Multi - Family Dwelling

4 - door apartment

- Types of Service: 230 V
- 2 wire, Line to Ground System
- Floor Area per unit: 80 sq. m
- Total Floor Area: 320 sq. m

Determine the branch circuit protection, size of conductor wires and the main header.

SOLUTION:

SOLUTION:

Assume that the dwelling unit is equipped with one 5.1 kW cooking unit; one unit laundry ckt. at 1.5 kW

A. Circuit - 1 For Lighting Load per unit (see plan)

1. By the area method, refer to Table, General Lighting Load by occupancy for dwelling units.

A. Circuit - 1 For Lighting Load per unit (see plan)

1. By the area method, refer to Table, General Lighting Load by occupancy for dwelling units.

80 sq. m x 24 watts per sq. m = 1920 watts

2. Compute for the Lighting Load. Divide:

1920 watts/230 volts = 8.35 A

3. Determine the size of the Branch Circuit conductor wire. Refer to Table. For 8.35 A load, use 2 pieces 2.0 mm

4. Determine the size of the conduit pipe. For number 14 AWG , TW wire use 13 mm minimum size of conduit pipe.

5. Determine the size or rating of the branch circuit protection. Refer again to Table. For 8.35 A load on a 2.0 mm

B. Circuit - 2 For Convenience Outlet Load

1. Solve for the total current load.

8 receptacles x 2 gang per outlet x 180 watts = 2880 watts

2. Solve for the appliance current load. Divide.

I = 2880 watts/230 volts = 12.52 A

3. Determine the size of the Branch Circuit conductor. Refer to Table 9.1 or 11.1. For a 12.52 A load, a 2.0 mm

4. But the National Electrical Code limits the size of convenience outlet wire to minimum of 3.5 mm

5. Determine the size of the conduit pipe. Refer to Table . For No. 12 TW wire, use 13 mm diameter pipe.

6. Find the Size of the Branch Circuit fuse protection. Refer to table. For 12.53 A non continuous load on convenience outlet, use 20 AT breaker.

C. Circuit - 3 Other Load

1. Laundry Circuit at 1500 watts per circuit (PEC provision)

1500 watts/230 volys = 6.52 A

2. Find the size of the branch circuit conductor. From Table, use 3.5 mm

3. Find the size of the conduit pipe. From Table, use 13 mm diameter pipe.

4. Find the size of the branch circuit fuse protection. From Table. The 6.52 A load on convenience outlet requires 20 A fuse or trip breaker.

D. Circuit - 4 Cooking Unit

1. Total Load is 5.1 kW = 5100 watts

2. Refer to Table Demand load for household. For electric range, apply 80% demand factor.

Total load x demand factor (Df)

5100 watts x .80 = 4080 watts

3. Compute for the line current load. Divide:

4080 watts/230 volts = 17.74 A

4. Find the size of the branch Circuit wire. Refer to Table 0.1 or 11.1. For 17.74 A line current, use 5.5 mm

5. Determine the size of the conduit pipe. From Table, for No.10 TW wire, use 20 mm diameter pipe.

6. Find the size of the branch circuit fuse protection. Refer to Table, for 17.74 A current load, use 30 A fuse or trip breaker.

E. Determine the sub-feeder per dwelling

1. Solve for the total connected load per dwelling.

Lighting load ........................................ 1920 watts

Convenience Load ,............................... 2880 watts

Other loads 5.1 + 1.5 kW........................ 6600 watts

TOTAL....................................11400 watts

2. Apply 80% demand factor (see Table)

TOTAL LINE CURRENT = (11400 watts x.80 df)/230 volts

= 39.65 A

3. Determine the Size of the sub-feeder and protection per dwelling for 39.65 A. For Table 9.1 or 11.1, use 8.0 mm

4. Find the size of the conduit pipe. For 8.0 mm

5. Determine the size or rating of the fuse protection. From Table, use 60 A molded Circuit breaker 2- wire 250 volts with solid bus.

F. Determine the Size of the Main Feeder

1. Solve for the Total connected Load on 4 dwelling units at 11400 watts each. Multiply:

11400 watts x 4 = 45600 watts

2. Refer to Table. For 4 dwelling units apply 45% demand factor. Multiply:

45600 watts x .45 = 20520 watts

3. Solve for the line current:

I = 20520 watts/230 volts = 89.22 A

4. Determine the Size of the Conductor wire. Refer to Table. For 89.22 A, use 2- 50 mm

COMMENT:

IT will be noted in Table, that the 89.22 A as computed does not exceed 80 % of the 120 allowable ampacity of 50 mm

5. Find the size of conduit pipe. Refer to Table. Use 38 nn diameter RSC or IMT pipe

6. Find the size or rating of the over-current protection. Refer to Table. Use 125 A safety switch,250 volts, 2 pole.

2. Compute for the Lighting Load. Divide:

1920 watts/230 volts = 8.35 A

3. Determine the size of the Branch Circuit conductor wire. Refer to Table. For 8.35 A load, use 2 pieces 2.0 mm

^{2}or No. 14 TW AWG copper wire4. Determine the size of the conduit pipe. For number 14 AWG , TW wire use 13 mm minimum size of conduit pipe.

5. Determine the size or rating of the branch circuit protection. Refer again to Table. For 8.35 A load on a 2.0 mm

^{2}wire conductor size, use 15 A fuse or trip breaker.B. Circuit - 2 For Convenience Outlet Load

1. Solve for the total current load.

8 receptacles x 2 gang per outlet x 180 watts = 2880 watts

2. Solve for the appliance current load. Divide.

I = 2880 watts/230 volts = 12.52 A

3. Determine the size of the Branch Circuit conductor. Refer to Table 9.1 or 11.1. For a 12.52 A load, a 2.0 mm

^{2}or No.14 TW AWG wire would be sufficient considering its 15 A ampacity that is bigger than 12.52 A as computed4. But the National Electrical Code limits the size of convenience outlet wire to minimum of 3.5 mm

^{2}or No. 12 AWG copper wire. The code must prevail. Use No. 12 TW.5. Determine the size of the conduit pipe. Refer to Table . For No. 12 TW wire, use 13 mm diameter pipe.

6. Find the Size of the Branch Circuit fuse protection. Refer to table. For 12.53 A non continuous load on convenience outlet, use 20 AT breaker.

C. Circuit - 3 Other Load

1. Laundry Circuit at 1500 watts per circuit (PEC provision)

1500 watts/230 volys = 6.52 A

2. Find the size of the branch circuit conductor. From Table, use 3.5 mm

^{2}or No. 12 TW copper wire, the minimum size for convenience outlet.3. Find the size of the conduit pipe. From Table, use 13 mm diameter pipe.

4. Find the size of the branch circuit fuse protection. From Table. The 6.52 A load on convenience outlet requires 20 A fuse or trip breaker.

D. Circuit - 4 Cooking Unit

1. Total Load is 5.1 kW = 5100 watts

2. Refer to Table Demand load for household. For electric range, apply 80% demand factor.

Total load x demand factor (Df)

5100 watts x .80 = 4080 watts

3. Compute for the line current load. Divide:

4080 watts/230 volts = 17.74 A

4. Find the size of the branch Circuit wire. Refer to Table 0.1 or 11.1. For 17.74 A line current, use 5.5 mm

^{2}or No.10 TW copper wire.5. Determine the size of the conduit pipe. From Table, for No.10 TW wire, use 20 mm diameter pipe.

6. Find the size of the branch circuit fuse protection. Refer to Table, for 17.74 A current load, use 30 A fuse or trip breaker.

E. Determine the sub-feeder per dwelling

1. Solve for the total connected load per dwelling.

Lighting load ........................................ 1920 watts

Convenience Load ,............................... 2880 watts

Other loads 5.1 + 1.5 kW........................ 6600 watts

TOTAL....................................11400 watts

2. Apply 80% demand factor (see Table)

TOTAL LINE CURRENT = (11400 watts x.80 df)/230 volts

= 39.65 A

3. Determine the Size of the sub-feeder and protection per dwelling for 39.65 A. For Table 9.1 or 11.1, use 8.0 mm

^{2}or No. 8 wire THW copper wire.4. Find the size of the conduit pipe. For 8.0 mm

^{2}wire, specify 25 mm diameter pipe.5. Determine the size or rating of the fuse protection. From Table, use 60 A molded Circuit breaker 2- wire 250 volts with solid bus.

F. Determine the Size of the Main Feeder

1. Solve for the Total connected Load on 4 dwelling units at 11400 watts each. Multiply:

11400 watts x 4 = 45600 watts

2. Refer to Table. For 4 dwelling units apply 45% demand factor. Multiply:

45600 watts x .45 = 20520 watts

3. Solve for the line current:

I = 20520 watts/230 volts = 89.22 A

4. Determine the Size of the Conductor wire. Refer to Table. For 89.22 A, use 2- 50 mm

^{2}TW copper wire or 2- 38 mm^{2}THW copper wire.COMMENT:

IT will be noted in Table, that the 89.22 A as computed does not exceed 80 % of the 120 allowable ampacity of 50 mm

^{2}TW copper wire or 125 ampacity of 38 mm^{2}THW copper wire. Therefore, any one of these two types of wire could be used for main feeder.5. Find the size of conduit pipe. Refer to Table. Use 38 nn diameter RSC or IMT pipe

6. Find the size or rating of the over-current protection. Refer to Table. Use 125 A safety switch,250 volts, 2 pole.

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ReplyDeletegood day sir,I'm an EE student doing some elecrical design for multi-family dwelling,can I ask if I have 20 units,what is the demand factor before computing for the total line current.

ReplyDelete