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The HK Polytechnic University Load Flow

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1 Load Flow Analysis

• A ‘load flow’ or ‘power flow’ is power system jargon for the steady-state solution

of a power system network subject to certain operational constraints, such as:

– Generation supplies the demand (load) plus losses.

– Bus voltage magnitudes remain close to rated values.

– Generators operate within specified real and reactive power limits.

– Transformer tap settings are within limits.

– Transmission lines and transformers are not overloaded.

• Load flow solution gives the nodal voltages and phase angles and hence the

power injection at all buses and power flows though transmission units such as

lines, cables and transformers.

• Load flow calculations are performed for power system planning, operational

planning and in connection with system operation and control.

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• Load flow studies are performed to investigate the following features of a power

system network:

1. Flow of MW and MVAr in the branches of the network.

2. Busbar (node) voltage.

3. Effect of the following changes on system loading:

(a) Rearranging circuits and incorporating new circuits.

(b) Temporary loss of generation and transmission circuits.(c) Injecting in-phase and quadrature boost voltages.

4. Optimum system running conditions and load distribution.

5. Minimising system losses.

6. Optimum rating and tap-range of transformers.

7. Improvements from change of conductor size and system voltage.

• Studies will normally be performed for various load conditions to ensure the

power network behaves properly under a wide range of operating conditions.

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2 Load Flow Problem

• Objectives: to determine the steady state operating conditions of (a) busbar

voltage, (b) generation, (c) branch power flows, and (d) circuit system loss.

• Conventional nodal or loop analysis is not suitable for load flow studies because

loads are normally given in terms of power rather than impedance. Also,

generators are considered as power sources, not voltage or current sources.

• Together with the power and voltage constraints, the load flow problem becomes

a nonlinear numerical problem formulated as a set of nonlinear algebraic

equations and the numerical solution must therefore be iterative in nature.

• A load flow solution of the power system requires mainly the following steps:

1. Formulation of the network equations (load flow equations).

2. Suitable mathematical technique for solution of the equations

(Gauss-Seidel, Newton-Raphson, and Fast Decoupled methods).

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3 Network Model Formulation

• In a power system, each bus is associated with 4 quantities:

1. real and reactive powers, P & Q.

2. bus voltage magnitude and angle, |V | & δ .

Among these 4 quantities , only 2 can be specified and the remaining 2 are

obtained through the load flow solution.

• Depending upon which quantities have been specified, the buses are classified

into three categories:

Bus Type Quantities Specified Quantities to be obtained

Load (PQ) bus P , Q |V |, δ

Generator (PV) bus P , |V | Q, δ

Slack (Swing) bus |V |

, δ P

,Q

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3.1 Bus Classification

1. Load or PQ bus: The complex power (P , Q) is specified. It is desired to find out

the voltage magnitude and phase angle through the load flow solution – loads.

2. Generation, PV or voltage control bus: The voltage magnitude and real

power are specified. Often limits to the value of the reactive power are giving as

well. It is required to find out the reactive power generation and the phase angle

of the bus voltage – capacitors, synchronous compensators and generators.

3. Slack, swing or reference bus: Bus voltage magnitude and angle are

specified, typically 1.0/0o, whereas its power P , Q are obtained through the

load flow to cover any power loss, which is not known precisely in advance of

the calculation, or mismatch of load and power generation – system frequency

control generators. This bus voltage angle will be taken as the reference. There

shall be only one such bus in a power system, and usually, the one with the

largest generation is assigned as the slack, swing or reference bus.

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3.2 Nodal Admittance Matrix

Load flow formulation can be established by using either the loop or bus frame of

reference.

loop: V = Z I where Z : impedance matrix V : voltage vector

bus: I = Y V Y : admittance matrix I : current vector

Generally, bus frame of reference in admittance form is preferred as :

1. data preparation is simple

2. its formation and modification is easy

3. the bus admittance matrix is a sparse matrix (i.e. most of its elements are zero)

– save computer memory and computational effort.

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Consider the nodal current at node 1,

I 1 = I 11 + I 12 + I 13= V 1y11 + (V 1 − V 2)y12 + (V 1 − V 3)y13

= V 1(y11 + y12 + y13) − V 2y12 − V 3y13

= V 1Y 11 + V 2Y 12 + V 3Y 13

where y11 is the shunt charging admittance

at bus 1 and y12 is the series admittance

between bus 1 and 2, and

Y 11 = y11 + y12 + y13Y 12 = −y12

Y 13 = −y13Similarly for node 2 and 3,

I 2 = V 1Y 21 + V 2Y 22 + V 3Y 23

I 3 = V 1Y 31 + V 2Y 32 + V 3Y 33

I

I I I I

I I

I

I 3

21 I

I I

33

12

11

13 21 23

22

3231

1 2

3

y12 = y21

Three-bus system

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Nodal current equations can be written in a matrix form:

I 1

I 2

I 3

=

Y 11 Y 12 Y 13

Y 21 Y 22 Y 23

Y 31 Y 32 Y 33

·

V 1

V 2

V 3

or I = Y V

or in compact form these equations can be written as:

I i =

3j=1

Y ij V j for i = 1, 2, 3

The above nodal current equations can be generalised to an n bus system:

I i =n

j=1

Y ij V j for i = 1, 2,. . . , n where Y ii =n

j=1

yij

Y ij = −yij

It can be shown that the nodal admittance matrix is a sparse matrix (only a few

number of elements are non-zero) for an actual power system.

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Consider the nodal admittance matrix for a 5 bus system.

Y =

Y 11 Y 12 0 Y 14 Y 15

Y 21 Y 22 Y 23 0 0

0 Y 32 Y 33 Y 34 0

Y 41 0 Y 43 Y 44 Y 45

Y 51 0 0 Y 54 Y 55

2

1

5

4

3

y12 = y21

y15 = y51

y45 = y54

y34 = y43

y23 = y32

y14 = y41

• For line, cable and tapped transformer:

Y ij = Y ji = −yij = −yji

• Y ij and Y ij are non-zero only if there is a connection between bus i and j.

• The diagonal element of each node is the sum of admittances connected to it.

• The off-diagonal element is the negated admittance between the nodes.

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3.3 Network Models

• Power network can be operating under balanced or unbalanced conditions. The

normal procedure for a load flow study is to assume a balanced system and to

use a single-phase representation equivalent to the positive sequence network.

• Shunt Branches – Rectors & Capacitor:

Shunt admittances are add to the diagonal elements, Y ii, corresponding to thenodes at which they are connected.

• Lines & Cables:

– Modelled as a π equivalent.

– Contribute to both the diagonal and

off-diagonal matrix elements

Y ii, Y jj , Y ij and Y ji .

yij

yii

y jj

ji

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• Tapped Transformers:

ji y

1 : a

V i V jaV i

I i I j

⇒

I ay

(a - a)y (1 - a)y2

i I j

I j = y (V j − aV i)

I i = −

aI j = ya2V i

−aV j

The equivalent circuit is just like an asymmetric π network.

I i = ((a2− a)y + ay)V i − ayV j = (yii + yij)V i − yijV j

I j = −ayV i + ((1 − a)y + ay)V j = −yijV i + (yjj + yij)V j

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3.4 Load Flow Equations

Recall the nodal current equations:

I i =

nj=1

Y ijV j for i = 1, 2, 3, . . . , n

= Y iiV i +

n

j=1,j=i

Y ijV j

or V i = 1

Y ii

I i −

nj=1,j=i

Y ijV j

= 1

Y ii

P i − jQi

V ∗i

−

nj=1,j=i

Y ijV j

with S ∗i = V

∗i I i = P i − jQi

The above load flow equations are nonlinear and can be solved by iterative methods

such as the Gauss-Seidel and Newton-Raphson methods.

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3.5 Power Calculations

The complex power S i delivered to bus i is:

S i = P i + jQi = V iI ∗i = V i

nj=1

Y ∗ijV

∗j

Express V i, V j and Y ij in polar coordinate (magnitude & angle):

P i + jQi = |V i|ejδi

nj=1

|Y ij |e−jθij |V j |e

−jδj = |V i|

nj=1

|V j ||Y ij|ej(δi−δj−θij)

The above complex equation can be expressed in the polar form:

P i = |V i|

nj=1

|V j ||Y ij | cos(δ i − δ j − θij) (1)

Qi = |V i|

nj=1

|V j ||Y ij | sin(δ i − δ j − θij) (2)

where Y ij = |Y ij |/θij

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LoadGen

From

other

buses

QiPi

QLiPLiQGiPGi

Bus i

Vi

(scheduled)

Pi(calculated)

(scheduled)

Qi(calculated)

• busbar voltage

Vi = |V i|/δ i

• net scheduled real power

Pi(scheduled) = PGi - PLi

• net scheduled reactive power

Qi(scheduled) = QGi - QLi

where PGi & QGi is generator power

PLi & QLi is load power

The power difference between the scheduled value (Pi(scheduled), Qi(scheduled)),

specified by the busbar generation and load, and the calculated value (Pi(calculated),

Qi(calculated)), derived from the best available busbar voltages and angles, is referred

to as the power mismatch (∆Pi, ∆Qi) where

∆Pi = Pi(scheduled) - Pi(calculated)

∆Qi = Qi(scheduled) - Qi(calculated)

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4 Gauss-Seidel (GS) Method

• The GS method is an iterative algorithm for solving a set of non-linear algebraic

equations.

• To start with, a solution vector is assumed. One of the equations is then used to

obtain the revised value of a particular variable by substituting in it the present

values of the remaining variable. The solution vector is immediately updated in

respect of this variable.

• The process is then repeated for all the variables thereby completing one

iteration. The iterative process is then repeated till the solution vector converges

within prescribed accuracy.

• The convergence is sensitive to the starting values assumed.

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Eg. Consider the following simple nodal equations:

2V 1 − 0.5V 2 − 1.5V 3 = I 1 = 1

−0.5V 1 + 1.25V 2 − 0.75V 3 = I 2 = −1.5

−1.5V 1 − 0.75V 2 + 2.25V 3 = I 3

where V 3 is 100 and V 2 is assumed to be 100 initially. Find V 1 and V 2.

In the iterations, the newly computed values are immediately used as soon as they are

obtained. 1V 1 = (1 + 150 + 50)/2 = 100.50001V 2 = (-1.5+75+50.25)/1.25 = 99.00002V 1 = (1 + 150 + 49.5)/2 = 100.25002V 2 = (-1.5+75+50.125)/1.25 = 98.9000

3V 1 = 100.2250 3V 2 = 98.8900

4V 1 = 100.2225 4V 2 = 98.8890

5V 1 = 100.22225 5V 2 = 98.8889

Hence, I 3 = -1.5 × 100.22225 - 0.75 × 98.8889 + 2.25 × 100 = 0.49995

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4.1 Load Flow Solution by Gauss-Seidel (GS) Method

• Recall the load flow equation: V i = 1

Y ii

P i− jQ

iV ∗i

−

nj=1,j=i

Y ijV j

(3)

• In GS method, the new calculated voltage V k+1i immediately replaces V ki and

is used in the solution of the subsequent equations. Hence, eqn (3) becomes:

V k+1i

= 1

Y ii

P i − jQi

(V ki

)∗ −

i−1j=1

Y ijV k+1j

−

nj=i+1

Y ijV kj

(4)

• For PV bus, Qi is unknown but can be calculated from power eqn (2).

• For slack bus, its load flow equation is excluded from the GS calculation as bothits voltage magnitude |V i| and angle δ i are specified while the 2 unknown

variables P i, Qi can be calculated from power eqn (1) and (2), i.e. there are

(n−1) load flow equations in total for a n bus system.

• Initial unknown voltage magnitude |V i| and angle δ i can be set up 1pu and 0o.

This is referred as the ‘flat start’ condition.

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4.2 Gauss-Seidel Solution Steps

1. Formulate the admittance matrix Y .

2. Separate out the slack, generator and load buses.

3. Assume any unknown bus voltage to, say, 1pu and 0o.

4. Start iteration process with first bus of the system (i=1).

5. Update Qi using equation (2) if i bus is a slack or PV bus.

Calculate the new bus voltage, V i, from the load flow equation (4).

6. Calculate the difference between old and new bus voltages.

∆V k+1i = V k+1i − V

ki

7. Using this new value of bus voltage in performing calculations for the next bus of

the system, except for the PV buses whose |V i| should remain constant.

8. Advance for the next bus of the system and repeat steps 5 to 7 until a new set of

values of bus voltages of all buses in the system is obtained – 1 GS iteration.

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9. Repeat the iterative process from step 4 to 8 until the difference ∆V i for all

buses is within a specified limit or tolerance.|∆V k+1i | < ǫ

where k is the iteration count and ǫ is the tolerance level.

4.3 Reactive Power Limits

For a generation bus, Qi should be checked for any limit violation.

Qimin ≤ Qi ≤ Qimax

Whenever there is a limit violation, Qi will be set to the limit and the bus type will be

switched to load, i.e. PQ, as it is not possible to keep the generator terminal voltage

to the specify voltage (V sp) while Qi is being limited.

When bus type switched, the bus voltage is also needed to be corrected to cater for

the Qi being limited. Once, Qi becomes within the limits, the bus type and terminal

voltage can be restored.

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4.4 Convergence Limits (Tolerance at Solution)

Usually 0.001, 0.0001 or 0.00001 (p.u. volts).

That is all ∆V k+1i must lie within this tolerance.

4.5 Acceleration Factors

• In practice, it is found that the process of convergence due to Gauss-Seidel

method is slow. A large number of iterations is required to obtain an accurate

solution.

• The rate of convergence can be increased by the use of acceleration factors.

V k+1iacc = V ki + α(V

k+1i − V

ki )

where α is known as acceleration factor and best values for the acceleration

factor α lie in the range of 1.1 to 1.6 with 1.4 is the most common used value

in practice.

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• Instead of computing the inverse of J , eqn (3) can be written as :

−F (x p) = J (x p)(x p+1 − x p) (5)

or ∆y p = J (x p)∆x p (6)

where F (x) = F (x p) + ∆y p = 0 (7)

and x p+1 = x p + ∆x p (8)

• There are 4 steps for each iteration :

1. compute ∆y p

2. compute J (x p)

3. solve for ∆x p by Gauss elimination and back substitution

4. compute x p+1

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5.3 Load Flow Solution by Newton-Raphson (NR) Method

First, rewrite the power flow equations (1) and (2) into an alternate form:

P i = Gii|V i|2 +

nj=1,j=i

|V i||V j ||Y ij| cos(δ i − δ j − θij) (9)

Qi = −Bii|V i|2 +

n

j=1,j=i

|V i||V j||Y ij | sin(δ i − δ j − θij) (10)

where Gii = |Y ii| cos(θii)

Bii = |Y ii| sin(θii)

Then, apply the Newton-Raphson method to form the following mismatch equation: ∆P i

∆Qi

=

∂P i∂δi

∂P i∂ |V i|

∂Qi∂δi

∂Qi∂ |V i|

∆δ i

∆|V i|

(11)

where ∆P i and ∆Qi are the power mismatch at bus i and

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∆P i = P i,scheduled − P i,calculated (12)

∆Qi = Qi,scheduled − Qi,calculated (13)

• P i,calculated and Qi,calculated:

obtained from the power flow eqn (1) and (2).

• P i,scheduled :

(a) Slack bus: P i,scheduled = P i,calculated and ∆P i = 0

(b) Otherwise (PV & PQ buses): P i,scheduled = P Gi − P Li.• Qi,scheduled:

(a) Slack & PV bus: Qi,scheduled = Qi,calculated and ∆Qi = 0(b) Otherwise (PQ bus): Qi,scheduled = QGi − QLi.

• To further improve the convergence, the mismatch equation can be rewritten as: ∆P i

∆Qi

=

∂P i∂δi

|V i| ∂P i∂ |V i|

∂Qi∂δi

|V i|∂Qi∂ |V i|

∆δ i∆|V i||V i|

=

H N

J L

∆δ i∆|V i||V i|

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• For a n bus system with m PQ buses, the mismatch equation becomes :

∆P 1

..

∆P n−1

∆Q1

..

∆Qm

=

H 1,1 .. H 1,n−1 N 1,1 .. N 1,m

.. H i,i .. .. N i,i ..

H n−1,1 .. H n−1,n−1 N n−1,1 .. N n−1,m

J 1,1 .. J 1,n−1 L1,1 .. L1,m

.. J i,i .. .. Li,i ..

J m,1 .. J m,n−1 Lm,1 .. Lm,m

∆δ1

..

∆δn−1

∆|V 1|

|V 1|

..

∆|V m|

|V m|

(14)

For i = j, H ii = ∂P i

∂δi= −Qi − Bii|V i|

2

N ii = |V i| ∂P i∂|V i|

= P i + Gii|V i|2

J ii = ∂Qi

∂δi= P i − Gii|V i|

2

Lii = |V i| ∂Qi∂|V i|

= Qi − Bii|V i|2

For i = j, H ij = ∂P i∂δj

= |V i||V j ||Y ij | sin(δi − δj − θij)

N ij = |V j | ∂P i∂|V j |

= |V i||V j ||Y ij | cos(δi − δj − θij)

J ij = ∂Qi

∂δj= −N ij

Lij = |V j |

∂Qi∂|V j | = H ij

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5.4 Steps in Newton-Raphson solution

1. Formulate the nodal admittance matrix Y .

2. Assume an initial set of bus voltages and set bus n as the reference bus.

3. Obtain the power injections P i and Qi for all i = 1,...(n − 1)

4. Obtain the power mismatches ∆P i and ∆Qi for all i = 1,...(n − 1)

5. Stop the iteration if all ∆P i and ∆Qi are within tolerance.

6. Obtain the Jacobian matrix elements using the best available voltage values.

7. Substitute the values obtained from steps (4) & (6) in equation (14). Solve this

linear simultaneous equation by a suitable method for vectors [∆δ ] and [∆|V i||V i| ].

8. Update δ i and |V i| for all i, i.e. δ k+1i = δ

ki + ∆δ i

V k+1i = V k

i (1 + ∆|V i||V i|

)

9. Goto step (3).

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5.5 Decoupled Load Flow (DFL)

An important characteristic of any practical electrical power transmission system

operating in steady state is strong interdependence between real powers and bus

voltage angles and reactive powers and voltage magnitudes.

If the P -δ and Q-V couplings are recognised to be much stronger than the P -V and Q-δ couplings the sub-matrices N and J can be ignored. Then separate

equations: [∆P ] = [H ] [∆δ ] (15)

[∆Q] = [L]

∆|V |

|V |

(16)

can be obtained and solved separately to give an approximate solution of |V | and δ .

Instead of the previous 2(n − 1) × 2(n − 1) matrix problem, there are two

(n − 1) × (n − 1) matrices to solve — save memory and easier to solve but take

more number of iterations to converge because of the approximation.

Techniques such as these are often used in on-line (very fast) load flow solutions

and in the starting (initial stage) of conventional full length load flows.

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5.7 Example for Newton Raphson and Fast Decoupled LF

✲

❄

V 1 = 1.05/0o V 2/δzser = 0.1 + j0.2

ysh = j0.15 ysh = j0.15

P L, QL0.1 + j0.2

B1 B2

Find V 2 by NR method with B1 as the slack bus and initial estimate for V 2 = 1/0o.

Power flow at B2: P 2 = |V 2|2

G22 + |V 1||V 2||Y 12| cos(δ 2 − θ12)

Q2 = −|V 2|2

B22 + |V 1||V 2||Y 12| sin(δ 2 − θ12)

Since B1 is the slack bus, only B2 mismatches are calculated. ∆P 2

∆Q2

=

H N

J L

∆δ 2∆|V 2||V 2|

=

∂P 2∂δ2

|V 2| ∂P 2∂ |V 2|

∂Q2∂δ2

|V 2|∂Q2∂ |V 2|

∆δ 2∆|V 2||V 2|

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Admittance matrix:

A =

yser + ysh −yser

−yser yser + ysh

=

2 − j3.85 −2 + j4

−2 + j4 2 − j3.85

where yser = 1

0.1+j0.2 = 2 − j4 and ysh = j0.15

i.e. Y 12 = −2 + j4 = 4.472/116.56o

Y 22 = 2 − j3.85 = G22 + jB22 ⇒ G22 = 2 and B22 = −3.85

H = ∂P 2

∂δ 2= −|V 1||V 2||Y 12| sin(δ 2 − θ12)

= −(1.05)(1.0)(4.472) sin(−116.56o) = 4.2

J = ∂Q2

∂δ 2= |V 1||V 2||Y 12| cos(δ 2 − θ12)

= (1.05)(1.0)(4.472) cos(−116.56o) = −2.1

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N = |V 2| ∂P 2

∂ |V 2| = 2|V 2|

2G22 + |V 1||V 2||Y 12| cos(δ 2 − θ12)

= 2|V 2|2

G22 + ∂Q2

∂δ 2= 2(2) − 2.1 = 1.9

L = |V 2|∂Q2

∂ |V 2| = −2|V 2|

2B22 + |V 1||V 2||Y 12| sin(δ 2 − θ12)

= −2|V 2|2

B22 − ∂P 2

∂δ 2= −2(−3.85) − 4.2 = 3.5

J 1

= 4.2 1.9

−2.1 3.5 P 12 = |V 2|

2G22 +

∂Q2

∂δ 2= 2 − 2.1 = −0.1

Q12 = −|V 2|

2B22 −

∂P 2

∂δ 2= 3.85 − 4.2 = −0.35

∆P 12 = P G − P L − P 2 = −0.1 + 0.1 = 0

∆Q12 = QG − QL − Q2 = −0.2 + 0.35 = 0.15

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0

0.15

=

4.2 1.9

−2.1 3.5

∆δ 12∆|V 1

2 |

|V 12 |

∆δ 12∆|V 1

2 |

|V 12 |

=

1

18.69

3.5 −1.9

2.1 4.2

0

0.15

=

−0.01525

0.0337

δ 12 = −0.01525 rad

|V

1

2 | = 1.0337 p.u.

Similarly for the second iteration:J 2

=

4.3080 2.0375

−2.2367 3.9199

∆P 22 = −0.00037

∆Q22 = −0.00596

δ 2 = δ 22 = −0.01475 rad

|V 2| = |V 22 | = 1.03243 p.u.

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35

✬

✫

✩

✪

Now find V 2 again by FD method instead of the NR method.

Recall: Y 12 = 4.472/116.56o

G22 = 2 B22 = −3.85P 12 = −0.1 ∆P

12 = 0

Q12 = −0.35 ∆Q12 = 0.15

[B′] = [−B22] = [3.85] [B′′] = [−B22] = [3.85]

From (17): ∆δ 2 = ∆P 2

−B22|V 2| = 0 rad

⇒ δ 12 = δ 2 + ∆δ 2 = 0 rad

From (18): ∆|V 2| = ∆Q2−B22

= 0.03896 p.u.

⇒ |V 12 | = |V 2| + ∆|V 2| = 1.03896 p.u.

Update the calculated power injection (P 2, Q2) and mismatch (∆P 2, ∆Q2) with

the latest bus voltage.

36

✬

✫

✩

✪

Repeat the above procedures, as shown below, until the solution converage or the

power mismatches are below the tolerance.

Iter P 2 Q2 ∆P 2 ∆Q2 ∆|V 2| ∆δ 2 |V 2| δ 2

1 -0.1 -0.35 0 0.15 0.0390 0 1.0390 0

2 -0.0229 -0.2078 -0.0771 0.0078 0.0020 -0.0193 1.0410 -0.0193

3 -0.1025 -0.1572 -0.0026 -0.0428 -0.0111 0.0006 1.0299 -0.0186

4 -0.1216 -0.2010 -0.0216 0.0010 0.0003 -0.0055 1.0301 -0.0132

5 -0.0977 -0.2122 -0.0023 0.0122 0.0032 -0.0006 1.0333 -0.0137

6 -0.0940 -0.1989 -0.0060 -0.0011 -0.0003 -0.0015 1.0330 -0.0153

7 -0.1010 -0.1966 0.0010 -0.0034 -0.0009 0.0003 1.0321 -0.0150

8 -0.1017 -0.2005 0.0017 0.0005 0.0001 0.0004 1.0323 -0.0146

9 -0.0996 -0.2009 -0.0004 -0.0002 0.0002 -0.0001 1.0325 -0.0147

10 -0.0996 -0.1998 0.0001 -0.0003 -0.0001 0.0000 1.0324 -0.0148

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THE HONG KONG

POLYTECHNIC UNIVERSITY FAX: (852) 2330 1544Department of Electrical Engineering Hung Hom, Kowloon, Hong Kong

EE4031 Power Systems

Tutorial on Power System Load Flow

1. Fig 1 shows a 4-bus system where all the transmission line series impedances are given

to a common base of 100 MVA while the shunt admittances of the lines are neglected.

Specifications at busbars are given in Table 1 and flat start conditions are assumed.

j0.2

j0.5

j0.25

j0.1 j0.33

1 2

34

LoadLoad

S 12S 14

S 21

S 24

S 23

Fig 1

Real Reactive Real Voltage VoltageBus Demand Demand Generation Magnitude Angle

(MW) (MVAr) (MW) (pu) (deg)

1 – – – 1.04 0

2 – – 100 1.02 –

3 80 60 – – –

4 90 50 – – –

Table 1

(a) Classify the type of each busbar.

(b) Determine the bus admittance matrix.

(c) Determine the initial power flows S 12, S 14, S 21, S 23 and S 24.

(d) Determine the initial power generations and mismatches at the bus 1 and 2.

(e) With justification, what should be the real power generation at bus 1 ?

(f) Recommend a solution method, with justifications, which is suitable for solving

this power flow problem.

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2. Fig 2 shows a single-line diagram of a 2-bus power system with parameters detailed in

Table 2. The series impedance of the line is given in per-unit on a common base of 100

MVA with shunt admittance neglected.

Z = j0.5

V2V = 1.02 01o

S = 50L2S = 30 + j10L1

SG1

21

Fig 2

Real Reactive Real Voltage VoltageBus Demand Demand Generation Magnitude Angle

(MW) (MVAr) (MW) (pu) (deg)

1 30 10 – 1.02 0

2 50 0 – – –

Table 2

(a) Name the slack bus and write down the bus admittance matrix Y .

(b) Based on the load flow equation given below :-

V i = 1

Y ii

P i − jQi

V ∗i−

n j=1,j=i

Y ijV j

Use Gauss-Seidel with flat start conditions to solve the load bus voltage V 2.

(c) With justification, what should be the reactive power generation at bus 1 ?

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3. Fig 3 shows a single-line diagram of a three-bus power system. All the transmission line

series impedances are given in per unit to a common base of 100 MVA while the shunt

admittances are neglected. Specifications at busbars are given in Table 3.

j0.1

j0.1 j0.1

1 2

3

200 MW

50 MVAr

Fig 3

Real Reactive Real Reactive Voltage VoltageBus Demand Demand Generation Generation Magnitude Angle

P L (MW) QL (MVAr) P G (MW) QG (MVAr) V (pu) δ (deg)

1 0 0 – – 1.01 0

2 200 50 0 0 – –

3 0 0 100 – 1.02 –

Table 3

(a) Classify each bus type and determine which of the variables V , δ , P and Q should

be treated as unknown.

(b) Write down the real power generation at bus 1 by inspecting the data.

(c) Write down the Jacobian matrix in terms of partial derivatives.

(d) Determine the bus admittance matrix.

(e) Given the power flow equations at bus i as follows

P i = |V i|2Gii +

n j=1,j=i

|V i||V j||Y ij| cos(δ i − δ j − θij)

Qi = −|V i|2Bii +

n j=1,j=i

|V i||V j||Y ij| sin(δ i − δ j − θij)

Derive the general equations for the diagonal coefficients of the Jacobian matrix

and hence find the diagonal coefficients of the Jacobian matrix for the first iter-

ation when the polar form of the Newton Raphson method is used with flat start

conditions.

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4. A single-line diagram of a three-bus power system is shown in Fig 4.

j0.4

j0.2 j0.1

1 2

3

200 MW50 MVAr

Fig 4

All the transmission line series impedances are given in per unit to a common base of

100 MVA while the shunt admittances are neglected. Specifications at busbars are givenin Table 4.

Real Reactive Real Reactive Voltage VoltageBus Demand Demand Generation Generation Magnitude Angle

P L (MW) QL (MVAr) P G (MW) QG (MVAr) V (pu) δ (deg)

1 0 0 – – 1.05 0

2 200 50 0 0 – –

3 0 0 100 – 1.02 –

Table 4

(a) Classify each bus type and determine which of the variables V , δ , P and Q should

be treated as unknown.

(b) Write down the bus admittance matrix [Y ].

(c) Using the Fast Decouple Load Flow (FDLF) convention :-

∆P

|V |

=

B′[∆δ ]

∆Q

|V |

= B

′′

∆|V |

|V |

write down the matrices [B′] and [B′′].

(d) Given the power flow equations at bus i as follows

P i = |V i|2Gii +

n j=1,j=i

|V i||V j ||Y ij| cos(δ i − δ j − θij)

Qi = −|V i|2Bii +

n j=1,j=i

|V i||V j ||Y ij | sin(δ i − δ j − θij)

carry out the first load flow iteration using the FDLF method.

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5. Fig 5 shows a single-line diagram of 3-bus system with parameters detailed in Table 5.

The series impedance of each transmission line is given in per-unit on a common base of

100 MVA with shunt admittance neglected.

j0.4

j0.4 j0.4

1 2

3

100 MW

-80 MVAr

100 MW

100 MW

60 MVAr

80 MW

-30 MVAr

Fig 5

Load Demand Specified Power Specified Voltage

Bus MW MVAr MW MVAr pu degree

1 100 0 – – 1.0 0

2 100 -80 80 -30 – –

3 100 60 0 0 – –

Table 5

(a) Name the slack bus and write down the bus admittance matrix Y .

(b) Based on the load flow equation given below :-

V i = 1

Y ii

P i − jQi

V ∗i−

n j=1,j=i

Y ijV j

Perform one iteration of the load flow using the Gauss-Seidel method with flat startconditions to calculate the appropriated voltages at bus 2 and 3.

(c) What should be the real power generation at bus 1 ?

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EE4031 Power Systems

Tutorial Solution on Power System Load Flow

1.

a) Bus Type1 Slack

2 Generator

3 Load

4 Load

b)

8.03 5 0 3.03

5 17 10 2

0 10 14 4

3.03 2 4 9.03

j j j

j j j jY

j j j

j j j j

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥

−⎣ ⎦

c) ]12

* *

12 1 1 2[ ( ) ser S V Y V V = −

1.04[ 5(1.04 1.02)]

0.104 pu 10.4 MVAr

j

j j

= −

= =

14 0.126 pu 12.6 MVAr S j j= =

21 0.102 puS j= −

23 0.204 puS j=

24 0.0408 puS j=

d) 1 12 14 23 MVAr S S S j= + =

2 21 23 24 14.28 MVAr S S S S j= + + =

Bus 1 – Slack mismatch = 0 MW

Bus 2 – PV bus P mismatch = Pg2 – Pl2 – Re(S2) = 100 MW

Q mismatch = 0 MW

e) No transmission loss Pg1 = 80 + 90 – 100 = 70 MW

f)

GS: simple, low memory usage, easy to implement NR: faster, better convergence, more reliable

FD: even faster & more reliable

2. a) slack bus : BUS1

⎥⎦

⎤⎢⎣

⎡

−

−=

22

22

j j

j jY

b) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−= − V Y V S Y V k

k

121*1

2

*

2

22

2

1, 5.022 −=−= S S L , 222 jY −= , 2 21 jY =

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4. a) Bus Type

1 Slack

2 PQ

3 PV

b) [ ]

7.5 2.5 5

[ ] 2.5 12.5 10

5 10 15

j j j

Y j j j G j

j j j

−⎡ ⎤⎢ ⎥= − = +⎢ ⎥⎢ ⎥−⎣ ⎦

B

c)

2

2 22 23 2

32 33 33

3

P

V B B

B B P

V

δ

δ

Δ⎡ ⎤⎢ ⎥ − − Δ⎡ ⎤ ⎡⎢ ⎥ = =⎢ ⎥ ⎢− − ΔΔ⎢ ⎥ ⎣ ⎦ ⎣⎢ ⎥⎣ ⎦

⎤⎥⎦

[ ] 222 2

22V Q

BV V

⎡ ⎤ ⎡ ΔΔ= − =

⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎤⎥

'12.5 10

10 15 B

−⎡⎡ ⎤ = ⎢⎣ ⎦ −⎣ ⎦

[ ] [ ]" 12.5 B =

d) Flat start conditions all angles are zero

i.e. no angle difference P2 = 0, P3 = 0

2 2 2

2000 2 p

100 schedule P P P u

−Δ = − = − = −

3 3 2

1000 1 pu

100 schedule P P P Δ = − = − =

2

3

2

12.5 101.0

1 10 15

1.02

δ

δ

−⎡ ⎤⎢ ⎥ Δ− ⎡ ⎤⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎢ ⎥ Δ−⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

⇒ 2

3

15 10 2 0.23081

10 12.5 0.98 0.088687.5

δ

δ

Δ − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢Δ −⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦

⎤⎥⎦

⇒ 2

3

0.2308rad

0.0886

δ

δ

−⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦

( ) ( ) ( )

2

2 2 22 2 1 21 2 1 21 2 3 23 2 3 23sin( ) sin( )

12.5 (1.0)(1.05)(2.5)sin 0.2308 1.0 1.02 10 sin 0.14222 2

0.152 pu

Q V B V V Y V V Y δ δ θ δ δ θ

π π

= − + − − + − −

⎛ ⎞ ⎛ = + − − + − −⎜ ⎟ ⎜

⎝ ⎠ ⎝

= −

⎞⎟ ⎠

⇒ 250

( 0.152) 0.348 pu100

Q −

Δ = − − = −

[ ]22 2

12.5Q V

V V

⎡ ⎤ ⎡Δ Δ=⎢ ⎥ ⎢

⎣ ⎦ ⎣

⎤⎥⎦

⇒ 20.348

0.028 pu12.5

V Δ = − = −

i.e. 2 21 0.972 puV V = + Δ =

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5. a) Slack bus - Bus 1

5 2.5 2.5

0.4, 2.5, 2.5 5 2.5

2.5 2.5 5

L L

j j j

Z j Y j Y j j j

j j j

−⎡ ⎤⎢ ⎥= = − = −⎢ ⎥

−⎢ ⎥⎣ ⎦

b)

Flat start:0 0 0

1 2 31 0 , 1 0V V V = ∠ ° = = ∠ °

2

3

1

2

1

3

1 0.8 0.8 0.3 0.2 0.5 pu

1 0.6 pu

1 0.2 0.5 0.2 5.52.5 2.5 1.1 0.04 1.1 2.08 pu

5 1 5

1 1 0.6 1 4.42.5 2.5 0.88 0.2 0.9 12.8 pu

5 1 5

S j j j

S j

j jV j j j

j j

j jV j j j

j j

= − + + − = − +

= − −

− − +⎡ ⎤= − − = = − = ∠ −⎢ ⎥− ⎣ ⎦

− + +⎡ ⎤= − − = = − − = ∠ −⎢ ⎥−

⎣ ⎦

°

°

c)

100 + 100 + 100 - 80 = 220 MW

ee4031 1 loadflow

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