The Power Flow Problem Scott Norr For EE 4501 April, 2017

Electric Concepts KVL: ∑ V = 0 (on closed path) Ohm’s Law: V = IR (V = IZ) Kirchoff: KCL: ∑ i = 0 (at any node) KVL: ∑ V = 0 (on closed path) Power: P = VI (S = VI*) = V2/R = I2R NOTE: i means electric current, j2 = -1 Electric Concepts

Previously: DC Circuits (RI=V) Resistance Matrix

AC Circuits: Phasor Analysis ZI = V (Thanks to Euler, Steinmetz) Impedance Matrix

DC Power: All electrical systems naturally seek an equilibrium point of lowest entropy Important to recognize that P󠄳V2 (try to find the proportional symbol in powerpoint sometime…..)

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 loads AC Power

On AC power systems, we don’t pre- determine 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 The Power Problem:

Consider an Example: 3 Node System

Unknowns: 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 Unknowns:

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 Balancing Power at Each Node:

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 pre-determined tolerance. This is called the Gauss-Seidel Solution Method Solve for Node Voltages and Angles:

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 A better Method:

Newton – Raphson Solution:

Powerflow Software is used by every electric utility in the world 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! Conclusions:

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, www.powerworld.com,2014 References: