EEE 103 - Load Flow Analysis [PDF]

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EEE 103 – Introduction to Power Systems

Power Flow Through Short Transmission Lines Let us consider a short transmission line. The single-phase equivalent circuit is shown below:

+•

Z = (r + jxL )L = |Z| ∠θ

I Vs = |Vs| ∠α s

IR V = |V | ∠0 R R

-• IS = I R = I

• +

• VS = ZI + VR

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EEE 103 – Introduction to Power Systems

We calculate for the current I and its conjugate I* :

v (VS ∠α ) − (VR ∠0 ) I = ( Z ∠θ ) v∗ (VS ∠ − α ) − (VR ∠0 ) I = (Z∠ −θ )

We calculate the single-phase complex power at the sending and receiving ends:

v v∗ SS = (VS ∠α ) ⋅ I v v∗ S R = (VR ∠0) ⋅ I

The direction of power flow will be inherent in the direction of the current I, i.e., SS is the supplied power when positive, and SR is the load power when positive. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Looking at the v sending-end v∗ complex power:

S S = (VS ∠α ) ⋅ I

v VS ∠ − α ) − (VR ∠0 ) ( S S = (VS ∠α ) ⋅ (Z∠ −θ ) v ' VS2 ( ' VS ⋅VR ( S S = ) ∠θ * − ) ∠ (α + θ ) * , + Z , + Z Getting the real and imaginary (reactive) components:

# VS2 $ # VS ⋅VR $ PS = ' ⋅ cos θ ( − ' ⋅ cos (α + θ ) ( * ) Z * ) Z # VS2 $ # VS ⋅VR $ QS = ' ⋅ sin θ ( − ' ⋅ sin (α + θ ) ( * ) Z * ) Z Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

If we assume the line reactance is much greater than the line resistance, i.e., xL >> rL, then we can neglect rL. This means θ = 90° and Z = X, which when we substitute in the previous equations yield:

" VS2 # " VS ⋅ VR # PS = & ⋅ cos 90° ' − & ⋅ cos (α + 90° ) ' ) ( X ) ( X " VS ⋅ VR # PS = & ⋅ sin α ' ( X ) " VS2 # " VS ⋅ VR # QS = & ⋅ sin 90° ' − & ⋅ sin (α + 90° ) ' ) ( X ) ( X " VS2 # " VS ⋅ VR # QS = & ' − & ⋅ cos α ' ) ( X ) ( X Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Looking at the receiving-end complex power:

v v∗ S R = (VR ∠0) ⋅ I

v VS ∠ − α ) − (VR ∠0 ) ( S R = (VR ∠0) ⋅ (Z∠ −θ ) 2 v ' VS ⋅VR ' ( V ( R SR = ) ∠ ( −α + θ ) * − ) ∠θ * + Z , + Z ,

Getting the real and imaginary (reactive) components: 2 $ % V ⋅ V V $ S R % R PR = ' ⋅ cos ( −α + θ ) ( − ' ⋅ cos θ ( ) Z * ) Z * 2 % $ VS ⋅VR % $ VR QR = ' ⋅ sin ( −α + θ ) ( − ' ⋅ sin θ ( ) Z * ) Z * Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

If we assume the line reactance is much greater than the line resistance, i.e., xL >> rL, then we can neglect rL. This means θ = 90° and Z = X, which when we substitute in the previous equations yield: 2 $ # VS ⋅ VR $ # VR PR = & ⋅ cos ( −α + 90° ) ' − & ⋅ cos 90° ' ( X ) ( X )

# VS ⋅ VR $ PR = & ⋅ sin α ' ( X ) 2 $ # VS ⋅ VR $ # VR QR = & ⋅ sin ( −α + 90° ) ' − & ⋅ sin 90° ' ( X ) ( Z ) 2 # $ V ⋅ V V # S R $ R QR = & ⋅ cos α ' − & ' ( X ) ( X ) Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Observations 1. 

Since we assumed that the transmission line consists of pure reactance, real power is not dissipated in the line and PS = PR.

2. 

If the transmission line resistance is non-negligible, we will have to use the “unsimplified” equations.

3. 

Maximum real power transfer occurs when α = 90°.

4. 

Real power transfer is more sensitive to the difference between phase angles of the supply voltage and the load voltage.

5. 

Reactive power transfer is more sensitive to the difference between magnitudes of the supply voltage and the load voltage. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

The Load Flow Problem Basic Electrical Engineering Solution How do you determine the voltage, current, power, and power factor at various points in a power system? Sending End

VS = ?

Line

1.1034 + j2.0856 ohms/phase ISR = ?

Receiving End

Solve for:·

VR = 13.2 kVLL

VOLTAGE DROP = VS - VR

Load 2 MVA, 3Ph 85%PF

1)  ISR = (SR/VR )* 2)  VD = ISRZL 3)  VS = VR + VD 4)  SS = VS·(ISR)*

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EEE 103 – Introduction to Power Systems

The Load Flow Problem Sending End

VS = ?

Line

1.1034 + j2.0856 ohms/phase ISR = ?

Receiving End

1)  ISR = (SR/VR )*

VR = 13.2 kVLL

S1φ = ( 2,000,000 / 3 )∠ cos −1 ( 0.85 ) = 666 ,666.67 ∠31.79 VA

Solve for:

Load 2 MVA, 3Ph 85%PF

2)  VD = ISRZL 3)  VS = VR + VD 4)  SS = VS·(ISR)*

VR = ( 13,200 / 3 )∠0 = 7621.02∠0 V ∗

% 666 ,666.67 ∠31.79 & I SR = ' ( = 87.48∠ − 31.79 A 7621.02∠0 ) * VD = ( 87.48∠ − 31.79 )(1.1034 + j2.0856 ) = 178.15 + j104.23 V VS = (7621.02 + j0 ) + (178.15 + j104.23 ) = 7,799.87 ∠0.77 V VS = 7,799.87 ∠0.77 /1000* 3 = 13.51 k V Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

The Load Flow Problem Load Flow From the Real World Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

How do you solve for: 1)  ISR = ?

Receiving End

VR = ? Load 2 MVA, 3Ph 85%PF

2)  VD = ? 3)  VR = ? 4)  SS = ? Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

The Load Flow Problem Load Flow of Distribution System Bus2

I23 , Loss23 = ?

Bus1 I12 , Loss12 = ?

Utility Grid

I24 , Loss24 = ? V1 = 67 kV P1 , Q1 = ?

V2 = ? P2 , Q2 = ?

How do you solve for the Voltages, Currents, Power and Losses?

Bus3 V3 = ? P3 , Q3 = ?

V4 = ? P4 , Q4 = ? Bus4 Lumped Load A 2 MVA 85%PF Lumped Load B 1 MVA 85%PF

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EEE 103 – Introduction to Power Systems

The Load Flow Problem Load Flow of Transmission and Subtransmission System G

G

Line 1

2

1 How do you solve for the Voltages, Currents and Power of a LOOP power system?

Line 2

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Line 3 3

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EEE 103 – Introduction to Power Systems

The Load Flow Problem !  How do you determine the voltage, current, and power flows, at various points in the power system, under existing conditions of normal operations? !  How do you determine the adequacy of the power system in meeting the demand during contingencies? !  How about if there are contemplated changes in the power system? How will you determine in advance the effects of: !  Growth or Addition of loads !  Addition or Decommissioning of generating plants !  Expansion of the transmission and distribution systems

before the proposed changes are implemented? Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

The Load Flow Problem ANSWER: THE LOAD FLOW STUDY! Load Flow Analysis simulates (i.e., mathematically determines) the performance of an electric power system under a given set of conditions. Load Flow (also called Power Flow) takes a snapshot of the electric power system at a given point in time.

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EEE 103 – Introduction to Power Systems

POWER SYSTEM MODELS FOR LOAD FLOW ANALYSIS

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EEE 103 – Introduction to Power Systems

Network Models !  The static components of the power system are modeled by the bus admittance matrix, [Ybus].

[YBUS] =

&Y11 $ $Y21 $ $Y31 $ $ $ $Yn 1 %

Y12 Y22 Y32  Yn 2

Y13 Y1 n # ! Y23 Y2 n ! ! Y33 Y3 n ! !   ! ! Yn 3 Ynn ! "

The number of buses (excluding the neutral bus) determines the dimension of the bus admittance, [Ybus]. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Generator Models 1. 

2. 

Voltage-controlled generating units to supply a scheduled active power P at a specified voltage magnitude V. The generators are equipped with voltage regulators to adjust the field excitation so that the units will supply or absorb a particular reactive power Q in order to maintain the voltage. Swing generating units to maintain the frequency at 60Hz in addition to the specified voltage. The generating unit is equipped with frequency following controller (quick-responding speed governor) and is assigned as the Swing Generator. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Bus Types !  The power system is interconnected through the busses. The busses must therefore be identified in the load flow model. ! Generators, shunt admittances, and loads are connected from their corresponding bus to the neutral bus. ! Transmission lines, transformers, and series impedances are connected from bus to bus.

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EEE 103 – Introduction to Power Systems

Bus Types !  To completely describe a particular bus, four quantities must be specified: ! Bus Voltage Magnitude, |VP| ! Bus Voltage Phase Angle, δP ! Bus Injected Active Power, PP ! Bus Injected Reactive Power, QP

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EEE 103 – Introduction to Power Systems

Swing Bus Swing Bus or Slack Bus The difference between the total load demand plus losses (both P and Q) and the scheduled generations is supplied by the swing bus. The voltage magnitude and phase angle are specified for the Swing Bus, also called the Slack Bus. P,Q + Swing Bus V∠δ G -

Specify: V, δ Unknown: P, Q

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EEE 103 – Introduction to Power Systems

Generator Bus Generator Bus (Voltage-Controlled) Bus PV Bus

or

The total real power Pp injected into the system through the bus is specified together with the magnitude of the voltage Vp at the bus. The bus voltage magnitude is maintained through reactive P,Q power injection. + Generator Bus V∠δ G -

Specify: P, V Unknown: Q, δ

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EEE 103 – Introduction to Power Systems

Load Bus Load Bus or PQ Bus The total injected power Pp and the reactive power Qp at Bus P are specified and are assumed constant, independent of the small variations in bus voltage. P,Q Load Bus

+

V∠δ -

Specify: P, Q Unknown: V, δ

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EEE 103 – Introduction to Power Systems

Summary of Bus Types Bus Type

Known Unknown Quantities Quantities

Swing

Vp, δp

Pp, Qp

Generator

Pp, Vp

Qp, δp

Load

Pp, Qp

Vp, δp

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EEE 103 – Introduction to Power Systems

Bus Types G Injected Powers:

2

1

PBUS = PGEN – PLOAD Line 2 QBUS = QGEN – QLOAD Bus No. 1 2 3

G

Line 1

Voltage Generation V (p.u.) δ P Q 1.0 0.0 * * 1.0 * 0.20 * * * 0 0

Line 3 3 Load

Remarks

P Q 0 0 0 0 0.60 0.25

Swing Bus Gen Bus Load Bus

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EEE 103 – Introduction to Power Systems

SOLUTIONS TO SIMULTANEOUS ALGEBRAIC EQUATIONS

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EEE 103 – Introduction to Power Systems

Numerical Methods !  Direct Methods ! Cramer’s Rule ! Matrix Inversion ! Gaussian Elimination Method ! Gauss-Jordan Reduction Method !  Iterative Methods ! Gauss Iterative Method ! Gauss-Seidel Iterative Method ! Newton-Raphson Method Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Iterative Methods An iterative method (root word: iterate) is a repetitive process for obtaining the solution of an equation or a system of equations. The solutions start from arbitrarily chosen initial estimates of the unknown variables from which a new set of estimates is determined. Convergence is achieved when the absolute mismatch between the current and previous estimates is less than some acceptable pre-specified precision index (the convergence index) for all variables.

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Given the system of algebraic equations,

a11 x1 + a12 x2 +  + a1n xn = y1 a21 x1 + a22 x2 +  + a2n xn = y 2 ↓

↓

↓

↓

a31 x1 + a32 x2 +  + ann xn = y3 In the above equation, the x’s are unknown. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method In general, the jth equation may be written as

1 n xj = ( b j − ∑i =1 a ji xi ) a jj

equation “a”

i≠ j

j = 1, 2, … n Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method In general, the Gauss iterative estimates are:

x1

k +1

x2

k +1

xn

k +1

a1n k y1 a12 k a13 k = − x2 − x3 − ... − xn a11 a11 a11 a11 a2n k y2 a21 k a23 k = − x1 − x3 − ... − xn a22 a22 a22 a22 an,n-1 y n an1 k an2 k k = − x1 − x2 − ... − x n -1 ann ann ann ann

where k is the iteration count Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method From an initial estimate of the unknowns (x10, x20, …xn0), updated values of the unknown variables are computed using equation “a”. This completes one iteration. The new estimates replace the original estimates. Mathematically, at the kth iteration,

x

k +1 j

1 n k = ( b j − ∑i =1 a ji x ) a jj i≠ j

equation “b”

i

j = 1, 2, … n Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method A convergence check is conducted after each iteration. The latest values are compared with their values respectively.

Δx = x k

k +1 j

−x

k j

equation “c”

j = 1, 2, … n The iteration process is terminated when:

max | Δx | < ε k j

(convergent)

k = maximum no. of iterations (non-convergent) Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method

Example:

4x1 − x2 + x3 = 4 x1 + 4x2 + x3 = 6 x1 + x2 + 3x3 = 5

Assume a convergence index of ε = 0.001 and use the following initial estimates:

a) x1 = x2 = x3 = 0.0 0

0

0

b) x1 = x2 = x3 = 0.5 0

0

0

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Solution: a) The system of equation must be expressed in standard form.

x1

1 = ( 4 + x2k - x3k ) 4

x2

1 = ( 6 - x1k - x3k ) 4

x3

1 = ( 5 - x1k - x2k ) 3

k +1

k +1

k +1

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 1 (k = 0):

1 ( 4 + 0 - 0 ) = 1.0 4 1 1 x2 = ( 6 - 0 - 0 ) = 1.5 4 1 x3 1 = ( 5 - 0 - 0 ) = 1.6667 3 Δx10 = 1 − 0 = 1 x1 1 =

Δx20 = 1.5 − 0 = 1.5 Δx30 = 1.6667 − 0 = 1.6667 max Δx30 = 1.6667 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 2 (k = 1):

1 x1 = ( 4 + 1.5 - 1.6667 ) = 0.958333 4 1 2 x2 = ( 6 - 1.0 - 1.6667 ) = 0.833333 4 1 x3 2 = ( 5 - 1.0 - 1.5 ) = 0.833333 3 Δx11 = 0.958325 − 1 = 0.041667 2

Δx21 = 0.833333 − 1.5 = 0.66667 Δx31 = 0.833333 − 1.6667 = −0.83334 max Δx31 = 0.83334 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 3 (k = 2):

1 x1 = ( 4 + 0.8333 - 0.8333 ) = 1.0 4 1 3 x2 = ( 6 - 0.9583 - 0.8333 ) = 1.0521 4 1 3 x3 = ( 5 - 0.9583 - 0.8333 ) = 1.0695 3 Δx12 = 1 − 0.958325 = 0.041667 3

Δx22 = 1.0521 − 0.833325 = 0.21877 Δx32 = 1.0695 − 0.8333 = 0.23617 max Δx32 = 0.23617 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 4 (k = 3):

1 x1 = ( 4 + 1.0521 - 1.0695 ) = 0.9956 4 1 4 x2 = ( 6 - 1.0 - 1.0695 ) = 0.9826 4 1 4 x3 = ( 5 - 1.0 - 1.0521 ) = 0.9826 3 Δx13 = 0.9956 − 1 = −0.0044 4

Δx23 = 0.9826 − 1.0521 = −0.0695 Δx33 = 0.9826 − 1.0695 = −0.0869 max Δx33 = 0.0869 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 5 (k = 4):

1 x1 = ( 4 + 0.9826 - 0.9826 ) = 1.0 4 1 5 x2 = ( 6 - 0.9956 - 0.9826 ) = 1.0054 4 1 5 x3 = ( 5 - 0.9956 - 0.9826) = 1.0073 3 Δx14 = 1 − 0.9956 = 0.0044 5

Δx24 = 1.0054 − 0.9826 = −0.0228 Δx34 = 1.0073 − 0.9826 = 0.0247 max Δx34 = 0.0247 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 6 (k = 5):

1 x1 = ( 4 + 1.0054 - 1.0073 ) = 0.9995 4 1 6 x2 = ( 6 - 1.0 - 1.0071 ) = 0.9982 4 1 6 x3 = ( 5 - 1.0 - 1.0054 ) = 0.9982 3 Δx15 = 0.9995 − 1 = −0.0005 6

Δx25 = 0.9982 − 1.0054 = −0.0072 Δx35 = o.9982 − 1.0073 = −0.0091 max Δx35 = 0.0091 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 7 (k = 6):

1 x1 = ( 4 + 0.9982 - 0.9982 ) = 1.0 4 1 7 x2 = ( 6 - 0.9995 - 0.9982 ) = 1.0006 4 1 7 x3 = ( 5 - 0.9995 - 0.9982) = 1.0008 3 Δx16 = 1 − 0.9995 = 0.0005 7

Δx 62 = 1.0006 − 0.9982 = 0.0024 Δx36 = 1.0008 − 0.9982 = 0.0026 max Δx36 = 0.0026 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method Iteration 8 (k = 7):

1 x1 = ( 4 + 1.0006 - 1.0008 ) = 0.9995 4 1 8 x2 = ( 6 - 1.0 - 1.0008 ) = 0.9998 4 1 8 x3 = ( 5 - 1.0 - 1.0008 ) = 0.9998 3 Δx17 = 0.9995 − 1 = −0.0005 8

Δx27 = 0.9998 − 1.0006 = −0.0008 Δx37 = 0.9998 − 1.0008 = −0.0010 max Δx37 = 0.0010 Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss Iterative Method The Gauss iterative method has converged at iteration 8. The method yields the following solution:

x1 = 0.9995 x2 = 0.9998 x3 = 0.9998

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EEE 103 – Introduction to Power Systems

GAUSS-SEIDEL ITERATIVE METHOD FOR A SYSTEM OF EQUATIONS

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EEE 103 – Introduction to Power Systems

Gauss-Seidel Iterative Method The Gauss-Seidel method is an improvement over the Gauss iterative method. As presented in the previous section, the standard form of the jth equation may be written as follows.

1 n xj = ( b j − ∑i =1 a ji xi ) a jj

j = 1, 2, … n

i≠ j

From an initial estimates (x10, x20,…xn0), an updated value is computed for x1 using the above equation with j set to 1.This new value replaces x10 and is then used together with the remaining initial estimates to compute a new value for x2. The process is repeated until a new estimate is obtained for xn. This completes one iteration. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss-Seidel Iterative Method Within an iteration, the most recent computed values are used in computing for the remaining unknowns. In general, at iteration k,

x

k+1 j

1 n = (b j − ∑ i =1 a ji x iα ) a jj i≠ j

j = 1, 2, … n where α = k if i > j = k + 1 if i < j After each iteration, a convergence check is conducted. The convergence criterion applied is the same with Gauss Iterative Method. Electrical and Electronics Engineering Institute University of the Philippines

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EEE 103 – Introduction to Power Systems

Gauss-Seidel Iterative Method

An improvement to the Gauss Iterative Method k +1

x1

k +1

x2

a1n k y1 a12 k = − x2 − ... − xn a11 a11 a11 a2n k y2 a21 k +1 = − x1 − ... − xn a22 a22 a22

ai,i-1 k +1 ai,i+1 k +1 ain k +1 yi aij k +1 x i = − xi − ... − xi-1 − xi+1 − xn aii aii aii aii aii k +1

k +1

xn

an,n-1 yn an1 k +1 k +1 = − x1 − ... − xn-1 ann ann ann

Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method Example: Solve the system of equations using the Gauss-Seidel method. Used a convergence index of ε = 0.001

4x1 − x2 + x3 = 4 x1 + 4x 2 + x3 = 6 x1 + x2 + 3x3 = 5 x1 = x2 = x3 = 0.5 0

0

0

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EEE 103 – Introduction to Power Systems

Gauss-Seidel Iterative Method Solution: a) The system of equation must be expressed in standard form.

x

k +1 1

x

k +1 2

x3k +1

1 = ( 4 + x2k - x3k ) 4 1 k +1 k = ( 6 - x1 - x3 ) 4 1 = ( 5 - x1k +1 - x2k +1 ) 3

Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method with x1 = x2 = x3 = 0.5 0

0

Iteration 1 (k =0):

0

1 x1 = ( 4 + 0.5 - 0.5 ) = 1.0 4 1 1 x2 = ( 6 - 1.0 - 0.5 ) = 1.125 4 1 1 x3 = ( 5 - 1.0 - 1.125 ) = 0.9583 3 Δx10 = 1 − 0.5 = 0.50 1

Δx20 = 1.125 − 0.50 = 0.625 Δx30 = 0.9583 − 0.50 = 0.4583 max | Δx20 | = 0.625 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method Iteration 2 (k = 1):

1 x1 2 = ( 4 + 1.125 - 0.9583 ) = 1.0417 4 1 x2 2 = ( 6 - 1.0417 - 0.9583 ) = 1.0 4 1 2 x3 = ( 5 - 1.0417 - 1.0 ) = 0.9861 3 Δx11 = 1.0417 − 1 = 0.0417

Δx21 = 1 − 1.125 = −0.125 Δx31 = 0.9861 − 0.9583 = 0.0323 max | Δx21 | = 0.125 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method Iteration 3 (k = 2):

1 x1 3 = ( 4 + 1.0 - 0.9861 ) = 1.0035 4 1 3 x2 = ( 6 - 1.0035 - 0.9891 ) = 1.0026 4 1 x3 3 = ( 5 - 1.0035 - 1.0026 ) = 0.9980 3 Δx12 = 1.0035 − 1.0417 = −0.0382

Δx22 = 1.0026 − 1 = 0.0026 Δx32 = 0.9980 − 0.9861 = 0.0119 max | Δx32 | = 0.0119 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method Iteration 4 (k = 3):

1 x1 4 = ( 4 + 1.0026 - 0.9980 ) = 1.0012 4 1 x2 4 = ( 6 - 1.0012 - 0.9980) = 1.0002 4 1 4 x3 = ( 5 - 1.0 - 1.0012 - 1.0002) = 0.9995 3 Δx13 = 1.0012 − 1.0035 = 0.0023

Δx23 = 1.0002 − 1.0026 = −0.0024 Δx33 = 0.9995 − 0.9980 = 0.0015 max | Δx23 | = 0.0024 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method Iteration 5 (k = 4):

1 x1 = ( 4 + 1.0002 - 0.9995 ) = 1.0002 4 1 x2 5 = ( 6 - 1.0002 - 0.9995) = 1.0001 4 1 5 x3 = ( 5 - 1.0002 - 1.0001) = 0.9999 3 Δx14 = 1.0002 − 1.0012 = −0.001 5

Δx24 = 1.0001 − 1.0002 = −0.0001 Δx34 = 0.9999 − 0.9995 = 0.0004 max | Δx 4 | = 0.001 < ε Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Iterative Method The Gauss-Seidel Iterative Method has converged after only 5 iterations with the following solutions:

x1 = 1.0002 x2 = 1.0001 x3 = 0.9999

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Gauss-Seidel Load Flow Linear Formulation of Load Flow Equations

The real and reactive power into any bus P is: Pp + jQp = Vp Ip*

or

(1)

Pp - jQp = Vp* Ip

where

Pp = real power injected into bus P Qp = reactive power injected into bus P Vp = phasor voltage of bus P Ip = current injected into bus P

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Gauss-Seidel Load Flow Equation (1) may be rewritten as: Pp - jQp _________ Ip = Vp*

(2)

From the Bus Admittance Matrix equation, the current injected into the bus are: Ip = Yp1V1 + Yp2V2 + … + YppVp + … + YpnVn

(3)

I1 = Y11V1 + Y12V2 + Y13V3 I2 = Y21V1 + Y22V2 + Y23V3 I3 = Y31V1 + Y32V2 + Y33V3 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow Substituting (3) into (2) Pp - jQp _________ = Y V + Y V + … + Y V + … + Y V p1 1 p2 2 pp p pn n * Vp P1 – jQ1 _________ = Y V + Y V + Y V 11 1 12 2 13 3 * V1

(4)

P2 – jQ2 _________ = Y V + Y V + Y V 21 1 22 2 23 3 * V2 P3 – jQ3 _________ = Y V + Y V + Y V 31 1 32 2 33 3 * V3 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow Solving for Vp in (4) P 1 – jQ1 _______ Y11V1 = - (___ + Y12V2 + Y13V3) V1*

1 V1 = Y11

& P1 − jQ1 # $ V * − Y12V2 − Y13V3 ! % " 1

P 2 – jQ2 _______ Y22V2 = - (Y12V2 + ___ + Y13V3) V2*

1 V2 = Y22

& P2 − jQ2 # − Y21V1 − Y13V3 ! $ V* % " 2

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Gauss-Seidel Load Flow P 3 – jQ3 _______ Y33V3 = - (Y13V1 + Y23V2 + ___) V3*

1 V3 = Y33

& P3 − jQ3 # − Y31V1 − Y32V2 ! $ V* % " 3

n P jQ 1 p p _______ ___ Vp = - Σ YpqVq Ypp Vp* q=1

(5)

q≠p

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Gauss-Seidel Load Flow Gauss-Seidel Load Flow Solution Generalizing the Gauss-Seidel Load Flow, the estimate for the voltage Vp at bus p at the kth iteration is: n P jQ 1 p p ___ _______ - Σ YpqVqα k+1 Vp = Ypp (Vpk)* q=1

(6)

q≠p

where, α = k α=k+1

if p < q if p > q

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Gauss-Seidel Load Flow !  Gauss-Seidel Voltage Equations of the form shown in (6) are written for all buses except for the swing bus. The solution proceeds iteratively from an estimate of all bus voltages !  For a Load Bus (Type 3) whose real power and reactive power are specified, the G-S voltage equation is used directly to compute the next estimate of the bus voltage. !  For a Generator Bus (Type 2) where the voltage magnitude is specified, an estimate of Qp must be determined first. This estimate is then compared with the reactive power limits of the generator. If it falls within the limits, the specified voltage is maintained and the computed Qp is inputted, in the Gauss-Seidel equation. Otherwise, the reactive power is set to an appropriate limit (Qmin or Qmax) and the bus is treated as a load bus in the current iteration. Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow Numerical Example Shown in the figure is a 3-bus power system. The line and bus data pertinent to the system are also given. The reactive limits of generator 2 are -50 MVARS and 50 MVARS. Use base power of 100 MVA. Solve the load flow problem using GaussSeidel iterative method with a 0.005 convergence index.

G

G

Line 1

2

1 Line 2

Line 3 3

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Gauss-Seidel Load Flow Branch Data Line No. Bus Code Impedance Zpq (p.u.) 1 2 3

1-2 1-3 2-3

0.08 + j0.24 0.02 + j0.06 0.06 + j0.18

Bus Data Bus Voltage Generation No. V (p.u.) δ P Q 1 1.0 0.0 * * 2 1.0 * 0.20 * 3 * * 0 0

Load

Remarks

P Q 0 0 0 0 0.60 0.25

Swing Bus Gen Bus Load Bus

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Gauss-Seidel Load Flow The Bus Admittance Matrix is:

[YBUS] = Y11 = 6.25 - j18.75

Y12 = -1.25 + j3.75

Y13 = -5 + j15

Y21 = -1.25 + j3.75 Y22 = 2.9167 - j8.75

Y23 = -1.6667 + j5

Y31 = -5 + j15

Y33 = 6.6667 - j20

Y32 = -1.6667 + j5

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Gauss-Seidel Load Flow Specified Variables: V1 = 1.0 δ1 = 0.0 V2 = 1.0

P2 = 0.2

P3 = -0.6

Q3 = -0.25

Note the negative sign of P and Q of the Load at Bus 3

Initial Estimates of Unknown Variables: δ20 = 0.0 V30 = 1.0 δ30 = 0.0 Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow Equations !  Bus 1: Swing Bus ( k +1)

V1

= 1.0∠0

for all iterations

!  Bus 2: PV Bus We must first estimate the Q2 for Bus 2 by: k ∗ 2

P2 − jQ2 = (V k

) ⋅ (Y

21

( k +1) 1

⋅V

(k ) 2

+ Y22 ⋅V

(k ) 3

+ Y23 ⋅V

)

Then we substitute the Q2 value (and only the Q2 value) to:

V2( k +1)

"$ 1 ') P2 − jQ2k = ⋅ Y22 ') (V ( k ) )∗ 2 .,

# % * − Y ⋅ V ( k +1) − Y ⋅ V ( k ) ( 21 1 23 3 ( * /

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Gauss-Seidel Load Flow Equations !  Bus 3: PQ Bus The Gauss-Seidel Equation for PQ Bus is straightforward:

V3( k +1)

"$ # % 1 ') P3 − jQ3 * ( k +1) ( k +1) ( = ⋅ − Y ⋅ V − Y ⋅ V 13 1 23 3 ∗ ( k ) ' ( Y33 ) (V ) * ., 3 /

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Gauss-Seidel Load Flow: Example Iteration 1: k = 0 k V Bus 1: 1 = 1.0∠0 k k ∗ Bus 2:

P2 − jQ2 = (V2 Y21

) ⋅ (Y

21

for all iterations ( k +1) 1

⋅V

(k ) 2

+ Y22 ⋅V

(k ) 3

+ Y23 ⋅V

)

V1k

% ( −1.25 + j3.75 ) ⋅ (1∠0 ) & V20 Q20 ( ∗ ' 0 P2 − jQ2 = (1∠0 ) ⋅ ' + ( 2.9167 − j8.75 ) ⋅ (1∠0 )( = 0 − j 0 ' + −1.6667 + j5 ⋅ 1∠0 ( V30 discard Y22 ) ( )( ) * V20

Y23

Q20 = 0

(This value is within limits.)

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Gauss-Seidel Load Flow: Example Bus 2: (iteration 1, continued)

V2( k +1)

"$ 1 ') P2 − jQ2k = ⋅ Y22 ') (V ( k ) )∗ 2 ., P2

# % * − Y ⋅ V ( k +1) − Y ⋅ V ( k ) ( 21 1 23 3 ( * / Q20

Y21 V1k

"$ 0.2 − j 0 % # '+ , − ( −1.25 + j3.75) ⋅ (1∠0 )( 1 ∗ 1 V2 = ⋅ '+- (1∠0 ) ,. Y23 V30 ( ( 2.9167 − j8.75) ' 0 ( V2 − ( −1.6667 + j5) ⋅ (1∠0 ) (0 '/ Y 22

V21 = 1.0071∠1.1705o → V21a = 1.0∠1.1705o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Bus 3: (iteration 1)

"$ # % 1 ') P3 − jQ3 * ( k +1) ( k +1) ( k +1) ( V3 = ⋅ − Y ⋅ V − Y ⋅ V 31 1 32 2 ∗ ( k ) ' ( Y33 ) (V ) * ., 3 / P3 Q3 Y31 V1k "$ −0.60 + j 0.25 % # '+ , − ( −5 + j15 ) ⋅ (1∠0 )( ∗ 1 , 1a( V31 = ⋅ '+- (1∠0 ) Y V . 32 2 (6.6667 − j 20) ' 0 ( o V 3 − − 1.6667 + j 5 ⋅ 1.0 ∠ 1.1705 '/ ( )( ) (0 Y33

V31 = 0.9799∠ − 1.0609o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Check for convergence: (iteration 1) 1

1 2

0 2

Δ V2 = V − V

o

o

= (1.0071∠1.1705 ) − (1.0∠0

)

= 0.0217 > 0.005

Δ1V3 = V31 − V30 = ( 0.9799∠ − 1.0609o ) − (1.0∠0o ) = 0.0272 > 0.005 Action: Continue iterating. 1a 2

V

1 o V = 0.9799 ∠ − 1.0609 = 1.0∠1.1705 3 o

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Gauss-Seidel Load Flow: Example Iteration 2: k = 1 Bus 2: P2 −

k ∗ 2

jQ2 = (V k

) ⋅ (Y

21

( k +1) 1

⋅V

(k ) 2

+ Y22 ⋅V

Y21

(k ) 3

+ Y23 ⋅V

)

V1k

! ( −1.25 + j3.75) ⋅ (1∠0o ) V21a " & ' P2 − jQ 12 = (1.0∠ − 1.1705) ⋅ &+ ( 2.9167 − j8.75) ⋅ (1.0∠1.1705o ) V31' & ' o ' & discard + − 1.6667 + j 5 ⋅ 0.9799 ∠ − 1.0609 )( )) Y22 ( ( V21a*

P2 − jQ 12 = 0.3024 − j 0.0160

Q21 = 0.0160

Y23 (This value is within limits.)

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Gauss-Seidel Load Flow: Example Bus 2: (iteration 2, continued)

V2( k +1)

"$ 1 ') P2 − jQ2k = ⋅ Y22 ') (V ( k ) )∗ 2 ., P2

# % * − Y ⋅ V ( k +1) − Y ⋅ V ( k ) ( 21 1 23 3 ( * / Q21

Y21 V1k

!$ 0.2 − j 0.0160 % o # − ( −1.25 + j3.75) ⋅ (1∠0 ) + ( ) * o 1 V22 = ⋅ *, 1.0∠ − 1.1705 V31 + Y23 ( 2.9167 − j8.75) *V 1a* − −1.6667 + j5 ⋅ 0.9799∠ − 1.0610o + ( )( )/ .2 Y22

V22 = 0.9965∠0.5648o → V22 a = 1.0∠0.5648o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Bus 3: (iteration 2)

V3( k +1)

"$ # % 1 ') P3 − jQ3 * ( k +1) ( k +1) ( = ⋅ − Y ⋅ V − Y ⋅ V 31 1 32 2 ∗ ( k ) ' ( Y33 ) (V ) * ., 3 / P3

Q3

Y31 V1k

!$ −0.60 + j 0.25 % o # − ( −5 + j15 ) ⋅ (1∠0 )+ ( ) * o 1 V32 = ⋅ *, 0.9799∠1.0610 - Y32 V22 a + (6.6667 − j 20) *V 1* o + 3 − − 1.6667 + j 5 ⋅ 1.0 ∠ 0.5648 ( )( )/ . Y33

V32 = 0.9791∠ − 1.2218o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Check for convergence: (iteration 2) o 0.9965 ∠ 0.5648 2 Δ V2 = V22 − V21a = − 1.0∠1.1705o

(

)

(

)

= 0.0111 > 0.005 o 0.9791 ∠ − 1.2218 ( ) 2 2 1 Δ V3 = V3 − V3 = − ( 0.9799∠ − 1.0610o ) = 0.0029 < 0.005

Action: Continue iterating.

V22 a = 1.0∠0.5648o V32 = 0.9791∠ − 1.2218o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Iteration 3: k = 2 Bus 2: P2 −

k ∗ 2

jQ2 = (V k

) ⋅ (Y

21

( k +1) 1

⋅V

(k ) 2

+ Y22 ⋅V

Y21

(k ) 3

+ Y23 ⋅V

)

V1k

! ( −1.25 + j3.75) ⋅ (1∠0 ) V 2 a " 2 & ' P2 − jQ 22 = (1.0∠ − 0.5648) ⋅ & + ( 2.9167 − j8.75) ⋅ (1.0∠0.5648o ) V32' & ' o ' & discard + − 1.6667 + j 5 ⋅ 0.9791 ∠ − 1.2218 )( )) Y22 ( ( V22 a*

Y23

P2 − jQ 22 = 0.2253 − j 0.0438

Q22 = 0.0438

(This value is within limits.)

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Gauss-Seidel Load Flow: Example Bus 2: (iteration 3, continued)

V2( k +1)

"$ 1 ') P2 − jQ2k = ⋅ Y22 ') (V ( k ) )∗ 2 ., P2

# % * − Y ⋅ V ( k +1) − Y ⋅ V ( k ) ( 21 1 23 3 ( * / Q21

Y21 V1k

!$ 0.2 − j 0.0438 % o # − ( −1.25 + j3.75 ) ⋅ (1∠0 ) + ( ) * o 1 V32 + Y23 V23 = ⋅ *, 1.0∠ − 0.5648 ( 2.9167 − j8.75) *V22 a* − −1.6667 + j5 ⋅ 0.9791∠ − 1.2218o + ( )( )/ . Y22

V23 = 0.9991∠0.4158o → V23a = 1.0∠0.4158o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Bus 3:

V3( k +1)

"$ # % 1 ') P3 − jQ3 * ( k +1) ( k +1) ( = ⋅ − Y ⋅ V − Y ⋅ V 31 1 32 2 ∗ ( k ) ' ( Y33 ) (V ) * ., 3 / P3

Q3

Y31 V1k

!$ −0.60 + j 0.25 % # − ( −5 + j15 ) ⋅ (1∠0 ) + ( ) * o 1 V23 a V33 = ⋅ *, 0.9799∠1.0610 - Y32 + (6.6667 − j 20) * V 2* o + − ( −1.6667 + j5 ) ⋅ (1.0∠0.4158 )/ . 3 Y33

V33 = 0.9790∠ − 1.2575o Electrical and Electronics Engineering Institute University of the Philippines

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Gauss-Seidel Load Flow: Example Check for convergence:

Δ V2

3

= V23 − V22 a =

o 0.9991 ∠ 0.4158 ( )

− (1.0∠0.5648o )

= 0.0027 < 0.005 o 0.9791 ∠ − 1.2575 ( ) 3 3 2 Δ V3 = V3 − V3 = − ( 0.9791∠ − 1.2218o ) = 0.0006 < 0.005

Action: Stop iterating. The solution has converged.

V23a = 1.0∠0.4158o V33 = 0.9790∠ − 1.2575o Electrical and Electronics Engineering Institute University of the Philippines

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Principles of Load Flow Control Generator Voltage & Power Control jX Ei∠δ

~

I

The complex power delivered to the bus (Generator Terminal) is

Vt∠0

& EiVt # Pt = $ sin δ ! % X "

& Ei ∠δ − Vt ∠0 # Pt + jQt = [Vt ∠0]I = [Vt ∠0]$ ! jX % " *

& EiVt Vt 2 # Qt = $ cos δ − ! X" % X

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Principles of Load Flow Control Generator Voltage & Power Control

& EiVt # Pt = $ sin δ ! % X "

& EiVt Vt 2 # Qt = $ cos δ − ! X" % X

Observations: 1.  Real Power is injected into the bus (Generator Operation), δ must be positive (Ei leads Vt) 2.  Real Power is drawn from the bus (Motor Operation), δ must be negative (Ei lags Vt) 3.  In actual operation, the numeric value of δ is small & since the slope of Sine function is maximum for small values, a minute change in δ can cause a substantial change in Pt Electrical and Electronics Engineering Institute University of the Philippines

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Principles of Load Flow Control Generator Voltage & Power Control

& EiVt # Pt = $ sin δ ! % X "

& EiVt Vt 2 # Qt = $ cos δ − ! X" % X

Observations: 4.  Reactive Power flow depends on relative values of EiCosδ and Vt 5.  Since the slope of Cosine function is minimum for small values of angle, Reactive Power is controlled by varying Ei •  Over-excitation (increasing Ei) will deliver Reactive Power into the Bus •  Under-excitation (decreasing Ei) will absorb Reactive Power from the Bus Electrical and Electronics Engineering Institute University of the Philippines

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Principles of Load Flow Control Capacitor Compensation ~ p

Ipq

The voltage of bus q can be expressed as

q + jQc

Observations:

PL - jQL

Eq = V p −

X pqQq Vp

−j

X pq Pq Vp

1.  The Reactive Power Qq causes a voltage drop and thus largely affects the magnitude of Eq 2.  A capacitor bank connected to bus q will reduce Qq that will consequently reduce voltage drop Electrical and Electronics Engineering Institute University of the Philippines

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Principles of Load Flow Control Tap-Changing Transformer a:1 q s

r

The π equivalent circuit of transformer with the per unit transformation ratio:

1 y pq a

p

Observation: The voltage drop in the transformer is affected by the transformation ratio “a”

1− a y pq 2 a

Electrical and Electronics Engineering Institute University of the Philippines

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Information from a Load Flow Study BASIC INFORMATION " " " " "

  Voltage Profile   Injected Power (Pp and Qp)   Line Currents (Ipq and Ipq)   Power Flows (Ppq and Qpq)   Line Losses (I2R and I2X)

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Information from a Load Flow Study

The bus voltages are: V1 = 1.0∠0° V2 = 0.9990∠0.4129° V3 = 0.9788∠-1.2560°

The power injected into the buses are: P1 - jQ1 = V1* [Y11V1 + Y12V2 + Y13V3 ] P1 - jQ1 = (1.0∠0) [(19.7642∠-71.5651°)(1.0∠0°) + (3.9528∠108.4349°)(0.9990∠0.4129°) + (15.8114∠108.4349°) (0.9788∠-1.25560°) = 0.4033 - j0.2272 Electrical and Electronics Engineering Institute University of the Philippines

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Information from a Load Flow Study

P2 - jQ2 = V2* [Y21V1 + Y22V2 + Y23V3 ] P2 - jQ2 = (0.999∠-0.4129°)[(3.9528∠108.4349°)(1.0∠0°) + (9.2233∠-71.5649°)(0.9990∠0.4129°) + (5.2705∠108.4349°)(0.9788∠-1.25560°) = 0.2025 - j0.04286 P3 - jQ3 = V3* [Y31V1 + Y32V2 + Y33V3 ] P3 - jQ3 = (0.9788∠1.256°) [(15.8114∠108.4349°)(1.0∠0°) + (5.2705∠108.4349°)(0.9990∠0.4129°) + (21.0819 ∠ -71.5650°)(0.9788∠-1.25560° ) = -0.600 + j0.2498 Electrical and Electronics Engineering Institute University of the Philippines

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Information from a Load Flow Study

Line Currents Ipq

Vp p

Iline

ypq

ypo

Vq

yqo

Iqp

q

The line current Ipq, measured at bus p is given by

I pq = I line + I po = y pq ( V p − Vq ) + y poV p Similarly, the line current Iqp, measured at bus q is

I qp = − I line + I qo = y pq ( Vq − V p ) + yqoVq Electrical and Electronics Engineering Institute University of the Philippines

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Information from a Load Flow Study The branch currents are:

I pq = I line = y pq ( V p − Vq )

I qp = − I line = y pq ( Vq − V p )

I12 = y12 [V1 - V2]

I21 = y12 [V2 – V1]

I13 = y13 [V1 – V3]

I31 = y13 [V3 – V1]

I23 = y23 [V2 – V3]

I32 = y23 [V3 – V2]

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Information from a Load Flow Study Power Flows The power flow (Spq) from bus p to q is

S

=P

pq

pq

+ jQ

pq

= V I* p

pq

The power flow (Sqp) from bus q to p is

S =P qp

qp

+ jQ = V I * qp

q qp

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Information from a Load Flow Study The branch power flows are: P12 – jQ12 = V1* I12

P21 – jQ21 = V2* I21

P13 – jQ13 = V1* I13

P31 – jQ31 = V3* I31

P23 – jQ23 = V2* I23

P32 – jQ32 = V3* I32

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 100 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Information from a Load Flow Study Line Losses The power loss in line pq is the algebraic sum of the power flows Spq and Sqp

Sloss = Ploss + jQloss = S pq + Sqp

= Vp I

* pq

− Vq I

= (V p + Vq )I

* pq

* pq

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 101 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Information from a Load Flow Study The line losses are: P12(Loss) – jQ12(Loss) = (P12 – jQ12) + (P21 – jQ21 )

P13(Loss) – jQ13(Loss) = (P13 – jQ13) + (P31 – jQ31 )

P23(Loss) – jQ23(Loss) = (P23 – jQ23) + (P32 – jQ32 )

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 102 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Information from a Load Flow Study Other Information: "   Overvoltage and Undervoltage Buses "   Critical and Overloaded Transformers and Lines "   Total System Losses

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 103 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies Sensitivity Analysis with Load Flow Study 1) Take any line, transformer or generator out of service. 2) Add, reduce or remove load to any or all buses. 3) Add, remove or shift generation to any bus. 4) Add new transmission or distribution lines. 5) Increase conductor size on T&D lines. 6) Change bus voltages. 7) Change transformer taps. 8) Increase or decrease transformer size. 9) Add or remove rotating or static var supply to buses. Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 105 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies ! Sensitivity Analysis Example IEEE 14-Bus System

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 106 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Sensitivity

Analysis Example

Removal of Line 4-5

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 107 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies ! Sensitivity Analysis Example IEEE 14-Bus System

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 108 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Sensitivity

Analysis Example

Removal of generator at Bus 2

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 109 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies ! Sensitivity Analysis Example IEEE 14-Bus System

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 110 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Sensitivity

Analysis Example

Removal of rotating VAR supply at bus 3

From 1.010 pu

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO EEE 103 AY2010-11 S2

111

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies ! Sensitivity Analysis Example IEEE 14-Bus System

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 112 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Sensitivity

Analysis Example

From 14.9 MW + 5.0 MVAR; V: from 1.035 pu Line 9-14: From 50% loading

Increase in P and Q at bus 14

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 113 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Analysis of existing conditions: ! Check for voltage violations (undervoltage/overvoltage). ! Check for transformer overloading/line overloading. ! Check for system losses.

!  Analysis for correction of power quality issues: ! Voltage adjustment at the delivery points ! Transformer tap changing ! Capacitor compensation: "  Compensation for Peak Loading "  Check for overvoltages during Off-Peak conditions "  Optimize capacitor allocation and capacitor switching Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 114 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Analysis for Expansion Planning: ! Construction of new substation ! Addition of capacity to existing substation ! Construction of new feeder segment ! Extension of existing feeder segment ! Addition of parallel feeder segment ! Replacement of conductors in existing feeder segments ! Conversion of entire feeder circuits from one voltage level to another voltage level ! Addition of generators Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 115 EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Uses of Load Flow Studies !  Contingency Analysis: ! Reliability of the Transmission, Subtransmission, and Distribution Systems Reliability denotes that not only is the power system working, but that it is working properly. That is, no physical and technical constraints must be violated – i.e., voltage must be well regulated and within acceptable range, load limits of the transformers and the lines must not be exceeded, and power balance must be satisfied.

!  System Loss Analysis: ! Identification of lossy components in the power system.

Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO 116 EEE 103 AY2010-11 S2