In Article " NonDwelling Buildings Load Calculations Part One " I introduced a List for ordinary NonDwelling Buildings Loads which was as follows:
 Lighting loads,
 Receptacles Loads,
 Kitchen Loads,
 Heating, Ventilation and air conditioning Loads (NonCoincident Loads),
 Motor Loads,
 Other Loads.
Again, but for above NonDwelling 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?
Important!!!
All design Calculations for Nondwelling 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 NonDwelling buildings
as permitted by NEC, Part IV. Optional Feeder and Service Load Calculations:

I explained the first four parts as follows:
1 Load Calculation for Lighting Loads of NonDwelling Building in Article " NonDwelling Buildings Load Calculations Part One ".
2 Load Calculation for Receptacle loads of NonDwelling Buildings in Articles:
3 Load Calculation for Kitchen Equipment Loads of Nondwelling Building in Article " NonDwelling Buildings Load Calculations Part Five ".
4 Load Calculation for Heating, Ventilation and Air Conditioning Systems Loads (NonCoincident Loads) of Nondwelling Building in Article " NonDwelling Buildings Load Calculations Part Six ".
Today, I will explain the Load Calculation for Motor and Other Loads of NonDwelling Building.
Fifth: Motor Loads
Important!!!
Regardless of the
type of structure—residential, commercial, or industrial, the Motors Loads will be calculated by in the same manner.
The motors can be fastened in place appliances or separate motors (air conditioning compressors, fan blower, etc.)

Rule#1: Largest Motor Load
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 voltamperes (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 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) :

Important!!!
Fullload currents for
3phase motors are in Table 430.250,
FullLoad Current for
TwoPhase Alternating Current Motors (4Wire) are in Table 430.249,
Fullload currents for
singlephase motors are in Table 430.248,
FullLoad Currents for DirectCurrent
Motors are in Table 430.247.

Important!!!
Kilovoltamperes (kVA) shall be considered
equivalent to kilowatts (kW) for Fastenedinplace Appliances.

Important!!!
If the Largest motor is air
conditioning compressor, 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#1:
For example, what is the minimum rating in amperes for conductors supplying a 2 hp, 230volt, singlephase motor?
Solution:
Find the fullload current in amperes for the given motor. As shown in Table 430.248, the current for a 2 hp, 230volt motor is 12 amperes.
Example#2:
What is the minimum rating in amperes for conductors supplying a 10 hp, 208volt, 3phase motor and a 7½ hp, 208volt, 3phase motor?
Solution:
The fullload current from Table 430.250 for a 10 hp, 208volt, 3phase motor is 30.8 amperes. The fullload current from the same table for a 7½ hp, 208volt, 3phase motor is 24.2 amperes.
Multiply the highest fullload current by 125 percent = 30.8 x 125 % = 38.5 A
Now add to this value to the fullload current of the other motor= 38.5 + 24.2 = 62.7 A
So, the minimum rating in amperes for conductors supplying above motors is 62.7 A
Example#3:
what is the minimum rating in amperes for conductors supplying a 5 hp, 230volt, 3phase motor; a 10 hp, 230volt, 3phase motor; a 15 hp, 230volt, 3phase motor; and a 10 hp, 230volt, singlephase motor?
Solution:
Fullload currents for 3phase motors are in Table 430.250, and fullload currents for singlephase motors are in 430.248.
The fullload current for a 5 hp, 230volt, 3phase motor is 15.2 amperes;
The fullload current for a 10 hp, 230volt, 3phase motor is 28 amperes;
The fullload current for a 15 hp, 230volt, 3phase motor is 42 amperes;
And the fullload current for a 10 hp, 230volt, singlephase motor is 50 amperes
Although the 15 hp motor has the largest amount of horsepower, the motor with the highest fullload current rating is the singlephase motor.
Multiply the motor with the highest fullload current rating by 125 percent = 50 × 125% = 62.5 A Now add to this value to the fullload currents of the other motors = 62.5 + 15.2 + 28 + 42 = 147.7 = 148 A
So, the minimum rating in amperes for conductors supplying these motors is 148 A
Example#4:
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, 240volt motor. How much load will this package unit add to a 240volt, singlephase 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 fullload current in amperes of a ½ hp, singlephase, 240volt 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
Rule#3: Motor Load
Balance
It is permissible to balance the motors as evenly as
possible between phases before performing motorload calculations.

Example#5:
What is the minimum rating in amperes for conductors supplying a 10 hp, 208volt, 3phase motor and three 3 hp, 120volt, singlephase motors?
Solution:
First, find the fullload currents for the motors.
The fullload current for a 10 hp, 208volt, 3phase motor is 30.8 amperes [Table 430.250].
Fullload currents for singlephase motors are in 430.248. Note that the currents listed in the 115volt column are permitted for system voltage ranges of 110 to 120 volts.
The fullload current for a 3 hp, 120volt motor is 34 amperes.
Next, balance the motors as evenly as possible between phases.
Connect the 3phase motor to each of three ungrounded (hot) conductors.
Because the 3 hp motors are 120volts, connect each motor to the grounded conductor and one ungrounded conductor. One motor will be on phase A, one motor on B and one on phase C. (see above image)
Although there are four motors total, there are only two motors on each phase
Because the motors are balanced between phases, the fullload current on each phase = 30.8 + 34 = 64.8A.
Because the single phase motor is the largest motor, multiply its amperes by 125 percent = 34 × 125% = 42.5 A
Now add to this value the fullload currents of the other motor on the same phase = 42.5 + 30.8 = 73.3 A = 73 A
So, the minimum rating in amperes for conductors supplying these motors is 73 A
Definition:
Continuous Load: A
load where the maximum current is expected to continue for 3 hours or more.

Important!!!
The loads under the following
categories are continuous loads :
While the Loads under the
following categories are Notcontinuous loads:

Rule#4: Continuous
Loads
When calculating a feeder or
service As per NEC Standard calculation method, the Loads classified as
Continuous Loads must be multiplied by 25 % and add it to the service
load calculation.

Sixth: All other Loads
Definition:
All Other Loads: are the
loads that don’t falling under one of the following loads categories:

Rule#5: Other Loads
All other Loads will
be calculated as follows:

Example#6:
A bank has the following loads:
 Parking Lot Lighting at 57 A, 120 V,
 Water Heater at 4 KW, 208 V.
Solution:
The two loads; Parking Lot Lighting and Water Heater are not falling under any category from the definition of All Other Loads.
So, these Loads will be calculated as follows:
Parking Lot Lighting is a continuous load, multiply its VA by 125% = 57 x 120 x 125% = 8,550 VA
The Water Heater is a Noncontinuous load, multiply its VA by 100% = 4 KW = 4,000 VA
The Service Load contribution of these loads = 8,550 + 4,000 VA = 12,550 VA
Rule#6: House
(Common Area) Loads in NonDwelling Buildings
As per NEC section 210.25(B), The systems,
equipment, and lighting for public or common areas are required to be
supplied from a separate “house load” panelboard.
This requirement permits
access to the branchcircuit disconnecting means without the need to enter
the space of any tenants. The requirement also prevents a tenant from turning
off important circuits that may affect other tenants.

In the next article, I will explain load calculations for special NonDwelling Buildings (Schools, Existing Installations and New Restaurants) by using NEC optional method. Please, keep following.
Aslam Walaicom
ReplyDeleteIt seem for me it's useful site for engineers I advice others to join, It is my first comment