Branch Circuit Design Calculations – Part Nine


In article " Receptacle Branch Circuit Design Calculations – Part Three ", I stated that a Receptacle in dwelling units may serve one of the following loads:

  1. General-use Receptacle Loads, 
  2. Small appliance Loads, 
  3. Laundry Load, 
  4. Cloth dryer Load, 
  5. Household cooking appliances load, 
  6. Fastened-in-place Appliance loads, 
  7. Heating and air conditioning loads, 
  8. 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




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 space-heating loads

As per NEC section 220.51, Fixed electric space-heating 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 240-volt, single-phase 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 space-heating watts are equivalent to volt-amperes (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 volt-amperes 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 one-family 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, 240-volt motor. How much load will this package unit add to a 240-volt, single-phase 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 full-load current in amperes of a ½ hp, single-phase, 240-volt 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 kilovolt-amperes. 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 full-load 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 one-family dwelling. How much load will this package unit add to a 240-volt, single-phase 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 full-load current rating plus the calculated load of the electric heat.
Multiply the motor’s full-load 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 kilovolt-amperes.



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.



Example#7:



What is the feeder load contribution for two 5-hp, 230-volt, 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 full-load current (FLC) of one motor. 
Full-load currents for single-phase, alternating-current 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 highest-rated 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:
  1. The service-entrance or feeder conductors have an ampacity of at least 100 amperes.




Important!!!
If the service-entrance ampacity calculated by the optional method is less than 100A, re-calculate 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:
  1. If No dwelling unit is supplied by more than one feeder
  2. Each dwelling unit is equipped with electric cooking equipment.
  3. Each dwelling unit is equipped with either electric space heating or air conditioning, or both.




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.



Rule#10: 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 Air-Conditioning service and feeder loads for dwelling units according to the type of dwelling unit as follows:

  1. A single dwelling unit, 
  2. A multifamily dwelling, 
  3. A two dwelling units,
  4. An existing dwelling unit.


First: for single dwelling units





Rule#11: Heating and Air-Conditioning Load as per NEC Optional calculation method

As per NEC section 220.82 (C), for Heating and Air-Conditioning Load, The largest of the following six selections (load in kVA) shall be included:

  1. 100 percent of the nameplate rating(s) of the air conditioning and cooling.
  2. 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electric heating.
  3. 100 percent of the nameplate rating(s) of the heat pump compressor and 65 percent of the supplemental electric heating for central electric space-heating systems. 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 space heating load.
  4. 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units.
  5. 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units.
  6. 100 percent of the nameplate ratings of electric thermal storage (ETS) and other heating systems where the usual load is expected to be continuous at the full nameplate value.




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 one-family 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 air-handler (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 volt-ampere (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 one-family 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 dual-fuel 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 geo-exchange, earth-coupled, ground-source or water-source 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 space-heating systems by 65 percent. 


Example#10:



A heat pump with supplemental electric heat will be installed in a one-family dwelling. The-2 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 space-heating load. 



Example#11:



A heat pump with supplemental electric heat will be installed in a one-family dwelling. The 3-ton 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 one-family 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 one-family 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 kilowatt-hours used during certain hours of the day and night. During off-peak hours, customers pay rates that are less than rates during peak hours. During off-peak 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 one-family 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#12: 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 one-family 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 Air-Conditioning service and feeder loads for dwelling units other than single dwelling unit. Please, keep following.




2 comments:

  1. Hello ENG. Ali,

    Rule#9 & Rule#10 are identical to each other. Feel it's a repetition.

    ReplyDelete

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