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Dwelling Units - NEC Standard Method Calculation
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Note: If a dwelling has more than one feeder, a separate load
calculation is needed for each feeder.
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Step-1 : Calculating general lighting and general receptacles
loads (except for small-appliance and Laundry Receptacles)
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Procedure
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Note
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Calculation
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From NEC Table 220.12,
for dwelling units, minimum general lighting load is 3 VA/ft2.
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Under any conditions, for
dwelling units, don’t use value less than 3 VA/ft2, there are no exceptions.
While the designer can choose a higher value based on the existing design
conditions
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The general lighting load
is calculated by multiplying the floor area (in ft2) of a dwelling unit by 3
VA/ft2.
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Calculate the floor area for each floor of Dwelling Unit in ft2.
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The floor area for each
floor shall be calculated from the outside dimensions of the dwelling unit.
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Floor area in ft2 =
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0
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ft2
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The calculated floor area
shall not include open porches, garages, or unused or unfinished spaces not
adaptable for future use (like some attics, cellars, and crawl spaces).
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General lighting and
general receptacles loads =
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0
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VA
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Step-2: Calculating Small-appliance branch circuits’ load
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Procedure
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Note
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Calculation
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Calculate the required
number of small-appliance branch ciruits in the dwelling unit
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As per NEC 210.1(C)(1),
In each dwelling unit, two or more 20-ampere small-appliance branch circuits
must be provided.
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The Small-appliance branch
circuits’ load, for dwelling units, is calculated by multiplying number of Small-appliance branch
circuits by 1,500 VA.
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the designer assign the
number of small-appliance branch ciruits based on the exisiting condition
(space dimension , number of required small appliances, etc.).
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As per 210.52(B)(1)Exception.2, An indvidual branch ciruit is
permitted for Refrigeration equipment at 1,500 VA. If you will apply this
rule,choose Number of Refergiator Circuits from F13
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2
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Don't exceed the
permissible loading of a 120 V, 20-
ampere branch circuit which is 2400 VA
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If a dwelling has more than one feeder, a separate load
calculation is needed for each feeder. It is not necessary to include
small-appliance branch circuit loads for feeders don’t supply such load. Does
feeder include small-appliance branch circuits?Select answer from F14
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N
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As per NEC section
220.52(A), each 2-wire small-appliance branch circuit load is calculated at
no less than 1,500 volt-amperes.
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Number of Small-appliance
branch circuits in the dwelling unit =
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0
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Ciruit
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0
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Small-appliance branch circuits’
load =
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0
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VA
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0
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Step-3: Calculating Laundry branch ciruit load
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Procedure
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Note
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Calculation
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Calculate the required number of laundry branch ciruits in the
dwelling unit
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As per NEC 210.11(C)(2),
In each dwelling unit, At least one 20-ampere branch circuit shall be
provided
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The Laundry branch circuits’ load,
for dwelling units, is calculated by multiplying number of Laundry branch circuits by 1,500
VA.
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As per NEC 220.52(B),
each 2-wire laundry branch circuit is calculated at no less than 1,500
volt-amperes.
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In multifamily dwelling
building, if Laundry facilities are provided on the premises and available to
all building tenants (as a common usage), Laundry branch circuits load will
not be added to each individual dwelling unit and will be added to a separate
“house load” panelboard. is Laundry Facility available to all building
tenants?Select answer from F21
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N
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A combination of clothes
washer and clothes dryer will be handled in calculations as it is a clothes
dryer, is there a combination? Select answer from F22
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Y
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Number of Laundry branch
circuits in the dwelling unit =
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1
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Ciruit
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1
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Laundry branch
circuits’ load =
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0
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VA
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Step-4: Applying Demand Factors from Table 220.42
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Sum Loads of Step-1,
Step-2 and Step-3
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Total Sum =
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0
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VA
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Calculate the demand of
The First 3,000 VA or Less at 100%
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demand of The First 3,000 VA or Less =
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0
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VA
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Calculate the demand of (
120,000 VA - 3,000 VA), if any, at 35%
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demand of ( 120,000 VA - 3,000 VA), if any =
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0
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VA
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Calculate the demand of
the reminder over 120,000 VA, if any, at 25%
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demand of the reminder over 120,000 VA, if any =
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0
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VA
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General Load for
Lighting, General Receptacles, Small Appliances and Laundry =
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0
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VA
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Step-5: Fastened in Place Appliances Load
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As per NEC section 220.53,
electric ranges, clothes dryers, space-heating equipment or air conditioning
equipment must not be included with the number of appliances that are
fastened in place. Also, All portable small Appliances for kitchen and others
are not Fastened-in-Place Appliances.
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Kilovolt-amperes (kVA)
shall be considered equivalent to kilowatts (kW)
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As per NEC section,
220.53, It shall be permissible to apply a demand factor of 75 % to the
nameplate rating load of four or more appliances fastened-in-place, that are served by the same feeder or
service in a one-family, two-family, or multifamily dwelling.
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In a multifamily
dwelling, the four or more fastened-in-place appliances do not have to be on
the same feeder for each dwelling unit. In this case, the 75% demand factor
will not apply to the feeder for each dwelling unit but it must be applied to
the multifamily dwelling service.
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As per NEC section
430.6(A)(1), Do not use the actual current rating marked on the nameplate.
When calculating motor loads, use the values given in Tables 430.247 through
430.250.
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Number of appliances
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Rating of appliances
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water heater
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Refrigerator
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Freezer
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dishwasher
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disposal
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Range hood
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microwave
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mini Refrigerator
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inst hot
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ironing center
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wine Clr
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Add more Appliances
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Add more Appliances
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Add more Appliances
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total number of appliances =
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0
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Total ratings of appliances
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0
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Total load =
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0
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VA
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Step-6: Clothes Dryers Load
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A clothes dryer is not a
requirement for a load calculation, Skip this step if there is no clothes
dryer.
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Kilovolt-amperes (kVA)
shall be considered equivalent to kilowatts (kW)
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Procedure
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Note
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Calculation
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Write the Nameplate Rating of
Cothes dryer
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As per NEC section
220.54, the load for household electric clothes dryers in a dwelling unit(s)
shall be either 5000 watts (volt-amperes) or the nameplate rating, whichever
is larger, for each dryer served.
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Write the Nameplate
Rating of Cothes dryer
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0
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VA
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A combination of clothes
washer and clothes dryer (see below image) will be handled in calculations as
it is a clothes dryer.
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5000 VA or the nameplate
rating, whichever is larger
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5000
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VA
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For single, two-family or
multi-family dwelling when each tenant has separate clothes dryer, Table
220.54 Demand Factors for Household Electric Clothes Dryers will be used
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Number of Clothes Dryers
=
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0
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In a multi-family
dwelling where there is a common Laundry area, use the full load of all
dryers without applying demand factor.
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Total Conneced load of
clothes dryers =
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0
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VA
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0
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0
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0
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0
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0
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0
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0
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0
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0
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0
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0
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0
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0
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Demand Factor =
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0
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Total Demand load of
clothes dryers =
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0
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VA
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Step-7: Household cooking appliances load
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We can skip the
calculation of Household Cooking Appliances Load if there are no cooking
appliances rated over 1.75 KW.
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Kilovolt-amperes (kVA)
shall be considered equivalent to kilowatts (kW)
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When the kilowatt rating
fraction is 0.5 or more, it must be rounded up to the next whole kilowatt
rating i.e. 14.5 KW up to a 15 KW and
When the fraction is less than 0.5, it can be dropped i.e. 14.4 KW dropped to
14 KW.
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The table 220.55 is not
applicable for ranges rated more than 27 kW because ranges rated more than 27
kW would not be considered household ranges.
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Case#1: Individual
Appliance
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Household cooking appliances Rating in KW=
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0
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Less that 3.5 KW Rating
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3.5-8.5 KW
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less than 12 KW
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12KW<rating<27KW
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0
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0
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0
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0
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Household cooking
appliances Demand load in KW=
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0
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Case#2: group of Appliances
with equal (same) ratings not over 12 KW
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Number of Household cooking appliances =
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0
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Household cooking appliances Rating in KW=
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0
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Put Demand/max. Demand
Factor from table 220.55
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Calculated demand load
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Less that 3.5 KW Rating
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0
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0
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3.5-8.5 KW
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0
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0
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less than 12 KW
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0
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Household cooking
appliances Demand load in KW=
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0
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Case#3: group of Appliances
with unequal ratings not over 12 KW
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group Rating #1
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group Rating #2
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group Rating #3
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Number of Household
cooking appliances =
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Household cooking
appliances Rating in KW=
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IF Less that 3.5 KW Rating, D.F =
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IF from 3.5 to 8.5 KW, D.F=
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IF less than 12 KW, Max. Demand =
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Calculated demand load
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0
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0
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0
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0
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0
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0
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Household cooking
appliances Demand load in KW=
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0
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Case#4: group of Ranges
with equal (same) ratings Over 12 kW through 27 KW
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Number of Household cooking appliances =
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Household cooking appliances Rating in KW=
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Put Demand/max. Demand
Factor from table 220.55
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Calculated demand load
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0
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Household cooking
appliances Demand load in KW=
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0
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Case#5: group of Ranges
with unequal ratings Over 8 3⁄4 kW through 27 kW
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total Number of Household cooking appliances =
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Max. Demand from Column
C, Table 220.55=
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Ranges Rating over 12 KW
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Ranges Rating Below 12 KW
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Number of Household
cooking appliances =
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Sum of Household cooking
appliances Rating in KW=
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the Average Rating =
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#DIV/0!
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Household cooking
appliances Demand load in KW=
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#DIV/0!
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Step-8:Heating and air conditioning loads
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the blower motor works
with both the heating and air conditioning system, it must be included in
both calculations.
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With a heat pump, the
compressor (and accompanying motors) and some or all of the electric heat can
be on at the same time. The load contribution of a heat pump is the air
conditioning system load plus the maximum amount of heat that can be on while
the air conditioner compressor is on.
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As per NEC section
430.6(A)(1), Do not use the actual current rating marked on the nameplate.
When calculating motor loads, use the values given in Tables 430.247 through
430.250.
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Noncoincident
Loads
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coincident
Loads
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Room air conditioners Load in VA at 100%
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FALSE
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Fixed electric space-heating Load in VA at 100%
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FALSE
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Central air conditioning Load in VA at 100%
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FALSE
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Central heating system Load in VA at 100%
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FALSE
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Heating and air
conditioning load in VA =
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0
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0
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Step-9: The Largest Motor
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As per NEC section
430.6(A)(1), Do not use the actual current rating marked on the nameplate.
When calculating motor loads, use the values given in Tables 430.247 through 430.250.
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Exceptions to 430.6(A)(1) :
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1- Motors built for low speeds
(less than 1,200 rpm) or high torques for multispeed motors.
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2- For equipment that employs a
shaded-pole or permanent-split capacitor-type fan or blower motor that is
marked with the motor type, use the full load current for such motor marked
on the nameplate of the equipment in which the fan or blower motor is
employed.
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3- For a listed
motor-operated appliance that is marked with both motor horsepower and
full-load current, use the motor full-load current marked on the nameplate of
the appliance.
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When calculating a feeder or
service As per NEC Standard calculation method, the largest motor must be
multiplied by 25 percent and add it to the service load calculation.
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If the motor is air
conditioning compressor, usually the air conditioning compressor is the
largest motor in dwelling units. in this case, multiply the load of one
compressor by 25 percent and add it to the service load calculation. But if
the heating load is larger than the air conditioning load, and because of
220.60 which states that” it is permissible to use only the larger of the
noncoincident loads” the air conditioning load will be omitted and the air
conditioning compressor will not be the largest motor in this case.
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VA of Largest Motor =
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Largest
Motor Additional Load =
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0
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VA
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TOTAL DEMAND LOAD =
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#DIV/0!
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VA
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