CHAPTER 5 BEE3143:POWER SYSTEM ANALYSIS- Power flow...
Transcript of CHAPTER 5 BEE3143:POWER SYSTEM ANALYSIS- Power flow...
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CHAPTER 5
BEE3143:POWER SYSTEM ANALYSIS- Power flow solution- Newton-Raphson Expected Outcomes Able to solve power flow solution using Newton-Raphson technique
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Newton-Raphson Power Flow Solution
• Newton-raphson method is found to be more practical and efficient for large power system.
• The number of iterations required to obtain a solution is independent of the system size, but more functional evaluations are required at each iteration.
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... Newton-Raphson Power Flow Solution Rewrite the current entering bus i in a
typical bus power system in figure below in terms of the bus admittance matrix
∑=
=n
jjiji VYI
1
The typical element Yij is
ijijijijijijijijij jBGYjYYY +=+=∠= θθθ sincos
The voltage at a typical bus i
(1)
Expressing Equation (1) in polar form
jij
n
jjiji VYI δθ +∠= ∑
=1
∑=
=
=−n
jjiji
iiii
VYV
IVjQP
1
*
*
yin
yi1
yi2
Ii
Vi
V1
V2
Vn
yi0
( )iiiiii jVVV δδδ sincos +=∠=
The complex conjugate of the power injected at bus i is
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...Newton-Raphson Power Flow Solution The complex conjugate of the power in polar form
ijij
n
jjiij
jij
n
jjijii
n
jjijiii
VVY
VYV
VYVjQP
δδθ
δθδ
−+∠=
+∠−∠=
=−
∑
∑
∑
=
=
=
1
1
1
*
Separating this equation into real and reactive parts
( )ijij
n
jjiiji VVYP δδθ −+= ∑
=1cos
( )ijij
n
jjiiji VVYQ δδθ −+−= ∑
=1sin
Equation (2) and (3) constitute a polar form of the power flow equations.
(2)
(3)
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...Newton-Raphson Power Flow Solution These Eqs. constitute a set of non-linear algebraic equations in terms of the independent variables, voltage magnitude in per-unit, and phase angle in radians. Expanding (2) and (3) in Taylor’s series about the initial estimate and neglecting all higher order terms results in the following set of linear equations.
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...Newton-Raphson Power Flow Solution
MISMATCHES
kn
k
kn
k
SCORRECTION
kn
k
kn
k
MATRIXJACOBIAN
k
n
nk
n
k
n
k
k
n
nk
n
k
n
k
k
n
nk
n
k
n
k
k
n
nk
n
k
n
k
Q
Q
P
P
V
V
VQ
VQ
JVQ
VQ
VP
VP
JVP
VP
QQJ
PPJ
PP
∆
∆
∆
∆
=
∆
∆
∆
∆
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
)(
)(2
)(
)(2
)(
)(2
)(
)(2
)()(
2
4
)(2
)(
2
2
)()(
2
2
)(2
)(
2
2
)()(
2
3
)(2
)(
2
2
)()(
2
1
)(2
)(
2
2
δ
δ
δδ
δδ
δδ
δδ
Newton-Raphson power flow equations:
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...Newton-Raphson Power Flow Solution
• From the previous equation, bus 1 is assumed to be slack bus. • The Jacobian matrix gives the linearized relationship between small
changes in voltage angle and voltage magnitude with the small changes in real and reactive power.
• Elements of Jacobian matrix are the partial derivatives of (2) and (3), evaluated at small changes in voltage angle and voltage magnitude
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...Newton-Raphson Power Flow Solution
• From the previous equation, bus 1 is assumed to be slack bus. • The Jacobian matrix gives the linearized relationship between small
changes in voltage angle and voltage magnitude with the small changes in real and reactive power.
• Elements of Jacobian matrix are the partial derivatives of (2) and (3), evaluated at small changes in voltage angle and voltage magnitude
• In short form, it can be written as:
∆∆
=
∆∆
VJJJJ
QP δ
43
21 (4)
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Procedure of NR
• For load buses, is calculated from (2) and calculated from (3); • is calculated from: • is calculated from: • For PV buses, is calculated from (2) and calculated from • The element of Jacobian matrix (J1, J2, J3 and J4) are calculated as
follows:
)(kiP )(k
iQ)(k
iP∆ )()( ki
schi
ki PPP −=∆
)()( ki
schi
ki QQQ −=∆)(k
iQ∆)(k
iP∆)(kiP
)()( ki
schi
ki PPP −=∆
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…Procedure of NR
The diagonal and off-diagonal elements of J1 are
∑≠
+−=∂∂
ijjiijijji
i
i YVVP )sin( δδθδ ( )ijijjiij
j
i VVYP
δδθδ
−+−=∂∂
sin
The diagonal and off-diagonal elements of J2 are
∑≠
+−+=∂∂
ijjiijijjiiiii
i
i YVYVVP )cos(cos2 δδθθ
( )ijijiijj
i VYVP δδθ −+=
∂∂ cos
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…Procedure of NR
The diagonal and off-diagonal elements of J3 are
∑≠
+−=∂∂
ijjiijijji
i
i YVVQ )cos( δδθδ
)cos( jiijijjij
i YVVQ δδθδ
+−−=∂∂
The diagonal and off-diagonal elements of J4 are
∑≠
+−−−=∂∂
ijjiijijjiiiii
i
i YVYVVQ )sin(sin2
||δδθθ
)sin(|| jiijiji
j
i YVVQ δδθ +−−=
∂∂
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…Procedure of NR
• The equation (4) is solved directly by optimally ordered triangular
factorization of Gaussian elimination • The new voltage and phase angle are computed from:
• The process is continued until the residuals and are less than specified accuracy
)(kiP∆ )(k
iQ∆
)()()1(
)()()1(
ki
ki
ki
ki
ki
ki
VVV ∆+=
∆+=+
+ δδδ
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