Power FlowRobert R. Krchnavek
Rowan UniversityGlassboro, New Jersey
Power Flow Calculations• Required for operation and planning.
• If all the voltages (magnitude and phase) could be measured, it would be a simple task to determine real and reactive power everywhere (linear system).
• Utilities measure voltage magnitude, real power and reactive power a certain buses.
• SCADA - supervisory control and data acquisition.
Power System – Terminology• Load Buses - P and Q are specified. These are
called PQ buses.
• Generator Buses - Voltage magnitude and real power, P, are specified. PV buses.
• Slack Bus - Voltage magnitude is specified, at 0˚ phase, and P and Q can be whatever! An “infinite” bus.
• P and Q are not specified. Neither is the voltage. Usually for transformers. Similar to PQ buses with P = Q = 0.
Example Power System(see textbook)
Example Power System
Example Power SystemAssuming a balanced 3-phase system, capable ofdelivering 100 MVA, determine the per-phase andper-unit quantities:
Vph =VLLp
3
Iph =1
3
MVA
Vph= Ibase
Zbase =Vbase
Ibase
199 kV
167 kA
1190.25 Ω
Ybase =1
Zbase
per-unit values
840 µΩ
Example Power SystemUsing a pi-circuit model for the transmission line, what are the line impedances (actual) and per-unit values?
For the 200 km long line:Zseries = 200⇥ (0.037 + |0.376) = (7.40 + |75.2) ⌦
! (0.0062 + |0.063) pu
Yshunt = �|200⇥ 4.5(10�6) f = 900 µf ! 1.07pu
Note: Be careful when working with susceptances and reactances and how they add.
Example Power SystemReconfiguring for node-voltage analysis:
Example Power SystemReconfiguring for node-voltage analysis:
Example Power SystemReconfiguring for node-voltage analysis:
Example Power SystemReconfiguring for node-voltage analysis:
Example Power SystemReconfiguring for node-voltage analysis:
Example Power SystemReconfiguring for node-voltage analysis:
Example Power SystemWrite the node-voltage equations:
A linear system of equations that can easily be solved numerically. Note, in the above equations, etc.
�I1 +V1 � V2
Z12+ V1Y1G +
V1 � V3
Z13= 0
I1 = V1Y1G +V1 � V2
Z12+
V1 � V3
Z13= V1Y1G + (V1 � V2)Y12 + (V1 � V3)Y13
I2 = V2Y2G + (V2 � V3)Y23 + (V2 � V1)Y12
I3 = V3Y3G + (V3 � V1)Y13 + (V3 � V2)Y23
Y12 = Y21
Example Power SystemThe node-voltage equations can be generalized:
Ik = VkYkG +X
m,m 6=k
(Vk � Vm)Ykm
Ik = Vk
0
@YkG +X
m,m 6=k
Ykm
1
A�X
m,m 6=k
VmYkm
Example Power SystemFor an arbitrarily large system with similar properties,we can create an Admittance Matrix.
First, some nomenclature:
Ykk = YkG +X
m,m 6=k
Ykm
and redefine in the second summation on the
previous page to be instead of .
This effectively buries the minus sign inside .
Ykm
� 1
Zkm
1
Zkm
Ykm
Example Power System2
6666664
I1I2······In
3
7777775=
2
6666664
Y11 Y12 ·· ·· ·· Y1n
Y21 Y22 ·· ·· ·· Y2n
·· ·· ·· ·· ·· ···· ·· ·· ·· ·· ···· ·· ·· ·· ·· ··Yn1 Yn2 ·· ·· ·· Ynn
3
7777775
2
6666664
V1
V2
······Vn
3
7777775
Ykk = YkG +X
m,m 6=k
1
Zkm
Ykm = � 1
Zkm
where
and
A linear system of equations that can easily be solved numerically.
Power Flow Equations
• In the matrix representation, the unknowns are the voltages (node-voltage analysis).
• The current injected at each node (I1, I2, etc.) is required. Unfortunately, these are not exactly known.
• PV buses - P is known and the magnitude of V.
• PQ buses - P and Q are known.
2
6666664
I1I2······In
3
7777775=
2
6666664
Y11 Y12 ·· ·· ·· Y1n
Y21 Y22 ·· ·· ·· Y2n
·· ·· ·· ·· ·· ···· ·· ·· ·· ·· ···· ·· ·· ·· ·· ··Yn1 Yn2 ·· ·· ·· Ynn
3
7777775
2
6666664
V1
V2
······Vn
3
7777775
Power Flow EquationsSk = Pk + |Qk = VkI
⇤k
I1 = Y11V1 +Y12V2 + · · ·
Ik =nX
m=1
YkmVm
Ykm = Gkm + |Bkm
Sk = Pk + |Qk = VkI⇤k = Vk
nX
m=1
Y⇤kmV⇤
m
= Vk
nX
m=1
(Gkm � |Bkm)V⇤m
Power Flow EquationsSk = Pk + |Qk =
nX
m=1
(Gkm � |Bkm)VkV⇤m
Pk + |Qk =nX
m=1
(Gkm � |Bkm)Vk ✓kVm �✓m
Pk + |Qk =nX
m=1
(Gkm � |Bkm)VkVm ✓k � ✓m
Pk + |Qk =nX
m=1
(Gkm � |Bkm)VkVme✓k�✓m
Pk + |Qk =
nX
m=1
(Gkm � |Bkm)VkVm [cos (✓k � ✓m) + | sin (✓k � ✓m)]
Power Flow Equations
nX
m=1,m 6=k
VkVm [Gkm cos (✓k � ✓m) +Bkm sin (✓k � ✓m)]
Pk = GkkV2k +
Qk = BkkV2k +
nX
m=1,m 6=k
VkVm [Gkm sin (✓k � ✓m)�Bkm cos (✓k � ✓m)]
Power Flow Equations
• Assume we have the following types of buses: one slack bus, PV-buses, and buses.
• equations where P is specified.
• equations where P and Q are specified.
• equations for Pk and/or Qk.
Consider an n-bus system:
nPV
nPQ
nPQ
1 + nPV + nPQ = n
nPV + nPQ
nPV + 2nPQ
Power Flow Equations
• Unknowns:
• unknown voltage magnitudes.
• unknown voltage phase angles.
• Total unknowns:
• Equal number of unknowns and equations.
nPQ
nPV + nPQ
nPV + 2nPQ
Power Flow EquationsP spk � Pk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0
Qspk �Qk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0
specified values of real and reactive power
All buses exceptthe slack bus.
All PQ buses.
Newton-Raphson Method
• Common method of solving a nonlinear system of equations.
• Relatively quick and likely to converge. Convergence usually occurs within 10 iterations regardless of matrix size.
• Other techniques are available, e.g., Gauss-Seidel.
Newton-Raphson Method
• c is a constant and f(x) is nonlinear.
• an initial guess of the solution.
• an adjustment in x gets closer to the solution. What should be?
• A Taylor series expansion is used to calculate a and a new x value.
c� f(x) = 0
x
(0)
c� f(x(0) +�x) ⇡ 0
f(a+ x) =NX
n=0
x
n
n!f
(n)(a) +RN
�x
�x
Newton-Raphson Methodf(a+ x) =
NX
n=0
x
n
n!f
(n)(a) +RN
f(x(0) +�x) = f(x(0)) +�x
df
dx
����x
(0)
+ · · ·
c� f(x(0) +�x) ⇡ 0 ) c�f(x(0)) +�x
df
dx
����x
(0)
�= 0
�x =c� f(x(0))
df
dx
���x
(0)
x
(1) = x
(0) +�x
Newton-Raphson Methodx
(1) = x
(0) +�x
�x =c� f(x(1))
df
dx
���x
(1)
x
(2) = x
(1) +�x
calculate a new :�x
and then a new x value:
and again, ……
Newton-Raphson Method
See Example 5.2 in the Textbook
and understand it !
N-R Method forPower Flow Equations
• Apply the Newton-Raphson method to the matrix of Pk and Qk equations.
• Pk and Qk are in estimates of voltage magnitudes and phases, or totally unknown voltages.
c� f(x) = 0
P spk � Pk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0
Qspk �Qk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0
N-R Method forPower Flow Equations
c� f(x(0)) = �x
df
dx
����x
(0)
P sp � PQsp �Q
�
| {z }(2nPQ+nPV )⇥1
=
2
4@P@✓
@P@V
@Q@✓
@Q@V
3
5
| {z }[J]
�✓�V
�
| {z }(2nPQ+nPV )⇥1
J is the Jacobian and consists of submatrices.
N-R Method forPower Flow Equations
nX
m=1,m 6=k
VkVm [Gkm cos (✓k � ✓m) +Bkm sin (✓k � ✓m)]
Pk = GkkV2k +
Recall:
@Pk
@✓k=
nX
m=1,m 6=k
VkVm
[�Gkm sin (✓k � ✓m) +Bkm cos (✓k � ✓m)]