EC&M’s Electrical Calculations Handbook - Chapter 2 pptx

... voltage - ( 120 120 ° = - 120 -6 0 ° ) = -( - 120 COS 60 -j 120 SIN 60) = -[ (-1 20 )(5) -j ( 120 )(.866)] = -[ -6 0 +j 103. 92] Phase a voltage: 120 = 120 COS 0 + j 120 SIN 0 = [ ( 120 )(1) +j ( 120 )(0)] 0 ° = ... Volt, 3-phase 4-wire wye Note 5 460 Volt, 3-phase 3-wire delta Note 4 600/347 Volt, 3-phase, 4-wire wye 24 00 Volt, 3-phase, 3-wire delta 24 00...

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... increase within the 3 12 Chapter Eleven 316 Chapter Eleven Figure 1 1-3 Solve for minimum conduit size given wire insulation type, wire size, and wire quantity. Figure 1 1 -2 Cross-sectional areas of ... so the cross-sec- tional area available within each type of raceway for the placement of conductors varies with the type of raceway. Figure 1 1 -2 on p. 315 shows the available cros...

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... Grounding Systems 21 5 Obtaining the System Grounding Point 21 7 Chapter 8. Lighting 22 1 The Lumen Method 22 1 The Point-by-Point Method 22 3 Indoor Lighting 22 9 Lighting Rules of Thumb 24 5 Contents vii RESISTOR CURRENT ... size 4 -2 7 Thermal damage curve for paper or rubber insulation 4 -2 8 Thermal damage curve for thermoplastic insulation 1 2- 1 Time-overcurrent character...

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... VECTOR 180 - 37.4° AMP 20 .9 -1 6.6 +j 12. 7 (-X, +J) (+X, +J) (-X, -J) (+X, -J) Step 3. Draw the resultant vector. +X +J -J (-X, -J) NEW RESULTANT VECTOR -1 6.6 +j 12. 7 180 ° - 37.4 ° AMP 20 .9 (-X, ... /1000 kVA = 58. 823 (58. 823 ) = (50.000) 2 + (kVAR) 2 kVA 2 = KW 2 + kVAR 2 (58. 823 ) 2 - (50.000) 2 = (kVAR) 2 30.986 = (kVAR) is the required ca...

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... 90 2 1 -2 5 7 0-7 7 1.08 1.05 1.04 2 6-3 0 7 8-8 6 1 1 1 3 1-3 5 8 7-9 5 0.91 0.94 0.96 3 6-4 0 9 6-1 04 0. 82 0.88 0.91 4 1-4 5 10 5-1 13 0.71 0. 82 0.87 4 6-5 0 11 4-1 22 0.58 0.75 0. 82 5 1-5 5 12 3-1 31 0.41 0.67 0.76 5 6-6 0 ... 125 100 150 120 170 135 00 145 115 175 135 195 150 000 165 130 20 0 155 22 5 175 0000 195 150 23 0 180 26 0 20 5 25...

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... 13.8-kV bus. Fault power is available to flow into this bus from 1. The electrical utility source 2. The 24 -megawatt (MW), 30-MVA generator 3. The 5000-hp motor 4. The 20 0-hp motor 5. The 40-hp ... form in Fig. 5 -2 . First, however, note that electrical short-circuit power can come from four sources in this circuit: 1. The 400,000-kVA electrical utility service 2. The 24 -MW, 3...

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... power conductors and generator overcurrent pro- 20 0 Chapter Six Gas-turbine generators arrive at the installation site prepackaged on skids. Figure 6 -2 Solve for generator overcurrent protection ... kilovoltampere rating is designed to be 125 percent of the generator kilowatt rating; and this is 20 2 Chapter Six Diesel engine generator set. Generator Sizing Calculations A generat...

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... such as a 460-volt (V), three-phase, three-wire delta system, an arc- ing fault that repeatedly restrikes can cause voltage “jack- ing” of up to six times the normal system phase-to-neutral voltage. ... minimum size of this conductor is shown in Table 25 0-1 22 of the National Electrical Code. See Fig. 7-3 for an example problem in sizing the equipment- grounding conductor. When feede...

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... necessary to use the point-by-point method. The Point-by-Point Method How to make point-by-point calculations Later in this chapter, lighting calculations within areas hav- ing reflective surfaces, ... perpendicular to the lamp. 24 8 Chapter Eight Figure 8-3 Zonal cavity calculation worksheet. 24 1 Figure 8 -2 Common lamps and their characteristics. 23 4 Footcandles ϭ The left...

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... transformer coil. 27 2 26 0 Figure 9-1 0 Table replicating part of NEC Table 45 0-3 (a), overcurrent protec- tion of transformers over 600 V. 27 4 Figure 9-1 2 Table replicating part of NEC Table 45 0-3 (b), overcurrent ... Fig. 9-1 5b. Note that the output voltage is either 24 0 V ϩ 24 V, or 26 4 V, or 24 0 V Ϫ 24 V, or 21 6 V, depending on whether the sec- ondary is c...

Ngày tải lên: 13/08/2014, 02:21