Commercial Electrical Load CalculationsKnowing how to correctly size loads in commercial applications is an essential skill for electriciansOct 1, 2004by Mike HoltNEC Trainer / Consultant, Mike Holt Enterprises

873 Articles

Even if you work with stamped drawings, you'll eventually need to do commercial load calculations in the field or on a

licensing exam. The NEC covers commercial calculations in Art. 220, but other articles also apply. For example, you must

know the definitions in Art. 100, be familiar with what Art. 210 says about continuous loads, and understand the

overcurrent protection requirements set forth in Art. 240.

Two items associated with this type of calculation repeatedly need clarification:

Voltage

The voltage to use for your calculations depends on the system design voltage. Thus when you calculate branch-circuit,

feeder, and service loads, you must use a nominal system voltage of 120V, 120/240V, 208Y/120V, 240V, 347V, 480Y/277V,

480V, 600Y/347V, or 600V unless otherwise specified (220.2) (Fig. 1 below).

Rounding

Refer to 200.2(B) to end the rounding mystery. When the ampere calculation exceeds a whole number by 0.5 or more,

round up to the next whole number. If the extra is 0.49 or less, round down to the next whole number. For, example,

round 29.5A up to 30A, but round 29.45A down to 29A.

Specific loads. Art. 220 doesn't cover all specific loads. For example, you'll find motors in Art. 430 and air conditioners in

Art. 440. To know if you should look in another Article, use the NEC index.

<b>Fig. 1.</b> Don’t make the mistake of using actual field measurements of system voltage in your calculations. Unless specified otherwise, loads shall be computed using the nominal system voltage such as 120V, 120/240V, 208Y/120V, 240V, 347V, 480Y/277V, 480V, 600Y/347V or 600V.

Art. 220 has specific requirements for most loads, including the following:

Dryers. Size the branch-circuit conductors and overcurrent protection device for commercial dryers to the appliance

nameplate rating. Calculate the feeder demand load for dryers at 100% of the appliance rating. If the dryers run

continuously, you must size the conductor and protection device at 125% of the load [210.19(A), 215.3, and 230.42]. Table

220.18 demand factors don't apply to commercial dryers.

Let's apply what we've just learned. What size branch-circuit conductor and overcurrent protection does the NEC require

for a 7kW dryer rated 240V when the dryer is in a multi-family dwelling laundry room (Fig. 2)?

I=P÷E

7,000W÷240V=29A

The ampacity of the conductor and overcurrent device must be at least 29A (240.4). Per Table 310.16, a 10 AWG

conductor at 60°C is rated 30A. Therefore, you must use a 30A breaker with a 10 AWG conductor.

<b>Fig. 2.</b> When determining proper branch-circuit protection and conductor size for a commercial clothes dryer, you must use a demand load of 100%. The reduced demand factors for multiple dryers (Table 220.18) don’t apply in a commercial setting.

Electric heat [424.3(B)]. Size branch-circuit conductors and the overcurrent protection device for electric heating to not

less than 125% of the total heating load, including blower motors. Calculate the feeder/service demand load for electric

heating equipment at 100% of the total heating load.

Kitchen equipment. Size branch-circuit conductors and overcurrent protection for commercial kitchen equipment per the

appliance nameplate rating.

To determine the service demand load for commercial kitchen equipment that has thermostatic control or intermittent

use, apply the demand factors from Table 220.20 to the total connected kitchen equipment load. The feeder or service

demand load can't be less than the sum of the two largest appliance loads. The demand factors of Table 220.20 don't

apply to space-heating, ventilating, or air-conditioning equipment.

Laundry equipment. Size these circuits to the appliance nameplate rating. You can assume a laundry circuit isn't a

continuous load and that commercial laundry circuits are rated 1,500VA — unless noted otherwise in the project drawings

or exam question.

Lighting. The NEC requires a minimum load per square foot for general lighting, depending on the type of occupancy

[Table 220.3(A)]. For the guestrooms of hotels, motels, hospitals, and storage warehouses, you can apply the general

lighting demand factors of Table 220.11 to the general lighting load.

Assume the general lighting load for commercial occupancies other than guestrooms of motels, hotels, hospitals, and

storage warehouses is continuous. Calculate it at 125% of the general lighting load listed in Table 220.3(A).

Receptacles. You don't do all receptacle load calculations the same way. The NEC has separate requirements, depending

on the application.

Multi-outlet receptacle assembly. For service calculations, consider every 5 feet (or less) of multi-outlet receptacle

assembly to be 180VA. When you can reasonably expect a multi-outlet receptacle assembly to power several appliances

simultaneously, consider each foot (or less) as 180VA for service calculations. Normally, a multi-outlet receptacle

assembly isn't a continuous load [220.3(B)(8)].

Receptacle VA load. The minimum load for each commercial or industrial general-use receptacle outlet is 180VA per strap

[220.3(B)(9)]. Normally, receptacles aren't continuous loads.

Number of receptacles permitted on a circuit. The maximum number of receptacle outlets permitted on a commercial or

industrial circuit depends on the circuit ampacity. To calculate that number, divide the VA rating of the circuit by 180VA

for each receptacle strap.

<b>Fig. 3.</b> The minimum load for each commercial general-use receptacle outlet is 180VA per strap. In this example, the 15A, 120V breaker could accommodate 1,800VA of load (120V x 15A = 1,800VA). Therefore, you could install a total of 10 receptacles on this circuit.

Let's work a sample problem. How many receptacle outlets are permitted on a 15A, 120V circuit (Fig. 3)?

Total circuit VA load for a 15A circuit:

120V×15A=1,800VA

Number of receptacles per circuit:

1,800VA÷180VA=10 receptacles

Receptacle sizing. The NEC permits 15A circuits in commercial and industrial occupancies, but some local codes require

a minimum 20A rating (310.5).

Receptacle service demand load. In other than dwelling units, you can add — to the lighting loads — receptacle loads

computed at not more than 180VA per outlet per 220.3(B)(9). You can also add fixed multi-outlet assemblies computed

per 220.3(B)(8). Both of these must adhere to the demand factors given in Table 220.11 or in Table 220.13.

Bank and office general lighting and receptacles. Calculate the receptacle demand load at 180VA for each receptacle

strap [220.3(B)(9)] if the number of receptacles is known, or 1VA for each square foot if the number of receptacles is

unknown [Table 220.3(A) Note b].

Signs. The NEC requires each commercial occupancy that's accessible to pedestrians to have at least one 20A branch

circuit for a sign [600.5(A)]. The load for the required exterior signs or outline lighting must be at least 1,200VA [220.3(B)

(6)]. A sign outlet is a continuous load. You must size the feeder load at 125% of the continuous load [215.2(A)(1) and

230.42].

The following question will allow you to practice what we've just covered. What's the demand load for one electric sign?

1,200VA×1.25=1,500VA

Neutral calculations. The neutral load is the maximum unbalanced demand load between the grounded (neutral)

conductor and any one ungrounded (hot) conductor — as determined by the calculations in Art. 220, Part B. This means

you don't consider line-to-line loads when sizing the grounded (neutral) conductor. What about load reduction? That

depends on certain factors, which we'll look at next.

Reduction over 200A. You can reduce the feeder/service net computed load for 3-wire, single-phase or 4-wire, 3-phase

systems that supply linear loads for that portion of the unbalanced load over 200A, by a multiplier of 70%.

To see how this would work for an actual installation, determine the neutral demand load for a balanced 400A, 3-wire,

120/240V feeder.

Total neutral load for 400A service:

First 200A at 100%: 200A×1.00=200A

Remainder at 70%: 200A×0.70=140A

Total demand load: 200A×140A=340A

Reduction not permitted. You can't reduce the neutral demand load for 3-wire, single-phase, 208Y/120V or 480Y/277V

circuits that consist of two line wires and the common conductor (neutral) of a 4-wire, 3-phase wye system. This is

because the common (neutral) conductor of a 3-wire circuit connected to a 4-wire, 3-phase wye system carries about the

same current as the phase conductors [310.15(B)(4)(b)].

<b>Fig. 4.</b> Sizing the grounded (neutral) conductor can be tricky. Just remember that you can’t reduce the neutral demand load for 3-wire, single-phase, 208Y/120V or 480Y/277V circuits that consist of two line wires and the common conductor (neutral) of a 4-wire, 3-phase system.

As proof of this theory, see the example in Fig. 4.

In addition, you can't reduce the neutral demand load for nonlinear loads supplied from a 3-phase, 4-wire, wye-connected

system, because they produce triplen harmonic currents that add on the neutral conductor. This situation can require the

neutral conductor to be larger than the ungrounded conductor load (220.22 FPN 2).

Knowing the correct way to do commercial load calculations makes you more valuable because you can play a key role in

the field design, inspection, and implementation process. It's one more skill that helps you do the job right the first time.

Kitchen Equipment Load CalculationsSep 9, 2014

by Mike HoltNEC Trainer / Consultant, Mike Holt Enterprises

873 Articles

When applying the demand factors of Table 220.56, the feeder or service demand load shall not be less than the sum of

_____.

a) the total number of receptacles at 180VA per receptacle outlet

b) the VA rating of all of the small-appliance branch circuits combined

c) the largest two kitchen equipment loads

d) the kitchen heating and air-conditioning loads

The correct answer is: c) the largest two kitchen equipment loads

Part III of Art. 220 focuses on feeder and service load calculation requirements. Section 220.56 addresses kitchen equipment installed in

locations other than dwelling units. As specifically stated in the last sentence of this section of the Code, "in no case shall the feeder or

service calculated load be less than the sum of the largest two kitchen equipment loads."

Cooking Equipment Calculations, Part II

Published: February 2004By Charles R. MillerSpecifications for calculating branch-circuit, feeder and service loads are in Article 220 of the National Electrical Code. This article is divided into four parts: I. General, II. Feeders and Services, III. Optional Calculations for Computing Feeder and Service Loads and IV. Method for Computing Farm Loads. Part I, though titled “General,” mostly contains specifications for computing branch circuits. After completing calculations in accordance with Part I, the results can either be used to size branch circuits, or can be added to other demand factors in Article 220 to size feeders or services. The computed load of a feeder or service must not be less than the load on the branch circuits supplied, as determined by Part I of Article 220, after any applicable demand factors permitted by Parts II, III or IV have been applied [220.10].

Part II includes a variety of demand factors for various types of occupancies. While general lighting demand factors are applicable in all occupancies, other demand factors are only applicable in certain occupancies. Some of these requirements only pertain to dwelling units and some only pertain to nondwelling units. Table 220.19 applies to cooking appliances in dwelling units, and also applies to household cooking appliances (rated over 1,750W) that are used in instructional programs. Last month’s In Focus began a new series covering cooking equipment calculations. This month, the discussion continues with a detailed explanation of cooking equipment demand factors.

If the rating of the range is higher than 8 3/4kW, but not higher than 12kW, the demand load can only be found in Column C. The numbers in Column C represent kilowatt ratings. For example, the demand load for one 12kW range is only 8kW. If the nameplate rating of the cooking appliance is more than 13/4kW to 83/4kW, the demand loads can be found in Columns A or B. The numbers in these columns are percentages and must be multiplied by the kilowatt rating to find the demand load. Column C is not just for equipment rated between 83/4kW and 12kW. Column C is for equipment not over 12kW. If the

kilowatt rating is within the parameters of Column A, the demand load can be calculated from either Column A or C. If the kilowatt rating is within the parameters of Column B, the demand load can be calculated from either Column B or C. Permission to use either column is stated in the first sentence of the third note under Table 220.19. In previous editions of the Code the columns were arranged in a different order. The column for equipment not over 12kW rating is now located on the right side of the table.

If the equipment is rated from 31/2 to 83/4kW, calculate the demand load from Column B, and then compare that demand with the demand found in Column C. It is permissible to use the lowest demand load. For example, what is the service demand load for one 8kW range? First, calculate the demand load from Column B by multiplying 8kW by 80 percent (8 x .80 = 6.4kW). Next, find the demand load in Column C, compare the two, and select the lowest. The demand load for one unit in Column C is 8kW. The lowest demand load permitted for one 8kW range is 6.4kW (See Figure 1). Do not drop numbers to the right of the decimal point because each tenth of a kilowatt is equal to 100W. For example, .4kW is 400W.

Do not make the mistake of thinking that the lowest demand can always be found in one particular column; it depends on the number of units as well as the kilowatt ratings. For example, what is the lowest demand load for 12 7kW ranges? Since there is more than one appliance, find the total load before multiplying by the demand factor percent in Column B. If all appliances are the same rating, multiply the number of units by the kilowatt rating of each (12 x 7 = 84kW). Next, find the demand factor percent in Column B for 12 appliances (12 appliances = 32 percent). Now compute the demand load by multiplying the total load by the demand factor percent (84 x .32 = 26.88kW). Finally, find the demand load in Column C, compare the two and select the lowest. The demand in Column C for six appliances is 27kW. The lowest demand load for 12 7kW ranges is 26.88kW or 26,880W (See Figure 2).

By adding one more 7kW range to the last example, the column with the lowest demand load changes. For example, find the lowest demand load for 13 7kW ranges. Compute the demand load in Column B by multiplying the total load by 32 percent (13 x 7 x .32 = 29.12kW). Now, find the demand load in Column C, compare the two and select the lowest. The demand in Column C for 13 appliances is 28kW. The lowest demand load for 13 7kW ranges is 28kW or 28,000W (See Figure 3).

Where the ratings of the cooking appliances fall under both Columns A and B, apply the demand factors for each column separately and add the results together. This specification is stated in the last sentence of the third note under Table 220.19. For example, what is the service demand load for 10 3kW ovens and 10 5kW counter-mounted cooking appliances (cooktops)? The demand for the ovens, under Column A, is 14.7kW (10 x 3 x .49 = 14.7). The demand load for the cooktops, under Column B, is 17kW (10 x 5 x .34 = 17). Add the results together to find the total demand load. The demand for these household cooking appliances is 31.7kW or 31,700W (See Figure 4).

When applying the demand factors in Columns A and B, remember to compare the total demand to the demand found in Column C and select the lowest. For example, what is the lowest demand load for 20 3kW ovens and 20 6.5kW cooktops? The demand for the ovens, under Column A, is 21kW (20 x 3 x .35 = 21). The demand load for the cooktops, under Column B, is 36.4kW (20 x 6.5 x .28 = 36.4). The total demand from Columns A and B combined is 57.4kW (21 + 36.4 = 57.4). Since these appliances are rated less than 12kW each, find the demand load from Column C to see if it is lower than Columns A and B. The total number of appliances is 40 (20 ovens and 20 cooktops). Where the number of appliances is from 26 to 40, add 15 to the number of appliances and the sum is the kilowatt demand load. Fifteen

added to 40 appliances is 55. Now compare the two columns and select the lowest demand. The combination of Columns A and B is 57.4kW, but the demand load from Column C is lower. The demand load for 20 3kW ovens and 20 6.5kW cooktops is 55kW or 55,000W (See Figure 5).

Next month’s In Focus, resuming with 220.19, will continue discussion of cooking equipment calculations. EC

MILLER, owner of Lighthouse Educational Services, teaches classes and seminars on the electrical industry. He is the author of “Illustrated Guide to the National Electrical Code” and NFPA’s “Electrical Reference.” He can be reached at 615.333-3336, charles@charlesRmiller.com

Article 220-Branch Circuit, Feeder and Service Calculations

Published: March 2004By Charles R. MillerCooking Equipment Calculations, Part III

220.19 Electric Ranges and Other Cooking Appliances

Article 220 of the National Electrical Code contains specifications for calculating branch-circuit, feeder and service loads. Part I, titled General, contains general requirements and specifications for computing branch-circuit loads. The branch-circuit loads are combined with other applicable demand factors in Parts II, III or IV to determine feeder and/or service loads. Some requirements in Part II are only applicable in nondwelling occupancies.

For example, receptacle loads in other than dwelling units must be calculated in accordance with 220.13. Some requirements only apply to dwellings units. For example, requirements in 220.16, 17 and 18 are applicable in dwelling units only. Demand loads in 220.19 apply to dwelling units unless meeting the stipulations in the fifth note to Table 220.19. Section 220.19 and Table 220.19 pertain to household electric ranges and other cooking appliances individually rated in excess of 1 3/4kW. Last month’s In Focus continued discussing provisions for calculating household cooking equipment. This month, the discussion continues with a detailed explanation of cooking equipment demand factors.

Up to this point, each household cooking appliance has been rated 12kW or less. Ranges over 12kW through 27kW must be calculated in accordance with either the first or second note under Table 220.19. When all the ranges are rated the same, use the first note. When they are not rated the same, use the second note. The demand loads in Column C are only applicable if the appliances are rated 12kW or less. For ranges individually rated more than 12kW, but not more than 27kW, the maximum demand in Column C must be increased 5 percent for each additional kilowatt of rating (or major fraction thereof) by which the rating of individual ranges exceeds 12kW (Note 1 under Table 220.19).

Note 1 applies if each range has the same kilowatt rating. The demand loads in Column C, whether one or more ranges, must be increased when the individual ranges are rated more than 12kW. The amount of increase is only 5 percent for each additional kilowatt over 12kW. After finding the demand load in Column C, increase that demand by the appropriate percentage. For example, what is the service demand load for one 14kW range? First, find the percentage by which Column C must be increased. A 14kW range exceeds 12kW by 2kW (14 - 12 = 2). Since Column C must be increased 5 percent for each

additional kilowatt of rating above 12, the maximum demand listed in Column C for one range must be increased by 10 percent (2 x 5 = 10). The increased amount is 0.8kW (8 x .10 = 0.8). This increased amount must be added to the original demand load (8 + 0.8 = 8.8). The service demand load for one 14kW range is 8.8kW (See Figure 1).

Regardless of the number, the first note applies if all the ranges are rated the same and are more than 12kW. For example, what is the demand load for 15 15kW ranges? A 15kW range exceeds 12kW by 3kW (15 - 12 = 3). The demand in Column C must be increased by 15 percent (3 x 5 = 15). Note: Do not increase the number of ranges by 15 percent. The increase applies to the maximum demand in Column C. The demand load for 15 ranges is 30kW. The increased amount is 4.5kW (30 x .15 = 4.5). Add the increased amount to the original demand (30 + 4.5 = 34.5). The demand load for fifteen 15kW ranges is 34.5kW (See Figure 2).

When applying note 1, the range ratings (before applying Column C) cannot include a fraction of a kilowatt. The fraction must either be dropped, or rounded up to the next whole kilowatt rating. When the fraction is less than .5, drop the fraction. For example, what is the service demand load for one 13.4kW range? Since the .4 is not a major fraction, drop the .4 and find the demand for one 13kW range. A 13kW range exceeds 12kW by 1kW (13 - 12 = 1). The demand in Column C must be increased by 5 percent (1 x 5 = 5). The increased amount is 0.4kW (8 x .05 = 0.4). Add the increased amount to the original demand (8 + 0.4 = 8.4). The service demand load for one 13.4kW range is 8.4kW (See Figure 3).

Round the kilowatt rating up to the next whole number when the kilowatt rating contains a fraction of .5 or more. For example, what is the service demand load for one 15.5kW range? Since the .5 is a major fraction, round the 15.5 up to a 16kW range and find the demand load. A 16kW range exceeds 12kW by 4kW (16 - 12 = 4). The demand in Column C must be increased by 20 percent (4 x .05 = .20). The increased amount is 1.6kW (8 x .20 = 1.6). Add the increased amount to the original demand (8 + 1.6 = 9.6). The service demand load for one 15.5kW range is 9.6kW (See Figure 4).

Apply this same principal to multiple ranges all of the same rating. For example, what is the service demand load for 40 13.6kW ranges? Since the .6 is a major fraction, round the 13.6 up to 14kW and find the demand. A 14kW range exceeds 12kW by 2kW (14 - 12 = 2). Since Column C must be increased 5 percent for each additional kilowatt of rating above 12, the maximum demand in Column C for 40 ranges must be increased by 10 percent (2 x .05 = .10). The total number of ranges is 40. Where the number of appliances is from 26 to 40, add 15 to the number of appliances and the sum is the kilowatt demand load. Fifteen added to 40 appliances is 55 (15 + 40 = 55). If the ranges were 12kW each, the demand load would be 55kW. Since the ranges are rated more than 12kW, the demand must be increased. The increased amount is 5.5kW (55 x .10 = 5.5). This increased amount must be added to the Column C demand load (55 + 5.5 = 60.5). The service demand load for 40 13.6kW ranges is 60.5kW (See Figure 5).

The first note under Table 220.19 applies to ranges over 12kW through 27kW, but this could be a little misleading. As previously mentioned, when the range rating contains a fraction of a kilowatt, it must either be dropped or rounded up. Fractions lower than 12.5kW are dropped, and therefore the first note does not apply. For example, what is the service demand load for 20 12.4kW ranges? Since the .4 is not a major fraction, drop the .4 and find the demand load. The demand in Column C for 20 12.4kW ranges is 35kW. Therefore, the first note under Table 220.19 actually applies to ranges 12.5kW through 27kW.

Next month’s In Focus continues the discussion of household cooking equipment computations in 220.19. EC

MILLER, owner of Lighthouse Educational Services, teaches classes and seminars on the electrical industry. He is the author of “Illustrated Guide to the National Electrical Code” and NFPA’s “Electrical Reference.” He can be reached at 615.333-3336, charles@charlesRmiller.com or www.charlesRmiller.com.

Article 220-Cooking Equipment Calculations, Part VI

Published: June 2004By Charles R. Miller220.19 Electric Ranges and Other Cooking Appliances

Specifications for calculating branch-circuit, feeder and service loads are in Article 220 of the National Electrical Code. This article is divided into four parts: I. General, II. Feeders and Services, III. Optional Calculations for Computing Feeder and Service Loads, and IV. Method for Computing Farm Loads. Part 1 contains general provisions and a number of branch-circuit calculations. After executing the calculation procedures in accordance with Part I, the results can either be used to size branch circuits, or can be added to other applicable demand factors permitted by Parts II, III or IV to size feeders or services. Part II contains specifications for calculating feeders and services. Optional calculations for computing feeder and service loads are in Part III. Part IV contains methods for computing farm loads.

While most of the procedures in Part II are used along with calculations from Part I, some feeder (or service) calculations will not require any calculations from Part I. For example, a panelboard is needed to supply power to 15 household ranges that will be used in culinary arts classes. The only load on this panel will be household electric ranges, and therefore the only calculation procedure needed is from 220.19 and Table 220.19. Last month’s Code In Focus continued a discussion of calculation methods for household electric ranges and other cooking appliances. This month, the discussion concludes the series on cooking equipment calculations.

While most of the provisions in 220.19 pertain to finding loads on feeders or services, the fourth note under Table 220.19 specifies how to calculate branch-circuit loads. As discussed last month, it is permissible to calculate the branch-circuit load for one range in accordance with Table 220.19. The second sentence in the fourth note states that the branch-circuit load for one wall-mounted oven or one counter-mounted cooking unit (cooktop) shall be the nameplate rating of the appliance. While the branch-circuit load for one range can be calculated at less than nameplate rating, the branch-circuit load for one oven or one cooktop cannot. For example, what size 75 C, copper conductors are required to feed an 8.6kW, 240V cooktop? Since the branch-circuit load must not be less than the nameplate rating, the branch-circuit load for one 8.6kW cooktop is 8.6kW or 8,600W. Although the load is as much as some ranges, because it is a cooktop, the branch-circuit load must be the nameplate rating. To determine the minimum conductor size, the branch-circuit ampacity must be known (Watts/volts = Amperes). The conductor ampacity must be rated for at least 36A (8,600/240 = 35.8 = 36). The minimum size 75 C, copper conductors required for an 8.6kW, 240V cooktop is 8 AWG (See Figure 1).

The branch-circuit load for one wall oven must be the nameplate rating of the appliance. Although not specifically mentioned, the branch-circuit load for a double wall oven is also the nameplate rating of the appliance. For example, what size 60 C, copper conductors are required to feed a 7.4kW, 240V double

wall oven? The branch-circuit load for this wall-mounted oven is 7.4kW or 7,400W. The conductor ampacity must be rated for at least 31A (7,400/240 = 30.8 = 31). The minimum size 60 C, copper conductors required for a 7.4kW, 240V double wall oven is 8 AWG (See Figure 2). Note, this example is asking for 60 C copper conductors, not 75 C.

The third sentence in the fourth note under table 220.19 contains the last branch-circuit load calculation. The branch-circuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens can be calculated by adding the nameplate rating of the individual appliances and treating the total as equivalent to one range. In this type of installation, the cooktop and oven(s) must be located in the same room and supplied from a single branch circuit. For example, what is the branch-circuit load for one 7.4kW, 240V cooktop and one 3.6kW, 240V wall oven? The household cooking appliances are supplied from a single branch circuit and located in the same room. The first step is to add the two appliances. Calculate the demand load as if calculating for one range at 11kW (7.4 + 3.6). By applying Table 220.19, an 11kW range has a maximum demand (in Column C) of 8kW or 8,000W (See Figure 3).

Where two wall-mounted ovens and one cooktop are supplied from a single branch circuit and located in the same room, the three appliances can be added together and treated as one range. For example, what is the branch-circuit load for one 7.4kW, 240V cooktop and two 3.6kW, 240V wall ovens? The household cooking appliances are supplied from a single branch circuit and located in the same room. The branch-circuit conductors will terminate in a junction box inside the cabinet under the cooktop. The first step is to add the three appliances. Calculate the branch-circuit demand load as if calculating for one range at 14.6kW (7.4 + 3.6 + 3.6). Because it is over 12kW, apply the first note under Table 220.19. The maximum demand in Column C must be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating exceeds 12kW. Since .6 is a major fraction, round the 14.6 up to 15kW. A 15kW range exceeds 12kW by 3kW (15 - 12 = 3). Since Column C must be increased 5 percent for each additional kilowatt of rating above 12, the maximum demand in Column C must be increased by 15 percent (3 x 5% = 15%). The increased amount is 1.2kW (8kW x 15% = 1.2kW). This increased amount must be added to the Column C demand load (8 + 1.2 = 9.2kW). The branch-circuit load for one 7.4kW, 240V cooktop and two 3.6kW, 240V wall ovens is 9.2kW or 9,200W.(See Figure 4). Derating the nameplate the load (in accordance with Table 220.19) is permitted only with a cooktop and one wall-mounted oven, or a cooktop and two ovens. Derating is not permitted when one branch circuit supplies two ovens and no cooktop.

Up to now, the demand load calculations in 220.19 have applied to household electric cooking equipment in dwelling units. Unless meeting the requirement in the last note under Table 220.19, load calculations for kitchen equipment in other than dwelling units must comply with 220.20 and Table 220.20. As stated in the fifth note, the demand loads in Table 220.19 also apply to household cooking appliances rated over 13/4kW and used in instructional programs. Household electric cooking equipment installed in other than dwelling units and not used in instructional programs must be calculated in accordance with 220.20.

Provided the equipment is household cooking equipment and used in instructional programs, the type of occupancy is not limited to dwelling units. For example, an existing high school classroom is being converted into a culinary arts classroom. Fifteen 12kW, 240V household electric ranges will be installed. A separate panelboard will be installed to supply the ranges. What is the load on the feeder for these ranges? Although the ranges are not in a dwelling, because the household cooking appliances will be

used in instructional programs, the load can be calculated from Table 220.19. The maximum demand in Column C for 15 12kW ranges is 30kW (See Figure 5). EC

MILLER, owner of Lighthouse Educational Services, teaches classes and seminars on the electrical industry. He is the author of “Illustrated Guide to the National Electrical Code” and NFPA’s “Electrical Reference.” He can be reached at 615.333-3336, charles@charlesRmiller.com or www.charlesRmiller.com.

Article 220-Branch Circuit, Feeder and Service Calculations, Part V

Published: May 2004By Charles R. Miller220.19 Electric Ranges and Other Cooking Appliances

Branch-circuit, feeder and service-load calculation provisions are in Article 220 of the National Electrical Code. While Part I is titled “General,” most of the first part pertains to branch-circuit calculations. Part II contains provisions and demand factors for calculating feeder and service loads. The provisions in Part I can be employed to size branch circuits, or can be added to other demand loads in Parts II, III or IV to determine feeders and services. The computed load of a feeder or service shall not be less than the total loads on the branch circuits supplied, as determined by Part I of Article 220, after any applicable demand factors permitted by Parts II, III or IV have been applied [220.10].

Computing feeder and service load calculations can sometimes be overwhelming. Breaking down load calculations into a series of steps can make a difficult task much easier. Different types of occupancies will require different steps. For example, a one-family dwelling will not have all of the same steps as a commercial building. One step included in most dwelling calculations, and some non-dwelling calculations is cooking equipment. Demand loads for electric ranges and other cooking appliances are in 220.19 and Table 220.19. Last month’s In Focus concluded with a discussion of the second note under Table 220.19. This month, the discussion begins with the third note.

In lieu of the method provided in Column C, it shall be permissible to add the nameplate rating of all household cooking equipment rated more than 1 3/4kW but not more than 8 3/4kW and multiply the sum by the demand factors specified in Column A or B for the given number of appliances, as per Note 3 under Table 220.19. If the cooking equipment is more than 1 3/4kW but less than 3 1/2kW, the demand can come from either Column A or Column C. If the equipment is 3 1/2kW to 8 3/4kW, the demand can come from either Column B or C. For example, what is the service demand load for two 4kW ovens and two 5kW cooktops? The four units have a combined rating of 18kW. The demand factor percent for four appliances in Column B is 50 percent. After applying the demand factor percent from Column B, the demand load is 9kW (18 ¥ .50 = 9). The demand load in Column C for four appliances is 17kW. Since it is permissible to use either column, select the lowest demand. The service demand for two 4kW ovens and two 5kW cooktops is 9kW (See Figure 1).

When calculating demand loads for household cooking equipment, if the kilowatt ratings fall in Columns A or B, compare them to Column C and select the lowest demand. Start by calculating the demand load from Column A or B. Then, compare that demand with the one from Column C. Finally, select the lowest demand load. For example, what is the service demand for six 8kW ranges? The demand load from Column B is 20.64kW (6 ¥ 8 ¥ .43 = 20.64). The demand from Column C for six appliances is 21kW. The lowest demand for six 8kW ranges comes from Column B (See Figure 2).

The demand loads from Columns A and B will not always be lower than the ones found in Column C. By adding one more 8kW range to the previous example, the column having the lowest demand changes. What is the service demand load for seven 8kW ranges? The demand load from Column B is 22.4kW (7 ¥ 8 ¥ .40 = 22.4). The demand from Column C for seven appliances is 22kW. While the lowest demand for six 8kW ranges comes from column B, the lowest demand for seven 8kW ranges comes from Column C (See Figure 3).

The second half of the third note provides instructions if the ratings of cooking appliances fall under both Column A and Column B. First, calculate the demand loads for the appliances that fall under Column A. Next, calculate the demand for appliances that fall under Column B. Finally, add the results together. For example, what is the service demand load for 30 3kW ovens and 15 5kW cooktops? The ovens are calculated from demand factors in Column A. The demand load from Column A is 27kW (30 ¥ 3 ¥ .30 = 27). The cooktops are calculated from demand factors in Column B. The demand load from Column B is 24kW (15 ¥ 5 ¥ .32 = 24). Add the results together to find the demand load. The service demand load for 30 3kW ovens and 15 5kW cooktops is 51kW (27 + 24 = 51). To make sure this is the lowest demand, find the kilowatt demand from Column C. Since all the appliances are rated less than 12kW, find the demand for all 45 appliances. Multiply 45 appliances by .75 and add 25. The demand in Column C for 45 appliances is 58.75kW (45 ¥ .75 = 33.75; 33.75 + 25 = 58.75). In this example, the lowest demand load is from Columns A and B (See Figure 4).

Up to this point, the purpose of all cooking equipment calculations was to size feeders and services. Nothing has been mentioned about sizing branch-circuits that feed cooking equipment. The fourth note under Table 220.19 specifies how to calculate branch-circuit loads. This note discusses several types of appliance installations. The first sentence stipulates how to calculate the branch-circuit load for one range. It shall be permissible to compute the branch-circuit load for one range in accordance with Table 220.19. For example, what size 75 C copper conductors are required to feed a 14kW, 240V range? First, calculate the branch-circuit demand load. Since this range exceeds 12kW, the demand in Column C must be increased 5 percent for each additional kilowatt of rating by which the rating exceeds 12kW. A 14kW range exceeds 12kW by 2kW (14 - 12 = 2). The maximum demand listed in Column C for one range must be increased by 10 percent (2 ¥ .5 = .10). The increased amount is .8kW (8 ¥ .10 = .8). This increased amount must be added to the original demand load (8 + .8 = 8.8). The branch-circuit demand for one 14kW range is 8.8kW or 8,800W. To determine the minimum conductor size, the branch-circuit ampacity must be known (watts/volts = amperes). The conductor ampacity must be rated at least 37A (8,800/240 = 36.6 = 37). The minimum size 75 C, copper conductors required for a 14kW, 240V range is 8 AWG (See Figure 5).

What size 75 C, copper conductors are required to feed an 8kW, 230V range? Calculate the demand load from Column B by multiplying 8kW by 80 percent (8 ¥ .80 = 6.4kW). Use the demand load of 6.4kW (or 6,400W), because it is lower than the Column C demand of 8kW. The conductor ampacity must be rated for at least 28A (6,400/230 = 27.8 = 28). The minimum size 75 C, copper conductors required for an 8kW, 230V range is 10 AWG (See Figure 6).

Next month’s In Focus continues the discussion of calculating branch-circuit loads for household cooking equipment. EC

MILLER, owner of Lighthouse Educational Services, teaches classes and seminars on the electrical industry. He is the author of “Illustrated Guide to the National Electrical Code” and NFPA’s “Electrical

Reference.” He can be reached at 615.333-3336, charles@charlesRmiller.com or www.charlesRmiller.com.