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A 6-pulse full-wave AC-DC converter is feeding power to a 10kW, 40V DC, RL load, as shown in Fig.13. The input supply system is 3-Phase, 415V, 50Hz.

2006-12-01 06:20:30 · 2 answers · asked by George 1 in Science & Mathematics Engineering

2 answers

KVA= Kw (more or less)

(400v now not 415v in the uk)

just add for power factor, (divide by0.8??)
add for efficiency (divide by 0.8??)

10,000/230/0.8/0.8/3 phases = 23A/phase

10,000/0.8/0.8 = 15.625kVA

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next sort out circuit protection for the equipment on the primary and secondary, you may need to increase the transformer so that it is safe.

probably go for 15kVA tx and circuit protection to suit. ( can see figure?)

the maths are a good guide but the manufacturer should be the most reliable.
hope this helps

2006-12-01 07:39:07 · answer #1 · answered by Mark G 2 · 0 0

Your kVA is actually going to be the area under the curve of your volt-amp profile of your alternating current.

I don't have your RL circuit (its in your textbook), but the impedence of this circuit will add a power factor, depnding on the magnitude of your R and your L.

2006-12-01 16:28:15 · answer #2 · answered by www.HaysEngineering.com 4 · 0 0

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