ee4031 1 loadflow
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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
34
✬
✫
✩
✪
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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The HK Polytechnic University Load Flow
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