Power Factor Correction 2

Question

Case1: True power = 300 W

Apparent power = 1817 VA

p.f. = 0.165

Case2: True power = 450 W

Apparent power = 1886

p.f. = 0.230

Case3: True power = 500 W

Apparent power = 1817

p.f. = 0.275

Required: Size of the capacitor that will make the p.f = 0.95

Please show complete solutions, thank you.

Answer 1

You need to learn how to do this yourself.  Power factor correction is rather easy if you draw right triangles and understand the concepts.

Look at this as a right triangle where the base is true power, the hypotenuse is apparent power, and the height is reactive power.

The angle of the triangle is Ф.  The cosФ = the power factor (PF)

Adding capacitors has the effect of shortening the height of the right triangle, decreasing the angle Ф.  The base of the triangle (watts) remains constant during your calculations.  Only the reactive and apparent values will change as you add capacitors.

The arccos(0.95) = 18.2°.

Draw a triangle with the true and apparant power values given, then compute how much you need to reduce the height to get the angle down to 18.2°.

The link below provides a good diagram of the power triangle.