Non-Dwelling Buildings Load Calculations- Part Six


In Article Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows: 


  1. Lighting loads, 
  2. Receptacles Loads, 
  3. Kitchen Loads, 
  4. Heating, Ventilation and air conditioning Loads (Non-Coincident Loads), 
  5. Motor Loads, 
  6. Other Loads. 


Again, but for above Non-Dwelling Buildings Loads, I will explain the following points:

  1. Where and how to distribute each type of load in a dwelling unit as per NEC code? 
  2. How to calculate its Demand load for feeder and service sizing calculations? 



Important!!!
All design Calculations for Non-dwelling Buildings will be as per NEC standard calculation method but I will explain design calculations as per NEC Optional calculation method only for the following Non-Dwelling buildings as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations:

  1. Schools,
  2. Existing Installations,
  3. New Restaurants.






I explained the Design Calculation for first type of Non-Dwelling Building loads which is Lighting loads in Article 
Non-Dwelling Buildings Load Calculations- Part One ",.


Also, I explained Design Calculation for second type; Receptacle loads for Non-Dwelling Buildings in Articles:




And I explained the third type of Non-dwelling Loads; Kitchen Loads in Article " Non-Dwelling Buildings Load Calculations- Part Five ".

Today, I will explain the Design Calculation of Heating, Ventilation and Air conditioning Loads (Non-Coincident Loads) in Non-Dwelling Building. 


You can review our course " Introduction to Heating, Ventilation and Air Conditioning Systems " for more information.




Forth: Heating, Ventilation and Air Conditioning Loads (Non-Coincident Loads)




Important!!!
Regardless of the type of structure—residential, commercial, or industrial, the Heating, Ventilation and Air Conditioning Systems will be calculated by NEC part III (standard method) in the following manner.




Definition:
A heat pump (see below image): 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.





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.




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.




Important!!!
Cord and plug space heaters are not a permanent fixed heaters, then it will not be included in load calculation in this step.
Fixed heating equipment, such as central heating systems, boilers, heating cable, and unit heaters (baseboard, panel, and duct heaters) will be included.





Important!!!
Fixed electric space heating shall be considered continuous load.



Example#1:


An Office 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 an Office Building that will have electric space heaters and window air conditioners?

The heat will consist of two offices with space heaters rated 1,500 watts each and three offices with space heaters rated 2,250 watts each.

The air conditioning will consist of four offices 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 Office Building, 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 Office Building is 11,040 VA
If the air conditioner compressors are the largest motors in this Office Building, 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 an Office. 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 will be the largest motor. 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.





In the next article, I will continue explaining load calculations for other types of loads in Non-Dwelling Buildings. Please, keep following.



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