angle - attn Don Kelly

Dec 04, 2006 2 Replies
F



>>>
>>
>>> Hi nice folks,
>>>
>>> I have a question for the Gurus please. The usual expression for
>>> the phase angle is angle = arc tan X/R. What is the expression for
>>> phase angle as related to real and apparent power, also as related to
>>> resistance and impedance? Cheers and TIA.
>>>
>>>
>>
>>arc tan X/R = arc cos R/Z for an impedance Basic trig
>>
>>However while phase angle is applied to impedances, strictly speaking it is
>>the relative angle between voltage and current =arc tan (reactive/real)
>>which is not necessarily the same.
>
>
>>>
>>
>>> Hi nice folks,
>>>
>>> I have a question for the Gurus please. The usual expression for
>>> the phase angle is angle = arc tan X/R. What is the expression for
>>> phase angle as related to real and apparent power, also as related to
>>> resistance and impedance? Cheers and TIA.
>>>
>>>
>>
>>arc tan X/R = arc cos R/Z for an impedance Basic trig
>>
>>However while phase angle is applied to impedances, strictly speaking it is
>>the relative angle between voltage and current =arc tan (reactive/real)
>>which is not necessarily the same.
>
>
>Thanks, but my primary interest was the expression for phase angle per se
>(not cos angle) between voltage and current as related to real and apparent
>power. Do you have the expression for this?

Dr. Kelly,



Oversight? TIA


Vote
Twitter Facebook LinkedIn
D

--------------

OK - real power/apparent power = power factor. If you look at the apparent power S(VA) real power P(W) and reactive "power" Q(VAR) and draw the power triangle.

S /| PF=cos (a) =P/S=P/root{P^2 +Q^2] or generally but not always R/root[R^2+X^2] / |Q a = angle =arccos(PF) = arctan Q/P = arcsin Q/S (1) /a_| P You have to use trig functions as available on most calculators to get the angle as the nice relationship is actually between the cosine and the power factor - not the angle and the pf.

PF angle Pf lead, angle is - PF lag, angle is +

1 0 0.8 36.87 degrees 0.6 53.12 0 90 Hence the relationship between pf and angle is as above angle = arccos(pf) etc That is it in the simplest form.
Vote
Twitter Facebook LinkedIn
F

Thanks. I keep the following for reference:

AC Electrical Circuits:

voltage E = 120 volts

resistance R = 10.8 ohms

current I = 10 amps

From which:

impedance Z = E / I = 12 ohms reactance X = Z - jR = 5.23 ohms

power P = I^2 * R = 1080 W

reactive power Q = I^2 * X = 523 VAR

apparent power S = E * I = 1200 VA

Phase Angle Between Voltage and Current:

angle = arc tan X / R = arc tan Q / P = 25.84 degrees

angle = arc cos R / Z = arc cos P / S = 25.84 degrees

angle = arc sin X / Z = arc sin Q / S = 25.84 degrees

Power Factor

power factor PF = cos angle = P / S = R / Z = 0.9

angle = arc cos PF = 25.84 degrees

Vote
Twitter Facebook LinkedIn

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required