Electricity - Power Factor / True, Apparent & Reactive Power - Educate ∞ Knowledge, Wisdom, Intelligence, Business, Commerce, Science, Technology, AI, ML, Software

Note : True power is also called as Real power or Active power, and is used interchangeably most of the time. Power fact...

Note: True power is also called as Real power or Active power, and is used interchangeably most of the time.

$PowerFactor={\frac{TruePower(kW)}{ApparentPower(kVA)}=Cos\theta}$

$TruePower(kW)={ApparentPower(kVA)\times{Cos\theta}}$

$TruePower(kW)={PowerFactor\times{ApparentPower(kVA)}}$

$ApparentPower(kVA)=\sqrt{TruePower(kW)^2+ReactivePower(kVAr)^2}$

$ApparentPower(kVA)={TruePower(kW)\times{Sin\theta}}$

$ReactivePower(kVAr)={ApparentPower(kVA)\times{Sin\theta}}$

$ReactivePower(kVAr)=\sqrt{{ApparentPower(kVA)^2}-{TruePower(kW)^2}}$

Power factor explained | Active Reactive Apparent Power correction

$V\underset{rms}{}=\frac{V\underset{peak}{}}{\sqrt{2}}$

$I\underset{rms}{}=\frac{I\underset{peak}{}}{\sqrt{2}}$

Real Power (P)

$P={V\underset{rms}{}}\times{I\underset{rms}{}}$

$P={I^2\times{R}}$

$P={\frac{V^2}{R}}$

Reactive Power (Q)

$Q={{I^2}\times{X}}$

$Q={\frac{V^2}{X}}$

Apparent Power (S)

$S={{V}\times{I}}$

$S={{I^2}\times{Z}}$

$S={\frac{V^2}{Z}}$

$S=\sqrt{{P^2}\times{Q^2}}}$

Where,

$V$ is Voltage/Potential Difference, in Volt
$I$ is Current, in Ampere
$P$ is Active/Real/True Power, in Watt
$R$ is Resistance, in Ohm
$Q$ is Reactive-Power, in VAr (Watt)
$X$ is Reactance, in Ohm
$S$ is Apparent-Power, in VA (Watt)
$Z$ is Impedence, in Ohm
$F\underset{p}{}$ is Power Factor, in Percentage

Example Calculations

Finding Power factor with given values of Voltage as 120V, Frequency as 60Hz, Resistance as 60 $\Omega$ , and Inductance as 160mH.

1) Power Factor (Fp) can be derived from equation,

$F\underset{p}{}={\frac{P}{S}}$

2.a) Active (Real/True) power (P) can be derived from equation,

$P={{I^2}\times{R}}$

2.b) Apparent power (S) can be derived from equation,

$S={{I^2}\times{Z}}$

3.a) Current (I) can be derived from equation,

$I={\frac{V}{Z}}$

3.b) Impedence (Z) can be derived from equation,

$Z={\sqrt{{R^2}+(X\underset{L}{}-X\underset{C}{})^2}$

4.a) Capacitive reactance (Xc) can be derived from equation,

$X\underset{C}{}={\frac{1}{2\pi{f}{C}}$

4.b) Inductive reactance (Xl) can be derived from equation,

$X\underset{L}{}={2\pi{f}{L}}$

Derivations start from bottoms up, with step 4(a & b), progressing to the top to finally at step 1.

4.b) Inductive reactance (Xl) for the given values of 60Hz, & 160mH is,
$X\underset{L}{}={2\pi{f}{L}}$
= 2(3.14)(60Hz)(0.160H)
= 60.139 $\Omega$

4.a) as there is no Capacitance present in the circuit the Capacitive Impedance value is zero,
$X\underset{C}{}={\frac{1}{2\pi{f}{C}}$
= 0 $\Omega$

3.b) Impedence (Z) for the given & derived values of 60 & 60.319 Ohms is,

$Z={\sqrt{{R^2}+(X\underset{L}{}-X\underset{C}{})^2}$
= $\sqrt{60\Omega^2+(60.319\Omega-0)^2}$
= 85.078 $\Omega$

3.a) Current (I) for the given & derived values of 60 Ohms & 60.319 Ohms is,

$I={\frac{V}{Z}}$
= $\frac{120V}{85.078\Omega}$
= 1.410A

2.b) Apparent power (S) for the given & derived values of 1.410A & 85.078 Ohms is,

$S={{I^2}\times{Z}}$
= ${(1.410A)^2}\times{85.078\Omega}$
= 169.143VA

2.a) Active power (S) for the given & derived values of 1.410A & 60 Ohms is,

$P={{I^2}\times{R}}$
= $(1.410A)^2\times{60\Omega}$
= 119.286W

1) Power Factor (Fp) can be derived from equation,

$F\underset{p}{}={\frac{P}{S}}$
     = ${\frac{119.286W}{169.143VA}}$
     = 0.705
     = 70.5%

The calculated Power factor is 70.5%, for the given values of 120V, 60Hz, 60 $\Omega$ , and 160mH.

Solution to the Problem: To improve the Power factor, we need to introduce a Capacitance to counteract the Reactive power (Inductive) of the circuit.

1) Inductive power (Xl) can be derived from equation,

$Q\underset{L}{}={I^2}\times{X\underset{L}{}}$
= $(1.410A^2)\times(60.319\Omega)$
= 119.920 VAR

2) Reactive power (Qc) can be derived from equation,

$Q\underset{C}{}=\frac{V^2}{X\underset{C}{}}$
$X\underset{C}{}=\frac{V^2}{Q\underset{C}{}}$
= $\frac{120V^2}{119.920VAR}$
= 120.080 $\Omega$

3) Capacitive Reactance (Xc) can be derived from equation,

$X\underset{C}{}=\frac{1}{2\pi{f}{C}}$
$C=\frac{1}{1\pi{f}X\underset{C}{}$
= $\frac{1}{2\pi(60Hz)(12.080\Omega)}$
= 22.090 $\mu{F}$

So, to correct & improve the Power factor, we need to connect 22.09uF Capacitance (Capacitor bank) in Parallel to the above given example circuit.

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Electricity - Power Factor / True, Apparent & Reactive Power

Note : True power is also called as Real power or Active power, and is used interchangeably most of the time. Power fact...