EE 158 ELECTRICAL SYSTEM DESIGN

Part VI Electrical System Design for Residential Occupancies

Prof. Rowaldo R. del Mundo Department of Electrical & Electronics Engineering

University of the Philippines Diliman

Classifications I. Classification of Dwelling Types According to PEC 1. Single-Family Dwelling 2. Two-Family Dwelling 3. Multi-Family Dwelling

Note: - A single-family dwelling or any individual apartment unit in a two-family or multi-family dwelling is designated as a dwelling unit.

Scope II. Scope of Premises Wiring System 1. Branch circuits in a dwelling unit 2. Feeders to individual dwelling units in a multi-family

dwelling 3. Service

Branch Circuit Design III. Branch Circuit Design for Dwellings 1. General Lighting and Receptacle Branch Circuits

- 24 VA/sqm of floor area - 15 A or 20 A branch circuit capacity - At least 2.0 mm2 for 15 A circuits or 3.5 mm2 for 20 A copper conductors

2. Small Appliance Branch Circuits - 1500 VA - 20 A Branch circuit - At least 3.5 mm2 copper wire

Branch Circuit Design 3. Laundry Branch Circuits

- 1500 VA - 20 A Branch circuit - At least 3.5 mm2 copper wire

4. Appliance Branch Circuits - Individual branch circuits for specific appliances - Branch circuit capacity and size of conductors depend on the full load current of the appliance.

5. Motor Branch Circuits - Apply rules for motors

Calculation Methods IV. Calculation Methods for Feeder and Service Design for

Single-Family and Multi-Family Dwelling Units 1. Standard Calculation Method 2. Optional Calculation Method

Standard Calculation Method V. Standard Calculation Method - SFDU & MFDU 1. General Lighting & Receptacle Loads

a. General lighting and receptacle loads in VA b. Small appliance loads in VA c. Laundry circuit load in VA d. Subtotal VA e. Apply demand factors to compute feeder load - 1st 3000 VA at 100% DF - Next 3001 to 120000 VA at 35% DF - Remainder over 120000 VA at 25% DF f. Subtotal(1) = Net Computed General Lighting & Receptacle Load in VA

Standard Calculation Method 2. Other Loads

- Subtotal (2): Apply specific demand factors for appliances (cooking appliances, clothes dryers, etc.).

3. Total Net Computed Load for Feeder = (1) + (2) Example:

The dwelling has a floor area of 145 sqm. It has the typical household appliances including one 1.5 HP room air-conditioning unit.

Standard Calculation Method Solution: 1. Total Load:

a. General lighting and Convenience Receptacle Load: (see Table 2.20.1.3(a)) Note: Table 2.20.1.3(a) includes the receptacle outlets of 20 A or less. 145 sqm x 24 VA/sqm x 1.25 = 4350 VA The computed load is 4350/230 = 18.91 A. One branch circuit of 20-A would be theoretically adequate. However, for flexibility and to allow for future needs, provide two 15 A or 20 A branch circuits for lighting and convenience outlets.

Standard Calculation Method b. Small Appliance Load: (see sec 2.20.2.7(a)) One 20 A circuit at 1500 VA = 1500 VA Provide one 20 A small appliance circuit. c. Laundry Circuit: (see Sec 2.20.2.7(b) and 3.3.2.7(b)). One 20 A circuit at 1500 VA = 1500 VA Provide one 20 A laundry circuit. d. Subtotal = 7350 VA e. Application of Demand factors: (see Table 3.3.2.2) First 3000 VA at 100% DF = 3000 VA Remainder at 35% DF = 4350VA x 0.35 = 1523 VA f. Net Computed Load = 4523 VA

Standard Calculation Method 2. Other Loads:

One 1.5 HP room ACU, 10 A x 230 V at 100% DF = 2300 VA Provide one 20 A room ACU circuit.

3. Total Net Computed Load = 6823 VA

Standard Calculation Method 4. Circuit requirement: Use five 20 A 2-wire branch circuits. 5. Feeder and Service Entrance:

Motor Loads: 2300 x 1.25 = 2875 VA Other Loads: 4523 VA Total = 7398 VA 7398 VA / 230 V = 32.17 A Use two 8 mm2 THW wires (40 A).

Standard Calculation Method Notes:

(1) See Sec 2.30.7.1(a) for the conductor and overcurrent protection ampere rating. (2) See Sec 2.30.4.3 for minimum service entrance conductor.

Standard Calculation Method 6. Service Equipment:

a. Maximum Current Rating of Protective Device: Inverse Time Circuit Breaker

b. Service Equipment Rating Use one 40 A trip, 240-V molded-case circuit breaker.

4523 VA20 A + 39.67 A230V

=

Standard Calculation Method 7. Diagram

Standard Calculation Method Example: Single-family dwelling unit, up to 50 sqm floor

area with load not exceeding 3680 VA.

The dwelling has a floor area of 50 sqm with typical appliance loads for small residential units.

Solution: 1. Total Load: a. General Lighting and Convenience Receptacle Load:

(see Table 2.20.1.3(a)) 50 sqm x 24 VA/sqm = 1200 VA

Standard Calculation Method b. Typical Appliance Load:

One electric flat iron = 1200 VA One television set = 80 VA One electric fan = 75 VA One radio = 20 VA Total load = 2575 VA

c. Total computed current = 2575 VA / 230 V = 11.2 A

Standard Calculation Method 2. Circuit requirement:

Use one 20 A 2-wire circuit (see Sec 2.1.2.5(a)). 3. Service Entrance Conductors:

Use two 3.5 mm2 TW copper wires (minimum) See Sec 2.20.3.2(b)(3) exception no. 3

Standard Calculation Method 4. Service Equipment: a. In case of three-wire single-phase service: (1) use one 30 A, 2 PST, 250 V safety switch with two 20 A

fuses; or (2) use one 20 A trip, 2-pole, 240 V enclosed molded-case

circuit breaker.

Standard Calculation Method b. In case of two-wire single-phase service: (1) use one 30 A, 1 PST, 250 V safety switch with one 20 A

fuse; or (2) use one 20 A trip, 1-pole, 240 V enclosed molded-case

circuit breaker

Standard Calculation Method Example: Single-family dwelling unit (more than six circuits).

The dwelling has a floor area of 145 sqm. It has the typical household appliances including one 8 kW electric range, two 1 HP room air-conditioning units, one 1.5 HP room air-conditioning unit, and one 1 HP water pump.

Standard Calculation Method Solution: 1. Total Load: a. General Lighting and Convenience Receptacle Load

(see Table 2.20.1.3(a)): Note: Table 2.20.1.3(a) includes the receptacle outlets of 20 A or less. 145 sqm x 24 VA/sqm x 1.25 = 4350 VA The computed load is 4350/230 = 18.91 A. One branch circuit of 20 A would be theoretically adequate. However for flexibility and to allow for future needs, provide two 15-A or 20-A branch circuits for lighting and convenience outlets.

Standard Calculation Method b. Small Appliance Load:

One 20 A circuit at 1500 VA = 1500 VA - Provide one 20 A small appliance circuit.

c. Laundry Circuit: One 20 A circuit at 1500 VA = 1500 VA - Provide one 20 A laundry circuit.

d. Subtotal = 7350 VA e. Application of Demand factors:

First 3000 VA at 100% DF = 3000 VA Remainder at 35% DF (4350VA x 0.35) = 1522 VA

Standard Calculation Method f. Other Loads:

One 8 kW electric range at 80% DF = 6400 VA - Provide one 40 A electric range circuit. Two 1 HP room ACU, 8 A x 230 V x 2 at 100% DF = 3680 VA - Provide two 20 A room ACU circuits. One 1.5 HP room ACU, 10 A x 230 V at 100% DF = 2300 VA - Provide one 20 A room ACU circuit. One 1 HP water pump, 8 A x 230 V at 100% DF = 1840 VA - Provide one 20 A water pump circuit

g. Total Net Computed Load = 18743 VA

Standard Calculation Method 2. Circuit requirement:

Use eight 20 A 2-wire branch circuits and one 40 A 2-wire branch circuit.

3. Service Entrance Conductors: Total Full Load Current:

Use two 30 mm2 THW wires.

18438 VA + 25% (2300 VA) 83 A230V

=

Standard Calculation Method 4. Service Equipment: a. Maximum Current Rating of Protective Device, with:

i. Non-time Delay Fuse [3000 VA + 1218 VA + 6400 VA + 3680 VA

+ 300% (2300 VA) + 1840 VA] 100 A230V

=

Standard Calculation Method ii. Inverse Time Circuit Breaker [3000 VA + 1218 VA + 6400 VA + 3680 VA

+ 250% (2300 VA) + 1840 VA] 95 A230V

=

Standard Calculation Method b. Service Equipment Rating

i. In case of three-wire single-phase service: (1) use one 100 A, 2 PST, 250 V safety switch with two 100 A fuses (renewable); or (2) use one 100 A trip, 2-pole, 240 V molded-case circuit breaker. ii. In case of two-wire single-phase service: (1) use one 100 A, 1 PST, 250 V safety switch with one 100 A fuse (renewable); or (2) use one 100 A trip, 1-pole, 240 V molded-case circuit breaker.

Standard Calculation Method 5. Diagrams a. three-wire Service: b. two-wire Service: Same as the figure above except that

the main and branch overcurrent protective devices are of the single-pole type.

Optional Calculation Method VI. Optional Calculation Method for Single Family Dwelling

Unit 1. Air-conditioning loads

- Apply 100% DF 2. Central electric space heating loads

- Apply 65% DF 3. Not less than 4 separately controlled space heating

- Apply 65% DF 4. Other Loads

- First 8 kVA at 100% DF - Remainder at 40% DF

Optional Calculation Method VII. Optional Calculation Method for Multi-Family Dwelling

Unit (sec 2.20.3.4) 1. Add the total computed load without applying the demand factors for each dwelling. 2. Apply demand factors to the total computed load of the multi-family dwelling. 3. Where two family dwelling units are supplied by a single feeder and the computed load based on the standard calculation method exceeds that for three identical units based on the optional calculation method, the lesser of the two loads shall be permitted.

Optional Calculation Method VIII. Conditions for Use of Optional Method 1. Service for one-family dwelling or feeder for individual dwelling unit of multi-family dwelling

a. Supplied by single 3-wire, 230 /115V service b. The ampacity of the service entrance or feeder

conductor is 100 A or more. 2. For services or feeders for multi-family dwellings, each individual dwelling must be:

a. Supplied by single feeder b. Equipped with electric cooking equipment c. Equipped with either electric heating or air conditioning, or both

Optional Calculation Method Example: Single Family Dwelling Unit, more than 150 sqm

floor area, 230 V single-phase service.

The dwelling has a floor area of 250 sqm. It has the typical household appliances including two 6 kVA package-type air-conditioning units, six 1.5 HP room air-conditioning units, two 4 kW wall-mounted ovens, one 12 kW electric range, one 8 kW electric range, one 5 kW clothes dryer, four 5 kW water heaters, and one 1.5 HP water pump.

Optional Calculation Method 1. Air Conditioning Load:

Two 6 kVA package type AC units = 12000 VA - Provide two 60 A package type ACU circuits. Six 2 HP room AC units, 12 A x 230 V x 6 = 16560 VA - Provide six 20 A room ACU circuits. Net computed AC loads (100% DF) = 28560 VA

2. Lighting and Convenience Receptacle Load: 250 sqm x 24 VA/sqm x 1.25 = 7500 VA Provide at least two 20 A (or three 15 A) lighting and convenience receptacle units. Note: The actual computed lighting load shall be used if the total VA exceeds that obtained from the 24 VA/sqm method.

Optional Calculation Method 3. Cooking Load: (see sec 2.20.2.11)

Two 4-kW wall mounted ovens, 4000 VA x 2 = 8000 VA 4000 VA / 230 V = 17.39 A Provide two 20 A wall mounted oven circuits. One 8-kW electric range = 8000 VA - 8000 VA / 230V = 34.78 A - Provide one 40 A electric range circuit. One 12-kW electric range = 12000 VA - Provide one 40 A electric range circuit.

Optional Calculation Method 4. Appliance Loads:

One 20 A small appliance circuit = 1500 VA - Provide one 20 A small appliance circuit. One laundry washing machine = 1500 VA - Provide one 20 A laundry circuit. One 5 kW clothes dryer = 5000 VA - Provide one 30-A clothes dryer circuit Four 5 kW water heaters = 20000 VA - Provide four 30 A water heater circuits

Optional Calculation Method One 1.5 HP water pump, 10 A x 230 V = 2300 VA - Provide one 20 A water pump circuit

5. Total other loads = 65800 VA Application of Demand Factors: - First 8KVA @ 100% = 8000 VA - Remaining 57800 VA @ 40% = 23120 VA Net Other Loads = 31120 VA

6. Net Computed Loads (All) 28560 VA + 31120 VA = 59 680 VA 59680 VA / 230 V = 259.5 A

Optional Calculation Method 7. Circuit Requirements:

Use thirteen 20 A 2-wire branch circuits, five 30 A 2-wire branch circuits, two 40 A 2-wire branch circuit, and two 60 A 2-wire branch circuits.

8. Service Entrance Conductors Total Full Load Current:

Use two 325 mm2 THW wires.

[66190 VA +25% (6000 VA)] 294 A 230 V

=

Optional Calculation Method 9. Service Equipment:

a. Maximum Current rating of Protective Device with: 1. Non-time Delay Fuse

2. Inverse Time Circuit Breaker [250% (6000 VA) + 6000 VA+ 19320 VA

+ 6000 VA + 16750 VA + 12120 VA] 327 A230V

=

[300% (6000) VA + 6000 VA + 19320 VA + 6000 VA + 16750 VA + 12120 VA] 340 A

230 V=

Optional Calculation Method b. Service Equipment Rating In case of ungrounded service: (a) use one 400 A, 2PST, 250 V safety switch with two 400 A fuses; or (b) use one 400 A trip, 2-pole 240 V molded-case circuit breaker.

Optional Calculation Method Example: Single-family dwelling unit, up to 150 sqm floor

area.

The dwelling has a floor area of 145 sqm. It has the typical household appliances including one 8 kW electric range, two 1 HP room air-conditioning units, one 1.5 HP room air-conditioning unit, and one 1 HP water pump.

Optional Calculation Method Solution: 1. Air conditioning Load (see Sec 4.40.1.6):

Two 1 HP room ACU, 8 A x 230 V x 2 = 3680 VA One 1.5 HP room ACU, 10 A x 230 V x 1 = 2300 VA

2. Other Loads: General lighting and Convenience Receptacle Load 145 sqm x 24 VA/sqm = 3480 VA One 20 A small appliance circuit = 1500 VA One 20 A laundry circuit = 1500 VA

Optional Calculation Method One 8 kW electric range = 8000 VA One 1 HP water pump, 8 A x 230 V = 1840 VA Total Other Loads = 16320 VA Application of Demand factors: First 8000 VA at 100% DF = 8000 VA Remainder at 40% DF (8320VA x 0.40) = 3328 VA Total net computed load = 17308 VA

Optional Calculation Method 3. Circuit requirement:

Use eight 20 A 2-wire branch circuits and one 40 A 2-wire branch circuit.

4. Service Entrance Conductors: Total Full Load Current: 17308 VA / 230 V = 76 A Use two 38 mm2 THW wires.

Optional Calculation Method 5. Service Equipment: a. Maximum Current Rating of Protective Device, with:

i. Non-time Delay Fuse

ii. Inverse Time Circuit Breaker

[3680 VA + 300% (2300 VA) + 8000 VA + 3328 VA] 95 A

230V=

[3680 VA + 250% (2300 VA) + 8000 VA + 3328 VA] 90 A

230 V=

Optional Calculation Method b. Service Equipment Rating:

i. In case of ungrounded service: (1) use one 200 A, 2 PST, 250 V safety switch with two 125 A fuses (renewable); or (2) use one 125 A trip, 2-pole, 240 V molded case circuit breaker. ii. In case of grounded service: (1) use one 200 A, 1 PST, 250 V safety switch with one 125 A fuse (renewable); or (2) use one 125 A trip, 1-pole, 240 V molded-case circuit breaker.

Optional Calculation Method 6. Diagram

Optional Calculation Method Example: Single-Family Dwelling Unit, more than 150 sqm

floor area, 230 V three-phase service.

Same conditions as previous example except that the two 6 kVA package-type air-conditioning units require a 3-phase supply.

Optional Calculation Method Solution: 1. Branch Circuit Arrangement:

A 3-phase fusible or circuit breaker type panelboard shall be provided for balanced distribution of single-phase loads.

2. Circuit Requirements: Use two 60 A 3-wire branch circuits, thirteen 20 A 2-wire branch circuits, five 30 A 2-wire branch circuits, and two 40 A 2-wire branch circuit.

Optional Calculation Method 3. Service Entrance Conductors

Total Full Load Current:

Use three 125 mm2 THW wires.

[66190 VA + 25% (6000 VA)] 170 A230 V 1.732

=

Optional Calculation Method 4. Service Equipment a. Maximum Current rating of Protective Device with:

Inverse Time Circuit Breaker

b. Service Equipment Rating:

Use one 250 A trip, 3-pole, 240 V molded-case circuit breaker.

[250% (6000 VA) + 6000 VA + 19320 VA + 6000 VA + 16750 VA + 12120 VA] 189 A

230 V 1.732=

Optional Calculation Method Example: Multi-family Dwelling Unit, 230 V single-phase

service.

A multifamily dwelling has 12 dwelling units with individual feeders to each dwelling unit. Each dwelling unit has a floor area of 90 sqm and typical household appliances including one 8 kW electric range and one 1 HP room air-conditioning unit.

Optional Calculation Method Solution: 1. Total Load for Each Dwelling Unit a. General Lighting and Convenience Receptacle Load:

Note: Table 2.20.1.3(a) includes the receptacle outlets of 20 A or less. 90 sqm x 24 VA/sqm = 2160 VA Provide two 20 A lighting and convenience receptacle circuits.

b. Small Appliance Load: One 20 A circuit = 1500 VA Provide one 20 A small appliance circuit.

Optional Calculation Method c. Laundry Circuit Load:

One 20 A at = 1500 VA - Provide one 20 A laundry circuit.

d. Subtotal = 5160 VA e. Application of Demand Factors: (see Table 3.3.2.2)

First 3000 VA @ 100% DF = 3000 VA Remainder @ 35% DF (2160 VA x 0.35) = 756 VA

Optional Calculation Method f. Other Loads:

One 8 kW electric range @ 80% DF = 6400 VA - Provide one 40 A electric range circuit

One 1 HP room ACU, 8 A x 230 V @ 100 % DF = 1840 VA - Provide one 20 A room ACU circuit. g. Total Net Computed Load = 1996 VA

Optional Calculation Method 2. Circuit Requirement for each dwelling unit:

Use five 20 A 2-wire branch circuits and one 40 A 2-wire branch circuit.

3. Service Feeder Conductors for each dwelling unit: Total Full Load Current: (see sec 6.6.2.4 and 6.7.1.7)

Use two 22 mm2 THW wires.

[11996 VA + 25% (1840 VA)] 54 A230 V

=

Optional Calculation Method 4. Service Equipment for each dwelling unit: a. Maximum Current rating of Protective Device with:

i. Non-time Delay Fuse

ii. Inverse Time Circuit Breaker

[3000 VA + 756 VA + 6400 VA + 300%(1840)] 68 A

230 V=

[3000 VA + 756 VA + 6400 VA + 250%(1840)] 64 A

230 V=

Optional Calculation Method b. Service Equipment Rating

i. In case of ungrounded service: (a) use one 100 A, 2PST, 250 V safety switch with two 100 A fuses; or (b) use one 90 A trip, 2-pole 240 V molded-case circuit breaker. ii. In case of grounded service, (a) use one 100 A, 1PST, 250 V safety switch with one 100 A fuse; or (b) use one 90 A trip, 1-pole, 240 V molded-case circuit breaker.

Optional Calculation Method 5. Total Load for 12 Dwelling Units: a. General lighting and convenience receptacle load,

2160 VA x 12 = 25920 VA b. Small appliance load, 1500 x 12 = 18000 VA c. Laundry circuit load, 1500 x 12 = 18000 VA d. One 8 kW electric range, 8000 x 12 = 96000 VA e. One 1 HP room ACU, 1840 x 12 = 22080 VA f. Total Computed Load = 180000 VA g. Application of Demand Factors:

180000 VA x 0.4 = 173800 VA

Optional Calculation Method 6. Main Service Entrance Conductors:

Total Full Load Current: 73800 VA / 230 V = 321 A Use two 325 mm2 or two sets of two 100 mm2 THW wires.

7. Main Service Equipment: a. Maximum Current Rating:

73800 VA / 230 V = 321 A

Optional Calculation Method b. Service Equipment Rating

i. In case of ungrounded service: (a) use one 600 A, 2PST, 250 V safety switch with two 500 A fuses; or (b) use one 500 A trip, 2-pole 240 V molded-case circuit breaker. ii. In case of grounded service, (a) use one 600 A, 1PST, 250 V safety switch with one 500 A fuse; or (b) use one 500 A trip, 1-pole, 240 V molded-case circuit breaker.