Branch Circuit Design Calculations – Part Ten

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:

• 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

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:

•  Receptacle Branch Circuit Design Calculations – Part One 

•  Receptacle Branch Circuit Design Calculations – Part Two

7- Heating and air conditioning loads – Part Two



In the previous Article " Heating and air conditioning loads – Part One ", I explained the following points:

1. Applied NEC Rules for Heating and air conditioning loads 
2. 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:

1. A multifamily dwelling, 
2. A two dwelling units, 
3. 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 air-conditioning load or the fixed electric space-heating 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 one-family 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 one-family dwellings. With the optional method one-family 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 space-heating load.

Example#1:


For a 30-unit 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 space-heating watts are equivalent to volt-amperes).



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 30-unit 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 12-unit 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 kilovolt-amperes).

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 12-unit 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 six-unit 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 kilovolt-amperes.


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 six-unit 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 small-appliance branch circuits, one 20A laundry branch circuit, fastened-in-place 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 general-use receptacle load for six units =3 VA x 150 ft2 x 6 = 18,900 VA 

Second: In accordance with 220.84(C)(2), The small-appliance 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 fastened-in-place 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 fastened-in-place 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 space-heating 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:

1. Not considering any house loads, the optional method feeder load calculation for six units is 97,918 VA.
2. 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 two-family dwelling will have 1,800 square feet of floor area, two 20-ampere (A) small-appliance branch circuits, one 20A laundry branch circuit, four fastened-in-place 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 two-family 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 small-appliance and laundry branch-circuit load at 1,500 VA for each circuit

The small-appliance and laundry branch-circuit load for each unit = 1,500VA X 3 = 4,500 VA The small-appliance and laundry branch circuit load for both units = 4,500VA X 2 = 9,000 VA 
The total general lighting load, including small-appliance 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 fastened-in-place appliance load for both units = 9,156 x 2 = 18,312 VA

Because there are more than three fastened-in-place appliances, it is permissible to apply a demand factor of 75 percent to this load.

After applying the Section 220.53 demand factor, the fastened-in-place 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 small-appliance and laundry branch-circuit load at 1,500 VA for each circuit

The small-appliance and laundry branch circuit load for each unit =1,500VA X 3 = 4,500 VA The small-appliance and laundry branch-circuit 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 fastened-in-place appliances, ranges and clothes dryers.

The load for the fastened-in-place appliances, ranges and clothes dryers in each unit = 9,156 + 12,000 + 5,000 = 26,156 VA

The load for the fastened-in-place 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 Air-Conditioning service and feeder loads for Existing dwelling units. Please, keep following.

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: 

• 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

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:

•  Receptacle Branch Circuit Design Calculations – Part One 

•  Receptacle Branch Circuit Design Calculations – Part Two

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. some other noncoincident loads examples are the standby motors/pumps.

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: 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: If No dwelling unit is supplied by more than one feeder Each dwelling unit is equipped with electric cooking equipment. 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.

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#10: 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: 100 percent of the nameplate rating(s) of the air conditioning and cooling. 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electric heating. 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. 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units. 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units. 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#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 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.