Power Flow Analysis

1. BEF43303
POWER SYSTEM ANALYSIS AND
PROTECTION
POWER FLOW ANALYSIS
WEEK 3

2. 3.0 CONTENTS
3.1 Power flow solution
3.2 Gauss-Seidel power flow solution
3.3 Line flows and losses

3. 3.1 POWER FLOW SOLUTION
• The system is assumed to be operating under balanced conditions.
• A single-phase model is used.
• Four quantities are associated with each bus:
1) Voltage magnitude 𝑉 ,
2) Phase angle 𝛿,
3) Real power 𝑃,
4) Reactive power 𝑄.

4. 3.1 POWER FLOW SOLUTION
• The system buses are classified into three types:
 Slack bus
• As reference, where 𝑉 and 𝛿 are specified. Makes up the losses in the
network.
 Load bus
• Called P-Q bus, where 𝑃 and 𝑄 are specified. 𝑉 and 𝛿 are unknown.
 Generator bus
• Called P-V bus, where P and V are specified. Q and 𝛿 are to be
determined.

5. 3.1 POWER FLOW EQUATION
• Consider a typical bus of a
power system network shown
in Figure 1.
• Transmission lines are
represented by their
equivalent model.
• Admittances are in per unit on
a common MVA base.
Figure 1

6. 3.1 POWER FLOW EQUATION
• Applying KCL to this bus:
• In general form:

7. 3.1 POWER FLOW EQUATION
• The real and reactive power at bus 𝑖 is:
• Substituting 𝐼𝑖 from KCL earlier:

8. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• The idea is to solve the set of non-linear equations in the previous
section for two unknown variables at each node.
• In this method, the equation is solved for 𝑉𝑖 and the iterative
sequence becomes:

9. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• Where:
• 𝑦𝑖𝑗 - actual admittance in per unit,
• 𝑃𝑖
𝑠𝑐ℎ
- net real power in per unit,
• 𝑄𝑖
𝑠𝑐ℎ
- net reactive power in per unit.
• Current entering bus 𝑖 – POSITIVE.
• Current leaving bus 𝑖 – NEGATIVE.
• 𝑃 and 𝑄 injected into the bus (generator buses) – POSITIVE.
• 𝑃 and 𝑄 flowing away from the bus (load bus) – NEGATIVE.

10. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• Solving for 𝑃𝑖 and 𝑄𝑖:

11. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• Substituting the derivation of the bus admittance matrix:

12. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• For P-Q buses:
 𝑃𝑖 and 𝑄𝑖 are known.
 Starting with initial estimate (1.0), the power flow equation is
solved for the real and imaginary components of voltage.
• For P-V buses:
 𝑃𝑖 and 𝑉𝑖 are known.
 First, the power flow equations is solved for 𝑄𝑖 and then is
used to solve for 𝑉𝑖 .

13. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• Since 𝑉𝑖 is specified, only the imaginary part of 𝑉
𝑖
(𝑘+1)
is retained.
• The real part is selected in order to satisfy the following condition:
• 𝑒𝑖
(𝑘+1)
and 𝑓𝑖
(𝑘+1)
are the real and imaginary components of the
voltage in the iterative sequence.

14. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• The convergence rate may be increased by using the acceleration
factor:
• The process is continued until the results are within a specified
accuracy:

15. 3.2 GAUSS-SEIDEL POWER FLOW
SOLUTION
• A voltage accuracy in the range of 0.00001 and 0.00005 pu is
satisfactory.
• In practice, the power mismatch is used as the stopping criteria.
• A typical power mismatch accuracy is 0.001 pu.

16. 3.3 LINE FLOWS AND LOSSES
• Following the iterative solution
of bus voltages, the next step is
the computation of line flow
and line losses.
• From Figure 2, the line current
𝐼𝑖𝑗, measured at bus 𝑖 and
defined positive in the
direction of 𝑖 → 𝑗 is given by:
Figure 2

17. 3.3 LINE FLOWS AND LOSSES
• Similarly, the line current 𝐼𝑗𝑖 measured at bus 𝑗 and defined positive
in the direction 𝑗 → 𝑖 is given by:
• The complex powers for both cases are:

18. 3.3 LINE FLOWS AND LOSSES
• The power loss in line 𝑖 − 𝑗 is the algebraic sum of the power flows
given as follows:

19. 3.3 LINE FLOWS AND LOSSES
Example 3.1
Figure 3 shows the one-line diagram of
a simple three-bus power system with
generation at bus 1. The magnitude of
voltage at bus 1 is adjusted to 1.05 pu.
The schedule loads at buses 2 and 3 are
as marked on the diagram. Line
impedance are marked in pu on a 100-
MVA base and the line charging
susceptances are neglected.
Figure 3

20. 3.3 LINE FLOWS AND LOSSES
Example 3.1
a) Using the Gauss-Seidel method, determine the phasor values of the voltage at
the load buses 2 and 3 (P-Q buses) accurate to four decimal places. (𝑉
2
(7)
=
0.9800 − 𝑗0.0600, 𝑉
3
(7)
= 1.0000 − 𝑗0.0500).
b) Find the slack bus real and reactive power. (𝑃1 = 4.095 𝑝𝑢, 𝑄1 = 1.890 𝑝𝑢).
c) Determine the line flows and line losses. Construct a power flow diagram
showing the direction of line flow. (𝑆𝐿12
= 8.5 𝑀𝑊 + 𝑗17.0 𝑀𝑉𝑎𝑟, 𝑆𝐿13
=
5.0 𝑀𝑊 + 𝑗15.0 𝑀𝑉𝑎𝑟, 𝑆𝐿23
= 0.8 𝑀𝑊 + 𝑗1.6 𝑀𝑉𝑎𝑟)

21. 3.3 LINE FLOWS AND LOSSES
Example 3.2
Figure 4 shows the one-line diagram of a simple three-bus power system with
generators at bus 1 and 3. The magnitude of voltage at bus 1 is adjusted to 1.05 pu.
Voltage magnitude at bus 3 is fixed at 1.04 pu with real power generation of 200
MW. A load consisting of 400 MW and 250 MVar is taken from bus 2. Line
impedance are marked in per unit on a 100 MVA base, and the line charging
susceptances are neglected. Obtain the power flow solution by the Gauss-Seidel
method including line flows and line losses. (𝑉2 = 0.97168∠ −2.6948°
pu, 𝑆3 =
2.0 + 𝑗1.4617 pu, 𝑉3 = 1.04∠ −0.498° pu, 𝑆1 = 2.1842 + 𝑗1.4085 pu, 𝑆𝐿12
= 8.39 +
𝑗16.79 MVA, 𝑆𝐿13
= 0.18 + 𝑗0.548 MVA, 𝑆𝐿23
= 9.85 + 𝑗19.69 MVA).

22. 3.3 LINE FLOWS AND LOSSES
Example 3.2
Figure 4

23. 3.3 LINE FLOWS AND LOSSES
Example 3.2
Figure 4 shows the one-line diagram of a simple three-bus power system with
generators at bus 1 and 3. The magnitude of voltage at bus 1 is adjusted to 1.05 pu.
Voltage magnitude at bus 3 is fixed at 1.04 pu with real power generation of 200
MW. A load consisting of 400 MW and 250 MVar is taken from bus 2. Line
impedance are marked in per unit on a 100 MVA base, and the line charging
susceptances are neglected. Obtain the power flow solution by the Gauss-Seidel
method including line flows and line losses. (𝑉2 = 0.97168∠ −2.6948°
pu, 𝑆3 =
2.0 + 𝑗1.4617 pu, 𝑉3 = 1.04∠ −0.498° pu, 𝑆1 = 2.1842 + 𝑗1.4085 pu, 𝑆𝐿12
= 8.39 +
𝑗16.79 MVA, 𝑆𝐿13
= 0.18 + 𝑗0.548 MVA, 𝑆𝐿23
= 9.85 + 𝑗19.69 MVA).

24. 3.3 LINE FLOWS AND LOSSES
Example 3.2
Figure 4

25. 3.3 LINE FLOWS AND LOSSES
Tutorial 3.1
In the power system network shown in Figure 5, bus 1 is a slack bus with 𝑉1 =
1.06∠0°pu and bus 2 is a load bus with 𝑆2 = 280 + 𝑗60 MVA. The line impedance
on a base of 100 MVA is 0.02 + 𝑗0.04 pu.
a) Using Gauss-Seidel method, determine 𝑉2. Use an initial estimate of 𝑉
2
(0)
=
1.0 + 𝑗0.0 and perform four iterations. (0.90 − 𝑗0.10 pu).
b) If after several iterations voltage at bus 2 converges to 𝑉2 = 0.90 − 𝑗0.10 pu
determine 𝑆1 and the real and reactive power loss in the line. (20 + 𝑗40 MVA).

26. 3.3 LINE FLOWS AND LOSSES
Tutorial 3.1
Figure 5

27. 3.3 LINE FLOWS AND LOSSES
Tutorial 3.2
Figure 6 shows the one-line diagram of a simple three-bus power system with generation at bus 1.
The voltage at bus 1 is 𝑉1 = 1.06∠0°pu. The scheduled loads on buses 2 and 3 are marked on the
diagram. Line impedances are marked in pu on a 100 MVA base. For the purpose of hand
calculations, line resistances and line charging susceptances are neglected.
a) Using Gauss-Seidel method and initial estimates of 𝑉
2
(0)
= 1.0 + 𝑗0.0 and 𝑉
3
(0)
= 1.0 + 𝑗0.0,
determine 𝑉2 and 𝑉3. Perform two iterations. (0.9089 − 𝑗0.0974 pu, 0.9522 − 𝑗0.0493)
b) If after several iterations the bus voltages converge to:
𝑉2 = 0.90 − 𝑗0.10 pu
𝑉3 = 0.95 − 𝑗0.05 pu
determine the line flows and line losses and the slack bus real and reactive power. (7.0 − 𝑗7.0,
𝑗60, 𝑗40, 𝑗10).

28. 3.3 LINE FLOWS AND LOSSES
Tutorial 3.2
Figure 6

29. 3.3 LINE FLOWS AND LOSSES
Tutorial 3.3
Figure 7 shows the one-line diagram of a simple three-bus power system with generation at buses 1 and 3. The
voltage at bus 1 is 𝑉1 = 1.025∠0°pu. Voltage magnitude at bus 3 is fixed at 1.03 pu with a real power
generation of 300 MW. A load consisting of 400 MW and 200 MVar is taken from bus 2. Line impedances are
marked in per unit on a 100 MVA base. For the purpose of hand calculations,
line resistances and line charging susceptances are neglected.
a) Using Gauss-Seidel method and initial estimates of 𝑉
2
(0)
= 1.0 + 𝑗0.0 and 𝑉
3
(0)
= 1.03 + 𝑗0.0 and keeping
𝑉3 = 1.03 pu, determine the phasor values of 𝑉2 and 𝑉3. Perform two iterations. (1.0001 − 𝑗0.0409 pu,
1.3671 pu, 1.0298 + 𝑗0.0216 pu).
b) If after several iterations the bus voltages converge to:
𝑉2 = 1.000571 − 𝑗0.0366898 pu
𝑉3 = 1.029706 + j0.0246 pu
determine the line flows and line losses and the slack bus real and reactive power. (100 + 𝑗90.51 MVA,
𝑗7.77 MVA, 𝑗1.25 MVA, 𝑗18.42 MVA).

30. 3.3 LINE FLOWS AND LOSSES
Tutorial 3.3
Figure 7