• 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)