Power factor (PF) is a ratio of real to reactive power at any electrical system. Lower PF means inefficient systems. Therefore, Electrical systems use capacitors to improve the power factor.
Today you’ll learn to choose a capacitor for improving the power factor of any system.
Let’s consider a system 400 V, 50 Hz system which power three loads L1, L2, and L3. The provided parameters are:
For L1 – 15 kW, θ = 60°,
For L2 – 30 kVAR, PF = 70%
For L3 – 5 HP, η = 80% and PF = 56%
The information provided in above case can be used to extract the information for drawing our power factor triangles.
Let’s understand the calculations. You can skip the three paragraphs below if you already mastered power factor triangle.
For the first load (L1), we used,
Tan 𝜃 = Q/P to calculate the reactive power.
For the second load, we first calculated the angle from PF by using InverseCos 𝜃 = PF and then applied Tan 𝜃 = Q/P for calculating the real power.
For the third load, we first converted HP to kW by using formula 1 HP = 745.6 kW and then used the efficiency formula to obtain actual power. Then from Tan 𝜃 = Q/P we obtained the kVAR.
Finally, we can add all triangles.
The kW and kVARs are simply added. Whereas the angle and reactive powers are calculated by using Tan 𝜃 = Perp/Base and basic geometry formula for the hypotenuse.
In our present case, the power factor is 0.58. Whereas the current drawn from the system is:
I = 154.60/400 V = 386.5 A.
Now if we improve the power factor, all other parameters except real power will change. Let’s assume that we want to improve our power factor to 0.9. The new triangle now is:
Size of the capacitor is 124.91 – 44.11 = 80.8 kVAR.
Using this capacitor in parallel to the system will reduce the current I from 386.5 A to 253 A.
Let’s watch another example in this video: