In Article " Non-Dwelling Buildings Load Calculations- Part One ", I introduced a List for ordinary Non-Dwelling Buildings Loads which was as follows:
- Lighting loads,
- Receptacles Loads,
- Kitchen Loads,
- Heating, Ventilation and air conditioning Loads (Non-Coincident Loads),
- Motor Loads,
- Other Loads.
Again, but for above Non-Dwelling Buildings Loads, I will explain the following points:
- Where and how to distribute each type of load in a dwelling unit as per NEC code?
- How to calculate its Demand load for feeder and service sizing calculations?
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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:
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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)
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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.
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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.
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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.
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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.
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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.
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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.
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Important!!!
Fixed electric
space heating shall be considered continuous load.
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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.
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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.
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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.
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Important!!!
Depending on the
design, the heating system and the air conditioning system might be
noncoincident loads.
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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.
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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.
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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.
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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.
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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.
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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
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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.
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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
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Important!!!
Noncoincident loads are not limited to heating and air
conditioning systems.
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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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