Example: 1
A 3 phase, 5 kW induction motor has a p.f (power factor) of 0.75 lagging. What size of capacitor in kvar is required to improve the p.f (power factor) to 0.90?
Solution
Motor input = P = 5 kW
Original p.f = cosθ1 = 0.75
Final p.f = cosθ2 = 0.90
θ1 = cos-1 = (0.75) = 41°.41; tan θ1 = tan (41°.41) = 0.8819
θ2 = cos-1 = (0.90) = 25°.84; tan θ2 = tan (25°.50) = 0.4843
Required Capacitor kVAR to improve p.f from 0.75 to 0.90
Required Capacitor kVAR = P (tan θ1 – tan θ2)
= 5kW (0.8819 – 0.4843)
= 1.99 kVAR
And rating of capacitors connected in each phase
1.99/3 = 0.663 kVAR
Example 2:
An Alternator is supplying a load of 650 kW at a p.f. (power factor) of 0.65. what size of capacitor in kVAR is required to raise the p.f (power factor) to unity. And how many more kW can the alternator supply for the same kVA loading when p.f. improved.
Solution
Supplying kW = 650 kW
Original p.f. = cosθ1 = 0.65
Final p.f. = cosθ2 = 1
θ1 = cos-1 = (0.65) = 49°.45; tan θ1 = tan (41°.24) = 1.169
θ2 = cos-1 = (1) = 0°; tan θ2 = tan (0°) = 0
Required capacitor kVAR to improve p.f from 0.75 to 0.90
Required capacitor kVAR = P (tan θ1 – tan θ2)
= 650kW (1.169– 0)
= 759.85 kVAR
Now, kVAi = kW/cosθi = 650 / 0.65 = 1000 kVA
kVAf = kVAi × cosθf = 1000 × 1 = 1000 kW
ΔkW = 1000 - 650 = 350 kW.
Example: 3
A Single phase 400V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of 0.6. The motor power factor has to be improved to 0.9 by connecting a capacitor in parallel with it. Calculate the required capacity of Capacitor in both kVAR and Farads.
Solution
Motor Input = P = V x I x cosθ
= 400V x 50A x 0.6
= 12kW
Actual p.f = cosθ1 = 0.6
Required p.f = cosθ2 = 0.90
θ1 = cos-1 = (0.60) = 53°.13; tan θ1 = tan (53°.13) = 1.3333
θ2 = cos-1 = (0.90) = 25°.84; tan θ2 = tan (25°.50) = 0.4843
Required capacitor kVAR to improve p.f from 0.60 to 0.90
Required capacitor kVAR = P (tan θ1 – tan θ2)
= 5kW (1.3333– 0.4843)
= 10.188 kVAR
We know that;
IC = V/ XC
Whereas XC = 1 / 2 π F C
IC = V / (1 / 2 π F C)
IC = V 2 F C
= (400) x 2π x (50) x C
IC = 125663.7 x C
And, kVAR = (V x IC) / 1000 ..… [kVAR =( V x I)/ 1000 ]
= 400 x 125663.7 x C
IC = 50265.48 x C … (ii)
Equating Equation (i) & (ii), we get,
50265.48 x C = 10.188C
C = 10.188 / 50265.48
C = 2.0268 x 10-4
C = 202.7 x 10-6
C = 202.7μF
Example 4
What value of capacitance must be connected in parallel with a load drawing 1 kW at 70% lagging power factor from a 208V, 60Hz source in order to raise the overall power factor to 90%.
Solution:
P = 1000W
Actual power factor = cosθ1 = 0.70
Desired power factor = cosθ2 = 0.90
θ1 = cos-1 = (0.70) = 45.57°; tan θ1 = tan (45.57°) = 1.0202
θ2 = cos-1 = (0.90) = 25.84°; tan θ2 = tan (25.84°) = 0.4843
Required capacitor kVAR to improve p.f from 0.70 to 0.90
Required capacitor kVAR = P (tan θ1 – tan θ2)
= 1 kW (1.0202– 0.4843)
= 0.5358 kVAR
C = VAR / (2 π f V2)
C = 535.8 / (2 π x 60 x 2082)
C = 32.87 μF (required Capacitance Value in Farads)
Important formulas
Power in Watts
kW = kVA x cosθ
kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power)
kW = √ ( kVA2– kVAR2)
kW = P = VI cosθ … (Single Phase)
kW = P =√3x V x I cosθ … (Three Phase)
Apparent Power in VA
kVA= √(kW2+ kVAR2)
kVA = kW/ cosθ
Reactive Power in VA
kVAR= √(kVA2– kW2)
kVAR = C x (2 π f V2)
Power factor (from 0.1 to 1)
Power Factor = cosθ = P / V I … (Single Phase)
Power Factor = cosθ = P / (√3x V x I) … (Three Phase)
Power Factor = cosθ = kW / kVA … (Both Single Phase & Three Phase)
Power Factor = cosθ = R/Z … (Resistance / Impedance)
XC = 1/ (2 π f C) … (XC = Capacitive reactance)
IC = V/ XC … (I = V / R)
Required Capacity of Capacitor in Farads/Microfarads
C = kVAR / (2 π f V2) in microfarad
Required Capacity of Capacitor in kVAR
kVAR = C x (2 π f V2)