CLEARING UP MULTIFAMILY DWELLING UNIT CALCULATIONS

-- 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.
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
- 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.
Applying the steps To see how these steps work, let’s do a sample calculation.
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
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.
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.
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 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 75ºC]. 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: - No more than one feeder per dwelling unit.
- Each dwelling 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.
- 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%.
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)
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.
1-888-NEC-CODE (1-888-632-2633)**