AC-Power factor of an RL circuit

Discussion in 'Homework Help' started by Rubberfrog, Nov 12, 2008.

  1. Rubberfrog

    Thread Starter New Member

    Nov 12, 2008
    7
    0
    Hello, im new to this forum and have used it in the past for studying, im wondering if anyone could give me help on a question I’ve been trying to figure out involving calculating power factor of an RL Circuit when the frequency has changed.

    A 240v 50hz appliance is rated at 2Kw and has a lag PF of 0.7
    Calculate the new PF at 60hz.

    Here’s what I’ve done so far:

    Sketched a Power triangle True power (P) = 2000w
    Apparent (S)= 2860VA (S= P/PF)

    Calculating I
    Vs=240v
    Apparent (S)=Vs x I

    I=S/Vs = 11.92A

    Finding Q (Reactive Comp.) by Pythagoras.
    b= root (c2-a2)
    Q= 2044va
    Finding Inductive Reactance
    Xl=Vs2/Q = 240squared/2044
    Xl=28.18

    Total Cct Impedance
    Z= Vs/I = 240/11.92= 20.13Ω

    Using XL = 2piFL
    Transpose to find L
    L= 632mH

    Cos θ = PF=0.7
    Cos-1PF= θ
    Θ=45.57degrees

    I have tried changing the Freq in the equation 2piFL =Xl giving two new values of Xl but I cant seem to be able to construct a new power triangle with just this info. Having got to this point I am not sure what my next step is any pointers would be of a great help.

    Cheers
    Rubberfrog
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Now, you have the inductance so you can calculate the new Xl for 60Hz. After you do it calculate the total impedance of the circuit. The arccos of the angle of this impedance is the new power factor. Another way to do it is to calculate again the total impedance at 60Hz, then find the current, then the real power dissipated by the real part of the impedance and finally divide the real power by the new apparent power to find the power factor.
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Rubberfrog,

    Hi RF. Love your name.

    Well, let's see. Let's just use the ratios. The resistance R is not going to change, but the reactance is going to increase by 6/5ths. So, we start with a normalized triangle with a hypotenuse of 1, a run of 0.7, and a rise of 0.71 . Then we increase the rise by 6/5ths or 0.71*6/5 = 0.86 . That makes a tangent of 0.86/0.7 = 1.22 . By any number of methods, it can be found that for a right triangle, an angle with a tan of 1.22 is the same as an angle with a cos of 0.63, which is the new PF.

    Ratch
     
    Last edited: Nov 21, 2008
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