = 122,100 / 360.3 = 339 amperes

 

Example 12A done another way that gives same answer.
(from Author)
This example illustrates a possible solution.  For installations such as this a legally licensed
and bonded professional electrical engineer should be consulted.

Ranges of unequal rating supplied by  3-phase 208Y/120 volt circuit.
6  ranges at 3 KW
10 ovens at 5 KW
14 ranges at 6 KW
45 ranges at 12 KW
25 ranges at 15 KW
1 3-phase range at 20 KW

This problem does not fit nicely within the rules of Section 220.55 or Table 220.55 and accompanying Notes.

Divide ranges to balance load and find how many are on each two phases.

Phases  AB                    AC                BC
           2 - 3 KW          2 - 3 KW      2 - 3 KW
           4 - 5 KW          3 - 5 KW      3 - 5 KW
           5 - 6 KW          5 - 6 KW      4 - 6 KW
          15 - 12 KW      15 - 12 KW  15 - 12 KW
           8 - 15 KW         8 - 15 KW     9 - 15 KW
             1 - 7 KW          1 - 7 KW     1 - 6 KW
_________________________________________
       35 - 366 KW        34 - 363 KW   34 - 359 KW

Step 1.  Find average value of one range using Note 2 of Table 220.5:

(30 x 12 + 45 x 12 + 25 x 15 + 1 x 20) / 101 = (360+ 540 + 375 + 20 ) / 101 = 1295 / 101 = 12.82

Note: 12 kw is used for values less than 12 kw per Table 220.55 note 2.

Step 2.  Next follow 220.55 second paragraph and example D5A in Annex D.

Maximum ranges connected between any two phases is 35. 

2 x 35 = 70

By table 220.55 column C demand is 25000 plus 3/4 KW for each range.

Demand is 25000 + 750 x 70 = 25000 + 52500watts = 77500

Increase this amount by 5 per cent since average range value is 12.82 KW

77500 + (0.05 x 77500)  = 77500 + 3875 = 81375

Step 3.  By example Annex D example D5A:

The demand load for each two phases is 81375 / 2 = 40688

Ans:  The total demand load is 3 x 40688= 122063 W

The line current is:

I = 122063 / (1.732 x 208) = 151125  / 360.3 =339 amperes

References:

 Annex D example D5A

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 (for 8 each 12 kw ranges) = 23,000 VA

Per phase demand = 23,000 VA ÷ 2 = 11,500 VA

Equivalent 3-phase load  (3 x 11500) = 34,500 VA

Table 220.55 Note 2

 “Over 8 3⁄4 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 8 3⁄4 kW and of different ratings, but none exceeding 27 kW, an   AVERAGE value of rating shall be calculated by adding together the ratings of all ranges to obtain the total connected load (using 12 kW for any range rated less than 12 kW) and dividing by the total number of ranges. Then the maximum demand in Column C shall be increased 5 percent for each kilowatt or major fraction thereof by which this average value exceeds 12 kW.”

Section 220.55 second paragraph:

“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.”

 

Example 13

Dwelling Unit - 3 phase range calculation supplied by 3-phase 4 wire 208/120 volt wye.      

Given:

            4-9kw

            4-13.8kw

            4-15kw

 Find:   Calculate Total Demand

37950 W is the answer for demand load.

Math:

Step 1.  Find average value of one range using Note 2 of Table 220.5:

(4 x 12 + 4 x 13.8 + 4 x 15) / 12 = (48+ 55.2 + 60 ) / 12 = 163.2 / 12 = 13.6

Note: 12 kw is used for 9 kw per Table 220.55 note 2.

Step 2.  Next follow 220.55 second paragraph and example D5A in Annex D.

Maximum ranges connected between any two phases is 4. 

2 x 4 = 8

By table 220.55 column C demand is 23000 for 8 ranges of 12 kw.  This must be increased 10 per cent since the average value of 13.6 is 1.6 kw above 12 kw.

Demand is 23000 + (0.10 x 23000) = 25300 watts.

Step 3.  By example Annex D example D5A:

The demand load for each two phases is 25300 / 2 = 12650

Ans:  The total demand load is 3 x 12650 = 37950 W

The line current is:

I = 37950 / (1.732 x 208) = 37950 / 360.3 = 105.3 amperes

References:

 Annex D example D5A

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 (for 8 each 12 kw ranges) = 23,000 VA

Per phase demand = 23,000 VA ÷ 2 = 11,500 VA

Equivalent 3-phase load  (3 x 11500) = 34,500 VA

Table 220.55 Note 2

 “Over 8 3⁄4 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 8 3⁄4 kW and of different ratings, but none exceeding 27 kW, an   AVERAGE value of rating shall be calculated by adding together the ratings of all ranges to obtain the total connected load (using 12 kW for any range rated less than 12 kW) and dividing by the total number of ranges. Then the maximum demand in Column C shall be increased 5 percent for each kilowatt or major fraction thereof by which this average value exceeds 12 kW.”

Section 220.55 second paragraph:

“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.”

Example 14

Commercial - 3 phase range calculation supplied by 3-phase 4 wire 208/120 volt wye.      

Given:

            4-9kw

            4-13.8kw

            4-15kw

 

 Find total load and apply demand factors given in Table 220.56

(4 x 9) + (4 x 13.8) + (4 x 15)  = (36+ 55.2 + 60 ) = 151.2 KW

Demand factor for 12 ranges from Table 220.56 is 65 per cent.

0.65 x 151.2 KW = 98.28 KW

Ans:  The total demand load is 98.28 KW

The line current is:

I = 98280 Watts / (1.732 x 208 Volts) = 98280 Watts / 360.3 = 272.8 amperes