In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads:
 Generaluse Receptacle Loads,
 Small appliance Loads,
 Laundry Load,
 Cloth dryer Load,
 Household cooking appliances load,
 Fastenedinplace Appliance loads,
 Heating and air conditioning loads,
 Motor loads.
I explained the first seven types in the following articles:
 Receptacle Branch Circuit Design Calculations – Part Four
 Receptacle Branch Circuit Design Calculations – Part Five
 Receptacle Branch Circuit Design Calculations – Part Six
 Receptacle Branch Circuit Design Calculations – Part Seven
 Branch Circuit Design Calculations – Part Eight
 Branch Circuit Design Calculations – Part Nine
 Branch Circuit Design Calculations – Part Ten
 Branch Circuit Design Calculations – Part Eleven
In the following paragraphs, I will explain Where and how to distribute each load outlets in a dwelling building as per NEC code.
You can review the following articles for more information:
8 Motor loads
You can review the following articles for more information:
8 Motor loads
8.1 Applied NEC Rules for Motor loads
There are many NEC rules that control the Motor Loads including:
 220.14(C) Motor Loads for All Occupancies
 220.18(A) Maximum Loads for circuits supplying MotorOperated and Combination Loads.
 220.50 Motors
 422.62 Appliances Consisting of Motors and Other Loads
 220.82 Optional Method  Dwelling Unit
 220.84 Optional Method  Multifamily Dwelling
 430.6 Ampacity and Motor Rating Determination
 430.22 Single Motor
8.2 Calculation of Motor loads
First: As per NEC Standard calculation method
The motors exist in dwelling units as fastened in place appliances or separate motors (air conditioning compressors, fan blower, etc.)
Rule#1: Motor Loads
as per NEC standard method
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.

Important!!!
Most electrical equipment is
rated in voltamperes (VA) or watt input. While motors traditionally have
been rated in horsepower output (Some motors are available with their output
ratings expressed in watts and kilowatts).

Rule#2: Motor Loads
as per NEC standard method
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.

Important!!!
Exceptions to 430.6(A)(1) :

Important!!!

Important!!!
Kilovoltamperes (kVA) shall be considered
equivalent to kilowatts (kW) for Fastenedinplace Appliances.

Important!!!
If the motor is one of the fastenedinplace appliances
, Apply NEC section, 220.53, which states that “ It shall be permissible to
apply a demand factor of 75 % to the nameplate rating load of four or more
appliances fastenedinplace, that are
served by the same feeder or service in a onefamily, twofamily, or multifamily
dwelling”.

Important!!!
As per NEC section 220.53,
electric ranges, clothes dryers, spaceheating 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 FastenedinPlace Appliances.

Important!!!
No derating is allowed when there are only one, two,
three fastenedinplace appliances.

Example#1:
The following appliances will be installed in a onefamily dwelling:
 A dishwasher rated 10 amperes at 120 volts;
 A ½ hp, 120volt kitchenwaste disposer;
 A 4.5 kW, 240volt water heater.
What is the total service load for these appliances?
Solution:
The load for these three appliances must be calculated at 100 percent.
The dishwasher load = 10 A × 120 V = 1,200 VA.
Since the kitchenwaste (or garbage) disposer is a motor, the load must be determined by multiplying the fullload current in Table 430.248 by the system voltage.
In accordance with Table 430.248, the fullload current of a ½hp, 120volt motor = 9.8 A
Note: Although the current value of 9.8 amperes is in the column titled “115 volts,” the currents listed in this column are permitted for system voltage ranges of 110 to 120 volts (see the last sentence of the text in Table 430.248).
The disposer load =9.8 A × 120 V = 1,176 VA
The load of the water heater = 4.5 × 1,000 = 4,500 VA
Since derating is not permissible for only three appliances,
The calculated load at 100 % = 1,200 + 1,176 + 4,500 = 6,876 VA
If the disposer motor is the largest motor, then
The calculated service load = 6,876 + disposer load x 25% = 7,170 VA
Example#2:
For the same case in example#, add a fourth appliance; a trash compactor rated 7.5 amperes at 120 volts, What is the total service load?
Solution:
The compactor load = 7.5 A × 120 V = 900 VA
The total before applying the demand factor = 1,200 + 1,176 + 900 + 4,500 = 7,776 VA
Next, because there are four appliances, multiply the load by the 75 percent demand factor.
The calculated service load for these appliances = 7,776 × 0.75 = 5,832 VA
If the disposer motor is the largest motor, then
The calculated service load = 5,832 + disposer load x 25% = 6,126 VA
Important!!!

Example#3:
A single feeder will supply two dwelling units. Each unit in this twofamily dwelling will have 1,800 square feet of floor area, two 20ampere (A) smallappliance branch circuits, one 20A laundry branch circuit, four fastenedinplace appliances with a total rating of 9,156 VA, a range rated 12,000 VA and an electric clothes dryer rated 5,000 VA. The heating and air conditioning system in each unit will consist of a compressor rated 4,200 VA, a blower motor rated 1,176 VA, a condenser fan motor rated 360 VA and electric heat rated 10,000 VA.
What is the feeder load calculation for this twofamily dwelling?
Solution:
Start by calculating the load as per NEC standard method as follows:
1 As per 220.12, Calculate the general lighting and general use receptacle load at 3 VA per square foot
The general lighting and general use receptacle load for each unit = 1,800 ft2 x 3 VA/ft2 = 5,400 VA
The general lighting and general use receptacle load for both units =5,400 VA X 2 = 10,800 VA
2 As per 220.52(A) and (B), calculate the smallappliance and laundry branchcircuit load at 1,500 VA for each circuit
The smallappliance and laundry branchcircuit load for each unit = 1,500VA X 3 = 4,500 VA
The smallappliance and laundry branch circuit load for both units = 4,500VA X 2 = 9,000 VA
The total general lighting load, including smallappliance and laundry branch circuits= 10,800 + 9,000 = 19,800 VA
3 Apply the Table 220.42 demand factors to the general lighting load.
The first 3,000 VA remain 3,000VA X 100% = 3,000VA
The remaining =19,800 – 3,000 = 16,800 VA at 35 percent = 16,800 VA X 35% = 5,880 VA
So, the general lighting load for both units = 3,000 + 5,880 = 8,880 VA
4 The total fastenedinplace appliance load for both units = 9,156 x 2 = 18,312 VA
Because there are more than three fastenedinplace appliances, it is permissible to apply a demand factor of 75 percent to this load.
After applying the Section 220.53 demand factor, the fastenedinplace appliance load = 18,312 X 75% = 13,734 VA
5 In accordance with 220.54, the electric clothes dryer load = 2 X 5,000 X100% = 10,000 VA
6 The maximum demand for two 12,000 VA ranges from Table 220.55 is 11,000 VA.
7 The heating load, with the blower motor, = 10,000 + 1,176 = 11,176 VA
This load is larger than the air conditioning load, and because of 220.60, it is permissible to use only the larger of the noncoincident loads.
Therefore, the total heating load for both units = 11,176 VA X 2 = 22,352 VA
8 As required by 220.50 and 430.24, this calculation must include 25 percent of the largest motor. Since the compressor was omitted, the largest motor is the blower motor.
25% of the largest motor = 1,176 X 25% = 294 VA
9 After applying all demand factors, the standard method load calculation for these two dwelling units = 8,880 + 13,734 + 10,000 + 11,000 + 22,352 + 294 = 66,180 VA
Don’t Forget…
For example#3, NEC optional method for three identical
units must be calculated, because In accordance with 220.85, where two
dwelling units are supplied by a single feeder and the calculated load as per
NEC standard method exceeds that for three identical units calculated in
accordance with NEC optional method, the lesser of the two loads shall be
permitted.

Example#4:
A heating/cooling package unit will be installed in a onefamily dwelling. The electric heater is rated 9.6 kW at 240 volts. The blower motor inside the package unit that circulates the air will be a ½ horsepower, 240volt motor. How much load will this package unit add to a 240volt, singlephase service? Assuming that the air conditioning load will be less than the heating load.
Solution:
The heat load= 9.6 KW × 1,000 = 9,600 watts
The fullload current in amperes of a ½ hp, singlephase, 240volt motor (from Table 430.248) = 4.9 A
The motor load = 4.9 A x 240 V = 1,176 VA
The service load for this package unit = 1,176 + 9,600 = 10,776 watts
Because the blower motor is the largest motor in the service load
The service load for this package unit = 10,776 + 1,176 x 25% = 11,070 watts
Second: As per NEC Optional calculation method
1 For single dwelling unit
Rule#3 Permanently Connected Motors
As per NEC section 220.82(B)(4), add the nameplate
ampere (A) or kilovolt ampere (kVA) rating of all permanently connected
motors to the general loads covered in 220.82(B). Include all motors that are
not already included with the general loads covered in 220.82(B)

Important!!!
Permanently connected motors must
be added to the general loads when calculating a feeder or service when
calculated by the NEC optional method. Although motors are listed separately,
some or all of the motors may have already been added to the load calculation
as fastenedinplace appliances.
For example, a kitchen waste
disposer could be added to an optional load calculation as a
fastenedinplace appliance or as a permanently connected motor. It is not
necessary to add motors that have already been added to the load calculation
as fastenedinplace appliances.

Important!!!
When calculating the feeder or
service load of a dwelling unit by the optional method, do not multiply the
fullload current of permanently connected motors by 125 percent.

Important!!!
Do not include heating and air
conditioning equipment in the list with permanently connected motors. Heating
and air conditioning equipment is covered in 220.82(C) and explained before
in articles:

Example#5:
What is the optional method service load calculation (before applying the demand factor) for a onefamily dwelling with the following permanently connected motors:
 Two automatic power attic roof ventilators rated horsepower (hp) at 115 volts (V) each
 A swimming pool pump rated 1 hp at 230V
Noting that the ampere rating of each motor is not known. These three motors have not been included with the general loads covered in 220.82(B).
Solution:
Step#1: find the fullload current rating in amperes for each motor.
Since the ampere ratings for these motors are not known, use the values given in Table 430.248 (which provides fullload currents in amperes for singlephase alternatingcurrent motors)
The fullload current for each of the automatic power attic roof ventilators = 4.4 A. The fullload current for the swimming pool pump = 10 A
Step#2: calculate voltamperes (VA) for each motor.
The load for each attic fan = 115 V x 4.4 A = 506 VA
The load for the swimming pool pump = 230 V x 10 A = 2,300 VA
The service load for these permanently connected motors (As per NEC optional calculation method) = 506 + 506 + 2,300 = 3,312 VA
2 For MultiFamily Dwelling
Rule#4: Permanently Connected Motors
As per NEC section 220.84(C)(4), add the nameplate
ampere (A) or kilovolt ampere (kVA) rating of all permanently connected
motors to other categories of loads included in 220.84(C). Include all motors
that are not already included with the item 220.84(C)(3).

Important!!!
For multifamily dwelling load
calculation, if there are any permanently connected motors that have not been
included, add the nameplate ampere or kilovoltampere rating of those motors
to the calculation.

Important!!!
Do not include motors that are part
of the air conditioning system or motors that are part of the fixed electric
spaceheating system. These motors will be included as 220.84(C)(5) requires and
as explained before in articles:

Important!!!
Include only permanently connected
motors that are supplied from dwelling units. Do not include permanently
connected motors that are supplied by house power.

Don’t Forget…
After calculation of feeder or
service load calculation for a Multifamily dwelling, apply the demand factors
of table 220.84.

Example#6:
The load of a multifamily dwelling will be calculated by the optional method. Two swimming pool pump motors and two automatic power attic roof ventilators will be powered by the house panel. Can these permanently connected motors be included with the connected loads to which Table 220.84 apply?
Solution:
Because these permanently connected motors are supplied by house power, do not add them to the loads that will be calculated in accordance with Table 220.84(Optional Calculations — Demand Factors for Three or More Multifamily Dwelling Units). These motors must be calculated in accordance with Part III of Article 220.
In the next Article, I will explain feeder and service calculations for complete dwelling units with solved examples. Please, keep following.
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