By (NFPA) National Fire Protection Association

Secure, effective, code-compliant electric installations are made uncomplicated with the newest ebook of this generally well known source. Like its hugely profitable prior variants, the NEC 2011 Handbook combines good, thorough, research-based content material with the instruments you want to construct an in-depth realizing of crucial issues. It offers the whole textual content of the up to date code rules along professional observation from code experts, delivering code motive, clarifications for brand new and up to date principles, and sensible, real-world suggestion on find out how to follow the code. New to the 2011 version are articles together with first-time Article 399 on October, Overhead Conductors with over six hundred volts, first-time Article 694 on Small Wind electrical structures, first-time Article 840 on Premises Powered Broadband Communications platforms, and extra. This successful mix has created a useful reference for these in or getting into careers in electric layout, install, inspection, and security.

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Sample text

5 x 1,000,000 = 764. 7200 Select number from Table, three-phase at 80% power factor, that is nearest but not greater than 764. This number is 745 which indicates the size of wire needed: No. 6. 73. For line to neutral voltage drop on a single phase system, divide single phase value by 2. Open Wiring The volt loss for open wiring installations depends on the separation between conductors. The volt loss is approximately equal to that for conductors in non-magnetic conduit. Section 310-15 offers a method to calculate conductor ampacity.

C. 732 Step 1. l. = Step 2. 5 Step 3. l. =1804A 25' - 500kcmil 6 Per Phase Service Entrance Conductors in Steel Conduit Fault X2 Step 4. 0349 6 x 22,185 x 480 Step 4. 4050 Step 5. M= Step 6. 0349 Step 5. M= Step 6. C. C. sym RMS = 49,803 + 7,216 = 57,019A LPS-RK-400SP Fuse ( fault X1) 50' - 500 kcmil Feeder Cable in Steel Conduit Fault X2 Motor Contribution System B M Fault X1 One-Line Diagram Available Utility Infinite Assumption Step 1. l. 732 Step 2. Multiplier = Step 3. C. 57 = 34,370A Step 4.

C. C. C. C. C. primary x Vprimary x (%Z) 100,000 x KVA trans Step 2. Calculate "M" (multiplier). Where: L = length (feet) of conductor to the fault. C = constant from Table of “C” values for conductors and busway. For parallel runs, multiply C values by the number of conductors per phase. I = available short-circuit current in amperes at beginning of circuit. M= 1 1+f Step 3. Calculate the short-circuit current at the secondary of the transformer. C. secondary = The L-N fault current is higher than the L-L fault current at the secondary terminals of a single-phase center-tapped transformer.

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Electrical Protection Handbook by (NFPA) National Fire Protection Association

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