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.