# Overall Power Factor of three motors for transformer voltage drop calculation

Hi

Suppose I have three motors connected to a single busbar which is fed by a single transformer and suppose that the current into the busbar (i.e. the total current of the transformer secondary) is known. If I wish to calculate the transformer voltage drop the I will obviously need the load power factor. If the power factors of each of the motors is known, can the powerfactor at the transformer secondary (i.e. the overall load powerfactor) be obtained by taking the average of the three motors.

E.g. if Motor A is assumed to have PF of 0.88, Motor B PF = 0.86 & Motor C PF = 0.84, is PF at busbar 0.86?

jim hardy
Gold Member
Dearly Missed
No.
The whole is equal to the sum of its parts not the average of its parts.

Make each motor current into rectangular form, real amps +jreactive amps ,

If this is a homework problem,
teacher probably wants you to solve some simultaneous equations to arrive at the answer.
Use phasors to make it a geometry problem that you can visualize

given only total current and power factor
you can draw a right triangle whose hypotenuse is total current and other two sides are real and imaginary amps.
what three smaller right triangles will add to give that one?

I think that'll work though haven't tried it myself.
Keep us posted?

old jim

Considering the following theoretical example, and based upon what you have said, does this look about right:

Motor A: Drawing 55A, PF = 0.88
Motor B: Drawing 62A, PF = 0.79
Motor C: Drawing 87A, PF = 0.82

Motor A:

Φ = cos-1(0.88) = 28.36°

Real current component = I * PF = 55*0.88 = 48.4A
Imaginary Current component = I * Sin(Φ) = 55*Sin(28.36) = 26.13A

Motor A Current is therefore 48.40 + j26.13A

Motor B:

Φ = cos-1(0.79) = 37.81°

Real current component = I * PF = 62*0.79 = 48.98A
Imaginary Current component = I * Sin(Φ) = 62*Sin(37.81) = 38.01A

Motor B Current is therefore 48.98 + j38.01A

Motor C:

Φ = cos-1(0.82) = 34.92°

Real current component = I * PF = 87*0.82 = 71.34A
Imaginary Current component = I * Sin(Φ) = 87*Sin(34.92) = 49.80A

Motor C Current is therefore 71.34 + j49.80A

Total Current

Total current = A+B+C = 48.40+j26.13 + 48.98+j38.01 + 71.34+j49.80 = 168.72+j113.94

Total Current = √(168.722+113.942) = 203.59A

Φ = tan-1(113.94/168.72) = 34.03°

Overall PF of three motor loads = cos(34.03) = 0.829

I hop this is correct as when I calculate the total current using KCL (current into busbar = current leaving busbar etc.) I get the following;

I = 55+62+87 = 204A

which is very close to the total value obtained using real and reactive component method for each motor?

jim hardy
Gold Member
Dearly Missed
and your orderly presentation is a delight to see.
Please excuse me for not checking the arithmetic in your polar to rectangular conversions , i know you are plenty capable
and i can hardly see the numbers on my slide rule anymore.......

That your total current comes out 203.59
versus 204 from just adding them without adjusting for their slightly different phase
is such a small difference that it comes as a bit of a surprise . It would be hard to discern that little bit of difference on my 6" slide rule.

still, that they differ makes the point.
I guess this wasn't homework - surely a textbook author would have chosen currents that made the point more emphatically ?
If you are curious make one motor have a leading PF (synchronous motor can do that)
and see what it does to result

Basic rule of thumb : Add in rectangular co-ordinates, multiply in polar.

Congratulations ! You done good ! I have vicariously enjoyed your success.
and it is really heartwarming to see such orderly work.
Nice job. I bet it felt good.

old jim