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 six types in the following articles:
 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
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
Definitions:
A heat pump (see below image)
is a device that acts as an air conditioner in the summer and as a heater in
the winter. Heat pumps look and function exactly like an air conditioner
except it has a reversible cycle.

7.1 Applied NEC Rules for Heating and air conditioning loads
There are many NEC rules that control the Heating and air conditioning loads including:
 220.50 Motors
 220.51 Fixed Electric Space Heating
 220.60 Noncoincident Loads
 220.82(C) Heating and air conditioning loads in Dwelling Unit
 220.83 Existing Dwelling Unit
 220.85 two family dwelling
 430.24 Several Motors or a Motor(s) and other Load(s)
 440.3 Other Articles
 440.6 Ampacity and Rating
7.2 Calculation of Heating and air conditioning loads
A For feeder and service calculation purposes
Important!!!
Most service load
calculations will include heating and/or air conditioning equipment, but not
all feeder load calculations will include these types of loads. If the feeder
will not supply power to heating and air conditioning equipment, calculate
just the general loads on this feeder. If a service will not supply heating
equipment calculate only the service for air condition only. If a service
will not supply power to heating and air conditioning equipment, ignore this
load in service load calculation.

First: As per NEC Standard calculation method
Rule#1: Service
load for Room air conditioners
The load for Room air conditioners shall be calculated at 100 % of its
ampere rating which may be indicated on its nameplate and will be used in
branch, feeder and service load calculations.

Rule#2: Service
load for Fixed electric spaceheating loads
As per NEC section 220.51,
Fixed electric spaceheating loads shall be calculated at 100 percent of the
total connected load. However, in no case shall a feeder or service load
current rating be less than the rating of the largest branch circuit
supplied.

Example#1:
A dwelling unit has seven wall heaters; each heater is rated 3,000 watts at 240 volts. How much load will these heaters add to a 240volt, singlephase service? Assuming that The air conditioning load will be less than the heating load.
Solution:
In accordance with 220.51, calculate the heaters service load at 100 %= 7 × 3,000 = 21,000 watts
Since the service voltage is known, the total current draw of the heaters can be calculated by dividing the total watts by 240 volts
The total current draw of the heaters = 21,000 ÷ 240 = 87.5 A = 88 A
Rule#3: Central air
conditioning and heating system Load
Central air conditioning and heating system Load shall be calculated at
100 % of its nameplate and will be a Noncoincident Load.

Rule#4:
Noncoincident
Loads
As per NEC section 220.60, where it is unlikely that two or more
noncoincident loads will be in use simultaneously, it shall be permissible to
use only the largest load(s) that will be used at one time for calculating
the total load of a feeder or service.

Important!!!
Depending on the
design, the heating system and the air conditioning system might be
noncoincident loads.

Example#2:
What is the service load contribution for a dwelling that will have electric space heaters and window air conditioners? The heat will consist of two rooms with space heaters rated 1,500 watts each and three rooms with space heaters rated 2,250 watts each.
The air conditioning will consist of four rooms with window air conditioning units rated 11.5 amperes at 240 volts. Assume spaceheating watts are equivalent to voltamperes (VA).
Solution:
In this dwelling, the heat load and the air conditioning load are noncoincident loads. The heat and the air conditioning will not be energized at the same time. Therefore, compare the total heat load to the total air conditioning load and omit the smaller of the two loads.
First: calculate the heat load The total heat load = 1,500 + 1,500 + 2,250 + 2,250 + 2,250 = 9,750 VA.
Second: calculate the air conditioning load
Start by finding the voltamperes of each unit (VA = E × I) (assume power factor or PF = 1.0).
The load of each air conditioner = 240 × 11.5 = 2,760 VA
The total air conditioner load = 2,760 VA × 4 = 11,040 VA
Since the air conditioner load is larger than the heat load, omit the heat load.
The service load contribution for the heating and air conditioning loads in this dwelling is 11,040 VA
If the air conditioner compressors are the largest motors in this dwelling, multiply the load of one compressor by 25 percent and add it to the service load calculation.
Rule#5:
the air handler (or blower motor) is
not a noncoincident load
Although the heating and air conditioning in package units and
split systems are noncoincident loads, the air handler (or blower motor) (or
evaporator motor) is not. Since the blower motor works with both the heating
and air conditioning system, it must be included in both calculations.

Example#3:
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 × 240 V = 1,176 VA
The service load for this package unit = 1,176 + 9,600 = 10,776 watts
Example#4:
A package unit has electric heat and air conditioning. The unit contains a compressor, a blower motor, a condenser fan motor and electric heat. The compressor draws 23 amperes at 240 volts, the blower motor draws 5 amperes at 240 volts and the condenser fan motor draws 2 amperes at 240 volts. The rating of the heat is 10 kW. Assume heating kilowatt rating is equivalent to kilovoltamperes. What is the service load contribution for this heating/cooling package unit?
Solution:
The load of the air conditioner compressor =23 A × 240 V= 5,520 VA The load of the blower motor = 5 A × 240 V = 1,200 VA
The load of the condenser fan motor = 2 A × 240 V = 480 VA
The total air conditioning load = 5,520 + 1,200 + 480 = 7,200 VA
Since the blower motor also works with the heat, the load of the blower motor must be added to the heat load. So, the heat load = 10 KW× 1,000 = 10,000 + 1,200 = 11,200 VA
Since the heat load is larger than the air conditioner load, omit the air conditioner load. The service load contribution for this package unit is 11,200 VA.
Rule#6:
Largest Motor in the feeder or service
load calculation
As per NEC sections 220.50 and 430.24, when calculating a feeder
or service, the largest motor must be multiplied by 125 percent.

Important!!!
Unless it is the largest motor in the feeder
or service load calculation, do not multiply the fullload current of the
motor by 125 percent.

Example#5:
In example#3, the blower motor in the heating/cooling package unit in the last example will be the largest motor in the onefamily dwelling. How much load will this package unit add to a 240volt, singlephase service?
Solution:
If the ½ hp blower motor is the largest motor in the calculation for the service, the ampacity must not be less than 125 percent of the fullload current rating plus the calculated load of the electric heat. Multiply the motor’s fullload current by 125 percent before adding it to the electric heat service load
The motor load = 4.9 A × 240 V = 1,176 VA The service load for this package unit = 1,176 x 1.25 + 9,600 = 11,070 watts
Rule#7:
A heat pump with supplementary heat is
not a noncoincident load
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.

Example#6:
What is the service load contribution for a heat pump with supplementary heat? The compressor draws 26.4 amperes at 240 volts, the blower motor draws 6.5 amperes at 240 volts and the condenser fan motor draws 3 amperes at 240 volts. The electric heat in this unit has a rating of 15 kW. Assume heating kilowatt rating is equivalent to kilovoltamperes.
Solution:
The compressor and all of the heat in this heat pump can be energized at the same time. The load of the air conditioner compressor = 26.4 × 240 = 6,336 VA
The load of the blower motor = 6.5 × 240 = 1,560 VA
The load of the condenser fan motor = 3 × 240 = 720
The total air conditioning load = 6,336 + 1,560 + 720 = 8,616 VA
Since the air conditioning system and all of the heat can be on at the same time, add the two together to get the service load contribution.
The service load contribution for this heat pump = 8,616 + 15,000 = 23,616 VA
Important!!!
Noncoincident loads are not limited to heating and air
conditioning systems. some other noncoincident loads examples are the standby motors/pumps.

Example#7:
What is the feeder load contribution for two 5hp, 230volt, single phase pump motors? The motors will be wired so only one motor is in operation at any time.
Solution:
Because these two motors are noncoincident loads, it is permissible to omit the load of one motor. Start by finding the fullload current (FLC) of one motor.
Fullload currents for singlephase, alternatingcurrent motors are in Table 430.248. From Table 430.248, The FLC of this motor is 28 amperes.
The load of one motor = 28 A × 230 V = 6,440VA
Although there are two motors, they will never operate at the same time. Therefore, only one motor must be added to the feeder calculation. The feeder load contribution = 6,440 VA
If this motor is the highestrated motor on this feeder, multiply the load of this motor by 25 percent, and add it to the feeder load calculation.
Second: As per NEC Optional calculation method
Rule#8: Application of NEC Optional calculation method
NEC Optional calculation method will be used if the
following condition is verified:

Important!!!
If the serviceentrance ampacity calculated by the
optional method is less than 100A, recalculate with using the standard
method.

Important!!!
In NEC Optional calculation method, for a
multifamily dwelling, Table 220.84 “Optional Calculations — Demand Factors
for Three or More Multifamily Dwelling Units” will be used if the following
conditions are verified:

Important!!!
The optional calculation can be used, provided all
of the conditions for using table 220.84 listed above are met. Otherwise, the
calculation for the multifamily dwelling is performed by using standard
calculation method.

Rule#9: Application of NEC Optional calculation method
NEC Optional calculation method is applicable only
for a single dwelling unit, an existing dwelling unit, a multifamily
dwelling, two dwelling units, a school, an existing installation and a new
restaurant.

I will explain the calculation of Heating and AirConditioning service and feeder loads for dwelling units according to the type of dwelling unit as follows:
 A single dwelling unit,
 A multifamily dwelling,
 A two dwelling units,
 An existing dwelling unit.
First: for single dwelling units
Rule#10: Heating
and AirConditioning Load as per NEC Optional calculation method
As per NEC section 220.82 (C), for Heating and
AirConditioning Load, The largest of the following six selections (load in
kVA) shall be included:

Case#1: If a dwelling has some type of heat other than electric
 In this case, calculate the air conditioning load only at 100 percent of the nameplate rating.
Example#8:
A central heating and air conditioning unit will be installed in a onefamily dwelling. The air conditioning system is electric, and the heating system is gas. The air conditioner compressor has a rating of 16.6 amperes (A) at 230 volts (V). The condenser fan motor has a rating of 2A at 115V and the airhandler (blower motor) has a rating of 3.2A at 115V. Calculating by the optional method, how much air conditioning load will be added to the general loads?
Solution:
First: calculating the voltampere (VA) load for each motor. The compressor load =16.6 A x 230 V = 3,818 VA
The condenser fan load = 2 Ax 115 V = 230VA
The air handler load = 3.2 A x 115 V = 368VA
Second: add the air conditioning loads to find the total The service load = 3,818 + 230 + 368 = 4,416 VA
Because air conditioning loads only exist, it will be calculated at 100 percent.
Important!!!
Where there is more than one air conditioning unit, the calculation method is the same as it is for one air conditioning unit.

Example#9:
Four window air conditioners will be installed in a onefamily dwelling. The heating system is not electric. Two units are rated 12.5A at 230V and the other two are rated 8.5 amperes at 230 volts. Calculating by the optional method, how much air conditioning load will be added to the general loads?
Solution:
Each of the larger air conditioners has a load = 12.5A x 230 V = 2,875 VA Each of the smaller window units has a load = 8.5A x 230 V = 1,955 VA
The total load for all four units = 2,875 + 2,875 + 1,955 + 1,955 = 9,660 VA
The multiplication factor for four air conditioning units is the same as it is for one unit at 100%
The service load for four air conditioning units = 9,660 VA x 100% = 9,660 VA
Case #2: Heat pumps equipped with or without electric supplemental heat
 Supplemental heat is sometimes referred to as auxiliary, backup or even emergency heat. A dualfuel heat pump is an electric heat pump and a gas furnace all in one. Dual fuel heat pumps can be fueled with natural gas or propane. Because geothermal heat pumps (sometimes referred to as geoexchange, earthcoupled, groundsource or watersource heat pumps) do not depend on the temperature of the outside air, they may or may not be equipped with supplemental heat.
 Heat pumps not equipped with supplemental electric heat are calculated exactly the same as the air conditioning equipment specified in case#1.
Case #3: When a heat pump is used with supplemental electric heat
 In this case, multiply the nameplate rating(s) of the heat pump compressor by 100 percent and multiply the supplemental electric heating for central electric spaceheating systems by 65 percent.
Example#10:
A heat pump with supplemental electric heat will be installed in a onefamily dwelling. The2 1/2 ton heat pump system has nameplate rating of 24A at 240V. This heat pump is equipped with 15 kilowatts (kW) of backup heat. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads?
Solution:
The heat pump load =24A x 240V = 5,760VA
The backup or supplemental heat= 15 kW x 1,000 = 15,000 VA
Multiply the backup heat by 65 percent. The backup or supplemental heat= 15,000 x 65% = 9,750 VA
The total heating and air conditioning load = 5,760 + 9,750 = 15,510 VA
Important!!!
Some heat pumps are designed so that only part of the electric heat will operate while the compressor is in operation. If the heat pump compressor is prevented from operating at the same time as the supplementary heat, it does not need to be added to the supplementary heat for the total central spaceheating load.

Example#11:
A heat pump with supplemental electric heat will be installed in a onefamily dwelling. The 3ton heat pump system has nameplate rating of 30A at 240V. Although this heat pump is equipped with 15 kW of backup heat, it is designed so that only 10 kW can operate while the compressor is running. The other 5 kW of electric heat will only operate when the compressor is in the off position. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads?
Solution:
The heat pump load = 30A x 240 V = 7,200 VA
Because only part of the backup heat can operate while the compressor is running, multiply the load of this part by 65 percent. The backup heat load = 10,000VA x 65% = 6,500 VA
The total heating and air conditioning load = 7,200 + 6,500 = 13,700 VA
Case#4: electric space heating by less than four separately controlled units
 In this case, 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units.
Example#12:
A total of three electric wall heaters will be installed in a onefamily dwelling. Two of the units are rated 3,000 watts (W) at 240 volts (V), and one unit is rated 4,800W at 240V. Each wall heater is separately controlled. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for these three electric space heaters?
Solution:
The total rating for all three heaters = 3,000 + 3,000 + 4,800 = 10,800 VA
Because there are less than four separately controlled space heating units, multiply the total rating by 65 percent. The total rating for all three heaters = 10,800VA x 65% = 7,020 VA
Case#5: electric space heating by four or more separately controlled units
 In this case, 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units.
Example#13:
Six separately controlled electric wall heaters will be installed in a onefamily dwelling. Two heaters are rated 1,500W at 120V, one heater is rated 2,000W at 240V, two heaters are rated 3,000W at 240V and one heater is rated 4,000W at 240V. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for these six electric space heaters?
Solution:
The total rating for all six heaters = 1,500 + 1,500 + 2,000 + 3,000 + 3,000 + 4,000 = 15,000 VA
Because there are at least four separately controlled space heating units, multiply the total rating by 40 percent. The total rating for all six heaters = 15,000 x 40% = 6,000 VA
Case#6: using electric thermal storage (ETS) and other heat systems
 Some electric utilities offer a discounted rate for kilowatthours used during certain hours of the day and night. During offpeak hours, customers pay rates that are less than rates during peak hours. During offpeak hours, electric elements are used to heat ceramic bricks. The stored heat is then released throughout the day when rates are higher. Many installations include multiple units placed throughout the house.
 Since all of the electric thermal storage units could be heating bricks at the same time, no demand factor can be applied to the units.
 In this case, 100 percent of the nameplate ratings of electric thermal storage (ETS) and other heat systems where the usual load is expected to be continuous at the full nameplate value.
Example#14:
Four electric thermal storage units will be installed in a onefamily dwelling. Two units are rated 5.4 kilowatts (kW) at 240V and two units are rated 7.2 kW at 240V. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for this dwelling?
Solution:
The total rating for all four units = 5.4 + 5.4 + 7.2 + 7.2 = 25.2 KVA
Because electric thermal storage system loads are calculated at 100 percent, 25.2 kVA will be added to service load.
Rule#11:
Noncoincident
Loads
As per NEC section 220.60, where it is unlikely that two or more
noncoincident loads will be in use simultaneously, it shall be permissible to
use only the largest load(s) that will be used at one time for calculating
the total load of a feeder or service.

Important!!!
Depending on the
design, the heating system and the air conditioning system might be
noncoincident loads.

Important!!!
Similarly, 220.82(C)
requires that only the largest of the six choices needs to be included in the
feeder or service calculation.

Example#15:
Three window air conditioners and five separately controlled electric wall heaters will be installed in a onefamily dwelling. One window unit is rated 11.5 amperes at 240V and the other two are rated 9 amperes at 240V. Two wall heaters are rated 2,000W at 240V, two heaters are rated 3,000W at 240V and one heater is rated 4,000W at 240V. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads?
Solution:
The larger air conditioner has a rating =11.5A x 240 V= 2,760 VA
Each smaller window unit has a rating = 9A x 240 V = 2,160 VA
The total load for all three units = 2,760 + 2,160 + 2,160 = 7,080 VA
Since the air conditioning load is calculated at 100 percent, the total air conditioning load = 7,080 x 100% = 7,080 VA
The total rating for all five heaters = 2,000 + 2,000 + 3,000 + 3,000 + 4,000 = 14,000 VA
Because there are at least four separately controlled space heating units, multiply the total rating by 40 percent. The total heating load = 14,000 x 40% = 5,600 VA
Because the heating and air conditioning systems in this example will not run simultaneously, it is only required to include the larger of the two loads. The larger load, after applying demand factors, is the air conditioning load. Add service load = air conditioning load = 7,080 VA
In the next Article, I will continue explaining the calculation of Heating and AirConditioning service and feeder loads for dwelling units other than single dwelling unit. Please, keep following.
شكر الله لك
ReplyDeleteHello ENG. Ali,
ReplyDeleteRule#9 & Rule#10 are identical to each other. Feel it's a repetition.