|Clearing Up Multifamily Dwelling Unit Calculations
dwelling unit service sizes needn't be confusing.
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
This is a six step
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
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:
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
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.
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
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, 1Ø systems. If it's over 400A, size the ungrounded conductors per
Table 310.16 to the calculated demand load.
Applying the steps
see how these steps work, let's do a sample calculation.
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)].
5,200VA (b) 40,590VA (c) 110,400VA (d) none of these
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);
Small-Appliance Circuits: 3,000VA
Laundry Circuit: 0VA
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 =
Next 117,000VA at 35%: 107,400VA x 0.35 = 37,590VA
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
Answer: (b) 240kW
Air-conditioning = 230V x 17A =
3,910VA x 40 = 156,400VA.
Heat =kW x 2 units = 6kW x 40 units =
We can omit the A/C, because it's smaller than the heat.
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?
(b) 134kVA (c) 7kVA (d) 5kVA
Answer: (a) 100kVA
Here are the
Waste Disposal: 40VA
Total Demand Load = 690VA x 20 units x 0.75 = 100,350VA
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
What is the feeder and service demand load for three ranges rated
(a) 15kW (b) 14kW (c) 17kW (d) 21kW
Answer: (c) 17kW
Refer to Table 220.19 Note 1-Over 12kW
"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
What is the minimum size service conductors for a multifamily
building that has a total demand load of 270kVA for a 120/208V, 3Ø 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
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
500 kcmil conductor has an ampacity of 380A [Table 310.16 at
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
No more than one feeder per dwelling unit.
unit has electric cooking equipment. The NEC does have an exception, if the
dwelling doesn't meet this condition).
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
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
Air-Conditioning versus Heat. Determine which load is larger,
air-conditioning or heat. Use the larger of the two at 100%.
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
Single phase: I =VA/E
Three phase: I =VA/(E x 1.732)
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, 1Ø
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.