Adopting Agency  BSC  BSCCG  SFM  HCD  DSA  OSHPD  DPH  

1  2  AC  SS  SS/CC  1  1R  2  3  4  5  
Adopt Entire Article  X  X  X  X  X  X  
Adopt entire Article as amended (amended sections listed below)  
Adopt only those sections that are listed below  X  
Article/Section  
Annex A  X  
Annex B  X 
This informative annex is not a part of the requirements of this NFPA document but is included for informational purposes only.
Selection of Conductors. In the following examples, the results are generally expressed in amperes (A). To select conductor sizes, refer to the 0 through 2000 volt (V) ampacity tables of Article 310 and the rules of 310.15 that pertain to these tables.
Voltage. For uniform application of Articles 210, 215, and 220, a nominal voltage of 120, 120/240, 240, and 208Y/120 V is used in calculating the ampere load on the conductor.
Fractions of an Ampere. Except where the calculations result in a major fraction of an ampere (0.5 or larger), such fractions are permitted to be dropped.
Power Factor. Calculations in the following examples are based, for convenience, on the assumption that all loads have the same power factor (PF).
Ranges. For the calculation of the range loads in these examples, Column C of Table 220.55 has been used. For optional methods, see Columns A and B of Table 220.55. Except where the calculations result in a major fraction of a kilowatt (0.5 or larger), such fractions are permitted to be dropped.
SI Units. For metric conversions, 0.093 m^{2} = 1 ft^{2} and 0.3048 m = 1 ft.
The dwelling has a floor area of 1500 ft^{2}, exclusive of an unfinished cellar not adaptable for future use, unfinished attic, and open porches. Appliances are a 12kW range and a 5.5kW, 240V dryer. Assume range and dryer kW ratings equivalent to kVA ratings in accordance with 220.54 and 220.55.
Calculated Load (see 220.40)
General Lighting Load 1500 ft^{2} at 3 VA/ft^{2} = 4500 VA
Minimum Number of Branch Circuits Required [see 210.11(A)]
General Lighting Load: 4500 VA ÷ 120 V = 38 A
This requires three 15A, 2wire or two 20A, 2wire circuits.
SmallAppliance Load: Two 2wire, 20A circuits [see 210.11(C)(1)]
Laundry Load: One 2wire, 20A circuit [see 210.11(C)(2)]
Bathroom Branch Circuit: One 2wire, 20A circuit (no additional load calculation is required for this circuit) [see 210.11(C)(3)]
Minimum Size Feeder Required [see 220.40]
General Lighting  4,500 VA 
Small Appliance  3,000 VA 
Laundry  1,500 VA 
Total  9,000 VA 
3000 VA at 100%  3,000 VA 
9000 VA — 3000 VA = 6000 VA at 35%  2,100 VA 
Net Load  5,100 VA 
Range (see Table 220.55)  8,000 VA 
Dryer Load (see Table 220.54)  5,500 VA 
Net Calculated Load  18,600 VA 
Net Calculated Load for 120/240V, 3wire, singlephase service or feeder
18,600 VA ÷ 240 V = 78 A
Sections 230.42(B) and 230.79 require service conductors and disconnecting means rated not less than 100 amperes.
Calculation for Neutral for Feeder and Service
Lighting and SmallAppliance Load  5,100 VA 
Range: 8000 VA at 70% (see 220.61)  5,600 VA 
Dryer: 5500 VA at 70% (see 220.61)  3,850 VA 
Total  14,550 VA 
Calculated Load for Neutral
14,550 VA ÷ 240 V = 61 A
Assume same conditions as Example No. Dl(a), plus addition of one 6A, 230V, room airconditioning unit and one 12A, 115V, room airconditioning unit,* one 8A, 115V, rated waste disposer, and one 10A, 120V, rated dishwasher. See Article 430 for general motors and Article 440, Part VII, for airconditioning equipment. Motors have nameplate ratings of 115 V and 230 V for use on 120V and 240V nominal voltage systems.
*(For feeder neutral, use larger of the two appliances for unbalance.)
From Example Dl(a), feeder current is 78 A (3wire, 240 V).
Line A  Neutral  Line B  

Amperes from Example D1(a)  78  61  78 
One 230V air conditioner  6  —  6 
One 115V air conditioner and 120V dishwasher  12  12  10 
One 115V disposer  —  8  8 
25% of largest motor (see 430.24)  3  3  2 
Total amperes per conductor  99  84  104 
Therefore, the service would be rated 110 A.
(see 220.82)
The dwelling has a floor area of 1500 ft^{2}, exclusive of an unfinished cellar not adaptable for future use, unfinished attic, and open porches. It has a 12kW range, a 2.5kW water heater, a 1.2kW dishwasher, 9 kW of electric space heating installed in five rooms, a 5kW clothes dryer, and a 6A, 230V, room airconditioning unit. Assume range, water heater, dishwasher, space heating, and clothes dryer kW ratings equivalent to kVA.
Air Conditioner kVA Calculation
6 A × 230 V ÷ 1000 = 1.38 kVA
This 1.38 kVA [item 1 from 220.82(C)] is less than 40% of 9 kVA of separately controlled electric heat [item 6 from 220.82(C)], so the 1.38 kVA need not be included in the service calculation.
General Load
1500 ft^{2} at 3 VA  4,500 VA 
Two 20A appliance outlet circuits at 1500 VA each  3,000 VA 
Laundry circuit  1,500 VA 
Range (at nameplate rating)  12,000 VA 
Water heater  2,500 VA 
Dishwasher  1,200 VA 
Clothes dryer  5,000 VA 
Total  29,700 VA 
Application of Demand Factor [see 220.82(B)]
First 10 kVA of general load at 100%  10,000 VA 
Remainder of general load at 40% (19.7 kVA × 0.4) 
7,880 VA 
Total of general load  17,880 VA 
9 kVA of heat at 40% (9000 VA × 0.4) =  3,600 VA 
Total  21,480 VA 
Calculated Load for Service Size
21.48 kVA = 21,480 VA
21,480 VA ÷ 240 V = 90 A
Therefore, the minimum service rating would be 100 A in accordance with 230.42 and 230.79.
Feeder Neutral Load in Accordance with 220.61
1500 ft^{2} at 3 VA  4,500 VA 
Three 20A circuits at 1500 VA  4,500 VA 
Total  9,000 VA 
3000 VA at 100%  3,000 VA 
9000 VA — 3000 VA = 6000 VA at 35%  2,100 VA 
Subtotal  5,100 VA 
Range: 8 kVA at 70%  5,600 VA 
Clothes dryer: 5 kVA at 70%  3,500 VA 
Dishwasher  1,200 VA 
Total  15,400 VA 
Calculated Load for Neutral
15,400 VA ÷ 240 V= 64 A
The dwelling has a floor area of 1500 ft^{2}, exclusive of an unfinished cellar not adaptable for future use, unfinished attic, and open porches. It has two 20A small appliance circuits, one 20A laundry circuit, two 4kW wallmounted ovens, one 5.1kW countermounted cooking unit, a 4.5kW water heater, a 1.2kW dishwasher, a 5kW combination clothes washer and dryer, six 7A, 230V room airconditioning units, and a 1.5kW permanently installed bathroom space heater. Assume wallmounted ovens, countermounted cooking unit, water heater, dishwasher, and combination clothes washer and dryer kW ratings equivalent to kVA.
Air Conditioning kVA Calculation
Total amperes = 6 units × 7 A = 42 A
42 A × 240 V ÷ 1000 = 10.08 kVA (assume PF = 1.0)
Load Included at 100%
Air Conditioning: Included below [see item 1 in 220.82(C)]
Space Heater: Omit [see item 5 in 220.82(C)]
General Load
1500 ft^{2} at 3 VA  4,500 VA  
Two 20A smallappliance circuits at 1500 VA each  3,000 VA  
Laundry circuit  1,500 VA  
Two ovens  8,000 VA  
One cooking unit  5,100 VA  
Water heater  4,500 VA  
Dishwasher  1,200 VA  
Washer/dryer  5,000 VA  
Total general load  32,800 VA  
First 10 kVA at 100%  10,000 VA  
Remainder at 40% (22.8 kVA × 0.4 × 1000) 
9,120 VA  
Subtotal general load  19,120 VA  
Air conditioning  10,080 VA  
Total  29,200 VA 
Calculated Load for Service
29,200 VA ÷ 240 V = 122 A (service rating)
Feeder Neutral Load, in accordance with 220.61
Assume that the two 44kVA wallmounted ovens are supplied by one branch circuit, the 5.1kVA countermounted cooking unit by a separate circuit.
1500 ft^{2} at 3 VA  4,500 VA 
Three 20A circuits at 1500 VA  4,500 VA 
Subtotal  9,000 VA 
3000 VA at 100%  3,000 VA 
9000 VA — 3000 VA = 6000 VA at 35%  2,100 VA 
Subtotal  5,100 VA 
Two 4kVA ovens plus one 5.1kVA cooking unit = 13.1 kVA. Table 220.55 permits 55% demand factor or 13.1 kVA × 0.55 = 7.2 kVA feeder capacity.
Subtotal from above  5,100 VA 
Ovens and cooking unit: 7200 VA × 70% for neutral load  5,040 VA 
Clothes washer/dryer: 5 kVA × 70% for neutral load  3,500 VA 
Dishwasher  1,200 VA 
Total  14,840 VA 
Calculated Load for Neutral
14,840 VA ÷ 240 V = 62
(see 220.82)
The dwelling has a floor area of 2000 ft^{2}, exclusive of an unfinished cellar not adaptable for future use, unfinished attic, and open porches. It has a 12kW range, a 4.5kW water heater, a 1.2kW dishwasher, a 5kW clothes dryer, and a 2^{1}/_{2}ton (24A) heat pump with 15 kW of backup heat.
Heat Pump kVA Calculation
24 A × 240 V ÷ 1000 = 5.76 kVA
This 5.76 kVA is less than 15 kVA of the backup heat; therefore, the heat pump load need not be included in the service calculation [see 220.82(C)].
General Load
2000 ft^{2} at 3 VA  6,000 VA 
Two 20A appliance outlet circuits at 1500 VA each  3,000 VA 
Laundry circuit  1,500 VA 
Range (at nameplate rating)  12,000 VA 
Water heater  4,500 VA 
Dishwasher  1,200 VA 
Clothes dryer  5,000 VA 
Subtotal general load  33,200 VA 
First 10 kVA at 100%  10,000 VA 
Remainder of general load at 40% (23,200 VA× 0.4)  9,280 VA 
Total net general load  19,280 VA 
Heat Pump and Supplementary Heat*
240 V × 24 A = 5760 VA
15 kW Electric Heat:
5760 VA + (15,000 VA × 65%) = 5.76 kVA + 9.75 kVA = 15.51 kVA
*If supplementary heat is not on at same time as heat pump, heat pump kVA need not be added to total.
Totals  
Net general load  19,280 VA 
Heat pump and supplementary heat  15,510 VA 
Total  34,790 VA 
Calculated Load for Service
34.79 kVA × 1000 ÷ 240 V = 145 A
Therefore, this dwelling unit would be permitted to be served by a 150A service.
A store 50 ft by 60 ft, or 3000 ft^{2}, has 30 ft of show window. There are a total of 80 duplex receptacles. The service is 120/240 V, single phase 3wire service. Actual connected lighting load is 8500 VA.
Calculated Load (see 220.40)
Noncontinuous Loads  
Receptacle Load (see 220.44)  
80 receptacles at 180 VA  14,400 VA 
10,000 VA at 100%  10,000 VA 
14,400 VA — 10,000 VA = 4400 at 50%  2,200 VA 
Subtotal  12,200 VA 
Continuous Loads  
General Lighting*  
3000 ft^{2} at 3 VA/ft^{2}  9,000 VA 
Show Window Lighting Load  
30 ft at 200 VA/ft [see 220.14(G)]  6,000 VA 
Outside Sign Circuit [see 220.14(F)]  1,200 VA 
Subtotal  16,200 VA 
Subtotal from noncontinuous  12,200 VA 
Total noncontinuous loads + continuous loads =  28,400 VA 
*In the example, the actual connected lighting load (8500 VA) is less than the load from Table 220.12, so the minimum lighting load from Table 220.12 is used in the calculation. Had the actual lighting load been greater than the value calculated from Table 220.12, the actual connected lighting load would have been used. 
Minimum Number of Branch Circuits Required
General Lighting: Branch circuits need only be installed to supply the actual connected load [see 210.11(B)].
8500 VA × 1.25 = 10,625 VA
10,625 VA ÷ 240 V = 44 A for 3wire, 120/240 V
The lighting load would be permitted to be served by 2wire or 3wire, 15 or 20A circuits with combined capacity equal to 44 A or greater for 3wire circuits or 88 A or greater for 2wire circuits. The feeder capacity as well as the number of branchcircuit positions available for lighting circuits in the panelboard must reflect the full calculated load of 9000 VA × 1.25 = 11,250 VA.
6000 VA × 1.25 = 7500 VA
7500 VA ÷ 240 V = 31 A for 3wire, 120/240 V
The show window lighting is permitted to be served by 2wire or 3wire circuits with a capacity equal to 31 A or greater for 3wire circuits or 62 A or greater for 2wire circuits.
Receptacles required by 210.62 are assumed to be included in the receptacle load above if these receptacles do not supply the show window lighting load.
Receptacle Load: 14,400 VA ÷ 240 V = 60 A for 3wire, 120/240 V
The receptacle load would be permitted to be served by 2wire or 3wire circuits with a capacity equal to 60 A or greater for 3wire circuits or 120 A or greater for 2wire circuits.
Minimum Size Feeder (or Service) Overcurrent Protection
Subtotal noncontinuous loads  12,200 VA  
Subtotal continuous load at 125% (16,200 VA × 1.25)  20,250 VA  
Total  32,450 VA 
32,450 VA ÷ 240 V = 135 A
The next higher standard size is 150 A (see 240.6).
Minimum Size Feeders (or Service Conductors) Required
For 120/240 V, 3wire system, 32,450 VA ÷ 240 V = 135 A Service or feeder conductor is 1/0 Cu in accordance with 215.3 and Table 310.15(B)(16) (with 75°C terminations).
An industrial multibuilding facility has its service at the rear of its main building, and then provides 480Y/277volt feeders to additional buildings behind the main building in order to segregate certain processes. The facility supplies its remote buildings through a partially enclosed access corridor that extends from the main switchboard rearward along a path that provides convenient access to services within 15 m (50 ft) of each additional building supplied. Two building feeders share a common raceway for approximately 45 m (150 ft) and run in the access corridor along with process steam and control and communications cabling. The steam raises the ambient temperature around the power raceway to as much as 35°C. At a tee fitting, the individual building feeders then run to each of the two buildings involved. The feeder neutrals are not connected to the equipment grounding conductors in the remote buildings. All distribution equipment terminations are listed as being suitable for 75°C connections. Each of the two buildings has the following loads:
Lighting, 11,600 VA, comprised of electricdischarge luminaires connected at 277 V
Receptacles, 22 125volt, 20ampere receptacles on generalpurpose branch circuits, supplied by separately derived systems in each of the buildings
1 Air compressor, 460 volt, three phase, 5 hp
1 Grinder, 460 volt, three phase, 1.5 hp
3 Welders, AC transformer type (nameplate: 23 amperes, 480 volts, 60 percent duty cycle)
3 Industrial Process Dryers, 480 volt, three phase, 15 kW each (assume continuous use throughout certain shifts)
Determine the overcurrent protection and conductor size for the feeders in the common raceway, assuming the use of XHHW2 insulation (90°C):
Calculated Load {Note: For reasonable precision, voltampere calculations are carried to three significant figures only; where loads are converted to amperes, the results are rounded to the nearest ampere [see 220.5(B)]}.
Noncontinuous Loads  
Receptacle Load (see 220.44) 22 receptacles at 180 VA  3,960 VA  
Welder Load [see 630.11(A), Table 630.11(A)]  
Each welder: 480V × 23A × 0.78 = 8,610 VA  
All 3 welders [see 630.11(B)] (demand factors 100%, 100%, 85% respectively)  
8,610 VA + 8,610 VA + 7,320 VA =  24,500 VA  
Subtotal, Noncontinuous Loads  28,500 VA  
Motor Loads (see 430.24, Table 430.250)  
Air compressor: 7.6 A × 480 V × √3 =  6,310 VA  
Grinder: 3 A × 480 V × √3 =  2,490 VA  
Largest motor, additional 25%:  1,580 VA  
Subtotal, Motor Loads  10,400 VA  
By using 430.24, the motor loads and the noncontinuous loads can be combined for the remaining calculation.  
Subtotal for load calculations, Noncontinuous Loads  38,900 VA  
Continuous Loads  
General Lighting  11,600 VA  
3 Industrial Process Dryers  15 kW each  45,000 VA 
Subtotal, Continuous Loads:  56,600 VA 
Overcurrent protection (see 215.3)
The overcurrent protective device must accommodate 125% of the continuous load, plus the noncontinuous load:
Continuous load  56,600 VA 
Noncontinuous load  38,900 VA 
Subtotal, actual load [actual load in amperes] [99,000 VA ÷ (480V × √3) = 119 A]  95,500 VA 
(25% of 56,600 VA) (See 215.3)  14,200 VA 
Total VA  109,700 VA 
Conversion to amperes using three significant figures: 109,700 VA / (480V × √3) = 132 A  
Minimum size overcurrent protective device: 132 A  
Minimum standard size overcurrent protective device (see 240.6): 150 amperes 
Where the overcurrent protective device and its assembly are listed for operation at 100 percent of its rating, a 125 ampere overcurrent protective device would be permitted. However, overcurrent protective device assemblies listed for 100 percent of their rating are typically not available at the 125ampere rating. (See 215.3 Exception.)
Ungrounded Feeder Conductors
The conductors must independently meet requirements for (1) terminations, and (2) conditions of use throughout the raceway run.
Minimum size conductor at the overcurrent device termination [see 110.14(C) and 215.2(A)(1), using 75°C ampacity column in Table 310.15(B)(16)]: 1/0 AWG.
Minimum size conductors in the raceway based on actual load [see Article 100, Ampacity, and 310.15(B)(3)(a) and correction factors to Table 310.15(B)(16)]:
95,500 VA / 0.7 / 0.96 = 142,000 VA
[70% = 310.15(B)(3)(a)] & [0.96 = Correction factors to Table 310.15(B)(16)]
Conversion to amperes:
142,000 VA / (480V × √3) = 171 A
Note that the neutral conductors are counted as currentcarrying conductors [see 310.15(B)(5)(c)] in this example because the discharge lighting has substantial nonlinear content. This requires a 2/0 AWG conductor based on the 90°C column of Table 310.15(B)(16). Therefore, the worst case is given by the raceway conditions, and 2/0 AWG conductors must be used. If the utility corridor were at normal temperatures [(30°C (86°F)], and if the lighting at each building were supplied from the local separately derived system (thus requiring no neutrals in the supply feeders), the raceway result (95,500 VA / 0.8 = 119,000 VA; 119,000 VA / (480V × √3) = 143 A, or a 1 AWG conductor @ 90°C) could not be used, because the termination result (1/0 AWG) based on the 75°C column of Table 310.15(B)(16) would become the worst case, requiring the larger conductor.
In every case, the overcurrent protective device shall provide overcurrent protection for the feeder conductors in accordance with their ampacity as provided by this Code(see 240.4). A 90°C 2/0 AWG conductor has a Table 310.15(B)(16) ampacity of 195 amperes. Adjusting for the conditions of use (35°C ambient temperature, 8 currentcarrying conductors in the common raceway),
195 amperes × 0.96 × 0.7 = 131 A
The 150ampere circuit breaker protects the 2/0 AWG feeder conductors, because 240.4(B) permits the use of the next higher standard size overcurrent protective device. Note that the feeder layout precludes the application of 310.15(A)(2) Exception.
Feeder Neutral Conductor (see 220.61)
Because 210.11(B) does not apply to these buildings, the load cannot be assumed to be evenly distributed across phases. Therefore the maximum imbalance must be assumed to be the full lighting load in this case, or 11,600 VA. (11,600 VA/277V = 42 amperes.) The ability of the neutral to return fault current [see 250.32(B) Exception(2)] is not a factor in this calculation.
Because the neutral runs between the main switchboard and the building panelboard, likely terminating on a busbar at both locations, and not on overcurrent devices, the effects of continuous loading can be disregarded in evaluating its terminations [see 215.2(A)(1) Exception No. 2]. That calculation is (11,600 VA ÷ 277V) = 42 amperes, to be evaluated under the 75°C column of Table 310.15(B)(16). The minimum size of the neutral might seem to be 8 AWG, but that size would not be sufficient to be depended upon in the event of a linetoneutral short circuit [see 215.2(A)(1), second paragraph]. Therefore, since the minimum size equipment grounding conductor for a 150 ampere circuit, as covered in Table 250.122, is 6 AWG, that is the minimum neutral size required for this feeder.
A multifamily dwelling has 40 dwelling units.
Meters are in two banks of 20 each with individual feeders to each dwelling unit.
Onehalf of the dwelling units are equipped with electric ranges not exceeding 12 kW each. Assume range kW rating equivalent to kVA rating in accordance with 220.55. Other half of ranges are gas ranges.
Area of each dwelling unit is 840 ft^{2}.
Laundry facilities on premises are available to all tenants. Add no circuit to individual dwelling unit.
Calculated Load for Each Dwelling Unit (see Article 220)
General Lighting: 840 ft^{2} at 3 VA/ft^{2} = 2520 VA
Special Appliance: Electric range (see 220.55) = 8000 VA
Minimum Number of Branch Circuits Required for Each Dwelling Unit [see 210.11(A)]
General Lighting Load: 2520 VA ÷ 120 V = 21 A or two 15A,
2wire circuits; or two 20A, 2wire circuits
SmallAppliance Load: Two 2wire circuits of 12 AWG wire [see 210.11(C)(1)]
Range Circuit: 8000 VA ÷ 240 V = 33 A or a circuit of two 8 AWG conductors and one 10 AWG conductor in accordance with 210.19(A)(3)
Minimum Size Feeder Required for Each Dwelling Unit (see 215.2)
Calculated Load (see Article 220):  
General Lighting  2,520 VA 
Small Appliance (two 20ampere circuits)  3,000 VA 
Subtotal Calculated Load (without ranges)  5,520 VA 
Application of Demand Factor (see Table 220.42)
First 3000 VA at 100%  3,000 VA 
5520 VA — 3000 VA = 2520 VA at 35%  882 VA 
Net Calculated Load (without ranges)  3,882 VA 
Range Load  8,000 VA 
Net Calculated Load (with ranges)  11,882 VA 
Size of Each Feeder (see 215.2)
For 120/240V, 3wire system (without ranges)
Net calculated load of 3882 VA ÷ 240 V = 16 A For 120/240V,
3wire system (with ranges)
Net calculated load, 11,882 VA ÷ 240 V = 50 A
Feeder Neutral
Lighting and SmallAppliance Load  3,882 VA 
Range Load: 8000 VA at 70% (see 220.61)  5,600 VA 
(only for apartments with electric range)  5,600 VA 
Net Calculated Load (neutral)  9,482 VA 
Calculated Load for Neutral
9482 VA ÷ 240 V = 39.5 A
Minimum Size Feeders Required from Service Equipment to Meter Bank (For 20 Dwelling Units — 10 with Ranges)
Total Calculated Load:  
Lighting and Small Appliance  
20 units × 5520 VA  110,400 VA 
Application of Demand Factor  
First 3000 VA at 100%  3,000 VA 
110,400 VA — 3000 VA = 107,400 VA at 35%  37,590 VA 
Net Calculated Load  40,590 VA 
Range Load: 10 ranges (not over 12 kVA) (see Col. C, Table 220.55, 25 kW)  25,000 VA 
Net Calculated Load (with ranges)  65,590 VA 
Net calculated load for 120/240V, 3wire system,
65,590 VA ÷ 240 V = 273 A
Feeder Neutral
Lighting and SmallAppliance Load  40,590 VA 
Range Load: 25,000 VA at 70% [see 220.61(B)]  17,500 VA 
Calculated Load (neutral)  58,090 VA 
Calculated Load for Neutral
58,090 VA ÷ 240 V = 242 A
Further Demand Factor [220.61(B)]
200 A at 100%  200 A 
242 A — 200 A = 42 A at 70%  29 A 
Net Calculated Load (neutral)  229 A 
Minimum Size Main Feeders (or Service Conductors) Required (Less House Load) (For 40 Dwelling Units — 20 with Ranges)
Total Calculated Load:  
Lighting and SmallAppliance Load 40 units × 5520 VA 
220,800 VA 
Application of Demand Factor (from Table 220.42)
First 3000 VA at 100%  3,000 VA 
Next 120,000 VA — 3000 VA = 117,000 VA at 35%  40,950 VA 
Remainder 220,800 VA — 120,000 VA = 100,800 VA at 25%  25,200 VA 
Net Calculated Load  69,150 VA 
Range Load: 20 ranges (less than 12 kVA)  
(see Col. C, Table 220.55)  35,000 VA 
Net Calculated Load  104,150 VA 
For 120/240V, 3wire system
Net calculated load of 104,150 VA ÷ 240 V = 434 A
Feeder Neutral
Lighting and SmallAppliance Load  69,150 VA 
Range: 35,000 VA at 70% [see 220.61(B)]  24,500 VA 
Calculated Load (neutral)  93,650 VA 
93,650 VA ÷ 240 V = 390 A
Further Demand Factor [see 220.61(B)
200 A at 100%  200 A 
390 A — 200 A = 190 A at 70%  133 A 
Net Calculated Load (neutral)  333 A 
[See Table 310.15(B)(16) through Table 310.15(B)(21), and 310.15(B)(2), (B)(3), and (B)(5).]
A multifamily dwelling equipped with electric cooking and space heating or air conditioning has 40 dwelling units.
Meters are in two banks of 20 each plus house metering and individual feeders to each dwelling unit.
Each dwelling unit is equipped with an electric range of 8kW nameplate rating, four 1.5kW separately controlled 240V electric space heaters, and a 2.5kW, 240V electric water heater. Assume range, space heater, and water heater kW ratings equivalent to kVA. Calculate the load for the individual dwelling unit by the standard calculation (Part III of Article 220).
A common laundry facility is available to all tenants [see 210.52(F), Exception No. 1].
Area of each dwelling unit is 840 ft^{2}.
Calculated Load for Each Dwelling Unit (see Part II and Part III of Article 220)
General Lighting Load:  
840 ft^{2} at 3 VA/ft^{2}  2,520 VA 
Electric range  8,000 VA 
Electric heat: 6 kVA (or air conditioning if larger)  6,000 VA 
Electric water heater  2,500 VA 
Minimum Number of Branch Circuits Required for Each Dwelling Unit
General Lighting Load: 2520 VA ÷ 120 V = 21 A or two 15A, 2wire circuits, or two 20A, 2wire circuits
SmallAppliance Load: Two 2wire circuits of 12 AWG [see 210.11(C)(1)]
Range Circuit (See Table 220.55, Column B):
8000 VA × 80% ÷ 240 V = 27 A on a circuit of three
10 AWG conductors in accordance with 210.19(A)(3) Space Heating: 6000 VA ÷ 240 V = 25 A
Number of circuits (see 210.11)
Minimum Size Feeder Required for Each Dwelling Unit (see 215.2)
Calculated Load (see Article 220):  
General Lighting  2,520 VA 
Small Appliance (two 20A circuits)  3,000 VA 
Subtotal Calculated Load (without range and space heating)  5,520 VA 
Application of Demand Factor
First 3000 VA at 100%  3,000 VA 
5520 VA — 3000 VA = 2520 VA at 35%  882 VA 
Net Calculated Load (without range and space heating) 
3,882 VA 
Range  6,400 VA 
Space Heating (see 220.51)  6,000 VA 
Water Heater  2,500 VA 
Net Calculated Load (for individual dwelling unit)  18,782 VA 
Size of Each Feeder
For 120/240V, 3wire system,
Net calculated load of 18,782 VA ÷ 240 V = 78 A
Lighting and Small Appliance  3,882 VA 
Range Load: 6400 VA at 70% [see 220.61(B)]  4,480 VA 
Space and Water Heating (no neutral): 240 V  0 VA 
Net Calculated Load (neutral)  8,362 VA 
Calculated Load for Neutral
8362 VA ÷ 240 V = 35 A
Minimum Size Feeder Required from Service Equipment to Meter Bank (For 20 Dwelling Units)
Total Calculated Load:  
Lighting and SmallAppliance Load  
20 units × 5520 VA  110,400 VA 
Water and Space Heating Load  
20 units × 8500 VA  170,000 VA 
Range Load: 20 × 8000 VA  160,000 VA 
Net Calculated Load (20 dwelling units)  440,400 VA 
Net Calculated Load Using Optional Calculation (see Table 220.84) 

440,400 VA × 0.38  167,352 VA 
167,352 VA ÷ 240 V = 697 A
Minimum Size Main Feeder Required (Less House Load) (For 40 Dwelling Units)
Calculated Load:  
Lighting and SmallAppliance Load  
40 units × 5520 VA  220,800 VA 
Water and Space Heating Load  340,000 VA 
40 units × 8500 VA  
Range: 40 ranges × 8000 VA  320,000 VA 
Net Calculated Load (40 dwelling units)  880,800 VA 
Net Calculated Load Using Optional Calculation (see Table 220.84)
880,800 VA × 0.28 = 246,624 VA
246,624 VA ÷ 240 V = 1028 A
Feeder Neutral Load for Feeder from Service Equipment to Meter Bank (For 20 Dwelling Units)
Lighting and SmallAppliance Load  
20 units × 5520 VA  110,400 VA 
First 3000 VA at 100%  3,000 VA 
110,400 VA — 3000 VA = 107,400 VA at 35%  37,590 VA 
Net Calculated Load  40,590 VA 
20 ranges: 35,000 VA at 70% [see Table 220.55 and 220.61(B)] 
24,500 VA 
Total  65,090 VA 
65,090 VA ÷ 240 V = 271 A
Further Demand Factor [see 220.61(B)]
First 200 A at 100%  200 A 
Balance: 271 A — 200 A = 71 A at 70%  50 A 
Total  250 A 
Feeder Neutral Load of Main Feeder (Less House Load) (For 40 Dwelling Units)
Lighting and SmallAppliance Load  
40 units × 5520 VA  220,800 VA 
First 3000 VA at 100%  3,000 VA 
Next 120,000 VA — 3000 VA = 117,000 VA at 35%  40,950 VA 
Remainder 220,800 VA — 120,000 VA = 100,800 VA at 25%  25,200 VA 
Net Calculated Load  69,150 VA 
40 ranges: 55,000 VA at 70% [see Table 220.55 and 220.61(B)]  38,500 VA 
Total  107,650 VA 
107,650 VA ÷ 240 V = 449 A
Further Demand Factor [see 220.61(B)]
First 200 A at 100%  200 A 
Balance: 449 — 200 A = 249 A at 70%  174 A 
Total  374 A 
All conditions and calculations are the same as for the multifamily dwelling [Example D4(a)] served at 120/240 V, single phase except as follows:
Service to each dwelling unit would be two phase legs and neutral.
Minimum Number of Branch Circuits Required for Each Dwelling Unit (see 210.11)
Range Circuit: 8000 VA ÷ 208 V = 38 A or a circuit of two 8 AWG conductors and one 10 AWG conductor in accordance with 210.19(A)(3)
Minimum Size Feeder Required for Each Dwelling Unit (see 215.2)
For 120/208V, 3wire system (without ranges),
Net calculated load of 3882 VA ÷ 2 legs ÷ 120 V/leg = 16 A
For 120/208V, 3wire system (with ranges),
Net calculated load (range) of 8000 VA ÷ 208 V = 39 A
Total load (range 4 lighting) = 39A + l6A = 55A
Reducing the neutral load on the feeder to each dwelling unit is not permitted [see 220.61(C)(1)].
Minimum Size Feeders Required from Service Equipment to Meter Bank (For 20 Dwelling Units — 10 with Ranges)
For 208Y/120V, 3phase, 4wire system,
Ranges: Maximum number between any two phase legs = 4 2 × 4 = 8.
Table 220.55 demand = 23,000 VA
Per phase demand = 23,000 VA ÷ 2 = 11,500 VA
Equivalent 3phase load = 34,500 VA
Net Calculated Load (total):
40.590 VA + 34,500 VA = 75,090 VA
75,090 VA ÷ (208 V)(1.732) = 208 A
Feeder Neutral Size
Net Calculated Lighting and Appliance Load & Equivalent Range Load:
40.590 VA + (34,500 VA at 70%) = 64,700 VA
Net Calculated Neutral Load:
64,700 VA ÷ (208 V)(1.732) = 180 A
Minimum Size Main Feeder (Less House Load) (For 40 Dwelling Units — 20 with Ranges)
For 208Y/120V, 3phase, 4wire system,
Ranges:
Maximum number between any two phase legs = 7 2 x7 = 14.
Table 220.55 demand = 29,000 VA
Per phase demand = 29,000 VA ÷ 2 = 14,500 VA
Equivalent 3phase load = 43,500 VA
Net Calculated Load (total):
69.150 VA + 43,500 VA = 112,650 VA
112,650 VA ÷ (208 V)(1.732) = 313 A
Main Feeder Neutral Size:
69.150 VA + (43,500 VA at 70%) = 99,600 VA
99,600 VA ÷ (208 V)(1.732) = 277 A
Further Demand Factor (see 220.61)
200 A at 100%  200.0 A 
277 A — 200 A = 77 A at 70%  54 A 
Net Calculated Load (neutral)  254 A 
All conditions and calculations are the same as for Optional Calculation for the Multifamily Dwelling [Example D4(b)] served at 120/240 V, single phase except as follows:
Service to each dwelling unit would be two phase legs and neutral.
Minimum Number of Branch Circuits Required for Each Dwelling Unit (see 210.11)
Range Circuit (see Table 220.55, Column B): 8000 VA at 80% ÷ 208 V = 31 A or a circuit of two 8 AWG conductors and one 10 AWG conductor in accordance with 210.19(A)(3)
Space Heating: 6000 VA ÷ 208 V = 29 A
Two 20ampere, 2pole circuits required, 12 AWG conductors
Minimum Size Feeder Required for Each Dwelling Unit
120/208V, 3wire circuit
Net calculated load of 18,782 VA ÷ 208 V = 90 A
Net calculated load (lighting line to neutral):
3882 VA ÷ 2 legs ÷ 120 V per leg = 16 amperes
Line to line = 14,900 VA ÷ 208 V = 72 A
Total load = 16.2 A + 71.6 A = 88 A
Minimum Size Feeder Required for Service Equipment to Meter Bank (For 20 Dwelling Units)
Net Calculated Load
167,352 VA ÷ (208 V)(1.732) = 465 A
Feeder Neutral Load
65,080 VA ÷ (208 V)(1.732) = 181 A
Minimum Size Main Feeder Required (Less House Load) (For 40 Dwelling Units)
Net Calculated Load
246,624 VA ÷ (208 V)(1.732) = 685 A
Main Feeder Neutral Load
107,650 VA ÷ (208 V)(1.732) = 299 A
Further Demand Factor [see 220.61(B)]
200 A at 100%  200.0 A 
299 A — 200 A = 99 A at 70%  69 A 
Net Calculated Load (neutral)  269 A 
Table 220.55, Column C, applies to ranges not over 12 kW. The application of Note 1 to ranges over 12 kW (and not over 27 kW) and Note 2 to ranges over 8^{3}/_{4} kW (and not over 27 kW) is illustrated in the following two examples.
Assume 24 ranges, each rated 16 kW.
From Table 220.55, Column C, the maximum demand for 24 ranges of 12kW rating is 39 kW. 16 kW exceeds 12 kW by 4.
5% × 4 = 20% (5% increase for each kW in excess of 12)
39 kW × 20% = 7.8 kW increase
39 + 7.8 = 46.8 kW (value to be used in selection of feeders)
Assume 5 ranges, each rated 11 kW; 2 ranges, each rated 12 kW; 20 ranges, each rated 13.5 kW; 3 ranges, each rated 18 kW.
5 ranges  × 12 kW =  60 kW (use 12 kW for range rated less than 12) 
2 ranges  × 12 kW =  24 kW 
20 ranges  × 13.5 kW =  270 kW 
3 ranges  × 18 kW =  54 kW 
30 ranges, Total kW =  408 kW 
408 ÷ 30 ranges = 13.6 kW (average to be used for calculation)
From Table 220.55, Column C, the demand for 30 ranges of 12kW rating is 15 kW + 30 (1 kW × 30 ranges) = 45 kW. 13.6 kW exceeds 12 kW by 1.6 kW (use 2 kW).
5% × 2 = 10% (5% increase for each kW in excess of 12 kW)
45 kW × 10% = 4.5 kW increase
45 kW + 4.5 kW = 49.5 kW (value to be used in selection of feeders)
Service conductors and feeders for certain dwellings are permitted to be sized in accordance with 310.15(B)(7).
With No Required Adjustment or Correction Factors. If a 175ampere service rating is selected, a service conductor is then sized as follows:
175 amperes × 0.83 = 145.25 amperes per 310.15(B)(7).
If no other adjustments or corrections are required for the installation, then, in accordance with Table 310.15(B)(16), a 1/0 AWG Cu or a 3/0 AWG Al meets this rating at 75°C (167°F).
With Required Temperature Correction Factor. If a 175ampere service rating is selected, a service conductor is then
175 amperes × 0.83 = 145.25 amperes per 310.15(B)(7).
If the conductors are installed in an ambient temperature of 40°C (104°F), the conductor ampacity must be multiplied by the appropriate correction factor in Table 310.15(B)(2)(a). In this case, we will use an XHHW2 conductor, so we use a correction factor of 0.91 to find the minimum conductor ampacity and size:
145.25/.91 = 159.6 amperes
In accordance with Table 310.15(B)(16), a 2/0 AWG Cu or a 4/0 AWG Al would be required.
If no temperature correction or ampacity adjustment factors are required, the following table includes conductor sizes calculated using the requirements in 310.15(B)(7). This table is based on 75°C terminations and without any adjustment or correction factors.
Conductor (AWG or kcmil)  

Service or Feeder Rating (Amperes)  Copper  Aluminum or CopperClad Aluminum 
100  4  2 
110  3  1 
125  2  1/0 
150  1  2/0 
175  1/0  3/0 
200  2/0  4/0 
225  3/0  250 
250  4/0  300 
300  250  350 
350  350  500 
400  400  600 
(see 240.6, 430.6, 430.22, 430.23, 430.24, 430.32, 430.52, and 430.62, Table 430.52, and Table 430.250)
Determine the minimum required conductor ampacity, the motor overload protection, the branchcircuit shortcircuit and groundfault protection, and the feeder protection, for three inductiontype motors on a 480V, 3phase feeder, as follows:
(a) One 25hp, 460V, 3phase, squirrelcage motor, nameplate fullload current 32 A, Design B, Service Factor 1.15
(b) Two 30hp, 460V, 3phase, woundrotor motors, nameplate primary fullload current 38 A, nameplate secondary fullload current 65 A, 40°C rise.
Conductor Ampacity The fullload current value used to determine the minimum required conductor ampacity is obtained from Table 430.250 [see 430.6(A)] for the squirrelcage motor and the primary of the woundrotor motors. To obtain the minimum required conductor ampacity, the fullload current is multiplied by 1.25 [see 430.22 and 430.23(A)].
For the 25hp motor,
34 A × 1.25 = 43 A
For the 30horsepower motors,
40 A × 1.25 = 50 A
65 A × 1.25 = 81 A
Motor Overload Protection Where protected by a separate overload device, the motors are required to have overload protection rated or set to trip at not more than 125% of the nameplate fullload current [see 430.6(A) and 430.32(A)(1)].
For the 25hp motor,
32 A × 1.25 = 40.0 A
For the 30hp motors,
38 A × 1.25 = 48 A
Where the separate overload device is an overload relay (not a fuse or circuit breaker), and the overload device selected at 125% is not sufficient to start the motor or carry the load, the trip setting is permitted to be increased in accordance with 430.32(C).
BranchCircuit ShortCircuit and GroundFault Protection The selection of the rating of the protective device depends on the type of protective device selected, in accordance with 430.52 and Table 430.52. The following is for the 25hp motor.
(a) NontimeDelay Fuse: The fuse rating is 300% × 34 A = 102 A. The next larger standard fuse is 110 A [see 240.6 and 430.52(C)(1), Exception No. 1]. If the motor will not start with a 110A nontimedelay fuse, the fuse rating is permitted to be increased to 125 A because this rating does not exceed 400% [see 430.52(C)(1), Exception No. 2(a)].
(b) TimeDelay Fuse: The fuse rating is 175% × 34 A = 59.5 A. The next larger standard fuse is 60 A [see 240.6 and 430.52(C)(1), Exception No. 1]. If the motor will not start with a 60A timedelay fuse, the fuse rating is permitted to be increased to 70 A because this rating does not exceed 225% [see 430.52(C)(1), Exception No. 2(b)].
Feeder ShortCircuit and GroundFault Protection
(a) Example using nontime delay fuse. The rating of the feeder protective device is based on the sum of the largest branchcircuit protective device for the specific type of device protecting the feeder: 300% × 34 A = 102 A (therefore the next largest standard size, 110 A, would be used) plus the sum of the fullload currents of the other motors, or 110 A + 40 A + 40 A = 190 A. The nearest standard fuse that does not exceed this value is 175 A [see 240.6 and 430.62(A)].
(b) Example using inverse time circuit breaker. The largest branchcircuit protective device for the specific type of device protecting the feeder, 250% × 34 A = 85. The next larger standard size is 90 A, plus the sum of the fullload currents of the other motors, or 90 A + 40 A + 40 A = 170 A. The nearest standard inverse time circuit breaker that does not exceed this value is 150 A [see 240.6 and 430.62(A)].
[see 215.2, 430.24, 430.24 Exception No. 1, 620.13, 620.14, 620.61, and Table 430.22(E) and 620.14]
Determine the conductor ampacity for a 460V 3phase, 60Hz ac feeder supplying a group of six elevators. The 460V ac drive motor nameplate rating of the largest MG set for one elevator is 40 hp and 52 A, and the remaining elevators each have a 30hp, 40A, ac drive motor rating for their MG sets. In addition to a motor controller, each elevator has a separate motion/operation controller rated 10 A continuous to operate microprocessors, relays, power supplies, and the elevator car door operator. The MG sets are rated continuous.
Conductor Ampacity Conductor ampacity is determined as follows:
(a) In accordance with 620.13(D) and 620.61(B)(1), use Table 430.22(E), for intermittent duty (elevators). For intermittent duty using a continuous rated motor, the percentage of nameplate current rating to be used is 140%.
(b) For the 30hp ac drive motor,
140% × 40 A = 56 A
(c) For the 40hp ac drive motor,
140% × 52 A = 73 A(I)
(d) The total conductor ampacity is the sum of all the motor currents:
(1 motor × 73 A) + (5 motors × 56 A) = 353 A
(e) In accordance with 620.14 and Table 620.14, the conductor (feeder) ampacity would be permitted to be reduced by the use of a demand factor. Constant loads are not included (see 620.14, Informational Note). For six elevators, the demand factor is 0.79. The feeder diverse ampacity is, therefore, 0.79 × 353 A = 279 A.
(f) In accordance with 430.24 and 215.3, the controller continuous current is 125% × 10 A = 13 A
(g) The total feeder ampacity is the sum of the diverse current and all the controller continuous current.
I_{total} = 279 A + (6 elevators × 12.5 A) = 354 A
(h) This ampacity would be permitted to be used to select the wire size.
See Figure D9.
FIGURE D9 Generator Field Control.
Determine the conductor ampacity for a 460V, 3phase, 60Hz ac feeder supplying a group of six identical elevators. The system is adjustablespeed SCR dc drive. The power transformers are external to the drive (motor controller) cabinet. Each elevator has a separate motion/operation controller connected to the load side of the main line disconnect switch rated 10 A continuous to operate microprocessors, relays, power supplies, and the elevator car door operator. Each transformer is rated 95 kVA with an efficiency of 90%.
Conductor Ampacity
Conductor ampacity is determined as follows:
(a) Calculate the nameplate rating of the transformer:
(b) In accordance with 620.13(D), for six elevators, the total conductor ampacity is the sum of all the currents.
6 elevators × 133 A = 798 A
(c) In accordance with 620.14 and Table 620.14, the conductor (feeder) ampacity would be permitted to be reduced by the use of a demand factor. Constant loads are not included (see 620.13, Informational Note No. 2). For six elevators, the demand factor is 0.79. The feeder diverse ampacity is, therefore, 0.79 × 798 A = 630 A.
(d) In accordance with 430.24 and 215.3, the controller continuous current is 125% × 10 A = 13 A.
(e) The total feeder ampacity is the sum of the diverse current and all the controller constant current.
I_{total} = 630 A + (6 elevators × 12.5 A) = 705 A
(f) This ampacity would be permitted to be used to select the wire size.
See Figure D10.
FIGURE D10 Adjustable Speed Drive Control.
(see 550.18)
A mobile home floor is 70 ft by 10 ft and has two small appliance circuits; a 1000VA, 240V heater; a 200VA, 120V exhaust fan; a 400VA, 120V dishwasher; and a 7000VA electric range.
Lighting and SmallAppliance Load
Lighting (70 ft × 10 ft × 3 VA per ft^{2})  2,100 VA  
Smallappliance (1500 VA× 2 circuits)  3,000 VA  
Laundry (1500 VA × 1 circuit)  1,500 VA  
Subtotal  6,600 VA  
First 3000 VA at 100%  3,000 VA  
Remainder (6600 VA — 3000 VA = 3600 VA) × 35%  1,260 VA  
Total  4,260 VA 
4260 VA ÷ 240 V = 17.75 A per leg
Amperes per Leg  Leg A  Leg B 

Lighting and appliances  18  18 
Heater (1000 VA ÷ 240 V)  4  4 
Fan (200 VA × 125% ÷ 120 V)  2  — 
Dishwasher (400 VA ÷ 120 V)  —  3 
Range (7000 VA × 0.8 ÷ 240 V)  23  23 
Total amperes per leg  47  48 
Based on the higher current calculated for either leg, a minimum 50A supply cord would be required.
For SI units, 0.093 m^{2} = 1 ft^{2} and 0.3048 m = 1 ft.
(see 552.47)
A park trailer floor is 40 ft by 10 ft and has two small appliance circuits, a 1000VA, 240V heater, a 200VA, 120V exhaust fan, a 400VA, 120V dishwasher, and a 7000VA electric range.
Lighting and SmallAppliance Load
Lighting (40 ft × 10 ft × 3 VA per ft^{2})  1,200 VA 
Smallappliance (1500 VA× 2 circuits)  3,000 VA 
Laundry (1500 VA × 1 circuit)  1,500 VA 
Subtotal  5,700 VA 
First 3000 VA at 100%  3,000 VA 
Remainder (5700 VA — 3000 VA = 2700 VA) × 35%  945 VA 
Total  3,945 VA 
3945 VA÷ 240 V = 16.44 A per leg
Amperes per Leg  Leg A  Leg B 

Lighting and appliances  16  16 
Heater (1000 VA ÷ 240 V)  4  4 
Fan (200 VA × 125% ÷ 120 V)  2  — 
Dishwasher (400 VA ÷ 120 V)  —  3 
Range (7000 VA × 0.8 ÷ 240 V)  23  23 
Totals  45  46 
Based on the higher current calculated for either leg, a minimum 50A supply cord would be required.
For SI units, 0.093 m^{2} = 1 ft^{2} and 0.3048 m = 1 ft.
(see Article 392)
D13(a) Multiconductor Cables 4/0 AWG and Larger
Use: NEC 392.22(A)(1)(a)
Cable tray must have an inside width equal to or greater than the sum of the diameters (Sd) of the cables, which must be installed in a single layer.
Example: Cable tray width is obtained as follows:
Cable Size Being Used  (OD) Cable Outside Diameters (in.) 
(N) Number of Cables 
SD = (OD) × (N) (Sum of the Cable Diameters) (in.) 

3conductor Type MC cable — 4/0 AWG  1.57  12  18.84 
The sum of the diameters (Sd) of all cables = 18.84 in., therefore a cable tray with an inside width of at least 18.84 in. is required.
Note: Cable outside diameter is a nominal diameter from catalog data.
D13(b) Multiconductor Cables Smaller Than 4/0 AWG
Use: NEC 392.22(A)(1)(b)
The sum of the crosssectional areas of all the cables to be installed in the cable tray must be equal to or less than the allowable cable area for the tray width, as indicated in Table 392.22(A), Column 1.
Table D13(b) from Table 392.22(A), Column 1
Inside Width of Cable Tray (in.)  Allowable Cable Area (in.^{2}) 

6  7.0 
9  10.5 
12  14.0 
18  21.0 
24  28.0 
30  35.0 
36  42.0 
Example: Cable tray width is obtained as follows:
Cable Size Being Used  (A) Cable CrossSectional Area (in.^{2}) 
(N) Number of Cables 
Multiply (A) × (N) (Which Is a Total Cable CrossSectional Area in in.^{2}) 

4conductor Type MC cable — 1 AWG  1.1350  9  12.15 
The total cable crosssectional area is 12.15 in.^{2}. Using Table D13(b) above, the next higher allowable cable area must be used, which is 14.0 in.^{2}. The table specifies that the cable tray inside width for an allowable cable area of 14.0 in.^{2} is 12 in.
Note: Cable crosssectional area is a nominal area from catalog data.
D13(c) Single Conductor Cables 1/0 AWG through 4/0 AWG
Use: NEC 392.22(B)(1)(d)
Cable tray must have an inside width equal to or greater than the sum of the diameters (Sd) of the cables. The cables must be evenly distributed across the cable tray.
Example: Cable tray width is obtained as follows:
Single Conductor Cable Size Being Used  (OD) Cable Outside Diameters (in.) 
(N) Number of Cables 
Sd = (OD) × (N) (Sum of the Cable Diameters) (in.) 

THHN — 4/0 AWG  0.642  18  11.556 
The sum of the diameters (Sd) of all cables = 11.56 in., therefore, a cable tray with an inside width of at least 11.56 in. is required.
Use: NEC 392.22(B)(1)(b)
The sum of the crosssectional areas of all the cables to be installed in the cable tray must be equal to or less than the allowable cable area for the tray width, as indicated in Table 392.22(B)(1), Column 1.
Table D13(d) from Table 392.22(B)(1), Column 1
Inside Width of Cable Tray (in.)  Allowable Cable Area (in.^{2}) 

6  6.5 
9  9.5 
12  13.0 
18  19.5 
24  26.0 
30  32.5 
36  39.0 
Example: Cable tray width is obtained as follows:
Cable Size Being Used  (A) Cable CrossSectional Area (in.^{2})  (N) Number of Cables  Multiply (A) × (N) (Which Is a Total Cable CrossSectional Area in in.^{2}) 

THHN — 500 kcmil  0.707  9  6.36 
The total cable crosssectional area is 6.36 in.^{2}. Using Table D13(d), the next higher allowable cable area must be used, which is 6.5 in.^{2}. The table specifies that the cable tray inside width for an allowable cable area of 6.5 in.^{2} is 6 in.
Note: Singleconductor cable crosssectional area from Chapter 9, Table 5.