Load Flow Analysis

Competency Training and Certification Program in Electric Power Distribution System Engineering

Certificate in

Power System Modeling and Analysis Training Course in

Load Flow Analysis

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Training Course in Load Flow Analysis

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Course Outline 1. The Load Flow Problem 2. Power System Models for Load Flow Analysis 3. Gauss-Seidel Load Flow 4. Newton-Raphson Load Flow 5. Backward/Forward Sweep Load Flow 6. Principles of Load Flow Control 7. Uses of Load Flow Studies U. P. National Engineering Center National Electrification Administration

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The Load Flow Problem

Basic Electrical Engineering Solution

Load Flow of Distribution System

Load Flow of Transmission and Subtransmission System

Load Flow of a Contemplated System

Load Flow of a Single Line

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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

Line

1.1034 + j2.0856 ohms/phase

Receiving End

Solve for: VS = ?

ISR = ?

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 = VSx(ISR)*

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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 = VSx(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 U. P. National Engineering Center National Electrification Administration

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

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

Receiving End

ISR = ?

How do you solve for:

VR = ?

Load 2 MVA, 3Ph 85%PF

1) ISR = ?

2) VD = ?

3) VR = ?

4) SS = ?

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The Load Flow Problem Load Flow of Distribution System Bus2 Bus1 I12 , Loss12 = ?

Utility Grid

V1 = 67 kV P1 , Q1 = ?

I23 , Loss23 = ?

Bus3 V3 = ?

P3 , Q3 = ?

I24 , Loss24 = ? V4 = ?

P4 , Q4 = ?

Bus4 V2 = ?

Lumped Load A P2 , Q2 = ?

2 MVA 85%PF Lumped Load B 1 MVA 85%PF

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

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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?

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Line 2

Line 3 3

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The Load Flow Problem Load Flow of a Contemplated System How about if there are contemplated changes in the System?

How will you determine in advance the effects of: • Growth or addition of new loads • Addition of generating plants • Upgrading of Substation • Expansion of distribution lines before the proposed changes are implemented?

Answer: LOAD FLOW ANALYSIS U. P. National Engineering Center National Electrification Administration

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The Load Flow Problem Load

Flow

Analysis

simulates (i.e., mathematically determine) the performance of an electric power system under a given set of conditions.

Load Flow (also called Power Flow) is a snapshot picture of the power system at a given point.

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The Load Flow Problem Load Flow of a Single Line Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

Receiving End

ISR = ?

VR = ?

Injected Power at Receiving End SR = VR x (ISR)*

Solving for the Current ISR = (SR / VR)* U. P. National Engineering Center National Electrification Administration

Load 2 MVA, 3Ph 85%PF

Voltage at Sending End VS = VR + Z x ISR

Voltage at Receiving End VR = VS - Z x SR*/VR* Competency Training & Certification Program in Electric Power Distribution System Engineering

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The Load Flow Problem Load Flow of a Single Line Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

Receiving End

ISR = ?

VR = ?

Load 2 MVA, 3Ph 85%PF

Converting Quantities in Per Unit Base Power = 1 MVA

VS(pu) = 13.2 /13.2 = 1/0

Base Voltage = 13.2 kV

SR(pu) = 2/cos-1(0.85) / 1

Base Impedance = [13.2]2/1

Zpu = (1.1034 + j2.0856)/174.24

= 174.24 ohms U. P. National Engineering Center National Electrification Administration

= 0.00633 + j0.01197 Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Load Flow Analysis

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The Load Flow Problem Load Flow of a Single Line Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

Receiving End

ISR = ?

VR = ?

Load 2 MVA, 3Ph 85%PF

VR(k) = VS - Z x [SR]* / [VR(k-1) ]* Let

VR(0) = 1/0

For k = 1

For k = 2

VR(1) = __________

VR(2) = __________

ΔV(1) = __________

ΔV(2) = __________

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The Load Flow Problem Load Flow of a Single Line Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

Receiving End

ISR = ?

VR = ?

Load 2 MVA, 3Ph 85%PF

VR(k) = VS - Z x [SR]* / [VR(k-1) ]* VR(2) = __________ For k = 3

For k = 4

VR(3) = __________

VR(4) = __________

ΔV(3) = __________

ΔV(4) = __________

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Training Course in Load Flow Analysis

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The Load Flow Problem Load Flow of a Single Line Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

Receiving End

ISR = ?

VR = ?

Load 2 MVA, 3Ph 85%PF

VS = __________

ISR = __________

VR = __________

SR = __________

VD = VS – VR

SS = VS x [ISR]*

VD = __________

SS = __________

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Power System Models for Load flow Analysis

Bus Admittance Matrix, Ybus

Network Models

Generator Models

Bus Types for Load Flow Analysis

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Power System Models for Load Flow Analysis The power system components are interconnected through the buses. The buses must therefore be identified in the load flow model.

Generators and loads are connected from bus to neutral.

Transmission lines and transformers are connected from one bus to another bus.

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Power System Models for Load Flow Analysis Network Models The static components (transformers and lines) are represented by the bus admittance matrix, Ybus

[YBUS] =

⎡ Y 11 ⎢ ⎢ Y 21 ⎢ ⎢ Y 31 ⎢ ⎢ M ⎢ ⎢Y n 1 ⎣

Y 12 Y 22 Y 32 M Yn2

Y 13 L Y 1 n ⎤ ⎥ Y 23 L Y 2 n ⎥ ⎥ Y 33 L Y 3 n ⎥ ⎥ M M ⎥ ⎥ Y n 3 L Y nn ⎥ ⎦

The number of buses (excluding the neutral bus) determines the dimension of the bus admittance, Ybus.

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Power System Models for Load Flow Analysis Network Models Line No. Bus Code Impedance Z pq (p.u.)

1 2 3

1 -2 1 -3 2 -3

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

Line 1 2

1 Line 2

Line 3

Set-up the Ybus

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Power System Models for Load Flow Analysis Network Models Compute the branch admittances to set up Ybus: 1 1 ______________ y12 = ____ = 1.25 - j3.75 = z12 0.08 + j0.24 1 1 ______________ y13 = ____ = 5 - j15 = z13 0.02 + j0.06 1 1 ______________ = 1.667 - j5 y23 = ____ = z23 0.06 + j0.18 U. P. National Engineering Center National Electrification Administration

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Power System Models for Load Flow Analysis Set-up the bus admittance matrix: Y11 = y12 + y13 = (1.25 - j3.75) + (5 - j15) = 6.25 - j18.75 = 19.7642 ∠ -71.5651° Y12 = -y12 = -1.25 + j3.75 = 3.9528 ∠ 108.4349° Y13 = -y13 = -5 + j15 = 15.8114 ∠ 108.4349° Y21 = Y12 = -y12 = -1.25 + j3.75 = 3.9528 ∠ 108.4349° U. P. National Engineering Center National Electrification Administration

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Power System Models for Load Flow Analysis Y22 = y12 + y23 = (1.25 - j3.75) + (1.6667 - j5) = 2.9167 - j8.75 = 9.2233 ∠ -71.5649° Y23 = -y23 = -1.6667 + j5 = 5.2705 ∠ 108.4349° Y31 = Y13 = -y13 = -5 + j15 = 15.8114 ∠ 108.4349° Y32 = Y23 = -y23 = -1.6667 + j5 = 5.2705 ∠ 108.4349° Y33 = y13 + y23 = (5 - j15) + (1.6667 - j5) = 6.6667 - j20 = 21.0819 ∠ -71.5650° U. P. National Engineering Center National Electrification Administration

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Power System Models for Load Flow Analysis Generator Models Voltage-controlled generating units to supply a scheduled active power (P) at a specified voltage (V). The generating units are equipped with voltage regulator to adjust the field excitation so that the units will operate at particular reactive power (Q) in order to maintain the voltage.

Swing generating units to maintain the frequency at 60Hz in addition to maintaining the specified voltage. The generating unit is equipped with frequency-following controller (very fast speed governor) and is assigned as Swing generator U. P. National Engineering Center National Electrification Administration

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Power System Models for Load Flow Analysis Bus Types for Load Flow Generators and loads are connected from bus to neutral.

Four quantities must be specified to completely describe a bus. These are:

Bus Bus Bus Bus

voltage magnitude, Vp voltage phase angle, δp injected active power, Pp injected reactive power, Qp

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Power System Models for Load Flow Analysis 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.

P,Q + Type 1: G V∠ δ Swing Bus U. P. National Engineering Center National Electrification Administration

Specify: V, δ Unknown: P, Q

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Power System Models for Load Flow Analysis Generator Bus (Voltage-Controlled) Bus or PV Bus 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. + Type 2: G V∠ δ Generator Bus U. P. National Engineering Center National Electrification Administration

Specify: P, V Unknown: Q, δ

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Power System Models for Load Flow Analysis 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 Type 3: Load Bus

+

V∠δ -

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Power System Models for Load Flow Analysis SUMMARY OF BUS TYPES Bus Type

Know n Q u a n t it ie s

Unknow n Q u a n t it ie s

Type1: S w in g

V p, δ p

P p, Q p

Type 2: G e n e ra to r

P p, V p

Q p, δ p

Type 3: Load

P p, Q p

V p, δ p

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Power System Models for Load Flow Analysis G

G

Line 1

2

1

Bus Types

Line 2

Line 3 3

Bus No. 1 2 3

Voltage Generation V (p.u.) δ P Q 1.0 0.0 * * 1.0 * 0.20 * * * 0 0 U. P. National Engineering Center National Electrification Administration

Load P Q 0 0 0 0 0.60 0.25

Remarks Swing Bus Gen Bus Load Bus

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Gauss-Seidel Load Flow

Linear Formulation of Load Flow Equations

Gauss-Seidel Load Flow Solution

Numerical Example

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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 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow Substituting (3) into (2) Pp - jQp _________ = Yp1V1 + Yp2V2 + … + YppVp + … + YpnVn Vp* P1 – jQ1 _________ = Y11V1 + Y12V2 + Y13V3 V1*

(4)

P2 – jQ2 _________ = Y21V1 + Y22V2 + Y23V3 V2* P3 – jQ3 _________ = Y31V1 + Y32V2 + Y33V3 V3* U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow Solving for Vp in (4) Y11V1 =

P 1 – jQ1 _______ V1*

1 V1 = Y11 Y22V2 =

⎡ P1 − jQ1 ⎤ − Y12V2 − Y13V3 ⎥ ⎢ V* ⎣ ⎦ 1

P 2 – jQ2 _______

1 V2 = Y22

- (___ + Y12V2 + Y13V3)

V2*

- (Y12V2 + ___ + Y13V3)

⎡ P2 − jQ2 ⎤ − Y21V1 − Y13V3 ⎥ ⎢ V* ⎣ ⎦ 2

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Gauss-Seidel Load Flow Y33V3 =

P3 – jQ3 _______ V3*

- (Y13V1 + Y23V2 + ___)

⎤ 1 ⎡ P3 − jQ3 V3 = − Y31V1 − Y32V2 ⎥ ⎢ * Y33 ⎣ V3 ⎦ 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 U. P. National Engineering Center National Electrification Administration

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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.

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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 zero and 50 MVARS, respectively.

Base power used is 100 MVA. Solve the load flow problem using Gauss-Seidel iterative method assuming 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 Z pq (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 U. P. National Engineering Center National Electrification Administration

Load P Q 0 0 0 0 0.60 0.25

Remarks Swing Bus Gen Bus Load Bus

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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 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow The Bus Admittance Matrix elements are: Y11 = 6.25 - j18.75 = 19.7642 ∠ -71.5651° Y12 = -1.25 + j3.75 = 3.9528 ∠ 108.4349° Y13 = -5 + j15 = 15.8114 ∠ 108.4349° Y21 = -1.25 + j3.75 = 3.9528 ∠ 108.4349° Y22 = 2.9167 - j8.75 = 9.2233 ∠ -71.5649° Y23 = -1.6667 + j5 = 5.2705 ∠ 108.4349° Y31 = -5 + j15 = 15.8114 ∠ 108.4349° Y32 = -1.6667 + j5 = 5.2705 ∠ 108.4349° Y33 = 6.6667 - j20 = 21.0819 ∠ -71.5650° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow Gauss-Seidel Equations Bus 1: Swing Bus V1

(k + 1 )

= 1∠ 0

for all iterations

Bus 2: Generator Bus Q2 must first be determined from: P2 - jQ2(k+1) = (V2(k))* [Y21V1(k+1) + Y22V2(k) + Y23V3(k)] then substitute it to: V2

(k + 1 )

(k + 1 ) ⎤ 1 ⎡ P2 − jQ 2 (k + 1 ) (k ) ⎢ = − Y 21 V 1 − Y 23 V 3 ⎥ * (k ) Y 22 ⎢ ⎥⎦ V ⎣ 2

(

)

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Gauss-Seidel Load Flow Bus 3: Load Bus

V3

(k + 1 )

⎤ 1 ⎡ P3 − jQ 3 (k + 1 ) (k + 1 ) ⎥ ⎢ = − Y 31 V 1 − Y 32 V 2 * Y 33 ⎢ V (k ) ⎥⎦ 3 ⎣

(

)

Iteration 1 (k = 0): V1 (1) = 1.0∠0° P2 - jQ2(1) = (1.0∠0°) [(-1.25 + j3.75)(1.0∠0°) + (2.9167 - j8.75)(1.0∠0°) + (-1.6667 + j5)(1.0∠0°) = 0.0 + j0.0 Q2(1) = 0.0 [This value is within the limits.]

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Gauss-Seidel Load Flow V2

(k + 1 )

V2

(1)

(k + 1 ) ⎡ ⎤ 1 P2 − jQ 2 (k + 1 ) (k ) ⎢ = − Y 21V 1 − Y 23 V 3 ⎥ * (k ) Y 22 ⎢ ⎥⎦ V 2 ⎣

(

)

1 0.2 - j0.0 ___________________ ___________ = 9.2233∠-71.5650 1.0∠0°

Y22

(k +1 )

V1 Y21 - (-1.25 +j3.75) (1.0∠0°) (k ) Y23 V3 - (-1.6667 + j5) (1.0∠0°)

P2 − jQ2

( k +1 )

(V ) (k )

*

2

= 1.0071∠1.1705° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow V3

(k + 1 )

⎤ 1 ⎡ P3 − jQ 3 (k + 1 ) (k + 1 ) ⎢ ⎥ = − Y 31V 1 − Y 32 V 2 * Y 33 ⎢ V (k ) ⎥⎦ 3 ⎣ P3 − jQ3

(

)

1 _____________________ V3 = 21.0819∠-71.5650 Y33 (k +1 ) Y31 V1 - (-5 +j15) (1.0∠0°) 1

-0.6 + j0.25 ____________ 1.0∠0°

(V )

k *

3

( k +1 ) Y32 V2 - (5.2705∠108.4349°)(1.0071∠1.1705°)

= 0.9816 ∠-1.0570° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow ΔV2 = V2(1) - V2(0) = 1.0071∠1.1705° - 1.0∠0° ⏐ΔV2⏐ = 0.0217

ΔV3 = V3(1) - V3(0) = 0.9816∠-1.0570° - 1.0∠0° ⏐ΔV3⏐ = 0.0259 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow Iteration 2 (k = 1): V1(2) = 1.0∠0° Let, V2(1) = 1.0∠1.1705° P2 - jQ2(2) = (1.0∠-1.1705°)[(-1.25 + j3.75)(1.0∠0°) + (9.2233∠-71.5649°)(1.0∠1.1705°) + (5.2705∠108.4349° )(0.9816∠-1.0570°) = 0.2995 - j0.0073 Q2 (2) = 0.0073

[This value is within the limits.]

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Gauss-Seidel Load Flow V2

(k + 1 )

V2

(2)

(k + 1 ) ⎡ ⎤ 1 P2 − jQ 2 (k + 1 ) (k ) ⎢ = − Y 21 V 1 − Y 23 V 3 ⎥ * ( ) k Y 22 ⎢ ⎥⎦ V2 ⎣

(

)

1 ___________________ = 9.2233 ∠ -71.5650

0.2 - j0.0073 ______________ 1.0 ∠ -1.1705°

- (-1.25 +j3.75) (1.0 ∠ 0°) - (5.2705 ∠ 108.4349° ) (0.9816 ∠ -1.0570°) = 0.9966 ∠ 0.5819° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow V3

(k + 1 )

⎤ 1 ⎡ P3 − jQ 3 (k + 1 ) (k + 1 ) ⎢ ⎥ = − Y 31 V 1 − Y 32 V 2 * ( ) k Y 33 ⎢ V ⎥⎦ 3 ⎣

(

)

-0.6 + j0.25 ___________________

1 _____________________ (2) V3 = 21.0819 ∠ -71.5650

0.9816 ∠ 1.0570°

- (-5 +j15) (1.0 ∠ 0°) - (5.2705 ∠ 108.4349°) (0.9966 ∠ 0.5819° ) = 0.9783 ∠ -1.2166° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow ΔV2 = V2(2) - V2(1) = 0.9966 ∠ 0.5819° - 1.0071 ∠ 1.1705° ⏐ΔV2⏐ = 0.0125

ΔV3 = V3(2) - V3(1) = 0.9783 ∠ -1.2166° - 0.9816 ∠ -1.0570° ⏐ΔV3⏐ = 0.004 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow Iteration 3 (k = 2): V1(2) = 1.0∠0° Let, V22 = 1.0 ∠ 0.5819° P2 - jQ22 = (1.0 ∠-0.5819°) [(-1.25 + j3.75)(1.0 ∠ 0°) + (9.2233 ∠ -71.5649° ) (1.0 ∠ 0.5819°) + (5.2705 ∠ 108.4349° ) (0.9783 ∠ -1.2166° ) = 0.2287 - j0.0472 Q22 = 0.0472

[This value is within the limits.]

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Gauss-Seidel Load Flow V2

(k + 1 )

V2

3

(k + 1 ) ⎡ ⎤ 1 P2 − jQ 2 (k + 1 ) (k ) ⎢ = − Y 21 V 1 − Y 23 V 3 ⎥ * ( ) k Y 22 ⎢ ⎥⎦ V2 ⎣

(

)

1 ___________________ = 9.2233 ∠ -71.5650

0.2 - j0.0472 ______________ 1.0 ∠ -0.5819°

- (-1.25 +j3.75) (1.0 ∠ 0°) - (5.2705 ∠ 108.4349° ) (0.9783 ∠ -1.2166° ) = 0.9990 ∠ 0.4129° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow V3

(k + 1 )

⎤ 1 ⎡ P3 − jQ 3 (k + 1 ) (k + 1 ) ⎢ ⎥ = − Y 31 V 1 − Y 32 V 2 * ( ) k Y 33 ⎢ V ⎥⎦ 3 ⎣

(

)

1 _____________________ 3 V3 = 21.0819 ∠-71.5650

-0.6 + j0.25 ___________________ 0.9783 ∠ 1.2166°

- (-5 +j15)(1.0∠0°) - (5.2705∠108.4349°)(0.9990∠0.4129°) = 0.9788∠-1.2560° U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow ΔV2 = V2(3) - V2(2) = 0.9990∠0.4129° - 1.0∠0.5819° ⏐ΔV2⏐ = 0.003 < 0.005

ΔV3 = V3(3) - V3(2) = 0.9788∠-1.2560° - 0.9783∠-1.2166° ⏐ΔV3⏐ = 0.0008 < 0.005

The solution has converged.

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Gauss-Seidel Load Flow 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 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow 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 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow 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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Gauss-Seidel Load Flow 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 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Load Flow 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

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Gauss-Seidel Load Flow Power FLOWS The power flow (Spq) from bus p to q is

S pq = Ppq − jQ pq = V p* I pq The power flow (Sqp) from bus q to p is

S qp = Pqp − jQqp = Vq* I qp

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Gauss-Seidel Load Flow 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 )

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Gauss-Seidel Load Flow Line Losses The power loss in line pq is the algebraic sum of the power flows Spq and Sqp

S loss = Ploss + jQloss = S pq + S qp

= VpI

* pq

â&#x2C6;&#x2019; VqI

= (V p + V q )I

* pq

* pq

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Gauss-Seidel Load Flow SUMMARY OF 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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Gauss-Seidel Load Flow OTHER INFORMATION Overvoltage and Undervoltage Buses Critical and Overloaded Transformers and Lines Total System Losses

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Newton-Raphson Load Flow

Non-Linear Formulation of Load Flow Equations

Newton-Raphson Load Flow Solution

Numerical Example

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Newton-Raphson Load Flow Non-Linear Formulation of Load Flow Equations The complex power injected into Bus p is

Pp − jQp = E*p I p

(1)

and the current equation may be written as Ip =

n

∑Y

pq

(2)

Eq

q =1

Substituting (2) into (1)

P p − jQ

p

= E

* p

•

n

∑Y E q =1

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pq

(3) q

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Newton-Raphson Load Flow Let E = V ∠ δ p p

E q = Vq ∠ δ q

p

Ypq = Ypq∠θ pq Substituting into equation (3),

Pp − jQ p =

n

∑VVY p

q

pq

∠ ( θ pq +δ q − δ p )

(4)

q =1

Separating the real and imaginary components

Pp =

n

∑VVY p

q

pq

co s( θ pq +δ q − δ p )

q =1

(5)

n

Q p = − ∑ V pV q Y pq sin( θ pq +δ q − δ p )

(6)

q =1

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Newton-Raphson Load Flow The formulation results in a set of non-linear equations, two for each Bus of the system.

Equations Pp are written for all Buses except the Swing Bus. Equations Qp are written for Load Buses only The system of equations may be written for

i number of buses minus the swing bus (n-1) j number of load buses

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Newton-Raphson Load Flow The system of equations may be written as

P1 = P1 ( δ 1 ,δ 2 ,...,δ i ,v1 ,v 2 ....,v j ) P2 = P2 ( δ 1 ,δ 2 ,...,δ i ,v1 ,v 2 ....,v j ) M M Pi = Pi ( δ 1 ,δ 2 ,...,δ i ,v 1 ,v 2 ....,v j )

(7)

Q1 = Q1 ( δ 1 ,δ 2 ,...,δ i ,v1 ,v2 ....,v j ) Q2 = Q2 ( δ 1 ,δ 2 ,...,δ i ,v1 ,v2 ....,v j ) M

M

Q j = Q j ( δ 1 ,δ 2 ,...,δ i ,v1 ,v2 ....,v j ) U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Equation (7) may be linearized using a First-Order Taylor-Series Expansion P1spec

= P1calc

+

∂P1 Δδ 1 ∂δ 1

+

∂P1 Δδ 2 ∂δ 2

+... +

∂P1 Δδ i ∂δ i

+

∂P1 ΔV1 ∂V1

+

∂P1 ΔV2 ∂V2

+... +

∂P1 ΔV j ∂V j

P2 spec

= P2calc

+

∂P2 Δδ 1 ∂δ 1

+

∂P2 Δδ 2 ∂δ 2

+... +

∂P2 Δδ i ∂δ i

+

∂P2 ΔV1 ∂V1

+

∂P2 ΔV2 ∂V2

+... +

∂P2 ΔV j ∂V j

+... +

∂Pi ΔV j ∂V j

M M M

M∂P

= Pi calc

+

Q1spec

= Q1calc

+

∂Q1 Δδ 1 ∂δ 1

Q2spec

= Q2calc

+

∂Q2 Δδ 1 ∂δ 1

M M M

Q

spec j

= Q

calc j

+

i

∂δ 1

M∂Q

j

∂δ 1

M

M

M

Δδ 2

+... +

+

∂Q1 Δδ 2 ∂δ 2

+... +

∂Q1 Δδ i ∂δ i

+

∂Q1 ΔV1 ∂V1

+

∂Q1 ΔV2 ∂V2

+... +

+

∂Q2 Δδ 2 ∂δ 2

+... +

∂Q2 Δδ i ∂δ i

+

∂Q2 ΔV1 ∂V1

+

∂Q2 ΔV2 ∂V2

+... +

i

∂δ 2

M ∂Q

Δδ1 +

j

∂δ 2

Δδ 2

M

+... +

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∂Q j ∂δ i

M

Δδ i

∂Pi + ΔV1 ∂V1

+

∂Q j ∂V1

M

∂Pi + ΔV2 ∂V2

M

∂Pi Δδ i ∂δ 2

Δδ 1 +

Pi spec

M ∂P

ΔV1 +

∂Q j ∂V2

M

ΔV2

+... +

∂Q1 ΔV j ∂V j ∂Q2 ΔV j ∂V j ∂Q j ∂V j

ΔV j

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Load Flow Analysis ⎡ P1 ⎢ ⎢ ⎢ spec ⎢ P2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ Pi s p e c ⎢ ⎢ ⎢ spec ⎢Q1 ⎢ ⎢ ⎢ Q spec ⎢ 2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ spec ⎣⎢ Q j spec

− P1

c a lc

− P2c a lc

M − Pi c a l c − Q 1c a l c − Q

ca lc 2

M − Q

ca lc j

⎡ ∂ P1 ⎤ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂ P2 ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂ Pi ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ∂Q1 ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ∂Q2 ⎥ ⎢ ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂Q j ⎥ ⎢ ⎦⎥ ⎣⎢ ∂ δ 1

71

∂ P1 ∂δ 2

L

∂ P1 ∂δ i

∂ P1 ∂V1

∂ P1 ∂V2

L

∂ P2 ∂δ 2

L

∂ P2 ∂δ i

∂ P2 ∂V1

∂ P2 ∂V2

L

M

M

M

M ∂ Pi ∂δ 2

L

∂ Pi ∂δ 2

∂ Pi ∂V1

∂ Pi ∂V2

L

∂Q1 ∂δ 2

L

∂Q1 ∂δ i

∂Q1 ∂V1

∂Q1 ∂V2

L

∂Q 2 ∂δ 2

L

∂Q2 ∂δ i

∂Q 2 ∂V1

∂Q2 ∂V2

L

M

M

M

M

M

∂Q

j

∂δ 2

L

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∂Q ∂δ i

j

∂Q

j

∂V1

∂Q

j

∂V2

L

∂ P1 ⎤ ⎡Δδ1 ∂V j ⎥ ⎢ ⎥ ⎥⎢ ∂ P2 ⎥ ⎢ Δδ2 ∂V j ⎥ ⎢ ⎥⎢ ⎥⎢ M ⎥⎢ M ⎥⎢ ⎥⎢ ∂ Pi ⎥ ⎢ ⎢Δδi ∂V j ⎥ ⎢ ⎥ ⎥⎢ ∂Q1 ⎥ ⎢ ΔV1 ∂V j ⎥ ⎢ ⎥⎢ ∂Q2 ⎥ ⎢ ⎥ ⎢ ΔV2 ∂V j ⎥ ⎢ ⎥⎢ ⎢ M ⎥⎥ ⎢ M ⎥⎢ ⎢ ∂Q j ⎥ ⎢ ⎥ ΔV j ∂ V j ⎥⎦ ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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Newton-Raphson Load Flow or simply

⎡ ∂P ⎡ ΔP ⎤ ⎢ ∂δ ⎢ΔQ ⎥ = ⎢ ∂Q ⎣ ⎦ ⎢ ⎣ ∂δ

∂P ⎤ ∂V ⎥ ⎡ Δδ ⎤ ∂Q ⎥ ⎢⎣ΔV ⎥⎦ ⎥ ∂V ⎦

∂P ⎤ ⎡ ∂P ⎡ ΔP ⎤ ⎢ ∂δ V ∂V ⎥ ⎡ Δδ ⎤ ⎥ ⎢ΔV ⎥ ⎢ΔQ ⎥ = ⎢ ∂Q ∂ Q ⎣ ⎦ ⎢ ⎥ ⎢⎣ V ⎥⎦ V ∂V ⎦ ⎣ ∂δ U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Newton-Raphson Load Flow Solution ⎡ΔP ⎤ ⎡ J1 ⎢ ⎥ ⎢ ⎢ ⎥= ⎢ ⎢ ⎥ ⎢ ⎣ ΔQ ⎦ ⎣ J 3

J2 ⎤ ⎥ ⎥ ⎥ J4 ⎦

⎡ Δδ ⎢ ⎢ ⎢ ΔV ⎢⎣ V

⎤ ⎥ ⎥ ⎥ ⎥⎦

n ⎧ ∂ Pp = ∑ V p V q Y p q s in ( θ p q + δ q − δ ⎪ q = 1 ,q ≠ p ⎪ ∂δ p J1 ⎨ ⎪ ∂ P p = − V V Y s in ( θ + δ − δ ) p q pq pq q p ⎪ ∂δ q ⎩ U. P. National Engineering Center National Electrification Administration

p

)

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Newton-Raphson Load Flow

J2

J3

∂ Pp ⎧ 2 V = P + V ⎪ p p p Y pp c o s θ pp ∂V p ⎪ ⎨ ⎪V ∂ Pp = V V Y c o s ( θ p q pq pq + δ q − δ p ) ⎪ q ∂V q ⎩ n ⎧ ∂Q p = ∑ V pV q Y pq c o s ( θ pq + δ q − δ ⎪ q = 1 ,q ≠ p ⎪ ∂δ p ⎨ ⎪ ∂ Q p = −V V Y c o s( θ p q pq pq + δ q − δ p ) ⎪ ∂δ q ⎩ U. P. National Engineering Center National Electrification Administration

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)

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Newton-Raphson Load Flow ⎧ ⎪V p ⎪ J4 ⎨ ⎪V ⎪ q ⎩

∂Q p ∂V p ∂Q p ∂Vq

= Q p − V p2 Y p p s in θ p q = − V p V q Y p q s in ( θ p q + δ q − δ

p

)

The solution of the load flow equations proceeds iteratively from the set of initial estimates. These estimates are updated after evaluating the Jacobian matrix.

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Newton-Raphson Load Flow At the kth iteration,

δ p( k +1 ) = δ p( k ) + Δδ p( k ) V p( k +1 ) = V p( k ) + ΔV p( k ) The process is terminated once convergence is achieved whrein

MAX Δ P ( k ) ≤ ε p

and

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Newton-Raphson 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 zero and 50 MVARS, respectively. Base power used is 100 MVA. Solve the load flow problem using Gauss-Seidel iterative method assuming a 0.005 convergence index.

G

G

Line 1

2

1 Line 2

Line 3 3

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Newton-Raphson Load Flow Branch Data Line No. Bus Code Impedance Z pq (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 U. P. National Engineering Center National Electrification Administration

Load P Q 0 0 0 0 0.60 0.25

Remarks Swing Bus Gen Bus Load Bus

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Newton-Raphson Load Flow The elements of the Bus Admittance Matrix are:

Y11 = 6.25 − j18.75 = 19.7642 ∠ − 71.5651 Y12 = − 1,25 + j 3.75 = 3.9528 ∠ 108.4349 Y13 = − 5 + j15 = 15.8114 ∠ 108.4349 Y21 = − 1.25 + j .375 = 3.9528 ∠ 108.4349 Y22 = 2.9167 − j 8.75 = 9.2233 ∠ − 71.5649 Y23 = − 1.6667 + j 5 = 5.2705 ∠ 108.4349 Y31 = − 5 + j15 = 15.811 4 ∠ 108.4349 Y32 = − 1.6667 + j 5 = 5.2705 ∠ 108.4349 Y33 = 6.6667 − j 20 = 21.0819 ∠ − 71.5650 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Bus 1: Swing Bus (Not included) Bus 2: Generator Bus (Compute for P2) Bus 3: Load Bus (Compute for P2 and Q2)

⎡ ∂P2 ⎡ ΔP2 ⎤ ⎢ ⎢ ⎥ ⎢ ∂δ2 ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ∂P3 ⎢ ΔP3 ⎥ = ⎢ ∂δ ⎢ ⎥ ⎢ 2 ⎢ ⎥ ⎢ ⎢⎣ΔQ3 ⎥⎦ ⎢ ∂Q3 ⎢⎣ ∂δ2

∂P2 ∂δ3 ∂P3 ∂δ3 ∂Q3 ∂δ3

⎤ ∂P2 ⎤ ⎡ V3 ⎥ ⎢ Δδ2 ⎥ ∂V3 ⎥ ⎢ ⎥ ⎥⎢ ⎥ ∂P3 ⎥ ⎢ V3 Δδ3 ⎥ ⎥ ∂V3 ⎥ ⎢ ⎥⎢ ⎥ ∂Q3 ⎥ ⎢ ΔV3 ⎥ ⎥⎢ ⎥ V3 ∂V3 ⎥⎦ ⎢⎣ V3 ⎥⎦

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Newton-Raphson Load Flow Specified Variables: V1 = 1.0 δ1 = 0.0 V2 = 1.0

P2 = 0.2

P3 = -0.6

Q3 = -0.25

Initial Estimates of Unknown Variables:

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Newton-Raphson Load Flow Compute Initial Power Estimates

P20 = V2V1Y21 cos( θ 21 + δ 1 − δ 2 ) + V2V2Y22 cos θ 22 + V2V3Y23 cos( θ 23 + δ 3 − δ 2 ) = ( 1.0 )( 1.0 )( 3.9528 )cos( 108.4349 + 0.0 + 0.0 ) + ( 1.0 )( 1.0 )( 9.2233 )cos( − 71.5649 ) + ( 1.0 )( 1.0 )( 5.2705 )cos( 108.4349 ) = 0.0

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Newton-Raphson Load Flow P30 = V3V1Y31 cos( θ 31 + δ 1 − δ 3 ) + V3V2Y32 cos( θ 22 + δ 2 − δ 3 ) + V3V3Y33 cos θ 33 = ( 1.0 )( 1.0 )( 15.8114 )cos( 108.4349 + 0.0 − 0.0 ) + ( 1.0 )( 1.0 )( 5.2705 )cos( 108.4349 + 0.0 + 0.0 ) + ( 1.0 )( 1.0 )( 21.0819 )cos( −71.5650 ) = 0.0

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Newton-Raphson Load Flow Q 30 = V 3V1Y31 sin( θ 31 + δ 1 − δ 3 ) + V 3V 2Y32 sin( θ 32 + δ 2 − δ 3 ) + V 3V 3Y33 sin θ 33 = ( 1.0 )( 1.0 )( 15.8114 ) sin( 108.4349 + 0.0 − 0.0 ) + ( 1.0 )( 1.0 )( 5.2705 ) sin( 108.4349 + 0.0 + 0.0 ) + ( 1.0 )( 1.0 )( 21.0819 ) sin( − 71.5650 ) = 0.0

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Newton-Raphson Load Flow Compute Power Mismatch

Δ P20 = 0 . 2 − 0 .0 = 0 . 2 Δ P30 = − 0 .6 − 0 .0 = − 0 .6 Δ Q 30 = − 0 .2 5 − 0 .0 = − 0 .2 5 Evaluate elements of Jacobian Matrix ⎡ ∂P ⎢ ∂δ J = ⎢ ⎢ ⎢ ∂Q ⎢⎣ ∂ δ

∂P ⎤ ∂δ ⎥ ⎥ ⎥ ∂Q ⎥ V ∂ V ⎥⎦ V

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Newton-Raphson Load Flow Elements of J1:

∂P2 = V2V1Y21 sin( θ 21 + δ 1 − δ 2 ) ∂δ 2 + V2V3Y23 sin( θ 23 + δ 3 − δ 2 ) = ( 1.0 )( 1.0 )( 3.9528 )sin( 108.4349 + 0.0 − 0.0 ) + ( 1.0 )( 1.0 )( 5.2705 )sin( 108.4349 + 0.0 − 0.0 ) = 8.75

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Newton-Raphson Load Flow Elements of J1:

∂P2 = −V2V3Y23 sin( θ 23 + δ 3 − δ 2 ) ∂δ 3 = −( 1.0 )( 1.0 )( 5.2705 ) sin( 108.4349 + 0.0 − 0.0 ) = −5

∂P3 = −V3V2 Y32 sin( θ 32 + δ 2 − δ 3 ) ∂δ 2 = −( 1.0 )( 1.0 )( 5.2705 ) sin( 108.4349 + 0.0 − 0.0 ) = −5 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Elements of J1:

∂P3 = V3V1Y31 sin( θ 31 + δ 1 − δ 3 ) ∂δ 3 + V3V2Y23 sin( θ 32 + δ 2 − δ 3 ) = ( 1.0 )( 1.0 )( 15.8114 )sin( 108.4349 + 0.0 − 0.0 ) + ( 1.0 )( 1.0 )( 5.2705 )sin( 108.4349 + 0.0 − 0.0 ) = 20

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Newton-Raphson Load Flow Elements of J2:

∂P2 V3 = V2V3Y23 cos(θ 23 + δ 3 + δ 2 ) ∂V3

= (1.0 )(1.0 )(5.2705) cos(108.4349 + 0.0 − 0.0 ) = −1.6667

∂P3 V3 = P3 + V32Y33 cos θ 33 ∂ V3 = 0.0 + ( 1.0 )2 ( 8.2233 )cos( − 71.5649 ) = 2.9167 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Elements of J3:

∂Q3 = −V3V2 Y32 cos( θ 32 + δ 2 − δ 3 ) ∂δ 2 = −( 1.0 )( 1.0 )( 5.2705 )cos( 108.4349 + 0.0 − 0.0 ) = 1.6667

∂ Q3 = V3V1Y31 cos( θ 31 + δ 1 − δ 3 ) ∂δ 3 + V3Y2Y32 cos( θ 32 + δ 2 − δ 3 ) = ( 1.0 )( 1.0 )( 15.8114 )cos( 108.4349 + 0.0 − 0.0 ) + ( 1.0 )( 1.0 )( 5.2705 )cos( 108.4349 + 0.0 − 0.0 ) = − 6.6667 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Elements of J4:

∂Q3 V3 = Q3 − V32Y33 sin θ 33 ∂V3 2 = 0.0 − (1.0) (21.0819 )sin (− 71.5649 ) = 20 In Matrix Form,

⎡ 8.75 ⎢ −5 ⎢ ⎢⎣1.6667

−5

−1.6667 ⎤ 20 2.9167 ⎥⎥ −6.6667 20 ⎥⎦

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Newton-Raphson Load Flow Solving for Gradients,

−5 −1.6667⎤ ⎡ Δδ2 ⎤ ⎡ 0.2 ⎤ ⎡ 8.75 ⎢ −0.6 ⎥ = ⎢ −5 ⎥ ⎢ Δδ ⎥ 20 2.9167 3 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎢⎣−0.25⎥⎦ ⎢⎣1.6667 −6.6667 20 ⎥⎦ ⎢⎣ΔV3 / V3 ⎥⎦

Δδ = 0.003984rad . = 0.2283 deg 0 2

Δδ = −0.02587rad. = −1.4822deg ΔV30 0 = −0.02145 ΔV3 = −0.02145 0 3

V3

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Newton-Raphson Load Flow Update Initial Estimates

δ 21 = δ 20 + Δδ 20 δ 21 = 0.0 + 0.2283 = 0.2283 δ 31 = δ 30 + Δδ 30 δ 31 = 0.0 − 1.4822 = −1.4822 V = V + ΔV 1 3

0 3

0 3

V31 = 1.0 − 0.02145 = 0.97855

Specified Variables

V11 = 1.0

δ 11 = 0.0 V21 = 1.0

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Newton-Raphson Load Flow Update Estimates of Injected Power

P21 = V2V1Y21 cos(θ 31 + δ1 − δ 3 ) + V2V2Y22 cos(θ 22 ) + V2V3Y23 cos(θ 23 + δ 3 − δ 2 ) = (1.0)(1.0)(3.9528) cos(108.4349 + 0.0 + 1.4822) + (1.0)(1.0)(9.2233) cos(− 71.5649) + (1.0)(0.97855)(5.2705) cos(108.4349 − 1.4822 − 0.2283) = 0.1975

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Newton-Raphson Load Flow Update Estimates of Injected Power

P31 = V3V1Y31 cos( θ31 + δ1 − δ 3 ) + V3V2Y32 cos( θ32 + δ 2 − δ 3 ) + V3V3Y33 cos( θ 33 ) = ( 0.97855 )( 1.0 )( 15.8114 )cos( 108.4349 + 0.0 + 1.4822 ) + ( 0.97855 )( 1.0 )( 5.2705 )cos(108.4349 + 0.2283 + 1.4822 ) + ( 0.97855 )( 0.97855 )( 21.0819 )cos( −71.5650 ) = −0.66633

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Newton-Raphson Load Flow Update Estimates of Injected Power

Q31 = −[V3V1Y31 sin( θ31 + δ1 − δ3 ) + V3V2Y32 sin( θ32 + δ2 − δ3 ) +V3V3Y33 sin( θ 33 )]

= - [(0.97855 )(1.0 )(15.8114 )sin(108.4349 + 0.0 + 1.4822 ) + (0.97855 )(1.0 )( 5.2705 )sin(108.4349 + 0.2283 + 1.4822 ) +(0.97855 )(0.97855 )( 21.0819 )sin( −71.5650 )] = −0.2375

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Newton-Raphson Load Flow Compute Power Mismatch Δ P2 = P2 ,s p − P2 ,c a lc = 0 .2 − 0 .1 9 7 5 = 0 .0 0 2 5

Δ P3 = P3 ,s p − P3 ,c a lc = − 0 .6 + 0 .6 6 3 3 = 0 .0 6 3 3

Δ Q 3 = Q 3 ,s p − Q 3 ,c a lc = − .0 .2 5 + 0 .2 3 7 5 = 0 .0 1 2 5 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Evaluate Elements of Jacobian Matrix Elements of J1:

∂P2 = VVY 2 1 21 sin( θ21 + δ1 − δ2 ) ∂δ2 +VV 2 3Y23 sin(θ23 + δ3 − δ2 ) = (1.0)(1.0)(3.9528)sin(108.4349 + 0.0 − 0.2283) + (1.0)(0.97855)(5.2705)sin(108.4349 − 1.4822 − 0.2283) = 8.6942

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Newton-Raphson Load Flow Elements of J1:

∂P2 = −V2V3Y23 sin(θ23 + δ3 − δ2 ) ∂δ3 = −(1.0 )(0.97855)(5.2705)sin(108.4349 − 1.4822 − 0.2283) = −4.9393 ∂P3 = −VV 3 2Y32 sin( θ32 + δ2 − δ3 ) ∂δ2 = −(0.97855)(1.0)(5.2705)sin(108.4349 + 0.2283 + 1.4822) = −4.8419

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Newton-Raphson Load Flow Elements of J1:

∂P3 = V3V1Y31 sin( θ31 + δ1 − δ3 ) ∂δ3 + V3V2Y23 sin( θ32 + δ2 − δ3 ) = (0.97855 )(1.0 )(15.8114 )sin(108.4349 + 0.0 + 1.4822 ) + (0.97855 )(1.0 )( 5.2705 )sin(108.4349 + 0.2283 + 1.4822 ) = 19.3887

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Newton-Raphson Load Flow Elements of J2:

∂P2 = V2V3Y23 cos(θ 23 + δ 3 + δ 2 ) V3 ∂V3

= (1.0 )(0.097855)(5.2705) cos(108.4349 + 1.4822 − 0.2283)

= −1.4842

∂P3 V3 = P3 + V32Y33 cos θ 33 ∂V3 = −0.6633 + ( 0.97855 )2 ( 21.0819 )cos( −71.5650 ) = 5.7205 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Elements of J3:

∂Q3 = −VV 3 2Y32 cos( θ32 + δ2 − δ3 ) ∂δ2 = −(0.97855)(1.0 )(5.2705)cos(108.4349 + 0.2283 + 1.4822) = 1.7762

∂Q3 = V3V1Y31 cos( θ31 + δ1 − δ 3 ) ∂δ 3 + V3Y2Y32 cos( θ32 + δ 2 − δ 3 ) = ( 0.97855 )( 1.0 )( 15.8114 )cos( 108.4349 + 0.0 + 1.4822 ) + ( 0.97855 )( 1.0 )( 5.2705 )cos( 108.4349 + 0.2283 + 1.4822 ) = −7.0470 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Elements of J4:

∂P3 V3 = Q3 − V32Y33 sinθ 33 ∂δ 3 = −0.2375 − ( 0.97855 )2 ( 21.0819 )sin( −71.565 ) = 18.9137 In Matrix Form,

⎡ 8.6942 −4.9393 −1.4842⎤ ⎢−4.8419 19.3887 5.7205 ⎥ ⎢ ⎥ ⎢⎣ 1.7762 −7.0470 18.9137 ⎥⎦ U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Solving for Gradients,

⎡ 0.0025 ⎤ ⎡ 8.6942 −4.9393 −1.4842⎤ ⎡ Δδ 2 ⎤ ⎢ 0.0633 ⎥ = ⎢−4.8419 19.3887 5.7205 ⎥ ⎢ Δδ ⎥ 3 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢⎣−0.0125⎥⎦ ⎢⎣ 1.7762 −7.0470 18.9137 ⎥⎦ ⎢⎣ΔV3 / V3 ⎥⎦

Δδ 21 = 0.0025 rad x 180 0 / πrad = 0.1458 0

Δδ 31 = 0.0038 rad x 180 0 / πrad = 0.2150 0

ΔV30 V3

= 0.0005

Δ V 31 = (0 .0005 )(0 .97855 ) = 0 .0005

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Newton-Raphson Load Flow Update Previous Estimates

δ 22 = δ 21 + Δδ 21 = 0.2283 + 0.1458 = 0.37410

δ 32 = δ 31 + Δδ 31 = − 1.4822 + 0.2150 = − 1.2672 0 V = V + ΔV 2 3

1 3

1 3

= 0.97855 + 0.0005 = 0.9791

Specified Variables

V11 = 1.0

δ 11 = 0.0 V21 = 1.0 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow Update Previous Estimates of Injected Power

P22 = V2V1Y21 cos( θ21 + δ1 − δ 2 ) + V2V2Y22 cos( θ22 ) + V2V3Y23 cos( θ23 + δ 3 − δ 2 ) = (1.0 )( 1.0 )( 3.9528 )cos( 108.4349 + 0.0 − 0.3741) + ( 1.0 )( 1.0 )( 9.2233 )cos( −71.5649 ) + ( 1.0 )( 0.9791)( 5.2705 )cos(108.4349 − 1.2672 − 0.3741) = 0.2018

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Newton-Raphson Load Flow Update Previous Estimates of Injected Power

P32 = V3V1Y31 cos( θ 31 + δ 1 − δ 3 ) + V3V2Y32 cos( θ 22 + δ 2 − δ 3 ) + V3V3Y33 cos θ 33 = ( 0.9791 )( 1.0 )( 15.8114 )cos( 108.4349 + 0.0 − 1.2672 ) + ( 0.9791 )( 1.0 )( 5.2705 )cos( 108.4349 + 0.3741 + 1.2672 ) + ( 0.9791 )( 0.9791 )( 21.0819 )cos( −71.5650 ) = −0.5995

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Newton-Raphson Load Flow Update Previous Estimates of Injected Power

Q32 = −[V3V1Y31 sin( θ31 + δ1 − δ3 )

+V3V2Y32 sin( θ32 + δ2 − δ3 ) + V3V3Y33 sinθ33 ]

= −[(0.9791)(1.0 )(15.8114 )sin(108.4349 + 0.0 − 1.2672 ) + (0.9791)(1.0 )( 5.2705 )sin(108.4349 + 0.37411 + 1.2672 ) +(0.9791)(0.9791)( 21.0819 )sin( −71.5650 )] = −0.2487

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Newton-Raphson Load Flow Compute Power Mismatch Δ P2 = P2 ,s p − P2 ,c a lc = 0 .2 − 0 .2 0 1 8 = 0 .0 0 1 8

Δ P3 = P3 ,s p − P3 ,c a lc = − 0 .6 + 0 .5 9 9 5 = 0 .0 0 0 5

Δ Q 3 = Q 3 ,s p − Q 3 ,c a lc = − .0 .2 5 + 0 .2 4 8 7 = 0 .0 0 1 3 U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Load Flow The solution of the Load Flow Problem is

V1 = 1.0∠0 0 V2 = 1.0∠0.37410 V3 = 0.9791∠ − 1.26720

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Backward/Forward Sweep Load Flow

Load Flow for Radial Distribution System

Procedure: Iterative Solution

Initialization

Solving for Injected Currents through the nodes

Backward Sweep

Forward Sweep

Solving for Injected Power

Solving for Voltage Mismatch

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Backward/Forward Sweep Load Flow Bus2 Bus1 I12 , Loss12 = ?

Utility Grid

I23 , Loss23 = ?

0.635 + j1.970 立

Bus3 V3 = ?

P3 , Q3 = ?

I24 , Loss24 = ? V4 = ?

P4 , Q4 = ?

0.4223 + j0.7980 立 Bus4 V2 = ?

Lumped Load A P2 , Q2 = ?

2 MVA 85%PF Lumped Load B 1 MVA 85%PF

0.131 + j1.595 立

V1 = 67 kV

P1 , Q1 = ?

Load Flow for Radial Distribution System

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Backward/Forward Sweep Load Flow Equivalent Circuit Bus2 V1 = 67 kV

Utility Grid

0.0364 +j 0.1131 pu

Bus1 0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458pu

1 + j0 pu

V2

Base Values

Bus3 V3

Bus4 V4

0.085 + j0.05267 pu

0.17 + j0.10536 pu

Sbase = 10 MVA Vbase1 = 67 kV

Base Z =13.22/10 =17.424立

Vbase2 = 13.2 kV U. P. National Engineering Center National Electrification Administration

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Backward/Forward Sweep Load Flow Iterative Solution 1. Solve Injected Currents by Loads 2. Solve Line Currents (Backward Sweep) 3. Update Voltages (Forward Sweep) 4. Solve for Injected Power 5. Solve for Power Mismatch Continue iteration by Backward-Forward Sweep until convergence is achieved After convergence, solve Iinj, Pinj, Qinj, PF, PLoss, QLoss U. P. National Engineering Center National Electrification Administration

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Backward/Forward Sweep Load Flow Initialization Bus2 V1 = 67 kV

Utility Grid

Bus1

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458 pu

1 + j0 pu

Initialize,

V2

V1(0) = 1/0

Bus3 V3

Bus4 V4

0.085 + j0.05267 pu

0.17 + j0.10536 pu

V2(0) = 1/0 V3(0) = 1/0 V4(0) = 1/0 U. P. National Engineering Center National Electrification Administration

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Backward/Forward Sweep Load Flow Solving for Injected Currents Bus2 V1 = 67 kV

Utility Grid

0.0364 +j 0.1131 pu

Bus1 0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458 pu

1 + j0 pu

Solve Injected Currents

by Loads

V2

Bus3 V3

Bus4 V4

0.17 + j0.10536 pu

0.085 + j0.05267 pu

I1(0) = 0 I2(0) = 0 I3(0) = S3* /[V3(0)]* = __________ I4(0) = S4* /[V4(0)]* = __________

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Backward/Forward Sweep Load Flow Backward Sweep Bus2 V1 = 67 kV

Utility Grid

0.0364 +j 0.1131 pu

Bus1 0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458 pu

1 + j0 pu

Solve Line Currents (Backward Sweep)

V2

Bus3 V3

Bus4 V4

0.17 + j0.10536 pu

0.085 + j0.05267 pu

I24(0) = I4(0) = _______ I23(0) = I3(0) = _______ I12(0) = 0 + I23(0) + I24(0) = _______

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Backward/Forward Sweep Load Flow Forward Sweep Bus2 V1 = 67 kV

Bus1

Utility Grid

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458 pu

1 + j0 pu

V2

Bus3 V3

Bus4 V4

0.085 + j0.05267 pu

0.17 + j0.10536 pu

Update Voltages

V1(1) = 1/0

(Forward Sweep)

V3(1) = V2(1) – [I23(0)][Z23] = ________

V2(1) = V1(0) – [I12(0)][Z12] = ________ V4(1) = V2(1) – [I24(0)][Z24] = ________

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Backward/Forward Sweep Load Flow Solving for Injected Power Bus2 V1 = 67 kV

Utility Grid

Bus1

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458 pu

1 + j0 pu

Solve Injected Power

V2

Bus3 V3

Bus4 V4

0.085+ j0.05267 pu

0.17 + j0.10536 pu

S1(1) = [V1(1)][I1(0)]* = ___________ S2(1) = [V2(1)][I2(0)]* = ___________ S3(1) = [V3(1)][I3(0)]* = ___________ S4(1) = [V4(1)][I4(0)]* = ___________

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Backward/Forward Sweep Load Flow Solving for Power Mismatch Bus2 V1 = 67 kV

Utility Grid

Bus1

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

V1

~

0.0242+j0.0458 pu

1 + j0 pu

V2

Bus3 V3

Bus4 V4

0.085 + j0.05267 pu

0.17 + j0.10536 pu

Solve ΔS1(1) = S1(sp) - S1(calc) = ____________ Power Mismatch ΔS2(1) = S2(sp) – S2(calc) = ____________ ΔS3(1) = S3(sp) – S3(calc) = ____________ ΔS4(1) = S4(sp) – S4(calc) = ____________ U. P. National Engineering Center National Electrification Administration

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Backward/Forward Sweep Load Flow Iterative Solution Iteration 2: Solve Injected Currents

by Loads

I1(1) = 0 I2(1) = 0 I3(1) = S3* /[V3(1)]* = __________ I4(1) = S4* /[V4(1)]* = __________

Solve Line Currents (Backward Sweep)

I24(1) = I4(1) = _______ I23(1) = I3(1) = _______ I12(1) = 0 + I23(1) + I24(1) = _______

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Backward/Forward Sweep Load Flow Iterative Solution Update Voltages

V1(2) = 1/0

(Forward Sweep)

V3(2) = V2(1) – [I23(1)][Z23] = ________

V2(2) = V1(1) – [I12(1)][Z12] = ________ V4(2) = V2(1) – [I24(1)][Z24] = ________

Solve Injected Power

S1(2) = [V1(2)][I1(1)]* = ___________ S2(2) = [V2(2)][I2(1)]* = ___________ S3(2) = [V3(2)][I3(1)]* = ___________ S4(2) = [V4(2)][I4(1)]* = ___________

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Backward/Forward Sweep Load Flow Iterative Solution Solve Power Mismatch

ΔS1(2) = S1(sp) - S1(calc) = ____________ ΔS2(2) = S2(sp) – S2(calc) = ____________ ΔS3(2) = S3(sp) – S3(calc) = ____________ ΔS4(2) = S4(sp) – S4(calc) = ____________

If Mismatch is higher than set convergence index, repeat the procedure (Backward-Forward Sweep) [Iteration 3]

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Backward/Forward Sweep Load Flow Iterative Solution Iteration 3: Solve Injected Currents

by Loads

I1(2) = 0 I2(2) = 0 I3(2) = S3* /[V3(2)]* = __________ I4(2) = S4* /[V4(2)]* = __________

Solve Line Currents (Backward Sweep)

I24(2) = I4(2) = _______ I23(2) = I3(2) = _______ I12(2) = 0 + I23(2) + I24(2) = _______

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Backward/Forward Sweep Load Flow Iterative Solution Update Voltages

V1(3) = 1/0

(Forward Sweep)

V3(3) = V2(2) – [I23(2)][Z23] = ________

V2(3) = V1(2) – [I12(2)][Z12] = ________ V4(3) = V2(2) – [I24(2)][Z24] = ________

Solve Injected Power

S1(3) = [V1(3)][I1(2)]* = ___________ S2(3) = [V2(3)][I2(2)]* = ___________ S3(3) = [V3(3)][I3(2)]* = ___________ S4(3) = [V4(3)][I4(2)]* = ___________

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Backward/Forward Sweep Load Flow Iterative Solution Solve Power Mismatch

ΔS1(3) = S1(3) - S1(2) ΔS2(3) = S2(3) – S2(2) = ____________ ΔS3(3) = S3(3) – S3(2) = ____________ ΔS4(3) = S4(3) – S4(2) = ____________

If Mismatch is lower than set convergence index, compute power flows

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Backward/Forward Sweep Load Flow Bus2 0.0364 +j 0.1131 pu

Bus1

Utility Grid

Bus3

0.0242+j0.0458 0.0075+j 0.0915 pu

Bus4 Lumped Load A 2 MVA 85%PF

VOLTAGE PROFILE Lumped Load B 1 MVA 85%PF

V1 = ________ V2 = ________

INJECTED POWER

V3 = ________

P1 + jQ1 = ________ + j ________

V4 = ________

P2 + jQ2 = ________ + j ________ P3 + jQ3 = ________ + j ________ P4 + jQ4 = ________ + j ________

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Backward/Forward Sweep Load Flow Bus2 0.0364 +j 0.1131 pu

Bus1

Utility Grid

Bus3

0.0242+j0.0458 pu 0.0075+j 0.0915 pu

POWER FLOW (P-Q)

Bus4 Lumped Load A 2 MVA 85%PF

P12 + jQ12 = ________ + j ________ P23 + jQ23 = ________ + j ________

Lumped Load B 1 MVA 85%PF

P24 + jQ24 = ________ + j ________ POWER FLOW (Q-P) P21 + jQ21 = ________ + j ________ P32 + jQ32 = ________ + j ________ P42 + jQ42 = ________ + j ________ U. P. National Engineering Center National Electrification Administration

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Backward/Forward Sweep Load Flow Bus2 0.0364 +j 0.1131 pu

Bus1

Utility Grid

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

Bus4

Branch Currents

Lumped Load A 2 MVA 85%PF

I12 = ________

Lumped Load B 1 MVA 85%PF

I23 = ________ I24 = ________

Bus3

POWER LOSSES I2R12 + jI2X12 = ________ + j ________ I2R23 + jI2X24 = ________ + j ________ I2R24 + jI2X24 = ________ + j ________

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Backward/Forward Sweep Load Flow Line sections in the radial network are ordered by layers away from the root node (substation bus).

1 2 4 7 13

8

8 14

21

22 27

Layer 1

3 5

9 15

23

Layer 2

6

10 11 17 16

12 18

32

20

19

24 28

Layer 3 Layer 4 26

25

29

30 33

34 35

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Layer 5 Layer 6 Layer 7 Layer 8

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Three-Phase Forward/ Backward Sweep Method The iterative algorithm for solving the radial system consists of three steps. At iteration k: Step 1: Nodal current calculation

⎡ I ia ⎤ ⎢I ⎥ ⎢ ib ⎥ ⎢⎣ I ic ⎥⎦

(k)

Where,

( ( (

⎡ S / V ( k −1 ) ⎢ ia ia( k −1 ) = ⎢ Sib / Vib ⎢ S / V ( k −1 ) ⎢⎣ ic ic

I ia , I ib , I ic

) ) )

⎤ ⎡Y * ia ⎥ ∗ ⎢ * Yib ⎥−⎢ ∗⎥ ⎢ ⎥⎦ ⎣ ∗

⎤ ⎡Via ⎤ ⎥⎢ ⎥ ⎥ ⎢Vib ⎥ *⎥ Yic ⎦ ⎢⎣Vic ⎥⎦

( k −1 )

Current injections at node i

S ia , S ib , S ic Scheduled power injections at node i V ia ,V ib ,V ic Voltages at node i Y ia ,Y ib ,Y ic

Admittances of all shunt elements at node i

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Three-Phase Forward/ Backward Sweep Method Step 2: Backward Sweep to sum up line section current Starting from the line section in the last layer and moving towards the root node. The current in the line section l is:

⎡ J la ⎤ ⎢J ⎥ ⎢ lb ⎥ ⎢⎣ J lc ⎥⎦ Where,

(k)

⎡ I ja ⎤ ⎢ ⎥ = − ⎢ I jb ⎥ ⎢ I jc ⎥ ⎣ ⎦

(k)

⎡ J ma ⎤ ⎥ ⎢ + ∑ ⎢ J mb ⎥ m∈M ⎢⎣ J mc ⎥⎦

(k )

J la , J lb , J jc

are the current flows on line section l

l and M

Is the set of line sections connected to node j

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Three-Phase Forward/ Backward Sweep Method Step 3: Forward Sweep to update nodal voltage Starting from the first layer and moving towards the last layer, the voltage at node j is:

⎡V ja ⎤ ⎢ ⎥ ⎢V jb ⎥ ⎢V jc ⎥ ⎣ ⎦

(k )

⎡Via ⎤ ⎥ ⎢ = ⎢Vib ⎥ ⎢⎣Vic ⎥⎦

(k)

⎡ zaa ,l ⎢ − ⎢ zab ,l ⎢⎣ zac ,l

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zab ,l zbb ,l zbc ,l

zac ,l ⎤ ⎡ J la ⎤ ⎥⎢ ⎥ zbc ,l ⎥ ⎢ J lb ⎥ zcc ,l ⎥⎦ ⎢⎣ J lc ⎥⎦

(k )

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Three-Phase Forward/ Backward Sweep Method After the three steps are executed in one iteration, the power mismatches at each node for all phases are calculated:

ΔS

(k ) ia

=V

(k ) ia

ΔS

(k ) ib

=V

(k ) ib

ΔS

(k ) ic

=V

(k ) ic

(I ) (I ) (I )

(k ) ∗ ia

* ia

− Y Via

2

− S ia

(k ) ∗ ib

* ia

− Y Vib

2

− S ib

(k ) ∗ ic

− Y Vic

2

− S ic

* ic

If the real and imaginary part (real and reactive power) of any of these power mismatches is greater than a convergence criterion, steps 1, 2 & 3 are repeated until convergence is achieved.

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Principles of Load Flow Control

Prime mover and excitation control of generators

Reactive Var Compensation (e.g., Capacitors)

Control of tap-changing and voltage regulating transformers

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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

⎡ E ∠δ − Vt ∠0 ⎤ Pt + jQt = [Vt ∠0]I * = [Vt ∠0]⎢ i ⎥ jX ⎦ ⎣

⎡ EiVt ⎤ Pt = ⎢ sinδ ⎥ ⎣ X ⎦ U. P. National Engineering Center National Electrification Administration

⎡ EiVt Vt 2 ⎤ Qt = ⎢ cosδ − ⎥ X⎦ ⎣ X Competency Training & Certification Program in Electric Power Distribution System Engineering

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Principles of Load Flow Control Generator Voltage & Power Control

⎡ EiVt ⎤ Pt = ⎢ sinδ ⎥ ⎣ X ⎦

⎡ EiVt Vt 2 ⎤ Qt = ⎢ cosδ − ⎥ 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 U. P. National Engineering Center National Electrification Administration

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Principles of Load Flow Control Generator Voltage & Power Control

⎤ ⎡ EiVt Pt = ⎢ sinδ ⎥ ⎣ X ⎦

⎡ EiVt Vt 2 ⎤ Qt = ⎢ cosδ − ⎥ 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 U. P. National Engineering Center National Electrification Administration

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Principles of Load Flow Control Capacitor Compensation ~ p

Ipq

The voltage of bus q can be expressed as

q + jQc PL - jQL

Eq = Vp â&#x2C6;&#x2019;

X pqQq Vp

â&#x2C6;&#x2019;j

X pq Pq Vp

Observations: 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 U. P. National Engineering Center National Electrification Administration

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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

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a −1 y pq a

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Uses of Load Flow Study

Sensitivity Analysis with Load Flow Study

Analysis of Existing Conditions

Analysis for Correcting PQ Problems

Expansion Planning

Contingency Analysis

System Loss Analysis

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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.

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Uses of Load Flow Studies 1) ANALYSIS OF EXISTING CONDITIONS • Check for voltage violations PGC: 0.95 – 1.05 p.u. (For Transmission) PDC: 0.90 – 1.10 p.u (For Distribution)* *Recommended 0.95 – 1.05 p.u.

• Check for branch power flow violations Transformer Overloads Line Overloads • Check for system losses Caps on Segregated DSL

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Uses of Load Flow Studies 2) ANALYSIS FOR CORRECTING PQ PROBLEMS â&#x20AC;˘ Voltage adjustment by utility at delivery point Request TransCo to improve voltage at connection point TransCo as System Operator will determine feasibility based on Economic Dispatch and other adjustments such as transformer tap changing and reactive power compensation

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Uses of Load Flow Studies 2) ANALYSIS FOR CORRECTING PQ PROBLEMS â&#x20AC;˘ Transformer tap changing Available Taps At Primary Side At Secondary Side Both Sides

Typical Taps

Tap 1: +5% Tap 2: +2.5% Tap 3: 0% (Rated Voltage) Tap 4: -2.5% Tap 5: -5%

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Uses of Load Flow Studies 2) ANALYSIS FOR CORRECTING PQ PROBLEMS •Capacitor compensation • Compensate for Peak Loading • Check overvoltages during Off-Peak • Optimize Capacitor Plan • System configuration improvement

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Uses of Load Flow Studies 3) EXPANSION PLANNING • • • • • • •

New substation construction Substation capacity expansion New feeder segment construction / extension Addition of parallel feeder segment Reconducting of existing feeder segment/ circuit Circuit conversion to higher voltage Generator addition

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Uses of Load Flow Studies 4) CONTINGENCY ANALYSIS Reliability analysis of the Transmission (Grid) and Subtransmission System 5) SYSTEM LOSS ANALYSIS Segregation of System Losses

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Competency Training & Certification Program in Electric Power Distribution System Engineering