Design_Project_1 - promoting a new energy policy by a...

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Electrical Engineering Department EE205 : Electric Circuits II First Design project Second Semester 2004-2005 ( 042 ) Due time : May 07, 2005 Three loads are connected in parallel across a 12.47 kV three-phase supply. Load 1: Inductive load, 60 kW and 660 KVAR. Load 2: Capacitive load, 240 kW at 0.8 power factor. Load 3: Resistive load of 60 kW. (a) Find the total complex power, power factor, and the supply current. (b) A Y-connected capacitor bank is connected in parallel with the loads. Find the total KVAR and the capacitance per phase in µ F to improve the overall power factor to 0.8 lagging. What is the new line current? Now SCECO utility company is removing its owned Y-connected capacitor bank and
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Unformatted text preview: promoting a new energy policy by a reduction in the energy rate from 0.2 SR/KWH to 0.15 SR/KWH provided that your load power factor is improved from 0.8 to a unity power factor. Design your capacitor bank to reduce your annual energy cost. Running your own capacitor bank would cost you 10% of its original cost annually as operational and maintenance cost. Remember that after five year of use, the installed capacitor bank has to be replaced by a new one. Will it be profitable as a project to install your own capacitor bank? Show all your calculations and assumptions Note: The cost of 1 KVAR reactive power is 1200 SR in the international market....
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This note was uploaded on 01/16/2010 for the course EE ee205 taught by Professor Profsyed during the Spring '09 term at King Fahd University of Petroleum & Minerals.

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