• 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)