 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 six types in the following articles:
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:
7 Heating and air conditioning loads – Part Two
In the previous Article " Heating and air conditioning loads – Part One ", I explained the following points:
 Applied NEC Rules for Heating and air conditioning loads
 Feeder and service calculation of Heating and air conditioning loads by two methods:
 First: As per NEC Standard calculation method
 Second: As per NEC Optional calculation method
The rules applied in second method differ from a dwelling unit type to another. I explained the rules for first dwelling type “A single dwelling unit” in above previous Article.
Today, I will continue explaining the rules for feeder and service calculation of Heating and air conditioning loads as per NEC Optional calculation method for other dwelling unit types which are:
 A multifamily dwelling,
 A two dwelling units,
 An existing dwelling unit,
Second: Multifamily Dwelling
Rule#1: Table 220.84 for Multifamily Dwelling
demand factors
As per NEC section 220.84, for Multifamily Dwelling,
the demand factors of Table 220.84 shall be applied to the larger of the
airconditioning load or the fixed electric spaceheating load.

Rule#2: Calculation
of feeder and service loads for air conditioning load in Multifamily Dwelling
With the optional
method multifamily dwelling load calculation, the air conditioning load is
calculated at 100 % of the nameplate rating.

Important!!!
The air
conditioning load is calculated the same way in multifamily dwellings as it
is in onefamily dwellings. In both types, the air conditioning load is
calculated at 100 percent of the nameplate rating.

Rule#3: Calculation of feeder and service loads for
electric space heating load in Multifamily Dwelling
With the optional
method multifamily dwelling load calculation, the load for space heating
units must be added to the calculation at the nameplate rating.

Important!!!
The electric space heating
load is not calculated the same way in multifamily dwellings as it is in
onefamily dwellings. With the optional method onefamily dwelling load
calculation, it is permissible to apply a demand factor to space heating
units. The demand factor depends on the number of units.

Don’t Forget…
When calculating a
multifamily dwelling by the optional method, use the larger of the air
conditioning loads or the fixed electric spaceheating load.

Example#1:
For a 30unit multifamily dwelling that have electric space heaters and window air conditioners as follows:
The heating for each unit will consist of two separately controlled wall heaters rated 2,250W each and one heater rated 1,500W.
The air conditioning for each unit will consist of two window air conditioning units rated 11.5A at 240V
What is the optional method service load calculation (before applying the Table 220.84 demand factor)? (Assume spaceheating watts are equivalent to voltamperes).
Solution:
First: calculate the heat load for each unit by adding the nameplate rating of heat units
The heat load for each unit = 2,250 + 2,250 + 1,500 = 6,000 VA So, the total heat load for 30 units = 6,000 VA x 30 = 180,000 VA
Second: calculate the air conditioning load for each unit by Multiplying volts, amperes and the number of air conditioners in each unit
The air conditioning load for each unit = 240V x 11.5A x 2 = 5,520 VA So, the total air conditioning load for 30 units = 5,520 VA x 30 = 165,600 VA
Third: compare between the heating load and the air conditioning load for all units
Since the heating load is larger than the air conditioning load, omit the air conditioning load.
The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning loads in this 30unit multifamily dwelling = 180,000 VA
Important!!!
Do not assume the
heating load will always be larger than the air conditioning load. The
heating system could be gas or oil. The dwelling could also be located in a
warm climate where the air conditioning load is larger than the heating load.

Example#2:
For a 12unit multifamily dwelling that has an identical central heating and air unit for each dwelling unit as follows:
Each package unit will contain a compressor, a blower motor, a condenser fan motor and electric heat. The compressor draws 23A at 240V, the blower motor draws 5A at 240V, and the condenser fan motor draws 2A at 240V. The rating of the electric heat is 5 kW.
What is the optional method service load calculation (before applying the Table 220.84 demand factor)? (Assume the heating kilowatt rating is equivalent to kilovoltamperes).
Solution:
First: calculate the air conditioning load for each unit and then for all units
The load of each air conditioner compressor =23A x 240 V = 5,520VA The load of each blower motor =5 A x 240 V = 1,200 VA
The load of each condenser fan motor = 2A x 240 V = 480VA
So, the air conditioning load for each unit = 5,520 + 1,200 + 480 = 7,200 VA And the total air conditioning load for 12 units = 7,200 VA x 12 = 86,400 VA
Second: calculate the heat load for each unit and then for all units
Since the blower motor also works with the heat, add the load of the blower motor to the heat load The load of blower motor = 5 x 1,000 VA = 5,000 VA
So, the heating load for each unit = 5,000 + 1,200 = 6,200 VA And the total heating load for 12 units = 6,200 x 12 = 74,400 VA
Third: compare between the heating load and the air conditioning load for all units
In this example, the air conditioning load is larger than the heating load; therefore, omit the heating load.
The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning system in this 12unit multifamily dwelling = 86,400 VA
Important!!!
Fixed electric
space heating is not limited to space heaters and electric strip heat that is
part of a package unit or furnace. Fixed electric space heating could also be
a heat pump with supplementary heat.

Rule#4: If the
Fixed electric space heating is a heat pump with supplementary heat
With a heat pump, the compressor (and accompanying motors) and some or
all of the electric heat can be energized at the same time.
So, The load contribution of a heat pump = the air conditioning system
load + the maximum amount of heat that can be on while the air conditioner
compressor is energized.

Example#3:
For a sixunit multifamily dwelling that has an identical heat pump for each dwelling unit as follows:
Each unit’s compressor draws 26.8A at 240V, the blower motor draws 5.8A at 240V, and the condenser fan motor draws 2.6A at 240V. The electric heat in this unit has a rating of 15 kW. The compressor and all of the heat in this heat pump can be energized at the same time.
What is the optional method service load calculation (before applying the Table 220.84 demand factor)? Assume the heating kilowatt rating is equivalent to kilovoltamperes.
Solution:
First: calculate the air conditioning and heat loads for each heat pump
The load of the air conditioner compressor = 26.8 A x 240 V= 6,432 VA The load of the blower motor = 5.8 A x 240 V= 1,392 VA
The load of the condenser fan motor = 2.6A x 240 V = 624 VA
So, the air conditioning load for each unit = 6,432 + 1,392 + 624 = 8,448 VA
Since the air conditioning system and all of the heat can be on at the same time, add the two together
The air conditioning and heat loads for each heat pump = 8,448 + 15,000 = 23,448 VA
Second: calculate the air conditioning and heat loads for all heat pumps
The total load for six units = 23,448VA x 6 = 140,688 VA
The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning system in this sixunit multifamily dwelling = 140,688 VA
Rule#5: house loads
of multifamily dwellings
Applying the Table
220.84 demand factor to house loads of multifamily dwellings is not
permitted. House loads must be calculated as per NEC standard method.

Rule#6: multifamily
dwelling buildings with multiple services and feeders
Sometimes, in large
multifamily dwelling buildings, multiple services and feeders may be
installed to supply power to different floors or different buildings; in this
case, it will be necessary to perform optional load calculation for each
feeder.
Generally, if the
number of units on the feeder is not the same as the number on the service, it will
be necessary to perform more than one load calculation.

Important!!!
If the total
service load for a multifamily dwelling is known, and there is a requirement to
supply the building by multiple feeders, do not just divide the service load
calculation by the number of feeders. Follow rule#6.

Example#4:
If the calculated services load for a 24 unit multifamily dwelling is 311,556 VA and there is a requirement to supply power for it by using four feeders, each feeder will supply power to six units.
Each unit in this multifamily dwelling will have 1,050 square feet of floor area, two 20 A smallappliance branch circuits, one 20A laundry branch circuit, fastenedinplace appliances rated 6,600 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 840 VA, a condenser fan motor rated 360 VA and electric heat rated 5,000 VA
What is the optional method feeder load calculation for six units?
Solution:
First: In accordance with 220.84(C)(1), The general lighting and generaluse receptacle load for six units =3 VA x 150 ft2 x 6 = 18,900 VA
Second: In accordance with 220.84(C)(2), The smallappliance branch circuit load for six units = 2 x 1,500VA x 6 = 18,000 VA
Third: In accordance with 220.84(C)(3), use the nameplate ratings for the fastenedinplace appliances, ranges and clothes dryers will be used:
The laundry branch circuit load for six units = 1,500VA x 6 = 9,000 VA.
The calculated load for the fastenedinplace appliances for six units = 6,600 VA X 6 = 39,600 VA
The calculated load for the ranges for six units = 12,000 VA X 6 = 72,000 VA
The calculated load for the clothes dryers for six units = 5,000 VA X 6 = 30,000 VA
Forth: In accordance with 220.84(C)(5), add to the load calculation the larger of the air conditioning load or the fixed electric space heating load
The fixed electric spaceheating load for six units = (electric heat load + blower motor load) X 6 = (5,000 + 840) X 6 = 35,040 VA The total air conditioning unit for six units = (compressor load + blower motor load + condenser fan motor load) x 6 = (4,200 VA + 840 VA +360 VA) x 6 = 32,400 VA
The air conditioning load is omitted because the heating load was larger.
Fifth: find the total connected load of the feeder
The total connected loads for a feeder supplying six units = 18,900 + 18,000 + 9,000 + 39,600 + 72,000 + 30,000 + 35,040 = 222,540 VA
Sixth: apply Table 220.84 demand factor (in below image) for six units is 44 percent.
The calculated load after applying the demand factor = 222,540 X 44% = 97,917.6 VA = 97,918 VA
Notes:
 Not considering any house loads, the optional method feeder load calculation for six units is 97,918 VA.
 Do not divide the service load calculation by the number of feeders to find the load of one feeder which will give incorrect answer as follows:
 The feeder load = service load / number of feeders = 311,556 VA / 4 = 77,889 VA
 While the correct calculated value for one feeder is 97,918 VA as calculated in example#4 above.
Third: Two family dwelling (that are supplied by a single feeder)
Rule#6: service and
feeder Calculation for Two family dwelling
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.

Important!!!
The optional method load calculation for
two dwelling units involves using the optional method load calculation procedures
for multifamily dwellings, but not for two dwelling units. Perform the load
calculation procedures in 220.84, but calculate these two dwelling units as
if there were three identical units.

Important!!!
In accordance with 220.85, it is necessary to calculate by both
methods and then select the lesser of the two loads.

Important!!!
Performing the optional method load calculation for two dwelling units
without performing the standard method load calculation is permissible, but
the result could be larger than the standard method load calculation.

Example#5:
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 optional method feeder load calculation for this twofamily dwelling?
Solution:
First: 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 (in below image) 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
Second: Perform the load calculation procedures in 220.84 (NEC optional method), but calculate these two dwelling units as if there were three identical units.
1 As per 220.84(C)(1), 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 three units = 5,400 VA X 3 = 16,200 VA
Remember, this optional calculation is based on three units, not two.
2 As per 220.84(C)(2), calculate the smallappliance and laundry branchcircuit load at 1,500 VA for each circuit
The smallappliance and laundry branch circuit load for each unit =1,500VA X 3 = 4,500 VA The smallappliance and laundry branchcircuit load for three units = 4,500 VA X 3 = 13,500 VA
3 In accordance with 220.84(C)(3), use the nameplate ratings for the fastenedinplace appliances, ranges and clothes dryers.
The load for the fastenedinplace appliances, ranges and clothes dryers in each unit = 9,156 + 12,000 + 5,000 = 26,156 VA
The load for the fastenedinplace appliances, ranges and clothes dryers for three units = 26,156 VA X 3 = 78,468 VA
4 In accordance with 220.84(C)(5), add to the load calculation the larger of the air conditioning load or the fixed electric space heating load.
In this example, the heating load is larger than the air conditioning load, therefore omit the air conditioning load. The heating load for each unit = 10,000 + 1,176 = 11,176 VA
The heating load for three units = 11,176VA X 3 = 33,528 VA
5 The total connected load for three units = 16,200 + 13,500 + 78,468 + 33,528 = 141,696 VA
6 After finding the total connected loads, apply the Table 220.84 demand factor for the number of dwelling units. The Table 220.84 demand factor for three units is 45 percent.
The calculated load after applying the demand factor =141,696 X 45% = 63,763 VA
Third: Compare between the results of standard and optional (as 3 units) loads and select the lesser of the two loads.
The result of the standard load calculation was 66,180 VA.
The result of the Optional load calculation (as 3 units) was 63,763 VA
So, the calculated feeder load for these two dwellings is 63,763 VA.
In the next Article, I will continue explaining the calculation of Heating and AirConditioning service and feeder loads for Existing dwelling units. Please, keep following.
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