Electrical Thumb Rules You MUST Follow // PART 2 (on photo: Osprey Lunch Among The Electrical Wires! by Kathleen Jackson via Flickr) Continued from first part: Electrical Thumb Rules You MUST Follow (Part 1) (electrical-engineering-portal/electrical-thumb-rules-you-must-follow-part-1?utm_source=Technical+Articles&utm_campaign=8390932fb8-RSS_EMAIL_CAMPAIGN&utm_medium=email&utm_term=0_ae069f0d91-8390932fb8-320848477) ** Useful Electrical Equations ------------------------------------------------------------ * For Sinusoidal Current: Form Factor = RMS Value/Average Value = 1.11 * For Sinusoidal Current: Peak Factor = Max Value/RMS Value = 1.414 * Average Value of Sinusoidal Current (I[av]) = 0.637 x Im (Im = Max.Value) * RMS Value of Sinusoidal Current (I[rms]) = 0.707 x Im (Im = Max.Value) * A.C Current = D.C Current/0.636. * Phase Difference between Phase = 360/ No of Phase (1 Phase=230/1=360°, 2 Phase=360/2=180°) * Short Circuit Level (electrical-engineering-portal/download-center/electrical-software/short-circuit-current-calculation-isc?utm_source=Technical+Articles&utm_campaign=8390932fb8-RSS_EMAIL_CAMPAIGN&utm_medium=email&utm_term=0_ae069f0d91-8390932fb8-320848477) of Cable in KA (I[sc]) = (0.094 x Cable Dia in Sq.mm) /√ Short Circuit Time (Sec) * Max.Cross Section Area of Earthing Strip (mm2) = √(Fault Current x Fault Current x Operating Time of Disconnected Device ) / K K = Material Factor, K for Cu = 159, K for Al = 105, K for steel = 58 , K for GI = 80 * Most Economical Voltage at given Distance = 5.5 x √ ((km/1.6) + (kw/100)) * Cable Voltage Drop (%) = (1.732 x current x (RcosǾ+jsinǾ) x 1.732 x Length (km) x 100) / (Volt(L-L) x Cable Run. * Spacing of Conductor in Transmission Line (mm) = 500 + 18 x (P – P Volt) + (2 x (Span in Length)/50). * Protection radius of Lighting Arrestor = √h x (2D-h) + (2D+L). Where h= height of L.A, D-distance of equipment (20, 40, 60 Meter), L=V x t (V=1m/ms, t=Discharge Time). * Size of Lighting Arrestor = 1.5x Phase to Earth Voltage or 1.5 x (System Voltage/1.732). * Maximum Voltage of the System = 1.1xRated Voltage (Ex. 66KV = 1.1 × 66 = 72.6KV) * Load Factor (electrical-engineering-portal/how-distribution-systems-control-customer-loads?utm_source=Technical+Articles&utm_campaign=8390932fb8-RSS_EMAIL_CAMPAIGN&utm_medium=email&utm_term=0_ae069f0d91-8390932fb8-320848477) = Average Power/Peak Power * If Load Factor is 1 or 100% = This is best situation for System and Consumer both. * If Load Factor is Low (0 or 25%) = you are paying maximum amount of KWH consumption. Load Factor may be increased by switching or use of your Electrical Application. * Demand Factor (electrical-engineering-portal/demand-factor-diversity-factor-utilization-factor-load-factor?utm_source=Technical+Articles&utm_campaign=8390932fb8-RSS_EMAIL_CAMPAIGN&utm_medium=email&utm_term=0_ae069f0d91-8390932fb8-320848477) = Maximum Demand / Total Connected Load (Demand Factor 1) Diversity factor should be consider for individual Load * Plant Factor (Plant Capacity) = Average Load / Capacity of Plant * Fusing Factor = Minimum Fusing Current / Current Rating (Fusing Factor>1). * Voltage Variation (1 to 1.5%) = ((Average Voltage – Min Voltage) x 100)/Average Voltage Ex: 462V, 463V, 455V, Voltage Variation= ((460 – 455) x 100)/455 = 1.1%. * Current Variation (10%) = ((Average Current – Min Current) x 100)/Average Current Ex: 30A,35A,30A, Current Variation = ((35-31.7) x 100)/31.7 = 10.4% * Fault Level at TC Secondary = TC (VA) x 100 / Transformer Secondary (V) x Impedance (%) * Motor Full Load Current = Kw /1.732 x KV x P.F x Efficiency [INS: :INS]
Posted on: Thu, 29 Aug 2013 12:28:11 +0000
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