t2ωsinθsin2

IVt2ωcos1θcos

2

IVθ)tω(I *tωtp mmmm

m SinCosVm

Average Power:

Power Factor:Power factor is often stated as percentage, e.g., 90% lagging (i.e., current lags voltage, inductive load) 60% leading (i.e., current leads voltage, capacitive load)

Reactive Power: The last term in power formula is the power flowing back and forth

between the source and the energy-storage elements. Reactive power is its peak power.

Apparent Power:

W θcosIVθcos2

IVP rmsrms

mm

angle powerthe called is θθθ ,θcosPF iv

Reactive) Amperes (Volt VAR θsinrms

Irms

Vθsin2

mImVQ

2rmsrms

22rmsrms

22rmsrms

22

rmsrms

IVθsinIVθcosIVQP

Ampere)-(Volt VA IVS

Note: 5kW load is different from 5kVA load.

Power FactorSuppose voltage & current phasors are:

Power dissipated in load Z is:

Average power: is the power factor

a lagging power factor (normal case: Current lags Voltage) a leading power factor (rare case: Current leads Voltage)

⇔𝑣 (𝑡 )=¿𝑉∨cos (𝜔𝑡+𝜃𝑉 )⇔𝑖 (𝑡 )=¿ 𝐼∨cos (𝜔𝑡+𝜃 𝐼)

¿12|𝑉||𝐼|cos (𝜃𝑉 −𝜃 𝐼 )+

12|𝑉||𝐼|cos ( 2𝜔𝑡+𝜃𝑉 +𝜃 𝐼 )

Power Factor CorrectionV = 230. Motor modelled as 5||7j Ω.

𝐼=𝑉

𝑅+𝑍𝐿¿ 46 − j 32.9 A¿𝟓𝟔 .𝟓∠−𝟑𝟔° 𝑨

𝑆=𝑉 𝐼∗=10 .6+ j7 .6 kVA kVA ⇒𝑐𝑜𝑠∅=𝑃

|𝑆|=𝑐𝑜𝑠36°=𝟎 .𝟖𝟏

Add a parallel capacitor of 300 μF:

𝑍𝐶=1

𝑗 𝜔𝐶=− 10 . 6 𝑗 Ω⇒𝐼𝐶=21. 7 𝑗 𝐴

𝐼=46 − 𝑗11 . 2 𝐴=𝟒𝟕∠−𝟏𝟒° 𝑨𝑆𝐶=𝑉 𝐼𝐶

∗=− 𝑗5𝑘𝑉𝐴𝑆=𝑉 𝐼∗=10 .6+ j2 .6 kVA=¿𝑐𝑜𝑠∅=

𝑃|𝑆|

=𝑐𝑜𝑠14°=𝟎 .𝟗𝟕

Average power to motor, P, is 10.6 kW in both cases. reduce from 56.5 to 47 A (-16%) Lower losses Effect of C: reduce VARs from 7.6 to 2.6 kVAR,

kVA

power factor from 0.81 to 0.97

Power factor correction – induction motor illustrated

Power Factor Correction – induction motor

• It is recommended that motors are run at their rated power in order to achieve the best power factor cosϕ

Changes in power factor for an induction motor as a function of motor loading.

• A 510 kVA synchronous generator for phase-shifting purposes.

• Today these applications for synchronous motors is rare.

• Sometimes they may still being used in older power stations with a rated power of more than 1000kVA.

Compensation of Reactive Power by Rotational Phase-Shifting Machines

P = AVERAGE POWER

Q = REACTIVE POWER

• Useful power – also known as ACTIVE POWER

• Converted to other useful form of energy – heat, light, sound, etc

• Power charged

• Power that is being transferred back and forth between load and source

• Associated with L or C – energy storage element – no losses

• Is not charged

• Inductive load: Q positive (current lags voltage, lagging PF ), Capacitive load: Q negative (current lead voltage, leading PF)

Active power Vs Reactive power

•Harmonics distortion in the grid have increased rapidly in recent years due primarily to the increasing application of non-linear semiconductor devices:Industrial: Rectifiers DC, arc furnaces, welding machines, motor drives, etc.Lighting: Fluorescent, Neon, Sodium lamps, Metal halide lamps, Mercury-

vapor lamps, etc. Consumer Electronics: PC, laptop charger, monitors, phase control

dimmers, heater, etc.

Harmonics Distortion - Introduction

Harmonics Distortion – Ex1: Industrial DC PS•3-phase rectifier with smoothing capacitor:

Better current wave form with 12-pulse rectifierLots of harmonic with current waveform

Harmonics Distortion – Ex2: LED Lighting, CFLsNote: new & more expensive light has internal PFC – reduce harmonic distortion.

•Single-phase rectifier with smoothing capacitor (R load)

Harmonics Distortion – Ex3: Consumer Electronics

Surge current may even create distortion on voltage if voltage source has high internal resistance.

• Any periodic waveform can be represented as a sum of: A sinusoidal term at the fundamental frequency Other sinusoidal terms (harmonics) having frequencies that

are multiples of the fundamental frequency.

Understanding Harmonics Distortion

a) The line rms current harmonics do not deliver any real power in Watts to the load, resulting in inefficient use of equipment capacity (i.e. low power factor).

b) Harmonics will increase conductor loss and iron loss in transformers, decreasing transmission efficiency and causing thermal problems.

c) The odd harmonics are extremely harmful to a three-phase system, causing overload of the unprotected neutral conductor.

d) Oscillation in power system should be absolutely prevented in order to avoid endangering the stability of system operation.

e) High peak harmonic currents may cause automatic relay protection devices to mis-trigger.f) Harmonics could cause other problems such as electromagnetic interference to interrupt

communication, degrading reliability of electrical equipment, increasing product defective ratio, insulation failure, audible noise, etc.

What is bad about high order harmonics?

• Defined as the ratio of real power (P) consumed by load to apparent power (S), which produces a figure from 0 to 1 that indicates the degree of distortion and phase shift in the current wave form.

• Intuitively, the overall PF is the product of the displacement PF and the distortion PF, where displacement PF consider only the fundamental and distortion PF consider higher order harmonics components of current wave form

Note: General definition of PF proposed by IEEE Std 1459-2000 has complicated mathematics formula, applied for any type of signal. All grid AC power supply has very low internal resistance, non-linear load has almost no effect on the voltage wave form. We can assume voltage wave form has no distortion, pure sin wave. The assumption allow us to simplify the math & think intuitively about PF.

Power Factor Generalized

•Displacement PF is the phase shift between voltage & fundamental component of current wave form:

•Distortion PF is calculated based on THD (total harmonics distortion)

Power factor – continue

𝑻𝑯𝑫=√∑𝒕=𝟐

∞

𝑰 𝒕𝟐

𝑰𝟏=√ 𝑰 𝒓𝒎𝒔

𝟐 − 𝑰𝟏𝒓𝒎𝒔𝟐

𝑰𝟏𝒓𝒎𝒔𝟐

𝑷𝑭=𝑷𝑺

=𝑷𝑭𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡∗𝑷𝑭 𝐷𝑖𝑠𝑡𝑜𝑟𝑡𝑖𝑜𝑛

•Both phase shift and distortion can reduce Power Factor:

Power Factor – continue

Passive PFC

Active PFC

PFC comparision

Input characteristic of PC power supplies with different PFC type: None, Passive, Active

Input line harmonics compare to IEC61000-3-2

For exams!

Practice Problems

A load operates at 20 kW, 0.8 pf lagging. The load voltage is 220∠0° V rms at 60 Hz. The impedance of the line is 0.09+j0.3 Ohm. Determine the voltage and power factor at the input to the line.

SS

If the power flow had actually been from network B to network A, the resultant signs on and would have been negative.