Last month’s article discussed the process to determine what load we will use to ensure the feeder conductors are sized appropriately. Remember, it’s not as simple as just going around and adding up nameplate values and picking a big enough wire size. Sure, this method might work, but it is likely going to give us a conductor that is way oversized and might make the difference between getting the project and having too high of a bid.

Now that we have an idea of how to determine the load our conductors must supply, let’s go through an example calculation to show how the process works. For this example, we will use a multibuilding industrial complex where there is a centrally located electrical service building and each of the manufacturing buildings are supplied by separate feeders. 

How to determine the load

Two new buildings will be added as part of an expansion, and we will run both 480/277V feeders in the same raceway to a convenient location to split and run into the respective buildings. Each feeder will contain a neutral conductor because the lighting will be 277V, but the connection to the equipment grounding conductor and grounding electrode system occurs at the service building. The overall length of each feeder is around 250 feet with about 200 feet in the shared raceway, and the ASHRAE Handbook lists the ambient temperature as 88°F.

Let’s look at the building’s equipment we are sizing the feeder to supply. Since we do not know the exact lighting load, we will be using Table 120.42(A) to determine that. The rest of the building contains, as follows:

• 50 receptacles at 120V and 20A circuits supplied by a separately derived system
• Two air compressor motors at 7.5 horsepower (HP), 460V, three-phase
• Four water pump motors at 1.5 HP, 460V, three-phase
• Three ceiling fan motors at 1 HP, 460V, three-phase
• Two industrial process water heaters at 20 kilowatts (kW), 480V, continuous
• Two industrial dryers at 10 kW, 480V, continuous
• One polycarbonate injection machine at 22.5 kW, 480V, noncontinuous
• The building is 10,000 sq. ft. 

First, we will determine the lighting load since that is still unknown. Table 120.42(A) lists the lighting demand for a manufacturing facility as 2.2 VA/sq. ft. 

Because our building is 10,000 sq. ft., that means we will use 22,000 VA for the lighting load. Table 120.45 lists no demand factors for manufacturing facilities, so we will use 100% of this and consider it a continuous load. 

However, the receptacle load might have a demand factor, since nondwelling receptacle loads above 10 kVA can have a demand factor of 50% applied. With 50 receptacles at 180 VA each, we fell just shy of the 10 kVA mark at 9,000 VA.

To figure out the load demand for the motors, we need to look up the FLC value for each motor in Table 430.250 and convert these values to VA so we can add them to the values we have.

Motor loads:

• 7.5 HP air compressors: 11A × 480V × 1.732 = 9,145 VA each
• 1.5 HP water pumps: 3A × 480V × 1.732 = 2,494 VA each
• 1 HP ceiling fans: 2.1A × 480V × 1.732 = 1,746 VA each
• Total motor load = 33,504 VA + 25% of largest motor = 35,790 VA

Keep in mind we only need to determine 125% of a single air compressor motor, not both. We can now add the motor load with the noncontinuous loads to avoid adding an additional 25% that will get applied to the continuous load. Granted, that would not pose a hazard but would likely require a larger size conductor than is ultimately necessary to supply this load.

The rest of the equipment on the list will simply add to what we have already determined. So now let’s look at adding up the continuous and noncontinuous loads.

Continuous loads:

• General lighting: 22,000 VA
• Industrial process water heaters: Two at 20 kW = 40,000 VA
• Industrial dryers: Two at 10 kW = 20,000 VA
• Total continuous load = 82,000 VA 
• Total amperes = 82,000 ÷ (480 × 1.732) = 99A

Noncontinuous and motor loads:

• Receptacle load = 9,000 VA
• Injection machine = 22,500 VA
• Total motor load = 35,790 VA
• Total noncontinuous and motor load = 67,290 VA 
• Total amperes = 67,290 VA ÷ (480 × 1.732) = 81A

Now that we have the total load supplied by the feeder, we can apply the comparison rule from Section 215.4. In 215.4(A)(1), we must determine the minimum size conductor based on the noncontinuous load plus 125% of the continuous load. This gives us a minimum ampacity of 204.75A, which, if we are using XHHW-2 copper conductors, would require a 4/0 AWG conductor from the 75°C column. Remember that while the insulation has a 90°C rating, Section 110.14(C) only allows us to use the 90°C ampacity for derating purposes unless we can verify that the terminations are listed for 90°C temperatures, which is not common.

Next, we will determine what size conductor is needed to supply the load (without adding 125% for continuous) after the application of adjustment and correction factors. Remember, the ambient temperature is 88°F, and there are two feeders in a shared raceway. However, here we can derate from the 90°C column if we are not using the conductor with a final derated ampacity that exceeds that of the 75°C column. 

We will count the neutrals as current-­carrying conductors, since the lighting will consist of nonlinear loads. Therefore, we need a conductor that is sufficient for 180A after applying a 70% ampacity adjustment for more than three current-carrying conductors in a raceway and an ambient temperature correction factor of 0.96. We can divide 180A by these two factors in decimal form to get a minimum conductor ampacity of 268A. From the 90°C column, a 250-kcmil copper conductor is required, and we just need to verify that we are not exceeding the 75°C rating of the conductor, which we are not. 

Since a 250-kcmil conductor is larger than the 4/0 AWG required by 215.4(A)(1), this will be the size of the conductor. However, depending on the size of the neutral conductor, this might result in a rather large raceway and difficulty pulling these feeders into a shared raceway. These are all considerations that need to be factored in when making the final decision for installation. Would it be more cost effective to run two separate raceways and be able to go with the smaller wire size? This is all part of the fun of building out the systems that power the world. 

Next month’s article will take this example and use it to fuel a discussion on feeder overcurrent protection and sizing the neutral and equipment grounding conductors. Until then, stay safe and remember to always test before you touch!

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