Schedule of load preparation is essential and a basic calculation for electrical engineers. In this process the proper sizing of conductors, overload protection and conduits are determined.
While there are different methods doing an electrical design but there is only one thing that cannot be altered  code requirements must be followed.
 This example emphasized the procedure rather than mimicking the actual loads of a residential unit.
 In this example the voltage drop and short circuit calculation is not included.
 The system voltage of this example is 220 VAC, 60 Hz.
Schedule of Loads
Ckt

Load

Ph

Rating
Per
outlet

No. of Outlets

VA

Volts

Amps

Wire

CB

Cond.

1

L.O.

1

100 VA

12

1,200

220

5.45

#14 TW

15 AT, 1P plugin

½” dia.

2

L.O.

1

100 VA

9

900

220

4.09

#14 TW

15 AT, 1P plugin

½” dia.

3

L.O.

1

100 VA

6

600

220

2.73

#14 TW

15 AT, 1P plugin

½” dia.

4

C.O

1

180 VA

10

1,800

220

8.18

#12 TW

20 AT, 1P plugin

¾” dia.

5

C.O

1

180 VA

12

2,160

220

9.82

#12 TW

30 AT, 1P plugin

¾” dia.

6

ACU

1

2.5 HP

1

2331

220

10.60

#10 TW

30 AT, 1P plugin

¾” dia.

7

ACU

1

2.5 HP

1

2331

220

10.60

#10 TW

30 AT, 1P plugin

¾” dia.

8

ACU

1

2.5 HP

1

2331

220

10.60

#10 TW

30 AT, 1P plugin

¾” dia.

9

Range Load

1

5000 W

1

5000

220

22.71

# 8 TW

80 AT, 1P plugin

1.0” dia.

Schedule of Loads
Schedule of loads are just a summary of data to easily identify and facilitate the necessary values and equipment rating to be used in any electrical installation. Any data given in the schedule of loads were backed by calculation based on a well settled electrical principles and code requirements.
The general rule: The cable and circuit breaker must always be coordinated.
Computations
Circuit 1:
I = 1,200 VA/ 220 V = 5.45 Ampere
Wire = 5.45 / 80% = 6.82 Amperes , Use 2.0 sqmm TW wire or #14 AWG [1]
Circuit Breaker = Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.
Circuit 2:
I = 900 VA/ 220 V = 4.09 Ampere
Wire = 4.09/ 80%= 5.11 Amperes , Use 2.0 sqmm TW wire or #14 AWG
Circuit Breaker = Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.
Circuit 3:
I = 600 VA/ 220 V = 2.72 Ampere
Wire = 2.72/ 80%= 3.41 Amperes , Use 2.0 sqmm TW wire or #14 AWG
Circuit Breaker = 6.82 Amperes, Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.
Circuit 4:
I = 1,800 VA/ 220 V = 8.18 Ampere
Wire = 8.18/ 80%= 10.23 Amperes , Use 3.5 sqmm TW wire or #12 AWG
Circuit Breaker = Use 20 A CB
Conduit = Use 3/4" diameter PVC conduit.
Main Feeder
By inspection:
Continuous loads = 9,963 VA or 45.29 A @ 220V (lighting loads and ACU)
Non Continuous = 8, 960 VA or 40.72 @ 220V (conv. outlet & range load)
Total Loads = 19, 923 VA
Main Feeder Current = (45.29 x 100% ) + (40.72 x 125%) = 96.19 Amperes [4]
Use 50 sqmm TW cable as main feeder or service entrance wire
Use 100 Ampere MCCB, 1 pole  10 kAIC*
note: 10 kAIC is just an assumed value, we need short circuit calculation to determine the right specs of the OCPD to be used in this example
Rules Applied:
1. NEC 2109a  Maximum to be served by branch circuit must not be less than 80% of the ampacity of the condutor
2. NEC 430 22 = The size of the wire supplying motorized load shall not be less than 125% of the rated full load current of the motor.
3. NEC 430 52 = The size of the branch circuit protection for motor loads shall not be greater than 250% of motor full load current for CB and 300% for nontime delay fuses on full voltage starting.
4. NEC 21022(C) = OverCurrent Protection Device shall be calculated as 100% of noncontinuous load + 125% of the continuous load.
References:
1. National Electrical Code 2011 (Handbook)
2. General Electric Circuit Breaker Catalogue
Computations
Circuit 1:
I = 1,200 VA/ 220 V = 5.45 Ampere
Wire = 5.45 / 80% = 6.82 Amperes , Use 2.0 sqmm TW wire or #14 AWG [1]
Circuit Breaker = Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.
Circuit 2:
I = 900 VA/ 220 V = 4.09 Ampere
Wire = 4.09/ 80%= 5.11 Amperes , Use 2.0 sqmm TW wire or #14 AWG
Circuit Breaker = Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.
Circuit 3:
I = 600 VA/ 220 V = 2.72 Ampere
Wire = 2.72/ 80%= 3.41 Amperes , Use 2.0 sqmm TW wire or #14 AWG
Circuit Breaker = 6.82 Amperes, Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.
Circuit 4:
I = 1,800 VA/ 220 V = 8.18 Ampere
Wire = 8.18/ 80%= 10.23 Amperes , Use 3.5 sqmm TW wire or #12 AWG
Circuit Breaker = Use 20 A CB
Conduit = Use 3/4" diameter PVC conduit.
Read: What are the Different Electric Current Symbols According to IEC Standard
Circuit 5:
I = 2,160 VA/ 220 V = 9.82 Ampere
Wire = 9.82/ 80% = 12.27 Amperes , Use 3.5 sqmm TW wire or #12 AWG
Circuit Breaker = Use 20 A CB
Conduit = Use 3/4" diameter PVC conduit.
Circuit 68:
VA = [ 2.5 HP x ( 746 Watts/ HP ) ] / 0.8 pf (assume 0.8 pf)
VA = 2331 VA
I = 2,331 VA/ 220 V = 10.60 Ampere
Wire = 10.60 x 125% = 13.24 Amperes , Use 3.5 sqmm TW wire or #12 AWG [2]
Circuit Breaker = 5.45 x 250% = 26.5 Amperes, Use 30 A Circuit Breaker [3]
Conduit = Use 3/4" diameter PVC conduit.
Note: since the breaker is 30 Ampere, we need to increase the cable size to 5.5 sqmm (rated 30 amps by NEC ) to maintain the coordination of cable and the circuit breaker.
Circuit 9:
VA = 5000 W / 1.0 pf (heating load is a resistive load w/ 100% pf)
VA = 5,000 VA
I = 5, 000 VA/ 220 V = 22.72 Ampere
Wire = 22.71 / 80% = 28.41 Amperes , Use 8.0 sqmm TW wire or #8 AWG
Circuit Breaker = Use 40 A Circuit Breaker
Conduit = Use 1.0" diameter PVC conduit.
Circuit 5:
I = 2,160 VA/ 220 V = 9.82 Ampere
Wire = 9.82/ 80% = 12.27 Amperes , Use 3.5 sqmm TW wire or #12 AWG
Circuit Breaker = Use 20 A CB
Conduit = Use 3/4" diameter PVC conduit.
Circuit 68:
VA = [ 2.5 HP x ( 746 Watts/ HP ) ] / 0.8 pf (assume 0.8 pf)
VA = 2331 VA
I = 2,331 VA/ 220 V = 10.60 Ampere
Wire = 10.60 x 125% = 13.24 Amperes , Use 3.5 sqmm TW wire or #12 AWG [2]
Circuit Breaker = 5.45 x 250% = 26.5 Amperes, Use 30 A Circuit Breaker [3]
Conduit = Use 3/4" diameter PVC conduit.
Note: since the breaker is 30 Ampere, we need to increase the cable size to 5.5 sqmm (rated 30 amps by NEC ) to maintain the coordination of cable and the circuit breaker.
Circuit 9:
VA = 5000 W / 1.0 pf (heating load is a resistive load w/ 100% pf)
VA = 5,000 VA
I = 5, 000 VA/ 220 V = 22.72 Ampere
Wire = 22.71 / 80% = 28.41 Amperes , Use 8.0 sqmm TW wire or #8 AWG
Circuit Breaker = Use 40 A Circuit Breaker
Conduit = Use 1.0" diameter PVC conduit.
Read: How to Select Proper Type of Miniature Circuit Breakers MCBs
Main Feeder
By inspection:
Continuous loads = 9,963 VA or 45.29 A @ 220V (lighting loads and ACU)
Non Continuous = 8, 960 VA or 40.72 @ 220V (conv. outlet & range load)
Total Loads = 19, 923 VA
Main Feeder Current = (45.29 x 100% ) + (40.72 x 125%) = 96.19 Amperes [4]
Use 50 sqmm TW cable as main feeder or service entrance wire
Use 100 Ampere MCCB, 1 pole  10 kAIC*
note: 10 kAIC is just an assumed value, we need short circuit calculation to determine the right specs of the OCPD to be used in this example
Rules Applied:
1. NEC 2109a  Maximum to be served by branch circuit must not be less than 80% of the ampacity of the condutor
2. NEC 430 22 = The size of the wire supplying motorized load shall not be less than 125% of the rated full load current of the motor.
3. NEC 430 52 = The size of the branch circuit protection for motor loads shall not be greater than 250% of motor full load current for CB and 300% for nontime delay fuses on full voltage starting.
4. NEC 21022(C) = OverCurrent Protection Device shall be calculated as 100% of noncontinuous load + 125% of the continuous load.
References:
1. National Electrical Code 2011 (Handbook)
2. General Electric Circuit Breaker Catalogue
11 comments
Thank you
Nice blog… Thanks for sharing very useful information about electrical circuits.
Learn Electronic Circuits
help:
question 1: circuit 4; how did you compute circuit breaker: 20.45 amps?
question 2: circuit 68; how did you compute circuit breaker: 5.45 amps?
Oops!? you multiplied the continous load by 100% and the noncontinous load by 125%.
in what chapter in the nec stated that 1 LO is rated 100VA?
..Hi, I'm just Wondering. Why did you always divide the AMPERAGE RESULT to 80% pf except Circuit 68 that you Multiply into 125%? thus this 80% pf different from 80% Demand Factor (DF)?
calculating voltage drop in the system. This code provides data of standard conductor properties that can be used in voltage drop calculation. CCTV Installers Melbourne
National Electrical Code 2011 (Handbook)
2. CESEEPS Red Book, Low Voltage Systems and Applications in the Industries
3. General Electric Circuit Breaker Catalogue http://www.electricaltest.cn/
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