Power System
Engineering
Eng. Nasser Al Mashaikhi
Unit 0: Course Outline
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Course Description
This
course
fundamental
will
deliver
understanding
the
of
main
electrical
engineering science and describing the
electrical network and the main components
installed in the network. Topics cover:
Electricity
Generation,
Transmission
&
Distribution Sector, Basic Laws, Network
design, Substations & components.
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✓ T O U N D E R S TA N D T H E E L E C T R I C A L S E C T O R I N O M A N A N D R O L E O F
EACH UTILITIES
✓ T O U N D E R S TA N D T H E P R I N C I P L E S A N D P R A C T I C E O F P O W E R S Y S T E M S
ENGINEERING
✓ T O U N D E R S TA N D P O W E R S Y S T E M C O M P O N E N T S A N D T Y P E O F
S U B S TAT I O N S
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OMAN ELECTRICAL STANDARDS
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Course Outcomes
Audience will:
Learn strong basics of Electrical Engineering and practical
implementation of Electrical fundamentals.
Understanding the main power system network in Oman
Learn different applications of commonly used electrical
components and operational international & local standards.
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Pre-assessment
Post-assessment
4 Days Program,
+ Practical Part
Course Duration & Assessment
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Unit 1: Electricity Generation,
Transmission & Distribution
Sector
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Dawn of the Electricity Reform
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Who Does
What Today
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Regulation
Responsible for regulating the
electricity sector and some
aspects of the water sector.
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Established by Article (19) of The
Law for the Regulation and
Privatization of the Electricity and
Related Water Sector promulgated
by Royal Decree 78/2004 on 1
August 2004 and amended by
Royal Decree 59/2009 and
47/2013.
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Has a legal personality and
enjoys financial and
administratively autonomy.
Thus, AER is reporting directly
to the Council of Ministers.
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NAMA Holding
❖To encourage electricity projects as well as electricity and Related Water projects which are financed by the private
sector;
❖To finance companies undertaking business in the electricity and Related Water sector whose capital is wholly-owned
by the State and that shall be within the framework of the general policy of the State, and shall safeguard the interest of
the State in these companies;
❖to undertake the measures which it considers imperative for achieving its objectives in the manner determined by the
Ministry of Oil and Gas;
❖To establish new companies or direct any company of its subsidiaries and that shall be for the purpose of securing or in
order to secure new Production Capacity or to manage, operate or maintain some of its assets or those which transfer
to it from any Electric Plant, Production Facility, Systems or those of such plants and facilities which are owned by a
Licensee whose License has been revoked pursuant to the provision of Article (121) of this Law, and all the aforesaid
shall be in the manner determined by the Ministry of Oil and Gas;
❖To provide central accounting services for the companies doing business in the electricity and Related Water sector
whose capital is wholly-owned by the State, and others of those who wish to receive such services for a charge;
❖To provide consultation to the Public Authority for Water, at its request, in relation to the future restructuring and
regulation of the unrelated water sector; and
❖To establish new companies on the direction of the Public Authority for Water, in coordination with the Ministry of
Finance, for the purpose of restructuring and privatizing the unrelated water sector.
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Power and Water Procurement
The Oman Power and Water Procurement Company (OPWP) member of
Nama Group is the single Buyer of power and water for all IPP/IWPP
projects within the Sultanate of Oman. As part of the OPWP
responsibility, the OPWP undertakes long term generation planning and
publishes a 7 year statement.
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Generation
✓OPWP purchases electricity and desalinated water in accordance with the
Power Purchase Agreements (PPAs) and Power and Water Purchase
Agreements (PWPAs) with the various generators and desalination
companies.
o
o
o
o
o
o
o
o
o
o
o
Al Rusail Power Company SAOG
United Energy Company SAOG
Al Kamil Power Company SAOG
Aqua Power Barka SAOG
Sohar Power Company SAOG
Dhofar Power Generation Company SAOC
SMN Barka Power SAOG
Al Batinah Power Company SAOG
Al Sawadi Power Company SAOG
Phoenix Energy Company SAOG
Simpecorp Salalah Power & Water Company SAOC
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Transmission
Oman Electricity
Transmission Co. (OETC)
member of Nama
Group owns and
operates a single
400/220/132 kV
transmission system to
serve the areas in the
Main Interconnected
System I(MIS) & Dhofar
Transmission System
(DTS).
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Distribution
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Reference
1. Oman Power System Companies
(https://www.researchgate.net/figure/Oman-power-system-companiesand-areas_fig7_270583843)
2. Authority of Electricity Regulation (https://www.aer.om/en/home)
3. Oman Electricity Transmission Company (www.omangrid.om)
4. Oman Power & Water Procurement (www.omanpwp.om)
5. Oman Electricity Sector: Features, Challenges and Opportunities for
Market Integration, Author: Shahid Hasan, Turki Al-Aqeel, Abdullah AlBadi, Yagyavalk Bhatt, Mohammed Al-Badi
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Unit 2: Basic Laws
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Basic Concepts
In electrical engineering, we are concerning to
transfer the energy from one point to another. To do
so, it is required an electric circuit. each components
in a circuit is called element.
An electric circuits is interconnection of electrical
elements
Elements are divided into Active and Passive Elements:
Active Elements are elements that can produce a power,
like: Generators, Batteries
Passive Elements are resistors, inductors or capacitors
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SI Units
Stands for: International
System of Units
SI selects some physical
quantities as its basis:
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Quantity
Basic Unit
Symbol
Mass
Kilogram
Kg
Length
Meter
m
Time
Second
s
Electric Current
Ampere
A
Temperature
Kelvin
K
Luminous Intensity
Candela
Cd
Amount of Substance
Mole
mol
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SI Units
One great advantage of SI unit is that it uses prefixes based on the
power of 10 to relate larger and smaller units to the basic unit.
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Multiplier Factor
Prefix
Symbol
1012
Tera
T
109
Giga
G
106
Mega
M
103
Kilo
K
10-3
Milli
m
10-6
Micro
µ
10-9
Nano
n
10-12
Pico
p
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Circuit Variables
Charge (Q): defines as electrical property of the atomic particles of
which matter consists, measured in coulombs (C).
Current (I) is the time (s) rate of change in charge (C), measured in
Amperes (A)
1 Ampere (A) = 1 Coulomb (C) /second (s)
Two type of current
◦ Direct Current (DC) is a current remain constant with time
◦ Alternating Current (AC) is a current varies (changing) with time
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War of Currents
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Circuit Variables
Voltage is the energy required to move a unit of charges through an
element, measured in Volt (V)
◦ Voltage (V) = Energy / Charge → V = W/Q
◦ Power is time rate of expending or absorbing energy, measure in watts (W)
◦ Power = Energy / time → P = W/t
◦ As per the equations above we can come up with this:
P = V I → W/Q . Q/t = W/t
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Ohm’s Law
Voltage across a conductor is directly proportional to the current provided
that temperature remain constant.
Voltage (V) = Current (I) x Resistance (R)
V
I
𝑉 = 𝐼𝑅
Conductor (R)
Figure 1: Simple formation of current flow in a conductor
Current is the flow of charges in a conductor from high potential to lower
potential.
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Resistance
The resistance that any piece of material has may be shown
experimentally depend on:
A. Length
B. Cross sectional Area
C. The material
D. Temperature
𝑙
⍴
𝑙
Resistance (R) = ⍴
𝐴
◦ where,
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Figure 2: Cylindrical shape of conductor
(l) is the length of the conductor in meter
(A) is the cross section area in m2
(⍴) is the resistivity of the material in Ω.m
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Resistance vs Temperature
The resistance of materials varies with temperature. For some metals, the
resistance increases with increasing in temperature.
R2=R1[1+ α1(θ2- θ1)]
where, (R1) is the initial resistance at temperature θ1
(θ1 ) is the initial temperature measured
(R2) is the final resistance at new temperature θ2
(θ2) is the new temperature measured
(α1) is temperature coefficient for a material
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Example 1
When the potential difference applied across a coil of copper wire at a
mean temperature of 20⁰C is 10 V a current of 1 A flows in it. After a
period of time the current falls to 0.95 A with a potential difference
unaltered. Determine the mean temperature of the coil. The
temperature coefficient of resistance of copper at 20⁰C as 4.28 x 10-3/K.
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Series Circuit
𝑅𝑇𝑂𝑇𝐴𝐿 = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑛
𝑣𝑠 = 𝑖𝑅𝑇𝑂𝑇𝐴𝐿
𝑣𝑅1 = 𝑖𝑅1
Figure 3: Series circuit
By voltage division expression for simple circuit:
𝑣𝑅1 =
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𝑅1
𝑣𝑠
𝑅1 + 𝑅 2
𝑣𝑅2 =
𝑅2
𝑣𝑠
𝑅1 + 𝑅 2
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Parallel Circuit
1
𝑅𝑇𝑂𝑇𝐴𝐿
=
1
1
1
+ + ⋯+
𝑅1 𝑅2
𝑅𝑛
𝑉𝐵 = 𝐼𝑅
𝐼1 =
Figure 4: Parallel circuit
𝑉𝐵
𝑅1
By Current division expression for simple circuit:
𝑖𝑅1 =
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𝑅2
𝐼
𝑅1 + 𝑅 2
𝑅1
𝑖𝑅2 =
𝐼
𝑅1 + 𝑅2
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Kirchhoff's Law
First Law: Kirchhoff Current Law (KCL) states that the algebraic sum of
all the currents leaving or entering a node is equal to ZERO.
𝑖1 + 𝑖2 − 𝑖3 − 𝑖4 = 0
Figure 9 : Node illustrate the KCL
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Kirchhoff's Law
Second Law: Kirchhoff Voltage Law (KVL) states that the algebraic sum of all
voltage drop or voltage rise in a loop is equal to ZERO.
−𝑣1 − 𝑣2 + 𝑣3 + 𝑣4 = 0
Figure 10 : Closed loop illustrate the KVL
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Example 3
Find the current in the circuit below:
Figure 11 : Example circuit
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Inductance
Inductance is the property of a circuit or coil that causes an
electromotive force to be set up due to a rate of change of current in
the circuit or coil. The device is called inductor and its unit is Henry (H)
Figure 12: Symbol for Inductor
𝑉𝐿 = 𝐼𝜔𝐿,
𝑋𝐿 = 2𝜋𝑓𝐿
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𝜔 = 2𝜋𝑓
Unit: Ohm (Ω)
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Capacitance
The property of a capacitor which determines how much charge can be
stored in it for a given potential difference between its terminals, equal
to the ratio of the charge stored to the potential difference and
measured in Farads (F)
Figure 13: Electric field between the
plates of a capacitor
𝐼𝐶 = 𝜔𝐶𝑉,
1
1
𝑋𝑐 =
=
𝜔𝐶 2𝜋𝑓𝐶
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Unit: Ohm (Ω)
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Impedance
An impedance (Z) vector consists of a real part (resistance, R) and an
imaginary part (reactance, X)
𝑍 = 𝑅 + 𝑗𝑋
𝑍 2 = 𝑅2 + 𝑋 2
𝑍=
𝑉
𝐼
Figure 14: RLC Series circuit
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Example 4:
Find the current (I) flows in the circuit above, if the voltage applied 10 V with 50 Hz
frequency, R= 5Ω, L= 20mH & C= 60μF
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Frequency & Period
■
If we plot an alternating quantity with a base of time, an example of the waveform in Fig. 1.
1. Period (T): is the time it takes the wave to
complete one cycle, unit is second (s)
2. Frequency (f): the number of cycles are
completed in a certain of time, unit is Hertz (Hz)
𝑓=
Figure 13. Waveform of an alternating quantity
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𝑇
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Sinusoidal waves
Alternating current and voltage can be expressed instantaneously in the
following function of time:
𝑣 = 𝑉𝑚 sin ω𝑡 = 𝑉𝑚 sin 2𝜋𝑓𝑡
𝑉𝑚
𝜋
2
𝜋
3𝜋
2
2𝜋
Figure. 14. Waveform of a sinusoidal alternating voltage
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RMS and mean values of AC
RMS stands for Root Mean Square
Consider a current 𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 flowing through a resistor R. at any instant,
the power p is given by:
𝑝 = 𝑖 2𝑅
RMS value can be derived from the above equation by:
𝐼𝑟𝑚𝑠 =
𝐼𝑚
√2
= 0.707 𝐼𝑚
Therefore, the effective value for the current is Irms
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Example 1
The equation relating the current in a circuit with time is:
𝑖 = 141.4 sin 377𝑡 Ampere
Find:
a. the r.m.s current
b. the frequency
c. the instantaneous value of the current when t is 3 ms.
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Power
Reactive Power (Q)
ϕ
Active Power (P)
𝑃2 + 𝑄2 = 𝑆 2
cos ϕ =
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𝑃
𝑆
Power Factor
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Power
The product of voltage and current in an AC circuit is termed the
apparent power:
𝑆 = 𝑉𝐼 , unit VAs
Therefore, the active power can be defined: 𝑃 = 𝑆 cos ϕ = 𝑉𝐼 cos ϕ
, in Watts
The reactive power Q, 𝑄 = 𝑉𝐼 sin ϕ , in VARs
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Example 2
1 Phase circuit takes a power of 4.2 KW at power factor 0.6 lag. Find
the value of the apparent power and the peak reactive power?
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Power Factor
Power factor is the ratio of active power (P) to apparent power (S).
It measures how effectively the active power is being used . A high
power factor signals efficient utilization of electrical power, while a
low power factor indicates poor utilization of electrical power .
Power factor = cos 𝜃 =
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𝑃
𝑆
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Example 3
An inductor coil is connected to supply of 250 V at 50 Hz and takes a current of
5 A. the coil dissipates 750 W. Calculate
◦ 1. the resistance and the inductance of the coil
◦ 2. the power factor of the coil
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Power Factor Correction
If we have an active power 100 KW & 150 KVA apparent power, the power factor will
be:
𝑃 100
𝑃. 𝐹 = =
= 66.6%.
𝑆 150
However, if the power factor improved by 90%, the new apparent power is:
𝑆=
𝑃
𝑃𝐹
=
100
90%
= 111 𝐾𝑉𝐴
Therefore, as Power Factor increase, you required less apparent power to deliver the
same power to do work.
One example of improving the power factor, is by installing a capacitor to the network
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Power Factor Correction
Example 4: Find the power factor in the figure below?
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Power Factor Correction
After adding 300 µF capacitor in parallel with the circuit, find new
power factor?
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Three phase system
The single phase AC circuit is suitable for most application but there are
fields of electrical engineering not well suited, power transmission and
electromechanical energy conversion by machines
Why do we use three phase system:
1. Three-phase lines transmit more power
2. Three-phase motors can start without the need for extra
equipment
3. three-phase generator produces more power than a single
phase generator
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Y Connection (Phase Voltage)
𝑉𝑎𝑛 = 𝑉𝑏𝑛 = 𝑉𝑐𝑛 = 𝑉𝑝ℎ = 𝑉
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Y connection (Line to Line Voltage)
𝑉𝑎𝑏 = √3 × 𝑉𝑝ℎ
𝑉𝑎𝑏 = 𝑉𝑐𝑎 = 𝑉𝑏𝑐
𝐿𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑝ℎ𝑎𝑠𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
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Delta Connection
𝑉𝐿 = 𝑉𝑝ℎ
𝐼𝐿 = 3 𝐼𝑝ℎ
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Power in Y connection
For single phase: 𝑃 = 𝑉𝐼 𝑐𝑜𝑠𝜃
For three phase: 𝑃 = 3 𝑉𝐼 𝑐𝑜𝑠𝜃
Therefore, in Y connection the real power is after considering the Line
voltage:
𝑷 = 𝟑 𝑽𝑳 𝑰𝑳 𝒄𝒐𝒔𝜽
■
For single phase: 𝑄 = 𝑉𝐼 𝑠𝑖𝑛𝜃
■
For three phase: 𝑄 = 3 𝑉𝐼 𝑠𝑖𝑛𝜃
■
Therefore, in Y connection the reactive power is after considering the Line
voltage:
𝑸 = 𝟑 𝑽𝑳 𝑰𝑳 𝒔𝒊𝒏𝜽
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Power in Δ connection
For single phase: 𝑃 = 𝑉𝐼 𝑐𝑜𝑠𝜃
For three phase: 𝑃 = 3 𝑉𝐼 𝑐𝑜𝑠𝜃
Therefore, in Δ connection the real power is after considering the Line
current:
𝑷 = 𝟑 𝑽𝑳 𝑰𝑳 𝒄𝒐𝒔𝜽
■
For single phase: 𝑄 = 𝑉𝐼 𝑠𝑖𝑛𝜃
■
For three phase: 𝑄 = 3 𝑉𝐼 𝑠𝑖𝑛𝜃
■
Therefore, in Δ connection the reactive power is after considering the Line
current:
𝑸 = 𝟑 𝑽𝑳 𝑰𝑳 𝒔𝒊𝒏𝜽
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References
1. Basic electrical engineering science, Author: I. McKenzie Smith & K. T.
Hosie, 1971 Longman
2. Circuit Analysis 1, Author: Steven T Karris, 2004
3. Basic Electrical Engineering, Author: V.HimaBindu, V.V.S Madhuri,
Chandrashekar.D
4. Basic electrical engineering science, I. McKenzie Smith & K. T. Hosie, 1971
Longman
5. Power Factor Correction Technical Data, Eaton
http://www.eaton.com/ecm/groups/public/@pub/@electrical/documents/
content/sa02607001e.pdf
6. Power System Analysis 1 SQU Slides, Dr. Mohammed Al Badi & Dr. Amer
Al Hinai
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Unit 3: Network Design
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Power System Structure
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Voltage Level in Oman Network
Step Up
Transformer
Step Down
Transformer
Step Down
Transformer
Step Down
Transformer
Load
G
…/132 KV
…/220 KV
…/400 KV
400/220/132/33 KV
11/0.433 KV
33/11 KV
Load
Generation
Transmission
33/0.433 KV
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Network design
❑ Relevant Legislation & Standards
❑ Assessment of Demand
❑ Selection and Sizing of Components
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Electrical
Standards
Local Standards:
✓Oman Electrical Standards (OES)
✓Oman Standards (OS)
International Standards
✓International Electro-technical Commission (IEC)
✓International Organization for Standardization (ISO)
Other Standards:
✓ British Standards (BS)
✓ American National Standards Institute (ANSI)
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Grid & Distribution Codes
◦ Grid Code referred to in the Sector Law and the Transmission
and Dispatch Licence granted by the Regulatory Authority to
the Oman Electricity Transmission Company SAOC
(“Transco”) under that Law. It contains rules in relation to
the planning, development, Connection to, Operation and
maintenance of and changes to Transco's Transmission
System.
◦ Distribution Code referred to in the Sector Law and the
licences granted by the Regulatory Authority to [Muscat
Distribution Company SAOC, Mazoon Distribution Company
SAOC, and Majan Distribution Company SAOC] (collectively,
Licensed Distributors) under that Law. It contains rules in
relation to the Connection to, Operation and maintenance of
and changes to the Distribution Systems of Licensed
Distributors.
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Assessment of demand
1.
Type and class of demand
a)
b)
c)
Industrial
Commercial
Residential
2.
Number of customers
3.
Size of individual demands
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Voltage
Thermal
Power Quality
Fault Level
Technical Constraints
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Voltage
Statutory voltage limits:
Consumers Connected at
Nominal voltage
Tolerance
HV
> 66 KV
±10%
MV
11 & 33 KV
±6%
LV
415/240 V
±6%
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Power Quality
❑ Voltage fluctuations
❑ Harmonics
❑ Phase unbalance
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Fault Level
❖ How much current can your circuit breaker
take?
❖How long can your assets carry that current
before they fail?
❖Is the fault level high enough to ensure
protection can differentiate between load
and a fault?
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Thermal
SOURCES OF HEAT GENERATION:
1.
EFFECTS OF SOLAR RADIATION
2.
THE INTERNAL RESISTANCE OF THE CONDUCTOR
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Unit 4: Network Stations
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Grid Station (400/220/132/33 KV)
The existing transmission system in northern Oman has three operating
voltages, i.e. 400KV, 220KV and 132KV.
Number of grid station in 2018 increased to 89 stations
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Primary Substation (33/11.5 KV)
After Grid Substation in the transmission side (132/33 KV), the network is connected in a
substation called distribution primary substation (33/11.5 KV)
The standard power rates for this substation in Oman (1, 3, 6, 10 & 20 MVA)
In-Door Substation
Out-Door Substation
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3/6 MVA Outdoor Substation
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Secondary Substation (11/0.433 KV)
Come after primary substation, this substation is near the customer side in residential areas or commercial. Two
types of substation depends on Transformer mounting either:
1. Pole Mounted Transformer (PMT), Transformer rated (100, 200 & 315 KVA)
2. Ground Mounted Transformer (GMT), Transformer rated (500 & 1000 KVA)
Ground Mounted Transformer (GMT)
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Pole Mounted Transformer (PMT)
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PMT Substation Layout
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GMT Substation Layout
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Unit 5: Network Components
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Classification of Tests
Some of the tests are performed during the early stages of development
and production(e.g. through Research and Development “R&D”),others
after production and installation.
Type tests are performed on each type of equipment before their
supply on a general commercial scale so as to demonstrate
performance characteristics meeting the intended application.
Routine tests are made by the manufacturer on every finished piece of
product to make sure that it fulfills the specifications.
Acceptance and commissioning tests are made by the purchaser.
Maintenance tests are usually carried out after maintenance or repair
of the equipment
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Energy Source
Fossil Fuel
◦ Oil
◦ Coal
◦ Natural Gas
Nuclear
Renewable Energy
◦ Solar
◦ Wind
◦ Water
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Generators
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Prime Movers for Electric Generators
Most of electric power is obtained by electromechanical energy conversion, whereby
mechanical energy is converted into electric energy by means of electric generators.
The source of mechanical energy is the prime mover which is directly coupled to the
generator.
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Transformer
Statistic component without any movement parts used for moving the power from
primary side to the secondary side with changing in the voltage and the current.
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Principle of operation
When the AC supply is given to the primary winding with a voltage of Vp, an alternating flux ϕ sets up
in the core of the transformer, which links with the secondary winding and as a result of it, an emf is
induced in it called Mutually Induced emf.
Physically, there is no electrical connection between the two windings, but they are magnetically
connected. Therefore, the electrical power is transferred from the primary circuit to the secondary
circuit through mutual inductance. The induced emf in the primary and secondary windings depends
upon the rate of change of flux linkage that is (N dϕ/dt) (Farady’s Law)
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EMF Calculation
𝐸=𝑁
𝑑ϕ
,
𝑑𝑡
𝐸=𝑁
𝑑
(ϕ sin ω𝑡)
𝑑𝑡 𝑚
Faraday’s Law
𝐸 = ω𝑁 (ϕ𝑚 cos ω𝑡)
𝐸𝑚𝑎𝑥 = ω𝑁 ϕ𝑚
𝐸𝑟𝑚𝑠 =
ω𝑁 ϕ𝑚
√2
=
2𝜋𝑓𝑁ϕ𝑚
√2
= 4.44𝑓𝑁ϕ𝑚
E is induced electromotive force
N is number of turns
f is frequency
Φm is flux measured in Weber
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Example
An Ideal 25 KVA transformer has 500 turns on the primary winding and
40 turns on the secondary winding. The primary winding is connected
to a 3 KV, 50 Hz supply. Calculate
1.the primary & secondary currents on full load
2.The secondary voltage
3.The maximum core flux
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Power Transformers
Step-Up Transformer: Used to step the voltage up (as the current is proportionally
step down) as much higher level so that power can be transmitted up to hundreds of
kilometers while conductor size and losses are kept down within practical limits.
Step-Down Transformer: Used to step the voltage down (current step up) as much
lower level so that power can be distributed safely to consumers and loads.
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Transformer Parts
Transformer structure consists of three main parts:
1. Primary Winding
2. Secondary Winding
3. Iron Core
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Transformer Parts
1.
Tap Changer
2.
Main Tank & Conservative Tank
3.
LV & HV Bushings
4.
Steel Structure
5.
Insulation Paper
6.
Oil
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Dry Type Transformer
Operation theory is same as oil
immersed transformer
1.
Used for distribution side!
2.
Indoor application mostly
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Cooling Method in the transformer
Oil immersed Transformer:
Dry Type Transformer
◦ ONAN
◦ ONAF
◦ OFAF
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◦ AN
◦ AF
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Transformers
How to distinguish between the HV Side and LV side in the transformer:
1. Bushing Size
2. Symbols (Capital & Small Letter)
3. Winding (inside the transformer)
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Transformer Losses
Transformer
Losses
Load Loss
Copper
Losses (I2R)
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No-load Loss
Eddy
Current
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Hysteresis
Loss
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Transformer Nameplate
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Transformers in Oman
◦ The main rates used in Oman in the distribution system as Oman Electrical
standards (OES 5, 5A & 6) for oil immersed:
1. 33/11 KV, 1, 3, 6, 10 & 20 MVA Power Transformers
2. 11/0.433 KV, 100, 200, 315, 500 & 1000 KVA Distribution Transformers
3. 33/0.433 KV, 100, 200, 315, 500 & 1000 KVA Distribution Transformers
Dry type rating:
1. 11/0.433 KV, 500, 630, 1000, 1600 & 2000 KVA Distribution Transformers
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Other types of Transformer
Instrument Transformers
1.Current Transformer
2.Voltage (Potential) Transformer
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Switchgear
Break, isolate and connect the electrical circuits in normal situation or
abnormal situation (e.g.. Faults, short circuits)
Indoor MV switchgear
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Outdoor MV Circuit Breaker
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Indoor Switchgear Type
Air Insulated Switchgear
Gas Insulated Switchgear
Insulation medium of the
switchgear interior is Air
Insulation medium of the
switchgear interior is SF6 Gas
Used for Voltage 33 KV & below
Prefer for High Voltages
Excellent overview, simple
handling and easy access.
Compact, multi-component
assembly.
Large dimensions due to clearance
requirement and poor dielectric
strength of air.
High outage time, Diagnosis of
internal fault and rectifying takes
very long time
Less Cost
Expensive
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11 KV Indoor Switchgear
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Air Insulated Switchgear
Compartments
LV Compartment
Busbar Compartment
Circuit Breaker Compartment
Cable Compartment
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Circuit Breaker Types
Oil Circuit Breaker
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Vacuum Circuit Breaker
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Vacuum Interrupter
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Circuit Breaker type
Sulphur Hexafluoride SF6 Circuit
Breaker
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Air Break Switch & Drop out Fuse
Air Break Switch
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Drop out Fuse
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Over Head Line Components
1.
Conductors
2.
Steel Towers
3.
Wooden & Concrete Poles
4.
Steel Channels
5.
Insulators
6.
Stay Set Materials
7.
Surge Arrestors
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Bare Conductor Types
Aluminum Conductor Steel Reinforced (ACSR)
All Aluminum Alloy Conductor (AAAC)
All Aluminum Conductor (AAC)
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Insulators
Insulator are made of Porcelain or Glass or
Polymeric (silicone rubber) material
The cone shape disk insulators have two
functions:
1. Increase the flashover distance between the
tower and conductor.
2. When rain falls, the insulators are
automatically cleaned.
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Underground Cables & Accessories
•Power Cables
•Cable Terminations and Joints
•Cable tiles and Caution tape
•Connecters & lugs
•Electrical Conduits and trays
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Cables Construction
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Cable components, materials and function
Conductor
◦ Current carrying component within the cable. The large cross area of
conductor, the large current flows.
◦ Material:
1. Copper
2. Aluminum
◦ Conductor type:
1. Solid Circular
2. Stranded
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Cable components, materials and function
Insulation
 The primary insulation for the conductor. As the voltage is higher, the insulation thickness shall
be increased
 Material:
1. Poly vinyl chloride (PVC)
2. Cross Linked Polyethylene (XLPE)
3. Ethylene Propylene Rubber (EPR)
Armoring
 Provide mechanical protection during installation and service. Non magnetic material shall be used
for Single core
 Material:
 Galvanized Steel Wire
 Aluminum Wire
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Cable components, materials and function
Outer Sheath (serving)
 Final coating of the cable.
 Material:
1. Poly vinyl chloride (PVC)
2. Medium Density Polyethylene (MDPE)
3. Low smoke and fume (LSF/LSOH)
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Cable sizing
There are five steps to calculating the right size of cable for a static load:
1.Calculate the load current (IL)
2.Select the type and current rating of the overcurrent device and should be more than load current
by 125%.
3.Apply the relevant correction factors to obtain the corrected load current (IL corrected). Correction
factors are applied to situations which inhibit a cable from dissipating its heat caused by the normal
flow of current through it.
4.The current-carrying capacity of the cable is then selected from the appropriate table of the
standard. The selected value should be at least equal to or slightly greater than the corrected load
current (IL corrected).
5.Calculate the voltage drop to ensure that it is not excessive. As per OES 4 the voltage drop is 2.5%
𝑉. 𝐷 =
𝑉𝑡×𝐼𝐿×𝐿
1000
Where, Vt is Tabulated Voltage Drop (mV/A.m), IL is Loadcurrent & L is cable length
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Example
A 21 KW load is to be supplied at 415 V and the power factor 0.8 by A
Copper conductor insulated with XLPE, 8 meters in length. The cable is
clipped on the surface through an area with ambient temperature of
50°C.
Calculate the minimum cable size required.
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Consumer Units
Meters (Electromechanical and Electronics)
Cut-out Fuses
Miniature Circuit Breaker (MCB)
Molded Case Circuit Breaker (MCCB)
Earth Leakage Circuit Breaker (ELCB)
Residual Current Circuit Breaker (RCCB)
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Reference
1.
Transformer Basics, URL: https://www.electronicstutorials.ws/transformer/transformer-basics.html
2.
Basic electrical engineering science, Author: McKenzie & Hosie
3.
Nuhas Cables Catalouge
4.
)‫ محمود جيالني‬.‫المرجع في التركيبات والتصاميم الكهربائية (د‬
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