### Non-Dwelling Buildings Load Calculations- Part Seven

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

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: Schools, Existing Installations, New Restaurants.

I explained the first four parts as follows:

Non-Dwelling Buildings Load Calculations- Part One ".

3- Load Calculation for Kitchen Equipment Loads of Non-dwelling Building in Article
" Non-Dwelling Buildings Load Calculations- Part Five ".

4- Load Calculation for Heating, Ventilation and Air Conditioning Systems Loads (Non-Coincident Loads) of Non-dwelling Building in Article " Non-Dwelling Buildings Load Calculations- Part Six ".

Today, I will explain the Load Calculation for Motor and Other Loads of Non-Dwelling Building.

 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 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 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) : Motors built for low speeds (less than 1,200 rpm) or high torques for multispeed motors. 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. 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 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, 230-volt, single-phase motor?

Solution:

Find the full-load current in amperes for the given motor. As shown in Table 430.248, the current for a 2 hp, 230-volt motor is 12 amperes.

Example#2:

What is the minimum rating in amperes for conductors supplying a 10 hp, 208-volt, 3-phase motor and a 7½ hp, 208-volt, 3-phase motor?

Solution:

The full-load current from Table 430.250 for a 10 hp, 208-volt, 3-phase motor is 30.8 amperes. The full-load current from the same table for a 7½ hp, 208-volt, 3-phase motor is 24.2 amperes.
Multiply the highest full-load current by 125 percent = 30.8 x 125 % = 38.5 A

Now add to this value to the full-load 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, 230-volt, 3-phase motor; a 10 hp, 230-volt, 3-phase motor; a 15 hp, 230-volt, 3-phase motor; and a 10 hp, 230-volt, single-phase motor?

Solution:

Full-load currents for 3-phase motors are in Table 430.250, and full-load currents for single-phase motors are in 430.248.

The full-load current for a 5 hp, 230-volt, 3-phase motor is 15.2 amperes;

The full-load current for a 10 hp, 230-volt, 3-phase motor is 28 amperes;

The full-load current for a 15 hp, 230-volt, 3-phase motor is 42 amperes;

And the full-load current for a 10 hp, 230-volt, single-phase motor is 50 amperes

Although the 15 hp motor has the largest amount of horsepower, the motor with the highest full-load current rating is the single-phase motor.

Multiply the motor with the highest full-load current rating by 125 percent = 50 × 125% = 62.5 A Now add to this value to the full-load 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, 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

 Rule#3: Motor Load Balance It is permissible to balance the motors as evenly as possible between phases before performing motor-load calculations.

Example#5:

What is the minimum rating in amperes for conductors supplying a 10 hp, 208-volt, 3-phase motor and three 3 hp, 120-volt, single-phase motors?

Solution:

First, find the full-load currents for the motors.

The full-load current for a 10 hp, 208-volt, 3-phase motor is 30.8 amperes [Table 430.250].

Full-load currents for single-phase motors are in 430.248. Note that the currents listed in the 115-volt column are permitted for system voltage ranges of 110 to 120 volts.

The full-load current for a 3 hp, 120-volt motor is 34 amperes.

Next, balance the motors as evenly as possible between phases.

Connect the 3-phase motor to each of three ungrounded (hot) conductors.

Because the 3 hp motors are 120-volts, 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 full-load 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 full-load 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 : General Lighting, Track Lighting, Show Window Lighting, Sign and Outline Lighting. While the Loads under the following categories are Not-continuous loads: General-Use Receptacles, Multioutlet Assemblies.

 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.

 Definition: All Other Loads: are the loads that don’t falling under one of the following loads categories: General lighting, Track lighting, Show window lighting, Sign and outline lighting, General-use receptacles, Multioutlet assemblies, Kitchen equipment, HVAC loads, Motor loads.

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 Non-continuous 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 Non-Dwelling 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 branch-circuit 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 Non-Dwelling Buildings (Schools, Existing Installations and New Restaurants) by using NEC optional method. Please, keep following.

#### 1 comment:

1. Aslam Walaicom
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