How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement

Post by: Omega delta electric




 How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement (Easiest way ever)

 Hi there! With a very important tutorial.. I hope you will find it very useful because I have already spent two days to prepare this article. I think all of those who have sent messages and mails about the topic will never ask again if they follow these simple methods to calculate the proper Size of Capacitor bank in kVAR and micro-farads for power factor correction and improvement in both single phase and three phase circuits. I think it’s too much..
 Now let’s begin…


Consider the following Examples.


 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 #1 (By Simple Table Method)


 Motor Input = 5kW
 From Table, Multiplier to improve PF from 0.75 to 0.90 is .398
 Required Capacitor kVAR to improve P.F from 0.75 to 0.90
 Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90
 = 5kW x .398
 = 1.99 kVAR
 And Rating of Capacitors connected in each Phase
 1.99/3 = 0.663 kVAR


 Solution # 2 (Classical Calculation Method)


 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 kVARHow to Calculate the Required Capacitor bank value in both kVAR and Farads?
 (How to Convert Farads into kVAR and Vice Versa)



 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.:


 (1) To find the required capacity of Capacitance in kVAR to improve P.F from 0.6 to 0.9 (Two Methods)


 Solution #1 (By Simple Table Method)


 Motor Input = P = V x I x Cosθ
= 400V x 50A x 0.6
 = 12kW


 From Table, Multiplier to improve PF from 0.60 to 0.90 is 0.849
 Required Capacitor kVAR to improve P.F from 0.60 to 0.90
 Required Capacitor kVAR = kW x Table Multiplier of 0.60 and 0.90
 = 12kW x 0.849
 = 10.188 kVAR


 Solution # 2 (Classical Calculation Method)


 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


 (2) To find the required capacity of Capacitance in Faradsto improve P.F from 0.6 to 0.9 (Two Methods)


 Solution #1 (Using a Simple Formula)


 We have already calculated the required Capacity of Capacitor in kVAR, so we can easily convert it into Farads by using this simple formula
 Required Capacity of Capacitor in Farads/Microfarads
 C = kVAR / (2 π f V2) in microfarad


 Putting the Values in the above formula
 = (10.188kVAR) / (2 x π x 50 x 4002)
 = 2.0268 x 10-4
 = 202.7 x 10-6
 = 202.7μF


 Solution # 2 (Simple Calculation Method)


 kVAR = 10.188 … (i)


 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 1kW at 70% lagging power factor from a 208V, 60Hz Source in order to raise the overall power factor to 91%.


 Solution:


 You can use either Table method or Simple Calculation method to find the required value of Capacitance in Farads or kVAR to improve Power factor from 0.71 to 0.97. So I used table method in this case.
 P = 1000W
 Actual Power factor = Cosθ1 = 0.71
 Desired Power factor = Cosθ2 = 0.97
 From Table, Multiplier to improve PF from 0.71 to 0.97 is 0.783
 Required Capacitor kVAR to improve P.F from 0.71 to 0.97
 Required Capacitor kVAR = kW x Table Multiplier of 0.71 and 0.97
 = 1kW x 0.783
 =783 VAR (required Capacitance Value in kVAR)
 Current in the Capacitor =


 IC = QC / V
 = 783 / 208
 = 3.76A


 And
 XC = V / IC
 = 208 / 3.76 = 55.25Ω
 C = 1/ (2 π f XC)
 C = 1 (2 π x 60 x 55.25)
 C = 48 μF (required Capacitance Value in Farads)


 Good to Know:
 Important formulas which is used for Power factor improvement calculation as well as used in the above calculation


 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)