Topic 1 AC power and power factor

Bài giảng ngành Điện: Giải các bài toán trong mạch điện điện thế thấp một pha và ba pha (Phần B) gồm các chủ đề sau: Solve problems in single and threephase low voltage circuits Part B Content Topic 1: AC power and power factor Topic 23: Multiphase systems Topic 4: Threephase STAR connection Topic 5: Threephase, Fourwire systems Topic 6: Threephase DELTA connection Topic 7 Interconnected Star and Delta Devices Topic 8 Energy and Power Requirements of AC Systems Topic 9: Harmonics

Solve problems in single and phase low voltage circuits

three-Topic 1: AC Power and Power Factor Part B

Alternating Current Principles - Power

Content

AC Power

 In AC circuits

 Resistive, Inductive, and/or Capacitive components

 Create different phase relationships that…

AC Power: Resistive Circuit

VR

VSƒ

IR

R

IV

AC Power: Resistive Circuit

 The power delivered to the circuit is continually pulsing, but

always positive.

 This means that energy is being converted from electrical energy into another form of energy eg heat, light, motion etc.

 Since this energy is being converted, it may be said that the

energy is CONSUMED by the circuit.

 This consumed energy is a type of power termed TRUE POWER (or sometimes active power).

 Symbol “P”, measured in WATTS (W).

AC Power: Resistive Circuit

TRUE POWER

P R = V R I R

Purely resistive circuits only

Where:

PR is the TRUE Power in Watts

VR is the Voltage across the resistive component in Volts

IR is the Current through the resistive component in Amps

Watts (W)

AC Power: Inductive Circuit

AC Power: Inductive Circuit

 Power is delivered to the circuit (positive pulse) and returned by the circuit (negative pulse).

 This means that electrical energy is NOT being consumed by the circuit, but simply OSCILLATES between the inductive

component and the supply.

 Since NO power is consumed, then the true power is zero, but there is still power oscillating in the circuit.

 This oscillating power is termed REACTIVE POWER.

 Symbol “Q”, measured in Volt-Amps-Reactive (VAR’s).

 Due to the inductive component, this power is referred to as a LAGGING reactive power.

AC Power: Inductive Circuit

AC Power: Capacitive Circuit

VSƒ

IC

IV

AC Power: Capacitive Circuit

out-of-phase with the power flowing in an inductive

referred to as a LEADING reactive power.

AC Power: Capacitive Circuit

REACTIVE POWER

Q C = V C I C

Purely capacitive circuits only

Where:

QC is the REACTIVE Power in VAR’s

VC is the Voltage across the capacitive component in Volts

IC is the Current through the capacitive component in Amps

Volt-Amps-Reactive

(VAR’s)

AC Power

 The total overall power flowing in an AC circuit is a

combination of TRUE Power and REACTIVE Power

 This total overall power flowing is called APPARENT POWER

 Symbol S, measured in Volt-Amps (VA)

AC Power: The POWER Triangle

P

True Power

Q

Reactive Power

S

Apparent Power

Ø

Phase Angle

 The power oscillating (stored then returned) in the circuit

 Measured in Volt-Amps-Reactive (VAR’s)

 Apparent Power (S)

 The total power flowing in the circuit

 Measured in Volt-Amps (VA)

 Phase Angle (Ø)

 This angle is the same as the phase angle between VS and IS

P True Power

Q Reactive Power

S Apparent Power

Ø

AC Power: Beer Analogy

Real Beer

Some froth

Real Beer

Lots

of froth

Power Factor

 Key Points

 Always a number between zero and one

 a pƒ of ONE is often referred to as UNITY power factor;

 For a given circuit, the pƒ may be LEADING or LAGGING, depending on the load

 in most circumstances, the pƒ will be lagging due to

inductive loads.

Power Factor Exercise

Motor 1 Motor 2 Motor 3 Rating: 5kW 5kW 5kW

Country: Germany Taiwan China

50 A

109 A25mm2

125 A

Comparison of Different Power Factors

 Best power factor

 Pƒ = 1

 Ø = 00E

QS

Comparison of Different Power Factors

Ø

Comparison of Different Power Factors

Comparison of Different Power Factors

 Worst case power

factor

 Pƒ = 0

 Ø = 900E

PQS

Power Factor

 A high power factor is:

 Approaching a value of 1 (eg 0.9)

 Represents a small phase angle (eg 25o)

 Requires a comparatively low supply current to operate the circuit.

 Represents a mostly RESISTIVE circuit

 A low power factor is:

 Approaching a value of zero (eg 0.2)

 Represents a large phase angle (eg 78o)

 Requires a comparatively high supply current to operate the circuit

 Represents a mostly REACTIVE circuit

AC Power

 Power Triangle Practical

AC Power

Effects of Low Power Factor

Effects of Low Power Factor

 The lower the pf, the higher the current required to

supply the same true power.

 For the electrical contractor, this higher current has the following implications

 Larger conductor sizes

 Higher rated circuit protection devices (fuses and circuit breakers)

 Higher rated switchgear

Effects of Low Power Factor

 For the energy distributor, the higher current results in:

 Higher losses in transmission conductors (more heating and higher voltage drops)

 Higher rated equipment for generation and transmission

 Larger supply transformers

 Lower dollar return for energy delivered (can only charge for kW, not kVA)

Power Factor

Supply Authority Requirements

Supply Authority Requirements

 No installation Pf shall be less than 0.8 lagging under

normal load conditions

 No installation shall have a leading power factor

 Lighting loads in excess of 240W total rating shall be

corrected to 0.8 lag pf.

Power Factor Correction

 A low power factor is generally due to lagging

REACTIVE POWER (inductive loads).

 By introducing leading REACTIVE POWER (eg capacitive loads) into a circuit or installation, the lagging power factor can be improved (phase angle will

decrease)

 Note: Improvement to UNITY pf is NOT cost effective.

Power Factor Correction

QLagging

QResultant

QLeading

Power Factor Correction

http://www.galco.com/circuit/images/Reactive.gif

Power Factor Correction

http://www.nokiancapacitors.com/images/03%20-%20products/low%20voltage/AutDetCapBank.jpg

Methods of Correcting Power Factor

 Short-term solution

 Ensuring that all motors and transformers are correctly loaded

 These devices produce their best (stated) pf when fully loaded

 Equivalent to increasing the true power consumed by the installation, thus reducing the phase angle (better pf)

Methods of Correcting Power Factor

 Expensive solution – fully

Methods of Correcting Power Factor

 Long-term solution – fixed correction value

 Adding capacitance to an installation (that is inductive)

 Capacitor banks or pf-corrected light fittings add leading reactive

power to an installation

Power Factor Exercises

Single Loads

Exercises: Single loads

 1 For a load of 2.1kW at pf of 0.75 lag connected to a 230V, 50Hz supply, determine the following:

 Iload

 Øload (phase angle)

 Draw a phasor diagram to scale showing VS, ILoad, and Øload

Exercises: Single loads

 1 For a load of 2.1kW at pf of 0.75 lag connected to a 230V, 50Hz supply, determine the following:

 Step 1: Find S

 S = P/pf = 2100/0.75 = 2800 VA

 Step 2: Find Iload

 Iload = S/V = 2800/230 = 12.2A

 Step 3: Find Øload

 Øload = cos-1(pf) = cos-1(0.75) = 41.40

 Phasor diagram next page

ILoad = 12.2A

VS = 230V ReferenceΦ=41.4o

Exercises: Single loads

 2 For a load of 750W at pf of 0.36 lag connected to a 230V, 50Hz supply, determine the following:

 Iload

 Øload (phase angle)

 Draw a phasor diagram to scale showing VS, ILoad, and Øload

VS = 230V ReferenceΦ=68.9o

Power Factor Exercises

Power Factor Correction

Exercises: Single loads

 Review:

 Neither of the previous two circuits complied with pf requirements.

 As such, they should undergo pf correction to alter them to comply with minimum requirements

 Energex specifies a minimum pf of 0.8 lag

 A phase angle for supply current of not less than 36.90E

 This can be achieved by using capacitors connected into parallel.

 Adding capacitance creates a Capacitor Current (IC)

 IC leads VS by 900E, thus causing Supply Current IS to be

‘corrected’ closer to the Supply Voltage

 The higher the capacitor current, the greater the ‘correction’

Exercises: Single loads

 For each of the two previous exercises,

complete the following:

 Exercise 1: Add a capacitor branch that draws 4A

 Exercise 2: Add a capacitor branch that draws 10A

ILoad = 12.2A

VS = 230V ReferenceΦ=41.4o

IS = 10.0AΦ=240 lag

Exercises: Single loads

 For a load of 2.4kW at pf of 0.4 lag connected to a 230V, 50Hz supply, determine the following:

 Iload

 Øload (phase angle)

 Draw a phasor diagram to scale showing VS, ILoad, and Øload

 After drawing the phasor diagram, perform power factor correction to correct this circuit to a power factor of 0.8 lag:

 Measure IS

 Measure IC

 Calculate QC

Exercises: Single loads Procedure

 Draw VS as ref

 Draw ILoad at Øload

 Step 5: Find ØCorrected

 Draw vert line up from ILoad

 Draw IS at ØCorrected

 Measure IS

 Measure IC

ILoad = 26.1A

VS = 230V ReferenceΦ=66.4o

Line for IS Corrected

ΦCorrected =36.9o Vertical line up from ILoad

Intersection point; then draw in IS Corrected

Measure value for IC

Draw IC phasor at

900 lead

IC=16.1A

IS Corrected=13.1A

Power Factor Correction

AS3000:2007 Requirements for the

Connection of Capacitors

Pf Correction AS3000:2007 Requirements

 General

likely to occur

 Rating of C/B’s, Switches, or Contactors

 CCC of supply conductors

 135% of the rated current of the capacitor; or

 The setting of the C/B

 Provision for discharge and control

Pf Correction AS3000:2007 Requirements

 Capacitors Clause 4.15 (cont.)

 Capacitors connected in parallel with individual appliances

 Permanent discharge path

 No switch or fuse between capacitor and appliance*

 Controlled by the control gear of appliance

 Capacitors NOT connected in parallel with individual appliances

 C/B must have overcurrent release

 Provision for discharge path

 No switch or fuse connected in discharge path*

Exercises: Multiple loads

 A 230V, 50Hz circuit consists of the following two loads:

 3.45kW heater with a unity pf

 2.415kW fan motor with a pf of 0.35 lag

 Determine the following:

 IHeater

 ØHeater

 IMotor

 ØMotor

 Draw a phasor diagram to scale showing VS, IHeater and IMotor

 Use phasor addition to find IS before correction

 Correct this circuit to a power factor of 0.8 lag and determine:

 IS After correction

 IC

 Calculate Capacitance required

VS = 230V ReferenceΦ=69.5o

Line for IS Corrected

Exercises: Multiple loads

 A 230V, 50Hz circuit consists of the following two loads:

 1.4kW brake solenoid with a pf of 0.2 lag

 2.0kW motor with a pf of 0.62 lag

 Determine the following:

 IS before correction

 Pf before correction

 If this circuit does not comply with supply authority

requirements, perform power factor correction to correct the circuit to 0.9 LAG Determine:

 IS After correction

 IC required

 Capacitance required

Exercises: Multiple loads

 IS before correction = 43.4A

 Pf before correction = 0.34 lag (70.00)

 IS After correction = 16.4A

 IC = 33.6A

AC Power

Power Factor Correction Practical