Power factor correction???

[shocked]

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.

 

[NormGA]

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

 

[230842]

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

 

[steviesparkie]

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

 

[DOWNTIME]

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?

 

[230842]

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