The Power Flow Problem

The Power Flow Problem
Scott Norr
EE 4501
April, 2015
Electric Concepts
• Ohm’s Law: V = IR
(V = IZ)
• Kirchoff: KCL: ∑ i = 0 (at any node)
∑ V = 0 (on closed path)
• Power: P = VI
(S = VI*)
= V2/R = I2R
• NOTE: i means electric current, j2 = -1
Previously: DC Circuits (RI=V)
Resistance Matrix
AC Circuits: Phasor Analysis
ZI = V (Thanks to Euler, Steinmetz)
DC Power:
All electrical
systems naturally
seek an equilibrium
point of lowest
Important to
recognize that
P ᴕ V2
(try to find the proportional
symbol in powerpoint
AC Power
• The complex power, S = VI* = P + jQ
• P is the average power (“real” power) in watts,
attributable to resistive loads
• Q is the reactive power (“imaginary” power) in
VAr, attributable to capacitive and inductive
The Power Problem:
• On AC power systems, we don’t predetermine the phase angles on the sources,
they are determined by the system
(additional unknowns to solve for!)
• Power is injected into nodes in the system via
sources and is removed at nodes via loads
(consumption points)
• Additionally, power is lost in the network
Consider an Example: 3 Node System
• At each bus (node) there are 4 parameters:
P, Q, V and Ө
• There are three types of buses:
– Load Buses: P, Q are known, V, Ө are unknown
– Generator Buses: P, V are known, Q, Ө are unknown
– Slack Bus: (unique) V, Ө are known, P, Q unknown
(this special generator node is allowed to accumulate errors
in the iterative solution of the system of equations)
So, for N nodes, 2N unknown node parameters
Balancing Power at Each Node:
• ∑Si = o
• SG-SL = Vi∑Ip*
• SG-SL = Vi∑Vp*Yp*
Can separate the real (P) from the
imaginary (Q)to form two equations
at each Bus
A system of 2N equations
Sparse, largely diagonalized matrices
Solve for Node Voltages and Angles:
• Vi new = (1/Yii)(Si/Viold - ∑Vp*Yp*)
• An iterative process, involving an initial
starting estimate and convergence to a predetermined tolerance.
• This is called the Gauss-Seidel Solution
A better Method:
• For analytic, complex differentiable systems,
can compute the low order terms of the Taylor
series and solve using Netwon’s method.
• In two variables, an iterative approach:
f1(x,y) = K = f(xo + Δxo, yo + Δyo)
g1(x,y) = L = g(xo + Δxo, yo + Δyo)
Computing the Taylor Series, and truncating it
yields an equation exploiting a Jacobian matrix
Newton – Raphson Solution:
• Powerflow Software is used by every electric
utility in the world. Many models contain 10,000
nodes or more.
• There are quite a few solution techniques that
are more efficient than the G-S and N-R methods
outlined here:
Fast-Decoupled N-R – decouples P,O from Q,V and
solves the two, smaller systems
Interior Point Newton - calculates a Hessian Mtx!
PowerWorld Simulator
• Stevenson, William D., Elements of Power
System Analysis, McGraw-Hill, 1982
• Tylavsky, Daniel, Lecture Notes #19, EEE 574,
Arizona State University, 1999
• PowerWorld Simulator,,2014