To Section O (By Jerald Tan Madroño) Electrical Floor - TopicsExpress



          

To Section O (By Jerald Tan Madroño) Electrical Floor plant Salahat Na Nag Tatanu kung ano ano at pano gawin Ang electrical plan sana makatulong To... kung mag tatanung kau nan dito si ..Junikz Zamora need namin help mo pre... Any comments for this article are welcome.... 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. 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 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
Posted on: Sat, 11 Oct 2014 09:07:09 +0000

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