Adam_151 said:6.71A ?
andy said:Adam_151 said:6.71A ?
where is the workin out?!
Nope ! 2/10 please see mebillynw said:2*pi*50*0.04 = 12.56ohms (Xl)
Vl / Xl = IL
100 / 12.56 = 7.96 amps
??
ambrougham said:So how about this one dragged up from the depths of memory of years (very much) gone by:
If a load draws a current of 10A from a 240V 50Hz supply with a power factor of 0.6 lagging, what value capacitance should be used to bring the power factor up to near unity ?
....... erhm, and heres hoping that I can dredge up a solution if needs be also
adam151 said:Now, I'm pretty sure the above is nonsense, but I'm not sure which part I'm making a false asumption in
Pensdown said:If I can remember back to my college days, I think thats why some suppliers insist on power factor correction in factories etc with a lagging power factor below about .7 All of their infrastructure acts as if it's taking the full 2544watts but the customer only pays for the 1440watts.
Adam_151 said:before correction it had a total imediance of 24ohms, the resistance and reactance were 8 ohms 22.6 ohms (approx), if you correct the power factor so take the reactance down to zero the best you can, the impedance is going to be about 22.6 ohms which means the current has actually risen to 10.6 , and the motor is giving a real power of 2544w
plugwash said:S=10*240=2400VA
P=2400*0.6=1440W
Q²=2400²-1440²=3686400
Q=1920
Xc*240=1920 <----- P = V²/Xc
Xc=8 <----- Xc = 30
Xc=1/(2 pi 50C)
100 pi C = 0.125
pi C = 0.00125
so the capacitor needs to be arround 400 (100) microfarads.
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