Contoh kiraan power factor

Dalam contoh kiraan ini, saya cuma dapatkan dari tempat lain, dan ditambah sedikit sebanyak untuk mudahkan pemahaman.

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 kVAR

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

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;

I_C = V/ X_C

Whereas X_C = 1 / 2 π F C

I_C = V / (1 / 2 π F C)

I_C = V 2 F C

= (400) x 2π x (50) x C

I_C = 125663.7 x C

And,

kVAR = (V x I_C) / 1000 … [kVAR =( V x I)/ 1000 ]

= 400 x 125663.7 x C

I_C = 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 =

I_C = Q_C / V

= 783 / 208

= 3.76A

And

X_C = V / I_C

= 208 / 3.76 = 55.25Ω

C = 1/ (2 π f X_C)

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)

X_C = 1/ (2 π f C) … (X_C = Capacitive reactance)

I_C = V/ X_C  … (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)