Docdownloader - once upon a time a few mistakes ago once upon a time a few mistakes ago PDF

Preview

Summary

once upon a time a few mistakes ago
once upon a time a few mistakes ago...

Description



ELECTRICAL RESIDENTIAL & COMMERCIAL CALCULATIO BONDING & GROUNDING & NEC GUIDELINE

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.



Dwellin llin g Unit Feede Feeder/S r/Se ervi ce Cond Conduct uctor or Ca Calcu lcu lat Dwe

How do you size conductors for r eside esidential ntial services and feede feeders? rs? Whether they're for journeymen, master electricians, or contractors, most electrical licensing exams requir requi calculate residential loads and service or feeders siz sizes es using one of two methods. The standard method, contained in Art. 220, Part II, involves more steps, but many people use it exclusively to avoid using the w method. However, most residential construction qualifies for the optional method in Art. 220, Part III, so it's understand both methods. In either case, you're free to exceed the NEC requirements — these are minim requirements, not design specifications. The standard method is where we'll start. It requires six sets of calculations for general lighting and recept small-appliance, small-appliance, and and laundry; laundry; air air conditioning conditioning versus versus heat; heat; appliances; appliances; clothes clothes dryer; dryer; cooking cooking equipment; equipment; conductor size. The following example should help illustrate how to apply these steps. What size service conductor does a 1,500 sq ft dwelling unit need, if it contains the following loads? The s 120/240V.



Disposal (940VA)



Dishwasher (1,250VA)



Trash compactor (1,100VA)



Water heater (4,500VA)



Dryer (4,000VA)



Cooktop (6,000VA)



Two ovens (each 3,000VA)

•

•

•

•

•

•

•

 Air conditioning (5 hp, 230V)

•



•

Three electric space heating units (each 3,000W)

General lighting and receptacles, small-appliance, and laundry laundry.. The NEC recognizes these circuits won't a ou may apply a demand factor to the total connected general lighting and ervice/feeder demand load, refer to Table 220.11 and follow these steps This website stores data such as

cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

ed load for general lighting and receptacles (3VA per sq ft) [Table 220.3 ,500VA, ,500VA, and and one one laundry laundry circuit circuit at at 1,500VA 1,500VA (220.16) (220.16) ((Fig. Fig. 1). 1).



 Fig. 1. 1.In In calculating conductor sizes fo unit, the NEC recognizes a homeowne all the appliances, turn on all the lights all receptacles at the same time.

General lighting/receptacles: 1,500 sq ft× 3VA=4,500VA Small-appliance Small-appliance circuits: circuits: 1,500VA×2 1,500VA×2 =3,000VA =3,000VA Laundry circuit: 1,500VA×1= 1,500VA Total connected load: load: 4,500VA+ 4,500VA+ 3,000VA+1,500VA=9,000VA 3,000VA+1,500VA=9,000VA First 3,000VA at 100%=3,000VA× 100%=3,000VA× 1.00=3,000VA Remainder at 35%=(9,000VA35%=(9,000VA- 3,000VA)×0.35=2,100VA 3,000VA)×0.35=2,100VA Total demand load=5,100VA Air-conditioning v versus ersus heat. heat. Because Because air-conditioning and heating heating loads aren't on simultaneously, you m smaller of the two loads (220.21). Calculate each of these at 100% (220.15) ((Fig. Fig. 2). 2). Air conditioning: 5 h hp, p, 230V VA=E×I (Table 430.148) 28 FL×230V=6,440 VA Heat: 3,000W×3 units=9,000W

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

Fig. 2.It 2.It should be obvious that the he cooling systems won’t be generated at time. Therefore, your calculations shou on the larger of the two loads.



3.When When determining load for for appliances, the Code allows you to use a 75% demand factor when fou Fig. 3. fastened-in-place appliances are on the same feeder. Appliances Appliances.. Per 220.17, you can use a 75% demand factor when four or more “fastened in place” applian as a dishwasher or waste disposal, are on the same feeder. Don't include clothes dryers, cooking equipm conditioning, or heat in this category ((Fig. Fig. 3). 3). Waste disposal: 940VA Dishwasher: 1,250VA Trash compactor: 1,100VA Water heater: 4,500VA Total Total connected connected appliance appliance load: load: 7,790VA× 7,790VA× 0.75=5,843VA 0.75=5,843VA

Fig. 3. 3.When When determining load for ap the Code allows you to use a 75% de factor when four or more fastened-in appliances are on the same feeder.

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

eder or service demand load for electric clothes dryers in a d dwelling welling unit e nameplate rating exceeds 5,000W, use that rating as the load. You can ric dryer provision. However, it's common to provide both gas and electr y electric will be omitted (Fig. (Fig. 4 4). ). Fig. 4.Electric 4.Electric clothes dryer demand accounted for as a separate load item



The service and feeder demand load for a 4kW dryer is 5,000W. Cooking equipment equipment.. For For household-cooking household-cooking appliances appliances rated higher higher than 1.75kW, you can use the dema listed in 220.19, Table and Notes 1, 2, and 3. All three of the the units in the example example are rated higher than 1.75kW and not higher than 8.75kW, so follow t instructions in Note 3. The two ovens are rated less than 3.5kW, so Table 220.19 Column A demand facto The cooktop is 6kW, so Column B demand factor applies ((Fig. Fig. 5). 5). Fig. Table 220.19 torated determin 5. load 5.See forSee cooling equipment high 1.75kW. 1.75kW.

Column A demand: 3kW×2 units× 0.75 demand factor=4.5kW Column B demand: 6kW×1 unit× 0.8=4.8kW Demand load=4.5kW+4.8kW= 9.3kW=9,300W

Feeder and service conductor size size.. 400A and less: For 33-wire, wire, 120/24 phase systems, size the feeder or service conductors conductors according to to Table 310.15(B)(6). For all others, use 310.16. 310.16. Size Size the the grounded grounded (neutral) (neutral) conductor conductor to to the the maximum maximum unbalanced unbalanced load load (220.22) (220.22) per per Table Table 310. Over Over 400A: 400A: Size Size the the ungrounded ungrounded and and grounded grounded (neutral) (neutral) conductors conductors per per Table 310.16. loads from steps 1 through 5.

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.



Optional method. method.You You can use the easier optional method found in 220.30 only when the total connected connecte served by a single 33-wire, wire, 120/240V or 208Y/120V set of service or feeder conductors with an ampacity o greater. Because this condition describes the typical residential service, the optional method is likely to ap can simplify the design process and save you time because you have s so o many fewer sets of calculations. General loads loads.. The calculated load shall not be less than 100% for the first 10kW, plus 40% of the remain following loads:



Small-appliance Small-appliance and and laundry laundry branch branch circuits: circuits: 1,500VA 1,500VA for for each each 20A 20A circuit. circuit.



General lighting and receptacles: 3VA per sq ft

•

•

 Appliances: The nam nameplate eplate VA rating o off all appliances and moto motors rs fastened in place (permanent (permanently ly

•

or on a specific circuit. Be sure to use the range and dr dryer yer at nameplate rating. HVAC HVAC.. Include the largest of the following:



100% of the nameplate rating of the air-conditioning equipment.



100% 100% of of the the heat-pump heat-pump compressors compressors and and supplemental supplemental heating, heating, unless unless the the controller controller prevents prevents sim sim operation of the compressor and supplemental heating.



100% of the nameplate ratings of electric thermal storage and other heating ssystems ystems where you ex usual load to be continuous at the full nameplate value. Don't configure such systems under any ot selection in this table.

•

•

•

•



65% of the nameplate the central electric space heating, heating, including integral integral supplementa supplement heat heat pumps pumps where where the the rating(s) controller controllerofprevents prevents simultaneous simultaneous operation operation of of including the the compressor compressor and and suppleme suppleme heating.



65% of the nameplate rating(s) of electric space heating, if there are less than four separately cont



40% of the nameplate rating(s) of electric space heating of four or more separately controlled units

•

•

Sizing service/feeder conductors. conductors. Now that we've seen how to determine residential loads, let's s size ize the service/feeder conductors. We'll use the same specifications that we used for the standard method so we compare apples to apples. Step 1:Determine 1: Determine general loads [230.30(B)]. Small appliance: 1,500VA×2 circuits = 3,000VA General lighting: 1,500 sq ft×3VA= 4,500VA

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

.

A)



Step 2:Compare 2:Compare air conditioner at 100% vs. heat at 65% [220.30(C)]. Air conditioner: 230V 230V×28A=6,440VA ×28A=6,440VA Heat [220.30(C)(5)]: 9,000 W×0.65 = 5,850 W (omit) Step 3:Calculate 3: Calculate service/feeder conductors per 310.15(B)(6). General loads=19,116VA Air conditioning=6,440V conditioning=6,440VA A Total demand load=25,556VA I=VA÷E=25,556VA ÷ 240V = 106.5A 310.15(B)(6) requires at least a 110A service with 3 AWG conductors. As you can see, in this this case the optional method method permitted a smal smaller ler service than the standard m method ethod of a service for a dwelling. Now that we've walked through the process of calculating residential services and feeders, you can see th is fairly easy. You need to calculate the loads first, and then move on to the service and feeder size. The N provides the requirements in Art. 220 and 230. Doing these calculations correctly can save you money du and construction, while providing safe homes for the families who occupy them.

Ap emand and f ac actt o r s f o r c o r r ect ec t l o ad c alc al c u l at atii o n s A p p l y d em A dwelling unit is a single unit that provides complete and independent living facilities, accordin according g to the N found in Art. 100 ((Fig. Fig. 1 ). ). Fig. 1.The 1.The definition of dwelling unit, as desc is found in Art. 100.

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.





•



•



•

Receptacles. Receptacles. You can use 15A or 20A receptacles on 20A circuits as long as there is more than on receptacle on the circuit. For these purposes, a duplex receptacle is considered to be ttwo wo receptac [210.21(B)(3)]. Continuous loads loads.. A continuous load is one in which the maximum current is expected to continue more, according to the Art. 100 definition. Fixed electric heating is one example of a continuous loa [424.3(B)]. When sizing branch circuit conductors and overcurrent devices for a continuous load, m load load by by 125% 125% [210.19(A)(1) [210.19(A)(1) and and 210.20(A)]. 210.20(A)]. Laundry rooms. rooms. A laundry area receptacle is required [210.52(F)], at least one of which must be wit wi washing [210.50(C)]. Any receptacle within 6 ft of the outside edge of a laundry sink must protectedmachine [210.8(A)(7)].

Required Required circu its.In its. In addition to the circuits required for dedicated appliances and those needed to serve lighting and receptacle load, a dwelling unit must have the following circuits:

 A minimum of two 20A, 120V small-appliance branch circuits for recep receptacles tacles in the kitchen, dining dining

•



•

breakfast room, pantry, or similar dining areas [220.11(C)(1)]. These circuits must not be used to se outlets, such as lighting outlets or receptacles from other areas [210.52(B)(2) Ex]. These circuits ar in the feeder/service calculation at 1,500VA for each circuit [220.52(A)]. One 20A, 120V branch circuit for the laundry receptacle(s). It can’t serve any other outlet(s), such a and can serve only receptacle outlets in in the laundry laundry area [210.52(F) and 210.11(C)(2)]. 210.11(C)(2)]. In your feed fee load load calculation, calculation, include include 1,500VA 1,500VA for for the the 20A 20A laundry laundry receptacle receptacle circuit circuit [220.52(B)], [220.52(B)], as as shown shown in in Fi

Fig. 2. 2.Per Per Sec. Sec. 210.11(C)(2), 210.11(C)(2), one 20A, 20A, 120V circuit is required for the laundry area rec recepta epta

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

ulations.Occupants ulations. Occupants don’t don’t use use all all loads loads simultaneously simultaneously under normal livin ctors” can be applied to many of the dwelling unit loads in order to size th



1.

2.

3.

General lighting VA load load.. When calculating branch circuits and feeder/service loads for dwellings, i minimum minimum 3VA 3VA per per sq sq ftft for for general general lighting lighting and and general-use general-use receptacles receptacles [220.12]. [220.12]. When When determinin determinin use the outside dimensions of the dwelling. Don’t include open porches, garages, or spaces not ad future use. Small appliance and laundry circuits circuits.. The 3VA per sq ft rule includes includes general general lighting and all all 15A an a 125V 125V general-use general-use receptacles, receptacles, but but doesn’t doesn’t include include small-appliance small-appliance or or laundry laundry circuit circuit receptacles. receptacles. TT you must calculate those at 1,500VA per circuit. See 220.14(J) for details. Number of branch circuits. circuits. Determine the number of branch circuits required for general general lighting lighting an use receptacles from the general lighting load and rating of the circuits [210.11(A)]. Although this is in Annex D, Example D1(a) of the NEC, let’s look at an another example.

Question: Question:What’s What’s the general lighting lighting and receptacle load for a 2,000-sq-ft dwelling unit that has 34 conve receptacles and 12 luminaires rated 100W each ((Fig. Fig. 3)? Fig. 3.Sample 3. Sample calculation showing how to fol rules rules in in Sec. Sec. 220.12 220.12 regarding regarding general general lighting lightin receptacles for a 2,000-sq-ft dwelling unit.

The calculation is pretty simple. 2,000 sq ft x 3VA = 6,000VA. No No additional additional load load is is required required for for general-use general-use receptacles receptacles and lighting outlets outlets because they they are included included

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

220.12 for dwelling units. See 220.14(J). le to determine the number of circuits required. are required for a 2,000-sq-ft dwelling unit? 00 sq ft x 3VA = 6,000VA :



Optional method for feeder and service load calculations calculations You can use the optional method [Art. 220, Part IV] only for dwelling units served by a single 120/240V or 3-wire set of service or feeder conductors with an ampacity of 100A or larger [220.82]. The optional metho of three calculation steps: 1. 2. 3.

General loads [220.82(B)] Heating and air-conditioning load [220.82(C)] Feeder/service conductors [310.15(B)(6)]

Step 1:General 1: General loads [220.82(B)]

The general calculated load must be at least 100% for the first 10kVA, plus 40% of the remainder of the fo loads: 1. 2. 3.

General lighting and receptacles: 3VA per sq ftft Small-appliance and laundry branch circuits: 1,500VA for each 20A, 120V small-appliance and lau circuit specified in 220.52. Appliances: The nameplate VA rating of all appliances and motors that are fastened in place (perm connected) or located on a specific circuit, not including heating or air-conditioning.

Be sure to calculate the range and dryer at their nameplate ratings. ratings. Step 2:Heating 2: Heating and air-conditioning air-conditioning load [220.82(C)] Include the larger of (1) through (6): 1. 2. 3.

4. 5. 6.

Air-conditioning equipment: 100% Heat-pump compressor without supplemental heating: 100% Heat-pump com compressor pressor and supple supplemental mental heating heating:: 100% of the nameplate rat rating ing of the heat-pump compressor and 65% of the supplemental electric heating for central electric space-heating system control circuit is designed so that the heat-pump compressor can’t run at the same time as the supp s up heat, omit the compressor from the calculation. Space-heating units (three o orr fewer separately controlled controlled un units): its): 65%. Space-heating units (f (four our or more separat separately ely controll controlled ed units): 40%. Thermal storage heating: 100%.

Step 3:Feeder/service 3: Feeder/service conductors [310.15(B)(6)]

This website stores data such as cookies to enable essential site functionality, as well as marketing, personalization, and analytics. You may change your settings at any time or accept the default settings.

ual dwelling units of one-family, two-family, and multi-family dwellings, us ,, single-phase, single-phase, 120/240V 120/240V service service or or feeder feeder conductors conductors (including (including neutra he main power feeder. Feeder conductors aren’t required to have an am nductors [215.2(A)(3)]. Size the neutral conductor to carry the unbalanc 310.15(B)(6) 310.15(B)(6) can’t can’t be be used used for for sizing sizing the the feeder feeder or or service conductors th g unit. ed conductors and the neutral conductor conductor using Table 310.16 for feeder/ do not fill all of the requirements for using Table 310 15(B)(6) Let’s try a



Step 1:General 1: General loads [220.82(B)] General lighting: 1,500 sq ft x 3VA = 4,500VA Small-appliance circuits: 1,500VA x 2 circuits = 3,000VA Laundry circuit: 1,500VA Appliances (nameplate): Cooktop: 6,000VA Disposal: 900VA Dishwasher: 1,200VA Dryer: 4,000V...