EC&M’s Electrical Calculations Handbook - Chapter 15 potx

... be placed i ᎏᎏ (1 ϩ i) n Ϫ 1 420 Chapter Fifteen Figure 1 5-2 3 Solve for the cash-flow diagram of a loan transaction at interest. Figure 1 5-2 4 Solve for the cash-flow diagram of a recurrent savings ... 1 ᎏᎏ 0.08 (1 ϩ i) n Ϫ 1 ᎏᎏ i 418 Chapter Fifteen Figure 1 5-2 2 Solve for the present value P of maintenance costs over the 2 0- year life of the generators. 376 Figure 1 5-2...

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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, ... (0.576 -j 0.17) x (0.576 -j 0.17) = V; Solve for V. 2. Then, multiply scalars together and add angles. V = (0.6 -1 6.4 ° ) x (0.6 -1 6.4 ° ) V = (0.6 x 0.6) (-1 6.4 ° +...

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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 ... REMARKS 31 0-1 6 OR 31 0-1 7 31 0-1 6 OR 31 0-1 7 UNDERGROUND FEEDER CABLE 31 0-1 6 OR 31 0-1 7 MOISTURE AND HEAT RESISTANT...

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... electrical systems against damage during short-circuit conditions is required by the National Electrical Code (Secs. 11 0-9 and 11 0-1 0). Short-Circuit Calculations 181 For example, in a three-phase ... 30-MVA generator 3. The 5000-hp motor 4. The motors connected to the 480-V bus Short-Circuit Calculations 191 Figure 5-1 Electrical power system one-line drawing for short-cir...

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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. ... enough equipment-grounding conductors, as shown in Fig. 7-4 . Since system grounding is done at the service-disconnect- ing means, the equipment-grounding conductor and the neu- tral (“groun...

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... characteristics 4-2 2 Heat losses of a copper bar from its dimensions and ampere load 8-1 Horizontal footcandle value using point method 4-1 1 Impedance of 600-V, 5-kV, and 1 5- kV copper cable 1-3 7 Impedance ... Solve for 1-3 0 Power factor from voltage, current, and true power 1 5-2 2 Present value of maintenance cost 1 4-2 Pull-box dimensions for an angle pull 1 4-1...

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... 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 2400 Volt, 3-phase, 3-wire delta 2400 Coltm 3-phase, 4-wire wye 4160/2400 Volt, 3-phase, 4-wire ... -j 104 = (180) 2 + (104) 2 Minus Phase b voltage - (120 120 ° = - 120 -6 0 ° ) = -( -1 20 COS 60 -j 120 SIN 60) = -[ (-1 20)(5) -j (120)(.866)] = -[ -6 0 +j 103.92]...

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... is 1000 kW and a 150 -hp motor is to be started meanwhile, the minimum kilowatt rating of the generator would be 1000 kW ϩ 150 kW, or 1150 kW, and the minimum kilovoltampere rat- ing of the generator ... the generator would be 0.5 ϫ [1000 kW ϩ 6 (150 )], or 950 kilovoltamperes (kVA). Select a generator set that meets or exceeds both the 1150 -kW and the 950-kVA requirements and that has a v...

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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, ... footcandles from the second luminaire. Sample point-by-point lighting calculations Refer to Fig. 8-1 for sample point-by-point lighting calcula- tions for both the nadir point and a po...

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... Figure 9-1 5 A standard two-winding transformer can be changed into an auto- transformer simply by interconnecting the primary coil with the secondary coil, as shown. 281 Figure 9-1 6 Three-phase ... coil. 272 260 Figure 9-1 0 Table replicating part of NEC Table 45 0-3 (a), overcurrent protec- tion of transformers over 600 V. 274 Figure 9-1 2 Table replicating part of NEC Table 4...

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