Mathcad - 02-23 - constant. Therefore, solve for the real...

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The plant probably would install 90 kVAR of capacitors (30 kVAR on each phase) to get the power factor above 0.93 Q cap 81.096 - kVAR = Q cap Q new Q old - = The difference between the old and new reactive power values has to be provided by the capacitors Q new 69.686kVAR = Q new S new 2 P load 2 - = and the reactive power will be S new 189.591 kVA = S new P load F pnew = After correction, the apparent power will be: Q old 150.782kVAR = Q old S load 2 P load 2 - = Then solve for the original reactive power and the new reactive power P load 176.32kW = P load S load F pold = Adding capacitors will change the reactive and apparent power but the real power will stay
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Unformatted text preview: constant. Therefore, solve for the real power Solution F pnew 0.93 = F pold 0.76 = S load 232 kVA = Given: kVAR kW kVA kW Define: P2-23. A machine shop has a three-phase electrical load consisting of many small motors. The total shop load is 232 KVA and the power factor is 0.76 lagging. The utility would like them to improve their power factor to greater than 0.93 lagging, but they must keep a lagging power factor. How many KVARs of capacitors will they require to achieve the minimum power factor....
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.

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