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Annex D Examples

  1. 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 exam‐ ples, 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 m2 = 1 ft2 and 0.3048 m = 1 ft.
    Example D1(a) One-Family Dwelling
    The dwelling has a floor area of 1500 ft2, exclusive of an unfin‐ ished cellar not adaptable for future use, unfinished attic, and open porches. Appliances are a 12-kW range and a 5.5-kW, 240- V 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 ft2 at 3 VA/ft2 = 4500 VA Minimum Number of Branch Circuits Required [see 210.11(A)] General Lighting Load: 4500 VA ÷ 120 V = 38 AThis requires three 15-A, 2-wire or two 20-A, 2-wire circuits.Small-Appliance Load: Two 2-wire, 20-A circuits [see 210.11(C)(1)]Laundry Load: One 2-wire, 20-A circuit [see 210.11(C)(2)]Bathroom Branch Circuit: One 2-wire, 20-A circuit (no addi‐ tional load calculation is required for this circuit) [see 210.11(C)(3)]Minimum Size Feeder Required [see 220.40]
    General Lighting 4,500 VASmall Appliance 3,000 VALaundry 1,500 VAimageTotal 9,000 VA3000 VA at 100% 3,000 VA9000 VA – 3000 VA = 6000 VA at 35% 2,100 VAimageNet Load 5,100 VARange (see Table 220.55) 8,000 VADryer Load (see Table 220.54) 5,500 VAimageNet Calculated Load 18,600 VA
    Net Calculated Load for 120/240-V, 3-wire, single-phase service or feeder18,600 VA ÷ 240 V = 78 ASections 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 Small-Appliance Load 5,100 VARange: 8000 VA at 70% (see 220.61) 5,600 VADryer: 5500 VA at 70% (see 220.61) 3,850 VAimageTotal 14,550 VA
    Calculated Load for Neutral14,550 VA ÷ 240 V = 61 A
    Example D1(b) One-Family Dwelling
    Assume same conditions as Example No. D1(a), plus addition of one 6-A, 230-V, room air-conditioning unit and one 12-A, 115-V, room air-conditioning unit,* one 8-A, 115-V, rated waste disposer, and one 10-A, 120-V, rated dishwasher. See Article 430 for general motors and Article 440, Part VII, for air- conditioning equipment. Motors have nameplate ratings of 115 V and 230 V for use on 120-V and 240-V nominal voltage systems.*(For feeder neutral, use larger of the two appliances for unbalance.)
    From Example D1(a), feeder current is 78 A (3-wire, 240 V). Feeder Neutral Load in Accordance with 220.61

    Line ANeutralLine B1500 ft2 at 3 VA
    4,500 VAAmperes from Example D1(a)786178Three 20-A circuits at 1500 VA
    4,500 VAOne 230-V air conditioner6—6
    Total9,000 VAOne 115-V air conditioner and1212103000 VA at 100%
    3,000 VAimage120-V dishwasherOne 115-V disposer — 8 8image25% of largest motor (see 430.24) 3 3 2Total amperes per conductor 99 84 104Therefore, the service would be rated 110 A.Example D2(a) Optional Calculation for One-Family Dwelling, Heating Larger Than Air Conditioning
    Calculated Load for Neutral15,400 VA ÷ 240 V= 64 A
    9000 VA – 3000 VA = 6000 VA at 35%
    2,100 VA
    Range: 8 kVA at 70%Subtotal5,100 VA5,600 VAClothes dryer: 5 kVA at 70% Dishwasher
    3,500 VA1,200 VATotal 15,400 VA
    (see 220.82)The dwelling has a floor area of 1500 ft2, exclusive of an unfin‐ ished cellar not adaptable for future use, unfinished attic, and open porches. It has a 12-kW range, a 2.5-kW water heater, a 1.2-kW dishwasher, 9 kW of electric space heating installed in five rooms, a 5-kW clothes dryer, and a 6-A, 230-V, room air- conditioning unit. Assume range, water heater, dishwasher, space heating, and clothes dryer kW ratings equivalent to kVA.Air Conditioner kVA Calculation6 A × 230 V ÷ 1000 = 1.38 kVAThis 1.38 kVA [item 1 from 220.82(C)] is less than 40% of9 kVA of separately controlled electric heat [item 6 from 220.82(C)], so the 1.38 kVA need not be included in the serv‐ ice calculation.General Load
    1500 ft2 at 3 VA 4,500 VATwo 20-A appliance outlet circuits at 1500 VA each 3,000 VA Laundry circuit 1,500 VARange (at nameplate rating) 12,000 VAWater heater 2,500 VADishwasher 1,200 VAClothes dryer 5,000 VAimageTotal 29,700 VA
    Application of Demand Factor [see 220.82(B)]
    First 10 kVA of general load at 100% 10,000 VAExample D2(b) Optional Calculation for One-Family Dwelling, Air Conditioning Larger Than Heating
    [see 220.82(A) and 220.82(C)]The dwelling has a floor area of 1500 ft2, exclusive of an unfin‐ ished cellar not adaptable for future use, unfinished attic, and open porches. It has two 20-A small appliance circuits, one 20-A laundry circuit, two 4-kW wall-mounted ovens, one 5.1-kW counter-mounted cooking unit, a 4.5-kW water heater, a 1.2-kW dishwasher, a 5-kW combination clothes washer and dryer, six 7-A, 230-V room air-conditioning units, and a 1.5-kW perma‐ nently installed bathroom space heater. Assume wall-mounted ovens, counter-mounted cooking unit, water heater, dish‐ washer, and combination clothes washer and dryer kW ratings equivalent to kVA.Air Conditioning kVA CalculationTotal amperes = 6 units × 7 A = 42 A42 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 ft2 at 3 VA 4,500 VATwo 20-A small-appliancecircuits at 1500 VA each 3,000 VARemainder of general load at 40% (19.7 kVA × 0.4)7,880 VA
    imageLaundry circuit 1,500 VATwo ovens 8,000 VAOne cooking unit 5,100 VATotal of general load 17,880 VA 9 kVA of heat at 40% (9000 VA × 0.4) = 3,600 VAimageTotal 21,480 VA
    Calculated Load for Service Size21.48 kVA = 21,480 VA 21,480 VA ÷ 240 V = 90 ATherefore, the minimum service rating would be 100 A in accordance with 230.42 and 230.79.Water heater 4,500 VADishwasher 1,200 VAWasher/dryer 5,000 VAimageTotal general load 32,800 VAFirst 10 kVA at 100%Remainder at 40%
    10,000 VA(22.8 kVA × 0.4 × 1000)
    9,120 VA
    image
    Air conditioningSubtotal general load19,120 VA10,080 VATotal29,200 VA
    Calculated Load for Service29,200 VA ÷ 240 V = 122 A (service rating)Remainder of general load at 40% (23,200 VA× 0.4)9,280 VA
    imageFeeder Neutral Load, in accordance with 220.61Assume that the two 4-kVA wall-mounted ovens are supplied by one branch circuit, the 5.1-kVA counter-mounted cooking unit by a separate circuit.
    1500 ft2 at 3 VA 4,500 VAThree 20-A circuits at 1500 VA 4,500 VAimageSubtotal 9,000 VA3000 VA at 100% 3,000 VA9000 VA – 3000 VA = 6000 VA at 35% 2,100 VAimageSubtotal 5,100 VA
    Two 4-kVA ovens plus one 5.1-kVA 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 VATotal net general load 19,280 VA
    Heat Pump and Supplementary Heat*240 V × 24 A = 5760 VA15 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.
    TotalsNet general load 19,280 VAimageHeat pump and supplementary heat 15,510 VA Total 34,790 VACalculated Load for ServiceOvens and cooking unit: 7200 VA × 70% for neutral load5,040 VA34.79 kVA × 1000 ÷ 240 V= 145 ATherefore, this dwelling unit would be permitted to beClothes washer/dryer: 5 kVA × 70% for neutral load 3,500 VA Dishwasher 1,200 VAimageTotal 14,840 VAserved by a 150-A service.
    Example D3 Store BuildingCalculated Load for Neutral14,840 VA ÷ 240 V = 62
    Example D2(c) Optional Calculation for One-Family Dwelling with Heat Pump (Single-Phase, 240/120-Volt Service)
    (see 220.82)The dwelling has a floor area of 2000 ft2, exclusive of an unfin‐ ished cellar not adaptable for future use, unfinished attic, and open porches. It has a 12-kW range, a 4.5-kW water heater, a 1.2-kW dishwasher, a 5-kW clothes dryer, and a 21∕2-ton (24-A) heat pump with 15 kW of backup heat.Heat Pump kVA Calculation24 A × 240 V ÷ 1000 = 5.76 kVAThis 5.76 kVA is less than 15 kVA of the backup heat; therefore, the heat pump load need not be included in the service calcu‐ lation [see 220.82(C)].General Load
    2000 ft2 at 3 VA 6,000 VA
    A store 50 ft by 60 ft, or 3000 ft2, has 30 ft of show window. There are a total of 80 duplex receptacles. The service is 120/240 V, single phase 3-wire service. Actual connected light‐ ing load is 8500 VA.Calculated Load (see 220.40)
    Noncontinuous LoadsReceptacle Load (see 220.44)80 receptacles at 180 VA 14,400 VA10,000 VA at 100% 10,000 VA14,400 VA − 10,000 VA = 4400 at 50% 2,200 VAimageSubtotal 12,200 VAContinuous LoadsGeneral Lighting*3000 ft2 at 3 VA/ft2 9,000 VAShow Window Lighting Load30 ft at 200 VA/ft [see 220.14(G)] 6,000 VAOutside Sign Circuit [see 220.14(F)] 1,200 VAimageSubtotal 16,200 VA Subtotal from noncontinuous 12,200 VAimageTotal noncontinuous loads +Two 20-A appliance outlet circuits at 1500 VA each3,000 VAcontinuous loads = 28,400 VALaundry circuit 1,500 VARange (at nameplate rating) 12,000 VAWater heater 4,500 VADishwasher 1,200 VAClothes dryer 5,000 VAimageSubtotal general load 33,200 VAFirst 10 kVA at 100% 10,000 VA(continues)
    *In the example, the actual connected lighting load (8500 VA) is less than the load from Table 220.12, so the minimum light‐ ing 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 RequiredGeneral Lighting: Branch circuits need only be installed to supply the actual connected load [see 210.11(B)].8500 VA × 1.25 = 10,625 VA10,625 VA ÷ 240 V = 44 A for 3-wire, 120/240 VThe lighting load would be permitted to be served by 2-wire or 3-wire, 15- or 20-A circuits with combined capacity equal to 44 A or greater for 3-wire circuits or 88 A or greater for 2-wire circuits. The feeder capacity as well as the number of branch- circuit positions available for lighting circuits in the panelboard must reflect the full calculated load of 9000 VA × 1.25 = 11,250 VA.Show Window6000 VA × 1.25 = 7500 VA7500 VA ÷ 240 V = 31 A for 3-wire, 120/240 VThe show window lighting is permitted to be served by 2-wire or 3-wire circuits with a capacity equal to 31 A or greater for 3- wire circuits or 62 A or greater for 2-wire 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.ReceptaclesReceptacle Load: 14,400 VA ÷ 240 V = 60 A for 3-wire, 120/240 VThe receptacle load would be permitted to be served by 2- wire or 3-wire circuits with a capacity equal to 60 A or greater for 3-wire circuits or 120 A or greater for 2-wire circuits.Minimum Size Feeder (or Service) Overcurrent Protection(see 215.3 or 230.90)
    Subtotal noncontinuous loads 12,200 VA
    Example D3(a) Industrial Feeders in a Common Raceway
    An industrial multi-building facility has its service at the rear of its main building, and then provides 480Y/277-volt 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 temper‐ ature 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 build‐ ings. 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 electric-discharge lumin‐ aires connected at 277 VReceptacles, 22 125-volt, 20-ampere receptacles on general- purpose branch circuits, supplied by separately derived systems in each of the buildings1 Air compressor, 460 volt, three phase, 5 hp1 Grinder, 460 volt, three phase, 1.5 hp3 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)Subtotal continuous load at 125% (16,200 VA × 1.25)
    32,450 VA ÷ 240 V = 135 A20,250 VA
    imageTotal 32,450 VADetermine the overcurrent protection and conductor size for the feeders in the common raceway, assuming the use of XHHW-2 insulation (90°C):Calculated Load {Note: For reasonable precision, volt-ampere calculations are carried to three significant figures only; where loads are converted to amperes, the results are rounded to theThe next higher standard size is 150 A (see 240.6).Minimum Size Feeders (or Service Conductors) Required [see 215.2, 230.42(A)]For 120/240 V, 3-wire 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).nearest ampere [see 220.5(B)]}.
    Noncontinuous LoadsReceptacle Load (see 220.44)22 receptacles at 180 VA 3,960 VAWelder Load [see 630.11(A), Table 630.11(A)]Each welder: 480V × 23A × 0.78 = 8,610 VAAll 3 welders [see 630.11(B)](demand factors 100%, 100%,85% respectively)8,610 VA + 8,610 VA + 7,320 VA = 24,500 VAimageSubtotal, Noncontinuous Loads 28,500 VA Motor Loads (see 430.24,Table 430.250)(continues)
    Air compressor: 7.6 A × 480 V × √3 = 6,310 VAGrinder: 3 A × 480 V × √3 = 2,490 VALargest motor, additional 25%: 1,580 VAimageSubtotal, Motor Loads 10,400 VABy 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 VAContinuous LoadsGeneral Lighting 11,600 VA3 Industrial Process Dryers 15 kW each 45,000 VAimageSubtotal, 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 VANoncontinuous load            38,900 VAmust be used. If the utility corridor were at normal tempera‐ tures [(30°C (86°F)], and if the lighting at each building were supplied from the local separately derived system (thus requir‐ ing 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 accord‐ ance 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 current-carrying conductors in the common raceway),195 amperes × 0.96 × 0.7 = 131 AThe 150-ampere 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. NoteSubtotal, actual load [actual load in amperes][99,000 VA ÷ (480V × √3) = 119 A]95,500 VAthat the feeder layout precludes the application of 310.15(A)(2) Exception.(25% of 56,600 VA) (See 215.3 ) 14,200 VAimageTotal VA 109,700 VAConversion to amperes using three significant figures: 109,700 VA / (480V × √3) = 132 AMinimum size overcurrent protective device: 132 AMinimum 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 125-ampere rating. (See 215.3 Exception.)Ungrounded Feeder ConductorsThe conductors must independently meet requirements for (1) terminations, and (2) conditions of use throughout the race‐ way run.Minimum size conductor at the overcurrrent device termina‐ tion [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 current- carrying 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 conductorsFeeder 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 contin‐ uous 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 line-to-neutral 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.
    Example D4(a) Multifamily Dwelling
    A multifamily dwelling has 40 dwelling units.Meters are in two banks of 20 each with individual feeders to each dwelling unit.One-half 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 ft2.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 ft2 at 3 VA/ft2 = 2520 VASpecial 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 15-A, 2- wire circuits; or two 20-A, 2-wire circuitsSmall-Appliance Load: Two 2-wire 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 VAimageSmall Appliance (two 20-ampere circuits) 3,000 VA Subtotal Calculated Load (without ranges) 5,520 VAApplication of Demand Factor (see Table 220.42)
    First 3000 VA at 100% 3,000 VA5520 VA – 3000 VA = 2520 VA at 35% 882 VAimageNet Calculated Load (without ranges) 3,882 VA Range Load 8,000 VAimageNet Calculated Load (with ranges) 11,882 VA
    Size of Each Feeder (see 215.2)For 120/240-V, 3-wire system (without ranges)Net calculated load of 3882 VA ÷ 240 V = 16 A For 120/240-V, 3-wire system (with ranges)Net calculated load, 11,882 VA ÷ 240 V = 50 ANet calculated load for 120/240-V, 3-wire system, 65,590 VA ÷ 240 V = 273 AFeeder Neutral
    Lighting and Small-Appliance Load 40,590 VA Range Load: 25,000 VA at 70% [see 220.61(B)] 17,500 VAimageCalculated Load (neutral) 58,090 VA
    Calculated Load for Neutral58,090 VA ÷ 240 V = 242 AFurther Demand Factor [220.61(B)]
    200 A at 100% 200 A242 A – 200 A = 42 A at 70% 29 AimageNet 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 Small-Appliance Load40 units × 5520 VΑ 220,800 VA
    Application of Demand Factor (from Table 220.42)
    First 3000 VA at 100% 3,000 VANext 120,000 VA – 3000 VA = 117,000 VA at 35% 40,950 VAFeeder NeutralRemainder 220,800 VA – 120,000 VA = 100,800 VAat 25%25,200 VA
    Lighting and Small-Appliance Load 3,882 VARange Load: 8000 VA at 70% (see 220.61) 5,600 VAimage(only for apartments with electric range) 5,600 VA Net Calculated Load (neutral) 9,482 VA
    Calculated Load for Neutral9482 VA ÷ 240 V= 39.5 AMinimum Size Feeders Required from Service Equipment to Meter Bank (For 20 Dwelling Units — 10 with Ranges)
    Total Calculated Load:Lighting and Small Appliance20 units × 5520 VΑ 110,400 VAimageApplication of Demand FactorFirst 3000 VA at 100% 3,000 VA110,400 VA – 3000 VA = 107,400 VA at 35% 37,590 VAimageNet Calculated Load 40,590 VA
    imageNet Calculated Load 69,150 VA Range Load: 20 ranges (less than 12 kVA)(see Col. C, Table 220.55) 35,000 VAimageNet Calculated Load 104,150 VAFor 120/240-V, 3-wire system
    Net calculated load of 104,150 VA ÷ 240 V = 434 AFeeder Neutral
    Lighting and Small-Appliance Load 69,150 VARange: 35,000 VA at 70% [see 220.61(B)] 24,500 VAimageCalculated Load (neutral) 93,650 VA
    93,650 VA ÷ 240 V = 390 AFurther Demand Factor [see 220.61(B)
    200 A at 100% 200 ARange Load: 10 ranges (not over 12 kVA) (see Col. C,25,000 VA390 A – 200 A = 190 A at 70% 133 AimageTable 220.55, 25 kW )                        Net Calculated Load (with ranges) 65,590 VANet 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).]

    Example D4(b) Optional Calculation for Multifamily Dwelling
    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 8- kW nameplate rating, four 1.5-kW separately controlled 240-V electric space heaters, and a 2.5-kW, 240-V electric water heater. Assume range, space heater, and water heater kW ratings equiv‐ alent 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 ft2.Calculated Load for Each Dwelling Unit (see Part II and Part III of Article 220)
    General Lighting Load:840 ft2 at 3 VA/ft2 2,520 VAElectric range 8,000 VARange 6,400 VASpace Heating (see 220.51) 6,000 VAWater Heater 2,500 VAimageNet Calculated Load (for individual dwelling unit) 18,782 VA
    Size of Each FeederFor 120/240-V, 3-wire system,Net calculated load of 18,782 VA ÷ 240 V = 78 AFeeder Neutral (see 220.61)
    Lighting and Small Appliance 3,882 VARange Load: 6400 VA at 70% [see 220.61(B)] 4,480 VASpace and Water Heating (no neutral): 240 V 0 VAimageNet Calculated Load (neutral) 8,362 VA
    Calculated Load for Neutral8362 VA ÷ 240 V = 35 ΑMinimum Size Feeder Required from Service Equipment to Meter Bank (For 20 Dwelling Units)
    Total Calculated Load:Lighting and Small-Appliance LoadElectric heat: 6 kVA (or air conditioning if larger)6,000 VA
    image20 units × 5520 VΑ 110,400 VAWater and Space Heating Load20 units × 8500 VΑ 170,000 VAElectric water heater 2,500 VA
    Minimum Number of Branch Circuits Required for Each Dwelling UnitGeneral Lighting Load: 2520 VA ÷ 120 V = 21 A or two 15-A, 2- wire circuits, or two 20-A, 2-wire circuitsSmall-Appliance Load: Two 2-wire 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 three10 AWG conductors in accordance with 210.19(A)(3) Space Heating: 6000 VA ÷ 240 V = 25 ANumber of circuits (see 210.11)Range Load: 20 × 8000 VΑ 160,000 VAimageNet 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 AMinimum Size Main Feeder Required (Less House Load) (For 40 Dwelling Units)
    Calculated Load:Lighting and Small-Appliance Load40 units × 5520 VΑ 220,800 VAMinimum Size Feeder Required for Each Dwelling Unit (see 215.2)Water and Space Heating Load 40 units × 8500 VΑ340,000 VA
    Calculated Load (see Article 220):General Lighting 2,520 VASmall Appliance (two 20-A circuits) 3,000 VAimageRange: 40 ranges × 8000 VΑ 320,000 VA Net Calculated Load (40 dwelling units) 880,800 VAimageNet Calculated Load Using Optional Calculation (see TableSubtotal Calculated Load (without rangeand space heating)
    Application of Demand Factor5,520 VA220.84)880,800 VA × 0.28 = 246,624 VA246,624 VA ÷ 240 V = 1028 AFeeder Neutral Load for Feeder from Service Equipment toFirst 3000 VA at 100% 3,000 VA5520 VA – 3000 VA = 2520 VA at 35% 882 VAimageNet Calculated Load 3,882 VA (without range and space heating)(continues)Meter Bank (For 20 Dwelling Units)
    Lighting and Small-Appliance Load20 units × 5520 VΑ 110,400 VA(continues)
    First 3000 VA at 100% 3,000 VA Net calculated load of 3882 VA ÷ 2 legs ÷ 120 V/leg = 16 A110,400 VA – 3000 VA = 107,400 VAat 35%37,590 VAFor 120/208-V, 3-wire system (with ranges),
    imageNet Calculated Load 40,590 VA20 ranges: 35,000 VA at 70% 24,500 VANet calculated load (range) of 8000 VA ÷ 208 V = 39 ATotal load (range + lighting) = 39 A + 16 A = 55 A[see Table 220.55 and 220.61(B)]
    65,090 VA ÷ 240 V = 271 AFurther Demand Factor [see 220.61(B)]
    imageTotal 65,090 VAReducing 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/120-V, 3-phase, 4-wire system,
    First 200 A at 100% 200 ABalance: 271 A – 200 A = 71 A at 70% 50 AimageTotal 250 A
    Feeder Neutral Load of Main Feeder (Less House Load) (For 40 Dwelling Units)
    Lighting and Small-Appliance Load40 units × 5520 VΑ 220,800 VAFirst 3000 VA at 100% 3,000 VANext 120,000 VA – 3000 VA = 117,000 VA at 35% 40,950 VARemainder 220,800 VA – 120,000 VA = 100,800 VA at 25,200 VA25%imageRanges: Maximum number between any two phase legs = 4 2 × 4 = 8.Table 220.55 demand = 23,000 VAPer phase demand = 23,000 VA ÷ 2 = 11,500 VA Equivalent 3-phase load = 34,500 VANet Calculated Load (total):40,590 VA + 34,500 VA = 75,090 VA75,090 VA ÷ (208 V)(1.732) = 208 AFeeder Neutral SizeNet Calculated Lighting and Appliance Load & EquivalentNet Calculated Load 69,150 VA Range Load:40 ranges: 55,000 VA at 70% [see Table 220.55 and 220.61(B)]38,500 VA40,590 VA + (34,500 VA at 70%) = 64,700 VA
    imageTotal 107,650 VA
    107,650 VA ÷ 240 V = 449 A
    Further Demand Factor [see 220.61(B)]First 200 A at 100%Balance: 449 – 200 A = 249 A at 70%
    200 A174 A
    Total374 A
    Example D5(a) Multifamily Dwelling Served at 208Y/120 Volts, Three Phase
    All conditions and calculations are the same as for the multi‐ family 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 120/208-V, 3-wire system (without ranges),Net Calculated Neutral Load:64,700 VA ÷ (208 V)(1.732) = 180 AMinimum Size Main Feeder (Less House Load) (For 40 Dwell‐ ing Units — 20 with Ranges)For 208Y/120-V, 3-phase, 4-wire system, Ranges:Maximum number between any two phase legs = 7 2 × 7 = 14.Table 220.55 demand = 29,000 VAPer phase demand = 29,000 VA ÷ 2 = 14,500 VA Equivalent 3-phase load = 43,500 VANet Calculated Load (total):69,150 VA + 43,500 VA = 112,650 VA112,650 VA ÷ (208 V)(1.732) = 313 AMain Feeder Neutral Size:69,150 VA + (43,500 VA at 70%) = 99,600 VA99,600 VA ÷ (208 V)(1.732) = 277 AFurther Demand Factor (see 220.61)
    200 A at 100% 200.0 A277 A – 200 A = 77 A at 70% 54 AimageNet Calculated Load (neutral) 254 A

    Example D5(b) Optional Calculation for Multifamily Dwelling Served at 208Y/120 Volts, Three Phase
    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 ATwo 20-ampere, 2-pole circuits required, 12 AWG conductorsMinimum Size Feeder Required for Each Dwelling Unit120/208-V, 3-wire circuitNet 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 ATotal load = 16.2 A + 71.6 A = 88 AMinimum Size Feeder Required for Service Equipment to Meter Bank (For 20 Dwelling Units)Net Calculated Load167,352 VA ÷ (208 V)(1.732) = 465 AFeeder Neutral Load65,080 VA ÷ (208 V)(1.732) = 181 AMinimum Size Main Feeder Required (Less House Load) (For 40 Dwelling Units)Net Calculated Load246,624 VA ÷ (208 V)(1.732) = 685 AMain Feeder Neutral Load107,650 VA ÷ (208 V)(1.732) = 299 AFurther Demand Factor [see 220.61(B)]
    200 A at 100% 200.0 A299 A – 200 A = 99 A at 70% 69 AimageNet Calculated Load (neutral) 269 A
    Example D6 Maximum Demand for Range Loads
    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 83∕4 kW (and not over 27 kW) is illustrated in the following two examples.
    1. Ranges All the Same Rating (see Table 220.55, Note 1)Assume 24 ranges, each rated 16 kW.From Table 220.55, Column C, the maximum demand for 24 ranges of 12-kW 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 increase39 + 7.8 = 46.8 kW (value to be used in selection of feeders)
    2. Ranges of Unequal Rating (see Table 220.55, Note 2)Assume 5 ranges, each rated 11 kW; 2 ranges, each rated 12 kW; 20 ranges, each rated 13.5 kW; 3 ranges, each rated18 kW.
      5 ranges × 12 kW = 60 kW (use12 kW for range
      rated less than 12)2 ranges× 12 kW =24 kW20 ranges× 13.5 kW =270 kW3 ranges× 18 kW =54 kW30 ranges, Total kW = 408 kW
      408 ÷ 30 ranges = 13.6 kW (average to be used for calcula‐ tion)From Table 220.55, Column C, the demand for 30 ranges of 12-kW 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 increase45 kW + 4.5 kW = 49.5 kW (value to be used in selection of feeders)
      Example D7 Sizing of Service Conductors for Dwell‐ ing(s)
      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 175-ampere 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).imageNWith Required Temperature Correction Factor. If a 175- ampere service rating is selected, a service conductor is then175 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 XHHW-2 conductor, so we use a correc‐
      tion factor of 0.91 to find the minimum conductor ampacity and size:145.25/.91 = 159.6 amperesIn 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)imageAluminum orMotor 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 full-load current [see 430.6(A) and 430.32(A)(1)].For the 25-hp motor, 32 A × 1.25 = 40.0 AFor the 30-hp motors, 38 A × 1.25 = 48 AWhere 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).Service or Feeder Rating100421103112521/0(Amperes)CopperCopper-CladAluminumBranch-Circuit Short-Circuit and Ground-Fault 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 25-hp motor.
      15012/0(a) Nontime-Delay Fuse: The fuse rating is 300% × 34 A =1751/03/0102 A. The next larger standard fuse is 110 A [see 240.6 and2002/04/0430.52(C)(1), Exception No. 1]. If the motor will not start with a2253/0250110-A nontime-delay fuse, the fuse rating is permitted to be2504/0300increased to 125 A because this rating does not exceed 400%300250350350350500400400600
      Example D8 Motor Circuit Conductors, Overload Protection, and Short-Circuit and Ground-Fault Protection
      (see 240.6, 430.6, 430.22, 430.23, 430.24, 430.32, 430.52, and430.62, Table 430.52, and Table 430.250)Determine the minimum required conductor ampacity, the motor overload protection, the branch-circuit short-circuit and ground-fault protection, and the feeder protection, for three induction-type motors on a 480-V, 3-phase feeder, as follows:
    1. One 25-hp, 460-V, 3-phase, squirrel-cage motor, name‐ plate full-load current 32 A, Design B, Service Factor 1.15
    2. Two 30-hp, 460-V, 3-phase, wound-rotor motors, name‐ plate primary full-load current 38 A, nameplate secondary full- load current 65 A, 40°C rise.Conductor Ampacity The full-load current value used to deter‐ mine the minimum required conductor ampacity is obtained from Table 430.250 [see 430.6(A)] for the squirrel-cage motor and the primary of the wound-rotor motors. To obtain the minimum required conductor ampacity, the full-load current is multiplied by 1.25 [see 430.22 and 430.23(A)].For the 25-hp motor, 34 A × 1.25 = 43 AFor the 30-horsepower motors, 40 A × 1.25 = 50 A65 A × 1.25 = 81 A[see 430.52(C)(1), Exception No. 2(a)].(b) Time-Delay 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 60-A time-delay 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 Short-Circuit and Ground-Fault Protection (a) Exam‐ ple using nontime delay fuse. The rating of the feeder protec‐ tive device is based on the sum of the largest branch-circuit protective device for the specific type of device protecting the feeder: 300% × 34 A = 102 A (therefore the next largest stand‐ ard size, 110 A, would be used) plus the sum of the full-load 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)].
    1. Example using inverse time circuit breaker. The largest branch-circuit potective device for the specific type of device protecting the feeder, 250% × 34 A = 85. The next larger stand‐ ard size is 90 A, plus the sum of the full-load currents of the other motors, or 90 A + 40 A + 40 A = 170 A. The nearest stand‐ ard inverse time circuit breaker that does not exceed this value is 150 A [see 240.6 and 430.62(A)].
      Example D9 Feeder Ampacity Determination for Gener‐ ator Field Control
      [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 460-V 3-phase, 60- Hz ac feeder supplying a group of six elevators. The 460-V 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 30-hp, 40-A, 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:
      1. In accordance with 620.13(D) and 620.61(B)(1), use Table 430.22(E), for intermittent duty (elevators). For intermit‐ tent duty using a continuous rated motor, the percentage of nameplate current rating to be used is 140%.
      2. For the 30-hp ac drive motor, 140% × 40 A = 56 A
      3. For the 40-hp ac drive motor, 140% × 52 A = 73 A(I)
      4. The total conductor ampacity is the sum of all the motor currents:(1 motor × 73 A) + (5 motors × 56 Α) = 353 Α
      5. In accordance with 620.14 and Table 620.14, the conduc‐ tor (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.
      6. In accordance with 430.24 and 215.3, the controller continuous current is 125% × 10 A = 13 A
      7. 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
      8. This ampacity would be permitted to be used to select the wire size.See Figure D9.
        Example D10 Feeder Ampacity Determination for Adjustable Speed Drive Control [see 215.2, 430.24, 620.13, 620.14, 620.61, and Table 430.22(E)]
        Determine the conductor ampacity for a 460-V, 3-phase, 60-Hz ac feeder supplying a group of six identical elevators. The system is adjustable-speed SCR dc drive. The power transform‐ ers 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 AmpacityConductor ampacity is determined as follows:
    1. Calculate the nameplate rating of the transformer:
      [D10]image3I      95 kVA  1000  133 A 460 V  0.90eff.
      imageMachine roomTo additional elevatorTo additional elevatorFeeder panelMotor controller
      620.13(A)
      Motion controller
      Generator field control system OperationcontrollerAC G MMG set620.13(A)Operating devicesCWTTo additional elevatorCar
      Machine room branch- circuit panel620.13(D) Feeder620.13(B)Machine room
      FIGURE D9 Generator Field Control.
    2. 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
    3. In accordance with 620.14 and Table 620.14, the conduc‐ tor (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.
    4. In accordance with 430.24 and 215.3, the controller continuous current is 125% × 10 A = 13 A.
    5. 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
    6. This ampacity would be permitted to be used to select the wire size.See Figure D10.
      Example D11 Mobile Home
      (see 550.18)A mobile home floor is 70 ft by 10 ft and has two small appli‐ ance circuits; a 1000-VA, 240-V heater; a 200-VA, 120-V exhaust fan; a 400-VA, 120-V dishwasher; and a 7000-VA electric range.

      imageMachine roomTo additional elevator Operation controllerOperating devicesMotion controllerAdjustable-speeddrive system, ac or dc620.13(A)MMotorcontrollerFeeder panelOptional power transformerCWTTo additional elevator CarTo additional elevator
      Machine room branch- circuit panel(see 552.47)Example D12 Park Trailer620.13(C)620.13(B)A park trailer floor is 40 ft by 10 ft and has two small appli‐ ance circuits, a 1000-VA, 240-V heater, a 200-VA, 120-V exhaust fan, a 400-VA, 120-V dishwasher, and a 7000-VA electric range.
      FeederLighting and Small-Appliance Load620.13(D)Lighting (40 ft × 10 ft × 3 VΑ per ft2) 1,200 VASmall-appliance (1500 VA× 2 circuits) 3,000 VALaundry (1500 VA × 1 circuit) 1,500 VAimageSubtotal 5,700 VAFirst 3000 VA at 100% 3,000 VARemainder (5700 VA – 3000 VA = 2700 VA) × 35% 945 VAimageTotal 3,945 VA3945 VA÷ 240 V = 16.44 A per leg
      Amperes per Leg
      Leg ALeg BLighting and appliances
      1616Heater (1000 VA ÷ 240 V)
      44Fan (200 VA × 125% ÷ 120 V)
      2—Dishwasher (400 VA ÷ 120 V)
      —3Range (7000 VA × 0.8 ÷ 240 V)
      2323
      Totals4546Machine room
      FIGURE D10 Adjustable Speed Drive Control.
      Lighting and Small-Appliance Load
      Lighting (70 ft × 10 ft × 3 VΑ per ft2) 2,100 VASmall-appliance (1500 VA× 2 circuits) 3,000 VALaundry (1500 VA × 1 circuit) 1,500 VAimageSubtotal 6,600 VAFirst 3000 VA at 100% 3,000 VA
      Based on the higher current calculated for either leg, a mini‐ mum 50-A supply cord would be required.For SI units, 0.093 m2 = 1 ft2 and 0.3048 m = 1 ft.
      Example D13 Cable Tray CalculationsRemainder (6600 VA – 3000 VA = 3600 VA) × 35%
      4260 VA ÷ 240 V = 17.75 A per leg1,260 VA
      imageTotal 4,260 VA(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.Lighting and appliances1818
      Heater (1000 VA ÷ 240 V)44(OD)(N)SD = (OD) × (N)Fan (200 VA × 125% ÷ 120 V)2—Cable SizeCable OutsideNumber of(Sum of the CableDishwasher (400 VA ÷ 120 V)—3Being UsedDiameters (in.)CablesDiameters) (in.)Range (7000 VA × 0.8 ÷ 240 V) 23 23
      3–conductor1.571218.84Total amperes per leg 47 48
      Based on the higher current calculated for either leg, a mini‐ mum 50-A supply cord would be required.For SI units, 0.093 m2 = 1 ft2 and 0.3048 m = 1 ft.
      Type MC cable — 4/0 AWG




















      Amperes per Leg Leg A Leg BExample: Cable tray width is obtained as follows:
      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 cross-sectional 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.
      imageTable D13(b) from Table 392.22(A), Column 1

      Sd = (OD) ×(N)Single

      (Sum of theConductor(OD)(N)CableCable SizeCable OutsideNumber ofDiameters)Being UsedDiameters (in.)Cables(in.)THHN —0.6421811.5564/0 AWG
      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.Note: Cable outside diameter from Chapter 9, Table 5.D.13(d) Single Conductor Cables 250 kcmil through 900 kcmilInside Width of Cable Tray (in.)Allowable Cable Area (in.2)Use: NEC 392.22(B)(1)(b)imageThe sum of the cross-sectional areas of all the cables to be6 7.09 10.512 14.02428.03035.03642.018 21.0installed 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
      image
      Example: Cable tray width is obtained as follows:Inside Width of Cable Tray (in.)Allowable Cable Area (in.2)
      imageCable Size Being Used
      1. Cable Cross- Sectional Area (in.2)
        (N)Number of Cables
        Multiply (A) ×
  2. (Which Is a Total Cable

Cross-Sectional Area in in.2)

6 6.5

image

9 9.5

12 13.0

18 19.5

24 26.0

30 32.5

36 39.0

4-conductor Type MC cable —

1 AWG

1.1350 9 12.15

Example: Cable tray width is obtained as follows:

Multiply (A) ×

The total cable cross-sectional 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.

image

Cable Size Being Used

(A)

Cable Cross- Sectional Area (in.2)

(N)

Number of Cables

(N) (Which Is a Total Cable

Cross-Sectional Area in in.2)

Note: Cable cross-sectional area is a nominal area from cata‐ log data.

D13(c) Single Conductor Cables 1/0 AWG through 4/0 AWG

THHN —

500 kcmil

0.707 9 6.36

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:

The total cable cross-sectional 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: Single-conductor cable cross-sectional area from Chap‐ ter 9, Table 5.

Informative Annex E Types of Construction

This informative annex is not a part of the requirements of this NFPA document but is included for informational purposes only.

Table E.1 contains the fire resistance rating, in hours, for Types I through V construction. The five different types of construction can be summarized briefly as follows (see also Table E.2):

Type I is a fire-resistive construction type. All structural elements and most interior elements are required to be noncombustible. Interior, nonbearing partitions are permitted to be 1 or 2 hour rated. For nearly all occupancy types, Type I construction can be of unlimited height.

Type II construction has three categories: fire-resistive, one- hour rated, and non-rated. The number of stories permitted for multifamily dwellings varies from two for non-rated and four for one-hour rated to 12 for fire-resistive construction.

Type III construction has two categories: one-hour rated and non-rated. Both categories require the structural framework and exterior walls to be of noncombustible material. One-hour

rated construction requires all interior partitions to be one- hour rated. Non-rated construction allows nonbearing interior partitions to be of non-rated construction. The maximum permitted number of stories for multifamily dwellings and other structures is two for non-rated and four for one-hour rated.

Type IV is a single construction category that provides for heavy timber construction. Both the structural framework and the exterior walls are required to be noncombustible except that wood members of certain minimum sizes are allowed. This construction type is seldom used for multifamily dwellings but, if used, would be permitted to be four stories high.

Type V construction has two categories: one-hour rated and non-rated. One-hour rated construction requires a minimum of one-hour rated construction throughout the building. Non- rated construction allows non-rated interior partitions with certain restrictions. The maximum permitted number of stories for multifamily dwellings and other structures is two for non- rated and three for one-hour rated.

Table E.1 Fire Resistance Ratings for Type I Through Type V Construction (hr)

Type IType IIType IIIType IVType V
4423322221110002112002HH111000
Exterior Bearing Walls aSupporting more than one floor, columns, or other bearing wallsSupporting one floor only Supporting a roof only
4
3
2
1
0b
2
2
2
1
0b
443321110b 0b222222110b 0b
Interior Bearing WallsSupporting more than one floor, columns, or other bearing wallsSupporting one floor only Supporting roofs only
4
3
2
1
0
1
0
2
1
0
33222111001100111100
ColumnsSupporting more than one floor, columns, or other bearing wallsSupporting one floor only Supporting roofs only
4
3
2
1
0
1
0
H
1
0
33222111001100H H1100
Beams, Girders, Trusses, and ArchesSupporting more than one floor, columns, or other bearing wallsSupporting one floor only Supporting roofs only
4
3
2
1
0
1
0
H
1
0
22222111001100H H1100
Floor/Ceiling Assemblies2221010H10
Roof/Ceiling Assemblies211∕211010H10
Interior Nonbearing Walls0000000000
Exterior Nonbearing Walls c0b0b0b0b0b0b0b0b0b0b

Source: Table 7.2.1.1 from NFPA 5000, Building Construction and Safety Code, 2012 edition. H: Heavy timber members.

aSee 7.3.2.1 in NFPA 5000.

bSee Section 7.3 in NFPA 5000.

cSee 7.2.3.2.12, 7.2.4.2.3, and 7.2.5.6.8 in NFPA 5000.

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