The most effective method to Calculate the Suitable Capacitor Size in Farads and kVAR for Power factor Improvement (Easiest way ever)

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Think about the accompanying Examples.

Model: 1

A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 slacking. What size of Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90?

Arrangement #1 (By Simple Table Method)

Engine 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

What’s more, Rating of Capacitors associated in each Phase

1.99/3 = 0.663 kVAR

Arrangement # 2 (Classical Calculation Method)

Engine input = P = 5 kW

Unique P.F = Cosθ1 = 0.75

Last 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

Furthermore, Rating of Capacitors associated in each Phase

1.99/3 = 0.663 kVAR

Tables (Capacitor measuring in kVAr and Farads for PF redress)

The accompanying tables have been set up to streamline kVAR estimation for control factor improvement. The size of capacitor in kVAR is the kW duplicated by factor in table to improve from existing force factor to proposed control factor. Check the others Examples beneath.

Table – from 0.01 to 0.09 (Click picture to enlarge)How to Calculate the Suitable Capacitor Size in Farads and kVARTable – from 0.10 to 0.30 (Click picture to enlarge)How-to-Calculate-the-Suitable-Capacitor-Size-in-Farads-amp-kVAR-for-Power-factor-Improvement-Easiest-way-everTable – from 0.31 to 0.49 (Click picture to enlarge)Calculate the Suitable Capacitor Size in Farads and kVAR for Power factor ImprovementTable – from 0.50 to 0.74 (Click picture to grow) strategies to ascertain the best possible Size of Capacitor bank in kVAR and miniaturized scale farads for control consider redress and improvement both single stage and three stage circuitsTable – from 0.75 to 1.00 (Click picture to enlarge)calculate the best possible Size of Capacitor bank in kVAR and small scale farads for control factor rectification The entire Table – from 0.10 to 1.0 (Click picture to enlarge)How to Calculate the Suitable Capacitor Size in Farads and kVAR for Power factor Improvement (Easiest way ever)

Model 2:

An Alternator is providing a heap 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 solidarity (1)? Furthermore, what number of more kW can the alternator supply for the equivalent kVA stacking when P.F improved.

Arrangement #1 (By Simple Table Method)

Providing kW = 650 kW

From Table 1, Multiplier to improve PF from 0.65 to solidarity (1) is 1.169

Required Capacitor kVAR to improve P.F from 0.65 to solidarity (1)

Required Capacitor kVAR = kW x Table 1 Multiplier of 65 and 100

= 650kW x 1.169

= 759.85 kVAR

We realize that P.F = Cosθ = kW/kVA . . .or on the other hand

kVA = kW/Cosθ

= 650/0.65 = 1000 kVA

At the point when Power Factor is raised to solidarity (1)

No of kW = kVA x Cosθ

= 1000 x 1 = 1000kW

Thus expanded Power provided by Alternator

1000kW – 650kW = 350kW

Arrangement # 2 (Classical Calculation Method)

Providing kW = 650 kW

Unique P.F = Cosθ1 = 0.65

Last 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

How to Calculate the Required Capacitor bank an incentive in both kVAR and Farads?

(Step by step instructions to Convert Farads into kVAR and Vice Versa)

Model: 3

A Single stage 400V, 50Hz, engine takes a stockpile current of 50A at a P.F (Power factor) of 0.6. The engine control factor must be improved to 0.9 by interfacing a capacitor in parallel with it. Compute the necessary limit of Capacitor in both kVAR and Farads.

Arrangement.:

(1) To locate the necessary limit of Capacitance in kVAR to improve P.F from 0.6 to 0.9 (Two Methods)

Arrangement #1 (By Simple Table Method)

Engine 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

Arrangement # 2 (Classical Calculation Method)

Engine Input = P = V x I x Cosθ

= 400V x 50A x 0.6

= 12kW

Genuine 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 locate the necessary limit of Capacitance in Faradsto improve P.F from 0.6 to 0.9 (Two Methods)

Arrangement #1 (Using a Simple Formula)

We have just determined the necessary Capacity of Capacitor in kVAR, so we can without much of a stretch believer it into Farads by utilizing this straightforward equation

Required Capacity of Capacitor in Farads/Microfarads

C = kVAR/(2 π f V2) in microfarad

Placing the Values in the above equation

= (10.188kVAR)/(2 x π x 50 x 4002)

= 2.0268 x 10-4

= 202.7 x 10-6

= 202.7μF

Arrangement # 2 (Simple Calculation Method)

kVAR = 10.188 … (I)

We realize that;

IC = V/XC

While 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

Furthermore,

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

= 400 x 125663.7 x C

IC = 50265.48 x C … (ii)

Likening Equation (I) and (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

Model 4

What estimation of Capacitance must be associated in parallel with a heap drawing 1kW at 70% slacking power factor from a 208V, 60Hz Source so as to raise the general power factor to 91%.

Arrangement:

You can utilize either Table technique or Simple Calculation strategy to locate the necessary estimation of Capacitance in Farads or kVAR to improve Power factor from 0.71 to 0.97. So I utilized table strategy for this situation.

P = 1000W

Genuine Power factor = Cosθ1 = 0.71

Wanted 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

What’s more,

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)

Great to Know:

Significant recipes which is utilized for Power factor improvement computation just as utilized in the above count

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)

Clear Power in VA

kVA= √(kW2+ kVAR2)

kVA = kW/Cosθ

Receptive 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 and 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)

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