AC Theory questions
- Thread starter billynw
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andy said:
Oh if I must, spoils it for everyone else though:
Reactance = 2xpi x frequency x value of inductor
= 6.28 x 50 x 0.04
= 12.56 ohms
As reactance opperatates at right angles to resistance, we have to apply pygothoras formula:
Impedance² = resistance² + reactance²
Impedance² = 8² + 12.56²
Impedance² = 64 + 157.75
Impedance² = 221.75
Impedance = root 221.75
Impedance = 14.89 ohms
Then its just a case of applying ohms law:
Voltage = Current x impedance
Current = Voltage / impedance
current = 100 / 14.89
current = 6.72A
Well at least I think thats right...
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Nope ! 2/10 please see mebillynw said:
You haven't taken into account the resistance of the coil - the calculation you have done is only correct for an inductor with zero resistance.
I = V / (R + jXL) which comes out as 6.7A /-57.5
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
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
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
S=10*240=2400VA
P=2400*0.6=1440W
Q²=2400²-1440²=3686400
Q=1920
Xc*240=1920
Xc=8
Xc=1/(2 pi 50C)
100 pi C = 0.125
pi C = 0.00125
so the capacitor needs to be arround 400 microfarads.
one thing I can't get my head round, taking the example plugwash solved:
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
Now, I'm pretty sure the above is nonsense, but I'm not sure which part I'm making a false asumption in
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
Now, I'm pretty sure the above is nonsense, but I'm not sure which part I'm making a false asumption in
adam151 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.
I stand to be corrected as I'm to lazy to get the book out
Pensdown said:
Sounds spot on to me .... they certainly don't enjoy putting in baby's arm sized cables to supply you particularly when you aren't paying for the power that you're appently using either so will therefore *encourage* you to install PFC .... one way or another !
However, I must say that 400uF wasn't the kind of figure I had in mind at stupid o'clock this morning
More like 100uF from a quicky calc when posting. Hmmmm, b*gger, maybe I *will* have to dredge up some more info from the darkest corners of ye olde memory and do the sums properly after all Supply Voltage: V = 240 /0
Power Factor: PF = cos(Phi)
Supply current: I = 10 /-Phi
Total Impedance: Z = V/I
Impedances in parallel: 1/ZT = 1/Z1 + 1/Z2 + 1/Z3 ....
Assuming a parallel combination of a resistance R with inductive reactance XL (lagging PF):
I/V = (10 /53.1) / (240 /0) = 1/R + 1/jXL --> R = 40 Ohms and XL = 30 Ohms
For unity PF, |XL| = |XC| --> 1/(2 * Pi * F * C) = 30
Therefore C = 106uF or should I say 100uF for cash ! ....................... maybe
Edited to add: OK, just checked this because what often seems like a good idea at stupid o'clock might not be at a more sensible time ! but I'm (almost 100%) certain it is correct.
i.e. with a parallel cct of R = 40 Ohms, L = 95.5mH and PFC C = 106uF, the supply current which would have been measured before PFC would be 10A (PF = 0.6) and the current measured after PFC would be 6A (PF = 1). Or do I need to go stand in the corner wearing the pointy hat for a while ?
Power Factor: PF = cos(Phi)
Supply current: I = 10 /-Phi
Total Impedance: Z = V/I
Impedances in parallel: 1/ZT = 1/Z1 + 1/Z2 + 1/Z3 ....
Assuming a parallel combination of a resistance R with inductive reactance XL (lagging PF):
I/V = (10 /53.1) / (240 /0) = 1/R + 1/jXL --> R = 40 Ohms and XL = 30 Ohms
For unity PF, |XL| = |XC| --> 1/(2 * Pi * F * C) = 30
Therefore C = 106uF or should I say 100uF for cash ! ....................... maybe
Edited to add: OK, just checked this because what often seems like a good idea at stupid o'clock might not be at a more sensible time ! but I'm (almost 100%) certain it is correct.
i.e. with a parallel cct of R = 40 Ohms, L = 95.5mH and PFC C = 106uF, the supply current which would have been measured before PFC would be 10A (PF = 0.6) and the current measured after PFC would be 6A (PF = 1). Or do I need to go stand in the corner wearing the pointy hat for a while ?
Adam_151 said:
nope because you put the capacitance in PARALELL the imaginary part of the Susceptance (reciprocal of impedance) cancels, hence the magnitude of the susceptance goes down and the magnitude of the impedance goes up.
what you are suggesting would only happen if you put the correction capacitor in series.
plugwash said:
Sorry for maybe appearing to be a smart@rse but this was starting to really wind me up ! Looking at doing it you're way, I think you have either dropped a clanger or I've lost the plot somewhere
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