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Power factor correction???

Power factor correction???

Power factor correction???

(OP)
Say two motors are connected in parallel across 240 v r.m.s, 50 hz supply. If one motor takes a current of 8A at a power factor of 0.8 lagging, and the other 5A at 0.6 power factor lagging, what is the total current taken from the supply and the corrected power factor.  

RE: Power factor correction???

Convert each to rectangular form and add, then change back to polar.  I get 12.9 A at 0.73 pf.

RE: Power factor correction???

  Problem can be solved in a different way than NormaGa's, which is absolutely correct. This deals whith apparent, real and reactive powers. You can add real (or active?) powers on one hand an reactive powers on the other. In your particular case, the power factor angles are:
   phi1 = acos 0.8 = 36'87º     sin phi1 = 0.6
   phi2 = acos 0.6 = 53'13º     sin phi2 = 0.8
 
 Apparent powers:
    Motor1   S1 = 240Vx8A = 1920 VA
    Motor2   S2 = 240Vx5A = 1200 VA
  Real (active) powers:
    Motor1   P1 = S1.cos(phi1) = 1920x0.8 = 1536 W
    Motor2   P2 = S2.cos(phi2) = 1200x0.6 = 720 W
                Total Real (Active)Power  = 2256 W
  Reactive powers:
    Motor1   Q1 = S1.sin(phi1) = 1920x0.6 = 1152 var
    Motor2   Q2 = S2.sin(phi2) = 1200x0.8 = 960 var
                Total Reactive Power      = 2112 var

  Total apparente power: sqrt(2256^2 + 2112^2) = 3090.3 VA
  Total current: S/V = 3090.3/240 = 12.87 A
  Total P.F. = P/S = 2256/3090.3 = 0.730

Julian

RE: Power factor correction???

agree with normGA
work out in polar-change to rectangular,add,convert back to polar
I=12.86A
pf=0.73
bear in mind normal uk supply is now quoted at 230V

RE: Power factor correction???

What happens when we are talking about 3-phase?, Does it matter if one machine has a power factor of 0.09 and another has one of 0.6, although the current  on both machines is the same as well as the voltage?

RE: Power factor correction???

To DOWNTIME: Three-phase installations can be managed as an aquivalent one-phase installation as follows:
  - Consider voltage line-to-neutral (or phase-to-phase voltage divided into square root of 3) as equivalent one-phase voltage
  - Consider one phase powers (three phase powers divided into three) as equivalente one-phase powers
  - Consider current line itself as equivalent one-phase current.
  - Power phactor is the same for three than for one-phase systems.

  Then you can deal whith these values an add currents as stated by NornaGa or steviesparkie , or add reals, and reactive powers as suggested by myself.

  If one machine has a power factor of 0.09 and another has one of 0.6, and the current on both machines is the same as well as the voltage, then the real or actual or active power
of both are:
  P1 = VxIx0.09    and     P2 = VxIx0.6  
     Let's say: P1/P2 = .09/.6 = 0.15

Julian

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