Open access peer-reviewed chapter - ONLINE FIRST

Power Flow Analysis

By Mohammed Albadi

Submitted: September 25th 2018Reviewed: December 7th 2018Published: March 1st 2019

DOI: 10.5772/intechopen.83374

Downloaded: 996

Abstract

Power flow, or load flow, is widely used in power system operation and planning. The power flow model of a power system is built using the relevant network, load, and generation data. Outputs of the power flow model include voltages at different buses, line flows in the network, and system losses. These outputs are obtained by solving nodal power balance equations. Since these equations are nonlinear, iterative techniques such as the Newton-Raphson, the Gauss-Seidel, and the fast-decoupled methods are commonly used to solve this problem. The problem is simplified as a linear problem in the DC power flow technique. This chapter will provide an overview of different techniques used to solve the power flow problem.

Keywords

  • power flow
  • load flow
  • iterative techniques
  • Newton-Raphson
  • Gauss-Seidel
  • fast-decoupled
  • DC power flow

1. Problem formulation

Power flow analysis is a fundamental study discussed in any power system analysis textbook such as [1, 2, 3, 4, 5, 6]. The objective of a power flow study is to calculate the voltages (magnitude and angle) for a given load, generation, and network condition. Once voltages are known for all buses, line flows and losses can be calculated. The starting point of solving power flow problems is to identify the known and unknown variables in the system. Based on these variables, buses are classified into three types: slack, generation, and load buses as shown in Table 1.

Bus typeVoltage (|Vi|δi)Real powerReactive power
MagnitudeAngleGenerationLoadNet (Pi)GenerationLoadNet (Q i)
Slack/SwingSpecifiedSpecifiedUnknownSpecifiedUnknownUnknownSpecifiedUnknown
Generator/Regulated/PVSpecifiedUnknownSpecifiedSpecifiedSpecifiedUnknownSpecifiedUnknown
Load/PQUnknownUnknownSpecifiedSpecifiedSpecifiedSpecifiedSpecifiedSpecified

Table 1.

Type of buses in the power flow problem.

The slack bus is required to provide the mismatch between scheduled generation the total system load including losses and total generation. The slack bus is commonly considered as the reference bus because both voltage magnitude and angles are specified; therefore, it is called the swing bus. The rest of generator buses are called regulated or PV buses because the net real power is specified and voltage magnitude is regulated. Most of the buses in practical power systems are load buses. Load buses are called PQ buses because both net real and reactive power loads are specified.

For PQ buses, both voltage magnitudes and angles are unknown, whereas for PV buses, only the voltage angle is unknown. As both voltage magnitudes and angles are specified for the Slack bus, there are no variables that must be solved for. In a system with n buses and g generators, there are 2(n-1)-(g-1) unknowns. To solve these unknowns, real and reactive power balance equations are used. To write these equations, the transmission network is modeled using the admittance matrix (Y-bus).

2. Admittance matrix and power flow equation

The admittance matrix of a power system is an abstract mathematical model of the system. It consists of admittance values of both lines and buses. The Y-bus is a square matrix with dimensions equal to the number of buses. This matrix is symmetrical along the diagonal.

Y=Y11Y1nYn1YnnE1

The values of diagonal elements (Yii) are equal to the sum of the admittances connected to bus i. The off-diagonal elements (Yij) are equal to the negative of the admittance connecting the two buses i and j. It is worth noting that with large systems, Y-bus is a sparse matrix.

Yii=j=0jinyijE2
Yij=Yji=yijE3

The net injected power at any bus can be calculated using the bus voltage (Vi), neighboring bus voltages (Vj), and admittances between the bus and its neighboring buses (yij) as shown in Figure 1.

Ii=Viyi0+ViV1yi1+ViV2yi2++ViVjyij

Figure 1.

Net injected power.

Rearranging the elements as a function of voltages, the current equation becomes as follows:

Ii=Viyi0+yi1+yi2+..+yijV1yi1V2yi2Vjyij
Ii=Vij=0jiyijj=1jiyijVj=ViYii+j=1jiYijVj

The power equation at any bus can be written as follows:

Si=Pi+jQ i=ViIi

Or

Si=PijQ i=ViIi

Substituting the expression on the current in Siequation results in the following formula:

Si=ViVij=0jiyijj=1jiyijVj=ViViYii+j=1jiYijVj

Real and reactive power can be calculated from the following equations:

Pi=ReViVij=0jiyijj=1jiyijVj=ReViViYii+j=1jiYijVj
Q i=ImViVij=0jiyijj=1jiyijVj=ImViViYii+j=1jiYijVj

Or

Pi=j=1nViVjYijcosθijδi+δjE4
Q i=j=1nViVjYijsinθijδi+δjE5

And the current (Ii) can be written as a function of the power as follows:

PijQ iVi=Vij=0jiyijj=1jiyijVj=ViYii+j=1jiYijVjE6

Example 1: Admittance matrix formation.

For the below 4-bus system in Figure 2, the admittance matrix is constructed by converting all impedances in the system into admittances as shown in Figure 3. Then, diagonal and off-diagonal elements are calculated using Eqs. (2) and (3).

Y=j7.5j4j2.50j4j7.750j2.5j2.50j4.5j20j2.5j2j4.5=7.590°490°2.590°0490°7.7590°02.590°2.590°04.590°290°02.590°290°4.590°

Figure 2.

Impedance diagram.

Figure 3.

Admittance diagram.

3. Gauss-Seidel technique

The Gauss-Seidel (GS) method, also known as the method of successive displacement, is the simplest iterative technique used to solve power flow problems. In general, GS method follows the following iterative steps to reach the solution for the functionfx=0:

  • Rearrange the function into the form x=gxto calculate the unknown variable.

  • Calculate the value gx0based on initial estimatesx0.

  • Calculate the improved valuex1=gx0.

  • Continue solving for improved values until the solution is within acceptable limitsxk+1xkϵ.

The rate of convergence can be improved using acceleration factors by modifying the step sizeα.

xk+1=xk+αgxk+1xkE7

In the context of a power flow problem, the unknown variables are voltages at all buses, but the slack. Both voltage magnitudes and angles are unknown for load buses, whereas voltage angles are unknown for regulated/generation buses.

The voltage Viat bus ican be calculated using either equations:

Vi=1j=1jiyijPischjQ ischVi+jyijVjE8
Vi=1YiiPischjQ ischVij=1jiYijVjE9

where yijis the admittance between buses iand j, Yijis the Y-bus element, Pischthe net scheduled injected real power, Q ischis the net scheduled injected reactive power, and Viis the conjugate of Vi. The net injected quantities are the sum of the generation minus load. Typically, the initial estimates ofVi=10°.

The iterative voltage equation is as follows:

Vik+1=1YiiPischjQ ischVikjiYijVjkork+1E10

Or

Vik+1=1j=0yijPischjQ ischVi+jiyijVjkork+1E11

Both real and reactive powers are scheduled for the load buses, and Eq. 6 is used to determine both voltage magnitudes and angles (Viδ) for every iteration (Vik+1).

For regulated buses, only real power is scheduled. Therefore, net injected reactive power is calculated based on the iterative voltages (Vik+1) using either equations:

Q ik+1=ImVikVikj=0nyijj=1jinyijVjkork+1E12
Q ik+1=j=1nVikVjkork+1Yijsinθijδik+δjkE13

where Viand Vjare the magnitudes of the voltage at buses iand j, respectively. δiand δjare the associated angles. yijis the admittance between buses iand j. Yijis the magnitude of the Y-bus element between the two buses; and θijis the corresponding angle.

Since the voltage magnitude (Vi) is specified at regulated/PV buses, Eqs. (8) or (9) will be used to determine the voltage angles only. To achieve this, two options can be used:

  1. When using the polar form (Viδi), discard the iterative voltage magnitude and keep the iterative angle.

  2. When using the rectangular form (Rei+jImi), discard the real part (Rei) and keep the imaginary part (Imi) of the iterative voltage. The new real part (Reinew) can be calculated from the specified magnitude (Vi) and the iterative imaginary part.

    Reinew=Vi2Imi2E14

The iterative process stops when the voltage improvement reaches acceptable limits: Vik+1Vikϵ.

Example 2: Gauss-Seidel power flow solution.

Figure 4 below shows a 3-bus system. Perform 2 iterations to obtain the voltage magnitude and angles at buses 2 and 3. Impedances are given on 100 MVA base.

Figure 4.

3-Bus power system.

Solution:

The admittance values of the transmission network and the injected power in per unit at buses 2 and 3 are calculated as shown in Figure 5. Note that net injected power at the load bus is negative while that of the PV bus is positive. Per units values are obtained by diving actual values (MW and MVAR) by the base (100 MVA).

Figure 5.

Power flow input data.

Iteration #1: assume V20=1.000°and V30=1.030°.

V21=1y21+y23P2schjQ2schV20+y21V1+y23V30=15j15+15j502+j0.51.000°+5j151.020°+15j501.000°
V21=1.0120j0.0260=1.01231.4717°

As Q3 is not given, it is calculated based on the latest available information using Eqs. (12) or (13).

Q31=ImV30V30j=0ny3jj=1jiny3jVjkork+1
Q31=ImV30V30y31+y32y31V1+y32V21
Q3[1]=Im{1.030°(1.030°(10j40+15j50)[(10j40)1.020°+(15j50)(1.0120j0.0260)])}=Im{1.7201j0.9373}=0.9373pu

Now that Q31is calculated, the voltage V31can be calculated:

V31=1y31+y32P3schjQ31V30+y31V1+y32V21=110j40+15j501.5j0.93731.030°+10j401.020°+15j501.0120j0.0260=1.0294j0.0022pu

Since the magnitude of V3is specified, we retain the imaginary part of V31and calculate the real part using Eq. (14).

Reinew=1.0320.00222=1.03pu

Therefore,

 V31=V31=1.03j0.0022=1.030.1226°pu

Iteration #2: considering V21=1.0120j0.0260and V31=1.03j0.0022.

V22=1y21+y23P2schjQ2schV21+y21V1+y23V31=15j15+15j502+j0.51.0120j0.0260+5j151.020°+15j501.03j0.0022V22=1.0115j0.0270=1.01191.5273°

Q32calculation is given below:

Q32=ImV31V31y31+y32y31V1+y32V22
Q3[2]=Im{(1.03j0.0022)((1.03j0.0022)(10j40+15j50)[(10j40)1.020°+(15j50)(1.0115j0.0270)])}=1.0000pu

The voltage V32is calculated as follows:

V32=1y31+y32P3schjQ32V31+y31V1+y32V22=110j40+15j501.5j11.030°+10j401.020°+15j501.0115j0.0270=1.0298j0.0030pu

Only imaginary value of the calculated V32is retained and the real part is calculated based on the retained imaginary values and the scheduled V3.

Reinew=1.0320.00302=1.03pu

Therefore,

V32=1.03j0.003=1.030.1644°pu

The iterative solution is presented in Table 2.

IterationV2kV3k
11.01231.4717°1.030.1226°
21.01191.5273°1.030.1644°
31.01191.5598°1.030.1846°
41.01191.5750°1.030.1941°
51.01181.5823°1.030.1986°
61.01181.5857°1.030.2008°

Table 2.

Gauss-Seidel iterative solution.

4. Newton-Raphson technique

The Newton-Raphson (N-R) technique, also known as the method of successive approximation, is based on Taylor’s expansion approximation. The unknown xin the function fx=ccan be determined using Taylor’s expansion approximation. Starting with an initial estimatex0, the deviation from the correct solution is x0. Applying Taylor’s expansion, the function can be written as follows:

fx0+x0=cE15
fx0+x0=fx0+fx0x0+12!fx0x02+13!fx0x03+=cE16

Assuming x0is small, the higher order terms (12!fx0x02+13!fx0x03+) are neglected and the function can be approximated by the first two terms.

fx0+x0fx0+fx0x0=cE17

Based on x0, the deviation from the correct solution can be iteratively calculated.

x0=cfx0fx0=fx0fx0E18

The improved solution can be calculated iteratively.

x1=x0+x0E19

The iterative process is stopped when the mismatch between scheduled and calculated value (fk=cfxk) is within acceptable limitsfkϵ.

In the context of power flow problems, unknown variables xare both voltage magnitude and angles (Viδi) at load buses, as well as voltage angles (δi) at regulated buses. The scheduled (specified) quantities (c) are both net real (Pisch) and reactive power (jQ isch) values at load buses and real power at generation buses as shown in Table 1. The iterative values of reactive power are calculated using Eqs. (12) or (13). Similarly, the iterative values of real power are calculated using Eqs. (20) or (21):

Pik+1=ReVikVikj=0nyijj=1jinyijVjkork+1E20
Pik+1=j=1nVikVjkork+1Yijcosθijδik+δjkE21

The Newton-Raphson power flow formulation can be written as follows:

xk=δikVik, c=PischQ isch, f=PikQ ik, and PischQ ischPikQ ik=PikQik.

When more variables are used, the derivative fis replaced by partial derivatives with respect to different variables. The partial derivative matrix is called the Jacobian matrix.

f=PiδiPiViQ iδiQ iVi=JJPVJJQVE22

Therefore, the Newton-Raphson power flow formulation can be solved using the below equation:

PischQ ischPikQ ik=PikQik=PiδikPiVikQ iδikQ iVikδikVikE23

To solve for the deviation, the inverse of the Jacobian matrix is required for every iteration.

δkVik=JkJPVkJkJQVk1PikQikE24

Then, new values are calculated:

δikVik=δik1Vik1+δikVikE25

The iterative process stops when the mismatch between calculated and scheduled quantities is within PikQikϵ.

Example 3: Newton-Raphson power flow solution.

Solve the power flow problem shown in Figure 3 using the Newton-Raphson technique. Perform two iterations.

Solution:

The first step in Newton-Raphson Power Flow technique is building Y-bus using Eqs. (2) and (3).

Y=15j555+j1510+j405+j1520j6515+j5010+j4015+j5025j90=57.0174.74°15.81108.43°41.23104.04°15.81108.43°68.0172.90°52.20106.70°41.23104.04°52.20106.70°93.4174.48°

Since the unknown variables are δ2, δ3, and V2; the scheduled quantities are P2sch, P3sch, and Q2sch, the following problem formulation can be written.

P2schP3schQ2schP2kP3kQ2k=P2δ2kP2δ3kP2V2kP3δ2kP3δ3kP3V2kQ2δ2kQ2δ3kQ2V2kδ2kδ3kV2k

To calculate the Jacobian matrix elements, P2, P3, and Q2equations are obtained using (4) and (5).

P2=j=1nV2VjY2jcosθ2jδ2+δj=V2V1Y21cosθ21δ2+δ1+V22Y22cosθ22+V2V3Y23cosθ23δ2+δ3
P2δ2=V2V1Y21sinθ21δ2+δ1+V2V3Y23sinθ23δ2+δ3
P2δ3=V2V3Y23sinθ23δ2+δ3
P2V2=V1Y21cosθ21δ2+δ1+2V2Y22cosθ22+V3Y23cosθ23δ2+δ3
P3=j=1nV3VjY3jcosθ3jδ3+δj=V3V1Y31cosθ31δ3+δ1+V3V2Y32cosθ32δ3+δ2+V32Y33cosθ33
P3δ2=V2V3Y32sinθ32δ3+δ2
P3δ3=V3V1Y31sinθ31δ3+δ1+V3V2Y32sinθ32δ3+δ2
P3V2=V3Y32cosθ32δ3+δ2
Q2=j=1nV2VjY2jsinθ2jδ2+δj=V2V1Y21sinθ21δ2+δ1V22Y22sinθ22V2V3Y23sinθ23δ2+δ3
Q3δ2=V2V1Y21cosθ21δ2+δ1+V2V3Y23cosθ23δ2+δ3
Q3δ3=V2V3Y23cosθ23δ2+δ3
Q3V2=V1Y21sinθ21δ2+δ12V2Y22sinθ22V3Y23sinθ23δ2+δ3

Iteration #1: assume V20=1.000°and V30=1.030°.

The calculated quantities:P21P31Q21=0.55000.56651.8000.

The scheduled quantities:P2schP3schQ2sch=21.50.5

The mismatch power matrix:∆P21∆P31Q21=21.50.50.55000.56651.8000=1.45000.93351.3000

The Jacobian matrix (J) elements for iteration # 1 are as follows:

J=66.851.519.4551.593.5215.4520.5515.4563.2

Newton-Raphson formulation is as follows:

1.45000.93351.3000=66.851.519.4551.593.5215.4520.5515.4563.2δ21δ31V21
δ21δ31V21=66.851.519.4551.593.5215.4520.5515.4563.211.45000.93351.3000=0.0279rad0.0033rad0.0123pu
δ21δ31V21=δ20δ30V20+δ21δ31V21
δ21δ31V21=001.00+0.0279rad0.0033rad0.0123pu=0.0279rad0.0033rad1.0123pu=1.5986°0.1891°1.0123pu

Iteration #2: Consider V21=1.01231.5986°and V31=1.030.1891°.

The calculated quantities: P22P32Q22=2.01091.52020.4621

The mismatch power matrix: ∆P22∆P32Q22=21.50.52.01091.52020.4621=0.01090.02020.0379

The Jacobian matrix elements are calculated as follows:

J=67.0751.7418.2652.5094.4914.1822.5116.9165.34
δ22δ32V22=67.0751.7418.2652.5094.4914.1822.5116.9165.3410.01090.02020.0379
δ22δ32V22=0.1272×103rad0.2154×103rad0.4810×103pu
δ22δ32V22=δ21δ31V21+δ22δ32V22
δ22δ32V22=0.0279rad0.0033rad1.0123pu+0.1272×103rad0.2154×103rad0.4810×103pu
δ22δ32V22==0.0277rad0.0035rad1.0118pu=1.5871°0.2005°1.0118pu

It is worth noting that the mismatch between calculated and scheduled quantities diminishes very quickly.

∆P21∆P31Q21=1.45000.93351.3000and∆P22∆P32Q22=0.01090.02020.0379

The iterative solution is presented in Table 3.

IterationV2kV3k
11.01231.5986°1.030.1891°
21.01181.5871°1.030.2005°
31.01181.5871°1.030.2005°

Table 3.

Newton-Raphson iterative solution.

5. Fast-decoupled technique

In high voltage transmission systems, the voltage angles between adjacent buses are relatively small. In addition, X/Rratio is high. These two properties result in a strong coupling between real power and voltage angle and between reactive power and voltage magnitude. In contrary, the coupling between real power and voltage magnitude, as well as reactive power and voltage angle, is weak. Considering adjacent buses, real power flows from the bus with a higher voltage angle to the bus with a lower voltage angle. Similarly, reactive power flows from the bus with a higher voltage magnitude to the bus with a lower voltage magnitude.

Fast-decoupled power flow technique includes two steps: (1) decoupling real and reactive power calculations; (2) obtaining of the Jacobian matrix elements directly from the Y-bus.

PQ=J00JQVδVE26

or

P=JδE27
Q=JQVVE28

Next step is to obtain Jand JQVfrom the Y-bus as flowing:

δ=B1PVE29
V=B1QVE30

Where the Band Bare relevant imaginary part of the Y-bus matrix elements. Bis related to the buses at which real power is scheduled (δis unknown) and Bis related to the buses at which reactive power is scheduled (Vis unknown).

Bij=YijsinθijE31

The fast-decoupled technique requires more iterations to converge compared to the Newton-Raphson power flow formulation, especially if X/Rratio is not high. Another advantage of this method is that the Jacobian matrix has constant term elements which are obtained and inverted once at the beginning of the iterative process.

Example 4: Fast-decoupled power flow solution

Solve the power flow problem shown in Figure 3 using fast-decoupled power flow technique. Perform two iterations.

Solution:

The first step in fast-decoupled power flow technique is obtaining Band Bfrom the Y-bus.

Y=15j555+j1510+j405+j1520j6515+j5010+j4015+j5025j90

Since the unknown voltage angles are δ2and δ3, elements for Bare obtained from the Y-bus (intersection of columns numbers 2 & 3 and rows numbers 2 & 3) as follows:

B=65505090

Since the unknown voltage magnitude is V2at bus 2, Bcontains one element only (intersection of the column and the row number 2):

B=65

The fast-decoupled power flow formulation becomes as follows:

δ2δ3=B1P2V2P3V3andV2=B1Q2V2

Or

δ2kδ3k=B1P2kV2k1P3kV3k1andV2k=B1Q2kV2k1

Iteration #1: assume V20=1.000°and V30=1.030°.

The calculated quantities:

P21P31Q21=0.55000.56651.8000

The mismatch power matrix:

∆P21∆P31Q21=21.50.50.55000.56651.8000=1.45000.93351.3000

Calculation of δ2and δ3:

δ21δ31=B1P21V20P31V30=6550509011.45001.000.93351.03=0.0254rad0.0041rad=1.4569°0.2324°
δ21δ31=δ20δ30+δ21δ31=00+1.4569°0.2324°=1.4569°0.2324°

Calculation of V2:

V21=6511.31.00=0.0200
V21=V20+V21=1.00+0.0200=1.020pu

Iteration #2: considering V21=1.0201.4569°and V30=1.030.2324°.

The calculated quantities:

P22P32Q22=1.66711.21330.0239

The mismatch power matrix:

∆P22∆P32Q22=21.50.51.66711.21330.0239=0.33290.28670.4761

Calculation of δ2and δ3:

δ22δ32=B1P22V21P32V31=6550509010.33291.020.28671.03
δ22δ32=0.0046rad0.0005rad=0.0300°0.0286°
δ22δ32=δ21δ31+δ22δ32=1.4569°0.2324°+0.0300°0.0286°=1.7213°0.2021°

Calculation of V2:

V22=6510.47611.02=0.0072
V22=V21+V22=1.020.0072=1.0128pu

The remaining five iterations are shown in Table 4. It shows that this method converges slower than Newton-Raphson method.

IterationV2kV3k
11.02001.4569°1.030.2324°
21.01281.7213°1.030.2021°
31.01111.6019°1.030.2029°
41.01181.5759°1.030.2026°
51.01191.5877°1.030.2027°

Table 4.

Fast-decoupled iterative solution.

6. DC power flow technique

DC power is an extension to the Fast-decoupled power flow formulation. In DC power flow method, the voltage is assumed constant at all buses; therefore, the(V,Q) equation is neglected. The (δ,P) equation can be further simplified to a linear problem that does not require iterative solution:

Bδ=PE32

Or

δ=B1PE33

Example 5: DC power flow

Solve the power flow problem shown in Figure 3 using the DC power flow technique.

Solution:

The first step in DC power flow technique is obtaining Bfrom the Y-bus.

Y=15j555+j1510+j405+j1520j6515+j5010+j4015+j5025j90

For this problem, Since the unknown voltage angles are δ2and δ3, elements for Bare obtained from the Y-bus as follows:

B=65505090
δ2δ3=655050901P2P3=65505090121.5=0.0313rad0.0007rad=1.7958°0.0428°

7. Slack bus power and losses calculations

The main objective of power flow calculations is to determine the voltages (magnitude and angle) for a given load and generation conditions. Once voltages are known for all buses, slack bus power, as well as line flows and losses, can be calculated. The slack bus real and reactive power are calculated using Eqs. (4) and (5), respectively. Overall system losses are the difference between generation and load.

SL=SGenSLoadE34

Specific branch losses are calculated using branch power flow. For example, the losses in the line i – j are the algebraic sum of the power flow.

SLij=Sij+SjiE35

Sijand Sjiare defined as follows:

Sij=ViIijE36
Sji=VjIjiE37

The current between buses Iijis a function of the voltages and the admittance between Viand Vj. It is worth noting that Iji=Iij

Iij=ViVjyijE38

Example 6: Slack bus power and losses

For the 3-bus system shown in Figure 3, the voltages at buses 2 and 3 were iteratively calculated: V1=1.020°, V2=1.01181.5871°and V3=1.030.2005°

  1. Calculate the slack bus power.

  2. Calculate the total system losses.

  3. Calculate individual branch losses.

Solution:

The polar form of the Y-bus is used.

Y=57.0174.74°15.81108.43°41.23104.04°15.81108.43°68.0172.90°52.20106.70°41.23104.04°52.20106.70°93.4174.48°

  1. The slack bus power

    The slack bus power can be calculated using real and reactive power equations:

    P1=j=1nV1VjY1jcosθ1jδ1+δj=V12Y11cosθ11+V1V2Y12cosθ12δ1+δ2+V1V3Y13cosθ13δ1+δ3=0.5195pu
    Q1=j=1nV1VjY1jsinθ1jδ1+δj=V12Y11sinθ11V1V2Y12cosθ12δ1+δ2V1V3Y13sinθ13δ1+δ3=0.4572pu

  1. The total system losses

The total real power losses can be calculated using the total net injected real a power at all buses:

PL=P1+P2+P3
PL=P1+P2+P3=0.51952+1.5=0.0195pu

Similarly, the reactive power losses can be calculated.

QL=Q1+Q2+Q3

However, Q3is unknown and can be calculated based on the given bus voltages:

Q3=j=1nV3VjY3jsinθ3jδ3+δj=V3V1Y31sinθ31δ3+δ1V3V2Y32sinθ32δ3+δ2V32Y33sinθ33=1.0220pu

Once Q2is calculated, the total reactive power losses are calculated:

QL=Q1+Q2+Q3=0.45720.5+1.0220=0.0648pu
  1. Branch losses

To calculate the losses in line 1–2, SL12is calculated by summing S12and S21. These flows are calculated as follows:S12=V1I12and S21=V2I21

I12=V1V2y12=1.020°1.01181.5871°5+j15=0.4635+j0.0121pu
I21=V2V1y21=I12=0.4635j0.0121pu
S12=V1I12=1.020°×0.4635j0.0121=0.4728j0.0123pu
S21=V2I21=1.01181.5871°×0.4635+j0.0121=0.4685+j0.0252pu
SL12=S12+S12=0.4728j0.0123+0.4685+j0.0252=0.0043+j0.0129pu

Similarly, power flow and losses in other branches are calculated.

Line 1–3:

S31=0.0456+j0.4494
S13=0.0467j0.4449
SL13=0.0011+j0.0045

Line 2–3:

S23=1.5315j0.5252
S32=1.5456+j0.5722
SL23=0.0141+j0.0470

Total losses:

SL=SL13+SL23+SL12=0.0195+j0.0648pu

The generation and load at different buses is shown in Figure 6.

Figure 6.

Power flow results.

8. Conclusions

Power flow analysis, or load flow analysis, has a wide range of applications in power systems operation and planning. This chapter presents an overview of the power flow problem, its formulation as well as different solution methods. The power flow model of a power system can be built using the relevant network, load, and generation data. Outputs of the power flow model include voltages (magnitude and angles) at different buses. Once nodal voltages are calculated, real and reactive power flows in different network branches can be calculated. The calculation of branch power flows enables technical loss calculation in different network branches, as well as the total system technical losses.

Power flow analysis is performed by solving nodal power balance equations. Since these equations are nonlinear, iterative techniques such as the Gauss-Seidel, the Newton-Raphson, and the fast-decoupled power flow methods are commonly used to solve this problem. In general, the Gauss-Seidel method is simple but converges slower than the Newton-Raphson method. However, the latter method required the Jacobian matrix formation of at every iteration. The fast-decoupled power flow method is a simplified version of the Newton-Raphson method. This simplification is achieved in two steps: 1) decoupling real and reactive power calculations; 2) obtaining of the Jacobian matrix elements directly from the Y-bus matrix. The DC power method is an extension to the fast-decoupled power flow formulation. In DC power flow method, the voltage is assumed constant at all buses and the problem becomes linear.

Acknowledgments

The author would like to acknowledge the financial support and thank the Rural Area Electricity Company and Sultan Qaboos University for sponsoring the publication of this chapter under project number CR/ENG/ECED/16/02.

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Mohammed Albadi (March 1st 2019). Power Flow Analysis [Online First], IntechOpen, DOI: 10.5772/intechopen.83374. Available from:

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