The Calculation of kVAr for 3-Phase Capacitor in Power Supply

What is the kVAr of a 3-phase capacitor needed to attain a PF = 0.95 in the lines with the motors?

The kVAr of a 3-phase capacitor needed to attain a PF = 0.95 in the lines with the motors is 10.85 kVAr.

Calculation Process:

Step 1: Finding the apparent power of the motors

We can find the apparent power of motor 1 as:

S1 = P1 / PF1

S1 = (40 hp * 0.746 kW/hp) / 0.85

S1 = 33.301 kW

Similarly, we can find the apparent power of motor 2 as:

S2 = P2 / PF2

S2 = (10 hp * 0.746 kW/hp) / 0.85

S2 = 8.325 kW

The total actual power P is:

P = P1 + P2

P = (40 hp * 0.746 kW/hp) + (10 hp * 0.746 kW/hp)

P = 34.96 kW

Step 2: Finding the reactive power of the motors

We can find the reactive power of motor 1 as:

Q1 = √(S1² - P1²)

Q1 = √(33.301 kW² - (40 hp * 0.746 kW/hp)²)

Q1 = 22.526 kVAR

Similarly, we can find the reactive power of motor 2 as:

Q2 = √(8.325 kW² - (10 hp * 0.746 kW/hp)²)

Q2 = 6.714 kVAR

Step 3: Finding the reactive power of the capacitor needed

The difference between Q1 and Q2 is the reactive power of the capacitor needed.

Qc = Q1 - Q2

Qc = 22.526 kVAR - 6.714 kVAR

Qc = 15.812 kVAR

Step 4: Using the formula Qc = P (tanφ1 - tanφ2)

We set tanφ2 = tan(18.19°) = 0.3249 to achieve PF = 0.95

Calculating for tanφ1 and then kVAr of the 3-phase capacitor needed:

Qc = P (tanφ1 - tanφ2)

Qc = 34.96 kW (0.7061 - 0.3249)

Qc = 10.85 kVAr

Therefore, the kVAr of a 3-phase capacitor needed to attain a PF = 0.95 in the lines with the motors is 10.85 kVAr.