LOAD-FLOW ANALYSIS
IN POWER SYSTEMS
Badrul H. Chowdhury
Professor
Electrical & Computer Engineering Department University of Missouri-Rolla
Introdu . . . . 11.1 Nomenclature . . . . 11.2 Developing Power Flow Equations . . . . 11.2 Power-Flow Solution . . . . 11.6 Example of N-R Solution . . . 11.12 Concluding Remarks . . . . 11.15 Bibliography . . . . 11.16
INTRODUCTION
The load-flow problem models the nonlinear relationships among bus power injections, power demands, and bus voltages and angles, with the network constants providing the circuit parameters. It is the heart of most system-planning studies and also the starting point for transient and dynamic stability studies. This section provides a formulation of the load-flow problem and its associated solution strategies. An understanding of the fundamentals of three-phase systems is assumed, including per-unit calculations, complex power relationships, and circuit-analysis techniques.
There are two popular numerical methods for solving the power-flow equations. These are the Gauss-Seidel (G-S) and the Newton-Raphson (N-R) Methods (Grainger and Stevenson, 1994; Elgerd, 1982; Glover and Sharma, 1994). The N-R method is superior to the G-S method because it exhibits a faster convergence characteristic. However, the N-R method suffers from the disadvantage that a “flat start” is not always possible since the solution at the beginning can oscillate without converging toward the solution. In or-der to avoid this problem, the load-flow solution is often started with a G-S algorithm fol-lowed by the N-R algorithm after a few iterations.
There is also an approximate but faster method for the load-flow solution. It is a varia-tion of the N-R method, called the fast-decoupled method, which was introduced by Stott and Alsac (1974). We will not be covering this method in this section.
NOMENCLATURE
SD complex power demand
SG complex power generation
S complex bus power PD real power demand, MW
PG real power generation
P real bus power
QD reactive power demand in
MVAR
QG reactive power generation
Q reactive bus power V bus voltage magnitude
bus voltage angle ˆV complex voltage B shunt susceptance yp shunt admittance
DEVELOPING POWER-FLOW EQUATIONS
A two-bus example, shown in Fig. 11.1, is used to simplify the development of the power-flow equations. The system consists of two busses connected by a transmission line. One can observe that there are six electrical quantities associated with each bus: PD,
PG, QD, QG,V, and . This is the most general case, in which each bus is shown to have
both generation and demand. In reality, not all busses will have power generation. The impedance diagram of the two-bus system is shown in Fig. 11.2. The transmission line is represented by a -model and the synchronous generator is represented by a source behind a synchronous reactance. The loads are assumed to be constant impedance for the sake of representing them on the impedance diagram. Typically, the load is represented by a constant power device, as shown in subsequent figures.
Figure 11.3 is the same as Fig. 11.2 but with the generation and demand bundled together to represent “bus power,” which represents bus power injections. Bus power is defined as
S1 SG1 SD1 (PG1 PD1) j(QG1 QD1) (1)
and
S2 SG2 SD2 (PG2 PD2) j(QG2 QD2) (2)
Also, injected current at bus 1 is
(3) Iˆ1IˆG1IˆD1 ys series admittance R series resistance X series reactance Zs series impedance XG synchronous reactance
Yii driving point admittance at bus i
Yij transfer admittance between
busses i and j Yij magnitude of Yij ij angle of Yij E synchronous machine-gener-ated voltage complex current Ybus bus admittance matrix
FIGURE 11.1 A two-bus power system.
FIGURE 11.2 Impedance diagram for the two-bus power system.
FIGURE 11.3 Bus powers with transmission line -model for the two-bus system.
and injected current at bus 2 is
(4) All quantities are assumed to be per unit. Then, since
S1 P1 jQ1 (P1 jQ1) (5) and, since S2Vˆ2Iˆ2*QP2 jQ2Vˆ2Iˆ*2 Q(P2 jQ2)Vˆ2*Iˆ2 (6) Iˆ1 Vˆ1* Q Iˆ1* Vˆ1 Q Iˆ1* Vˆ1 Iˆ2IˆG2IˆD2
Let us define current flows in the circuit as shown in Fig. 11.4. Therefore, at bus 1
(7) (8) where Y11 sum of admittances connected at bus 1 yp ys (9)
Y12 negative of the admittance between busses 1 and 2 ys (10)
Similarly, at bus 2
(11) (12)
Y22 sum of all admittances connected at bus 2 y p ys (13)
Y21 negative of the admittance between busses 2 and 1 y s Y12 (14)
Hence, for the two-bus power system, the current injections are
(15)
In matrix notation,
Ibus YbusVbus (16)
The two-bus system can easily be extended to a larger system. Consider an n-bus sys-tem. Figure 11.5a shows the connections from bus 1 of this system to all the other busses. Figure 11.5b shows the transmission line models. Equations (5) through (16) that were
Vˆ 1 ˆ V2 Y11 Y12 Y21 Y22 I1 I2 Iˆ2Y21Vˆ1Y22Vˆ2 Iˆ2 (ys)Vˆ1 (ypys)Vˆ2 Vˆ2yp (Vˆ2Vˆ1)ys Iˆ2Iˆ2 Iˆ2 Iˆ1Y11Vˆ1Y12Vˆ2 Iˆ1 (ypys)Vˆ1 (ys)Vˆ2 Vˆ1yp (Vˆ1Vˆ2)ys Iˆ1Iˆ1 Iˆ1derived for the two-bus system can now be extended to represent the n-bus system. This is shown next. 1 yp12 yp13 yp1n ( )ys12 ( )ys13 ( )ys1n (yp12 yp13 yp1n ys12 ys13 ys1n) ys12 ys13 ys1n (17) Y11 Y12 Y13 Y1n (18) where Y11 (yp12 yp13 yp1n ys12 ys13 ys1n) (19)
sum of all admittances connected to bus 1
Y12 yS12; Y13 yS13; Y1n ys1n (20)
(21) n j1 YijVˆj Iˆ1 Vˆn Vˆ3 Vˆ2 Vˆ1 Iˆ1 Vˆn Vˆ3 Vˆ2 Vˆn Vˆ2 Vˆ1 Vˆ3 Vˆ1 Vˆ2 Vˆ1 Vˆ1 Vˆ1 Vˆ1 IˆFIGURE 11.5a Extending the analysis to an n-bus system.
Also, extending the power Eq. (5) to an n-bus system,
P1 jQ1 I1 (22)
Equation (22) can be written for any generic bus i:
Pi jQi i 1, 2, . . . , n (23)
Equation (23) represents the nonlinear power-flow equations. Equation (15) can also be rewritten for an n-bus system:
(24)
or
Ibus YbusVbus (25)
where
bus admittance matrix (26)
POWER-FLOW SOLUTION
Let us take a generic bus as shown in Fig. 11.6. As mentioned earlier, each bus has six quantities or variables associated with it. They are V ,, PG, QG, PD, and QD. Assuming
that there are n busses in the system, there would be a total of 6n variables. Ybus
Y11 Y21 Yn1 Y12 Y22 Yn2 … … … Y1n Y2n Ynn Iˆ1 Iˆ2 Iˆn Y11 Y21 Yn1 Y12 Y22 Yn2 … … … Y1n Y2n Ynn ˆ V1 ˆ V2 ˆ Vn … n j1 YijVˆj Vˆi* n j1 Y1jVˆj Vˆ1* Vˆ1*The power-flow Eq. (23) can be resolved into the real and reactive parts as follows:
Pi Real i 1, 2, . . . , n (27)
Qi Imag i 1, 2, . . . , n (28)
Thus, there are 2n equations and 6n variables for the n-bus system. Since there cannot be a solution in such case, 4n variables have to be prespecified. Based on parameter specifi-cations, we can now classify the busses as shown in Table 11.1.
We will now describe the methods used in solving the power-flow equations.
Gauss-Seidel (G-S) Algorithm for Power-Flow Solution Note that the power-flow equations are
(23) (29) (30) (31) (32) Also, from (29), (33) and Pi Re
Vˆi*YiiVˆi n j1, j i Vˆi*YijVˆj Q Vˆi PijQi Vˆi* n j1, j i YijVˆj Yii Q YiiVˆi PijQi Vˆi* n j1, j i YijVˆj Q Vˆi*YiiVˆi (PijQi) n j1, j i Vˆi*YijVˆj Vˆi*YiiVˆi n j1, j i Vˆi*YijVˆj PijQiVˆi* n j1 YijVˆj i 1, 2, . . . , n Vˆi* n j1 YijVˆj Vˆi* n j1 YijVˆjTABLE 11.1 Bus Classifications
Bus classification Prespecified variables Unknown variables Slack or swing V ,, PD, QD PG, QG
Voltage-controlled V , PG, PD, QD , QG
(34)
where Yijare elements of the Ybus.
The G-S Algorithm
Step 0. Formulate and Assemble Ybusin Per Unit
Step 1. Assign Initial Guesses to Unknown Voltage Magnitudes and Angles
V 1.0, 0
Step 2a. For Load Buses, Find form Eq. (32)
where k iteration no. For voltage-controlled busses, find using (34) and (32) to-gether. That is, find Qifirst.
Then
However,Vi is specified for voltage-controlled busses. So,
In using Eqs. (32) and (34), one must remember to use the most recently calculated values of bus voltages in each iteration. So, for example, if there are five busses in the system being studied, and one has determined new values of bus voltages at busses 1 – 3, then during the determination of bus voltage at bus 4, one should use these newly calcu-lated values of bus voltages at 1, 2, and 3; busses 4 and 5 will have the values from the previous iteration.
Step 2b. For Faster Convergence, Apply Acceleration Factor to Load Buses
(35)
where acceleration factor.
Step 3. Check Convergence
Re Re [ ] (36)
That is, the absolute value of the difference of the real part of the voltage between successive iterations should be less than a tolerance value . Typically, 104, and also,
Imag Imag (37)
That is, the absolute value of the difference of the imaginary value of the voltage should be less than a tolerance value .
[Vˆi(k)] [Vˆi(k1)] Vˆi( k) [Vˆi( k1)] (V(k) i V( k) i,acc) V(k) i,acc V(k1) i,acc Vi ( k1) V i , spec i,calc (k1) Vi(k1)
(PijQi)/Vˆi*(k) n j1, j i YijVˆj* Yii Qi(k1) ImagVˆi*(k)Vi(k)Yii n j1, j i YijVj(k)Vˆi Vi (k1)(P i jQi) Vi*(k) n j1, ji YijVˆj(k) Yii Vˆi Qi ImagVˆi*YiiVˆi n j1, j i Vˆi*YijVˆj
If the difference is greater than tolerance, return to Step 3. If the difference is less than tolerance, the solution has converged; go to Step 4.
Step 4. Find Slack Bus Power PGand QGfrom Eqs. (27) and (28) Step 5. Find All Line Flows as Described in the Next Section
Computing Line Flows. As the last step in any power-flow solution, one has to find the line flows. This is illustrated by the two-bus system shown in Fig. 11.7. Line current, ij,
at bus i is defined positive in the direction i : j.
(38) Let Sij, Sjibe line powers defined positive into the line at bus i and j, respectively.
(39) (40) The power loss in line (i j) is the algebraic sum of the power flows determined from (39) and (40).
SLij Sij Sji (41)
Newton-Raphson (N-R) Method for Power-Flow Solution
The Newton-Raphson method enables us to replace the nonlinear set of power-flow equa-tions of (23) with a linear set. We will show this after the basis for the method is explained. The Taylor series expansion of a function f(x) of a single variable, x, around the point (x a) is given by
f(x) f(a) (xa) (42)
where value of the derivative evaluated at x a. The series converges if lim n 0.
n: f x
a f xa (xa)n 2! 2f x2 a (xa)n n! nf xn n SjiPjijQjiVˆjIˆji*VˆjVˆj*Vˆi*ys* Vj 2ypi* SijPijjQijVˆiIˆij*VˆiVˆi*Vˆj*
ys* Vi 2ypi* IˆijIˆsIˆpi (VˆiVˆj)ysVˆi ypi Iˆ FIGURE 11.7 A two-bus system illustrating line-flow
If (x a) 1 then we can neglect the higher-order terms and write (42) as
f(x) f(a) (x a) (43)
For a function of n variables, one can expand around the point: (x1 a1), (x2 a2),
(xn an) with (xk ak) 1 and k 1, 2, . . . , n. Then, Eq. (42) becomes
f(x1, x2, . . . , xn) f(a1, a2, . . . , an) (x1 a1) (x2 a2)
… (xn an) (44)
Let us consider a set of nonlinear equations, each a function of n variables: f1(x1, x2, . . . , xn) y1 f2(x1, x2, . . . , xn) y2 (45) fn(x1, x2, . . . , xn) yn or fk(x1, x2, . . . , xn) yk k 1, 2, . . . , n
Assume initial values and some correction,xk, which when added to yield .
When are close to the solution, xk, the are small.
Using the approximate Taylor’s series, we have
fk(x1, x2, . . . , xn) fk x1 x2
xn yk k 1, 2, . . . , n (46)
or, in matrix form,
(47)
or
[U](0) [J](0)[X](0) (48)
where [J] is the Jacobian matrix.
[X] ([J](0))1[U](0) (49)
x1 x2 xn y1f1(x1 (0), x2 (0), . . . , xn(0)) y2f2(x1 (0), x2 (0), . . . , xn(0)) ynfn(x1 (0), x2 (0), . . . , xn(0)) fk xnx(0) n fk x2 x(0) 2 fk x1x(0) 1 (x1(0), x2(0), . . . , xn (0)) xks xk(0) xk(1) xk(0) xk(0) f xn an f x2 a2 f x1 a1 f xa x1(0) x2(0) … xn(0) x1(0) x2(0) … xn(0) x1(0) x2(0) … xn(0) fn xn fn x2 fn x1 f2 xn f2 x1 f2 x1 f1 xn f1 x2 f1 x1To continue iteration, find [X](1)from
[X](1) [X](0) [X](0) (50)
Generally,
[X](k1) [X](k) [X](k) (51)
where k iteration number.
The Newton-Raphson Method Applied to Power-Flow Equations The N-R method is typically applied on the real form of the power-flow equations:
Pi Vi Vk yik cos (kiik) fip (52)
Qi Vi Vk yik sin (kiik) fiq
i 1, . . . , n
(53)
Assume, temporarily, that all busses, except bus 1, are of the “load” type. Thus, the unknown parameters consist of the (n 1) voltage phasors, , . . . , . In terms of real variables, these are:
Angles 2,3, . . . ,n (n 1) variables
Magnitudes V2, V3, . . . , Vn (n 1) variables
Rewriting (47) for the power-flow equations,
Vˆn Vˆ2
n k1 n k1 fnq 3 (0) fnq 2 (0) f3 q 3 (0) f3q 2 (0) f2 q 3 (0) f2q 2 (0) fnp 3 (0) fnp 2 (0) f3 p 3 (0) f3 p 2 (0) f2 p 3 (0) f2 p 2 (0) … … … … … … … … … … … … fnq V2 (0) fnq n (0) f3q V2 (0) f3q n (0) f2q V2 (0) f2q n (0) fnp V2 (0) fnp n (0) f3 p V2 (0) f3 p n (0) f2 p V2 (0) f2 p n (0) fnq Vn (0) f3q Vn (0) f2q Vn (0) fnp Vn (0) f3p Vn (0) f2p Vn (0) ∆P2(0) ∆P3(0) ∆Pn(0) ∆Q2(0) ∆Q(0)3 ∆Qn(0) ∆2(0) ∆3(0) ∆n(0) ∆V2(0) ∆V3(0) ∆Vn(0) (54)Before proceeding any further, we need to account for voltage-controlled busses. For every voltage-controlled bus in the system, delete the corresponding row and column from the Jacobian matrix. This is done because the mismatch element for a voltage-controlled bus is unknown.
Writing Eq. (54) in matrix form,
(55) where U(0) vector of power mismatches at initial guesses
J(0) the Jacobian matrix evalutated at the initial guesses
X(0) the error vector at the zeroth iteration
The N-R Algorithm
Step 0. Formulate and Assemble Ybusin Per Unit
Step 1. Assign Initial Guesses to Unknown Voltage Magnitudes and Angles for a Flat Start
V 1.0, 0
Step 2. Determine the Mismatch Vector U for Iteration k Step 3. Determine the Jacobian Matrix J for Iteration k Step 4. Determine Error Vector X from Eq. (55)
Set X at iteration (k 1): X(k1) X(k) X(k). Check if the power mismatches are
within tolerance. If so, go to Step 5. Otherwise, go back to Step 2.
Step 5. Find Slack Bus Power PGand QG from Eqs. (27) and (28)
Step 6. Compute Line Flows Using Eqs. (39) and (40) and the Total Line Losses from Eq. (41)
EXAMPLE OF N-R SOLUTION
Consider the three-bus system shown in Fig. 11.8. Known quantities are also shown. Given:Vˆ11.00p.u.,V2 1.0 p.u., P2 0.6 p.u., P3 0.8 p.u., Q3 0.6.
U(0)J(0) X(0)
Step 0. Ybus Step 1. Step 2. P2 f2p y21 V2 V1 cos (1221) y22 V2 V2 cos (2222) y23 V2 V3 cos (3223) (56) Q2 f2q y21 V2 V1 sin (1221) y22 V2 V2 sin (2222) y23 V2 V3 sin (3223) (57) P3 f3p y31 V3 V1 cos (1331) y32 V3 V2 cos (2332) y33 V32cos (3333) (58) Q3 f3q y31 V3 V1 sin (1331) y32 V3 V2 sin (2332) y33 V32sin23 (59)
The specified bus powers are: P2S 0.6, P3S 0.8, Q3S 0.6. The calculated bus
powers at this iteration are: 2.1.1 cos (0 0 90) 6.1.1 cos (90°) 4.1.1 cos (0 0 90) 0, 5.1.1 cos 90 4.1.1 cos 90 9.12cos (90) 0,
(5.1.1 sin 90 4.1.1 sin 90° 9.12 sin (90°)) 0. Therefore, the mismatches are
P2 0.6; P3 0.8; Q3 0.6. Step 3. y21 V1 V2 sin (221) y23 V2 V3 sin (2323) (60) y23 V2 V3 sin (3223) (61) f2q 3 f2p 2 Q0 3 P03 P02 X(0)
(0) 2 (0) 3 V2 0 V 3 0 0 0 1.0 1.0 j7 j2 j5 j2 j6 j4 j5 j4 j9 f3q V2 f3p V2 , f2p V2 , f2q 3 f2q V3 , f2q V2 , f2q 2,No need to evaluate these since bus 2 is a voltage-controlled bus.
y23 V2 cos (3223) (62)
Therefore, at the initial guesses,
6 4 0
y32 V3 V2 sin (2332) 4 (63)
[y31 V3 V1 sin (331) y32 V3 V2 sin (3232)] 9 (64)
[y31 V1 cos (1331) y32 V2 cos (2332)
2y33 V3 cos 33] 0 (65)
[y32 V3 V2 cos (2332)] 0 (66)
[y31 V3 V1 cos (3131)
y32 V3 V2 cos (3232)] 0 (67)
[y31 V1 sin (1331) y32 V2 sin (2332)
2y33 V3 sin 33] 9 (68)
After the row and column corresponding to bus 2 are eliminated:
Step 4.
To solve the preceding equations, one can resort to inversion of the Jacobian matrix. However, computationally, it is more efficient to apply a numerical technique such as the
0.6
0.8 0.6 6 4 0 4 9 0 0 0 0 2 3 V3 P2(0) P3(0) Q2(0) Q3(0) 6 4 0 0 4 9 0 0 0 0 10 4 0 0 4 9 2 3 V2 V3 f3q V3 0 f3q d2 0 f3q 2 0 f3p V3 0 f3p 3 0 f3p 2 0 f2p V3 f2p 3 0 f2p 2 0 f2p V3Gaussian elimination technique. The latter can be found in any textbook dealing with numerical analysis. This technique is applied next.
Divide by 6 By back substitution, V3(1) V2(0) V3 V3 ; 1.0 0.067 3 0.063 ; 0.933 2 0.1 0.667 3 2 0 0.063 0.063 0.058 ; 2 0 0.058 0.058
Continue further iterations until convergence is achieved.
CONCLUDING REMARKS
The two solution strategies described here comprise the basic steps in a load-flow solu-tion. The reader should be reminded that, occasionally, an off-nominal transformer, a capacitor, or other network devices also have to be modeled. Most of these models can be represented in the bus admittance matrix. Another practical consideration that one needs to bear in mind is that all generators have upper and lower limits of reactive power generation. Hence, if during a load flow iteration it is found that any one of the generators is violating its limits, then that particular bus where the generator is located is said to have lost voltage control and, thus, should be treated as a load bus in subse-quent iterations.
As is obvious from the two methods, computer-based analysis is essential for obtain-ing accurate load-flow solutions of any realistically sized power system. A
computer-0.6 9 0.067 0.1
0.063 0.6 1 0 0 0.667 1 0 0 0 9 2 3 V3 0.1 0.1 0.6 1 0 0 0.667 1.583 0 0 0 9 2 3 V3 Divide by 1.583 0.1 0.2 0.6 1 1 0 0.667 2.25 0 0 0 9 2 3 V3Add this row to row 1 0.6
0.8 0.6 6 4 0 4 9 0 0 0 0 2 3 V3 Divide by 4based analysis typically utilizes many numerical techniques, such as optimal ordering and sparsity techniques, in order to reduce memory and storage requirements. There are several excellent load-flow programs available that are widely used by engineers in util-ity companies for frequent system studies. While industry-grade load-flow software tends to be very expensive, there are now many educational versions of load-flow soft-ware available that are inexpensive and quite adequate for classroom use or for studying small-scale systems.
BIBLIOGRAPHY
Elgerd, O. I. 1982. Electric Energy Systems Theory — An Introduction, 2nd ed. New York: McGraw-Hill.
Glover, J. D., and M. Sharma. 1994. Power System Analysis, and Design, 2nd ed. Boston: PWS Publishing.
Grainger, J. J., and W. D. Stevenson. 1994. Power System Analysis. New York: McGraw-Hill. Stott, B., and O. Alsac. 1974. “Fast Decoupled Load Flow,” IEEE Transactions on Power Apparatus
