SINGLE FAMILY DWELLING
Type of Service - 230 V
Single Phase 2 wire 60 Hz
Line to ground Current system
PROBLEM:
A single family dwelling is to be circuited with the following requirements as shown on the figure above. Determine the:
SOLUTION
A. Circuit - 1 for lighting load
1. From the Figure above, determine the number of lighting outlets. By direct counting, there are 8 light outlets.
The National Electrical Code provides that:
"100 watts shall be the maximum load for each household lighting outlet."
Adopting the 100 watts per lighting outlet we have:
8 outlets x 100 = 800 watts
2. Determine the Total Current load
800 watts/230 volts = 3.48 amperes
3. Determine the size of conductor wire for circuit - 1 . Refer to Table 9.1 or 11.1. use 2 pieces 2.0 mm2 or No. 14 TW copper wire having an ampacity of 15 amperes that is much largr than the 3.48 amperes computed maximum load.
4. Determine the size of the Conduit Pipe. Refer to Table . The smallest diameter of a conduit pipe that could accommodate up to 3 pieces of No.14 TW conductor wire is 13 mm diameter. therefore, specify 13 mm diameter conduit pipe.
5. determine the size or rating of the fuse protective device. Refer to Table . Use 15 amperes fuse
The National Electrical Code provides that:
"Ampacity of the connected load shall not exceed 80% of the amperage capacity of the conductor and the fuse."
In Table , the maximum ampacity load of a or 2.0 mm2 No.14 AWG copper wire is 15 amperes. 80 % of 15 is 12, the maximum allowable load of the circuit sufficient enough to carry the 3.48 amperes computed load for a maximum 100 watts per light outlet. therefore the use of 2.0 mm2 or No.14 TW is safe.
B, Circuit - 2 For Small Appliance load
SOLUTION
The National Electrical Code provides that:
"For each single receptacle shall be considered at no less than 180 watts rating."
It simply mean that, each convenience outlet, is considered to have a maximum load of not less than 180 watts per plug or gang. thus:
1. From Figure above, there are 6 convenience receptacles for small appliance load. considering that there are two plug outlet, the total number of plug will be:
6 outlets x 2 plug = 12 pieces
2. Solve for the Total Estimated Load
12 x 180 watts per outlet = 2,160 watts
3. Determine the Maximum Expected Current Load:
2,160 watts / 230 volts = 9.39 amperes
4. Determine the Size of the Conductor wire . Refer to Table 9.1 or 11.1. For 9.39 amperes, use 2 pieces 3.5 mm2 or No. 12 TW copper wire for Circuit no.2
5. Determine the Size of the conduit pipe. for the 2 - No. 12 TW wire, refer to Table . Use 13 mm conduit pipe.
6. Determine the Over Current fuse protection. Refer to Table . under the column of fuse and breaker rating, the 20 amperes fuse can safely carry a maximum load of 16 amperes the 80 % of 20 amperes fuse can safely carry a maximum load of 16 amperes the 80% of 20 amperes load permitted by the National Electrical Code on No. 12 circuit wire.
COMMENT:
1. On convenience outlet receptacle, the National Electrical Code provides that, " Each single receptacle shall be considered at no less than 180 watts rating."
2. Examining the values given on Table , the 2.0 mm2 or No. 14 AWG TW copper wire has an allowable ampacity rating of 15 amperes. Granting that only 80 % of this 15 amperes is considered the derated value, still 12 amperes is very much larger than the 9.36 amperes computed as maximum load for 6-convenience outlet. Why specify a bigger 3.5 mm2or No. 12 AWG conductor wire?
3. Although the 2.0 mm2or No. 14 AWG wire conductor could safely carry the 9.36 amperes computed load, considering its 15 amperes ampacity rating,yet, we cannot do so because the Code specifically mandated the use of 3.5 mm2or No. 12 AWG copper wire as the minimum size for all types of convenience outlet wiring except, for an appliance with limited load wherein a 2.0 mm2or No.14 AWG wire is permitted.
C. Circuit - 3 for other loads
SOLUTION
1. Examining Figure above, other loads are:
1 - unit electric stove at 1.1 kw = 1,1oo watts
1 - unit water heater at 2.5 kw = 2,500 watts
Total Load..........3,600 watts
2. Compute for the current load. Divide:
3,600 watts / 230 volts = 15.65 amperes
3. Determine the size of the service conductor wire. Refer to Table 9.1 or 11.1.
4. For the 15.65 amperes load, use 2 pieces 3.5mm2or No.12 AWG TW copper wire
5. Determine the Size of the Conduit Pipe (if reqd) to Table. Two pieces No. 12 AWG wire can be accommodated comfortably in a 13 mm diameter conduit pipe. Specify 13 mm diameter conduit pipe.
6. Determine the size or rating of the Over-current Protection. Refer to Table . For the 15.65 amperes load use 20 amperes fuse rating.
COMMENTS:
The fuse rating is 20 amperes. Granting that it will be derated at 80 % x 20, the 16 amperes derated value is still higher than the computed load of 15.65 amperes. Therefore, the 20 amperes fuse over current protection is accepted.
FINDING THE SIZE OF SERVICE ENTRANCE
The size of service entrance being the supply conductor and equipment for delivering energy from the electricity supply to the wiring system of the building, is also computed based on the total load supplied by the branch circuit. Continuing the solution, we have the following:
1. Solve for the total load of circuit 1 to circuit 3.
Total current load = Total connected load / voltage rating
= (800 W + 2,160 W + 3,600 W) / 230 V
= 28.52 amperes
2. Apply 80% demand factor as permitted by the National Electrical Code.
28.52 x .80 = 22.8 amperes
3. Find the Size of the Service Wire. Refer 22.8 amperes to Table . Use 2- 8.0 mm2or No.8 TW copper wire
4. Determine the size of conduit pipe for the service wire. Refer Table , for No.8 TW copper wire, use 20 mm diameter conduit pipe.
COMMENT:
1. A demand factor of 80% was applied considering that not all receptacles and outlets are being used simultaneously.
2. These type of loads are classified as non-continuous load. From Table 9.1, the 5.5 mm2or
No.10 AWG copper wire conductor has 30 amperes ampacity which is bigger than 22.8 amperes as computed. However, we do not specify the use of No.10 AWG wire because the code limits the use of 8.0 mm2or No.8 AWG, conductor as minimum size for Service Entrance.
3. The National Electrical Code on Service Entrance provides that:
" Service entrance shall have sufficient ampacity to carry the building load. They shall have the adequate mechanical strength and shall not be smaller than 8.0 mm2or 3.2 mm diameter except for installation to supply limited load of a single branch circuit such as small poly-phase power, controlled water heaters and the like and they shall not be smaller than 3.5 mm2or 2.0 mm diameter copper or equivalent.
THE MAIN DISCONNECTING MEANS OR SAFETY SWITCH
Find the total computed load
Circuit - 1 ........ 3.48 amperes
Circuit - 2 ........9.39 amperes
Circuit - 3 ........15.65 amperes
TOTAL........28.52 amperes
2. Use 2 pieces 30 amperes fuse parallel connection 60 amperes 2 pole single throw (PST) 250 volts safety switch
3. Provide 2-double branch circuit cut out with two 15 and 2-20 amperes fuse respectively.
Type of Service - 230 V
Single Phase 2 wire 60 Hz
Line to ground Current system
PROBLEM:
A single family dwelling is to be circuited with the following requirements as shown on the figure above. Determine the:
- size of the branch circuit wire for lighting outlets
- size of the conduit pipes
- size or rating of the fuse protective device
SOLUTION
A. Circuit - 1 for lighting load
1. From the Figure above, determine the number of lighting outlets. By direct counting, there are 8 light outlets.
The National Electrical Code provides that:
"100 watts shall be the maximum load for each household lighting outlet."
Adopting the 100 watts per lighting outlet we have:
8 outlets x 100 = 800 watts
2. Determine the Total Current load
800 watts/230 volts = 3.48 amperes
3. Determine the size of conductor wire for circuit - 1 . Refer to Table 9.1 or 11.1. use 2 pieces 2.0 mm2 or No. 14 TW copper wire having an ampacity of 15 amperes that is much largr than the 3.48 amperes computed maximum load.
4. Determine the size of the Conduit Pipe. Refer to Table . The smallest diameter of a conduit pipe that could accommodate up to 3 pieces of No.14 TW conductor wire is 13 mm diameter. therefore, specify 13 mm diameter conduit pipe.
5. determine the size or rating of the fuse protective device. Refer to Table . Use 15 amperes fuse
The National Electrical Code provides that:
"Ampacity of the connected load shall not exceed 80% of the amperage capacity of the conductor and the fuse."
In Table , the maximum ampacity load of a or 2.0 mm2 No.14 AWG copper wire is 15 amperes. 80 % of 15 is 12, the maximum allowable load of the circuit sufficient enough to carry the 3.48 amperes computed load for a maximum 100 watts per light outlet. therefore the use of 2.0 mm2 or No.14 TW is safe.
B, Circuit - 2 For Small Appliance load
SOLUTION
The National Electrical Code provides that:
"For each single receptacle shall be considered at no less than 180 watts rating."
It simply mean that, each convenience outlet, is considered to have a maximum load of not less than 180 watts per plug or gang. thus:
1. From Figure above, there are 6 convenience receptacles for small appliance load. considering that there are two plug outlet, the total number of plug will be:
6 outlets x 2 plug = 12 pieces
2. Solve for the Total Estimated Load
12 x 180 watts per outlet = 2,160 watts
3. Determine the Maximum Expected Current Load:
2,160 watts / 230 volts = 9.39 amperes
4. Determine the Size of the Conductor wire . Refer to Table 9.1 or 11.1. For 9.39 amperes, use 2 pieces 3.5 mm2 or No. 12 TW copper wire for Circuit no.2
5. Determine the Size of the conduit pipe. for the 2 - No. 12 TW wire, refer to Table . Use 13 mm conduit pipe.
6. Determine the Over Current fuse protection. Refer to Table . under the column of fuse and breaker rating, the 20 amperes fuse can safely carry a maximum load of 16 amperes the 80 % of 20 amperes fuse can safely carry a maximum load of 16 amperes the 80% of 20 amperes load permitted by the National Electrical Code on No. 12 circuit wire.
COMMENT:
1. On convenience outlet receptacle, the National Electrical Code provides that, " Each single receptacle shall be considered at no less than 180 watts rating."
2. Examining the values given on Table , the 2.0 mm2 or No. 14 AWG TW copper wire has an allowable ampacity rating of 15 amperes. Granting that only 80 % of this 15 amperes is considered the derated value, still 12 amperes is very much larger than the 9.36 amperes computed as maximum load for 6-convenience outlet. Why specify a bigger 3.5 mm2or No. 12 AWG conductor wire?
3. Although the 2.0 mm2or No. 14 AWG wire conductor could safely carry the 9.36 amperes computed load, considering its 15 amperes ampacity rating,yet, we cannot do so because the Code specifically mandated the use of 3.5 mm2or No. 12 AWG copper wire as the minimum size for all types of convenience outlet wiring except, for an appliance with limited load wherein a 2.0 mm2or No.14 AWG wire is permitted.
C. Circuit - 3 for other loads
SOLUTION
1. Examining Figure above, other loads are:
1 - unit electric stove at 1.1 kw = 1,1oo watts
1 - unit water heater at 2.5 kw = 2,500 watts
Total Load..........3,600 watts
2. Compute for the current load. Divide:
3,600 watts / 230 volts = 15.65 amperes
3. Determine the size of the service conductor wire. Refer to Table 9.1 or 11.1.
4. For the 15.65 amperes load, use 2 pieces 3.5mm2or No.12 AWG TW copper wire
5. Determine the Size of the Conduit Pipe (if reqd) to Table. Two pieces No. 12 AWG wire can be accommodated comfortably in a 13 mm diameter conduit pipe. Specify 13 mm diameter conduit pipe.
6. Determine the size or rating of the Over-current Protection. Refer to Table . For the 15.65 amperes load use 20 amperes fuse rating.
COMMENTS:
The fuse rating is 20 amperes. Granting that it will be derated at 80 % x 20, the 16 amperes derated value is still higher than the computed load of 15.65 amperes. Therefore, the 20 amperes fuse over current protection is accepted.
FINDING THE SIZE OF SERVICE ENTRANCE
The size of service entrance being the supply conductor and equipment for delivering energy from the electricity supply to the wiring system of the building, is also computed based on the total load supplied by the branch circuit. Continuing the solution, we have the following:
1. Solve for the total load of circuit 1 to circuit 3.
Total current load = Total connected load / voltage rating
= (800 W + 2,160 W + 3,600 W) / 230 V
= 28.52 amperes
2. Apply 80% demand factor as permitted by the National Electrical Code.
28.52 x .80 = 22.8 amperes
3. Find the Size of the Service Wire. Refer 22.8 amperes to Table . Use 2- 8.0 mm2or No.8 TW copper wire
4. Determine the size of conduit pipe for the service wire. Refer Table , for No.8 TW copper wire, use 20 mm diameter conduit pipe.
COMMENT:
1. A demand factor of 80% was applied considering that not all receptacles and outlets are being used simultaneously.
2. These type of loads are classified as non-continuous load. From Table 9.1, the 5.5 mm2or
No.10 AWG copper wire conductor has 30 amperes ampacity which is bigger than 22.8 amperes as computed. However, we do not specify the use of No.10 AWG wire because the code limits the use of 8.0 mm2or No.8 AWG, conductor as minimum size for Service Entrance.
3. The National Electrical Code on Service Entrance provides that:
" Service entrance shall have sufficient ampacity to carry the building load. They shall have the adequate mechanical strength and shall not be smaller than 8.0 mm2or 3.2 mm diameter except for installation to supply limited load of a single branch circuit such as small poly-phase power, controlled water heaters and the like and they shall not be smaller than 3.5 mm2or 2.0 mm diameter copper or equivalent.
THE MAIN DISCONNECTING MEANS OR SAFETY SWITCH
Find the total computed load
Circuit - 1 ........ 3.48 amperes
Circuit - 2 ........9.39 amperes
Circuit - 3 ........15.65 amperes
TOTAL........28.52 amperes
2. Use 2 pieces 30 amperes fuse parallel connection 60 amperes 2 pole single throw (PST) 250 volts safety switch
3. Provide 2-double branch circuit cut out with two 15 and 2-20 amperes fuse respectively.
MULTI-ground system and line to line service
The protection of branch circuit is tapped to the hot line of live wire. The grounded line being in neutral zero voltage is not protected with fuse. this is one advantage of the MULTI-GROUND SYSTEM being adopted by the electric cooperative implemented by the RURAL ELECTRIFICATION PROGRAM of the government. The branch circuit and cutout should be doubled because the engaged voltage in the line is only 230 V while the other is zero being grounded ( see figure)
Other electric service system on the other hand, are classified as LINE TO LINE SERVICE wherein the engaged voltage is 115/230 volts which requires FUSE PROTECTIO FOR BOTH LINES.
NOTE:
The quantity of materials is subject to change depending upon the area and the choice of the designing engineers. For open onstallation, conduit pipe can be changed to split knobs or PDX wires.
NOTE:
The quantity of materials is subject to change depending upon the area and the choice of the designing engineers. For open onstallation, conduit pipe can be changed to split knobs or PDX wires.
SINGLE FAMILY DWELLING
Type of Service - 115/230 V
Single Phase 3 wire 60 Hz
Line to Line system
SOLUTION:
Examining the lighting plan of the above figure, there are 19 lighting outlets. Split the 19 outlets into two circuits A and B.
A. Circuit - 1 Lighting Load (10 light outlets)
1. The PEC provides that 100 watts be the maximum load per light outlet. thus, for 10 light outlets at 100 watts, multiply:
10 outlets x 100 watts = 1000 watts
2. Compute The Current Load
1000 watts/230 volts = 4.35 amperes
3. Find the size of Branch circuit wire. Refer to Table 9.1 or 11.1. For 4.35 amperes, use 2.0 mm2 TW copper wire.
4. Find the rating of overcurrent protection. Refer to Table. for 4.35 A, use 15 amperes trip breaker.
5. Determine the size of conduit pipe. Refer to Table , for No. 14 TW copper wire, use 13 mm conduit pipe.
B. Circuit - 2 Lighting Load (9 light outlets)
1. For 9 outlets, find the Total load in watts.
9 outlets x 100 watts per outlet = 900 watts
Divide : 900 watts/230 volts = 3.91 amperes
2. Determine the Size of the Branch circuit Wire. Refer to Table 9.1 or 11.1. For 3.91 A load, use 2.0 mm2or No.14 TW copper wire.
3. Determine the size of the conduit pipe. Refer to Table. For 2 pieces No. 14 TW copper wire, use the 13 mm minimum size of conduit pipe.
4. Determine the size or rating of the overcurrent Protection. Refer to Table. For 3.91 A load, use 15 A load fuse of trip breaker.
C. Circuit - 3 For small appliance load
Section 3.3.1.2 of the the PEC specif 180 watts load limit per convenience outlet. Thus,
1. Find the number of appliance load outlet and the current load.
6 outlets x 2 gang per outlet x 180 watts
12 x 180 = 2,160 watts
Divide: 2160 watts/230 volts = 9.39 A
2. Determine the Size of the Service Wire conductor. Refer to Table 9.1 or 11.1. For the 9.39 A load, specify the minimum wire gauge for convenience outlet.
2 pieces 3.5 mm2or No.12 TW copper wire
3. Determine the Size of the conduit pipe. Refer to Table. For 2 pieces No.12 TW copper wire. Use 13 mm diameter conduit pipe.
4. Solve for the Size or Rating of the Over current Protection. Refer to Table. For 9.39 A on No. 12 TW copper wire specify:
20 A fuse or trip breaker.
D. Circuit- 4 for Small Appliance Load
1. The load of circuit 4 is identical with circuit . Use the same size of wire, conduit, and wire protection rating.
E. Circuit - 5 for Range Load
1. Range load (appliance rating) at 8.0 kW = 8000 watts
2. Solve for the Line current.
8000 watts / 230 V= 34.78 A
3. Refer to Table , apply 80% demand load factor .
34.78 x .80df = 27.82 A
4. Determine the Size of the Branch Circuit wire. Refer to Table 9.1 or 11.1. For the 27.82 A, use 8.0 mm2 or No. 8 TW copper wire.
5. Determine the Size of Conduit pipe. Refer to Table. For 2 pieces No.8 wire use 200 mm diameter pipe.
6. Find the Size or Rating of the Fuse or Trip Breaker. Refer to Table. For appliance load, use 40 A fuse or trip breaker.
F. Circuit - 6 for Water heater Load
1. one unit water heater at 2.5 kW = 2500 watts
2. The current load will be:
2500 watts/230 volts = 10.86 A
3. Solve for the Size of the Branch circuit wire. Refer to Table 9.1 or 11.1 . For 10.86 A convenience outlet use 2 pcs 3.5 mm2 or No. 12 TW copper wire.
4. Determine the Size of the conduit pipe. Refer to Table. For 2 - No.12 Tw copper wire, use 13 mm conduit pipe.
5. Find the Size or Rating of the Over Current Protection. For the 10.86 A load, use 20 A fuse fuse or trip breaker.
G. Circuit 7 and 8 with 1 - unit ACU each
1. One unit ACU at 1.5 Hp is
1.5 Hp x 746 = 1119 watts
Article 6.7 of the Philippine Electrical Code provides that: "BRANCH CIRCUIT CONDUCTOR SUPPLYING A MOTOR SHALL HAVE AN AMPACITY NOT LESS THAN 125% OF THE FULL LOAD CURRENT."
2. Current Load: 1119 watts/230 V = 4.86 A
4.86 A x 125% = 6.07 A
3. Find the Size of the Branch circuit service wire. Refer to Table. The 6.7 A can be served by a 2.0 No.14 TW copper wire, but the Code limits the size of convenience outlet to No. 12 AWG copper wire. Specify No.12 THW copper wire for circuit 7 and 8.
4. Find the Size of the conduit pipe. Refer to Table. For two No.12 wire, use 13 mm conduit pipe.
5. Find the Size and Rating of the Branch Circuit Protection. The Code on branch circuit protection for a single motor provides that" It shall be increased by 250% of the full load current of the motor." thus,
4.86 x 250% = 12.15 A. From Table for a continuous load use 2-30 AT breaker.
Type of Service - 115/230 V
Single Phase 3 wire 60 Hz
Line to Line system
SOLUTION:
Examining the lighting plan of the above figure, there are 19 lighting outlets. Split the 19 outlets into two circuits A and B.
A. Circuit - 1 Lighting Load (10 light outlets)
1. The PEC provides that 100 watts be the maximum load per light outlet. thus, for 10 light outlets at 100 watts, multiply:
10 outlets x 100 watts = 1000 watts
2. Compute The Current Load
1000 watts/230 volts = 4.35 amperes
3. Find the size of Branch circuit wire. Refer to Table 9.1 or 11.1. For 4.35 amperes, use 2.0 mm2 TW copper wire.
4. Find the rating of overcurrent protection. Refer to Table. for 4.35 A, use 15 amperes trip breaker.
5. Determine the size of conduit pipe. Refer to Table , for No. 14 TW copper wire, use 13 mm conduit pipe.
B. Circuit - 2 Lighting Load (9 light outlets)
1. For 9 outlets, find the Total load in watts.
9 outlets x 100 watts per outlet = 900 watts
Divide : 900 watts/230 volts = 3.91 amperes
2. Determine the Size of the Branch circuit Wire. Refer to Table 9.1 or 11.1. For 3.91 A load, use 2.0 mm2or No.14 TW copper wire.
3. Determine the size of the conduit pipe. Refer to Table. For 2 pieces No. 14 TW copper wire, use the 13 mm minimum size of conduit pipe.
4. Determine the size or rating of the overcurrent Protection. Refer to Table. For 3.91 A load, use 15 A load fuse of trip breaker.
C. Circuit - 3 For small appliance load
Section 3.3.1.2 of the the PEC specif 180 watts load limit per convenience outlet. Thus,
1. Find the number of appliance load outlet and the current load.
6 outlets x 2 gang per outlet x 180 watts
12 x 180 = 2,160 watts
Divide: 2160 watts/230 volts = 9.39 A
2. Determine the Size of the Service Wire conductor. Refer to Table 9.1 or 11.1. For the 9.39 A load, specify the minimum wire gauge for convenience outlet.
2 pieces 3.5 mm2or No.12 TW copper wire
3. Determine the Size of the conduit pipe. Refer to Table. For 2 pieces No.12 TW copper wire. Use 13 mm diameter conduit pipe.
4. Solve for the Size or Rating of the Over current Protection. Refer to Table. For 9.39 A on No. 12 TW copper wire specify:
20 A fuse or trip breaker.
D. Circuit- 4 for Small Appliance Load
1. The load of circuit 4 is identical with circuit . Use the same size of wire, conduit, and wire protection rating.
E. Circuit - 5 for Range Load
1. Range load (appliance rating) at 8.0 kW = 8000 watts
2. Solve for the Line current.
8000 watts / 230 V= 34.78 A
3. Refer to Table , apply 80% demand load factor .
34.78 x .80df = 27.82 A
4. Determine the Size of the Branch Circuit wire. Refer to Table 9.1 or 11.1. For the 27.82 A, use 8.0 mm2 or No. 8 TW copper wire.
5. Determine the Size of Conduit pipe. Refer to Table. For 2 pieces No.8 wire use 200 mm diameter pipe.
6. Find the Size or Rating of the Fuse or Trip Breaker. Refer to Table. For appliance load, use 40 A fuse or trip breaker.
F. Circuit - 6 for Water heater Load
1. one unit water heater at 2.5 kW = 2500 watts
2. The current load will be:
2500 watts/230 volts = 10.86 A
3. Solve for the Size of the Branch circuit wire. Refer to Table 9.1 or 11.1 . For 10.86 A convenience outlet use 2 pcs 3.5 mm2 or No. 12 TW copper wire.
4. Determine the Size of the conduit pipe. Refer to Table. For 2 - No.12 Tw copper wire, use 13 mm conduit pipe.
5. Find the Size or Rating of the Over Current Protection. For the 10.86 A load, use 20 A fuse fuse or trip breaker.
G. Circuit 7 and 8 with 1 - unit ACU each
1. One unit ACU at 1.5 Hp is
1.5 Hp x 746 = 1119 watts
Article 6.7 of the Philippine Electrical Code provides that: "BRANCH CIRCUIT CONDUCTOR SUPPLYING A MOTOR SHALL HAVE AN AMPACITY NOT LESS THAN 125% OF THE FULL LOAD CURRENT."
2. Current Load: 1119 watts/230 V = 4.86 A
4.86 A x 125% = 6.07 A
3. Find the Size of the Branch circuit service wire. Refer to Table. The 6.7 A can be served by a 2.0 No.14 TW copper wire, but the Code limits the size of convenience outlet to No. 12 AWG copper wire. Specify No.12 THW copper wire for circuit 7 and 8.
4. Find the Size of the conduit pipe. Refer to Table. For two No.12 wire, use 13 mm conduit pipe.
5. Find the Size and Rating of the Branch Circuit Protection. The Code on branch circuit protection for a single motor provides that" It shall be increased by 250% of the full load current of the motor." thus,
4.86 x 250% = 12.15 A. From Table for a continuous load use 2-30 AT breaker.
CALCULATING THE AMPACITY OF THE SERVICE
ENTRANCE CONDUCTOR AND THE MAIN DISCONNECTING MEANS
1. Find The total current load of circuit 1 to circuit 8:
lighting load Ct. 1 and Ct.2 ....................1900 watts
small appliance load Ct. 3 and Ct. 4.......4320 watts
other loads Ct.5 and Ct 6.,......................10500 watts
TOTAL LOAD(except the ACU)....16720 watts
2. From Table , OPTIONAL CALCULATION for dwelling Unit, apply demand factor.
for the first 10000 w at 100%(df)...........10000 watts
subtract: 16720 - 10000 = 6720 watts
for other load, multiply by 40%
6720 x .4 .........................................2688 watts
Aircon unit at 100% demand factor
2-units at 1119 watts........................2238 watts
TOTAL .....................14, 926 watts
ENTRANCE CONDUCTOR AND THE MAIN DISCONNECTING MEANS
1. Find The total current load of circuit 1 to circuit 8:
lighting load Ct. 1 and Ct.2 ....................1900 watts
small appliance load Ct. 3 and Ct. 4.......4320 watts
other loads Ct.5 and Ct 6.,......................10500 watts
TOTAL LOAD(except the ACU)....16720 watts
2. From Table , OPTIONAL CALCULATION for dwelling Unit, apply demand factor.
for the first 10000 w at 100%(df)...........10000 watts
subtract: 16720 - 10000 = 6720 watts
for other load, multiply by 40%
6720 x .4 .........................................2688 watts
Aircon unit at 100% demand factor
2-units at 1119 watts........................2238 watts
TOTAL .....................14, 926 watts
TOTAL CONNECTED LOAD PLUS 25% OF THE LARGEST MOTOR
1. Ampere I = 14.926w + (25% of 1.119w)/230 V
= 63.37 A
2. Find the Size of Main feeder and the Neutral line.
- Use 2 - 3.8 mm2 TW copper wire
3. The Neutral Conductor of a 3 - wire line to line supply system shall have an ampacity of not less than 70% of the ungrounded(live wire) conductor or TWO TRADE SIZE SMALLER SIZE THAN THE UNGROUNDED CONDUCTOR. (PEC specs) Therefore use 1- 22 mm2 Tw copper wire for the Neutral line.
4. Determine the Size of the Conduit pipe. Refer to Table , use 32 mm diameter pipe.
5. For main Breaker, refer to Table . Use 2 - 100 A 2 - wires 250 V, 2 pole molded air circuity breaker.
COMMENT:
The total computed load is 63.37 A. The 30 mm2 copper wire could be used considering its 90 A capacity. However, The NEC provides that:
" If the computed load exceeds 10000 watts, the conductor and overcurrent protection shall be rated not less than 100 A.
THEREFORE USE 2 - 38 mm2 TW WIRE FOR THE MAIN FEEDER AND 2 - 100 A FOR THE MAIN BREAKER.
= 63.37 A
2. Find the Size of Main feeder and the Neutral line.
- Use 2 - 3.8 mm2 TW copper wire
3. The Neutral Conductor of a 3 - wire line to line supply system shall have an ampacity of not less than 70% of the ungrounded(live wire) conductor or TWO TRADE SIZE SMALLER SIZE THAN THE UNGROUNDED CONDUCTOR. (PEC specs) Therefore use 1- 22 mm2 Tw copper wire for the Neutral line.
4. Determine the Size of the Conduit pipe. Refer to Table , use 32 mm diameter pipe.
5. For main Breaker, refer to Table . Use 2 - 100 A 2 - wires 250 V, 2 pole molded air circuity breaker.
COMMENT:
The total computed load is 63.37 A. The 30 mm2 copper wire could be used considering its 90 A capacity. However, The NEC provides that:
" If the computed load exceeds 10000 watts, the conductor and overcurrent protection shall be rated not less than 100 A.
THEREFORE USE 2 - 38 mm2 TW WIRE FOR THE MAIN FEEDER AND 2 - 100 A FOR THE MAIN BREAKER.
Small Family Dwelling
Type of Service - 230 volts; two wire
Line to Ground system
Type of Service - 230 volts; two wire
Line to Ground system
A single family dwelling with a floor area of 80 square meters has the following receptacles and outlets load.
LIGHTING:
7 pcs. - 40 watts fluorescent lamps
2 pcs. - 20 watts Incandescent lamps
CONVENIENCE OUTLET:
1 - Electric Iron .............................................. 1000 watts
1 - Electric stove.............................................. 1100 watts
2 - Electric fan ................................................ 500 watts
1 - 7 cu. ft Refrigerator .................................... 175 watts
1 - Portable stereo ........................................... 100 watts
1 - 20" TV set ................................................... 300 watts
SOLUTION:
A. Circuit 1 - Lighting Load by the Area method
1. Determine the wattage required per square meter area. From, the wattage required per square meter for dwelling units is 24 watts. Multiply:
80 sq.m x 24 watts = 1920 watts
2. Determine the current load. Divide:
1920 watts/ 230 volts = 8.35 A
3. Compute the actual lighting load. Multiply:
7 - Fluorescent lamps x 40 watts = 280 watts
2 - Incandescent bulb x 60 watts = 120 watts
TOTAL............ 400 watts
4. Solve the actual current load. Divide:
400 watts/230 volts = 1.74 A
5. Determine th Size of the Branch Circuit wire. From , the 1.74 A is very small load to be carried by 2.0 mm2 or No. 14 TW copper wire. Therefore, the No. 14 wire is safe.
6. Determine the Size of the conduit pipe. Refer to table, for 2- No.14 wire, use 13 mm conduit pipe.
7. Determine the size or rating of the branch circuit protection. Refer to table. For 2.0 mm2 or No.14 copper wire conductor, use 15 A fuse or trip breaker.
B. Circuit - 2 For small appliance load
1. Solve for the total appliance current load.
LOAD CURRENT = ( 1000 + 1100 + 500 + 175 + 300 + 100) / 230 volts
= 3175 watts/230 volts
= 13.81 A
2. Determine the size of the Branch circuit wire conductor. Refer to Table. For a convenience load of 13.81 A specify 3.5 mm2 or No. 12 TW copper wire, the minimum size required for convenience outlet.
3. Find the size of the conduit pipe. Refer to Table, for 2 pieces No.12 TW copper wire, use 13 mm diameter pipe.
4. Find the Size or rating of the Protection device. See Table, for 13.81 A, use 1 - 20 A fuse.
COMMENT:
It is interesting to note that only one 20 A fuse protection was used because the current is a LINE TO GROUND OR MULTI-GROUND SYSTEM where one line is zero voltage being grounded. Unlike the LINE TO LINE SYSTEM of current supply, it is necessary to provide 2 fuses to protect the two line branch circuit.
FINDING THE SIZE OF THE SERVICE ENTRANCE OR FEEDER
1. Get the sum total of connected load. Add:
Lighting Load.................................... 1920 watts
Small appliance load ......................... 3175 watts
TOTAL.................................. 5095 watts
2. Solve for the total connected load current. Divide:
5095 watts/230 volts = 22.15 A
3. Find the size of Service Entrance. Refer to Table. For 22.15 A, use No. 8 TW copper wire, the minimum size for service entrance.
4. For Main Protection, use 1-safety switch, 2 pole, 2 wires, 250 volts.
Under the preceding set-up, one safety switch could supply both lighting and convenience outlet at different branch circuit without the use of fuse cutout. This is only applicable to the line to ground or multi-ground system being used by the electric cooperative.
LIGHTING:
7 pcs. - 40 watts fluorescent lamps
2 pcs. - 20 watts Incandescent lamps
CONVENIENCE OUTLET:
1 - Electric Iron .............................................. 1000 watts
1 - Electric stove.............................................. 1100 watts
2 - Electric fan ................................................ 500 watts
1 - 7 cu. ft Refrigerator .................................... 175 watts
1 - Portable stereo ........................................... 100 watts
1 - 20" TV set ................................................... 300 watts
SOLUTION:
A. Circuit 1 - Lighting Load by the Area method
1. Determine the wattage required per square meter area. From, the wattage required per square meter for dwelling units is 24 watts. Multiply:
80 sq.m x 24 watts = 1920 watts
2. Determine the current load. Divide:
1920 watts/ 230 volts = 8.35 A
3. Compute the actual lighting load. Multiply:
7 - Fluorescent lamps x 40 watts = 280 watts
2 - Incandescent bulb x 60 watts = 120 watts
TOTAL............ 400 watts
4. Solve the actual current load. Divide:
400 watts/230 volts = 1.74 A
5. Determine th Size of the Branch Circuit wire. From , the 1.74 A is very small load to be carried by 2.0 mm2 or No. 14 TW copper wire. Therefore, the No. 14 wire is safe.
6. Determine the Size of the conduit pipe. Refer to table, for 2- No.14 wire, use 13 mm conduit pipe.
7. Determine the size or rating of the branch circuit protection. Refer to table. For 2.0 mm2 or No.14 copper wire conductor, use 15 A fuse or trip breaker.
B. Circuit - 2 For small appliance load
1. Solve for the total appliance current load.
LOAD CURRENT = ( 1000 + 1100 + 500 + 175 + 300 + 100) / 230 volts
= 3175 watts/230 volts
= 13.81 A
2. Determine the size of the Branch circuit wire conductor. Refer to Table. For a convenience load of 13.81 A specify 3.5 mm2 or No. 12 TW copper wire, the minimum size required for convenience outlet.
3. Find the size of the conduit pipe. Refer to Table, for 2 pieces No.12 TW copper wire, use 13 mm diameter pipe.
4. Find the Size or rating of the Protection device. See Table, for 13.81 A, use 1 - 20 A fuse.
COMMENT:
It is interesting to note that only one 20 A fuse protection was used because the current is a LINE TO GROUND OR MULTI-GROUND SYSTEM where one line is zero voltage being grounded. Unlike the LINE TO LINE SYSTEM of current supply, it is necessary to provide 2 fuses to protect the two line branch circuit.
FINDING THE SIZE OF THE SERVICE ENTRANCE OR FEEDER
1. Get the sum total of connected load. Add:
Lighting Load.................................... 1920 watts
Small appliance load ......................... 3175 watts
TOTAL.................................. 5095 watts
2. Solve for the total connected load current. Divide:
5095 watts/230 volts = 22.15 A
3. Find the size of Service Entrance. Refer to Table. For 22.15 A, use No. 8 TW copper wire, the minimum size for service entrance.
4. For Main Protection, use 1-safety switch, 2 pole, 2 wires, 250 volts.
Under the preceding set-up, one safety switch could supply both lighting and convenience outlet at different branch circuit without the use of fuse cutout. This is only applicable to the line to ground or multi-ground system being used by the electric cooperative.
what is the distance between each convenience outlet? and the distance from the floor to the convenience outlet?
ReplyDeleteEvry 3m less for distance of each oulet ... and height of convenience outlet from finish floor.level , 450mm
DeleteThis is very helpful specially circuit breaker here in the Philippines
ReplyDeletehow do you got the 63.37amp. for your service entrance wire
ReplyDeleteThis is one of the best samples....Thank you!
ReplyDeletethanks for the info.. but where can i find those tables??
ReplyDeletethanks for the info.. but where can i find those tables??
ReplyDeletethanks for the info.. but where can i find those tables??
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ReplyDeleteTy.
ReplyDeleteThe National Electrical Code set 100 watts as the max. for lighting, why we need to use 100 in the computation of total watts, what if I'm using 20 pcs. of 5W LED bulb, is using 100 as multiplier, it is not an over design? as the actual is only 100W not 2000W
ReplyDeleteI think over design is better than exact because if ever you missed something in your calculation you wont get wrong. Or if you have plans for additional loads, you dont need to change your wires. Besides it is PEC standards for us to follow.
DeleteI think they should change this rule. In my place, they will have to require you to buy your own transformer if they see that your load is high. As to how high, they don't really tell you how high. As-built plan may be better, or apply for a construction light before proceeding with const activities.
DeleteThe Electrical Code has provisions for minimum safety requirements. Using 100VA per lighting outlet is much safer than using the wattage of the LED. Practicality is also considered here. How if the owner in the future changed the LED lamp into a much higher wattage? This will prompt him also to replace or upgrade the wire/cable. FYI...the Electrical Code specifies that total calculated loads shall be based on the total ampere ratings of such units and not on the total watts of the lamps. I hope this answer helps.
DeleteWHAT IS THE DEMAND FACTOR FOR A SIMPLE STUDIO TYPE APARTMENT. TOTAL WATTAGE IS 3480 WATTS PER UNIT
ReplyDeleteHow many units?
Delete5-small appliance duplex outlets, 3- light bulbs and 1-1hp ACU. of the total wattage, what will be the demand factor? hope to hear from you and it will be greatly appreciated if someone share their knowledge on this matter
ReplyDeleteWhy not use a THHN wire?
ReplyDeleteIs TW more ideal?
TW is a minimum requirement. THHN is better.
DeleteThis comment has been removed by the author.
ReplyDeleteHi sir. In a standard circuit of LP how many lights is the limit per circuit? and Where can I find it in PEC.
ReplyDeleteThanks.
Thanks. Its very helpful
ReplyDeleteHi.. is there any standard minimum VA rating for every equipment? For instance, a 1Hp ACCU shall be rated at 1840VA?.. TIA
ReplyDeleteI have 129.22A. what will be the service entrance conductor? thank you
ReplyDeleteUSE: 150AT CB, 2-50mm2 THHN and 1-14mm2 THHN ground wire in 50mm diameter conduit.
DeleteI graduated electrical engineering, it is very helpful to me for preparing for a job and have a expirience in computaion of load. Keep it up mam/sir!
ReplyDeleteCan I ask why did you not use power factor for the ACU load?
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ReplyDeleteFrom PEC:
ReplyDelete(i) Receptacle Outlets.
Except as covered in 2.20.2.5(j) and (k), receptacle outlets shall be calculated at not less than 180 volt-amperes for each single or for each multiple receptacle on one yoke. A single piece of equipment consisting of a multiple receptacle comprised of four or more receptacles shall be calculated at not less than 90 volt- amperes per receptacle. This provision shall not be applicable to the receptacle outlets specified in 2.10.1.11(c)(1) and (c)(2).
Please clarify the 180va per outlet.
per PEC 180w for each yoke, in my understanding 2 gang outlet = 180 or 1-3 gang computed 180w
DeleteHow many lighting outlets maximum in one branch circuit?
ReplyDeleteHiw to compute MPB for 2 panel board., 1 at ground floor & 1 at 2nd floor
DeleteIf it says minimum requirement, does it also mean putting thicker wire is much better? For instance, 3.5mm thhn wire for lights and 5.5mm thhn wire for outlets?
ReplyDeleteYes it is much better based from the safety perspective. But if you are installing for example branch circuit for convenience outlet then it is not. In terms of economy and practicality it is not advisable to increase the size of wire. Using a larger size will cost you more when in fact the smaller size is allowed by the code as long as it meets the minimum calculation. A 5.5sqmm wire also requires a larger conduit, thus increasing the cost. There are also some electrical device that a larger wire size would not fit or will be difficult to install.
DeletePLEASE DISPLAY THE FLOOR PLAN
ReplyDeleteHow to find or calculate the loop for lighting?
ReplyDeleteSolve for this problem,A1.5 air conditioning unit is to be circuited, determine the ff
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ReplyDeleteThis guide enables a more straightforward way of determining conductor sizes, conduit, and the ratings of protective devices for Residential Electrical Estimates and provides useful tools to ensure rapid accurate estimation.
ReplyDelete