Overview
•  Review complex power, real power,
reactive power
•  Introduce one-line diagram (circuit)
•  Introduce the power flow problem
Introduction to the Power
Flow Problem
Smith College, EGR 325
Sept 15, 2015
– Must employ a numerical solution (not
an analytic solution)
– (Why?)
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Power System Diagrams
Memories of Circuit Theory
•  Circuit vs. one-line diagram
•  Kirchhoff’s Laws
–  Equations
–  The underlying physics represented by the
equations (conservation laws?)
•  For circuits problems, typically we are…
–  Given what information?
–  Solving for what?
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4
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Power System Diagrams
•  Circuit vs. one-line diagram
16.8 MW
16.0 MW
6.4 MVR
0.0 MVR
44.94 kV
16.8 MW
40.0 kV
16.0 MW
6.4 MVR
16.0 MVR
Generators
are shown as
circles
Transmission lines
are shown as a single
line
16.0 MVR
Arrows are
used to
show loads
New York + Northeast
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2
Power Flow Analysis v. Circuits
200kV-1MV
Transmission
transformer
•  In circuit analysis, we know (or calculate)
–  V, I, Z; and ultimately the power, P
High voltage
transmission line
•  In power system analysis we do not know
–  The complex voltages (magnitude & phase)
–  The complex current injections
Transmission
transformer
Power plant
Service
transformer
120/240V
•  Instead, we do know
15kV-25kV
Distribution line
Distribution
transformer
–  the complex power being demand at each load
–  estimates for the power ‘injected’ by generators
–  estimates for the voltage magnitudes at generators
•  Therefore
–  we need to go beyond Ohm’s law
–  we must use ‘power flow’ equations
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Phasors: Power & Phase Angle
Complex Power S = P + jQ
•  Instantaneous – at a given moment in time
•  Average – over a specified time period
•  Phasor domain (SSS)
•  Writing power in phasor notation:
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V = Vm∠θ v = Vm eθv
P = Vm I m cos(θ v − θ i )
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θi
p(t ) = v(t )i(t ) = Vm I m cos(ωt + θv ) cos(ωt + θi )
•  Moving to complex power (with RMS):
I = I m∠θ i = I m e
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1
p(t) = Vm I m cos(θ v − θ i ) + Vm I m cos(2ω t + θ v + θ i )
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2
VI =
VI* =
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S = Re{VI*} + jIm{VI*} = P + jQ
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Concept: “Real” Power, P
Reactive (Imaginary?) Power
S = P + jQ
•  Reactive power, Q
–  Oscillates within system
–  Energy stored in, and oscillating between,
capacitance and inductance
•  P is what we buy from the power company
•  Most customers do not pay for Q
•  To maximize revenue, power companies
would like to generate _____?
•  Voltage and (stored) reactive power allow
real power to flow
–  How would they operate or design the system
to do this?
–  A role for the power factor?
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Reactive Power Analogy
•  Reactive power
–  Energy stored in capacitance and inductance
–  Supports the electromagnetic fields along
transmission lines
–  Cannot be transmitted long distances
•  Analogy
–  Inflatable firefighters’ water hoses
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For Power Systems Analysis:
Complex Power – Phasor Domain
IMPORTANT
θ is the power factor angle
•  “Power flow” refers to how the power is
moving through the system
–  At all times in the actual power system, and in
any simulation, the total power flowing into
any bus (node) *must* equal zero (KCL)
S ≡V I
*
V
I
θ
I = I∠ − θ
S = V I * =V (cos 0 + j sin 0) ⋅ I (cos θ + j sin θ )
S = V I cosθ + j V I sin θ
S = P+ j Q
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Power Flow Equations
Real Power
Reactive Power
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Power Flow Equations
•  Using KCL, we can write the current
flow into the power system from any
node i
Ii = IGi – IDi = ΣIik
•  Since I = YV (Ohm’s Law, where Y = Z-1),
we know: (subscripts in the summation?)
Ii = IGi – IDi = ΣYikVk
•  What do “i” and “k” represent?
•  Turning to the power injection at node i,
we can write
Si = ViIi* = Vi(ΣYikVk)* = ViΣYik* Vk*
•  We want to be able to calculate Pi and
Qi separately, so we need to expand
the expression for Si into real and
reactive components.
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Power Flow Equations
Power Flow Equations
•  Use the following expressions to
expand the equation for complex
power
•  Finally, we can write
Si = ViΣYik* Vk* = Pi + jQi
= Σ |Vi||Vk| ejθik (Gik – jBik)
= Σ |Vi||Vk|(cosθik + jsinθik) (Gik – jBik)
Ø  Yik = Gik + jBik
Ø  Vi = |Vi|<θi = |Vi|ejθ
•  Separate this into real & reactive parts
•  ejθ = cosθ + jsinθ
Ø  θik = θi – θk
Ø  Si = Pi + jQi
Pi = Σ Vi Vk (Gik cosθ ik + Bik sin θ ik ) = PGi − PDi
Qi = Σ Vi Vk (Gik sin θ ik − Bik cosθ ik ) = QGi − QDi
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Belgian High Voltage Network
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Power Flow Analysis
•  For power systems, we know
–  The system topology (the circuit diagram)
–  The impedance of each line, R, X, B
–  The load at each load bus (S = P + jQ)
–  The capability of each generator (P, V)
“i” takes on each bus
number in turn, 1 through 53.
•  We want to know
–  The output of each generator (S = P + jQ)
–  The voltage at each bus (V = V∠θ)
–  The power flow on each line (Pflow)
If “i” is bus 41, for example, then
“k” takes on numbers 35, 36,
37, 40, 46, and 47 à all buses
directly connected to bus 41.
There will be a set of power flow equations,
Pi and Qi, for each bus “i”
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Real Power Flow Equations
Power Flow Summary
•  How many equations and how many
unknowns?
•  Numerical methods
•  What is the purpose of power flow
analysis?
•  What is the process?
•  What data do we have and seek?
•  How is our understanding improved
by performing a power flow?
– Lack of convergence
– Slack bus
•  Definition
•  Mathematical and physical role
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Power Factor Review
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Power Factor Review 2
•  For inductive loads
•  What is the power factor?
•  In words and mathematically
•  Power factor = cos(θv - θi)
•  The phase angle (θv - θi) is defined as the
power factor angle
•  Note that this is also the impedance angle –
do you see why?
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•  The current lags the voltage, showing
that θi is less than (or more negative
than) θv
•  Therefore, (θv - θi) > 0
•  The power factor is said to be lagging
•  The load is said to be lagging
•  Reactive power, Q > 0 à An inductor
consumes Q
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Power Factor Review 3
•  For capacitive loads
•  The current leads the voltage, showing
that θi is greater than θv
•  Therefore, (θv - θi) < 0
•  The power factor is said to be leading
•  The load is said to be leading
•  Reactive power, Q < 0 à A capacitor
generates Q
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Summary
SIEPAC – Central America
•  Introduction to the power flow
problem
– Review of complex power (and
complex numbers)
– Develop the power flow equations
– Recap of power factor
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http://www.google.com/url?sa=t&source=web&cd=2&ved=0CCIQFjAB&url=http%3A%2F%2Fwww6.miami.edu%2Fhemisphericpolicy%2FMartin_Central_America_Electric_Int.pdf&rct=j&q=central%20america%20transmission%20siepac&ei=6kCPTd7eKmV0QHxjbWfCw&usg=AFQjCNHOxI1aHVoKhEQ7vJ5nk_SCxsC-zQ&cad=rja
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