ELE-B7 Power Systems Engineering
Load Flow (Gauss-Seidel Method )
Dr. Ramadan El-Shatshat
Classification of buses:
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Dr. A.M. Gaouda
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Different buses at the network can be classified as:
1.
2.
3.
PQ bus
The Load Buses (PQ bus)
The Generator Bus (PV bus)
The Slack or Swing Bus
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PQ bus
Slack
Bus
PV bus
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PV bus
PV bus
PV bus
1.
PV bus
PQ bus
The Load Buses (PQ bus)
A non-generator bus. The active and reactive powers are specified at this
bus. The voltage magnitude and phase angle are unknown.
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Pig and Qig
Bus i
Pid and Qid
Pi and Qi are known & | Vi | and δ i are unknown
Generators Power:
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Delivered Power:
Pig = 0 and Qig = 0
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Pid and Qid are known
In practice, the load real power is known from measurement, load forecasting
or historical record and the reactive power is assumed based on 0.85 pf.
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Pi ,sch = Pig − Pid = − Pid
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and
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Qi ,sch = Qig − Qid = −Dr.UAE
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2.
The Generator Bus (PV bus)
The bus is also known as “Voltage controlled bus” because
the voltage magnitude can be kept constant. At this bus the
net active power and the voltage magnitude are specified.
The reactive power and the voltage phase angle are
unknown.
Pi and | Vi | are known & Qi and δ i are unknown
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Dr. A.M. Gaouda
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G1
Bus i
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Pi,sch
| Vi |
NOTE: There are certain buses without generators may have voltage controlled
capability. At these buses the real power generation is zero.
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3.
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The Slack or Swing Bus
Because the system losses are not known precisely before completing the
power flow solution, it is not possible to specify the real power injected at
every bus. Hence, The real power of one of the generator buses is allowed to
swing. The swing bus supplies the slack between the scheduled real power
generation and the sum of all loads and system losses. The voltage angle of the
slack bus serves as a reference, δ i = 0
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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| Vi | and δ i are known & Pi and Qi are unknown
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NOTE:
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Real Power losses = Total generation − Total load
N
PL =
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N
N
∑P −∑P = ∑P
gi
i =1
di
i =1
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i
i =1
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In the load flow problem, we select the slack bus at which the
power Pg is not scheduled.
After solving the load flow problem, the difference (Slack) between
the total specified power (P) going into the system at all other buses
and the total output (P) plus the losses (I2R) are assigned to the
slack bus.
For this reason a generator bus must be selected as a slack bus.
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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Voltage of the swing bus is selected as a reference. Generally, the
bus of the largest generator is selected as swing bus and numbered
as bus 1.
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Solution of Non-Linear Equations
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The two load flow equations are:
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n
PDr.i A.M.
=|Gaouda
Vi |
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Qi =| Vi |
∑| Yip || V p | cos ( δ i − δ p − γ ip )
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p =1
n
∑| Yip || V p | sin ( δ i − δ p − γ ip )
p =1
These equations provide the calculated value of net real power and net reactive
power entering bus ‘i’. The equations are non-linear and only a numerical
solution is possible. There are different methods could be implemented to solve
these equations. Among those is the Gauss-Seidel method.
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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Gauss-Seidel Method
Consider a system of non-linear equations having “n” unknowns x1 , x 2 , .........., x n
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f 1 ( x1 , x 2 , .........., x n )
f 2 ( x1 , x 2 , .........., x n )
.. ..
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.. ..Dr. A.M...Gaouda..
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f n ( x1 , x 2 ,ELEC572,
..........
n )
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Rearranging, then
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x i = f i ( x1 , x 2 , .........., x n )
1≤ i ≤ n
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Eq. 1
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Assuming initial values,
x1o , x 2o , .........., x no
Substituting the initial values in Eq. 1, then
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first iteration
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first variable
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x = f ( x , x , .........., x )
1
1
1
1
o
1
o
2
o
n
x 21 = f 21 ( x11 , x 2o , .........., x no )
x = f ( x , x , x .........., x no )
1
3
1
3
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1
1
1
2
o
3
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Or in general
All values are initial values
o
o
xDr.1o A.M.
, xGaouda
2 , .........., x n
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x 1 = x 11 from previous step
and all other values are
initial values
x 2o , .........., x no
x 1 = x 11 & x 2 = x 21
and
o
x 3 , .........., x no
x i1 = f i 1 ( x11 , x 21 , ....., x io ,....., x no )
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Where
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x
x
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is the first approximation
using
the
initial
assumed
values.
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i
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i
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The kth approximation of
K
th
iteration
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xi
is:
x = f i ( x , x , .....x
i th variable
k
i
1
k
1
k
2
k
i −1
The changes in the magnitude of each variable
the previous iteration is:
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,x
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k −1
i
x
,x
k −1
i+1
....., x
k −1
n
)
k
k −1
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from
its
value
at
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x
Δ x i = x ik − x ik − 1
If Δ x i < ε then the solution has converged.
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Where,
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ε
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is a small value ( for exmple : ε = 0.001 )
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EXAMPLE:
For the following equation, find an accurate value for x up to 5 decimal places.
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2 x − log( x ) = 7
SOLUTION:
Using Gauss-Seidel
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x = 0.5 ( 7 + log x )
x =1
x 1 = 0.5( 7 + log 1 ) = 3.5
x = 3.5
x 2 = 0.5 ( 7 + log 3.5 ) = 3.772034
x 2 = 3.772034
x 3 = 0.5( 7 + log 3.772034 ) = 3.788287
o
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1
x 3 = 3.788287
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1 st iteration
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2 nd iteration
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x 4 = 0.5( 7 + log 3.788287 ) = 3.789221
x 5 = 3.789274
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x = 3.789278
6
ε = 0.000004
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EXAMPLE:
For the following equations, find an x and y after 4 iterations.
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x = 0.7 sin x + 0.2 cos y
&
y = 0.7 cos x − 0.2 sin y
SOLUTION:
Using Gauss-Seidel, assuming initial values
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x o = y o = 0.5 ( rad )
x 1 = 0.7 sin x o + 0.2 cos y o
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x 1 = 0.7 sin 0.5 + 0.2 cos 0.5
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x 1 = 0.51111
y 1 = 0.7 cos 0.51111 − 0.2 sin 0.5
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y = 0.51465
1
x 2 = 0.516497
x 3 = 0.520211
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x = 0.522520
4
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y 2 = 0.510241
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y 3 = 0.509722
y 4 = 0.509007
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Gauss-Seidel Method for Load Flow Analysis
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Advantages
1.
2.
3.
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Simplicity
Small computer memory requirement
Less computational time per iteration
Disadvantages
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1.
2.
3.
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Slow rate of convergence, and therefore large number of iterations.
Increase in the number of iterations as the number of system buses
increases.
The speed of convergence is affected by the selected slack bus.
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I - G-S Method when PV buses are absent
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Assuming a power system in which the voltage controlled buses are absent. If the
system has n buses, then; one bus will be considered as a slack bus and the other n-1
buses are load buses (PQ-buses).
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For the Slack or Swing Bus:
| Vi | and δ i = 0 are known & Pi and Qi are unknown
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The swing bus voltage is taken as a reference. It is voltage magnitude is known and
its phase shift angle is set equal to zero.
For (n-1) Load Buses (PQ bus):
Pi and Qi are known & | Vi | and δ i are unknown
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Using Gauss-Seidel Method, we assume the initial values for the magnitude and
phase shift angle of (n-1) buses. These values are updated at each iteration.
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For an ‘n’ bus system
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I bus = Ybus Vbus
….. Eq. 1
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For the ith bus of an ‘n’ bus system, the current
entering this bus is:
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
I i = Yi1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn
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….. Eq. 2
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n
I i = Yii Vi + ∑ Yip V p
….. Eq. 3
p=1
p≠ i
⎛
⎞
n
⎟
1 ⎜
Vi =
⎜ I i − ∑ Yip V p ⎟
Yii ⎜
p=1
⎟
≠
p
i
⎝
⎠
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….. Eq. 4
In power systems, power is known rather than currents. The complex power
injected into the ith bus is:
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Si = Pi + jQi = Vi I *i
….. Eq. 5
S i* = Vi* I i
….. Eq. 6
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OR
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Ii =
Substituting in Eq. 4
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Pi − jQi
….. Eq. 7
Vi*
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⎞
⎛
n
⎟
⎜
1 Pi − jQi
Vi =
− ∑ Yip V p ⎟
⎜
*
Yii ⎜ Vi
p=1
⎟
p≠ i
⎠
⎝
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Dr. A.M. Gaouda
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….. Eq. 8
Rearranging in GS
method
“Vi” is moved to
the left hand side
of the equation.
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Since bus 1 is the slack bus “reference”, then Vi represents n-1 set of
equations for i= 2, 3, …., n. These equations will be solved using G-S
method for the unknowns V2, V3, …..Vn.
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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NOTES:
1.
Eq. 8 can be written as:
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NOTE
The values for P and Q
are the scheduled
values for PQ Bus.
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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n Y
1 Pi − jQi
ip
Vi = *
− ∑ Vp
Vi
Yii
p = 1 Yii
….. Eq. 9
n
Ki
Vi = * − ∑ Lip V p
Vi p = 1
….. Eq. 10
p≠ i
P − jQi
Ki = i
Yii
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p≠ i
Lip =
and
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Yip
Yii
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The values for Ki and Lip are computed once in the beginning and used in every
iteration.
2.
The voltages at all the buses in a power system are close to 1.0 pu.
Therefore, we can start the G-S iteration process assuming initial values for the
voltages equal to 1.0 and making zero angle.
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V2o = V3o = ...... Vno = 1∠0
3.
At each step in the iteration process use the most updated values for the
voltages to compute the new values for the bus voltages.
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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n
Ki
Vi = * − ∑ Lip V p
Vi
p=1
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….. Eq. 11
p≠ i
Ki
Vi = * −
Vi
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Therefore, for the
Vi k + 1
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(kth+1)
i −1
∑L
p=1
ip
Vp −
n Dr. A.M. Gaouda
∑L
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p= i + 1
ip
Vp
iteration,
Ki
=
−
k *
( Vi )
i −1
( k +1 )
L
V
∑ ip p −
p=1
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….. Eq. 12
The most updated voltage
values are from the
previous iteration
n
k
L
V
∑ ip p
p= i + 1
….. Eq. 13
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The most updated voltage
values are from the same
iteration
for i = 1 ,2 ,..........n
The iteration process is continuous till the convergence occurs, i.e.;
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| ΔVi k + 1 | = | Vi k + 1 | − | Vi k | ⟨ ε
….. Eq. 14
for i = 1 ,2 ,..........n
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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Dr. A.M. Gaouda
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4. The current and complex power at
ith
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bus are:
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1
2
3
I i = Yi1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn
And
Or
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Si = Pi + jQi = Vi I *i
Linear
Network
Ii
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
i
Vi
n
Reference
S i* = Pi − Qi = Vi* I i
Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Pi = Re{ Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )}
Qi = − Im{ Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )}
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The two equations are known as the rectangular form of the load flow equations.
They provide the calculated value of net real power and net reactive power
entering bus ‘i’.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
EXAMPLE:
For the three bus system. Write the expression
for the bus voltages using GS method.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
SOLUTION:
The system contains 3 buses, (n=3).
i- Select bus 1 as a slack bus “reference”.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
G1
2
1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
P2 , Q 2
| V1 | = 1 and δ 1 = 0
ii- Buses 2 and 3 are load buses.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
| V1 | = 1
δ1 = 0
3
P3 , Q3
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
P2 , P3 , Q2 and Q3 are known
V2 ,V3 ,δ 2 and δ 3 are unknown
3
K2
V2 = * − ∑ L2 p V p
V2 p = 1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
p≠ 2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3 Y
1 P2 − jQ2
2p
V2 = *
−∑
Vp
V2
Y22
p = 1 Y22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
p≠ 2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
NOTE
----------The load flow problem
is solve when the
mismatch is equal to
zero.
Calculated values
=
Dr. A.M. Gaouda
UAE University
Scheduled values
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y23 ⎤
1 P2 − jQ2 ⎡ Y21
V3 ⎥
V2 = *
− ⎢ V1 +
Y22 ⎦
V2
Y22
⎣ Y22
….. Eq. 15
Y32 ⎤
1 P3 − jQ3 ⎡ Y31
− ⎢ V1 +
V2 ⎥
V3 = *
Y33 ⎦
V3
Y33
⎣ Y33
….. Eq. 16
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
NOTE
The values for P and Q
are the scheduled
values
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Using GS method, select the initial values for the unknowns as:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V = V = 1∠0
o
2
o
3
Start the first iteration
The most updated voltage
value is the initial value
Y23 o ⎤
1 P2 − jQ2 ⎡ Y21
V =
V3 ⎥
− ⎢ V1 +
o *
( V2 )
Y22
Y22
⎣ Y22
⎦
1
2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
….. Eq. 17
The most updated voltage
value is from this iteration
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y32 1 ⎤
1 P3 − jQ3 ⎡ Y31
− ⎢ V1 +
V =
V2 ⎥
o *
( V3 )
Y33
Y33 ⎦
⎣ Y33
1
3
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
….. Eq. 18
Dr. A.M. Gaouda
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ELEC572, 04/05
Dr. A.M. Gaouda
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ELEC572, 04/05
Start the second iteration
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y23 1 ⎤
1 P2 − jQ2 ⎡ Y21
V3 ⎥
V =
− ⎢ V1 +
1 *
Y22 ⎦
( V2 )
Y22
⎣ Y22
2
2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y32 2 ⎤
1 P3 − jQ3 ⎡ Y31
− ⎢ V1 +
V2 ⎥
V =
1 *
Y33
( V3 )
Y33
⎣ Y33
⎦
2
3
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
| ΔV22 | = | V22 | − | V21 | ⟨ ε
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
| ΔV32 | = | V32 | − | V31 | ⟨ ε
If Eqs. 21, 22 are not satisfied then start a new iteration.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The most updated voltage
value is from this iteration
….. Eq. 20
for i = 1 ,2 ,..........n
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
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ELEC572, 04/05
….. Eq. 19
Dr. A.M. Gaouda
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ELEC572, 04/05
Compare the results for convergence
| ΔVi k + 1 | = | Vi k + 1 | − | Vi k | ⟨ ε
The most updated voltage
value is from previous
iteration
….. Eq. 21
….. Eq. 22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
EXAMPLE
1:
UAE University
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
ELEC572, 04/05
For the system shown in the figure, the line impedances are as indicated in per
unit on 100MVA base.
A. Using Gauss-Seidel method find the bus voltages after 7 iterations.
B. Using the bus voltages find the Slack bus real and reactive power.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
2
0.02 + j 0.04 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
G
0.01 + j 0.03 pu
110.2
MVAR
V1 = 1.05 ∠0 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
138.6
MW
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
0.0125 + j 0.025 pu
o
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
256.6
MW
45.2
MVAR
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
Formulation of the Bus Admittance Matrix
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
0.02 + j 0.04 pu
y12 =
1
= 10 − j 20 = y 21
0.02 + j 0.04 Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
y13 =
= 10 − j 30 = y 31
0.01 + j 0.03
y 23 =
Ybus
0.01 + j 0.03 pu
V1 = 1.05 ∠0 o pu
138.6
⎡Y11 Y12 Y13 ⎤
⎡ y12 + y13
= ⎢⎢Y21 Y22 Y23 ⎥⎥ = ⎢ − y 21
⎢
⎢Y31 Y32 Y33 ⎥
⎢⎣ − y 31
⎢
⎥
Ybus =
0.0125 + j 0.025 pu
110.2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
MW
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
MW
MVAR
3
1
= 16 − j 32 = y 32
0.0125 + j 0.025
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
G
256.6
45.2
MVAR
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
− y12
y 21 + y 23
− y 32
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤
⎢ − 10 + j 20 26 − j 52 − 16 + j 32 ⎥
Dr. A.M. Gaouda
⎢
⎥
UAE University
ELEC572, 04/05
⎢⎣ − 10 + j 30 − 16 + j 32 26 − j 62 ⎥⎦
− y13 ⎤
− y 23 ⎥⎥
y 31 + y 32 ⎥⎦
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Classification of buses:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Bus 1: Slack Bus
0.02 + j 0.04 pu
G
V1 = 1.05∠0 pu
o
V1 = 1.05∠0 o pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
0.01 + j 0.03 pu
138.6
MW
V2 ,V3 ,δ 2 and δ 3 are unknown
P2 ,d = 256.6 MW
P3 ,d = 138.6 MW
110.2
MVAR
3
P2 , P3 , Q2 and Q3 are known
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
256.6
MW
0.0125 + j 0.025 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Buses 2 and 3: Load Buses (PQ bus)
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q2 ,d = 110.2 MVAR
Q3 ,d = 45.2 MVAR
45.2
MVAR
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Pig and Qig
G
Pi ,sch = Pgi − Pdi
&
Dr. A.M. Gaouda
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ELEC572, 04/05
Qi ,sch = Q gi − Qdi
Dr. A.M. Gaouda
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ELEC572, 04/05
S i ,sch = Pi ,sch + jQi ,sch
Dr. A.M. Gaouda
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ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d )
Bus i
Pid and Qid
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
A.M. Gaouda
( P2 , g − P2 ,d ) + j ( Q2 , g − Q2Dr.UAE
,d )University
=
pu
ELEC572, 04/05
Base MVA
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 2 ,sch
S 2 ,sch
( 0 − 256.6 ) + j ( 0 − 110.2 )
Dr. A.M. Gaouda
=
pu
UAE University
100
MVA
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Reminder
The bus admittance matrix
is
⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤
⎥
⎢
26 − j 52 − 16 + j 32 ⎥
Dr. A.M. Gaouda⎢ − 10 + j 20
UAE University ⎢
− 10 + j 30 − 16 + j 32 26 - j 62 ⎦⎥
ELEC572, 04/05⎣
S 2 ,sch = − 2.566 − j 1.102 pu
S 3 ,sch = − 1.386 − j 0.452 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Using GS method, select the initial values for the unknowns as:
Start the first iteration
V2o = V3o = 1∠0
Y23 o ⎤
1 P2 ,sch − jQ2 ,sch ⎡ Y21
1
V2 =
V3 ⎥
− ⎢ V1 +
o *
( V2 )
Y22
Y22
Y22
⎣
⎦
− jQ
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S *i ,sch = Pi ,sch
1
Dr. A.M. GaoudaV 2
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Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
i ,sch
⎤
− 2.566 + j 1.102 ⎡ − 10 + j 20
− 16 + j 32
−⎢
=
1.05 +
1.0 ⎥
*
26 − j 52
26 − j 52
( 1.0 )
⎦
⎣ 26 − j 52
1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
OR, to simplify the calculations, we have:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Reminder
The bus admittance matrix
is
Yip
1 Pi − jQi
− ∑ Vp
Vi = *
Vi
Yii
p = 1 Yii
n
p≠ i
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
n
K2
V2 = * − ∑ L2 p V p
V2 p = 1
[
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
p≠ 2
K2
o
V =
−
L
V
+
L
V
21 1
23 3
( V2o )*
1
2
⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤
⎥
⎢− +
⎢ 10 j 20 26 − j 52 − 16 + j 32 ⎥
⎢⎣ − 10 + j 30 − 16 + j 32 26 - j 62 ⎥⎦
]
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The values for Ki and Lip are computed once in the beginning and used in every
iteration.
Y21
L21 =
Y22
P − jQ2
K2 = 2
Y22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L21 = -0.3846
K2 = -0.0367 - j0.031
V1 = 1.05 ∠0 o pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
and
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V = 0.9825 − j 0.310
1
2
and
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y23
L23 =
Y22
L23 = -0.6154
V3o = 1∠0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
[
K3
1
L
V
L
V
V31 =
−
+
31 1
32 2
( V3o )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
P3 − jQ3
K3 =
Y33
and
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V1 = 1.05 ∠0 o pu
and
Y32
Y33
V21 = 0.9825 − j 0.310
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
K 2 , K 3 , L21 , L23 , L31 , L32 constants will be the same.
K2
1
V =
−
L
V
+
L
V
21 1
23 3
( V21 )*
[
]
[
]
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L32 =
L32 = -0.5310 - 0.0354i
UAE
University − j 0.0353
V31 = ELEC572,
1.0011
04/05
Start the second iteration
V32 =
and
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
2
Y31
Y33
L31 = - 0.4690 + 0.0354i
K3 = -0.0142 - j0.0164
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L31 =
]
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
K3
2
−
L
V
+
L
V
31 1
32 2
( V31 )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
VDr.22A.M.=Gaouda
0.9816 − j 0.0520
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V32 = 1.0008 − j0.0459
Dr. A.M. Gaouda
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ELEC572, 04/05
Start the third iteration
Dr. A.M. Gaouda
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ELEC572, 04/05
V23 =
Dr. A.M. Gaouda
UAE University
K 2 , K 3 ,Dr.UAE
LA.M.
,Gaouda
L23 , L31 , L32 constants will be the same.ELEC572,
21
04/05
University
[
ELEC572, 04/05
K2
2
L
V
L
V
−
+
21 1
23 3
( V22 )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
]
[
K3
3
V =
−
L
V
+
L
V
31 1
32 2
( V32 )*
3
3
Start the fourth iteration
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
= 1.0004 - j0.0488
K 2 , K 3 ,Dr.LA.M.
L23 , L31 , L32 constants will be
the same.
21 ,Gaouda
Dr. A.M. Gaouda
[
UAE University
ELEC572, 04/05
]
V24 =
K2
3
−
L
V
+
L
V
21 1
23 3
( V23 )*
V34 =
K3
4
−
L
V
+
L
V
31 1
32 2
( V33 )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
]
= 0.9808 - j0.0578
[
]
UAE University
ELEC572, 04/05
= 0.9803 - j0.0594
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
= 1.0002 - j0.0497
After 7 iterations,
V27 = 0.9800 − j0.0600
= 0.98183∠ − 3.5035 o pu
Dr. A.M. Gaouda
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
UAE University
ELEC572, 04/05
V = 1.0000 − j 0.0500 = 0.00125∠ − 2.8624 pu
7
3
o
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
B. Using the bus voltages find
the Slack bus real and reactive power.
UAE University
Dr. A.M. Gaouda
ELEC572, 04/05
UAE University
ELEC572, 04/05
1
2
0.02 + j 0.04 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
G
0.01 + j 0.03 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
256.6
MW
0.0125 + j 0.025 pu
110.2
MVAR
V1 = 1.05∠0 o pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
138.6
MW
45.2
MVAR
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V1 = 1.05 + j 0.0 o pu
V2 = 0.9800 − j0.0600 = 0.98183∠ − 3.5035 o pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
o
Dr. A.M.=
Gaouda
∠
−
V3 = 1.0000 − j 0.0500
0
.
00125
2
.
8624
pu
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Using the rectangular form of the load flow equations,
then the net active and reactive powers at 1th bus are:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Pi = Re{ V1* ( Y11 V1 + Y12 V2 + Y13 V3 )}
Reminder
The bus admittance matrix
is
⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤
⎥
⎢
26 − j 52 − 16 + j 32 ⎥
Dr. A.M. Gaouda⎢ − 10 + j 20
UAE University ⎢
− 10 + j 30 − 16 + j 32 26 - j 62 ⎦⎥
ELEC572, 04/05⎣
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qi = − Im{ V1* ( Y11 V1 + Y12 V2 + Y13 V3 )}
P1 − jQ1 = V1* ( Y11 V1 + Y12 V2 + Y13 V3 )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Pi − jQi =
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
4.0938 - j1.8894
1
P1 = 4.0938 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
G
Dr. A.M. Gaouda
UAE University
0.01 + j 004/05
.03 pu
ELEC572,
P1 = 409.38 MVA
256.6
MW
0.0125 + j 0.025 pu
110.2
V1 = 1.05∠0 o pu
Base MVA=100
Q1 = 188.94 MVA
2
0.02 + j 0.04 pu
Q1 = 1.8894 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
MVAR
3
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
138.6
MW
45.2
MVAR
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Problem 1:
The line impedances are as indicated in per unit on 100MVA base. Using GaussSeidel method:
1. Classify each bus
2. find bus admittance matrix
2. find bus 2 voltage after the first iteration.
3. find bus 1 real and reactive power.
NOTE: select the initial value for bus 2 voltage as:
V2o = 1∠ − 22.0169 o
SG 2 =
0.25 + j 0.6459 pu
G1
1
S D1
G2
V1 = 1.0∠0 o pu
V2
Z L = j 0.5 pu
2
S D2 =
1.0 + j 0.5 pu
Ybus =
-j2.0
j2.0
j2.0
-j2.0
S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d )
S 2 ,sch = - 0.7500 + j0.1459 pu
Using GS method, select the initial values for the unknowns as:
V1 = 1.0∠0 o pu
V2o = 1∠ − 22.0169 o
Start the first iteration
V21
=
1
P2 ,sch − jQ2 ,sch
( V2o )*
Y22
⎡ Y21
−⎢
V1
⎢⎣ Y22
⎤
⎥
⎥⎦
V21 = 1∠ − 22.0238 o = 0.9271 - j0.3749
Power at bus 1
P1 − jQ1 = V1* ( Y11 V1 + Y12 V2 )
P1 + jQ1 = 0.7500 +j0.1459
II. Modifying G-S Method when PV buses are present
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Assuming a power system has n buses, then; one bus will be considered as a
slack bus and the other buses are load buses (PQ-buses) and voltage controlled
buses (PV-buses). Let the system buses be numbered as:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
i=1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Salck bus
i = 2 , 3 , ..... , m
PV − buses
i = m + 1 , m + 2 , ...... , n
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
PQ − buses
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
For the voltage controlled buses,
Pi and | Vi | are known & Qi and δ i are unknown
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
| Vi | = | Vi | Specified
Qi ,min ⟨ Qi ⟨ Qi ,max
….. Eq. 23
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
….. Eq. 24
The second requirement for the voltage controlled bus may be violated if the
bus voltage becomes too high or too small. It is to be noted that we can control
the bus voltage by controlling the bus reactive power.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Therefore, during any iteration, if the PV-bus reactive power
violates its limits then set it according to the following rule.
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qi ⟩ Qi ,max
set Qi = Qi ,max
Qi ⟨ Qi ,min
set Qi = Qi ,min
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qi ,min ⟨ Qi ⟨ Qi ,max
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
NOTE
For PQ − bus
And treat this bus as PQ-bus.
Pi and Qi are known
& | Vi | and δi are unknown
Load flow solution when PV buses are present
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
a. Calculate Qi
In the polar form,
n
Qi =| Vi |
For the
(kth+1)
iteration,
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
( k +1 )
i
Q
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
= | Vi |speci
+ | Vi |speci
∑| Yip || V p | sin ( δ i − δ p − γ ip )
p =1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
i −1
( k +1 )
(k )
( k +1 )
|
Y
|
|
V
|
sin
(
δ
−
δ
− γ ip )
∑ ip p
i
p
p=1
n
(k )
(k )
(k )
|
Y
|
|
V
|
sin
(
δ
−
δ
− γ ip )
∑ ip p
i
i
p= i
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
For p = 1 to ( i − 1 ), use | V | & δ p of ( k + 1 ) iteration
Dr. A.M. Gaouda
UAE University
p
ELEC572, 04/05
th
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
For p = i to n , use | V p | & δ p of ( k th ) iteration
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Set | Vi | = | Vi | speci
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
In the rectangular form,
Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qi = − Im{ Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )}
b. Check Qik+1 to see if it is within the limits
Qi ,min ⟨ Qi ⟨ Qi ,max
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Case 1: If the reactive power limits are not violated,
calculate Vi k + 1
Vi
k +1
Ki
=
−
k *
( Vi )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
i −1
∑ Lip V
p=1
( k +1 )
p
−
n
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
•
Use the most
updated value of Qi
to calculate Ki .
k
L
V
∑ ip p = |Vik + 1 |∠δ i
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
p= i + 1
k +1
•
New Voltage
A.M. Gaouda
magnitudeDr.
and
UAE University
ELEC572, 04/05
angle are obtained
Use
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
| Vi | speci and δ ik + 1 For the PV-bus voltage.
Reset the magnitude
|Vi
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
k +1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
| = |Vi |Speci
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Voltage magnitude is known for PV
bus, therefore the new calculated
magnitude will not be used.
∠δ ik + 1
Vi k + 1 = |Vi |Speci
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Only the calculated angle
will be updated and used.
Case 2: If the reactive power limits are violated,
Qik + 1 ⟩ Qi ,max
Or
set Qik + 1 = Qi ,max
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qik + 1 ⟨ Qi ,min
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
set Qik + 1 = Qi ,min
Consider this bus as a PQ-Bus, calculate bus voltage Vi k + 1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Vi k + 1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Ki
=
−
k *
( Vi )
Vi
i −1
( k +1 )
L
V
∑ ip p −
k +1
p=1
= |Vi
n
k
L
V
∑ ip p
p= i + 1
k +A.M.
1 Gaouda
k + 1 ∠δ Dr.
iUAE University
|
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The PV-bus becomes PQ-bus
and both Voltage magnitude
and angle are calculated
and
Dr. A.M.
Gaouda
UAE University
used
ELEC572, 04/05
EXAMPLE:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Each line has an impedance of 0.05+j0.15
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
Line Data for the 5 buses Network
4
From
Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
5
G1
Dr. A.M. Gaouda
UAE University
ELEC572, 2
04/05
G
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
To
Dr. A.M.
Gaouda
Bus
UAE University
ELEC572, 04/05
R
X
1
2
0.0500
0.1500
2
3
0.0500
0.1500
2
4
0.0500
0.1500
3
4
0.0500
0.1500
1
5
Dr. A.M. Gaouda
0.0500
0.1500
UAE University
4
5
0.0500
ELEC572, 04/05
0.1500
The shunt admittance is neglected
Bus Data for the the 5 buses Network Before load flow solution
Load
MW
Bus
No.
Bus
code
VoltDr. A.M. GaoudaVolt
UAE University
Mag.ELEC572,
Angle
04/05
1
Slack
1.0200
0
100
50
?
?
0
2
PV
1.0200
?
0
0
200
?
3
PQ
?
?
50
20
PQ
?
?
PQ
?
?
Dr. A.M. Gaouda
4
UAE University
ELEC572, 04/05
5
Load
MVAR
Gen.Dr. A.M.Gen.
Gaouda
UAE University
MW ELEC572,
MVAR
04/05
Q
Min.
Q
Max.
Inject
MVAR
0
0
20
60
0
0
0
0
0
0
50
Dr. A.M. Gaouda
UAE University
20
ELEC572, 04/05
0
0
0
0
50
20
0
0
0
0
Dr. 0
A.M. Gaouda
UAE University
ELEC572,
04/05
0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
For the ‘5’ bus system
Construct the bus admittance matrix Ybus
Find Q2 ,δ 2 ,V3 ,V4 and V5
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qmax = 0.6 pu
Qmin = 0.2 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
4
2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
G1
G2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
5
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
SOLUTION:
3
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
A.M. Gaouda
4 Dr.
UAE University
ELEC572, 04/05
Ybus Construction
y =
1
1
Dr. A.M. Gaouda = 2 − j 6
=
UAE University
z 0.05 +ELEC572,
j 0.15
04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
Y11 = y12 + y15 = 4 − j 12
G2
G1
5
1
Y22 = y 21 + y 23 + y 25 = 6 − j 18
Y = y + y 34 = 4 − j 12
Dr. A.M. Gaouda
UAE University
33ELEC572, 04/05
32
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y44 = y43 + y 45 = 4 − j 12
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Ybus =
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
-2.0 + J6.0
6.0 -J18.0
-2.0 + J6.0
0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y15 = − y15 = −2 + j 6
Y13 = Y14 = 0
Y55 = y51 + y52 + y54 = 6 − j 18
4.0 - J12.0
-2.0 + J6.0
0
0
-2.0 + J6.0
Y12 = − y12 = −2 + j 6
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
0
ELEC572, 04/05
0
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
-2.0 + J6.0
-2.0 + J6.0
0
-2.0 + J6.0
6.0 -J18.0Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The net scheduled power injected at each bus is:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
4
Bus
No.
Load
MW
Bus
code
Volt
Mag.
Volt
Angle
Load
MVAR
1
Dr. A.M. Gaouda
UAE University
2
ELEC572, 04/05
Slack
1.0200
0
100
PV
1.0200
?
0
3
PQ
?
?
50
20
4
PQ
?
?
50
5
PQ
?
?
50
Gen.
MW
Gen.
MVAR
Q
Min.
Q
Max.
Inject
MVAR
2
G1
G2
5
1
Dr.50A.M. Gaouda ?
UAE University 200
0
ELEC572, 04/05
?
0
0
0
?
20
60
0
0
0
0
0
0
20
0
0
0
0
0
20
0
0
0
0
0
S 1 ,sch = ( P1 , g − P1 ,d ) + j ( Q1 , g − Q1 ,d )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 1 ,sch = ( P1 , g − 1.0 ) + j ( Q1 , g − 0.5 )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 2 ,sch = ( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d )
S 2 ,sch = ( 2.0 − 0 ) + j ( Q2 , g − 0 )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3 ,sch
S
= ( 0 − 0.5 ) + j ( 0 − 0.2 )
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 3 ,sch = − 0.5 − j 0.2
S 4 ,sch = − 0.5 − j0.2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 5 ,sch = − 0.5 − j 0.2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
The
known values are:
UAE University
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
4.0 - J12.0
ELEC572, 04/05
V1 = 1.02∠0 o
-2.0 + J6.0
0
0
-2.0 + J6.0
|V2 |spec = 1.02
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The bus admittance matrix is
-2.0 + J6.0
6.0 -J18.0
-2.0 + J6.0
0
-2.0 + J6.0
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
0
0
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
-2.0 + J6.0
-2.0 + J6.0
0
-2.0 + J6.0
6.0 -J18.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q2 ,min = 0.2
Q2 ,max = 0.6
and
Using GS method, select the initial values for the unknowns as:
V3o =V4o = V5o = 1∠0o
and
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
δ 2o = 0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Start the first iteration
Bus 2 is PV Bus
Q2 ,min ⟨ Q2 ⟨ Q2 ,max
Check Q2 is within the limits
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )
Q21 = − Im{ V2* ( Y21 V1 + Y22 V2o + Y23 V3o + Y24 V4o + Y25 V5o )}
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q = 0.2448
1
2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q2 ,min ⟨ Q2 ⟨ Q2 ,max
i .e .;
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
0.20 ⟨ 0.2448 ⟨ 0.6
The reactive power limits are not violated,
Calculate:
[
K2
1
o
o
o
−
L
V
+
L
V
+
L
V
+
L
V
V2 =
21 1
23 3
24 4
25 5
( V2o )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
]
The values for Ki and Lip are computed once in the beginning and used in every
iteration.
Y
L21 = 21
Y22
P2 − jQ2
K2 =
Y22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y
L23 = 23
Y22
Y
L24 = 24
Y22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
S 2 ,sch = 2.0 + j 0.2448
L21 = -0.3333
K2 = 0.0456 + j0.0959
L23 = -0.3333
V = 1.0555∠5.1113
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Reset the magnitude
Therefore,
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
2
o
|V | = |V2 |Speci = 1.02
1
2
δ 21 = 5.1113o
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V21 = 1.02∠5.1113o
L24 = 0.0
L25 =
Y25
Y22
L25 = -0.3333
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Voltage magnitude is
known and fixed for a
PV bus, therefore the
new calculated
magnitude willDr.not
be
A.M. Gaouda
UAE University
used. ELEC572, 04/05
Bus 3 is PQ Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V31 =
P3 − jQ3
K3 =
Y33
[
K3
1
o
o
−
L
V
+
L
V
+
L
V
+
L
V
31 1
32 2
34 4
35 5
( V3o )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L31 =
K3 = -0.0275 - j0.0325
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y31
Y33
L31 = 0.0
L32 =
Y32
Y33
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L34 =
L32 = -0.5000
L35 =
L34 = -0.5000
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Bus 4 is PQ Bus
[
K4 =
P4 − jQ4
Y44
K4 = -0.0275 - j0.0325
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y
L41 = 41
Y44
L41 = 0.0
Y42
L42 =
Y44
V = 0.9631∠ - 1.5489
1
4
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L43 =
Y43
Y44
L43 = -0.5000
L42 = 0.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y35
Y33
L35 = 0.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
K4
1
1
o
L
V
L
V
L
V
L
V
V =
−
+
+
+
41
1
42
2
43
3
45
5
( V4o )*
1
4
]
Y34
Y33
V31 = 0.9806∠0.7559o
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
o
]
L45 =
Y45
Y44
L45 = -0.5000
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Bus 5 is PQ Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
[
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
K5
1
3
1
L
V
L
V
L
V
L
V
V =
−
+
+
+
51 1
52 2
53 3
54 4
( V5o )*
1
5
K5 = -0.0183 - 0.0217i
L53 = 0.0
L52 = -0.3333
L51 = -0.3333
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V51 = 0.9812∠ - 0.0031o
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
]
L54 = -0.3333
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Start the second iteration
Bus 2 is PV Bus
Check Q2 is within the limits
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
0.2 ⟨ Q2 ⟨ 0.6
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q22 = − Im{ V21* ( Y21 V1 + Y22 V21 + Y23 V31 + Y24 V41 + Y25 V51 )}
Q22 = 0.0290
The reactive power limits are violated
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q2 ⟨ Qi ,min
And treat this bus as PQ-bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
set Q2 = Qi ,min = 0.2
S 2 ,sch = 2.0 + j 0.2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Use the most
updated value of
Q2 to calculate the
constant K2
Dr. A.M. Gaouda
UAE University
All Buses 2, 3, 4 and 5 are PQ Buses. Find the bus voltages using GS method
ELEC572, 04/05
Problem 2:
A. The line impedances are as indicated in per unit on 100MVA base.
The line charging susceptances are neglected. Using Gauss-Seidel method find
the power flow solution of the system. Ignoring the limits of Q3.
1
2
0.02 + j 0.04 pu
G
400
MW
0.0125 + j0.025 pu
0.01 + j 0.03 pu
250
MVAR
V1 = 1.05 ∠0 o pu
3
| V3 | = 1.04
200
MW
G
1
1
y12 =
= 10 − j 20 = y 21
0.02 + j 0.04
1
y13 =
= 10 − j 30 = y 31
0.01 + j 0.03
y 23 =
2
0.02 + j 0.04 pu
G
250
V1 = 1.05 ∠0 o pu
MVAR
3
| V3 | = 1.04
200
MW
Ybus
⎡Y11 Y12 Y13 ⎤
⎡ y12 + y13
⎥
⎢
= ⎢Y21 Y22 Y23 ⎥ = ⎢ − y 21
⎢
⎢Y31 Y32 Y33 ⎥
⎢⎣ − y 31
⎥
⎢
Ybus =
MW
0.0125 + j 0.025 pu
0.01 + j 0.03 pu
1
= 16 − j 32 = y 32
0.0125 + j 0.025
400
− y12
y 21 + y 23
− y 32
⎡ 20 − j 50 − 10 + j 20 − 10 + j 30 ⎤
⎢ − 10 + j 20 26 − j 52 − 16 + j 32 ⎥
⎢
⎥
⎢⎣ − 10 + j 30 − 16 + j 32 26 − j 62 ⎥⎦
G
− y13 ⎤
− y 23 ⎥⎥
y 31 + y 32 ⎥⎦
Classification of buses:
1
2
0.02 + j 0.04 pu
Bus 1: Slack Bus
G
250
V1 = 1.05 ∠0 pu
MVAR
3
Bus 2: Load Bus (PQ bus)
| V3 | = 1.04
200
MW
V2 and δ 2 are unknown
P2 and Q2 are known
S 2 ,sch =
( P2 , g − P2 ,d ) + j ( Q2 , g − Q2 ,d )
S 2 ,sch =
Base MVA
pu
( 0 − 400 ) + j ( 0 − 250 )
pu
100
S 2 ,sch = − 4 − j 2.5 pu
Bus 3: Voltage Controlled Bus (PV bus)
|V3 | and Pg ,3 are known
0.0125 + j 0.025 pu
0.01 + j 0.03 pu
o
V1 = 1.05∠0 o pu
Q3 ,sch and δ 3 are unknown
P3 ,sch = 2.0 pu
400
MW
G
Using GS method, select the initial values for the unknowns as:
V1 = 1.05∠0 o pu
V2o = 1∠0
| V3 | = 1.04
Start the first iteration
Bus 2 is PQ Bus
n
K2
V2 = * − ∑ L2 p V p
V2 p = 1
p≠ 2
[
K2
o
L
V
L
V
V =
−
+
21 1
23 3
( V2o )*
1
2
K2 = - 0.0692 - j0.0423
L21 = - 0.3846
]
L23 = -0.6154
V21 = 0.9746 - j0.0423
δ 3o = 0 o
Bus 3 is PV Bus
Calculate and Check Q3 is within the limits
Q3 ,min ⟨ Q3 ⟨ Q3 ,max
Q31 = − Im{ V3* ( Y31 V1 + Y32 V21 + Y33 V3o )}
Q31 = j1.1600
K3
1
1
V3 =
−
L
V
+
L
V
31 1
32 2
( V3o )*
[
K3 = 0.0274 + j0.0208
L31 = -0.4690 + j0.0354
]
L32 = -0.5310 - j0.0354
V31 = 1.0378 - j0.0052 = 1.0378∠ - 0.2854o
Reset the magnitude
|V31 | = |Vi |Speci = 1.04
V31 = 1.04∠ - 0.2854o
V31 = 1.0400 - j0.0052
Voltage magnitude is
fixed for a PV bus,
therefore the new
calculated magnitude
will not be used.
Start the second iteration
Bus 2 is PQ Bus
K 2 , L21 , L23 are constants and will be the same.
[
K2
1
L
V
L
V
V =
−
+
21 1
23 3
( V21 )*
V22 = 0.9711 - j0.0434
2
2
Bus 3 is PV Bus
Calculate and check Q3 is within the limits
]
Q3 ,min ⟨ Q3 ⟨ Q3 ,max
Q32 = − Im{ V3* ( Y31 V1 + Y32 V22 + Y33 V31 )}
Q32 = j1.3881
[
K3
2
V =
−
L
V
+
L
V
31 1
32 2
( V31 )*
2
3
]
L31 and L32 are constants and will be the same.
K 3 is changed as Q3 change
P3 − jQ32
K3 = 0.0305 + j0.0194
K3 =
Y33
V32 = 1.0391 - j0.0073 = 1.0391∠ - 0.4028o
Reset the magnitude
V32 = 1.04∠ - 0.4028o = 1.0400 - j0.0073
EXAMPLE:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Each line has an impedance of 0.05+j0.15
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
Line Data for the 5 buses Network
4
From
Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
5
G1
Dr. A.M. Gaouda
UAE University
ELEC572, 2
04/05
G
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
1
To
Dr. A.M.
Gaouda
Bus
UAE University
ELEC572, 04/05
R
X
1
2
0.0500
0.1500
2
3
0.0500
0.1500
2
4
0.0500
0.1500
3
4
0.0500
0.1500
1
5
Dr. A.M. Gaouda
0.0500
0.1500
UAE University
4
5
0.0500
ELEC572, 04/05
0.1500
The shunt admittance is neglected
Bus Data for the the 5 buses Network Before load flow solution
Load
MW
Bus
No.
Bus
code
VoltDr. A.M. GaoudaVolt
UAE University
Mag.ELEC572,
Angle
04/05
1
Slack
1.0200
0
100
50
?
?
0
2
PV
1.0200
?
0
0
200
?
3
PQ
?
?
50
20
PQ
?
?
PQ
?
?
Dr. A.M. Gaouda
4
UAE University
ELEC572, 04/05
5
Load
MVAR
Gen.Dr. A.M.Gen.
Gaouda
UAE University
MW ELEC572,
MVAR
04/05
Q
Min.
Q
Max.
Inject
MVAR
0
0
20
60
0
0
0
0
0
0
50
Dr. A.M. Gaouda
UAE University
20
ELEC572, 04/05
0
0
0
0
50
20
0
0
0
0
Dr. 0
A.M. Gaouda
UAE University
ELEC572,
04/05
0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
For the ‘5’ bus system
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Construct the bus admittance matrix Ybus
Find Q2 ,δ 2 ,V3 ,V4 and V5
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Qmax = 0.6 pu
Qmin = 0.2 pu
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
SOLUTION:
3
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
A.M. Gaouda
4 Dr.
UAE University
ELEC572, 04/05
Ybus Construction
y =
1
1
Dr. A.M. Gaouda = 2 − j 6
=
UAE University
z 0.05 +ELEC572,
j 0.15
04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
2
Y11 = y12 + y15 = 4 − j 12
G2
G1
5
1
Y22 = y 21 + y 23 + y 25 = 6 − j 18
Y = y + y 34 = 4 − j 12
Dr. A.M. Gaouda
UAE University
33ELEC572, 04/05
32
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y44 = y43 + y 45 = 4 − j 12
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Ybus =
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
-2.0 + J6.0
6.0 -J18.0
-2.0 + J6.0
0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y15 = − y15 = −2 + j 6
Y13 = Y14 = 0
Y55 = y51 + y52 + y54 = 6 − j 18
4.0 - J12.0
-2.0 + J6.0
0
0
-2.0 + J6.0
Y12 = − y12 = −2 + j 6
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
0
ELEC572, 04/05
0
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
-2.0 + J6.0
-2.0 + J6.0
0
-2.0 + J6.0
6.0 -J18.0Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The net scheduled power injected at each bus is:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
3
4
Bus
No.
Load
MW
Bus
code
Volt
Mag.
Volt
Angle
Load
MVAR
1
Dr. A.M. Gaouda
UAE University
2
ELEC572, 04/05
Slack
1.0200
0
100
PV
1.0200
?
0
3
PQ
?
?
50
20
4
PQ
?
?
50
5
PQ
?
?
50
Gen.
MW
Gen.
MVAR
Q
Min.
Q
Max.
Inject
MVAR
2
G1
G2
1
5
?
0
?
0
20
20
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr.50A.M. Gaouda ?
UAE University 200
0
ELEC572, 04/05
0
0
20
60
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
The
known values are:
UAE University
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
4.0 - J12.0
ELEC572, 04/05
-2.0 + J6.0
0
0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
The bus admittance matrix is
-2.0 + J6.0
6.0 -J18.0
-2.0 + J6.0
0
-2.0 + J6.0
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
0
0
0
-2.0 + J6.0
4.0 -J12.0
-2.0 + J6.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
-2.0 + J6.0
-2.0 + J6.0
0
-2.0 + J6.0
6.0 -J18.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Using GS method, select the initial values for the unknowns as:
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Start the first iteration
Bus 2 is PV Bus
Q2 ,min ⟨ Q2 ⟨ Q2 ,max
Check Q2 is within the limits
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Pi − jQi = Vi* ( Yi 1 V1 + Yi 2 V2 + ...... + Yii Vi + ...Yin Vn )
Q21 = − Im{ V2* ( Y21 V1 + Y22 V2o + Y23 V3o + Y24 V4o + Y25 V5o )}
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q =
1
2
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Q2 ,min ⟨ Q2 ⟨ Q2 ,max
i .e .;
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
0.20 ⟨ 0.448 ⟨ 0.6
The reactive power limits are not violated,
Calculate:
[
K2
1
o
o
o
L
V
L
V
L
V
L
V
V2 =
−
+
+
+
21 1
23 3
24 4
25 5
( V2o )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
]
The values for Ki and Lip are computed once in the beginning and used in every
iteration.
P2 − jQ2
K2 =
Y22
Y21
L21 =
Y22
K2 = 0.0456 + j0.0959
L21 = -0.3333
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L23 =
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L23 = -0.3333
V21 = 1
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Reset the magnitude
Therefore,
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y
L24 = 24
Y22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L24 = 0.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
|V21 | =
δ 21 = 5
V21 = 1
Y23
Y22
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y25
L25 =
Y22
L25 = -0.3333
Voltage magnitude is
known and fixed for a
PV bus, therefore the
new calculated
magnitude will not be
A.M. Gaouda
used. Dr.
UAE University
ELEC572, 04/05
Bus 3 is PQ Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V31 =
P3 − jQ3
K3 =
Y33
[
K3
1
o
o
L
V
L
V
L
V
L
V
−
+
+
+
31 1
32 2
34 4
35 5
( V3o )*
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L31 =
K3 = -0.0275 - j0.0325
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y31
Y33
L31 = 0.0
L32 =
Y32
Y33
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L34 =
L32 = -0.5000
L35 =
L34 = -0.5000
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Bus 4 is PQ Bus
[
K4 =
P4 − jQ4
Y44
K4 = -0.0275 - j0.0325
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y
L41 = 41
Y44
L41 = 0.0
Y42
L42 =
Y44
V = 0.9631∠ - 1.5489
1
4
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
L43 =
Y43
Y44
L43 = -0.5000
L42 = 0.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Y35
Y33
L35 = 0.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
K4
1
1
o
L
V
L
V
L
V
L
V
V =
−
+
+
+
41
1
42
2
43
3
45
5
( V4o )*
1
4
]
Y34
Y33
V31 = 0.9806∠0.7559o
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
o
]
L45 =
Y45
Y44
L45 = -0.5000
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Bus 5 is PQ Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
[
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
K5
1
3
1
L
V
L
V
L
V
L
V
V =
−
+
+
+
51 1
52 2
53 3
54 4
( V5o )*
1
5
K5 = -0.0183 - 0.0217i
L51 = -0.3333
L52 = -0.3333
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
V51 = 0.9812∠ - 0.0031o
L53 = 0.0
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
]
L54 = -0.3333
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Start the second iteration
Bus 2 is PV Bus
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
Dr. A.M. Gaouda
UAE University
ELEC572, 04/05
EXAMPLE:
Each line has an impedance of 0.05+j0.15
Line Data for the 5 buses Network
3
4
2
5
G1
G2
1
Bus
nl
Bus
nr
R
1
2
0.0500
0.1500
2
3
0.0500
0.1500
2
4
0.0500
0.1500
3
4
0.0500
0.1500
1
5
0.0500
0.1500
4
5
0.0500
0.1500
X
The shunt admittance is neglected
Bus Data for the the 5 buses Network Before load flow solution
Load
MW
Bus
No.
Bus
code
Volt
Mag.
Volt
Angle
Load
MVAR
Gen.
MW
Gen.
MVAR
Q
Min.
1
Slack
1.0200
0
100
50
?
?
0
2
PV
1.0200
?
0
0
200
?
3
PQ
?
?
50
20
0
4
PQ
?
?
50
20
5
PQ
?
?
50
20
Q
Max.
Inject
MVAR
0
0
20
60
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
For the ‘5’ bus system
Construct the bus admittance matrix Ybus
Find Q2 ,δ 2 ,V3 ,V4 and V5
Qmax = 0.6 pu
Qmin = 0.2 pu
SOLUTION:
Solution
Ybus Construction
3
4
2
G1
G2
1
5
Bus
nl
Bus
nr
R
1
2
0.0500
0.1500
2
3
0.0500
0.1500
2
4
0.0500
0.1500
3
4
0.0500
0.1500
1
5
0.0500
0.1500
4
5
0.0500
0.1500
X
Solution
3
4
2
G1
G2
1
5