By Mike Holt, NEC Consultant

Your commercial application may have special requirements for calculating the load.

In Part 1, we looked at specific types of commercial loads. Let’s use some examples to show yet another difference in demand factors. If you were to watch daily operations at a marina [555.12], you’d see why its demand factors differ from those for, say, hotel rooms. As you can imagine, the usage is very different. The demand factors must change to make sense for the application.

The NEC permits a demand factor for the marina receptacle outlets that provide shore power for boats. Table 555.12 demand factors are based on the number of receptacles on the feeder. The receptacles must also be balanced between the phases to determine the number of receptacles on any given line.

It’s helpful to look at an example problem. What’s the demand load for a 120/240V, single-phase service required for a marina with twenty 30A, 250V receptacles and twenty 20A, 125V receptacles?

(a) 200A      (b) 420A      (c) 560A      (d) 720A

Answer: (c) 560A

The demand load is based on the number of receptacles, so distribute the receptacles equally on each line to determine the connected load per line. All the 250V receptacles connect between L1 and L2 but for the 125V receptacles, half are put on L1 and the other half on L2.

      Line 1          Line 2

Twenty 30A, 250V     20 x 30A     20 x 30A

Twenty 20A, 125V     10 x 20A     10 x 20A

Thirty receptacles      30-800A     30-800A

30 receptacles = 0.70 demand factor [Table 555.12]

Demand Load Per Line = Connected Load Per Line x Demand Factor

The connected load of Line 1 (or Line 2 since they are equal) = (20 x 30A) + (10 x 20A) = 800A per line

Demand Load per Line = 800A x 0.70

Demand Load per Line = 560A per line

Optional method

Let’s move on to yet another wrinkle: the optional method. You can use this to calculate the service load for a new restaurant that has electric space heating or electric air-conditioning (or both) [220.88].

Step 1: Determine the total connected load. Add the nameplate rating of all loads at 100 percent; include both the air-conditioning and heat load [Table 220.88 Note].

Step 2: Apply the demand factors from Table 220.88 to the total connected load calculated in Step 1.

Question: What is the service conductor ampacity required for an all-electric restaurant (120/208V, three-phase) that has a total connected load of 200 kVA?

(a) 420A      (b) 444A      (c) 520A      (d) 600A

Answer: (b) 444A

Total Connected Load 200 kVA

First 200 kVA at 80% 200 kVA x 0.80 160 kVA

Total Demand Load 160 kVA

I = VA/(E x 1.732)

I = 160,000 VA/(208V x 1.732)

I = 160,000 VA/360V

I = 444A

To determine the ungrounded service conductor size if installed parallel in two raceways:

444A/2 raceways = 222A

4/0 AWG rated 230A at 75degrC, Table 310.15(B)(16)

Welders

Commercial applications also include special loads, not just special occupancies. A welders is a very specialized load. Electric welders operate by either creating an electric arc between two surfaces or heating a metal rod until it melts. Each technique produces high momentary current draws.

The supply conductors for arc welders must have an ampacity at least that of the welder nameplate rating [630.11(A)]. If the nameplate rating isn’t available, the supply conductors must have an ampacity at least that of the rated primary current as adjusted by the multiplier in Table 630.11(A), based on the duty cycle of the welder.

Question: A nonmotor-generator arc welder has a primary current rating of 30A with a duty cycle of 30 percent. What's the minimum size branch-circuit conductor for the welder?

(a) 12 AWG      (b) 10 AWG      (c) 8 AWG       (d) 6 AWG

Answer: (a) 12 AWG

Demand Load = Primary Rating x Multiplier

[Table 630.11(A)]

Demand Load = 30A x 0.55

Demand Load = 16.50A

12 AWG rated 20A at 60degrC [110.14(C)(1)(a), Table 310.15(B)(16)]

Groups of welders

Conductors that supply a group of resistance welders must have an ampacity that’s at least the sum of the value determined using 630.31(A)(2) for the largest welder in the group, plus 60 percent of the values determined for all remaining welders [630.31(B)].

Question: What’s the minimum size conductors that supply five 30A resistance welders with a duty cycle of 30 percent?

(a) 4 AWG      (b) 3 AWG      (c) 2 AWG      (d) 1 AWG

Answer: (a) 4 AWG

Demand Load = Primary Rating x Multiplier

[Table 630.31(A)(2)] x Welder Percentage [630.31(B)]

Welder 1: 30A x 0.55 = 16.50A x 100% 16.50A

Welder 2: 30A x 0.55 = 16.50A x 60% 9.90A

Welder 3: 30A x 0.55 = 16.50A x 60% 9.90A

Welder 4: 30A x 0.55 = 16.50A x 60% 9.90A

Welder 5: 30A x 0.55 = 16.50A x 60% Total Demand Load 56.10A

Conductor size: 4 AWG rated 70A at 60degrC [110.14(C)(1)(a), Table 310.15(B)(16)]

Putting it all together

This example problem helps put various calculation pieces together in a typical light industrial situation. Question: Calculate the service size for a light industrial manufacturing building with the following loads:

• Lighting, 11,600 VA, comprised of electric-discharge luminaries operating at 277V.
• Twenty-two 20A, 125V receptacle outlets on general-purpose branch circuits, supplied by a separately derived system (since system for building is 480V three-phase).
• Three nonmotor generator arc welders (23A, 480V, 60% duty).
• Air compressor, 460V, three-phase, 5 hp. ¥ Grinder, 460V, three-phase, 1½ hp.
• Three continuous use industrial process dryers, 480V, three-phase, 15 kVA each.

(a) 100A      (b) 110A      (c) 125A      (d) 150A

Answer: (d) 150A

 

Step 1: Determine the noncontinuous loads and motor loads.

The noncontinuous and motor loads can be combined [430.24].

(1) Noncontinuous Loads

(a) Receptacle Load [220.44]

22 receptacles at 180 VA 3,960 VA

(b) Welder Load [630.11(A) and Table 630.11(A)]

A multiplication factor of 0.78 is allowed for

60% duty cycle welders.

Each welder: 480V x 23A x 0.78 = 8,600 VA

Welder demand factors for 3 welders: [630.11(B)]

First welder 100%

Second welder 100%

Third welder 85%

8,600 VA x 100% = 8,600 VA

8,600 VA x 100% = 8,600 VA

8,600 VA x 85% = 7,320 VA

Total Welder Demand Load = 24,510 VA

Total Noncontinuous Loads (receptacles and welders) = 28,470 VA

(2) Motor Loads [430.24 and Table 430.250]

(a) Air Compressor: 5 hp FLC = 7.60A

[Table 430.250] 480V x 7.60A x 1.732 = 6,318 VA

(b) Grinder: 1½ hp FLC = 3A

[Table 430.250] 3A x 480V x 1.732 = 2,494 VA

(c) Largest Motor, Additional 25 percent: 6,310 VA x 25% = 1578

Total Motor Loads = 10,390 VA

Total Noncontinuous Loads and the Motor Loads [430.24] 38,860 VA

Step 2: Determine the continuous loads.

(1) General Lighting = 11,600 VA

(2) Three Industrial Process

Dryers,15 kVA each = 45,000 VA

Total Load 56,600 VA

Total Continuous Load at 125% = 56,600 VA x 1.25 = 1,4150 VA

Total Continuous Load = 70,750 VA

 

Step 3: Determine the overcurrent protection [215.3].

The overcurrent device must be sized at 125 percent of the continuous loads, plus the noncontinuous loads:

(1) Noncontinuous Loads 38,860 VA

(2) Continuous Loads 70,750 VA

Total VA 109,610 VA

 

Step 4: Convert to amperes to size the overcurrent device:

I = VA/(E x 1.732)

I = 109,610 VA/(480V x 1.732)

I = 109,610 VA/831V

I = 132A

Overcurrent Device, next size up [240.4(B)] =   150A [240.6]

Conductors = 1/0 AWG. These have an ampacity of 150A at 75degrC [110.14(C)(1)(b) and Table 310.15(B)(16)]. Keep in mind that parallel conductors must comply with 310.10(H).

Keeping commercial calculations code compliant

With so many application-specific requirements, commercial calculations could become confusing. Some final thoughts here can help you avoid that.

The calculations used to size feeders and service conductors for commercial installations don’t include the same demand factors allowed for dwelling units. The nature of nondwelling occupancies doesn’t create the same kinds of diversity found in homes and apartment buildings.

Article 220 isn’t the only Article that contains service calculation requirements. For example, you’ll also find them in Article 550 (mobile home parks), Article 555 (marinas), Article 630 (electric welders), and Article 430 (motors). When performing commercial service calculations, be careful to look at all components of the installation to determine if some other Article applies to your installation.

You can apply certain demand factors to commercial services. When dealing with individual branch circuits and feeders for continuous loads, remember to use a 125 percent multiplier for the ungrounded conductors. This carries over to the continuous duty portions of loads when determining the actual feeder and service conductors for commercial occupancies.

You can size the neutral conductor at 100 percent of the continuous and noncontinuous loads for branch circuits and feeders when it’s not connected to an overcurrent device [210.19(A)(1) Ex 2 and 215.2(A)(1) Ex 2]. Optional calculations are allowed for some commercial installations, so be certain to see if your application is one of those.