How To Prepare Schedule of Loads

Introduction:

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.

Sample Electrical Plan:

Sample Electrical Plan

• The diagram describe above is simple example of an electrical plan where the actual number of outlets in lighting and convenience outlet are known.
• 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, Line-Neutral.

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 plug-in	½” dia.
2	L.O.	1	100 VA	9	900	220	4.09	#14 TW	15 AT, 1P plug-in	½” dia.
3	L.O.	1	100 VA	6	600	220	2.73	#14 TW	15 AT, 1P plug-in	½” dia.
4	C.O	1	180 VA	10	1,800	220	8.18	#12 TW	20 AT, 1P plug-in	¾” dia.
5	C.O	1	180 VA	12	2,160	220	9.82	#12 TW	30 AT, 1P plug-in	¾” dia.
6	ACU	1	2.5 HP	1	2331	220	10.60	#10 TW	30 AT, 1P plug-in	¾” dia.
7	ACU	1	2.5 HP	1	2331	220	10.60	#10 TW	30 AT, 1P plug-in	¾” dia.
8	ACU	1	2.5 HP	1	2331	220	10.60	#10 TW	30 AT, 1P plug-in	¾” dia.
9	Range Load	1	5000 W	1	5000	220	22.71	# 8 TW	80 AT, 1P plug-in	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.  


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 plugin type 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 plugin type 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 plugin type 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 = 20.45 Amperes, Use 20 A plugin type Circuit Breaker
Conduit = Use 3/4" 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 plugin type Circuit Breaker
Conduit = Use 3/4" diameter PVC conduit.

Circuit 6-8:

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 plugin type Circuit Breaker [3]
Conduit = Use 3/4" diameter PVC conduit.  

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 plugin type Circuit Breaker
Conduit = Use 1.0" 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 210-9a - 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 non-time delay fuses on full voltage starting.

4. NEC 210-22(C) = Over-Current Protection Device shall be calculated as 100% of non-continuous load + 125% of the continuous load. 

References:

1. National Electrical Code 2011 (Handbook)
2. CESEEPS Red Book, Low Voltage Systems and Applications in the Industries
3. General Electric Circuit Breaker Catalogue

Any comments for this article are welcome....