the gauss seidel method of load flow analysis is explained with detailed flowchart and algorithmic steps with an design problem

1. Gauss-Seidel Method of Load Flow
Analysis
Contents:
 Algorithm
 Flowchart
 Problems
 Advantages & Disadvantages
Kongunadu College of Engineering & Technology Gauss-Seidel Method Prepared by,
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1

2. Algorithm
Step 1:
Assume a flat voltage profile 1+j0 for all buses except slack bus
Step 2:
Assume a suitable value of convergence criterion ε
Step 3:
Set iteration count k=0 and assume V1
0 V2
0 V3
0 ………. Vn
0 except
slack bus.
Step 4:
Set bus count p=1
Step 5:
Check for slack bus. If it is slack bus then go to step-12,
otherwise go to next step.
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3. Contd…
Step 6: Check for generator bus. If it is generator bus go to next step,
otherwise go to step 9
Step 7:
Set |Vp
k|=|Vp|spec calculate the reactive power by,
𝑄 𝑝,𝑐𝑎𝑙
𝑘+1
=(-1) im 𝑉𝑃
𝑘 ∗
× 𝑌𝑝𝑞 𝑉𝑞
𝑘+1𝑝−1
𝑞=1 + 𝑌𝑝𝑞 𝑣 𝑞
𝑘𝑛
𝑞=𝑝
If the calculated reactive power is within specified limits then consider
this bus as generator bus and set 𝑄 𝑃=𝑄 𝑝,𝑐𝑎𝑙
𝑘+1
and go to next step.
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4. Contd…
If calculated Q violates the specified limit then treat this bus as
load bus
if Qp,cal
k+1 < Qp,min then Qp = Qp,min
Qp,cal
k+1 > Qp,max then Qp = Qp,max
go to step-9
Step 8:
For generator bus the voltage magnitude is constant. The
phase of bus voltage calculated by,
𝑉𝑝,𝑡𝑒𝑚𝑝
𝑘+1 1
𝑌𝑝𝑝
𝑃 𝑝−𝑄 𝑝
𝑉𝑝
𝑘 ∗ − 𝑌𝑝𝑞 𝑉𝑞
𝑘+1𝑝−1
𝑞=1 − 𝑌𝑝𝑞 𝑣 𝑞
𝑘𝑛
𝑞=𝑝+1
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5. Contd…
Step 9: For the load bus the value of voltage can be calculated by,
𝑉𝑝
𝑘+1
=
1
𝑌𝑝𝑝
𝑃 𝑝−𝑄 𝑝
𝑉𝑝
𝑘 ∗ − 𝑌𝑝𝑞 𝑉𝑞
𝑘+1𝑝−1
𝑞=1 − 𝑌𝑝𝑞 𝑣 𝑞
𝑘𝑛
𝑞=𝑝+1
Step 10: An acceleration factor α can be used for faster convergence.
Vp,acc
k+1 = Vp
k+α(Vp
k+1- Vp
k)
Then set,
Vp
k+1 = Vp,acc
k+1 α=1.6
Step 11:
Calculate, ∆Vp
k+1=Vp
k+1- Vp
k
Step 12:
Repeat steps 5 to 11 until all the bus voltages have been calculated .
Continue until bus count is n.
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6. Contd…
Step 13:
Find the largest of the absolute value of
change in voltage. |∆Vmax|<ε then move to next step.
Otherwise increment the iteration count and go to
step-4.
Step-14
Calculate the line flows and slack bus power
using bus voltages.
6
Kongunadu College of Engineering & Technology Gauss-Seidel Method Prepared by,
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7. Flowchart: 7
Kongunadu College of Engineering & Technology Gauss-Seidel Method Prepared by,
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8. 8
Kongunadu College of Engineering & Technology Gauss-Seidel Method Prepared by,
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9. The above algorithm has been explained in the next slide with an
example.
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10. 1. The figure given below shows a power system.
Bus 1:Slack bus V Specified=1.05p.u
Bus 2:PV bus 𝑽 =1.2p.u Pg=3 p.u
Bus 3: PQ bus PL=4p.u QL=2p.u
Carry out one iteration of load flow solution by G-S method. Take
Q limits of generator 2 as 0 < Q < 3.
Step 1:Form 𝒀 𝒃𝒖𝒔 matrix
𝑌𝑏𝑢𝑠 =
3 − 𝑗9
−2 + 𝑗5
−1 + 𝑗4
−2 + 𝑗5
5 − 𝑗14
−3 + 𝑗9
−1 + 𝑗4
−3 + 𝑗9
4 − 𝑗13
Step 2: Initial voltages are considered as 1 p.u for all buses.
𝑉2
0
= 1.2+j0 𝑉3
0
= 1+j0
Bus 1 is slack bus so its voltage remains constant for all iterations.
𝑉1
0
=𝑉1
1
= 𝑉1
2
= 𝑉1
3
= 1.05+j0
Step 3: Start iteration count, for first iteration set k=0
Step 4: Bus 1 is slack bus p=1 ∴ 𝑽 𝟏
𝟏
= 1.05+j0
Step 5: Bus 2 is generator bus(p=2)
For generator bus calculate the phase of the voltage
To calculate phase angle of bus voltage estimate the reactive power at bus 2 by
using.
𝑸 𝒑,𝒄𝒂𝒍
𝒌+𝟏
=(-1) im 𝑽 𝑷
𝒌 ∗
× 𝒀 𝒑𝒒 𝑽 𝒒
𝒌+𝟏𝒑−𝟏
𝒒=𝟏 + 𝒀 𝒑𝒒 𝒗 𝒒
𝒌𝒏
𝒒=𝒑
Substitute k=0 and p=2
𝑄2
1
=(-1) im 𝑉2
0 ∗
× 𝑌2𝑞 𝑉𝑞
11
𝑞 =1 + 𝑌2𝑞 𝑣 𝑞
03
2
𝑄2
1
=(-1) im 𝑉2
0 ∗
∗ 𝑌21 𝑉1
1
+ 𝑌22 𝑉2
0
+ 𝑌23 𝑉3
0
= (-1) im 1.2 − 𝑗0 ∗ (−2 + 𝑗5) 1.05 + 𝑗0 + 5 − 𝑗14 1.2 + 𝑗0 + −3 +
𝑗9 (1 + 𝑗0)
=(-1)im 𝑗1.08 − 𝑗3.06 ∴ 𝑸 𝟐
𝟏
= 3.06 p.u 0 < Q < 3.
10
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11. The Q-limit is violated which is greater than the upper limit. Set 𝑄2 = 3 𝑝. 𝑢 Therefore the
bus acts as load bus. ∴ 𝑽 𝟐
𝟎
= 1+j0
For load bus calculate the value of voltage magnitude and phase angle,
𝑉𝑝 ,
𝑘+1
=
1
𝑌𝑝𝑝
𝑃 𝑝 −𝑗𝑄 𝑝
𝑉𝑝
𝑘 ∗ − 𝑌𝑝𝑞 𝑉𝑞
𝑘+1𝑝 −1
𝑞 =1 − 𝑌𝑝𝑞 𝑣 𝑞
𝑘𝑛
𝑞 =𝑝 +1
𝑉2
1
=
1
𝑌22
𝑃2−𝑗𝑄 2
𝑉2
0 ∗ − 𝑌21 𝑉1
1
− 𝑌23 𝑣3
0
=
1
5−𝑗14
3+𝑗3
1−𝑗0
− −2 + 𝑗5 (1 + 𝑗0) – −3 + 𝑗9 (1 − 𝑗0)
𝑉2
1
=1.023 + j0.063p.u =1.02 3.54°p.u
Step 6: Bus 3 as load bus,
𝑉3
1
=
1
𝑌33
𝑃3−𝑗 𝑄3
𝑉3
0 ∗ − 𝑌31 𝑣1
1
− 𝑌32 𝑣2
1
=
1
4−𝑗 13
−4+𝑗2
1−𝑗0
− −1 + 𝑗4 (1.05 + 𝑗0) – −3 + 𝑗9 (1.023 + 𝑗0.063)
𝑉3
1
= 0.803 – j0.194 p.u = 0.826 −13.60 p.u
Result:
The voltages at the end of first iteration is,
𝑉1
1
= 1.05+j0 p.u
𝑉2
1
=1.023 + j0.063p.u
𝑉3
1
= 0.803 – j0.194 p.u
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12. Advantages and disadvantages of G-S method.
Advantages:
 Simple calculation and less programming task.
 Less memory requirement.
 Useful for small systems.
Disadvantages:
 Requires large number of iterations to reach convergence
 Not suitable for large systems
 Convergence time increases with size of the system.
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13. References:
1. Hadi Saadat, ‘Power System Analysis’, Tata McGraw Hill Education Pvt.
Ltd., New Delhi, 21st reprint, 2010.
2. Kundur P., ‘Power System Stability and Control, Tata McGraw Hill Education
Pvt. Ltd., New Delhi, 10th reprint, 2010.
3. Pai M A, ‘Computer Techniques in Power System Analysis’, Tata Mc Graw-Hill
Publishing
Company Ltd., New Delhi, Second Edition, 2007.
4. J. Duncan Glover, Mulukutla S. Sarma, Thomas J. Overbye, ‘ Power System
Analysis & Design’, Cengage Learning, Fifth Edition, 2012.
5. Olle. I. Elgerd, ‘Electric Energy Systems Theory – An Introduction’, Tata
McGraw Hill Publishing Company Limited, New Delhi, Second Edition, 2012.
6. C.A.Gross, “Power System Analysis,” Wiley India, 2011.
7. M.Jeraldin Ahila “Power System Analysis”, Lakshmi Publications, Chennai,
Eleventh Edition 2017.
8. Other Web Resources
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Kongunadu College of Engineering & Technology Gauss-Seidel Method Prepared by,
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