CLEARING UP MULTIFAMILY DWELLING UNIT CALCULATIONS


Calculating multifamily dwelling unit service sizes needn’t be confusing.

-- Mike Holt

Multifamily service calculations (Figure 10-1) can seem confusing, especially since you have a choice of methods-standard (Article 220, Part II) and optional (Article 220, Part III). Breaking down each method into its component steps clears up the calculation confusion. But, which of the two methods should you use?

On electrical exams, use the standard method unless the question specifies the optional method. In the field, the optional method is usually preferable-if the dwelling qualifies for it. The optional method has fewer steps and results in a smaller service. In either case, the calculations differ somewhat from the single-family dwelling calculations.

Standard method

This is a six step process.

Step 1: Determine the general load. This consists of the general lighting and receptacles (3VA per sq ft), small appliance (3,000VA) and the laundry (1,500VA) circuit loads of all dwelling units [Table 220.11]. You can omit the laundry load (1,500VA), if a central laundry facility is available to all building occupants [210.52(F) Ex. 1].

Because these circuits will not all be on (loaded) simultaneously, 220.16 permits you to apply Table 220.11 demand factors to the total connected load as follows:

  • First 3,000VA at 100% demand
  • Next 117,000VA at 35% demand
  • Remainder at 25% demand

Step 2: Determine air-conditioning versus heat [220.15, 220.21]. Because the air-conditioning and heating loads don’t run simultaneously, you can omit the smaller of the two loads. Observe these two requirements:

  • Calculate the air-conditioning demand load at 100%-in other words, assume the air-conditioning runs at maximum.
  • Calculate electric space-heating loads at 100% of the total connected load [220.15]-in other words, assume all the space heaters run simultaneously.

Step 3: Determine the appliance demand load [220.17]. You can use a demand factor of 75% for four or more appliances fastened in place. These are such appliances as a dishwasher, waste disposal, trash compactor, or water heater-not space-heating equipment [220.15], clothes dryers [220.18], cooking appliances [220.19] or air-conditioning equipment.

Step 4: Determine the clothes dryers load [220.18]. Use 5,000W or the nameplate rating-whichever is greater. Then, adjust that per the demand factors in Table 220.18. You can omit the dryer load if the dwelling units do not have hook-ups for electric dryers. Don’t calculate laundry room dryers per this method.

Step 5: Determine the cooking equipment loads [220.19]. For household cooking appliances rated over 13/4kW, calculate feeder and service loads per the demand factors of 220.19, Table and Notes.

Step 6: Size the feeder and service conductors. First, add up all of your loads. How large the conductors must be depends on whether the load is greater than 400A. If it’s 400A or less, size the ungrounded conductors per Table 310.15(B)(6) for 120/240V, 1 systems. If it’s over 400A, size the ungrounded conductors per Table 310.16 to the calculated demand load.

Applying the steps

To see how these steps work, let’s do a sample calculation.

Step 1: Determine the general load.

What is the demand load for a 20-unit apartment building? Each apartment is 840 sq ft., and there is a central laundry facility provided on the premises for all tenants [210.52(F)].

(a) 5,200VA (b) 40,590VA (c) 110,400VA (d) none of these

Answer: (b) 40,590VA

How did we get that answer? We add up the loads and apply the demand factor, this way:

General Lighting: (840 sq ft x 3VA); 2,520VA
Small-Appliance Circuits: 3,000VA
Laundry Circuit: 0VA

Total connected load for one unit: 5,520VA

Demand Factor per Table 220.11: 5,520VA x 20 units = 110,400VA

First 3,000VA at 100%: 3,000VA x 1.00 = 3,000VA
Next 117,000VA at 35%: 107,400VA x 0.35 = 37,590VA

Total Demand Load = 3,000VA + 37,590VA = 40,590VA

Step 2: Determine air-conditioning versus heat.

What is the net computed air-conditioning versus heat load for a 40-unit multifamily building with A/C (3-hp, 230V) and two baseboard heaters (3kW) in each unit?

(a) 160kW (b) 240kW (c) 60kW (d) 50kW

Answer: (b) 240kW

Air-conditioning = 230V x 17A = 3,910VA
3,910VA x 40 = 156,400VA.
Heat =kW x 2 units = 6kW x 40 units = 240kW

We can omit the A/C, because it’s smaller than the heat.

Step 3: Determine the appliance demand load

What is the appliance demand load for a 20-unit multifamily building where each unit contains a 940VA waste disposal, a 1,250VA dishwasher, and a 4,500VA water heater?

(a) 100kVA (b) 134kVA (c) 7kVA (d) 5kVA

Answer: (a) 100kVA

Here are the numbers:

Waste Disposal: 40VA
Dishwasher: 1,250VA
Water Heater: 4,500VA

Total Demand Load = 690VA x 20 units x 0.75 = 100,350VA

Step 4: Determine the clothes dryers load

Each unit of a 10-unit multifamily dwelling building contains a 4.5kW electric clothes dryer. What is the feeder and service dryer demand load for the building?

(a) 5kW (b) 25kW (c) 60kW (d) none of these

Answer: (b) 25kW

Demand Load =

5kW x 10 units = 50kVA.
50kVA x 50% = 25kW

Step 5: Determine the cooking equipment loads

What is the feeder and service demand load for three ranges rated 15.6kW each?

(a) 15kW (b) 14kW (c) 17kW (d) 21kW

Answer: (c) 17kW (closest answer)

Refer to Table 220.19 Note 1-Over 12kW

(a) “Column C” demand load = 14kW (3 units).

(b) The average range (15.6kW) exceeds 12kW by 3.6kW. Increase “Column C” demand load (14kW) by 20%. Thus: 14kW x 1.2 = 16.8kW.

Step 6: Size the feeder and service conductors

What is the minimum size service conductors for a multifamily building that has a total demand load of 270kVA for a 120/208V, 3 system, if the service conductors are run in parallel?

(a) Two 300 kcmil per phase (b) Two 350 kcmil per phase
(c) Two 500 kcmil per phase (d) Two 600 kcmil per phase

Answer: (c) Two 500 kcmil per phase

To get this answer, refer to Table 310.15(B)(6)

I =VA/(E x 1.732)

I = 270,000VA/(208V x 1.732) = 750A

Amperes per parallel set: 750A/2 raceways = 375A per phase.

500 kcmil conductor has an ampacity of 380A [Table 310.16 at 75C].

You can use an 800A protection device for two sets of 500 kcmil conductors (380A x 2) [240.4(B)].

The optional method

This is a three-step process. To use this method, the dwelling must meet three conditions:

  1. No more than one feeder per dwelling unit.
  2. Each dwelling unit has electric cooking equipment. The NEC does have an exception, if the dwelling doesn’t meet this condition).
  3. Each dwelling unit has electric space heating, air conditioning, or both.

Step 1: Determine the total connected load. You need to add up the following loads:

  • General Lighting. 3VA per sq ft.
  • Small-Appliance and Laundry Branch Circuit. 1,500VA for each 20A small-appliance and laundry branch circuit.
  • Appliances. The nameplate VA rating of all appliances and motors fastened in place (permanently connected), but not the air-conditioning or heating load. Use the range and dryer nameplate ratings!
  • Air-Conditioning versus Heat. Determine which load is larger, air-conditioning or heat. Use the larger of the two at 100%.

Step 2: Determine demand load.

You determine the net computed demand load by applying the demand factor from Table 220.32 to the total connected load (Step 1). You can covert the net computed demand load (kVA) to amperes by:

Single phase: I =VA/E
Three phase: I =VA/(E x 1.732)

Step 3: Determine feeder and service conductor size. As with the standard method, how you do this depends on whether the load is greater than 400A. If it’s 400A or less, size the ungrounded conductors per Table 310.15(B)(6) for 120/240V, 1 systems up to 400A. If it’s over 400A, size the ungrounded conductors per Table 310.16-based on the calculated demand load.

As you can see, these calculations aren’t hard if you take them one step at a time. Should you do these if you are using stamped drawings? Yes. Even stamped drawings can contain errors-doing these calculations is a good way to ensure the conductor sizes meet or exceed NEC requirements. If your drawings use the optional method, you should also ensure the dwelling qualifies for that method.

Now that you’ve seen how these calculations work, you can head off service conductor sizing mistakes before construction and inspection. That eliminates delays and expensive rework. More importantly, you will have done your work with the sure knowledge the service conductors are safe for every family in that multifamily dwelling.

 

Copyright © 2003 Mike Holt Enterprises,Inc.
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