Mathcad - 02-25 - (b) The shop owner would like to expand...

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S syn 99.162 kVA = S syn P in F psyn = F psyn 0.8 = Since the new motors operate at 0.8 power factor, the apparent power will be P in 79.33kW = P in P out η = P out 100 hp = For the new motors, the power out is 100 hp or 74.6 kW (2 times 50 hp), so the power in is the power out divided by efficiency S old 166.5 151.337i + kVA = S old S load e j θ old = θ old acos F pold ( 29 = b. We can find the total load with the new motors by adding the two loads together as complex numbers I L 270.633A = I L S load 3 V L = a. The line current can be found from the three-phase power formula Solution η 94% = V L 480 V = F pold 0.74 = S load 225 kVA = Given: kVAR kW kVA kW Define: P2-25. A machine shop is fed by a 300 KVA, 480 Volt, three-phase transformer bank that has a load consisting of many small motors. The total shop load is 225 KVA and the power factor is 0.74 lagging. (a) What is the line current in the feeder from the transformer to the facility?
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Unformatted text preview: (b) The shop owner would like to expand the services provided to customers. This requires adding two 50 HP motors that will run at rated load. One possibility is to use synchronous motors and run them at a 0.8 leading power factor. The synchronous motors are 94% efficient. Would the total load fit within the size of the existing service? SHOW ALL YOUR WORK!! syn acos F psyn ( 29-= syn 36.87-deg = S newmot S syn e j syn = S newmot 79.33 59.497i-kVA = So the total load is the old plus the new S tot S old S newmot + = S tot 245.83 91.839i + kVA = S tot 262.425kVA = The total load is 262.4 kVA, so the transformer is big enough! F ptot cos arg S tot ( 29 ( 29 = F ptot 0.937 =...
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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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Mathcad - 02-25 - (b) The shop owner would like to expand...

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