Commercial Loads - Part 2

By Mike Holt – taken from my NEC Exam Preparation textbook, Unit 11

Calculating commercial receptacle loads, and a look at the optional calculation method for commercial loads.

Commercial Receptacle Load Calculations
The multioutlet receptacle assembly, such as a plug strip, is common in commercial applications. For each 5 ft (or fraction thereof) of multioutlet receptacle assembly, use 180 VA in your feeder/service calculations. This is assuming that it’s unlikely for the appliances plugged into this assembly to operate simultaneously [220.14(H)].
If you expect several appliances to operate simultaneously from the same multioutlet receptacle assembly, consider each foot (or fraction of a foot) as 180 VA for feeder/service calculations. A multioutlet receptacle assembly isn’t generally considered a continuous load.
Try this sample problem. What’s the calculated load for 10 work stations, each of which has 10 ft of multioutlet receptacle assembly (not used simultaneously) and 3 ft of multioutlet receptacle assembly (used simultaneously)?
(a) 5 kVA         (b) 6 kVA         (c) 7 kVA         (d) 9 kVA
Answer: (d) 9 kVA.
10 stations with 10 ft per station = 100 ft of multioutlet
     assembly not simultaneously used
10 stations with 3 ft per station = 30 ft of multioutlet assembly
     simultaneously used
100 ft/5 ft per section =              20 sections x 180 VA             3,600 VA
30 ft/1 ft per section =                30 sections x 180 VA           + 5,400 VA
Calculated Load                                                                         9,000 VA
This example shows how to size the feeder to accommodate this load.
First find the VA allowed per circuit:
120V x 20A = 2,400 VA for noncontinuous loads
Then divide by 180 VA to find how many 180 VA sections a 20A circuit can serve:
2,400 VA/180 VA = 13
Each work bench requires 2 – 180 VA sections for the ten ft section, and 3 – 180 VA sections for the 3 ft section, so that is 5 – 180VA sections per workbench. At 13 sections per circuit, a 20A branch circuit can serve two tables.

Receptacle VA load
Receptacles are generally not considered continuous loads.
The load for a general-use receptacle outlet in a non-dwelling occupancy is 180 VA per strap [220.14(I)]. The maximum number of receptacle outlets permitted on a commercial or industrial circuit depends on the circuit ampacity. Calculate the number of receptacles per circuit by dividing the VA rating of the circuit by 180 VA for each receptacle strap (also called a yoke).
Based on the Article 100 definition, a duplex receptacle is two receptacles on the same yoke. For the purposes of this calculation, a single receptacle or a duplex receptacle each count as 180 VA [220.14(I)].

Receptacle feeder/service calculated load
Calculate receptacle loads at not less than 180 VA per outlet (strap) per 220.14(I), and fixed multioutlet assemblies per 220.14(H). According to 220.44, you can add these calculated loads to the lighting loads and apply the lighting load demand factors given in Table 220.42. Alternatively, you can use the demand factors for receptacles given in Table 220.44.
The Table 220.44 receptacle demand factors are as follows:

• First 10 kVA at 100 percent demand factor.
• Remainder over 10 kVA at 50 percent demand factor.

Calculate the receptacle load using 180 VA for each single or multiple receptacle on one yoke, or strap [220.14(I)].
The receptacle calculated load for office buildings and banks is the larger calculation of (1) or (2):
(1) Determine the receptacle calculated load at 180 VA per receptacle yoke [220.14(I)], then apply the demand factor from Table 220.44.
(2) Determine the receptacle calculated load at 1 VA per sq ft.
It’s common not to know the exact number of receptacles that will eventually be installed in an office building or bank. The main structure is built first, then individual office space that’s rented out to each tenant will often have a custom installation, or a new tenant will remodel the space to suit his or her needs. A calculation of 1 VA per square foot allows a generic feeder/service demand for general receptacles.
What is the receptacle calculated load for an 18,000 sq ft bank/office building containing one hundred sixty 15A and 20A, 125V receptacles (straps)?
(a) 15,400 VA  (b) 19,400 VA  (c) 28,800 VA  (d) 142 kVA
Answer: (b) 19,400 VA
[220.14(K)(1)]
160 receptacles (straps) x 180 VA [220.14(I)] = 28,800 VA
Total Receptacle Load =        28,800 VA
First 10,000 at 100%               -10,000 VA     x 1.00  10,000 VA
Remainder at 50%                   18,800 VA       x 0.50 + 9,400 VA
    [Table 220.44]        
Receptacle Calculated Load                                      19,400 VA
Compare this to the 1 VA per sq ft method [220.14(K)(2)]
18,000 x 1 VA per sq ft = 18,000 VA (smaller answer, omit)

Sign circuits
The NEC requires each commercial occupancy accessible to pedestrians to have at least one 20A branch circuit for a sign [600.5(A)]. The load for the required exterior sign or outline lighting must be a minimum of 1,200 VA [220.14(F)]. A sign outlet is a continuous load and the feeder/service conductor must be sized at 125 percent of the continuous load [215.2(A)(1) and 230.42].
What is the feeder/service conductor calculated load for one electric sign?
(a) 1,200 VA    (b) 1,500 VA    (c) 1,920 VA    (d) 2,400 VA
Answer: (b) 1,500 VA
Feeder/Service Calculated Load = 1,200 VA x 1.25
Feeder/Service Calculated Load = 1,500 VA
Commercial/industrial vs. residential
You’ve probably noticed that receptacle calculations for commercial/industrial applications differ from those that apply to residential applications. The differences exist because in residential locations, the receptacles are generally placed much closer together for convenience purposes but used in a diverse manner, so that all receptacles are not heavily loaded during all periods of time. In commercial occupancies, there are less rules governing receptacle placement, so they may be placed as needed, but may be called into use more often and for longer periods of time. In dwelling units, the receptacle load is included in the general lighting load VA calculated according to Table 220.12 and is then subject to the demand factors of Table 220.42. The 180 VA per receptacle strap allowance does not apply to dwelling unit calculations.

In Part 1, we looked at how to calculate commercial loads using the standard method. You can save time by using the optional method. The optional method calculations are located in  Part IV of Article 220. The optional method calculations vary according to the type of building:

• New dwelling units: 220.82.
• Existing dwelling units: 220.83.
• Multifamily dwellings: 220.84.
• Schools: 220.86.
• New Restaurants: 220.88.

All-Electric Restaurant
If a restaurant has electric space heating, electric air-conditioning, or both, you can use the optional method The optional method consists of two steps:

1. Determine the total connected load. Add the nameplate rating of all loads at 100 percent, including both the air-conditioning and heating load [Table 220.88 Note].
2. Apply the demand factors from Table 220.88 to the total connected load calculated in Step 1.

An example will help illustrate how the optional method works for restaurants. What’s the calculated load for an all-electric restaurant (120/208V, three-phase) that has a total connected load of 300 kVA?
(a) 420A          (b) 472A          (c) 520A          (d) 600A
Answer: (b) 472A
Total Connected Load             300 kVA
First 200 kVA at 80%              -200 kVA         x 0.80                160 kVA
Next 201-325 kVA at 10%        100 kVA          x 0.10              +  10 kVA
Total Calculated Load                                                               170 kVA
I = VA/(E x 1.732)
I = 170,000 VA/(208V x 1.732)
I = 170,000 VA/360V
I = 472A
Paralleling two conductors per phase: 472A/2 raceways = 236A
250 kcmil is rated 255A at 75ºC: 255A x 2 conductors (in parallel)  = 510A
The minimum neutral size allowed when paralleling conductors is 1/0 AWG [250.24(C)(2) and 310.4].
What size grounding electrode conductor do you need if the service uses two sets of 250 kcmil conductors in parallel?
(a) 4 AWG       (b) 1/0 AWG    (c) 2/0 AWG    (d) 3/0 AWG
Answer: (b) 1/0 AWG
First, find the equivalent area of the parallel conductors:
            250 kcmil x 2 conductors  = 500 kcmil [Table 250.66, Note 1]
500 kcmil requires a 1/0 AWG grounding electrode conductor [Table 250.66].
The largest grounding electrode conductor to a ground rod is 6 AWG. The largest to a concrete encased electrode (Ufer) is 4 AWG [250.66(A) and 250.66(B)].

Not All-Electric Restaurant
What if the restaurant isn’t all-electric? Try calculating the load for a not all-electric restaurant (120/208V, three-phase) that has a total connected load of 300 kVA. What’s the right answer?
(a) 420A          (b) 460A          (c) 520A          (d) 695A
Answer: (d) 695A
Total Connected Load              300 kVA
First 200 kVA at 100%            – 200 kVA        x 1.00              200 kVA
201-325 kVA at 50%                 100 kVA         x 0.50             +  50 kVA
Total Calculated Load                                                             250 kVA
I = VA/(E x 1.732)
I = 250,000 VA/(208V x 1.732)
I = 250,000 VA/360V
I = 694A
Paralleling two conductors per phase: 694A/2 raceways = 347A
500 kcmil is rated 380A at 75ºC: 380A x 2 = 760A.
When a calculation result does not correspond to a standard overcurrent protection size, we are allowed to round up to the next standard size, as long as it does not exceed 800A [240.4(B)]. This would allow the use of an 800A overcurrent device [240.6(A)]
The minimum neutral size when paralleling conductors is 1/0 AWG [250.24(C)(2) and 310.4]
Fortunately, you don’t have to use both the standard and optional methods and then pick the one that is the larger. You are allowed to use either method, so you can save time by using the optional method instead ofthe standard method.

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