EET308
Chapter 3: Power Flow Solution
Gauss Seidel
Introduction
• Power flow studies (load flow) is an important part of
power system analysis.
• Necessary for planning, economic scheduling and
control of an existing system as well as planning its
future expansion.
• The problem consists of determining the magnitudes
and phase angle of voltages at each bus and active
and reactive power flow in each line.
Introduction
• In solving power flow problem, the system is assumed
to be operating under balanced conditions and a
single-phase model is used.
• Four quantities are associated with each bus
i) voltage magnitude, ïVï
ii) phase angle, d
iii) real power, P
iv) reactive power, Q
Introduction
• The system buses are generally classified into three types:1. Slack bus (swing bus or reference bus) – one generator at
slack bus - is taken as reference where the magnitude
voltage (|V|) and phase angle of the voltage (d) are
specified. This bus makes up the difference between
scheduled loads and generated power that caused by the
losses in the network.
2. Load buses (P-Q buses) – the reactive (Q) and active powers
(P) are specified at these buses. The magnitude and the
phase angle of the bus voltages are unknown.
3. Regulated buses (Generator buses or P-V buses) – known as
voltage-controlled buses. The real power (P) and voltage
magnitude (|V|) are specified. The phase angles of the
voltages and reactive power to be determined.
Power Flow Equation
• Consider bus of a power system network as shown in figure
below. Transmission lines are represented by equivalent p
models where impedances have been converted to per unit
admittances on a common MVA base.
• Applying KCL to this bus results in
!" = $"% &" + $"( &" − &( + $"* &" − &* + ⋯ + $", &" − &,
= $"% + $"( + $"* + ⋯ + $", &" − $"( &( − $"* &* − ⋯ − $", &,
,
,
= &" - $". − - $". &. 0 ≠ 2
./%
./(
Power Flow Equation
The real and reactive power at bus i is
!" + $%" =
'" ("∗
0
!" − $%"
→ (" =
'"∗
0
!" − $%"
= '" , 1"- − , 1"- ''"∗
-./
$≠4
-.2
From this relationship, the mathematical formulation of the
power flow problem results in a system of algebraic nonlinear equations which must be solved by iterative
techniques.
Gauss-Seidel Power Flow Solution
• In power flow study, it is essential to solve the set of non-linear
equations for two unknown variables at each node.
• In Gauss-Seidel method, !" is solved and the iterative
sequences become
!"
(+,-)
%"&'( − 1)"&'(
(+)
∑
+
#
!
"$ $
∗(+)
!"
=
∑ #"$
1≠6
where,
#"$ shown in lowercase letters is the actual admittance in per unit
%"&'( and )"&'( are the net real and reactive powers in per unit
Gauss-Seidel Power Flow Solution
• In KCL, current entering bus i was assumed positive. Thus, for
buses where real and reactive powers are injected into the
bus, such as generator buses, !"#$% and &"#$% have positive
values.
• For load buses where real and reactive powers are flowing
away from the bus, !"#$% and &"#$% have negative values.
4
[()*]
!"
= ℜ ."∗ ( ." ( 0 5"1 − 0 5"1 .1 (
123
4
[()*]
&"
4
12*
4
= −ℑ ."∗ ( ." ( 0 5"1 − 0 5"1 .1 (
123
7≠9
12*
7≠9
Gauss-Seidel Power Flow Solution
• Rewriting the power flow equation in terms of the bus
admittance matrix (Y-bus).
• Since the off-diagonal elements of the bus admittance matrix
!"#$ shown by uppercase letters, !%& = −)%& and the diagonal
elements are !%% = ∑ )%& ,the equation becomes
Gauss-Seidel Power Flow Solution
• !"" includes the admittance to ground of life charging
susceptance and any other fixed admittance to ground.
• Since both components (V and d) are specified for slack bus,
there are 2 (n-1) equations which must be solved iteratively.
• Under normal operating conditions, |V| of buses are in
neighborhood of 1.0 per unit or close to the |V| of the slack
buses.
• |V| of load buses are lower than the slack bus value, depending
on reactive power demand whereas the scheduled voltage at
the generator buses are higher.
Gauss-Seidel Power Flow Solution
• The d of the load buses are below the reference angle in
accordance to power demand whereas d of the generator
buses may be above the reference value depending on
the amount of real power flowing into the bus.
• Thus, for Gauss-Seidel method, an initial voltage estimate
of 1.0+j0.0 for unknown voltage is satisfactory and the
converged solution correlates with actual operating
states.
Gauss-Seidel Power Flow Solution
• For P-Q buses (load buses), !"#$% and &"#$% are known. Starting
()*+)
with an initial estimate, '"
is solved for the real and
imaginary components of voltage.
• For P-V buses (regulated buses), !"#$% and '" are specified,
()*+)
()*+)
&"
is solved, then used in '"
.
• Since '" is specified, only imaginary part of '"
And its real part is selected in order to satisfy
[)*+]
-"
()*+)
-"
=
()*+)
'"
[#$%] 2
− ."
()*+)
is retained.
[)*+] 2
where
and ."
are the real and imaginary components
()*+)
of the voltage '"
in the iterative sequence.
Gauss-Seidel Power Flow Solution
• The rate of convergence is increased by applying acceleration
factor (α) to the approximate solution obtained from each
iteration.
!"
($%&)
= !"
($)
($)
!" +,-
+*
− !"
($)
• α is the acceleration factor and is in the range of 1.3 to 1.7.
• The process is continued until changes in the real and
imaginary components of bus voltages between successive
iterations are within specified accuracy.
($%&)
− /"
($%&)
− 2"
/"
2"
($)
≤1
($)
≤1
Gauss-Seidel Power Flow Solution
• A voltage in accuracy is in the range of 0.00001 to
0.00005 per unit is satisfactory.
• In practice, the method for determining the completion of
a solution is based on an accuracy index set up on the
power mismatch.
• The iteration continues until the magnitude of the largest
element in the DP and DQ columns in less than specified
value.
• Normally, power mismatch accuracy is 0.001 per unit.
• After solving for bus voltages and angles, power flows and
losses in the network branches are calculated.
Line Flows and Losses
• Consider the line connecting the two buses, bus i and bus j in
transmission line model below. The line current, Iij measured at
bus i and defined positive in the direction.
Example 1
Figure below shows the single-line diagram of three-bus power
system with generation at bus 1. The scheduled loads at buses 2
and 3 are as marked on the diagram. Line impedances are marked
in per unit on 100 MVA base and the line charging susceptances
are neglected.
Example 1
a) Using Gauss-Seidel method, determine the phasor values of
the voltage at load buses 2 and 3 (P-Q buses) to 4 decimal
places.
b) Find slack bus real and reactive power.
c) Determine the line flows and line losses.
d) Construct a power flow diagram showing the direction of line
flow.
Solution (E1)
a) Line impedances are converted to admittances
!"#
1
=
= 10 − *20,
0.02 + *0.04
!". = 10 − *30,
!#. = 16 − *32
Solution (E1)
At the P-Q buses, the complex loads expressed in per units are
!"#$%
!2#$%
256.6 + -110.2
=−
= −2.566 − -1.102 01
100
138.6 + -45.2
=−
= −1.386 − -0.452 01
100
Bus 1 is taken as reference bus (slack bus). Starting from an initial
estimate of
(8)
6"
(=)
6"
(8)
= 1.0 + -0.0 :;< 62 = 1.0 + -0.0
>"#$% − -?"#$%
(8)
+
A
6
+
A
6
="
=
"2
2
∗(8)
6"
=
A=" + A"2
−2.566 + -1.102
+ 10 − -20 1.05 + -0 + 16 − -32 1.0 + -0
1.0 − -0
=
(26 − -52)
= 0.9825 − -0.0310 01
Solution (E1)
(3)
!2
42567 − ,82567
(3)
+
;
!
+
;
!
32
3
"2
"
∗(:)
!2
=
;32 + ;"2
−1.386 + ,0.452
+ 10 − ,30 1.05 + ,0 + 16 − ,32 0.9825 − ,0.0310
1.0 − ,0
=
(26 − ,62)
= 1.0011 − ,0.0353
For the 2nd iteration,
(")
!"
−2.566 + ,1.102
0.9825 + ,0.0310 + 10 − ,20 1.05 + ,0 + 16 − ,32 1.0011 − ,0.0353
=
(26 − ,52)
= 0.9816 − ,0.0520
and
(")
!2
−1.386 + ,0.452
1.0011 + ,0.0353 + 10 − ,30 1.05 + ,0 + 16 − ,32 0.9816 − ,0.052
=
(26 − ,62)
= 1.0008 − ,0.0459
Solution (E1)
The process is continued and a solution is converged with an accuracy
of 5×10%& per unit in seven iterations as given below.
The final solution is
Solution (E1)
b) The slack bus is
or the slack bus real and reactive powers are
!" = 4.095 )* = 409.5 +, and -" = 1.890 )* = 189.0 +012.
c) To find the line flows, first find the line currents are computed. With
the line charging capacitors neglected, the line currents are
Solution (E1)
The line flows are
Solution (E1)
The line losses are
d) The power flow diagram is shown below:-
Example 2
Figure below shows the single-line diagram of three-bus power system
with generator at buses 1 and 3. The magnitude of voltage at bus 1 is
adjusted to 1.05 per unit. Voltage magnitude at bus 3 is fixed at
1.04 per unit with a real power generation at 200 MW. A load consisting
of 400 MW and 250 Mvar is taken from bus 2. Line impedances are
marked in per unit on a 100 MVA base and the line charging
susceptances are neglected. Obtain the power flow solution by GaussSeidel method including line flows and line losses.
Solution (E2)
Line impedances are converted to admittances
Bus 1 is taken as reference bus (slack bus). Starting from an initial
estimate of
Solution (E2)
Bus 3 is a regulated bus where voltage magnitude and real power are
specified. For the voltage-controlled bus, the first reactive power is
The value of !"($) is used as !"&'( for the computation of voltage at bus 3.
The complex voltage at bus 3, denoted by )'"($) , is calculated
Solution (E2)
Since !" is held constant at 1.04 pu, only the imaginary part of !#"(%) is retained,
i.e, '"(%) = −0.005170, and its real part is obtained from
(%)
1
(%)
= 1.039987 − 50.005170
/" = (1.04)1 − 0.005170
Thus
!"
= 1.039987
For the 2nd iteration,
(1)
!1
(1)
617#8 − 5917#8
(%)
+
<
!
+
<
!
%1
%
1"
"
∗(%)
!1
=
<%1 + <1"
−4.0 + 52.5
0.97462 + 50.042307 + 10 − 520 1.05 + 16 − 532 1.039987 − 50.005170
=
26 − 552
= 0.971057 − 50.043432
9" = −ℑ !"∗ % !" % <%" + <1" − <%" !% − <1" !1 1
= −ℑ{ 1.039987 + 50.005170 [ 1.039987 − 50.005170 26 − 562
− 10 − 530 1.05 + 50 − 16 − 532 0.971057 − 50.043432 ]}
= 1.38796
Solution (E2)
The value of !"($) is used as !"&'( for the computation of voltage at bus 3.
The complex voltage at bus 3, denoted by )'"($) , is calculated
($)
)'"
6"&'( − 5!"&'(
($)
+
:
)
+
:
)
"8 8
"$ $
∗(8)
)"
=
:"8 + :"$
2.0 − 51.38796
+ 10 − 530 1.05 + 16 − 532 0.971057 − 50.043432
1.039987 + 50.00517
=
26 − 562
= 1.03908 − 50.00730
Since )" is held constant at 1.04 pu, only the imaginary part of )'"($)is
retained, i.e, *"($) = −0.00730, and its real part is obtained
($)
from
1" = (1.04)$ − 0.00730
or
)" = 1.039974 − 50.00730
($)
$
= 1.039974
Solution (E2)
The process is continued and a solution is converged with an accuracy
of 5×10%& per unit in seven iterations as given below.
The final solution is
'( = 0.97168∠ − 2.6948° 45
67 = 2.0 + 91.4617 45
'7 = 1.04∠ − 0.498° 45
6: = 2.1842 + 91.4085 45
Solution (E2)
Line flows and line losses are computed as below
The power flow diagram is as below
Exercise 1
Figure below shows the single-line diagram of a three-bus power system with
generation at bus 1. The voltage at bus 1 is V1 = 1.025Ð0° per unit. The
scheduled loads on bus 2 and bus 3 are marked on the diagram. Line
impedances are marked in per unit on a 100 MVA base. Line resistances and
line charging susceptances are neglected.
a) By using Gauss-Seidel method and initial estimates of V2(0) = 1.0Ð0° and
V3(0) = 1.3285Ð19.8°, determine V2 and V3. Perform calculation for two
iterations.
2
b) After several iterations, the bus V1 = 1.025Ð0° 1
j0.05
350 MW
voltages converge to
Determine the line flows and
losses and the slack bus real and
reactive power.
200 Mvar
Slack Bus
j0.015
300 MW
250 Mvar
3
j0
.0
5
!" = 0.9346 − +0.1159 ./
!0 = 0.9724 − +0.0604 ./
Exercise 2
Figure below shows the single-line diagram of three-bus power system with
generation at bus 1 and bus 3. The voltage at bus 1 is V1 = 1.025Ð0° per unit.
The voltage magnitude at bus 3 is fixed at 1.05 per unit with a real power
generation of 250 MW. The scheduled load on bus 2 is marked on the
diagram. Line impedances are marked in per unit on a 100 MVA base. Line
resistances and line charging susceptances are neglected.
V1 = 1.025Ð0°
j0.1
1
j0.4
2
250 MW
150 Mvar
Slack Bus
j0.2
| V3 | =1.05
j0.1
.2
j0
By using Gauss-Seidel method
and initial estimates of
V2(0) = 1.0Ð0° and
V3(0) = 1.05Ð0°, determine V2
and V3. Perform calculation for
one iteration.
P3 = 250 MW
3