9 Load Flow Analysis

__________________________________________________________ Introduction In the early days of the development of electrical power engineering, generators were physically close to the load they supplied. Central generating stations were established in many towns and cities for the purpose of supplying power to local customers. By the 1920s many different companies operated such generating stations each with their own standard for supply. In some cities a direct current supply was available, in others alternating current at 12.5 Hz, 25 Hz, or 50 Hz might be generated. At this time there was no need for load flow analysis as the electrical power systems were small, isolated and generation matched load. The lack of a standard supply severely hampered the development of the domestic appliance market. In 1926, a British Act of Parliament established the Central Electricity Board with the specific objective of standardizing the supply voltage and frequency. The Board also oversaw the interconnection of the most efficient generating stations by a 132 kV high-voltage network. Interconnection between generating stations and controlling the real and reactive powers flowing along the interconnectors then became a priority. This required an understanding of the basics of load flow analysis. In 1948 the British Electricity Supply Industry was nationalized and two organizations were established in England and Wales (1) the area boards which were mainly concerned with distribution and customer services and (2) the generating boards which were responsible foe the generation of electricity and the operation of the high voltage transmission network. In Scotland two boards were established, the North of Scotland Hydro Electricity Board and the South of Scotland Electricity Board. There boards were responsible for all aspects of generation, transmission and distribution within their geographical areas. By the 1980s the electrical network in the United Kingdom was fully interconnected with transmission voltages increased to 275 kV and 400 kV (the super grid). Electrical power generation was concentrated at large power stations directly connected to either the 275 kV or 400 kV network and the direction of real power flow was from the H-V network down to the distribution network (132 kV, 33 kV, 11 kV etc). Load flow analysis of this complex interconnected network had become an essential tool to aid electrical power systems engineers in the daily running of the power system. At the present time (2001) the British electricity supply industry has been privatized the move away from concentrated large scale coal, oil, and nuclear generation to small scale distributed generation using renewable energy sources has led to an increase in the number of load flow studies performed. Generation is now distributed at all voltage levels within the power system and out with the direct control of a single grid control centre.

341

Load Flow

The load flow calculation is used to ensure that the steady state voltages and currents flowing the interconnected network are within the continuous ratings of the equipment for both normal and emergency operating conditions. Such calculations are performed on computer as hand calculations are too time consuming and impractical for large systems. In this Chapter we will consider relatively simple power systems with two, three, or four bus-bars to illustrate the techniques used in practical load flow analysis. In a real load flow study an analysis involving perhaps 2000 or more bus-bars is not uncommon. Learning Objectives To understand the term load flow analysis as applied to an interconnected power system

To understand the factors which influence the flow of real and reactive powers over an inter-connector To understand that the flow of both real power and reactive power contribute to the real power losses in the transmission network To understand the principle aims and objectives of a load flow study To understand that the load flow problem is non-linear and cannot be solved by classical circuit analysis To understand how to construct the bus admittance matrix for a simple power system To understand the Gauss-Seidel Method as applied to the iterative solution of simultaneous equations To understand the three different types of bus bar in a load flow study. To understand which quantities are known and which are unknown for each type at the beginnings of the calculation To understand how to calculate the voltages in a power system using the Gauss-Seidel Method To understand how to calculate the real and reactive powers on a generator bus bar once all the bus bar voltages have been found.

342

Load Flow

9.1 Power and VAr flow in an Inter-connector

Consider the two bus-bar power system shown in Figure 9.1. Two generators and their local loads are connected by a short transmission line which interconnects them. Without the transmission line connecting the bus-bars 1 and 2, the generator G1 could only supply the local load L1 while the generator G2 could only supply the load L2. By interconnecting the bus-bars, it is possible for the generators to share both real and reactive powers. The question then arises, how much real power and reactive power is transmitted over the inter-connector?

Figure 9.1 Two interconnected generating stations

If we consider the injection of real power P1 and reactive power Q1 at bus-bar 1and the delivery of real power P2 and reactive power Q2 at bus-bar 2, we can derive equations of the real and reactive power flows along the transmission line.

Figure 9.2 Directions for P and Q at each end of the interconnector

343

Load Flow

Figure 9.3 Phasor diagram showing voltages at each end of the interconnector Assuming the inter-connector can be regarded as a short transmission line we can neglect the shunt capacitance and leakage resistance to earth. The current flowing along the line per phase is therefore given by:

Z

VVI 21 −= (9.1)

where Z is the series impedance per phase of the line Z = (R + jX) and V1 and V2 are the phasors representing the phase voltages on bus-bar 1 and bus-bar 2 as shown in Figure 9.3. Note that the angle between the phasors V1 and V2 which is denoted by δ is referred to as either the transmission angle, the load angle, or the power angle. (It is unfortunate that three different names are in common use to describe the same quantity.) The complex power per phase at bus-bar 1 is given by:

∗

∗∗ −==+=

ZVVVIVjQPS 21

1*

1111 (9.2)

If the line resistance is small then in the limit as R tends to zero, then and

. The expression for the complex power per phase SjxZ →

jxZ −→∗1 can now be rewritten

as:

jx

eVVVjx

VVVVZ

VVVS

jδ21

21

*21

*11

*

*2

*1

11 −

−=

−−

=−

= (9.3)

Expanding using Euler's Rule yields: δ+δ=δ sinjcose j

344

Load Flow

jx

sinδVVjcosδVVVS 2121

21

1 −

−−=

Rearranging this equation gives:

xcosδVVV

jx

sinδVVS 21

2121

1

−+= (9.4)

Equation (9.4) and equation (9.2) are both expressions for the complex power S1. Hence we can write the following.

x

cosδVVVj

xsinδVV

jQPS 212

121111

−+=+= (9.5)

By inspection of equation (9.5) it becomes apparent that we can write the following equations for the real power and the reactive power per phase injected into bus-bar 1 by generator 1:

xsinVV

P 211

δ= Watts/phase (9.6)

x

cosδVVVQ 21

21

1

−= VArs/phase (9.7)

where V1 is the voltage per phase at bus-bar 1 V2 is the voltage per phase at bus-bar 2 X is the reactance of the transmission line per phase δ is the transmission angle In deriving equations (9.11) and (9.12) we assumed that the series resistance of the line was zero. This implies that real power loss in the line is zero, so the real power injected at bus-bar 1 must equal the real power received at bus-bar 2 i.e.:

x

sinVVPPP 21

21

δ=== (9.8)

Also if the reactive power injected in the transmission line at bus 1 is

345

Load Flow

x

cosδVVVQ 21

21

1

−= (9.9)

then by symmetry we see that the reactive power at bus 2 is

x

VcosδVVQ

2221

2

−= (9.10)

Here we note that . This is due to the reactive power required to magnetize the series reactance of the transmission line.

21 QQ ≠

Having derived equations (9.8 – 9.10) we can now attempt to understand the important mechanism whereby the real power flow and the reactive power flow may be controlled in both magnitude and direction.

9.1.1 Control of Real Power Flow over inter-connector The reader must remember that the transmission angle δ was defined as the phase angle between V1 and V2. Note that the voltages and reactance must be given in per-phase values to yield per-phase values of power. The reactance of the line X is a fixed parameter. If we assume automatic excitation control of the generators at each end of the line so as to maintain a constant terminal voltage magnitude, i.e. |V1| and |V2| are fixed, then equation (9.8) shows that the only way in which P can be made to change is to change the transmission angle δ. Figure 9.4 shows how equation (9.8) can be represented by the power-delta curve for fixed magnitudes of V1 and V2. When the transmission angle δ is zero, no power is transferred over the inter-connector. If the phase angle of V1 is advanced relative to V2, the value of δ increases so the power flow increases up to a maximum value at δ=90o given by

XVV

P 21max = (9.11)

If we were to attempt to increase δ further beyond 90o the power transfer will decrease. At this point the transmission collapses; the two generators at each end of the inter-connector would fall out of synchronism, this means that generator G1 and the load on bus-bar 1 would run at a different frequency from generator G2 and the load on bus-bar 2. A transmission angle of δ = 90o corresponds to the transmission limit or static stability limit of the line. In practice, δ is normally limited to 300 to

346

Load Flow

ensure system stability during faults and other transients. (A 3rd power limit exists, this is the level at which the line heating limit is reached, this is much higher than Pmax)

Figure 9.4 Power-Delta Curve for a short inter-connector

If the transmission angle δ is increased in the negative sense (the phase angle of V1 now lags behind V2) then figure 9.4 shows that the power flow becomes negative. The direction of power transmission has now reversed. Note the important difference between ac and dc transmission. In a dc system the sending-end voltage magnitude must exceed that of the receiving-end voltage in order for power to flow in the proper direction. In an ac system the voltage magnitudes do not determine the power direction but the voltage phase difference does. By definition real power flows from leading to lagging phasor. This is illustrated in diagrams 9.5 and 9.6.

Figure 9.5 Real power flow from bus1 to bus 2

Figure 9.6 Real power flow from bus 2 to bus 1

347

Load Flow

Example 9.1

A 132 kV three-phase transmission line is 50 miles long and has a reactance of 40 Ω per phase. If the line is to be operated as a “flat line” at rated voltage (i.e. the sending voltage magnitude is equal to the received voltage magnitude), determine (a) the maximum possible steady state power flow and (b) the transmission angle δ if the real power flow is 200 MW.

Solution 9.1 Part a) As this is a three phase transmission line, the given voltage is the rated line to line

voltage. Working per phase, the rated phase voltage is therefore kV76.2kV3

132= .

The sending voltage is therefore 76.2 kV and the received voltage is 76.2 kV since this is a flat line. The maximum steady state power that can be transmitted per phase is given by equation (9.11)

MW14540

1076.21076.2XVV

P33

21max =

⋅×⋅==

The maximum steady state three-phase power is therefore:

MW435MW1453Pmax =×= Part b)

If the real power flow is 200 MW, the power per phase is MW66,673MW200

=

The voltage per phase at the sending and receiving ends is still 76.2 kV, the required transmission angle is found from equation (9.8)

o

336

21

27.3δ

sinδ40

1076.21076.21066.67

sin

=

⋅×⋅×

=×

=x

VVP

δ

The transmission angle required to transfer 200 MW over the line is therefore 27.3o. Example 9.2 Repeat Example 9.1, using a per-unit calculation.

348

Load Flow

Solution 9.2

Part a) Choose 200 MVA as Sbase and 132 kV as Vbase and calculate Zbase.

Ω87.1210200

)10(132S

)(VZ 6

23

base

2base

base =⋅⋅

==

The reactance of the 40 Ω transmission line in per unit is therefore

pu0.45987.12

40Zpu ==

The voltage at the sending end is 132 kV or 1 pu. The voltage at the receiving end is also 1 pu. The maximum power transfer is given by equation (9.11)

pu2.1780.459

1.01.0XVV

P 21max =

×==

Part b)

If the real power flow is 200 MW, in per unit this is 1.0pu200MVA200MW

=

The required transmission angle is found from equation (9.8)

o

21

27.3δ

sinδ0.459

0.10.11.0

sin

=

⋅×

=

=x

VVP

δ

Examples 9.1 and 9.2 show that the equations derived above for real and reactive powers along an inter-connector are valid for both per phase and per unit calculations.

9.1.2 Control of Reactive Power over inter-connector In our discussions on the control of the real power flow over the inter-connector we noted that in practice the transmission angle δ is usually limited to values less than 30o, to ensure stability during transients and other disturbances. In general we can

349

Load Flow

therefore assume that the transmission angle is relatively small and cos equal to one (cos 0

δo = 1, cos 30o = 0.866) Equations (9.9) and (9.10) can therefore be simplified:

( )21121

2121

21

1 VVxV

xVVV

xcosδVVV

Q −=−

=−

= (9.12)

( 212

2221

2221

2 VVx

Vx

VVVx

VcosδVVQ −=

−=

−= ) (9.13)

The factor inside the parenthesis in the equations for Q1 and Q2 shows us that the reactive power flow tends to be proportional to the difference in voltage magnitudes. By definition reactive power tends to flow in the direction of lowest voltage amplitude. The larger the voltage difference, the stronger the reactive power flow. This reactive power flow tendency is perhaps more clearly seen by considering the average reactive power flow:

x2VV

2QQQ

22

2121

ave

−=

+≡ (9.14)

Remember that Q1 is not equal to Q2 since reactive power is absorbed by the series reactance of the transmission line.

Figure 9.7 shows the phasor relationship for a small power flow from 1 to 2, and a large reactive power flow from 2 to 1.

Figure 9.7 Phasor relationship

Example 9.3

Consider again the 132 kV transmission line in examples 9.1 and 9.2. Calculate the reactive power flow in the line for a real power flow of 200 MW with the following voltage profiles:

350

Load Flow

Case a) 1.0puVV 21 ==

Case b) 1.0puV1.2pu,V 21 ==

Case c) 1.2puV1.0pu,V 21 == Solution 9.3

In example 9.2 using 200 MVA and 132 kV as base quantities, the per unit impedance of the 40 Ω transmission line was shown to be 0.459pu. Case a) The transmission angle corresponding to 1pu power was calculated to be 27.3o.in example 9.2. Using equations (9.9) and (9.10), the per-unit reactive powers can be calculated.

( ) ( ) ( ) ( ) pu0.243

0.45927.3cos1.01.01.0

xcosδVVV

Q2

212

11 =

⋅−=

−=

( ) ( ) ( ) ( ) pu0.2430.459

1.027.3cos1.01.0x

VcosδVVQ

22221

2 −=−⋅

=−

=

Note that when the voltage profile of the line is kept flat, reactive power flows into the line from both ends. Also note that the simplified equations for Q1 and Q2 have not been used for this calculation. Case b) First we must calculate the transmission angle δ. Using equation (9.8)

o

21

22.5δ

sinδ0.459

0.12.11.0

sin

=

⋅×

=

=x

VVP

δ

Hence using equations (9.9) and (9.10), the new per-unit reactive powers can be calculated.

( ) ( ) ( ) ( ) pu0.721

0.45922.5cos1.21.01.2

xcosδVVV

Q2

212

11 =

⋅−=

−=

351

Load Flow

( ) ( ) ( ) ( ) pu0.2360.459

1.022.5cos1.01.2x

VcosδVVQ

22221

2 =−⋅

=−

=

Case c) The transmission angle δ will be the same as in Case b), i.e. δ = 22.5o. Using equations (9.9) and (9.10) as before:

( ) ( ) ( ) ( ) pu0.2360.459

22.5cos1.01.21.0x

cosδVVVQ

221

21

1 −=⋅−

=−

=

( ) ( ) ( ) ( ) pu-0.7210.459

1.222.5cos1.21.0x

VcosδVVQ

22221

2 =−⋅

=−

=

In this case the reactive power flow is from bus 2 to bus 1. The results of these calculations are shown in graphical form in Figure 9.8. In is interesting to note that by changing the voltage levels of the two bus-bars, a strong effect is noted in the reactive power flow but not effect whatsoever on the real power flow. There is however a slight effect on the transmission angle δ. This example confirms the claim that the magnitude of the reactive power flow depends on the voltage magnitude and that the reactive power flows towards the low voltage bus-bar.

Figure 9.8 Varying reactive power flows

352

Load Flow

Real Power Losses in an Inter-connector The real and reactive power flows shown in Figure 9.8 which refers to Example 9.3 show a substantial reactive power loss on the line. This reactive power is consumed by the series reactance of the line. Equations (9.8), (9.9), and (9.10) were derived on the assumption of zero resistance and the real power losses were zero. The effect of the shunt capacitance of the line was also neglected. In practice a transmission line will have a certain series resistance which will cause a real power loss which can be expressed as:

2IRP =Ω (9.15)

This power loss is of much more practical significance that the reactive power loss. Due to the distributed transmission line series impedance and shunt capacitance the voltage, current, and powers vary along the length of the line. Let Vave, Iave, Pave, and Qave represent average values measured at the mid point of the line. Theses variable are related by the following equation: *

aveaveaveave IVjQP ⋅=+ The complex conjugate of the average current can therefore be written as

ave

aveave*ave V

jQPI

+=

and

*ave

aveaveave V

jQPI

−=

Multiplying these equations yields

2ave

2ave

2ave

ave

aveave*

ave

aveave2ave

*aveave

VQP

VjQP

VjQP

III+

=+

⋅−

==⋅

Substituting this expression into equation (9.15) gives the following approximate loss formula:

2ave

2ave

2ave2

ΩV

QPRIRP

+=≅ (9.16)

353

Load Flow

The formula is important because it shows us that the real and the reactive line powers contribute equally to the real power loss in the line. From the point of view of power loss we should therefore minimize the reactive power flow in the line.

Example 9.4

A three-phase 140kV transmission line has the following parameters per phase: resistance = 9.10 Ω, inductive reactance = j50.5 Ω, total shunt admittance = j334 µΩ-1. a) Using the π model of the line calculate the sending end conditions if the

magnitude of the receiving end voltage is 140 kV and the load complex power is S = 100 + j60 MVA.

b) Use the sending and receiving end complex powers to calculate the power loss in the line and show that equation (9.16) can be used to estimate the approximate line loss.

Solution 9.4

Part a) The solution of this problem requires a five step approach similar to that used in Chapter 5. Step 1 Draw the per-phase circuit diagram and mark all voltages, currents,

real and reactive powers. Step 2 Calculate the receiving end current, remembering to work per phase

The complex power at the load is:

354

Load Flow

j20.0033.333

60j3

100*IVS 222 +=+=⋅= MVA per phase

If V2 is the reference phasor then, ( )kVj0.080.82kVj0.03

140V2 +=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

The received current is therefore given by

( )Aj247.4412.480.82

j20.0033.33I2 −=+

= /phase

Step 3 Find the sending end voltage The current Ish2 is given by ( ) Aj13.510j1671080.82YVI 63

22sh2 =⋅×=⋅= − /phase

The line current IL is now calculated by adding the load current and the shunt current.

( ) ( )Aj233.9412.4j13.5j247.4412.4III sh22L −=+−=+= /phase The voltage drop across the series branch is ( ) ( ) ( ) ( ) kVj18.7015.5650.59.10233.9412.4jXRI∆V L +=+⋅−=+⋅= /phase The sending end voltage is then calculated ( ) ( ) kVj18.7096.39j18.7015.5680.82∆VVV 21 +=++=+= /phase Step 4 Find the sending-end current The current Ish1 is given by ( ) ( ) ( )Aj16.13.110j16710j18.7096.39YVI 63

11sh1 +−=⋅⋅⋅+=⋅= − /phase The sending end current therefore equals ( )Aj217.8409.3III Lsh11 −=+= /phase Step 5 Find the sending end power

( ) ( )( ) 6

3*111

1064.2838.35

8.2173.4091070.1839.96

⋅+=

−⋅⋅+=⋅=

j

jjIVS

355

Load Flow

The sending complex power is therefore 35.38 MW and 28.64 MVAr per phase. The total three phase real and reactive powers are 106.14 MW and 85.93 MVAr.

Part b)

The sending real power is 106.14 MW and the received real power is 100.00MW. The difference between these must be the power lost in transmission, i.e 6.14 MW. To use equation (9.16) the average real power, reactive power and voltage must be found. The magnitude of the received voltage per phase is 80.82 kV. The magnitude of the sending voltage per phase is

( ) kV98.181018.7096.3910j18.7096.39V 32231 =⋅+=+=

The average voltage is therefore

kV89.52

98.1880.82Vave =+

= per phase

The average real power per phase is

MW34.352

35.3833.33Pave =+

= /phase

The average reactive power per phase is

MW24.322

28.6420.00Qave =+

= /phase

The power loss per phase in the transmission line is therefore

( ) ( )

( )MW2.01

1089.51024.321034.359.10

VQPRIRP 23

2323

2

222

Ω =⋅

⋅+⋅=

+=≅ /phase

The total three-phase power loss is therefore 2.01 x 3 = 6.03 MW.

356

Load Flow

The use of the average voltage, real power and reactive power to calculate the loss returns a value of 6.03 MW, the accurate calculation returns a value of 6.14 MW. The error in using the approximation is less that 2%.

Tutorial Questions

9.1.1 A 140 kV three-phase transmission line has a reactance of 50 Ω per phase. If the line is to be operated

as a “flat line” at rated voltage (i.e. the sending voltage magnitude is equal to the received voltage magnitude), determine (a) the maximum possible steady state power flow and (b) the transmission angle δ if the real power flow is 100 MW.

9.1.2 A 220 kV, three-phase transmission line has a series reactance of 140 Ω per phase and negligible shunt admittance. Calculate the transmission angle δ and the reactive power flow in the line for a real power flow of 40 MW with the following voltage profiles:

Case a) 1.0puVV 21 ==

Case b) 1.0puV1.20pu,V 21 ==

9.1.3 A three-phase 140kV transmission line has the following parameters per phase: resistance = 9.10 Ω,

inductive reactance = j50.5 Ω, total shunt admittance = j334 µΩ-1. The sending end voltage is kept constant at 145 kV. The three-phase sending end power equals 120 MW at a power factor of 0.8 lagging.

Using the π model of the line calculate:

a) the current at each end of the line b) the voltage at the receiving end of the line c) the three-phase power at the receiving end of the line d) the efficiency of the transmission.

9.1.4 Show that the power loss in a transmission line can be estimated using the following equation:

2

22

LV

QPRP +=

where R is the series resistance of the line per phase P is the average of the sending and receiving powers per phase Q is the average of the sending and receiving reactive powers per phase V is the average of the sending and receiving voltage magnitudes per phase

Use the calculated real and reactive powers and voltages per phase at the sending and receiving ends of the transmission line in question 9.1.3. to estimate the power loss per phase and efficiency of the three-phase transmission.

357

Load Flow

9.2 The Load Flow Problem In the previous section, we have seen how the real power flowing over an inter-

connector is a function of the difference in voltage phase angle between the two ends of the inter-connector. The reactive power flow is a function of the difference in voltage magnitude between the two ends of the inter-connector.

In practice electrical power systems have many inter-connections and calculating the real and reactive power flows in the system is not a simple task. The purpose of load flow analysis is to determine the actual real and reactive power flow in a given system, and more importantly how to control these power flows. The primary aims and objectives of any load flow study are as follows:

1. Find the real and reactive power flows in the transmission lines based on certain

assumptions concerning the nature of the loads and the generators on the system.

2. Calculate the voltages on all bus-bars in the power system. 3. Check that no cable, transmission line, transformer, or generator is overloaded. 4. Find the optimum configuration of the power system to minimise transmission

losses. Any electrical engineer will immediately identify the load flow problem as an electrical circuits problem. The standard approach to solving such problems is to first represent the sources in the network as either voltage or current sources. Network equations are then written down using Kirchhoff’s Laws in which either the voltages or the currents take on the role of unknown quantities. The loads are represented by impedances, and if these are assumed known and constant, the resultant equations are linear and can be easily solved. In this problem the source e.m.f. is known as is the circuit impedances. In an electrical power system, the loads do not behave in a simple manner and cannot generally be represented by constant impedances. The e.m.f.s of each generator is not known and cannot be measured since it is internal to the generator. In the load flow problem we therefore have to establish a system of equations which describe the network in terms of quantities which are easily measured. These quantities are real powers, reactive powers and voltage magnitudes. A load flow study therefore involves network equations written in terms of voltages and power, not voltages and currents as in a classical circuits problem. This difference results in the load flow equations being non-linear. This eliminates the possibility of a

358

Load Flow

direct solution to the load flow equations in most cases. We can however always arrive at an iterative numerical solution to solve this system of non-linear equations. Load flow analysis of a power system containing hundreds of bus-bars and transmission lines is therefore a rather complex task always performed using a computer program and beyond the scope of the introductory text. It is possible however to consider relatively simple power systems to illustrate the techniques required for the iterative solution of the load flow problem. Consider the three bus-bar power system shown in Figure 9.10, a generator is connected to each of the first two buses and an electrical load is connected to the third bus. If the voltages in bus-bars 1 and 2 where known, then the real and reactive power transferred over the transmission line linking bus-bar 1 and bus-bar 2 could be calculated using the relationships derived in the previous section.

Figure 9.10 Simple three bus-bar power system

The real and reactive power demanded at the load bus-bar is known, as are the voltages at the generator bus-bars 1 and 2. The three transmission lines interconnecting the buses contain both resistance and reactance, therefore there must be losses in the system. The two generators must jointly supply the total load (including losses) whilst staying within their individual power limits. Each line must operate within its normal limits, i.e. δ the transmission angle across each line should not be too large. The voltage on the load bus-bar 3 must be within acceptable limits. No part of the power system must be over-loaded. The first step in performing the load flow analysis is to take the single line diagram in figure 9.10 and form the bus admittance matrix. This matrix contains all the information required to describe the power system.

359

Load Flow

9.2.1 The Bus Admittance Matrix In order to form the bus admittance matrix, the generators in the network are represented as current sources, the loads are represented as current sinks, and the transmission lines interconnecting each bus-bar are represented by their π equivalent circuit. This allows us to redraw the single line diagram shown in Figure 9.10 as the network shown in Figure 9.11. Normally we would represent each line using its series impedance Z and shunt admittance Y. In this case it is more convenient to convert the series impedance Z to its equivalent admittance Y.

22 XRjXR

jXRjXR

jXR1

jXR1

Z1Y

+−

=−−

⋅+

=+

== (9.17)

In this way we work with admittances only, having converted all impedances to admittances.

Figure 9.11 Admittance Diagram

In Figure 9.11, the current source at bus-bar 1 represents generator 1, the current source at bus-bar 2 represents generator 2, and the current sink on bus-bar 3 represents the load. The transmission line linking bus-bars 1 and 3, is represented by the three admittances Ya, Ye, and Yf, the line linking bus-bars 3 and 2 is represented by the three admittances Yb, Yg, and Yh, the line linking bus-bars 1 and 2 is represented by the three admittances Yc, Yd, and Yi.

360

Load Flow

In circuit analysis we are used to using Ohm’s Law to relate the voltage across an element to the current through the element and the impedance of the element: ZIV ⋅=The current can be calculated from

YVZVI ⋅== (9.18)

where Z1Y = is the admittance of the circuit element.

If we apply Kirchoff’s current law at each bus-bar in turn, sum of the currents flowing into a bus-bar must equal the sum of the currents flowing away from the bus-bar, the following bus-bar voltages are obtained:

( ) ( ) ( )( ) ( ) (( ) ( ) ( )gf3b23a133

ih2c12b322

ed1a31C211

YYVYVVYVVIYYVYVVYVVIYYVYVVYVVI

++−+−=−++−+−= )++−+−=

(9.19)

Rearranging the equations:

( )( )

( ) 3gfba2b1a3

3b2ihcb1c2

3a2c1edca1

VYYYYVYVYIVYVYYYYVYI

VYVYVYYYYI

++++−−=−−++++−=

−−+++= (9.20)

These equations can be written in matrix format:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 3

2

1

33

23

13

32

22

12

31

21

11

3

2

1

VVV

YYY

YYY

YYY

III

Bus Injector Current Vector

[Ibus]

Bus Admittance Matrix

[Ybus]

Bus Voltage Vector

[Vbus]

361

Load Flow

The bus admittance matrix is square, sparse and symmetrical. A diagonal element

kkY is called the self-admittance of bus k and is found by summing the primitive admittances of all lines and transformers connected to bus k, plus the admittances of all shunt connections between bus k and ground.

An off diagonal element is the mutual admittance of the line or transformer between buses k and m. It is equal to the negative of the admittance of the line or transformer between buses k and m. It is therefore zero if there is no connection from bus k to bus m.

kmY

The formation of the admittance, matrix for the power systems to be studied is the first stage of the load flow analysis. In should be noted that in deriving the bus admittance matrix we included the shunt admittance representing the capacitance to ground of the transmission lines; in practice this admittance is often neglected. This simplifies the formation of the bus admittance matrix without introducing a significant error.

Example 9.5

For the three bus-bar system shown below, form the bus admittance matrix.

Solution 9.5 In this question, the shunt admittance of the line is neglected. The impedances in the diagram are in per unit. As this is a three bus-bar power system, the bus admittance matrix will be a 3 x 3 matrix.

362

Load Flow

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

332313

322212

312111

bus YYYY

YYYYYY

It is easier to start with the off diagonal elements in the bus admittance matrix. An off diagonal element Ykm is the mutual admittance of the line or transformer connected between buses k and m. It is equal to the negative of the admittance of the line or transformer between buses k and m.

( )( )( )

( ) puj2)1(0.40.2j0.40.2

j0.40.2j0.40.2

j0.40.21

j0.40.21YY 222112 +−=

++−

=−−

+−

=+

−==

( )( )( )

( ) puj4)2(0.20.1j0.20.1

j0.20.1j0.20.1

j0.20.11

j0.20.11YY 223223 +−=

++−

=−−

+−

=+

−==

0YY 1331 ==

Elements on the main diagonal of the bus admittance matrix Ykk are the self admittance of bus k formed by summing all the primitive admittances connected to bus k. ( )j2-1-YY 1211 == ( ) ( ) ( )j63j42j21YYY 232122 −=−+−=−−= ( )j42YY 3233 −=−= The bus admittance matrix of this three bus-bar power system is therefore:

[ ]( ) ( ) ( )( ) ( ) (

( ) ( ) ( ) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+−−+−

+−−=

j42j420j42j63j21

0j21j21Ybus )

In practice an interconnected power system will have more than three busbars and the admittance matrix must be derived from the system single line diagram. As a power system will contains transformers an many different voltage levels, the load flow analysis will always be performed in per unit. It is therefore necessary to take the system line diagram and per-unit impedances and convert these to admittances on a common base before forming the bus admittance matrix. This is best illustrated by example.

363

Load Flow

Example 9.6

For the six bus power system shown below, derive the per unit bus admittance matrix. The reactance of each transmission line is 20 Ω, and the generators and transformers are rated as follows: Item MVA rating Voltage rating Impedance on rated MVA and Voltage G1 20 MVA 12 kV X = 1.2 pu G2 60 MVA 13.8 kV X = 1.4 pu G3 50 MVA 13.2 kV X = 1.4 pu T1 25 MVA 12/69 kV X = 0.08 pu T2 75 MVA 13.8/69 kV X = 0.18 pu T3 60 MVA 69/13.2 kV X = 0.14 pu T4 75 MVA 69/13.8 kV X = 0.16 pu Use a Sbase of 100 MVA and a voltage base of 12 kV in the circuit of generator 1.

Solution 9.6

We are told to use a Sbase of 100 MVA and a voltage base of 12kV at generator 1. As the transformer T1 has a voltage ratio of 12kV: 69kV, the voltage base on bus bar 5 is 69 kV (remember the voltage base changes in proportion to the turns ratio of each transformer). Working through the system the voltage base at each bus bar is found to be as follows: Bus No 1 2 3 4 5 6 Voltage Base 12 kV 13.8 kV 13.2 kV 13.8 kV 69 kV 69 kV

364

Load Flow

The generator impedances are given in the question, but as we represent the generator by a current source in the load flow model this information is not required. The impedance of each transformer is given on its MVA rating and kV rating. As the kV rating corresponds to the base voltages in the table above we only need to convert the given per unit impedance to our chosen MVA rating.

For transformer 1: pu0.3225

100(0.08)XT1 ==

For transformer 2: pu0.213375

100(0.16)XT2 ==

For transformer 3: pu0.233360

100(0.14)XT3 ==

For transformer 4: pu0.213375

100(0.16)XT4 ==

The impedance of each transmission line is given as 20 Ω per phase. To convert these to per unit we must find the base impedance at 69 kV and 100 MVA.

( ) Ω

2

47.6110100

1069SV

Z 6

3

base

2base

base =⋅

⋅==

The impedance of each transmission line in per unit is therefore pu0.4247.61

20= .

Now that we have the per unit impedance for each element in the network we can construct the admittance matrix using the rules outlined above. The bus admittance matrix is square, sparse and symmetrical. A diagonal element

kkY is called the self-admittance of bus k and is found by summing the primitive admittances of all lines and transformers connected to bus k, plus the admittances of all shunt connections between bus k and ground. An off diagonal element kmY is the mutual admittance of the line or transformer between buses k and m. It is equal to the negative of the admittance of the line or transformer between buses k and m. It is therefore zero if there is no connection from bus k to bus m. Using these rules:

365

Load Flow

-j13.736j4.762-j4.688--j4.286Y-Y--YY

-j12.575j4.762-j4.688--3.125Y-Y--YY

-j4.688-YY

-j4.286-YY

-j4.688-YY

-j3.125-YY

)( j4.762j0.42

1j0.42

1jX

1jX1

YY

j4.688j0.2133

1jX1

YY

j4.286j0.2333

1jX

1YY

j4.688j0.2133

1jX

1YY

j3.125j0.32

1jX

1YY

56463666

56525155

4644

3633

2522

1511

L2L16556

6446

T36336

T25225

T15115

parallel in are lines ion transmiss twoThe

======

========

=−−=−−==

=−=−==

=−=−==

=−=−==

=−=−==

4

The bus admittance matrix can now be formed:

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

=

j13.736j4.762j4.688j4.28600j4.762j12.57500j4.688j3.125j4.6880j4.688000j4.28600j4.28600

0j4.68800j4.68800j3.125000j3.126

Ybus

Note that the bus admittance matrix is symmetrical about the main diagonal; this is a useful check to ensure that the matrix is correct. Students should ensure that they understand the formation of the bus admittance matrix.

366

Load Flow

Tutorial Questions

9.2.1 The impedances of the transmission lines in a four bus-bar power system are listed below.

Line Impedance 1 – 2 0.20 + j0.6 1 – 3 0.10 + j0.3 2 – 3 0.30 + j0.9 2 – 4 0.10 + j0.3 3 – 4 0.20 + j0.6

Form the bus-admittance matrix.

9.2.2 The single line diagram of power system is shown below. The reactance of the transmission line is 30 Ω. The generators and transformers are rated as follows:

Item MVA rating Voltage rating Impedance on rated MVA and Voltage

G1 20 MVA 12 kV X = 1.20 pu G2 60 MVA 13.8 kV X = 1.40 pu G3 20 MVA 13.2 kV X = 1.20 pu T1 25 MVA 12/69 kV X = 0.08 pu T2 60 MVA 69/13.8 kV X = 0.14 pu T3 75 MVA 13.2/69 kV X = 0.14 pu

Using an Sbase of 50 MVA and a voltage base of 12 kV in the circuit of generator 1, form the bus admittance matrix.

367

Load Flow

The Jacobi and Gauss-Seidel Iterative Methods Consider the following set of simultaneous equations:

5.00.20.10.12.03.04.01.10.11.08.06.0

=++=++=++

zyxzyxzyx

To solve these equations using an iterative procedure we proceed as follows. Inspect the equations and choose the equation with the largest coefficient multiplying the variable x. In this case this is the second equation, which is divided throughout by the coefficient of x, this gives:

1.12.0

1.13.0

1.14.0

1.11.1

=++ zxx

or 18182.027273.03664.0 =++ zyx Now inspect the remaining equations to find the one with the highest coefficient multiplying the variable y. In this case it is the first equation which is divided throughout by this coefficient to give:

8.0

18.01.0

8.08.0

8.06.0

=++ zyx

or 25.1125.075.0 =++ zyx The third remaining equation is now used, the equation is divided by the coefficient of z which yields:

25.0

22

21

21

=++ zyx

or 25.05.05.0 =++ zyx Writing these three new equations together:

368

Load Flow

25.05.05.025.1125.075.018182.027273.036364.0

=++=++=++

zyxzyxzyx

These three equations can be solved using the Jacobi Iteration scheme. An initial guess is made for the solution of the set of equations (x0, y0, z0) and a set of updated estimates are found (x1, y1, z1) using the following equations in turn:

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )nynxnz

nznxnynznynx

5.05.025.01125.075.025.11

27273.036364.018182.01

−−=+−−=+

−−=+

When the new values of x(n+1), y(n+1), z(n+1) are obtained these are then used to calculate x(n+2), y(n+2), z(n+2) and so on. The solution continues until no further improvement in the values returned is obtained. Using this technique with an initial guess for the solution of (0,0,0) it takes 55 iterations for the solution to converge as shown below.

369

Load Flow

In practice the Jacobi Iteration method is not used because the rate of convergence may be improved by using the new values as soon as they become available. The previous equations then become:

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )15.015.025.01

125.0175.025.1127273.036364.018182.01

+−+−=+−+−=+

−−=+

nynxnznznxny

nznynx

Using this improved technique with an initial guess for the solution of (0,0,0) it takes only 11 iterations for the solution to converge as shown below

This improved iterative scheme known as the “Gauss Siedal” method converges to a solution 5 times faster that the simple Jacobi method.

It is important to realize that the convergence of the iterative scheme is not guaranteed, a poor choice of the initial guess may prevent the scheme converging on a solution. The advantage of the Jacobi and Gauss Siedal methods is that they are straight forward to program on a computer and no not require a large quantity of computer memory. The Gauss Seidel method was the first to be applied to the load flow problem.

370

Load Flow

Tutorial Questions

9.2.3 Find one root (solution) to the following quadratic equation using the Gauss-Seidel iteration scheme.

Use an initial guess of x(0) = 1. x2 – 6x + 2 = 0

Use the quadratic formula 2a

4acbbx2 −±−

= to find both roots of the equation. Why does the

Gauss-Seidel scheme return only one solution ? 9.2.4 Solve the following set of equations in x and y

08.109.10.3

2 =−+

=+−

xyxy

using the Gauss-Seidel iteration scheme. Use an initial guess of x(0) = 1 and y(0) = 1.

Classification of bus-bars in a load flow study In a load flow study, each bus bar can be classified as being one of three types.

1. Load Bus or P-Q bus At the load bus k, the net real and reactive power demands are specified or scheduled, i.e.

( ) ( dkgkdkgk

spk

spk

spk

QQjPPjQPS

−+−=+=

) (9.22)

where:

k.busatdemandpowerreactiveandrealQ,P

k.busatgenerationpowerreactiveandrealQ,P.kbusatpowerreactiveandrealspecifiedQ,P

dkdr

gkgk

spk

spk

=

==

2. Generator Losses or P-V bus

Generators are connected to P-V buses. At these buses, the real power generation and the voltage magnitude are specified, i.e.

371

Load Flow

spmm

dmGmspm

VV

PPP

=

−= (9.23)

Vm

sp is the specified voltage magnitude at bus m. The voltage magnitude at a P-V bus is fixed at its specified or controlled values provided that the reactive power generated is within operational limits. Otherwise, the voltage is allowed to seek its proper level, and the bus is converted into a P-Q bus by specifying the reactive power at the bus.

3. Slack Bus or Swing Bus

As the system losses are not known precisely before the load flow analysis is performed, it is not possible to specify real power injected at every bus. Hence the real power of one of the generator buses is allowed to swing, and it supplies the differences between the scheduled real power generation and the sum of all loads and system loses. The swing bus voltage magnitude is specified and its voltage phase angle is usually chosen as the system reference and set equal to zero, i.e.

0s

spss

0VV

=δ=

(9.24)

Figure 9.12 Four Bus Power System

Consider the four bus power system shown in Figure 9.12. All the load bus bars in the system are P-Q buses. The power demanded at each load bus-bar is specified and the voltage at each load bus is unknown. There are two generator buses on the system,

372

Load Flow

one of them is a P-V bus where the generator power and terminal voltage is specified. The second generator bus is the slack or swing bus, its voltage is specified but its power is unknown. These known and unknown quantities are summarised in the table below.

Bus No

Type Known quantities Unknown quantities

1 Slack δ,V1 11 Q,P 2 Load

22 Q,P 22 δ,V 3 Load

33 Q,P 33 δ,V 4 Generator

44 , VP 44 δ,Q

At the end of the load flow study all the unknown quantities in this table will have been found. Equations to be solved in the load flow study

We have seen that the power system can be represented by following the matrix equation:

[ ] [ ] [ ]busbusbus VYI =

At bus k, the phase current kI injected into the bus is given by:

kNkN3k22k11k IVY.....YVYVY =+++ (9.25)

where:

system thein buses ofnumber total

matrix. admittance bus thefromelement

m busat tagephasor vol

===

NYV

km

m

Impedance The specified, or scheduled complex power injection at bus k may be expressed in terms of the current injected into the bus and the bus voltage phasor as follows:

*kk

spk

spk

spk IVjQPS =+= (9.26)

Rearranging this equation:

373

Load Flow

K

spspk*

kV

jQPI

+=

form which the current injected at bus k is found to be:

*K

spspk

kV

jQPI

−= (9.27)

Substituting this into equation (9.25) gives kI in terms of the bus-bar voltages and the elements of the bus admittance matrix.

*k

spk

spk

NkN3k32k21k1V

jQPV....YVYVYVY

−=+++ (9.28)

Moving all terms except kkk YV to the RHS and dividing the resultant by kkY gives

the following expression for the phasor voltage . kV

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡+++++−

−= NNkmkm22k11k*

k

spk

spk

kkk YY...VY...VYVY

VjQP

Y1V (9.29)

for km ≠

Remember the swing bus voltage is to be taken as the reference phasor, its voltage magnitude is specified and its phase angle is set to zero. The above equation for the voltage at bus k is used to write a system of (N-1) simultaneous algebraic equations relating the phasor voltage at the individual buses with the corresponding power injections at the bus and phasor voltages at all the buses. These equations are coupled through the elements of the bus admittance matrix. For example, consider the four bus-bar power system shown in Figure 9.12. The voltage equations to be solved in the load flow analysis of this system are as follows:

374

Load Flow

( )

( )

( )⎥⎦

⎤⎢⎣

⎡⋅+⋅+⋅−

−=

⎥⎦

⎤⎢⎣

⎡⋅+⋅+⋅−

−=

⎥⎦

⎤⎢⎣

⎡⋅+⋅+⋅−

−=

=

334224114*4

SP4

SP4

444

443223113*3

SP3

SP3

333

442332112*2

SP2

SP2

222

o11

VYVYVYV

jQPY1V

VYVYVYV

jQPY1V

VYVYVYV

jQPY1V

0VV

The voltage on the generator bus bar 1 is known since this generator is the slack bus, its voltage magnitude is specified and its phase angle set to zero. The voltage equations for the load bus bars 2 and 3 use elements from the bus admittance matrix and the specified real and reactive power at each bus. The voltage equation for the voltage on the generator bus bar 4 cannot be solved immediately since the reactive power on this bus is unknown (remember that on a generator bus the voltage magnitude and the reactive power is specified). Rearranging equation (9.28), the reactive power of the generator can be estimated:

( ) spk

spkNkN3k32k21k1k

k

spk

spk

NkN3k32k21k1

*k

spk

spk

NkN3k32k21k1

jQPV....YVYVYVYV

V

jQPV....YVYVYVY

V

jQPV....YVYVYVY

+=+++

+=+++

−=+++

For a load bus (sink) the reactive power would be given by the imaginary part of the expression on the LHS of the equation above. For a generator (source) the reactive power generated with therefore be the negative of the imaginary part of the expression of the LHS of the above equation, i.e. the calculated reactive power on the generator bus is given by:

( )NkN3k32k21k1kcalcK V....YVYVYVYVQ +++−= Im

This value is substituted into the equation for V4. The load flow problem can be solved by finding the solution to these three simultaneous equations for the voltages V2, V3, and V4. These equations can be solved using the Gauss-Seidel iterative technique.

375

Load Flow

Gauss-Seidel Solution of the Load Flow Problem The Gauss-Seidel method is an iterative technique for solving a system of non linear algebraic equations, such as that formed by the (N-1) voltage equations derived for the load flow analysis of an N bus power system. Using this technique, the phasor voltage at a bus is found by using the latest computed values of the phasor voltage at the other buses. This is identical to the procedure used to solve the three simultaneous equations in the example above. The Gauss-Seidel method applied to the load flow problem, produces the following general result for the voltage at each bus k:

( )( )⎥⎥⎦

⎤

⎢⎢⎣

⎡−

−= ∑

=

β+w

k#m,1mmkmi

k

spk

spk

kk

1ik VY

V

jQPY

1V (9.30)

kmkm

for1iβforiβ

<>

+==

where i represents the iteration count.

In the solution of these equations, the most recently computed values for each voltage are used. The solution procedure by passes the swing because the phasor voltage of the swing bus is already known. For a P-Q or load bus, the procedure is applied directly to obtain an improved estimate of the phasor voltage in as much as and are both given. For a P-V or generator bus, the bus voltage is specified; hence, is not directly available. Therefore an estimate is computed by using the current estimates of the phasor voltages.

spkP

spkQ kV

spkQ calc

kQ

( )[ ]nknmkm22k11k

*km

calck VY...VY...VYVYVIQ +++−= (9.31)

This value of reactive power is used to replace sp

kQ and then is recalculated. Then

the magnitude of kV

kV is reset to its specified value spkV , but the new value of its phase

angle is retained. The Gauss-Seidel method has the attractions of simplicity, comparatively good performance, and non-storage of previous values. It has a very reliable convergence characteristic, but the rate of convergence is slow.

The iterative process is said to have converged when the process no longer yields any improvement on the solution. At this point, the phasor voltages at all the busses have been found and may be used to drive other information about the steady-state operating characteristics of the power system. For example, the real and reactive

376

Load Flow

powers flowing through any transformer or transmission line, with series impedance Z and shunt admittance Y, connected between buses k and m may be expressed as:

*

kmk

kkmkmkm V2Y

ZVVVjQPS ⎟

⎠⎞

⎜⎝⎛ +

−=+= (9.32)

Having read the above text, the reader may feel that the load flow problem and its solution is a very complex task. By approaching the problem in a systematic manner the calculations become very straight forward, this is best illustrated by an example.

Example 9.7

For the power system shown below, bus 1 is selected as the slack or swing bus and its voltage is set to V1=(1.0 + j0.0) pu. The chosen power base is 100 MVA. Generator 2 delivers a real power of 0.75 pu at a voltage of 1.02 pu. The loads on buses 3 and 4 are Sd3 = (0.40 + j 0.30) pu and Sd4 = (0.80 + j 0.60) pu respectively. The impedance parameters for the transmission lines referred to the given power base and a voltage base of 115 kV are: Line 1 - 2 Z = (0.01 + j0.05) pu Y = j0.30 pu Line 2 – 3 Z = (0.03 + j0.15) pu Y = j0.90 pu The transformer is connected between buses 3 and 4, and its reactance is 0.10 pu. From a flat start, perform one iteration of the Gauss-Seidel iterative technique to find the voltages at buses 2, 3, and 4.

Solution 9.7

First we need to calculate the elements in the bus admittance matrix as these are required to form the iterative voltage equations. All impedances and admittances

377

Load Flow

parameters are given in per unit. Using the rules for building the bus admittance matrix the following are calculated.

j10YY

85.416.0j10j0.45j6.411.28Y2

j0.9YY

78.425.56j0.45j0.15j6.411.28j19.233.8462

j0.92

j0.3YYY

78.619.46j0.15j19.233.8462

j0.3YY

j10j0.1

1YY

101.36.54j6.411.28j0.150.031

YY

0YY

0YY

0YY

101.319.61j19.233.8460.050.01

j0.05)(0.01j0.050.011

YY

3444

o342333

o232122

o1211

4334

o3223

4224

4114

3113

o222112

−=−=

−∠=−+−=−+−=

−∠=++−+−=++−−=

−∠=+−=+−=

=−

==

∠=+−=+−

==

======

∠=+−=+

−−=

+−

==

As bus 1 is the swing bus the voltage on bus 1 is known, V1 = 1.0∠0o. The voltage on bus 2 is found from the following equation iterative equation

⎥⎦

⎤⎢⎣

⎡⋅−⋅−⋅−

−= (0)

424(0)323121(0)*

2

22

22

12 VYVYVY

VjQP

Y1V

where o(0)

4(0)3

o(0)2

o1 01.0VV,01.02V,01.0V ∠==∠=∠=

A flat start is where the voltage on the swing bus is set at 1 pu and an angle of zero degrees, the voltage on the generator at bus 2 is set to its specified magnitude at an angle of zero degrees, and the completely unknown voltages on the load bus bars are set to 1 pu at an angle of zero degrees as shown above. The iterative equation for V2 cannot be solved since the reactive power at this generator bus is unknown (remember only real power and voltage magnitude is specified at a generator bus). The reactive power at the generator bus is calculated using equation (9.31).

378

Load Flow

( )[ ]

( ) ( ) ( ) ( ) (( ) ( )

pu0.62

001.0101.36.5401.0278.425.5601.0101.319.61

01.02Im

VYVYVYVY*VImQ

oo

o0oo*o

42432322212122

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+∠⋅∠+

∠⋅−∠+∠⋅∠∠−=

)⋅+⋅+⋅+⋅−=

This value of reactive power is now substituted into the iterative equation for V2:

( )( ) (

( ) (o

oo

oo*o

o

(0)424

(0)323121(0)*

2

22

22

(1)2

1.61.007

01.0101.36.54

01.0101.319.6101.02j0.620.75

78.425.561

VYVYVYV

jQPY1V

∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠⋅∠−

∠⋅∠−∠

+

−∠=

⎥⎦

⎤⎢⎣

⎡⋅−⋅−⋅−

−=

)

)

Thus the 1st iteration returns a value of V2 equal to 1.007 ∠1.6o

We can now perform the 1st iteration for the voltage on bus bar 3. The scheduled real and reactive powers on bus 3 are calculated using equation (9.22).

pu0.3QQQ

pu0.4PPP

D3G33

D3G33

−=−=−=−=

The iterative equation for the voltage on bus 3 is:

( )( ) (

( ) ( )pu0.81.009

01.0j10

1.61.007101.36.54001.0j0.30.4

85.416.01

VYVYVYV

jQPY1V

o

o

oo*o

o

(0)434

(1)232131(0)

3

33

33

(1)3

−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠⋅−

∠⋅∠−−∠

+−

−∠=

⎥⎦

⎤⎢⎣

⎡⋅−⋅−⋅−

−=

)

Notice that the updated value for V2 is used in this calculation i.e. 1.007 ∠1.6o and not the initial value of 1.0 ∠0o. It is a requirement of the Gauss-Seidel technique that updated values of voltages are used as soon as they become available. In a similar way the voltage at bus bar 4 can be calculated.

379

Load Flow

pu0.6QQQ

pu0.8PPP

D4G44

D4G44

−=−=−=−=

( )

( ) ( )o

o

ojj

j

VYVYVYV

jQPY

V

7.5953.0

8.0009.1100000.1

6.08.010

1

1

*

)1(343

)1(242141)*0(

4

44

44

)1(4

−∠=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−∠⋅−−−

∠

+−−

=

⎥⎦

⎤⎢⎣

⎡⋅−⋅−⋅−

−=

After the first iteration the calculated bus bar voltages are therefore:

o(1)4

o(1)3

o(1)2

o(1)1

5.70.953V

0.81.009V1.61.007V

01.0V

−∠=

−∠=

∠=

∠=

These updated voltages are used as the input to the next iteration and the whole process is repeated until additional iterations fail to produce a change in the calculated voltages. At this point the solution is said to have converged and the analysis is almost complete. Once all the bus bar voltages are found, the currents in the net work can be calculated and the powers flowing along all transmission lines can be found. Transmission losses can then be calculated.

380

Load Flow

Tutorial Questions

9.2.5 For the two busbar power system shown below:

i) Form the bus admittance matrix (all impedances shown are in per-unit) ii) If the load on bus 2 is S = (0.5 + j0.3) derive an iterative expression using the Gauss-

Seidel method for the voltage bus 2 iii) Find the voltage on bus 2 iv) Find the real and reactive powers on the generator bus 1 v) Estimate the efficiency of the transmission line.

9.2.6 For the three busbar power system shown below:

i) Form the bus admittance matrix (all impedances shown are in per-unit) ii) For the given loads on buses 2 and 3 find the voltages on buses 2 and 3 using the

Gauss-Seidel method. Assume a flat start. iii) Find the apparent power injected into the system by the generator on bus 1

9.2.7 The three busbar power system of question 9.2.6 is modified by the addition of a transmission line

between buses 2 and 3 of series impedance Z = (0.1 + j0.3). i) Modify the bus admittance matrix to allow for this extra line ii) Derive an iterative expression for the voltages on buses 2 and 3.

Perform one iteration of the Gauss-Seidel method to update the voltages on buses 2 and 3 from a flat start.

381

Load Flow

Final Comments on Load Flow Analysis Many students find the concepts introduced in this chapter difficult to grasp at the first attempt for the following reasons 1. The load flow problem is not a difficult problem to solve, but it requires the

power system description to be set up in a very specific way using the bus-admittance matrix. The rules for constructing the matrix are easy to remember and with practice become routine.

2. The need to specify real and reactive powers on load bus bars, and voltage and

power on generator bus bars is very different from classical circuit analysis where loads are represented by impedances supplied at specified voltages. This results in the load flow problem being non-linear.

3. As the load flow problem is non-linear iterative techniques must be used to

solve for the system voltages. A common technique described in the text is the Gauss-Seidel Method. Other techniques such as the Newton-Raphson algorithm and Stott’s algorithm provide faster solutions but are more mathematically intensive and beyond the scope of this text.

4. In practice the load flow calculation will involve many hundreds, perhaps

thousands of bus bars in an interconnected power system. The load flow calculation will then be performed using a computer program. This chapter has provided a introduction to the terminology and techniques used in a load flow study.

382

Load Flow

Chapter Summary

To understand the term load flow analysis as applied to an interconnected power system The load flow calculation is used to ensure that the steady state voltages and currents flowing the interconnected network are within the continuous ratings of the equipment for both normal and emergency operating conditions. Such calculations are performed on computer as hand calculations are too time consuming and impractical for large systems. To understand the factors which influence the flow of real and reactive powers over an inter-connector The power flow between two bus bars is a function of the transmission angle δ across the transmission line and is not a function of voltage magnitude. Real power flows from leading

to lagging phasor. For a loss less line x

sinVVPPP 21

21

δ=== . The reactive power flow

is a function of the difference in voltage magnitude. The larger the difference in magnitude, the stronger the reactive power flow. Reactive power flows from the high voltage bus-bar to the low voltage bus-bar. The reactive power flow at 1 into the line is

xcosδVVV

Q 212

11

−= and the reactive power out of the line at 2 is

xVcosδVV

Q2

2212

−=

To understand that the flow of both real power and reactive power contribute to the real power losses in the transmission network The power loss in a transmission link can be estimated from the line average power,

averages reactive power, and average voltage thus 2ave

2ave

2ave2

ΩV

QPRIRP

+=≅

To understand the principle aims and objectives of a load flow study

383

Load Flow

To find the real and reactive power flows in the transmission lines based on certain assumptions concerning the nature of the loads and the generators on the system. To calculate the voltages on all bus-bars in the power system. Check that no cable, transmission line, transformer, or generator is overloaded. To find the optimum configuration of the power system to minimise transmission losses.

To understand that the load flow problem is non-linear and cannot be solved by classical circuit analysis A load flow study network equations written in terms of voltages and power, not voltages and currents as in a classical circuits problem. This difference results in the load flow equations being non-linear. This eliminates the possibility of a direct analytical solution to the load flow equations in most cases. To understand how to construct the bus admittance matrix for a simple power system The bus admittance matrix is square, sparse and symmetrical. A diagonal element kkY is called the self-admittance of bus k and is found by summing the primitive admittances of all lines and transformers connected to bus k, plus the admittances of all shunt connections between bus k and ground. An off diagonal element kmY is the mutual admittance of the line or transformer between buses k and m. It is equal to the negative of the admittance of the line or transformer between buses k and m. It is therefore zero if there is no connection from bus k to bus m.

To understand the Gauss-Seidel Method as applied to the iterative solution of simultaneous equations The Gauss-Seidel Method is used to solve sets of simultaneous equations. The equations are rewritten in the form .......)n(Y)n(X)1n(X ++=+ , one equation for each unknown. Initial guesses are made for each unknown which are then updated by substituting these values in the iterative equations. Convergence is not guaranteed and an intelligent first guess is required.

To understand the three different types of bus bar in a load flow study. To understand which quantities are known and which are unknown for each type at the beginnings of the calculation At a Load Bus or P-Q bus the net real and reactive power demands are specified and the voltage magnitude and phase angle are unknown. Generators are connected to P-V buses at which the real power generation and voltage magnitude are specified; the reactive power and voltage angle are unknown. The swing bus voltage magnitude is specified and its

384

Load Flow

voltage phase angle is usually chosen as the system reference and set equal to zero; the power on the swing bus is unknown.

To understand how to calculate the voltages in a power system using the Gauss-Seidel Method The Gauss-Seidel method applied to the load flow problem, produces the following general result for the voltage at each bus k:

( )( )⎥⎥⎦

⎤

⎢⎢⎣

⎡−

−= ∑

=

β+w

k#m,1mmkmi

k

spk

spk

kk

1ik VY

V

jQPY

1V

kmkm

for1ifori

<>

+==

ββ

where i represents the iteration count.

To understand how to calculate the real and reactive powers on a generator bus bar once all the bus bar voltages have been found. For a generator the reactive power generated is given by:

( )NkN3k32k21k1kcalcK V....YVYVYVYVQ +++−= Im

385

Load Flow

Chapter 9 Tutorial Solutions 9.1.1 (a) The voltage given in the question is the line to line voltage. This calculation must be

performed working per phase.

kV80.8kV3

1403

VV Lph ===

For flat line operation kV80VV 21 == per phase.

Power transfer is sinδXVV

P 21= per phase

Maximum power transfer occurs when 1sinδ = , i.e. o90δ =

( ) ( ) MW/ph130

501080.81080.8

XVV

P33

21max =

⋅⋅==

Maximum three-phase power is 3 x 130 MW = 390 MW (b) If the power flow is 100 MW, then the power per phase is given by

MW/ph333.333

100=

Power transfer is sinδXVV

P 21= per phase

( ) ( )sinδ50

1080.81080.81033.3333

6 ⋅⋅=⋅

solving for δ yields 14.78o

The transmission angle is 14.78o

386

Load Flow

9.1.2

The voltage profile is given in per unit, so perform all calculations in per unit.

Choose 220 kV as Vbase, 40 MVA as Sbase and workout Zbase

( ) ( ) Ω1210104010220

SV

Z 6

23

base

2base

base =⋅⋅

==

The impedance of the line in per unit is therefore pu.11501210140

=ΩΩ

Case (a) pu1.0V1 = and pu1.0V2 =

pu1.04040MW40P ===

Power transfer is sinδXVV

P 21= per phase

( ) ( )sinδ

0.1151.01.01.0 ⋅

=

solving for the transmission angle δ yields 6.6o

The reactive power at the sending end is:

( ) ( ) ( )

pu0.0570.115

cos6.61.01.01.0X

cosδVVVQ

o2

212

11

=

⋅⋅−=

⋅−=

The reactive power at the receiving end is

( ) ( ) ( )

0.057pu0.115

1.0cos6.61.01.0X

VcosδVVQ

2o

2221

2

−=

−⋅=

−⋅=

Case (b) pu1.2V1 = and pu1.0V2 =

pu1.04040MW40P ===

387

Load Flow

Power transfer is sinδXVV

P 21= per phase

( ) ( )sinδ

0.1151.01.21.0 ⋅

=

solving for the transmission angle δ yields 5.49o

The reactive power at the sending end is:

( ) ( ) ( )

pu2.130.115

cos5.491.01.21.0X

cosδVVVQ

o2

212

11

=

⋅⋅−=

⋅−=

The reactive power at the receiving end is

( ) ( ) ( )

pu 1.690.115

1.0cos5.491.01.2X

VcosδVVQ

2o

2221

2

+=

−⋅=

−⋅=

The results for cases (a) and (b) are summarized in the table below:

1V 2V δ P1 P2 Q1 Q2

Case 1 1.0 1.0 6.60 1.0 1.0 0.057 -0.057 Case 2 1.2 1.0 5.49 1.0 1.0 2.13 1.69

388

Load Flow

The case (a) the reactive power required to energize the series impedance of the line comes from both ends of the line. In case (b) there is a strong reactive power flow from bus bar 1 to bus bar 2. This example clearly shows that raising the sending end voltage increases the reactive power flow in the system.

9.1.3

This problem is very similar to Example 9.4. Instead of the receiving end conditions being specified, the sending end is specified. All information should be transferred to the per phase equivalent circuit and the voltages and currents are then calculated using straight forward circuit analysis.

Sending voltage = 145 kV line to line or kV83.73

145= per phase.

Sending three phase power is 120 MW at 0.8 pf

Sending three phase apparent power is MVA1500.8120

=

Sending three phase reactive power is MVA90120150 22 =−

Per phase values P = 40 MW Q = 30 MVAr S = 40 MVA The complex power per phase is ( )MVAj3040jQPS 111 +=+=

389

Load Flow

Taking V1 as the reference phasor, ( )kVj0.083.7V1 += The complex power is also , hence the complex conjugate of the current can be calculated, then the current I

*111 IVS ⋅=

1 can be found:

( ) ( )

( )Aj358477I

Aj3584771010

83.7j4040

VSI

1

3

6

1

1*1

−=

+=+

==

The current flowing into the shunt capacitance at the sending end can be found from the shunt admittance and the sending end line voltage: Aj13.9710j1671083.7YVI 63

11sh1 =⋅×⋅=⋅= −

The line current can now be calculated by applying Kirchhoff’s 1st Law 11shL III =+

( ) ( )Aj371477j13.97j358477III sh11L −=−−=−= The voltage drop across the series impedance of the line can now be calculated

( )( ) ( )

( ) kVj207122307518735j3376j240884340

j50.59.10j371477jXRI∆V LL

+=+−+=

+⋅−=+⋅=

The sending voltage is ∆VVV 21 += , hence we can write

( )Vj2071260625

j207122307583700∆VVV 12

−=−−=

−=

It is convenient to calculate the magnitude of the received voltage:

( ) ( ) kV642071260625V 222 =+= per phase

The current flowing into the shunt capacitance at the receiving end can be found from the shunt admittance and the receiving end line voltage: ( ) ( )Aj10.1244.310j1672071260625YVI 6

22sh2 +=⋅×−=⋅= −j

390

Load Flow

The received current I2 can now be calculated by applying Kirchhoff’s 1st Law

2sh2L III += ( ) ( ) ( )Aj360.8473.5j10.123.45j371477III sh2L2 −=+−−=== As V2 and I2 are now known the complex power S2 can be determined:

( ) (( )( ) VA10j1236.17

107.47j9.80j21.828.7j360.8473.5j2071260625

IVS

6

6

*222

⋅+=

⋅+−+=

+⋅−=⋅=

)

This is the complex power per phase. The three-phase real power is therefore MW108.5MW6.1733 =× and the three-phase reactive power is MVAr36MVAr123 =×

The sending real power is 120 MW and the receiving power is 108.5 MW. The efficiency of the transmission is therefore:

0.90120

108.5PinPoη === or 90%.

Summary Sending current = (477 – j 358) A per phase Sending voltage = 83.7 kV per phase = 145 kV line to line Sending complex power = (120 +j 90) MVAr (three phase) Received current = (474 – j 360) A per phase Received voltage = 64 kV per phase = 110 kV line to line Received complex power =(108.5 +j 36) MVA (three phase)

9.1.4

The equation for the approximate power loss in a transmission line was derived in the main text. Using the summary table at the end of the solution to 9.1.3 the average per phase real power, reactive power and voltage can be determined.

391

Load Flow

( )

( )

( ) kV73.86483.721V

MVAr21.0012.030.021Q

MW28.0836.174021P

ave

ave

ave

=+=

=+=

=+=

( ) ( )( )

MW/phase3.151073.8

1021.001028.089.1V

QPRlossPower 23

2626

2

22

=⋅

⋅+⋅=

+=

Total three phase loss is 3 x 3.15 MW = 9.45 MW.

92%or0.92120

9.45120Efficiency =−

=

The approximate calculation is seen to give a reasonable estimate of the power loss in the transmission line.

9.2.1

From the given line impedances a single line diagram of the power system can be drawn. This can aid the formation of the bus admittance matrix. It is easier to form the off diagonal elements first, i.e. the mutual admittances. Ykn is the negative of the admittance connecting bus k to bus n.

392

Load Flow

j00YY

j1.50.5j0.160.2

1YY

j3.01.0j0.30.1

1YY

j3.01.0j0.30.1

1YY

j1.00.333j0.90.3

1YY

j1.50.5j0.60.2

1YY

1441

4334

4224

3113

3223

2112

+==

+−=+

−==

+−=+

−==

+−=+

−==

+−=+

−==

+−=+

−==

The main diagonal elements can now be found, i.e. the self admittances. Ykk is the sum of all the primitive admittances connected to bus k.

j4.51.5j1.5)(0.5j3.0)(1.0j0)(0YYYYj5.51.833j1.5)(0.5j1.0)(0.333j3.0)(1.0YYYYj5.51.833j3.0)(1.0j1.0)(0.333j1.5)(0.5YYYY

j4.51.5j0)(0j3.0)(1.0j1.5)(0.5YYYY

43424144

34323133

24232122

14131211

−=−+−++=−−−=−=−+−+−=−−−=−=−+−+−=−−−=

−=++−+−=−−−=

The bus admittance matrix can now be formed

( ) ( ) ( ) ( )( ) ( ) ( ) (( ) ( ) ( ) (( ) ( ) ( ) ( ) ⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+−+−++−−+−+−+−+−−+−++−+−−

=

j4.51.5j1.50.5j3.01.0j00j1.50.5j5.51.833j1.00.333j3.01.0j3.01.0j1.00.333j5.51.833j1.50.5

j00j3.01.0j1.50.5j4.51.5

Ybus

))

9.2.2

We are told to use an Sbase of 50 MVA and a voltage base of 12 kV at generator 1. Redraw the system showing the voltage bases in each part of the network. These are found using the transformer turns ratio.

393

Load Flow

The impedance of each transformer is given on its own MVA rating and kV rating. As the kV rating corresponds to the base voltages in the diagram above we need only convert the given per unit impedance to our chosen MVA base.

For transformer 1: pu0.162550(0.08)XT1 =⎟

⎠⎞

⎜⎝⎛=

For transformer 2: pu0.11676050(0.14)XT2 =⎟

⎠⎞

⎜⎝⎛=

For transformer 3: pu0.10667550(0.16)XT3 =⎟

⎠⎞

⎜⎝⎛=

The impedance of the transmission line is given as 30 Ω per phase. To convert to per unit find base impedance at 69 kV and 50 MVA.

( ) ( ) Ω95.22

10501069

SV

Z 6

3

base

2base

base =⋅⋅

==

The impedance of the transmission line in per unit is therefore pu0.31595.22

30= .

Now that the per unit impedance of every element in the network is known we construct the bus admittance matrix. Starting with the off diagonal elements, i.e. the mutual admittances between buses.

394

Load Flow

j3.174j0.315

1YY

j9.375j0.1066

1YY

j8.57j0.1167

1YY

j6.25j0.16

1YY

5445

4334

5225

4114

=−==

=−==

=−==

=−==

The self admittance of each bus can now be calculated

j11.74j3.174j8.57YYYj18.8j3.174j9.375j6.25YYYY

j9.375YYj8.57YYj6.25YY

545255

45434144

3433

2522

1411

−=−−=−−=−=−−−=−−−=

−=−=−=−=−=−=

The bus admittance matrix can now be formed.

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−

=

j11.72j3.1740j8.570j3.174j18.8j9.3750j6.25

0j9.375j9.37500j8.5700j8.570

0j6.2500j6.25

Ybus

395

Load Flow

9.2.3

To solve the equation using an iterative approach it must be written in an alternative form.

can be written as 026xx 2 =+− 2x61

31x +=

The iterative scheme to solve this equation is therefore:

2(n)1)(n x

61

31x +=+

Initial guess 1x(0) =

Iteration 1 ( ) 0.5161

31x(1) 2 =+=

Iteration 2 ( ) 0.3750.561

31x(2) 2 =+=

Iteration 3 ( ) 0.35680.37561

31x(3) 2 =+=

Iteration 4 ( ) 0.35340.356861

31x(4) 2 =+=

To two decimal places the solution is 0.35 The quadratic formula

2

2862

214662a

4acbbx22 ±

=⋅⋅−±

=−±−

=

gives the solutions as x = 0.354 and x = 5.64 The Gauss-Seidel iterative technique finds the root closest to the initial guess. A guess close to 5 would find the 2nd root of the equation. The success of the Gauss-Seidel method in finding the roots of the equation is not guaranteed. An intelligent initial guess is usually required to aid the convergence towards a solution.

396

Load Flow

9.2.4 The equations to be solved are:

01.8xy01.93xy

2 =−+

=+−

These equations can be rewritten as:

2x1.8y

0.63333yx

−=

+=

This suggest the following iterative scheme:

2

1)(n1)(n

(n)1)(n

x1.8y

0.63333

yx

++

+

−=

+=

Using the initial guess of 1x(0) = and 1x(y) = , the solution proceeds as follows:

( )

( )

( )

9045.0)0463.0(8.1

9436.06333.03

9311.0

9311.0)9321.0(8.1

9321.06333.03

8975.08975.09500.08.1

9500.06333.03

9502.0

9502.09218.08.1

9218.06333.03

8656.0

8656.09666.08.1

9666.06333.031

2)5(

)5(

2)4(

)4(

2)3(

)3(

2)2(

)2(

2)1(

)1(

=−=

=+=

=−=

=+=

=−=

=+=

=−=

=+=

=−=

=+=

y

x

y

x

y

x

y

x

y

x

The subsequent iterations are listed (0.9348, 0.9261), (0.9420, 0.9126), (0.9375, 0.9210), (0.9403, 0.9158), (0.9385, 0.9192), (0.9397, 0.9169), (0.9389, 0.9184). The Gauss-Seidel Method slowly approaches the correct solution of (0.93926,0.9178).

397

Load Flow

9.2.5 I) Form the bus admittance matrix using the rules for mutual and self admittances.

( ) ( )

( )( )0.40.2

j4.02.0YY

j4.02.0j0.20.1

1YY

2122

1211

2112

jYY −=−=−=−=

+−=+

−==

So the bus admittance matrix is therefore

( ) ( )( ) ( ⎥

⎦

⎤⎢⎣

⎡−+−+−−

=j4.02.0j4.02.0j4.02.0j4.92.0

Ybus )

ii) The load on bus bar 2 is S = 0.5 + j 0.3 pu. The scheduled real and reactive powers on

bus 2 are therefore and 0.5Psp2 −= 0.3Qsp

2 −= The iterative equation for the voltage on bus 2 is

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅−

−=+

112*(i)2

sp2

sp2

22

1)(i2 VY

V

jQPY1V

Bus bar 1 is the swing or slack bus, the voltage on bus 1 is therefore equal to 1 pu at an

angle of zero degrees. Substituting known values:

( ) (

( )

)

1.0V

j0.070.11

j0.01.0j4.02.0V

j0.30.5j4.0)(2.0

1V

*(i)2

*(i)2

1)(i2

+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⋅+−−

+−−

=+

This equation is the iterative equation for the voltage on bus bar 2. Initial guess for V2 is 1 pu at an angle of zero degrees. The solution proceeds as follows:

398

Load Flow

9j0.060.8676V

j0.0700.8682V

j0.0680.8710V

j0.070.89V

j0.01.0V

(4)2

(3)2

(2)2

(1)2

(0)2

−=

−=

−=

−=

+=

So after four iterations the solution has converged to puj0.07)(0.86V2 −= .

iii) The real and reactive powers on 1 can be determined using the following equation:

( )( ) ( )( ) ( )(( )( )

j0.42)(0.56j3.58)1.44(j4.02.0

j0.70.86j4.02.0j0.01.0j4.02.0j0.01.0

VYVYVjQP*

212111*

111

−=+−+−=

−+−++−+=

⋅+⋅=−

)

The generator on bus bar 1 is therefore delivering 0.56 pu real power and 0.42 pu reactive power to the network.

iv) Input power on bus 1 is 0.56 pu. Output power on bus 2 is 0.5 pu.

Efficiency = 9.056.05.0=

9.2.6

Form the bus admittance matrix using the rules for mutual and self admittances.

( ) ( )

( )

( ) ( ) (( )( )j1.50.5YY

j3.0)1YYj4.51.5j1.50.5j31YYY

0YY

j1.50.5j0.6)(0.2

1YY

j3.01.0j0.30.1

1YY

3133

2122

131211

3223

3113

2112

−=−=−=−=

−=−+−=−−===

+−=+

−==

+−=+

−==

)

So the bus admittance matrix is therefore

399

Load Flow

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−−+−

+−+−−=

j1.50.50j1.50.50j31j31

j1.50.5j31j4.51.5Ybus

)

ii) The load on bus bar 2 is S = 0.2 + j 0.1 pu. The scheduled real and reactive powers on

bus 2 are therefore and 0.2Psp2 −= 0.1Qsp

2 −= The iterative equation for the voltage on bus 2 is

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅−⋅−

−=+

323112*(i)2

sp2

sp2

22

1)(i2 VYVY

V

jQPY1V

Bus bar 1 is the swing or slack bus, the voltage on bus 1 is therefore equal to 1 pu at an

angle of zero degrees. Substituting known values gives the following iterative procedure for the voltage on bus 2.

( ) ( ) (

( )

)

0.1V

05.005.0

j0.01.0j31V

j0.10.2-j3-1

1V

*(i)2

*(i)2

1)(i2

+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⋅+−−

+=+

j

Initial guess for V2 is 1 pu at an angle of zero degrees. The solution proceeds as follows:

9j0.0490.94441V

j0.0500-0.94444V

j0.4972-0.94475V

j0.0500-0.94999V

j0.01.0V

(4)2

(3)2

(2)2

(1)2

(0)2

−=

=

=

=

+=

So after four iterations the solution has converged to puj0.05)(0.94V2 −= . The load on bus bar 3 is S = 0.2 + j 0.1 pu. The scheduled real and reactive powers on bus 3 are therefore and 0.2Psp

3 −= 0.1Qsp3 −=

The iterative equation for the voltage on bus 3 is

400

Load Flow

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅−⋅−

−=+

332131*(i)3

sp3

sp3

33

1)(i3 VYVY

V

jQPY1V

Substituting known values gives the following iterative procedure for the voltage on bus

3.

( ) ( )(

( )

)

0.1V

1.01.0

0.00.15.15.0V

1.02.0j1.5-0.5

1V

*(i)3

*(i)3

1)(i3

+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡++−−

+−=+

j

jjj

Initial guess for V3 is 1 pu at an angle of zero degrees. The solution proceeds as follows:

j0.100000.87419V

j0.999190.87429V

j1.000000.87500V

j0.975600.87804V

j0.100000.89999V

j0.01.0V

(5)3

(4)3

(3)3

(2)3

(1)3

(0)3

−=

−=

−=

−=

−=

+=

So after five iterations the solution has converged to puj0.10)(0.87V3 −= .

iii) The real and reactive powers on 1 can be determined using the following equation:

( )

( ) ( )( ) ( )(( )( )

2558.04186.01.0874.05.15.0

j0.005-0.94440.30.1j0.01.0j4.5-1.5j0.01.0

VYVYVjQP

*

312212111*

111

jjj

jVY

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−+

+−+++=

⋅+⋅+⋅=−

)

The generator on bus bar 1 is therefore delivering 0.418 pu real power and 0.255 pu reactive power to the network.

9.2.7 i) A line is added linking bus 2 and bus 3 so the mutual admittance Y23 is no longer zero

( ) ( )j31j0.30.1

1YY 3223 +−=+

−==

401

Load Flow

The self admittance of buses 2 and 3 both change due to the addition of the line 2-3

j4.5)(1.5

j3.0)(1.0j1.5)(0.5YYY 323133

−=−+−=

−−=

j6.0)(2.0

j3.0)(1.0j3.0)(1.0YYY 232122

−=−+−=

−−=

The modified bus admittance matrix is now

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+−+−−+−+−+−−

=j4.51.5j31j1.50.5j31j62j31j1.50.5j31j4.51.5

Ybus

)

ii) The iterative equation for the voltage on bus 2 is

( ) ( )( ) ( )

( ) 0.5V0.5V

j0.0250.025

Vj31j0.01.0j31V

j0.10.2j6)(2

1

VYVYV

jQPY1V

(i)3*(i)

2

(i)3*(i)

2

323112*(i)2

sp2

sp2

22

1)(i2

+⋅+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅+−−++−−

+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅−⋅−

−=+

The iterative equation for the voltage on bus 3 is

( ) ( )( ) ( )

( ) 0.3333V0.6666V

j0.0330.033

Vj31j0.01.0j1.55.0V

j0.10.2j4.5)(1.5

1

VYVYV

jQPY1V

(i)2*(i)

3

(i)2*(i)

3

232131*(i)3

sp3

sp3

33

1)(i3

+⋅+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅+−−++−−

+−−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅−⋅−

−=+

402

Load Flow

The expression for the voltage on bus 2 is now a function of the voltage on bus 3 and visa versa. These two equations must now be iterated together to find both voltages.

iii) Assuming a flat start, the first iteration for V2 yields

( )

( )( ) ( )

j0.2500)-(0.9750

0.5j0.01.00.5j0.01.0j0.0250.025V

0.5V0.5V

j0.0250.025V

(1)2

(i)3*(i)

2

1)(i2

=

++⋅++−−

=

+⋅+−−

=+

This updated value of V2 is now used to estimate V3.

( )

( )( ) ( )

j0.0496)-(0.9493

0.3333j0.025-0.97500.6666j0.01.00j0.0330.033V

0.3333V0.6666V

j0.0330.033V

(1)3

(i)2*(i)

3

1)(i3

=

+⋅++−−

=

+⋅+−−

=+

A further 7 iterations would be required for the solution to converge to ( )j0.06220.9272V2 −= ( )j0.07440.9116V3 −=

403