Dwelling Unit Calculations

The service neutral load for household electric clothes dryers is permitted to be calculated at _____ percent of the load as determined by 220.54.

70 Response Feedback: 220.61(B)(1)

If the total demand load is 30 kW for a 120/240V dwelling unit, what's the copper conductor size?

2 AWG Response Feedback: [310.12] I = VA/E I = 30,000 VA/240V I = 125A minimum service disconnect rating Service Conductor: 125A × 0.83=104A, 2 AWG, rated 115A [Table 310.16]

What's the general lighting load and general-use receptacle load for a 2,100 sq ft home that has 74 receptacle outlets and nine recessed luminaires rated at 100 W each?

6,300 VA Response Feedback: (2,100 sq ft @ 3 VA) No additional load is required for general use receptacles and lighting outlets [Table 220.12, Note a].

What's the feeder or service demand load for a waste disposer (940 VA), dishwasher (1,250 VA), and a water heater (4,500W)?

6,690 VA Response Feedback: Waste Disposer 940 VA Dishwasher 1,250 VA Water Heater 4,500 VA6,690 VA

What's the feeder or service demand load for an 11.50 kW range?

8 kW Response Feedback: [Table 220.55, Column C]: 8 kW

What's the general lighting load for a 2,700 sq ft dwelling unit?

8,100 VA Response Feedback: [220.12] 2,700 sq ft @ 3 VA = 8,100 VA

What's the feeder or service demand load for one air conditioner (28A, 230V) and three baseboard heaters (3 kW each)?

9,000 VA Response Feedback: Air Conditioner: 5 hp, 230V FLC = 28A [Table 430.248] Air Conditioner VA = (V × A) Air Conditioner VA = 240V × 28A Air Conditioner VA = 6,720 VA* (omit) Heat: [220.51] 3,000 × 3 = 9,000 VA *Use the larger of the two loads [220.60].

What's the feeder or service demand load for one 6 kW and two 3 kW cooking appliances?

9.3 kW Response Feedback: Table 220.55 and Table 220.55 Note 3]Column B: 6 kW x 0.80 4.80 kW Column A: 3 kW x 2 units 6 kW x 0.75 + 4.50 kW Calculated Load 9.30 kW

Household cooking appliances rated 1kW can have the feeder and service loads calculated according to the demand factors of Table 220.55.

False

A feeder or service dryer load is required even if the dwelling unit doesn't contain an electric dryer.

False Response Feedback: 220.54 requires a dryer load only if the dryer is installed in thedwelling unit.

A load demand factor of 75 percent is permitted for _____ or more appliances fastened in place such as waste disposers, dishwashers, trash compactors, water heaters, and so forth.

Four Response Feedback: 220.53

What size service conductors are required if the demand load for a dwelling unit equals 120A, and the service disconnect is rated 125A?

2 AWG Response Feedback: Service Conductor = Service Rating multiplied by 83% Service Conductor = 125A × 0.83 Service Conductor =104A [310.12] 2 AWG rated 115A at 75°C [Table 310.16]

What's the dwelling unit service neutral load for one 9 kW range?

5.6 kW Response Feedback: The neutral load is based on the feeder calculated load derived from Table 220.55 Column C, which is 8 kW for one unit. The neutral calculated load is based on 70% of the feeder calculated load of 8 kW [220.61(B)(1)]. Range Neutral Load = Range Load × 70% [220.61(B)(1)] Range Neutral Load = 8 kW × 0.70 Range Neutral Load = 5.60 kW

The feeder or service neutral load for household cooking appliances, such as electric ranges, wall-mounted ovens, or counter-mounted cooking units, is permitted to be calculated at _____ percent of the load as determined by 220.55.

70 Response Feedback: 220.61(B)(1)

What's the feeder or service demand load for a 13.60 kW range?

8.8 kW Response Feedback: The Column C value (8 kW) must be increased 5% for each kW or major fraction of a kW (0.50 kW or larger) over 12 kW [220.55 Note 1]. 13.60 kW - 12 kW = 1.60 kW (round up to 2) 2 x 5% = 10%, an increase of 10% of the Column C value results in 110%, or a 1.10 multiplier Calculated Load = Column C Value x Multiplier Calculated Load = 8 kW x 1.10 Calculated Load = 8.80 kW

The NEC recognizes that the general lighting and receptacles, and small-appliance and laundry circuits won't all be on or loaded at the same time and permits a(n) _____ to be applied to the total of these loads. (Use the correct code terminology)

Demand Factor Response Feedback: [Article 100, 220.42, and Table 220.42]

The NEC requires a minimum of 3 VA for each square foot of living space for the required general lighting and general-use receptacles for dwelling units. The dimensions for determining the area are computed from the outside of the building and don't include _____.

Open porches, garages, and spaces not adaptable for future use Response Feedback: 220.12

The feeder or service demand load for electric clothes dryers located in dwelling units must not be less than _____.

The greater of 5,000W or the nameplate rating Response Feedback: 220.54

The NEC doesn't require a separate circuit for the washing machine, but does require a separate circuit for the laundry room receptacle or receptacles, of which one can be for the washing machine.

True Response Feedback: 210.11(C)(2)

The 3 VA for each square foot of living space for general lighting includes receptacles rated 15A and/or 20A, but not the small-appliance or laundry circuit receptacles.

True Response Feedback: 220.12 and 220.14(J)

Because the air-conditioning and heating loads aren't on at the same time they're called "noncoincidental," and it's permissible to omit the smaller of the two loads.

True Response Feedback: 220.60

What size single-phase, 3-wire feeder is required for a 110A service of which 36A consist of 240V loads? Size the conductors based on 75°C terminals in accordance with 110.14(C)(1)(b).

Two 3 AWG and one 4 AWG Response Feedback: [110.14(C), 220.61, 310.12(A), and Table 310.16

When sizing the feeder or service conductor according to the optional method we must:

a. Determine the total connected load of the general lighting and receptacles, small-appliance and laundry branch circuits, and the nameplate VA rating of all appliances and motors. Determine the demand load by applying the following demand factors to the total connected load: The first 10 kVA at 100 percent, and the remainder at 40 percent. Determine the air-conditioning versus heat demand load.

What size service conductors are required if the demand load for a dwelling unit equals 140A, and the service disconnect is rated 150A?

1 AWG Response Feedback: Service Conductor = Service Rating multiplied by 83% Service Conductor = 150A × 0.83 Service Conductor =125A [310.12] 1 AWG rated 130A at 75°C [Table 310.16]

Each dwelling unit must have a feeder load consisting of _____ for the 20A laundry circuit.

1,500 VA Response Feedback: 220.52(B)

What's the total connected load before demand factors for general lighting and receptacles, small-appliance, and laundry circuits for a 6,540 sq ft dwelling unit? (Note: This is asking for total connected load, not the demand load. This calculation is the calculated load before we apply the demand factors of Table 220.42)

24,120 VA Response Feedback: 220.14(J) and 220.52] General Lighting and Receptacles:6,540 sq ft x 3 VA 19,620 VA Small-Appliance Circuits: 1,500 VA x 2 3,000 VA Laundry Circuit: 1,500 VA x 1 1,500 VA Total Connected Load 24,120 VA

When sizing the feeder or service, each dwelling unit must have a minimum feeder load of _____ for the two small-appliance branch circuits.

3,000 VA Response Feedback: 220.52(A)

What's the demand load for an air conditioner (18A, 230V) and electric space heating (4 kW)?

4,140 VA Response Feedback: Air Conditioner VA Load = 18A × 230V Air Conditioner VA Load = 4,140 VA Use the larger of the two, which is the air-conditioning [220.60].

What's the feeder neutral load for one 6 kW household dryer?

4.2 kW Response Feedback: [220.61(B)(1)] The neutral load is based on the feeder calculated load derived from Table 220.54, which is the nameplate for one unit. The neutral calculated load is based on 70% of the feeder calculated load of the nameplate [220.61(B)(1)]. Dryer Neutral Load = Dryer Nameplate × 0.70 Dryer Neutral Load = 6 kW × 0.70

What's the feeder or service demand load for two 3 kW cooking appliances?What's the feeder or service demand load for two 3 kW cooking appliances?

4.5 kW Response Feedback: Table 220.55 applies to cooking appliances (household) over 1 3/4 kW.

What's the demand load for a 4.50 kW dryer?

5 kW Response Feedback: 220.54

What's the feeder or service demand load for a 4 kW dryer?

5 kW Response Feedback: The dryer load must not be less than 5,000 VA or the nameplate rating if greater than 5 kW (for standard calculations) [220.14(B) and 220.54]. This doesn't apply to optional calculations [220.82].

What's the net demand load for the general lighting, small-appliance, and laundry circuits for a 1,500 sq ft dwelling unit?

5,100 VA Response Feedback: [220.52(A), 220.52(B), Table 220.12, and Table 220.42] General Lighting 1,500 sq ft × 3 VA = 4,500 VA 2 Small-Appliance Circuits (1,500 × 2) 3,000 VA Laundry 1,500 VA 9,000 VA -3,000 VA at 100% = 3,000 VA 6,000 VA at 35% = 2,100 VA Total Calculated Load 5,100 VA Note: 15A and 20A receptacles are considered part of the general lighting load (3 VA) [Table 220.12].

What's the feeder or service demand load for one 6 kW cooking appliance?

4.8 kW Response Feedback: [Table 220.55 Column B]Calculated Load = Total Nameplate Rating x DemandFactor [Table 220.55]Calculated Load = 6 kW x 0.80 [Table 220.55 B]Calculated Load = 4.80 kW

What size service conductor is required for a 2,000 sq ft dwelling unit containing the following loads? Use the standard method for this calculation.- Dishwasher 1,000 VA- Waste Disposer 1,000 VA- Water Heater 4,500W- Dryer 5,500W- Range 14,000W- Air-Conditioning 20A, 240V- Electric Heating (one control unit) 9,600W

1 AWG Response Feedback: Selected Answer: d. 1/0 AWG Answers: a. 2/0 AWG b. 1 AWG c. 3/0 AWG d. 1/0 AWG Response Feedback: Step 1: General Lighting and Receptacles(2,000 sq ft × 3 VA) [220.12] 6,000 VA Small-Appliance Circuits (1,500 VA × 2) [220.52(A)] 3,000 VA Laundry Circuit (1,500 VA × 1) [220.52(B)] +1,500 VA Total Connected Load 10,500 VA First 3,000 VA at 100% [Table 220.42] -3,000 VA x 1.00 3,000 VA Remainder at 35% 7,500 VA x 0.35 +2,625 VA Demand Load 5,625 VA Step 2: Appliance Demand Load [220.53]Dishwasher 1,000 VA Waste Disposer 1,000 VA Water Heater +4,500 W Demand Load 6,500 VA 6,500 VA The demand factor of 75% doesn't apply, since there are only three appliances Step 3: Dryer Demand Load [220.54]5,500W = 5,500W Step 4: Cooking Equipment Demand Load [220.55]Step a: Determine the column C value. 14 kW exceeds 12 kW by 2 kW. The Column C value (8 kW) must be increased 5% for each kW over 12 kW [Table 220.55, Note 1]. 2 × 5% = 10% increase of the Column C value, resulting in 110%, or a 1.10 multiplier. Step b: Determine the demand load. Demand Load = Table 220.55 Column C kW Value × Note 1 Multiplier Demand Load = 8 kW × 1.10 Demand Load = 8,800 W 8,800 W Step 5: Air-Conditioning versus Heat DemandLoad [220.60]A/C VA Load = Volts × Amperes A/C VA Load = 240V × 20A A/C VA Load = 4,800 VA 4,800 VA/1,000 = 4.80 kVA, omit [220.60] Heat = 9.60 kW 9,600 W Step 6: Service Size [310.15(B)(7)]Step 1. General Lighting Demand Load 5,625 VA Step 2. Appliance Demand Load 6,500 VA Step 3. Dryer Demand Load 5,500 W Step 4. Cooking Equipment Demand Load 8,800 W Step 5. Heating Demand Load +9,600 W Total Demand Load 36,025 VA Service Size in Amperes = VA Demand Load/System Volts Service Size in Amperes = 36,025 VA/240V Service Size in Amperes = 150A, 150A service [240.6(A)] Service Conductor = 83% of Service Rating [310.15(B)(7)] Service Conductor = 150A × 83% 124.50A, 1 AWG rated 130A at 75oC [Table 310.15(B)(16)]

What's the feeder or service demand load for a waste disposer (940 VA), dishwasher (1,250 VA), a trash compactor (1,100 VA), and a water heater (4,500W)?

5,843 VA Response Feedback: Waste Disposer 940 VA Dishwasher 1,250 VA Trash Compactor 1,100 VA Water Heater + 4,500 VA Connected Load 7,790 VA Calculated Load = Connected Load x Demand Factor[220.53]Calculated Load = 7,790 VA x 0.75

What's the demand load for a dwelling unit that has one dishwasher (1.50 kVA) and one water heater (4 kW)?

5.5 kVA Response Feedback: [220.53] Dishwasher 1,500 VA Water Heater 4,000 W Calculated Load at 100% 5,500 VA Calculated Load in kVA 5,500 VA /1,000 = 5.50 kVA

What's the feeder or service demand load for a 5.50 kW dryer?

5.5 kW Response Feedback: The dryer load must not be less than 5,000 VA or the nameplate rating if greater than 5 kW (for standard calculations) [220.14(B) and 220.54]. This doesn't apply to optional method calculations

The feeder and service neutral load is the maximum unbalanced calculated load between the neutral conductor and any one ungrounded conductor as determined by Article 220. Since 240V loads can't be connected to the neutral conductor, 240V loads aren't considered for sizing the feeder grounded conductor.

True Response Feedback: 220.61

A 1,500 sq ft dwelling unit contains the following loads:- Waste Disposal (1 kVA)- Dishwasher (1.50 kVA)- Water Heater (5 kW)- Dryer (5.50 kW)- Cooktop (6 kW)- Two Ovens (each 3 kW)- Air-Conditioning (3 hp, 230V)- Two separately controlled 5 kW space-heating unitsUse the optional method of calculation to determine what size aluminum conductors are required for the 120/240V, single-phase service.

1 AWG AL Response Feedback: [220.82] Step 1: Determine the total connected load.General Lighting 1,500 sq ft × 3 VA 4,500 VA Small-Appliance Circuits 1,500 VA × 2 circuits 3,000 VA Laundry Circuit + 1,500 VA 9,000 VA Appliances (nameplate ratings) Waste Disposer 1,000 VA Dishwasher 1,500 VA Water Heater 5,000 VA Dryer 5,500 VA Cooktop 6,000 VA Ovens 3,000 VA × 2 + 6,000 VA Connected Load 34,000 VA Step 2: Determine the calculated load.Connected Load 34,000 VA First 10,000 VA at 100% 10,000 VA at 100% = 10,000 VA Remainder at 40% 24,000 VA at 40% = +9,600 VA Calculated Load 19,600 VA Step 3: Determine the air-conditioning versus heat.Air-conditioning or heat-pump compressors at 100% versus 65% of one, two, or three separately controlled electric spaceheating units [220.82(C)]. Air-Conditioning: 3 hp, 230V FLC = 17A [Table 430.248] Air-Conditioning VA = V × A Air-Conditioning VA = 240V × 17A Air-Conditioning VA= 4,080 VA, [220.60] Heat [220.82(C)(4)]: 10,000 VA × 0.65 = 6,500 VA Use the heat since it's the larger of the two. Step 4: Determine the total calculated load.From Step 2 19,600 VA From Step 3 + 6,500 VA Total Calculated Load 26,100 VA I = VA/E I = 26,100 VA/240V I = 109A minimum service disconnect rating is 110A [240.6(A)]. Feeder/Service Conductors: 1 AWG AL rated 100A (110A × 0.83 =91A minimum conductor ampacity) [Table 310.12].

What's the feeder or service demand load for an air conditioner (28A, 230V) and electric space heater (9.60 kW with a 3A, 240V blower motor)?

10,320 VA Response Feedback: Air Conditioner: 5 hp, 230V FLC = 28A [Table 430.248] Air Conditioner VA = (V × A) Air Conditioner VA = (240V × 28A) Air Conditioner VA = 6,720 VA* (omit) Heat: 9,600 VA + (3A × 240V) = 10,320 VA *Use the larger of the two loads [220.60].

A minimum of _____ 20A small-appliance branch circuit(s) is(are) required for receptacle outlets in the kitchen, dining room, breakfast room, pantry, and similar areas.

2 Response Feedback: 210.11(C)(1) and 210.52(B)(1)