Branch Circuit Design Calculations – Part Twelve


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 seven 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:

 




    8- Motor loads


    8.1 Applied NEC Rules for Motor loads


    There are many NEC rules that control the Motor Loads including:


    • 220.14(C) Motor Loads for All Occupancies 
    • 220.18(A) Maximum Loads for circuits supplying Motor-Operated and Combination Loads. 
    • 220.50 Motors 
    • 422.62 Appliances Consisting of Motors and Other Loads 
    • 220.82 Optional Method - Dwelling Unit 
    • 220.84 Optional Method - Multifamily Dwelling 
    • 430.6 Ampacity and Motor Rating Determination 
    • 430.22 Single Motor 


    8.2 Calculation of Motor loads



    First: As per NEC Standard calculation method



    The motors exist in dwelling units as fastened in place appliances or separate motors (air conditioning compressors, fan blower, etc.)




    Rule#1: Motor Loads as per NEC standard method

    When calculating a feeder or service As per NEC Standard calculation method, the largest motor must be multiplied by 25 percent and add it to the service load calculation.




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




    Rule#2: Motor Loads as per NEC standard method

    As per NEC section 430.6(A)(1), Do not use the actual current rating marked on the nameplate. When calculating motor loads, use the values given in Tables 430.247 through 430.250.














    Important!!!
    Exceptions to 430.6(A)(1) :
    1. Motors built for low speeds (less than 1,200 rpm) or high torques for multispeed motors.
    2. For equipment that employs a shaded-pole or permanent-split capacitor-type fan or blower motor that is marked with the motor type, use the full load current for such motor marked on the nameplate of the equipment in which the fan or blower motor is employed.
    3. For a listed motor-operated appliance that is marked with both motor horsepower and full-load current, use the motor full-load current marked on the nameplate of the appliance.




                                             
    Important!!!
    • Full-load currents for 3-phase motors are in Table 430.250,
    • Full-Load Current for Two-Phase Alternating- Current Motors (4-Wire) are in Table 430.249,
    • Full-load currents for single-phase motors are in Table 430.248,
    • Full-Load Currents for Direct-Current Motors are in Table 430.247.





    Important!!!
    Kilovolt-amperes (kVA) shall be considered equivalent to kilowatts (kW) for Fastened-in-place Appliances.



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




    Important!!!
    As per NEC section 220.53, electric ranges, clothes dryers, space-heating equipment or air conditioning equipment must not be included with the number of appliances that are fastened in place.
    Also, All portable small Appliances for kitchen and others are not Fastened-in-Place Appliances.




    Important!!!
    No derating is allowed when there are only one, two, three fastened-in-place appliances.




    Example#1:

    The following appliances will be installed in a one-family dwelling:

    • A dishwasher rated 10 amperes at 120 volts; 
    • A ½ hp, 120-volt kitchen-waste disposer; 
    • A 4.5 kW, 240-volt water heater. 

    What is the total service load for these appliances?



    Solution:

    The load for these three appliances must be calculated at 100 percent.

    The dishwasher load = 10 A × 120 V = 1,200 VA.

    Since the kitchen-waste (or garbage) disposer is a motor, the load must be determined by multiplying the full-load current in Table 430.248 by the system voltage.

    In accordance with Table 430.248, the full-load current of a ½-hp, 120-volt motor = 9.8 A

    Note: Although the current value of 9.8 amperes is in the column titled “115 volts,” the currents listed in this column are permitted for system voltage ranges of 110 to 120 volts (see the last sentence of the text in Table 430.248).

    The disposer load =9.8 A × 120 V = 1,176 VA

    The load of the water heater = 4.5 × 1,000 = 4,500 VA

    Since derating is not permissible for only three appliances,

    The calculated load at 100 % = 1,200 + 1,176 + 4,500 = 6,876 VA


    If the disposer motor is the largest motor, then

    The calculated service load = 6,876 + disposer load x 25% = 7,170 VA




    Example#2:

    For the same case in example#, add a fourth appliance; a trash compactor rated 7.5 amperes at 120 volts, 
    What is the total service load? 

    Solution: 


    The compactor load = 7.5 A × 120 V = 900 VA

    The total before applying the demand factor = 1,200 + 1,176 + 900 + 4,500 = 7,776 VA

    Next, because there are four appliances, multiply the load by the 75 percent demand factor.

    The calculated service load for these appliances = 7,776 × 0.75 = 5,832 VA

    If the disposer motor is the largest motor, then

    The calculated service load = 5,832 + disposer load x 25% = 6,126 VA





    Important!!!
    • If the motor is air conditioning compressor, usually the air conditioning compressor is the largest motor in dwelling units. in this case, multiply the load of one compressor by 25 percent and add it to the service load calculation.
    • But if the heating load is larger than the air conditioning load, and because of 220.60 which states that” it is permissible to use only the larger of the noncoincident loads” the air conditioning load will be omitted and the air conditioning compressor will not be the largest motor in this case.





    Example#3:

    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 feeder load calculation for this two-family dwelling?



    Solution: 

    Start by calculating the load as per NEC standard method as follows: 
    1- As per 220.12, Calculate the general lighting and general use receptacle load at 3 VA per square foot 

    The general lighting and general use receptacle load for each unit = 1,800 ft2 x 3 VA/ft2 = 5,400 VA

    The general lighting and general use receptacle load for both units =5,400 VA X 2 = 10,800 VA

    2- As per 220.52(A) and (B), calculate the 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 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





    Don’t Forget…
    For example#3,  NEC optional method for three identical units must be calculated, because In accordance with 220.85, where two dwelling units are supplied by a single feeder and the calculated load as per NEC standard method exceeds that for three identical units calculated in accordance with NEC optional method, the lesser of the two loads shall be permitted.




    Example#4:


    A heating/cooling package unit will be installed in a 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 x 240 V = 1,176 VA

    The service load for this package unit = 1,176 + 9,600 = 10,776 watts

    Because the blower motor is the largest motor in the service load

    The service load for this package unit = 10,776 + 1,176 x 25% = 11,070 watts



    Second: As per NEC Optional calculation method



    1- For single dwelling unit





    Rule#3 Permanently Connected Motors

    As per NEC section 220.82(B)(4), add the nameplate ampere (A) or kilovolt ampere (kVA) rating of all permanently connected motors to the general loads covered in 220.82(B). Include all motors that are not already included with the general loads covered in 220.82(B)




    Important!!!
    Permanently connected motors must be added to the general loads when calculating a feeder or service when calculated by the NEC optional method. Although motors are listed separately, some or all of the motors may have already been added to the load calculation as fastened-in-place appliances.

    For example, a kitchen waste disposer could be added to an optional load calculation as a fastened-in-place appliance or as a permanently connected motor. It is not necessary to add motors that have already been added to the load calculation as fastened-in-place appliances.




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




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








    Example#5:

    What is the optional method service load calculation (before applying the demand factor) for a one-family dwelling with the following permanently connected motors:

    • Two automatic power attic roof ventilators rated horsepower (hp) at 115 volts (V) each 
    • A swimming pool pump rated 1 hp at 230V 

    Noting that the ampere rating of each motor is not known. These three motors have not been included with the general loads covered in 220.82(B).



    Solution:

    Step#1: find the full-load current rating in amperes for each motor.

    Since the ampere ratings for these motors are not known, use the values given in Table 430.248 (which provides full-load currents in amperes for single-phase alternating-current motors)

    The full-load current for each of the automatic power attic roof ventilators = 4.4 A.
    The full-load current for the swimming pool pump = 10 A 

    Step#2: calculate volt-amperes (VA) for each motor. 

    The load for each attic fan = 115 V x 4.4 A = 506 VA

    The load for the swimming pool pump = 230 V x 10 A = 2,300 VA

    The service load for these permanently connected motors (As per NEC optional calculation method) = 506 + 506 + 2,300 = 3,312 VA



    2- For Multi-Family Dwelling




    Rule#4: Permanently Connected Motors

    As per NEC section 220.84(C)(4), add the nameplate ampere (A) or kilovolt ampere (kVA) rating of all permanently connected motors to other categories of loads included in 220.84(C). Include all motors that are not already included with the item 220.84(C)(3).




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




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









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




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






    Example#6:

    The load of a multifamily dwelling will be calculated by the optional method. Two swimming pool pump motors and two automatic power attic roof ventilators will be powered by the house panel. Can these permanently connected motors be included with the connected loads to which Table 220.84 apply?



    Solution: 

    Because these permanently connected motors are supplied by house power, do not add them to the loads that will be calculated in accordance with Table 220.84(Optional Calculations — Demand Factors for Three or More Multifamily Dwelling Units). These motors must be calculated in accordance with Part III of Article 220. 



    In the next Article, I will explain feeder and service calculations for complete dwelling units with solved examples. Please, keep following.




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