Intro + Per Unit

1. BEF 43303
BEF 43303
POWER SYSTEMS ANALYSIS
POWER SYSTEMS ANALYSIS
POWER SYSTEMS ANALYSIS
POWER SYSTEMS ANALYSIS
AND PROTECTION
AND PROTECTION
1
SEM I 2013/2014
SEM I 2013/2014

2. Course Overview
Course Overview
Synopsis:
The subject deals with topics related to the
power system analysis and protection: per-unit
power system analysis and protection: per unit
system, power flow analysis, analysis of balanced
and unbalanced faults, power system stability and
application power system control overcurrent
application, power system control, overcurrent
protection, differential protection and application,
distance protection and application. Overall, this
course focuses on analysis of power system and
course focuses on analysis of power system and
the protection schemes for power system
network.
2

3. Course Outcome
At the end of this course the student
At the end of this course the student
should be able to:
1. Analyze power system faults based on balance
1. Analyze power system faults based on balance
and unbalanced faults techniques. (PLO4-
CTPS-C4)
2. Demonstrate power flow analysis using related
software. (PLO3-CS-P4)
3 Express the suitable protection schemes based
3. Express the suitable protection schemes based
on power system requirement. (PLO11-SD-A3)
3

4. Lecture Plan
WEEKS CONTENTS
1 Per-Unit System
2 Power Flow Analysis
3 Analysis of Balanced Fault
4 Analysis of Unbalanced Fault
5 Power System Stability
5 Power System Stability
6 Application of Power System Stability
7 Load Frequency and Automatic Generation Controls
8 Reactive Power and Voltage Controls
9 Non-Directional Overcurrent and Earth Fault Protection
4

5. L t Pl
Lecture Plan
WEEK CONTENTS
10 Directional Overcurrent and Earth Fault Relay
11 Differential Protection Scheme
12 Differential Protection Application
13 Distance Protection Scheme
14 Distance Protection Application
Syllabus details
5

6. A t
Assessments
Oral 5 %
Oral 5 %
Test 1 15 % (week 4)
Test 2 15 % (week 8)
Test 2 15 % (week 8)
Assignment 15 %
Final exam 50 %
Final exam 50 %
Total 100%
WITH WISDOM WE EXPLORE 6

7. This Week Lecture Plan
This Week Lecture Plan
CHAPTER CONTENTS
1 Per-Unit System (3 Hours)
• Introduction
• Vectors
• Operators
• Convention Used for Voltage Direction
B Q titi d P U it S t
• Base Quantities and Per-Unit System
• Transferring Per-Unit Quantities from One Set
of Base Values
7

8. Representation of Electric Power System
One-Line Diagram (OLD)
 Definition:
◦ A diagram showing the interconnection of
various components of a balanced three phase
various components of a balanced three- phase
power system by standard symbols on a single
phase basis.
R
3-phase system One line diagram
R
Y
B
=
8
3-phase system One-line diagram

9. S
Advantages of OLD
 Simplicity
 1-Φ represents all 3-Φs of the balanced system
 The equivalent circuits of the components are replaced
 The equivalent circuits of the components are replaced
by their standard symbols
 The completion of the circuit through the neutral is
p g
omitted
9

10. Symbols for One-Line Diagram
Machine or rotating armature
Or Machine or rotating armature
2-winding power transformer
Or
Or
3-winding power transformer
Or
Power Circuit Breaker (CB)
Load
Or
Power Circuit Breaker (CB)
(oil/liquid) (OCB)
Air CB (ACB)
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skip next page

11. Busbar
3-phase, 3-wire
delta connection
Transmission line
delta connection
3-phase wye,
neutral ungrounded
Fuse
neutral ungrounded
3-phase wye,
neutral grounded
A
Current
transformer
(CT)
neutral grounded
Ammeter
V
or Potential
transformer
(PT or VT)
Ammeter
Voltmeter
11
( )

12. Impedance (Z) and Reactance (X) Diagram
 Impedance (Z = R + jX) diagram is converted from
OLD showing the equivalent circuit of each
t f th t It i d d t l l t th
component of the system. It is needed to calculate the
performance of a system under normal and abnormal
conditions i.e. load conditions (Load Flow (LF) studies)
th f f lt/ h t i it (f lt
or upon the occurrence of a fault/short circuit (fault
analysis studies).
 Reactance (jX) diagram is further simplified from Z
diagram by omitting all static loads, all Rs, the
magnetizing I (Im) of each transformer, and the
capacitance (C) of the transmission line. It is applied
only to fault calculations, and not to LF studies.
 Z and X diagrams sometimes called the Positive-
sequence diagram.
sequence diagram.
12

13. Z and X Diagrams
Example: OLD of an EPS
Load B
T2
T1
Load B
Load A
WITH WISDOM WE EXPLORE 13

14. Z diagram corresponding to the OLD
g p g
E1 E2 E3
Gen.
3
Load
B
Transformer
T2
Transmission
Line
Transformer
T1
Load
A
Generators
1 and 2
14

15. X diagram corresponding to the OLD
E1 E2 E1
Generators
1 and 2
Transmission
Line
Transformer
T2
Gen.
3
Transformer
T1
15

16. Per - unit (P.U) Representation
 Common quantities used in power system analysis (PSA) are
voltage (V) (in kV), current (I) (in kA), voltamperes (in kVA or
MVA), and impedance (in Ω). It is very cumbersome to convert
I t diff t lt l l i PS h i t V
Is to different voltage levels in a PS having two or more V
levels.
 P.U. representation is introduced in such a way that the various
h i l titi d d i l f ti
physical quantities are expressed as a decimal fraction or
multiples of base quantities and is defined as:
actual quantity
Quantity in per-unit
base value quantity

Example:
For instance, if a base voltage of 275 kV is chosen, actual
voltages of 247 5 kV 275 kV and 288 75 kV becomes 0 90
16
voltages of 247.5 kV, 275 kV, and 288.75 kV becomes 0.90,
1.00, and 1.05 per-unit.

17. For 1-Φ systems: The formula relates the various quantities
for 1-Φ system:
(1- )
( )
base kVA (in kVA)
Base I (in A)
base V (in kV)
LN


LN
1φ
V
voltage
base
kV
voltage,
base
kVA
base
A
current,
Base 
( )
2
( )
base V (in V)
Base Z (in ohms)
base I (in A)
(base V ) (in kV)
B Z (i h )
LN
LN

LN
A
current,
base
V
voltage,
base
impedance
Base 
2
LN )
kV
voltage,
(base
impedance
Base  ( )
(1- )
(1- ) (1- )
Base Z (in ohms)
base MVA (in MVA)
Base power (in kW) base kVA (
LN

 

 in kVA)
Base power (in MW) base MVA

1φ
MVA
base
impedance
Base 
Base P, MW1φ = Base MVA1φ
B Q MVAR1 B MVA1 (3- ) (1- )
Base power (in MW) base MVA
 

 
)
( l
Base Q, MVAR1φ = Base MVA1φ
 
 2
.
)
(
base
base
base
u
p
V
VA
Z
Z
actual
Z
Z 




18.  F 3 Φ t
 For 3-Φ systems:
The formula relates the various quantities for 3-Φ system:
(3- )
( )
base kVA
Base I (in A)
3 X base V (in kV)
LL


LL
3φ
kV
voltage,
base
X
3
kVA
base
A
current,
Base 
( )
2
( )
(3- )
(base V ) (in kV)
Base Z (in ohms)
base MVA
Base power (in kW) base kVA
LL
LL


3φ
2
LL
MVA
base
)
kV
voltage,
(base
impedance
Base 
(3- ) (3- )
(3- ) (3- )
Base power (in kW) base kVA
Base power (in MW) base MVA
 
 


3φ
3φ
3φ
3φ
MVA
base
MW
power,
Base
kVA
base
kW
power,
Base


 
 2
)
( base
u
p
VA
Z
actual
Z
Z 



18
 2
.
base
base
u
p
V
Z


19. Example:
The base impedance and base voltage for a given
power system are 10Ω and 400V, respectively.
Calculate the base kVA and the base current.
Calculate the base kVA and the base current.
Solution:
F m Ohm’ l
A
40
400

From Ohm’s law,
Base current = A
40
10
kVA
X
16
400
40

Base kVA =
Base current
19
kVA
16
1000

Base kVA
19

20. Example:
The base current and the base voltage of a
The base current and the base voltage of a
345kV system are chosen to be 3000A and 300
kV, respectively. Determine the base impedance
and the per unit voltage for the system
and the per-unit voltage for the system.
Solution:
Solution:
Base impedance = 


100
10
300 3
345
Base impedance 
100
3000
1.15pu
300
345

Per-unit voltage =
20

21. Example:
A 3-Φ, Y-connected system is rated at 100 MVA and 132 kV.
Express 80 MVA of 3-Φ apparent power (S) as a p.u. value
referred to:
(a) the 3-Φ system MVA as base and
(b) the 1-Φ system MVA as base.
(a) For the 3-Φ base,
Base MVA = 100 MVA = 1 p.u.
and Base kVLL = 132 kV = 1 p.u.
LL p
so p.u. MVA = 80/100 = 0.8 p.u.
(b) For the 1-Φ base,
Base MVA = 100/3 MVA = 33.33 MVA = 1 p.u.
and Base kV = 132/√3 = 76 21 kV = 1 p u
21
and Base kV = 132/√3 = 76.21 kV = 1 p.u.
so p.u. MVA = (1/3)*(80/33.333) = 0.8 p.u.

22. Changing the Base of P.U. Quantities
 The Z of individual generators and transformers are
generally in terms of % or p.u. quantities based on
their ratings given by manufacturer.
their ratings given by manufacturer.
 For PSA, all Zs must be expressed in p.u. on a
common system base. Thus, it is necessary to
f
convert the p.u. Zs from one base to another
(common base, for example: 100 MVA).
 P.U. Z of a circuit
element
2
(actual Z in ) * (base MVA)
(base V) in kV


 The equation shows that p.u. Z is directly proportional
to the base MVA and inversely proportional to the
22
square of the base V.

23.  Therefore, to change from old base p.u. Z to new
2
   
, g p
base p.u. Z, the following equation applies:
2
old new
new old
new old
base kV base MVA
P.U. Z P.U. Z
base kV base MVA
   
    
   
Example 1:
The reactance X” of a generator is given as
0.20 p.u. based on the generator’s nameplate
rating of 13.2 kV, 30 MVA. The base for
calculations is 13.8 kV, 50 MVA. Find X” on this
calculations is 13.8 kV, 50 MVA. Find X on this
new base.
2
13 2 50
   
23
13.2 50
x" 0.20 0.306 p.u.
13.8 30
   
 
   
   

24. Example 2:
Calculate the p.u. impedance of a synchronous
motor rated 200 kVA, 13.2 kV and having reactance
of 50 ohm
of 50 ohm
Example 3:
The primary and secondary sides of a single
phase 1 MVA, 4kV/2kV transformer have a
p ,
leakage reactance of 2 ohm each. Find p.u. X if
the transformer referred to primary and secondary
side
side.
24

25. A l i f PS bl l i lifi d b
Analysis of PS problems are greatly simplified by
using single-line Z diagram in which system
parameters are expressed in p.u. The steps to
compute p.u. values are summarized as follows:
compute p.u. values are summarized as follows:
 Step 1: Select a common volt-ampere base for
the entire power system and a voltage base for one
part of the system
part of the system.
 Step 2: Compute voltage bases for all parts of the
PS by correlating the transformation ratios of the
y g
transformer banks.
25

26.  Step 3: Convert p.u. values (which is provided by the
Step 3: Convert p.u. values (which is provided by the
nameplate of the equipment) to the common system
volt-ampere base and the applicable voltage base. In
case the parameters are provided in actual ohmic
case the parameters are provided in actual ohmic
values, compute base Z for the part of the PS in which
the equipment is connected and calculate the p.u.
values.
values.
 Step 4: Draw a single-line diagram of the PS indicating
values of all parameters in p u Proceed to analyze the
values of all parameters in p.u. Proceed to analyze the
PS.
 Step 5: Convert to actual values wherever required
 Step 5: Convert to actual values wherever required.
26

27. Example 4:
Draw the reactance diagram of system shown
below. Assume reactance for the transmission line is
60 ohm and select the generator rating as base in
60 ohm and select the generator rating as base in
the generator circuit
6.6/66 kV
27

28. Example 5:
Example 5:
A 30 MVA 13.8 kV 3-Φ generator has a sub-transient
reactance (Xd’’) of 15%. The generator supplies two
motors over a tr. line having transformers at both
motors over a tr. line having transformers at both
ends, as shown in OLD below. The motors have rated
inputs of 20 MVA and 10 MVA, both 12.5 kV with x” =
20%. The 3-Φ transformer T1 is rated 35 MVA, 13.2/116
,
(∆/Y) kV with leakage reactance (Xl) of 10%. 3-Φ
transformer T2 is rated at 10 MVA, 116/12.5 (Y/∆) kV
with Xl of 10%. Series X of the tr. line is 80 Ω. Draw the
X diagram with all Xs marked in p.u. Select the
generator rating as base in the generator circuit.
28