Power System Analysis: Modeling, Power Flow, and Stability, Study Guides, Projects, Research of Electrical Engineering

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Power System Analysis
K. Tomsovic V. Venkatasubramanian
School of Electrical Engineering and Computer Science
Washington State University
Pullman, WA
1. Introduction
The interconnected power system is often referred to as the largest and most complex
machine ever built by humankind. This may be hyperbole, but it does emphasize an
inherent truth: the complex interdependency of different parts of the system. That is,
events in geographically distant parts of the system may interact strongly and in
unexpected ways. Power system analysis is concerned with understanding the operation
of the system as a whole. Generally, the system is analyzed either under steady-state
operating conditions or under dynamic conditions during disturbances.
Electric power is primarily transmitted as a three phase signal. That is, three AC current
currents are sent that are out of phase by 120o but of equal magnitude. Such balanced
currents sum to zero and thus, obviate the need for a return line. If the voltages are
balanced as well, the total power transmitted will be constant in time, which is a more
efficient use of equipment capacity. For large scale systems analysis, the assumption is
usually made that the system is balanced. Each phase can be then analyzed independently
greatly simplifying computations. In the following, the implicit assumption is that three
phase systems are being used.
2. Steady-State Analysis
In steady state analysis, any transients from disturbances are assumed to have settled
down and the system state is unchanging. Specifically, system load, including
transmission system losses, are precisely matched with power generation so that the
system frequency is constant, e.g., 60 Hz in North America. Perhaps, the foremost
concern during steady-state is economic operation of the system but reliability is also
important as the system must be operated to avoid outages should disturbances occur.
The primary analysis tool for steady-state operation is the so-called power flow analysis,
where the voltages and power flow through the system is determined. This analysis is
used for both operation and planning studies and throughout the system at both the high
transmission voltages and the lower distribution system voltages.
The power system can be roughly separated into three subcomponents: generation,
transmission and distribution, and load. The transmission and distribution network
consists of power transformers, transmission lines, capacitors, reactors and protection
devices. The vast majority of generation is produced by synchronous generators. Loads
consist of a large number of, and a diverse assortment, of devices, from home appliances
and lighting to heavy industrial equipment to sophisticated electronics. As such,
modeling the aggregate effect is a challenging problem in power system analysis. In the
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Power System Analysis

K. Tomsovic V. Venkatasubramanian

School of Electrical Engineering and Computer Science

Washington State University

Pullman, WA

1. Introduction

The interconnected power system is often referred to as the largest and most complex machine ever built by humankind. This may be hyperbole, but it does emphasize an inherent truth: the complex interdependency of different parts of the system. That is, events in geographically distant parts of the system may interact strongly and in unexpected ways. Power system analysis is concerned with understanding the operation of the system as a whole. Generally, the system is analyzed either under steady-state operating conditions or under dynamic conditions during disturbances.

Electric power is primarily transmitted as a three phase signal. That is, three AC current currents are sent that are out of phase by 120o^ but of equal magnitude. Such balanced currents sum to zero and thus, obviate the need for a return line. If the voltages are balanced as well, the total power transmitted will be constant in time, which is a more efficient use of equipment capacity. For large scale systems analysis, the assumption is usually made that the system is balanced. Each phase can be then analyzed independently greatly simplifying computations. In the following, the implicit assumption is that three phase systems are being used.

2. Steady-State Analysis

In steady state analysis, any transients from disturbances are assumed to have settled down and the system state is unchanging. Specifically, system load, including transmission system losses, are precisely matched with power generation so that the system frequency is constant, e.g., 60 Hz in North America. Perhaps, the foremost concern during steady-state is economic operation of the system but reliability is also important as the system must be operated to avoid outages should disturbances occur. The primary analysis tool for steady-state operation is the so-called power flow analysis, where the voltages and power flow through the system is determined. This analysis is used for both operation and planning studies and throughout the system at both the high transmission voltages and the lower distribution system voltages.

The power system can be roughly separated into three subcomponents: generation, transmission and distribution, and load. The transmission and distribution network consists of power transformers, transmission lines, capacitors, reactors and protection devices. The vast majority of generation is produced by synchronous generators. Loads consist of a large number of, and a diverse assortment, of devices, from home appliances and lighting to heavy industrial equipment to sophisticated electronics. As such, modeling the aggregate effect is a challenging problem in power system analysis. In the

following, the appropriate models for these components in the steady-state are introduced.

2.a Modeling

2.a.1 Transformers

A transformer is a device used to convert voltage levels in an AC circuit. They have numerous uses in power systems. To begin, it is more efficient to transmit power at high voltages and low current than low voltage and high current. Conversely, lower voltages are safer and more economic for end use. Thus, transformers are used to step-up voltages from the generators and then used to step-down the voltage for end use. Another wide use of transformers is for instrumentation so that sensitive equipment can be isolated from the high voltages and currents of the transmission system. Transformers may also be used as means of controlling real power flow by phase-shifting.

Transformers function by the linkage of magnetic flux through a core of ferromagnetic material. Figure 2.1a illustrates a magnetic core with a single winding. When a current I is supplied to the first set of windings, called the primary windings, a magnetic field, H ,

will develop and magnetic flux, φ, will flow in the core. Ampère’s Law relates the

enclosed current to the magnetic field encountered on a closed path. If we assume that H is constant throughout the path then

Hl = NI (2.1)

where l is the path length through the core and N is the number of windings on the core so that NI is the enclosed current by the path referred to as the magnetomotive force (mmf).

I (^) I

Figure 2.1 a) flux flows through core from first winding, b) flux is linked to a second set of windings

The magnetic field is related to the magnetic flux by the properties of the material, specifically, the permeability. If we assume a linear relationship, i.e., neglecting

hysteresis and saturation effects, then the flux density B or the flux φ is

l

NI

B = μ H = μ or l

NI

φ = μ A (2.2)

approximately modeled by a shunt resistor. Saturation is an important non-linear effect that results in additional losses and the creation of odd order harmonics in the current and voltage signals. Since in the steady-state system analysis, only the 60 Hz component of the currents and voltages are considered, saturation effects are typically ignored.

An equivalent circuit for the transformer model described above is shown in Figure 2.2.

Figure 2.2 Transformer circuit model

The main difficulty with this model as it now stands is the ideal transformer component. Carrying this component around in the calculations creates unnecessary complexity. Further from engineering point of view, the voltages and currents in the system are most easily seen relative to their rated values. Thus, most system analysis is done on a normalization called the per unit system. In the per unit system, a system power base is established and the rated voltages at each point in the network are determined. All system variables are then given relative to this value. These base quantities for the currents can be found as

B

B B V

S
I = (2.7)

and for impedances

B

B B

B B (^) S

V
I
V
Z

2 = = (2.8)

This normalization has the great added advantage of reducing the need to represent the ideal transformer in the circuit. One must simply keep track of the nominal base voltage in each part of the network.In this way, the equivalent transformer model is as given in Figure 2.3. Note, phase-shifting and off-nominal transformer ratios result in asymmetric circuits and require some additional manipulation in the per unit framework. Those details are omitted for brevity.

Figure 2.3 Simplified transformer circuit model under per unit system

2.a.2 Transmission line parameters

As mentioned previously, electric power is transmitted in three phases. This accounts for the common site of three lines, or for dual circuits six lines, seen strung between transmission towers. Typically, a high voltage transmission line has several feet of spacing between the three conductors. The conductors themselves are stranded wire for improved mechanical properties, as well as electrical properties. If the currents will be large, several conductors may be strung per phase. This improves cooling compared to using one large conductor. This geometry is important as it impacts the electrical properties of the line.

To begin, as current flows in each conductor, a magnetic field develops. Adjacent lines then may induce voltages in nearby conductors through mutual induction (as we saw for transformers only now the coupling is not as tight). This interaction largely determines the inductance seen by the respective phase currents. To understand this phenomenon, consider a single line of radius r and infinite length with some current flow, I , as sketched in Figure 2.4. Similar to the transformer development, we will apply Ampère’s Law to characterize the magnetic field. The magnetic field at some distance x from the line can be found by assuming that the field is constant at all points equal distance from the line.

Then the closed path is a circle with circumference 2 π x , which gives

I

x Figure 2.4 Infinite transmission line

2 π xH = I or x

I
H

If x is less than the line radius, the closed path will not link all of the current. Assuming an equal distribution of current throughout the wire, then

D
D D

Figure 2.5 End view of equally spaced phase conductors

In practice, the phase conductors may not be equally spaced. This results in unbalanced conditions due to the imbalance in mutual inductance. High voltages transmission lines with such a layout can be transposed so that on average the distance between phases is equal canceling out the imbalance. The equivalent distance of separation between phases can then be found as the geometric mean of this spacing. Similarly, if several conductors are used per phase an equivalent conductor radius can be found as the geometric mean.

Transmission lines also exhibit capacitive effects. That is, whenever a voltage is applied to a pair of conductors separated by a non-conducting medium, charge accumulates leading to capacitance. Similar to the previous development for inductance, we can determine the capacitance based on Gauss’s Law. For a point P at a distance x from a conductor with charge q , the electric flux density D is

x

q D

Assuming a homogeneous medium, the electric field density E is related to D by the

permitivity ε of the dielectric, which in this case will be assumed to be that of free space.

x

q E

Integrating E over some path (a radial path is chosen for simplicity) yields the voltage difference between the two end points.

2

1 0 0

12 2 2 ln

2

1^ R

q R dx x

q V

R

R πε^ πε

= (^) ∫ = (2.18)

Now consider a three phase transmission line again with each line spaced equally by the distance D. Superposition holds so that the voltage arising from each of the charges can be added. To find the voltage from phase to ground arising from each of the conductors, assume first, a balanced system with q (^) a + qb + qc = 0 , and two, a neutral located at some

far distance R from phase a so

r

D

q D

R

q D

R

q r

R

V (^) an qa b c a ln 2

ln ln ln 2

Now recalling that capacitance is simply the ratio of charge to voltage, the capacitance from phase a to ground per unit length of line will be

r

V^ D

q C an

a an ln

Again if the conductors are not evenly spaced, transposition results in an equivalent geometric mean distance and using bundled conductors per phase can also be accommodated by using a geometric mean.

Finally, conductors have finite resistances that depend upon the temperature, the frequency of the current, the conductor material, and so on. For most systems analysis problems, these can be based on values provided by manufacturers or compiled into tables for commonly used conductors and typical ambient conditions.

2.a.3 Transmission line circuit models

Transmission lines may be classified based on their total length. If the line is around 50 miles or less, a so-called “short line,” capacitance can be neglected and the series inductance and resistance can be modeled as lumped parameters. Figure 2.6 depicts the short line model per phase. The series resistance and inductance is simply the per unit distance parameters times the line length so at 60 Hz for line length l the line impedance is

Z = Rl + j Ll = R + jX

Figure 2.6 Short line model

For lines longer than 50 miles, up to around 150 miles, capacitance can no longer be neglected. A reasonable circuit model is to simply split the total capacitance evenly with each half represented as a shunt capacitor at each end of the line. This is depicted as the π-circuit model in Figure 2.7. Again, the total capacitance is simply the per unit distance capacitance times the line length so

Y = j Canl = jB

where the prime indicates the modified circuit values arising from a long line.

2.a.4 Generators

Three phase synchronous generators produce the overwhelming majority of electricity in the modern power systems. Synchronous machines operate by applying a DC excitation to a rotor that, when mechanically rotated, induces a voltage in the armature windings due to changing flux linkage. The per phase flux for a balanced connection can be written as

λ = Kf If sin θ m (2.30)

where If is the field current, θ m is the angle of the rotor relative to the armature and Kf is a

constant that depends on the number of windings and the physical properties of the machine. The machine may have several poles so that the armature will “see” multiple rotations for each turn of the rotor. So for example, a four pole machine appears

electrically to be rotating twice as fast as two pole machine. For a machine rotating at ω m

rad/s with p poles, the electric frequency is

p

ω s = ω m (2.31)

with ω s the desired synchronous frequency. If the machine is rotated at a constant speed

Faraday’s Law tells us the induced voltage will be

ω sin (ω θ 0 )

= = K I t + dt

d V (^) f f s s (2.32)

If a load is applied to the armature windings, then current will flow and the armature flux will link with the field. This effectively puts a mechanical load on the rotor and so power input must be matched to this load in order to maintain the desired constant frequency. Some of the armature flux “leaks” and does not link with the field. In addition, there are winding resistive losses but those are commonly neglected. The circuit model shown in Figure 2.8 is a good representation for the synchronous generator in the steady state. Note that most generators are operated at some fixed terminal voltage with a constant power output. Thus, for steady state studies the generator is often referred to as a PV bus since the terminal node has fixed power P and voltage V.

Figure 2.8 Simple synchronous generator model

2.a.5 Loads

Modeling power system loads remains a difficult problem. The large number of different devices that could be connected to the network at any given time renders precise modeling intractable. Broadly speaking, loads may vary with voltage and frequency. In the steady-state frequency is constant so one only needs to be concerned with the voltage. For most steady-state analysis, a fixed, i.e., constant over an allowable voltage range, power consumption model can be used. Still, some analysis requires consideration of voltage effects to be useful and then the traditional exponential model can be used to represent real power consumption P and reactive power consumption Q

P PV^ a = 0 (2.33)

Q QV^ b = 0 (2.34)

where the voltage V is normalized to some rated voltage. The exponents a and b can be 0, 1 or 2 where they could represent constant power, current or impedance loads, respectively. Alternatively, they can represent composite loads with a generally ranging between 0.5 and 1.8 and b ranging between 1.5 and 6.

2.b Power flow analysis

Power flow equations represent the fundamental balancing of power as it flows from the generators to the loads through the transmission network. Both real and reactive power flows play equally important roles in determining the power flow properties of the system. Power flow studies are amongst the most significant computational studies carried out in power system planning and operations in the industry. Power flow equations allow us to compute the bus voltage magnitudes and their phase angles, as well as the transmission line current magnitudes. In actual system operation, both the voltage and current magnitudes need to be maintained within strict tolerances, for meeting consumer power quality requirements and for preventing overheating of the transmission lines, respectively. The difficulty in computing the power-flow solutions arises from the fact that the equations are inherently nonlinear arising from the balancing of power

jx

∠ − V ∠ δ

I (2.35)

Next, the complex power S delivered to the load bus can be calculated as

( ) ( ) x

V

x

V δ π 2 2 ∠ π 2

S = VI* = (2.36)

Therefore, we get the real and reactive power balance equations to be

x

V V
Q

x

V
P

δ cos δ

sin −^2 +

After setting Q =0 in (2.37), we can simplify the two equations in (2.37) into a quadratic equation in V^2 as follows

V^4 − V^2 + x^2 P^2 = 0 (2.38)

Therefore, given any real power load P , the corresponding power-flow solution for the bus voltage V can be solved from (2.38). We note that for nominal load values, there are two solutions for the bus voltage V and they are the positive roots of V^2 in the next equation

2 1 1 4 x^2 P^2 V

Equation (2.39) implies that there exist two power-flow solutions for load values P < Pmax where Pmax = 1/(2 x ), and there exist no power-flow solutions for P > Pmax. A qualitative plot of the power-flow solutions for the bus voltage V in terms of different real power loads P is shown in Figure 2.10.

From the plot and from the analysis thus far, we can make the following observations:

  1. The dependence of the bus voltage V on the load P is very much nonlinear. It has been possible for us to compute the power-flow solutions analytically for this simple system. In the large power system with hundreds of generators delivering power to thousands of loads, we have to solve for thousands of bus voltages and their phase angles from large coupled sets of nonlinear power-flow equations, and the computation is a nontrivial task.
  2. Multiple power-flow solutions can exist for a specified loading condition. In Figure 2.10, there exist two solutions for any load P < Pmax. Among the two solutions, the solution on the upper locus with voltage V near 1 pu is considered the nominal solution. For the solutions on the lower locus, the bus voltage V may be unacceptably low for normal operation. The lower voltage solution also requires higher line current in order to deliver the specified load P , and the line current values can become unacceptably high. In general, for any specified

loading condition, we would like to locate the power-flow solution that has the most acceptable values of voltages and currents among the multiple power-flow solutions. In this example of a single generator delivering power to a single load, there exist two power-flow solutions. In a large power system, there may exist a very large number of possible power-flow solutions.

  1. Once the bus voltage V has been computed from (2.39), the bus voltage phase

angle δ can be computed from (2.37). Then, the line current phasor I can be

solved from (2.35). Specifically, we would like to ensure that the magnitude of the line current I stays below the thermal limit of the transmission line for preventing potential damage to the expensive transmission line.

  1. Power-flow solutions may fail to exist at high loading conditions, for instance, when P > Pmax in Figure 2.10. The loading value Pmax beyond which power-flow solutions do not exist is called the static limit in the power literature. Since the power-flow solutions denote the steady state operating conditions in our formulation, lack of power-flow solutions implies that it is not possible to transfer power from the generator to the load in a steady state fashion, and the dynamic interactions of the generators and the loads become significant. Operating the power system at loading conditions beyond the static limit may lead to catastrophic failure of the system.

V

P

Pmax

Figure 2.10 Qualitative plot of the power-flow solutions

2.b.2 Power-flow problem formulation

In this subsection, we will construct the power-flow equations in a structured manner using the admittance matrix Ybus representation of the transmission network. The admittance matrix Ybus is assumed to be known for the system under consideration. Let us first look at the complex power-balance at any bus, say bus i , in the network.

equations (2.43) and (2.44), and our aim in the rest of this section is to develop algorithms for solving this problem.

Let us consider a purely load bus first, that is, with PGi = QGi = 0. In this case, the loads PLi and QLi are assumed to be known either from measurements or from load estimates, and the bus voltage variables Vi , and δi are the unknown variables. Purely load buses with no generation support are called “PQ buses” in the power-flow studies since both real- power injection Pi and reactive power Qi have been specified at these buses.

Typically, every generator in the system consists of two types of internal controls, one for maintaining the real power output of the generator, and the other for regulating the bus voltage magnitude. In power-flow studies, we usually assume that both these control mechanisms are operating perfectly and so the real power output PGi and Vi are maintained at their specified values. Again, the load variables PLi and QLi are also assumed to be known. This leaves the generator reactive output QGi and the voltage phase angle δi as the two unknown variables for the bus. In terms of injections, the real power injection Pi and the bus voltage Vi are then the specified variables, and thus the generator buses are normally denoted “PV buses” in power-flow studies.

In reality, the generator voltage control for keeping the bus voltage magnitude at a specified value becomes inactive when the control is pushed to the extremes, say when the reactive output of the generator becomes either too high or too low. This voltage control limitation of the generator can be represented in the power-flow studies by keeping track of the reactive output QGi. When the reactive generation QGi becomes larger than a prespecified maximum value say QGi,max or goes lower than a prespecified minimum value QGi,min , the reactive output is assumed to be fixed at the limiting value QGi,max or QGi,min respectively, and the voltage control is disabled in the formulation. That is, the reactive power QGi becomes a known variable, either at QGi,max or QGi,min and the voltage Vi then becomes the unknown variable for bus i. In power-flow terminology, we say that the generator at bus i has reacted its reactive limits and hence, bus i has changed from a PV bus to a PQ bus. Owing to space limitations, we will not discuss generator reactive limits in any more detail in this section.

In addition to PQ buses and PV buses, we also need to introduce the notion of a “slack bus” in the power-flow formulation. Note that power conservation demands that the real power generated from all the generators in the network must equal the sum of the total real power loads and the line losses on the transmission network.

∑ =^ ∑ + ∑∑ i j

losses i

L i

PG (^) i Pi P ij (2.45)

The line losses associated with any transmission line in turn depends on the line resistance and the line current magnitude. As stated earlier, one of the main objectives of the power-flow studies to compute the line currents, and as such, the line current values are not known at the beginning of a power-flow computation. Therefore, we do not know the actual values for the line losses in the transmission network. Looking at (2.45), we

need to assume that at least of one of the variables PGi or PLi should be a free variable for satisfying the real power conservation. Traditionally, we assume that one of the generations is a “slack” variable and such a generator bus is denoted the “slack bus”. At the slack bus, we specify both the voltage Vi and the angle δi. The power injections Pi and Qi are the unknown variables. Again, by tradition, we set the voltage at slack bus to be the rated voltage or at 1 pu, and the phase angle to be at zero.

Like in standard text books, slack bus is defined in this section to be the first bus in the network with V 1 = 1 and the angle δ 1 =0. Assuming the number of generators to be NG , the buses 2 through NG +1 are set to be the PV buses. The remaining buses NG +2 through N are then the PQ buses.

2.b.3. Gauss-Seidel algorithm:

Let us consider a set of simultaneous linear equations of the form A x = b , where A is an n X n matrix, and, x and b are n X 1 vectors. Clearly, there exists a unique solution to the problem when the matrix A is invertible and the solution is given by x = A-^1 b. When the matrix size is very large, it may not be possible to compute the inverse of the matrix A for finding the solution and there exist other numerical techniques. Gauss-Seidel algorithm is one such classical algorithm that tries to arrive at the solution x = A-^1 b iteratively by starting from an approximate initial condition say x^0. The iteration for the solution xk+ from the previous iterate xk^ proceeds as follows.

b a x a x i n a

x j i ji

k ij j

k i ij j ii

k i for^1 ,^2 ,...,

= −∑ −∑ < >

(2.46)

Here aij denotes the ( i,j )-the entry of the matrix A as usual. It can be shown that the iterative solution xk^ converges to the exact solution A-^1 b for any initial condition x^0 , provided the matrix A satisfies certain “diagonal dominance” properties. The details are limited here to save space and they can be seen in standard numerical analysis text-books.

In the previous section, we have formulated the power-flow problem as a set of simultaneous nonlinear equations (2.42) and as such, it is not obvious how the Gauss- Seidel algorithm can be applied for solving these equations. The trick here is to visualize the power-balance equations to be arising from the network admittance equations Ybus Vbus = Ibus. The matrix Ybus takes over the role of the matrix A in the linear equations. We will be solving for the bus voltage vector Vbus. The current injections Ibus are not known per se. The current injections are in fact dependent on the bus voltages. As we see next, they can also be computed iteratively from the power injections Si by using the relationship Ii = Si *^ / V i *. For a PQ bus, the injection Si is a specified variable and hence is known. For PV buses, only the real power injection Pi is known while the reactive injection Qi is evaluated first using the latest estimate of bus voltages Vbus.

An outline of the Gauss-Seidel algorithm for solving the power-flow equations (2.42) is presented next. Let us start with an initial condition for the bus voltages Vbus^0 and we would like to compute the iterate Vbus k+1^ from the previous iterate Vbus k. Recall that bus 1

nonlinear equations into linear equations at each step, and by solving the linearized equations exactly.

Suppose we want to solve the nonlinear equations F ( x ) = 0 where x is a n X 1 vector, and

F : ℜ n^ → ℜ n is a smooth nonlinear function. We have been given an initial condition x^0. Then, for computing the estimate xk +1^ from xk , we first linearize the functions F ( x ) at xk^ as follows

( k )

x

k (^) x x x

F

F x F x k

The solution to the linearized equations (2.b.1.17) is defined as the iteration estimate xk+.

( k

x

k k Fx x

F

x x k

1 1

= − )^ (2.52)

Note that the linearization (2.51) will be a good approximation if the estimate xk^ is close to the true solution say x*^ since F ( x )= 0. The NR algorithm stated in (2.52) can be proved to converge to the true solution _x_ , when the initial condition x^0 is sufficiently close to x*. On the other hand, for initial conditions away from x* , the approximation (2.51) becomes poorly justified, and the iterations can quickly diverge away from x*. When the iterations converge, owing to the linearized nature of the algorithm, the norm of the error decreases to zero in a “quadratic” fashion. Roughly speaking, the ratios of the error norm from one iteration to the square of the error norm in the previous iteration tends to be a constant. An example would be that the error norms decrease from 0.1 in one iteration, to 0.01 in the next iteration, to 0.0001 in the following iteration. Therefore, given good initial conditions, the NR algorithm can typically get to an accurate solution estimate within a few iterations.

Let us apply the NR algorithm for solving the power-flow equations (2.43) and (2.44). We will solve for the unknown variables among the bus voltage magnitudes Vi and angles δi first. That is, we define the vector x as consisting of all the PV and PQ bus angles, and all the PQ bus voltages. PV bus voltages are known and hence, they are not included in x.

T

x = δ 2 , ...,δ N G + 1 ,δ NG + 2 ,...,δ N , VNG + 2 ,..., VN (2.53)

The corresponding power-flow equations are as follows

N j N j j

N N j N j

N j N j j

N N j N j

N j Nj j

N N j N j

N j N j j

N N j N j

N j N j j

N N j N j

j j j

j j

N N

N N

N N

N N

N N

Q Y V V
Q Y V V
P Y V V
P Y V V
P Y V V
P Y VV

Q q x

Q q x

P p x

P p x

P p x

P p x

F x

G G G G

G G G G

G G G G

G G

G G

G G

, ,

2 2 , 2 2 2 ,

, ,

2 2 , 2 2 2 ,

1 1 , 1 1 1 ,

2 2 , 2 2 2 ,

2 2

2 2

1 1

2 2

sin

sin

cos

cos

cos

cos

The entries of the F function in (2.54) are the differences between the specified power injections and the computed power injections from the current power-flow solutions, and these are usually denoted as the real and reactive power mismatches at the different buses. In the power-flow problem, then we want to find a solution that makes the power mismatches in (2.54) to be zero.

Suppose an initial condition x^0 has been specified. Then, the NR algorithm for solving the power-flow equations (2.54) proceeds iteratively as follows:

Newton-Raphson iterations:

(1) Compute the power mismatches F ( xk ) for step k from (2.54). If the mismatches are within desired tolerance values, the iterations stop.

(2) Compute the power-flow Jacobian xk

k x

F
J

=. Owing to the nice structure of the

equations in (2.54), explicit formulas can be derived for the entries of the Jacobian matrix, and the Jacobian for step k can be evaluated by substituting the current values of xk^ into these formulas. (3) Compute the correction factors ∆ xk^ from (2.52) by solving a set of simultaneous linear equations

J k^ ∆ xk = − F ( xk ) (2.55)

The Jacobian matrix Jk^ is extremely sparse even for very large power systems, which facilitates the solution of ∆ xk^ in (2.55). (4) Evaluate xk+1^ from xk^ by adding the correction factors ∆ xk.

x k^ +^1 = xk +∆ x^ k (2.56)