Faculty of Electrical EngineeringUniversiti Teknologi Malaysia

Mechanical and Electrical Systems SKAA 2032

Power Supply (AC and DC)

Alternating voltage and current

• Electricity is produced by generators at power station.

• Electricity is then distributed by a vast network of transmission lines called National Grid System.

• It is easier and cheaper to generate AC than DC.• It is more convenient to distribute AC than DC

since the voltage can be readily altered using transformer.

• Whenever DC is needed, devices called rectifiers are used for conversion.

Alternating voltage and current

Power socket

Rectifier

Generation of Single Phase

• An electric current can be induced in a circuit by a changing magnetic field – Faraday’s Law

• The direction of the induced current is such that the induced magnetic field always opposes the change in the flux – Len’z Law

• Direction of current for generator – Fleming’s right hand rule.

Single phase

• Single phase electricity is generated by rotating a single turn coil through a magnetic field.

• The shape of the waveform produced by a generator (i.e. the alternator) is in the form of sine wave.

• Wires used:– Live conductor (yellow)– Neutral conductor (blue)– Earth conductor (green) –connected from neutral via a

protective gear to earth

Single phase

Single phase system

A general expression for the sinusoid is given by: v(t) = Vm sin (wt + q)

whereVm is the amplitude or peak valueω is the angular frequency radian/s given by ω=2πftf is the frequency in hertz (Hz)t is the time in second (s)T is the period in second, given by T=1/f θ is the phase angle in degree

Single phase system

v(t)Vm

-Vm

t

w

2T

T1f

f2wThe angular frequency in radians/second

Single phase system

• A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes.

• We can transform a sinusoid from sine to cosine form or vice versa using this relationship:

cos ωt = sin (ωt + 90o)

sin ωt = cos (ωt - 90o)

Single phase system

Example 1.1

Find the amplitude, phase angle, angular frequency, period and frequency of the sinusoidal waveform

(a) v(t) = 12 cos (50t + 10o)

(b) v(t) = 5 sin (4πt - 60o)

(a) (12V, 10o, 50rads/sec, 0.126 sec., 7.937 Hz)

(a) (5V, -60o, 4π rads/sec, 0.5 sec., 2 Hz)

Single phase system

• Sinusoids are easily expressed in terms of phasors. • A phasor is a complex number that represents the

amplitude and phase of a sinusoid. v(t) = Vm cos (ωt + θ)

q mVV

Time domain

Phasor domain

Time domain Phasor domain)cos( qwtVm qmV

om 90V q)sin( qwtVm

)cos( qwtmI qmIo

m 90I q)sin( qwtmI

Single phase system

Instantaneous and Average Power• The instantaneous power is the power at any

instant of time: p(t) = v(t) i(t)

• Where v(t) = Vm cos (ωt + θv) i(t) = Im cos (ωt + θi)

• Using the trigonometric identity, gives )2cos(

21)cos(

21)( ivmmivmm tIVIVtp qqwqq

Single phase system

The average power is the average of the instantaneous power over one period.

T

0dtt

T1 )(pP

)cos( ivmmIV21

qqP

p(t)

t

)cos( ivmmIV21

qq

mmIV21

Single phase system

• The effective value is the root mean square (rms) of the periodic signal.

• The average power in terms of the rms values is given by

Where

)cos( ivP qq rmsrmsIV

2V

V mrms

2I

I mrms

Single phase system

Example 1.2An ac voltage of a sinusoidal waveform has a peak value of 300 V. What is the rms value of this voltage?(212.1 V)

Example 1.3What is the peak voltage of 120 V rms?(169.7)

Single phase system

Example 1.4An alternating current of sinusoidal waveform has a r.m.s value of 10A. What are the peak values of this current over one cycle?

(14.14A & -14.14A)

Single phase system

Example 1.5An alternating voltage can be represented by v=141.4 sin 377t. Determine: (a) r.m.s. voltage (b) frequency (c) the instantaneous voltage when t = 3 ms

(100V, 60Hz, 127.8V)

Single phase system

Apparent Power, Reactive Power and Power Factor

The apparent power is the product of the rms values of voltage and current.

The reactive power is a measure of the energy exchange between the source and the load reactive part.

rmsrmsIVS

)sin( ivQ qq rmsrmsIV

Single phase system

The power factor is the cosine of the phase difference between voltage and current.

The complex power:

)cos( ivfactor Power qqSP

iv

jQPqq

rmsrms IV

Single phase system

rmsrmsIVS

)sin( ivQ qq rmsrmsIV

)cos( ivrmsrmsIVP qq True or active power:

Apparent power:

Watts (W)

reactive volt·amperes (var)

Reactive power:

S Q

Pθv–θi

volt·amperes (VA)

Three phase system

Three phase system

• A three-phase electricity is generated when three coils are placed 120° apart, and the whole rotated in a magnetic field.

• The result is three independent supplies of equal phase voltage - distinguished by 120° phase angle.

• The convention adopted to identify the phase voltages: R-red, Y-yellow, B-blue.

• The standard phase sequence is R, Y, B.

Generation of Three-phase

• Suppose three similar loops of wire with terminals R-R’, Y-Y’ and B-B’ are fixed to one another at angles of 120o and rotating through a magnetic field.

R

R1

B

B1

Y

Y1

N S

Three phase system

• Three conductors (lines) to carry the three phase supply, colored red, yellow and blue.

• A fourth conductors called the neutral, connected through protective device to earth.

• The three phase system is usually connected using: – star connection (sources i.e. alternators)– delta connection (transformers, motors and other

loads)

Generation of Three-phase

The instantaneous e.m.f. generated in phase R, Y and B:

vR = VR sin wt vY = VY sin (wt -120o) vB = VB sin (wt -240o) = VBsin (wt +120o)

v(t)

wt

vR

vY vB

Generation of Three-phase

Phase sequences:(a) RYB or positive sequence

120o

-120o

120o VR

VY

VB

w

o)rms(YY 120VV

o)rms(RR 0VV

o)rms(B

o)rms(BB

120V

240VV

VR leads VY, which in turn leads VB.This sequence is produced when the rotor rotates in the counterclockwise direction.

Generation of Three-phase

(b) RBY or negative sequence

o)rms(BB 120VV

o)rms(RR 0VV

o)rms(Y

o)rms(YY

120V

240VV

120o

-120o

120o VR

VB

VY

w

VR leads VB, which in turn leads VY.This sequence is produced when the rotor rotates in the clockwise direction.

Star Connection

Three wire systemR

Y

B

ZR

Y B

Star Connection

Four wire system

VRN

VBN VYN

ZR

R

BN

Y

Star Connection of Load

Z1

Z3

Z2

R

B

Y

NLoad

Z3

R

Y

B

Load

N

Delta Connection

R

Y

B

Y

B

R

Delta Connection of Load

Zc

Za

Zb

R

B

Y

Load

Za

R

Y

B

Load

Star Connection

N

R

Y

B

VRY

VYB

VBR

VYN

VBN

VRN

IR

IY

IB

Phase voltages (line-to-neutral voltages):

240

120

0

phaseBN

phaseYN

phaseRN

V

V

V

V

V

V

# Reference: VRN

# Positive sequence.

Line voltages (line-to-line voltages):

RNBNBR

BNYNYB

YNRNRY

VVVVVVVVV

Star Connection

N

R

Y

B

VRY

VYB

VBR

VYN

VBN

VRN

IR

IY

IB

Line currents, Iline:

BYR III ,,

Phase currents are equal to their line currents:

linephase II

linephase

linephase

V

II

V

Star Connection – Line Voltages

N

R

Y

B

VRY

VYB

VBR

VYN

VBN

VRN

IR

IY

IB

303

120j1200j0

1200

seph

oooophase

phasephase

YNRNRY

aV

V

VVVVV

)sin()(cos()sin(cos

The two other can be calculated using similar method.

Star Connection - Line voltages

RNBNBR VVV

1503

2103

903

303

phase

phaseBR

phaseYB

phaseRY

V

VV

VV

VV

BNYNYB VVV

YNRNRY VVV

Star connection - Vector diagram

30°

-120°

VBR VRY

VYB

VYN

VRN

VBN

-VYN

• Phasor diagram is used to visualize the system voltages• Star system has two type of voltages: Line-to-neutral, and line-to-line.• The line-to-neutral voltages are shifted with 120o

• The line-to-line voltage leads the line to neutral voltage with 30o

• The line-to-line voltage is times the line-to-neutral voltage

Star connection - Distribution

Typical distribution voltage of 415/240V, 3 phase 4 wires system

Delta Connection

VRY

VBR

VYB

VRY

R

Y

BV

YB

VBR

Phase voltages are equal to the line voltages# Reference: IRY

# Positive sequence. linephase VV

Delta Connection

Phase currents:

240II

120II

0II

phaseBR

phaseYB

phaseRY

linephase

linephase

II

VV

VRY

VBR

VYB

VRY

R

Y

B

VYB

VBR

R

YB

Delta Connection – Line Currents

30I3

120j1200j0I

120I0IIII

seph

oooophase

phasephase

BRRYR

a

sin(cos)sin(cos

The two other can be calculated using similar method.

Delta Connection – Line Currents

BRRYR III

RYYBY III

YBBRB III

90I3

270I3I

1503I

30I3I

phase

phaseB

phaseY

phaseR

V

Delta Connection – Vector Diagram

-30°

120°

IY I R

IB

I YB

IRY

I BR

-I BR

TNB Supply System

Voltage 3 phase, 50 Hz

The main transmission and substation network are: - 275 kV - 132 kV - 66 kV

The distribution are: - 33 kV - 22 kV - 11 kV - 6.6 kV - 415 volts - 240 volts (single phase) drawn from 415 volts 3 phase (phase voltage), between line (R, Y, B) and Neutral (N)

TNB Supply System

The low voltage system (415/240 V) is 3-phase four wire.The low voltage system is a mixture of overhead lines and under ground cables.

The high voltage and extra high voltage system is 3-phase three wire Configuration. Overhead line and under ground cable system are used.

Supply Method (two types of premises)1. Single consumer such as private dwelling house, workshop, factory, etc.

a. Single phase, two wire, 240 V, up to 12 kVA max demandb. Three phase, four wire, 415 V, up to 45 kVA max demandc. Three phase, four wire, C. T. metered 415 V, up to 1,500 kVA max demand

TNB Supply System2. Multi tenanted premises, such as high rises flats, commercial, office blocks, etc

- Low Voltage

a. Three phase, three wires, 6,600 and 11,000 V for load of 1, 500 kVA max demand and above, whichever voltage is available

b. Three phase, three wires, 22,000 and 33,000 V for load of 5,000 kVA max demand and above, whichever voltage is available

c. Three phase, three wires, 66,000 V, 132,000 V and 275,000 for exceptionally large load of above 20 MVA max demand

Three phase, four wire, C.T. metered 415 V, up to 1,500 kVA maxdemand

- High Voltage and Extra High Voltage

Standby Supply

• Standby generator(s) may be used by the applicant at their premises, subject to compliance with the relevant laws.

• The generators shall remain a separate system from TNB distribution system and the applicant shall declare to TNB on the safe installation of the generator(s).

• This may be used in place of TNB’s supply source through a suitable, approved changeover facility.

• The Energy Commission and other relevant authorities govern the usage of generators and standby supply.

• This may be used in place of the TNB’s supply source through a suitable, approved change over facility under emergency conditions.