Mechanical and Electrical Systems
SKAA 2032
Power Supply (AC and DC)
Faculty of Electrical Engineering
Universiti Teknologi Malaysia
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
Rectifier
Power socket
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)
where
Vm is the amplitude or peak value
ω is the angular frequency radian/s given by ω=2πft
f 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
2
T
w
1
f 
T
w  2f
The angular frequency in radians/second
t
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)
Time domain
V  Vm q
Phasor domain
Time domain
Phasor domain
Vm cos( wt  q )
Vm sin( wt  q )
Vm q
Vm q  90o
I m cos( wt  q )
I m q
I m sin( wt  q )
I m q  90o
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
1
1
p (t )  Vm I m cos(q v  q i )  Vm I m cos(2wt  q v  q i )
2
2
Single phase system
The average power is the average of the
instantaneous power over one period.
1 T
P   p( t ) dt
T 0
1
P  Vm I m cos( q v  q i )
2
p(t)
1
Vm I m
2
1
Vm I m cos( q v  q i )
2



t
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
P  VrmsIrms cos( qv  qi )
Where
Vrms 
Vm
I rms 
Im
2
2
Single phase system
Example 1.2
An 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 1.3
What is the peak voltage of 120 V rms?
(169.7)
Single phase system
Example 1.4
An 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 1.5
An 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.
S  Vrms I rms
The reactive power is a measure of the energy
exchange between the source and the load reactive
part.
Q  VrmsIrms sin( qv  qi )
Single phase system
The power factor is the cosine of the phase
difference between voltage and current.
P
Power factor   cos( qv  qi )
S
The complex power:
 P  jQ
 Vrms I rms qv  qi
Single phase system
True or active power:
P  Vrms I rms cos(q v  q i ) Watts (W)
Apparent power:
S  Vrms I rms volt·amperes (VA)
S
Q
θv–θi
Reactive power:
Q  VrmsIrms sin( qv  qi )
reactive volt·amperes (var)
P
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 RR’, Y-Y’ and B-B’ are fixed to one another at angles of
120o and rotating through a magnetic field.
R
B1
Y1
N
S
Y
B
R1
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
v(t)
vR

vY

vB
wt
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)
Generation of Three-phase
Phase sequences:
(a) RYB or positive sequence
VB
VR  VR ( rms) 0o
w
120o
120o
VR
-120o
VY
VY  VY( rms)   120o
VB  VB ( rms)   240o
 VB ( rms) 120o
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
VY
VR  VR ( rms) 0o
w
VB  VB( rms)   120o
120o
120o
VR
-120o
VY  VY ( rms)   240o
 VY ( rms) 120o
VB
VR leads VB, which in turn leads VY.
This sequence is produced when the rotor rotates in
the clockwise direction.
Star Connection
Three wire system
R
ZR
ZY
Y
B
Z
B
Star Connection
Four wire system
R
ZR
VRN
V BN
V YN
ZY
Y
N
B
Z
B
Star Connection of Load
R
R
2
Z
Z1
Y
Y
Load
Z3
B
N
Z1
B
N
Z2
Z3
Load
Delta Connection
R
R
Y
Y
B
B
Delta Connection of Load
R
Load
R
Zc
Z
c
Zb
Y
B
Za
Y
Zb
Za
B
Load
Star Connection
IR
Phase voltages (line-to-neutral
voltages):
R
VRN
VRN  V phase 0
VRY
N
V YN
VYN  V phase   120
IY
VBR
Y
VBN
VYB
IB
B
# Reference: VRN
# Positive sequence.
VBN  V phase   240
Line voltages (line-to-line
voltages):
VRY  VRN  VYN
VYB  VYN  VBN
VBR  VBN  VRN
Star Connection
IR
R
Line currents, Iline:
VRN
I R , IY , I B
VRY
N
V YN
IY
VBR
Y
VBN
Phase currents are equal to
their line currents:
I phase  I line
VYB
IB
B
I phase  I line
Vphase  Vline
Star Connection – Line Voltages
VRY  VRN  VYN
 Vphase 0  Vphase   120

 Vphase (cos0o  j sin0o )  (cos(120o )  j sin (120o )

 3 Vphase  30
IR
R
The two other can be
calculated using similar
method.
VRN
VRY
N
V YN
IY
VBR
Y
VBN
VYB
IB
B
Star Connection - Line voltages
VRY  VRN  VYN
VRY  3 Vphase  30
VYB  VYN  VBN
VYB  3 Vphase   90
VBR  VBN  VRN
VBR  3 Vphase   210
 3 Vphase 150
Star connection - Vector diagram
• 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
VBR
VBN
VRY
30°
-120°
VYN
VYB
-VYN
VRN
Star connection - Distribution
Typical distribution voltage of 415/240V, 3 phase 4 wires system
Delta Connection
R
V
V
RY
BR
Y
V
V
B
V
RY
V
BR
YB
YB
# Reference: IRY
# Positive sequence.
Phase voltages are equal
to the line voltages
V phase  Vline
Delta Connection
R
V
R
V
RY
BR
V
Y
V
Y
B
RY
V
BR
YB
B
Phase currents:
V
YB
Vphase  Vline
I phase  I line
I RY  I phase 0
I YB  I phase   120
I BR  I phase   240
Delta Connection – Line Currents
I R  I RY  I BR
 I phase 0  I phase 120

 I phase (cos0o  j si n0o )  (cos120o  j si n120o
 3 I phase   30
The two other can be calculated using
similar method.

Delta Connection – Line Currents
I R  I RY  I B R
I R  3 I phase   30
I Y  I YB  I RY
I Y  3 Vphase   150
I B  I B R  I YB
I B  3 I phase   270
 3 I phase  90
Delta Connection – Vector Diagram
IB
I BR
120°
I RY
-30°
IY
I YB
-I BR
IR
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 demand
b. Three phase, four wire, 415 V, up to 45 kVA max demand
c. Three phase, four wire, C. T. metered 415 V, up to 1,500 kVA max
demand
TNB Supply System
2. Multi tenanted premises, such as high rises flats, commercial,
office blocks, etc
- Low Voltage
Three phase, four wire, C.T. metered 415 V, up to 1,500 kVA max
demand
-
High Voltage and Extra High 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
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.