PSAS Lab Manual

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PARUL INSTITUTE OF ENGINEERING & TECHNOLOGYDepartment of Electrical Engineering Power System Analysis

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PRACTICAL NO.1

Aim: Introduction to FEMM

1. Introduction

Finite Element Method Magnetics (FEMM) is a finite element package for solving 2D planar and

axisymmetric problems in low frequency magnetics and electrostatics. The program runs under

runs under Windows 95, 98, ME, NT, 2000 and XP. The program can be obtained via the FEMM

home page athttp://femm.foster-miller.com.

The package is composed of an interactive shell encompassing graphical pre-and postprocessing;

a mesh generator; and various solvers. A powerful scripting language, Lua 4.0, is integrated with

the program. Lua allows users to create batch runs, describe geometries parametrically, perform

optimizations, etc. Lua is also integrated into every edit box in the program so that formulas can

be entered in lieu of numerical values, if desired. (Detailed information on Lua is available from

http://www.lua.org.) There is no hard limit on problem sizemaximum problem size is limited

by the amount of available memory. Users commonly perform simulations with as many as amillion elements.

The purpose of this document is to present a step-by-step tutorial to help new users get "up and

running" with FEMM. In this document, the solution for the field of an air-cored coil is

considered.

2. Model Construction and Analysis

This will take you through a step-by-step process to analyze the magnetic field of an aircoredsolenoid sitting in open space. The coil to be analyzed is pictured in Figure 1. The coil has an

inner diameter of 1 inch; an outer diameter of 3 inches; and an axial length of 2 inches. The coil

is built out of 1000 turns of 18 AWG copper wire. For the purposes of this example, we will

consider the case in which a steady current of 1 Amp is flowing through the wire.
http://femm.foster-miller.com/http://femm.foster-miller.com/http://femm.foster-miller.com/http://femm.foster-miller.com/


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In FEMM, one models a slice of the axisymmetric tric problem. By convention, the r = 0 axis is

understood to run vertically, and the problem domain is restricted to the region where r 0. In

this convention, positive-valued currents flow in the into-the-page direction.

Figure 1: Air-cored coil to be analyzed in first example.

2.1 Create a New Model

Run the FEMM application by selecting femm 4.0 from the femm 4.0 section of the Start Menu.

The default preferences will bring up a blank window with a minimal menu bar. Select New

from the main menu. A dialog will pop up with a drop list allowing you to select the type of new

document to be created. Select the Magnetics Problem entry and hit the OK button. A new blank

magnetics problem will be created, and a number of new toolbar buttons will appear.

2.2 Set Problem Definition

The first task is to tell the program what sort of problem is to be solved. To do this, select

Problem from the main menu. The Problem Definition dialog will appear. Set Problem Type to

Axisymmetric. Make sure that Length Units is set to Inches and that the Frequency is set to 0.

When the proper values have been entered, hit the OK button.


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2.3 Draw Boundaries

The first task is to draw boundaries for the solution region. Fundamentally, finite element solversmesh and find a solution over a finite region of space that contains the objects of interest. In this

case, we will choose our solution region to be a sphere with a radius of 4 inches.

First, we can adjust the view so that it will contain the entire solution region. Select

View|Keyboard off of the main menu to bring up a dialog that allows you to specify the extents

of the visible screen. In this dialog, specify Bottom to be -4, Left to be 0, Right to be 4, and Top

to be 4 and hit the OK button. The screen will then be rescaled to the smallest rectangle that

contains the specified region.

Next, node points need to be defined that bound the sphere. To draw these node points, select the

Operate on nodes button from the from the tool bar (this is the farthest button on the left with a

small black box :). Place nodes at the top and bottom of the sphere, at (0, 4) and (0-4), and at the

origin (0,0). One can place nodes either by moving the mouse pointer to the desired location and

pressing the left mouse button, or by pressing the key and manually entering the point

coordinates via a popup dialog.

Select the Operate on segments toolbar button (second button from the left with a blue line: ). To

select a node to be the endpoint of a line, click near the desired endpoint with the left mouse

button. Draw a line down the axis of symmetry by selecting the point at (0,-4) and then the point

at (0, 4). A line will appear linking the nodes as soon as the second point is selected.

Select the Operate on arc segments toolbar button (third button from the left with a blue arc :).

Draw an arc down the axis of symmetry by selecting the point at (0,-4) and then the point at (0,

4). A dialog will appear asking you for some attributes of the arc. In FEMM, arc are

approximated by a series of small, straight lines. The Max. Segment specifies the coarseness

with which the arc is divided into sections. Enter 2.5 into this edit box to get a fairly fine

representation of the arc. Put 180 in the Arc Angle edit box to denote that a half circle is being

drawn.


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2.4 Draw Coil

Now, the coil itself can be drawn. Switch back to Nodes mode by pressing the Operate on nodestoolbar button. Place nodes at (0.5,-1), (1.5,-1), (1.5,1) and (0.5,1) defining the extents of the

coil. Select the Operate on segments toolbar button so that lines can be drawn connecting the

points. By selecting the nodes defining the coil in sequence, one obtains lines between each of

the nodes and result in a large connected box.

2.5 Place Block Labels

Now click on the Operate on Block Labels toolbar button denoted by concentric green circles .

Place a block label in the coil region, and place one in the air outside the coil region. Like node

points, block labels can be placed either by a click on the left mouse button, or via the

dialog. The program uses block labels to associate materials and other properties with various

regions in the problem geometry. Next, we will defined some material properties, and then we

will go back and associate them with particular block labels.

NOTE: If snap-to-grid is enabled then it may be sometimes be difficult to place the block label

in the empty space. If this is the case, disable snap-to-grid by de-selecting the tool bar button

with the point and arrow.

2.6 Add materials to the model

Select Properties|Materials Library off of the main menu. The drag-and-drop Air from Library

Materials to Model Materials to add it to the current model. Go into the Copper AWG Sizes

folder and drag 18 AWG into Model Materials. Click on OK.

2.7 Add a "Circuit Property" for the coil

Select Properties|Circuits off of the main menu. On the dialog that appears, push the Add

Property button to create a new circuit property. Name circuit by replacing the new circuit name

with Coil. Specify that the circuit property is to be applied to a wound region by selecting the

Series radio button. Enter 1 as the Circuit Current. The j edit box denotes the imaginary


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component of the current, which is used in time harmonic problems to denote the phase of the

current. In this case, the problem is magnetostatic, so the imaginary component is ignoredjust

put zero in the j edit box. Click on OK for both the Circuit Property and Property Definition

dialogs.

2.8 Associate properties with block labels.

Right click on the block label node in the air region outside the coil. The block label will turn

red, denoting that it is selected. Press space to open the selected block label (Instead of

pressing the space bar, one can use the Open up Properties Dialog toolbar button). A dialog will

pop up containing the properties assigned to the selected label. Set the Block type to Air.

Uncheck the Let Triangle choose Mesh Size checkbox and enter 0.1 for the Mesh size. The mesh

size parameter defines a constraint on the largest possible elements size allowed in the associated

section. The mesher attempts to fill the region with nearly equilateral triangles in which the sides

are approximately the same length as the specified Mesh size parameter. When the Let Triangle

choose Mesh Size box is checked, the mesher is free to pick its own element size, usually

resulting in a somewhatcoarse mesh. Click on OK. The block label will then be labeled as Air,

and a circle will appear about the block label indicating the approximate mesh size in the

associated region.

Repeat the same for the block label node inside the coil region, changing the mesh size to 0.1.

However, set this Block type to Copper. We want to assign currents to flow in this region, so

select the Coil circuit from the In Circuit drop list. The Number of turns edit box will become

activated if a series-type circuit is selected for the region (e.g. the Coil property that was

previously defined). Enter 1000 as the number of turns for this region, denoting that the region if

filled with 1000 turns wrapped in a counter-clockwise direction (i.e. positive turns in the right-hand-screw rule sense). Click on OK.

NOTE: If we wanted to denote that the turns are wrapped in a counter-clockwise direction

instead, we could have specified the number of turns to be1000.


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2.9 Create Boundary Conditions

Select Properties|Boundary from the menu bar, then click on the Add Property button. Replacethe name New Boundary with ABC and change the BC Type to Mixed. The ABC name is meant

to denote that we are creating an "asymptotic boundary condition" that approximates the

impedance of an unbounded, open space. In this way, we can model the field produced by the

coil in an unbounded space while still only modeling a finite region of that space. When the

Mixed boundary condition type is selected the c0 coefficient and c1 coefficient boxes will

become enabled. These entries are meant to represent coefficients in a boundary condition of the

form:

where A is magnetic vector potential, r is the relative magnetic permeability of the region

adjacent to the boundary, o is the permeability of free space, and n represents the direction

normal to the boundary. For our asymptotic boundary condition, we need to specify:

where R is the outer radius of a spherical problem domain. To enter these values into the dialog

box, enter 0 as the c1 coefficient and 1/(uo*4*inch) as the c0 coefficient. The Lua scripting

language processes the contents of each edit box automatically when the dialog is closed,

substituting the numerical value of the permeability of free space for uo and 0.0254 for inch and

evaluating the result.


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To assign this boundary condition, switch to operate on arc segments mode. Select the arc

defining the outer boundary by clicking on the arc with the left mouse button and push the space

bar to open the arc's properties for editing. Select ABC from the Boundary cond. drop list and

click on OK. You have now defined enough boundary conditions to solve the problem, since a

zero potential is automatically applied along the r = 0 line for axisymmetric problems.

You have now completed modeling the coil. The finished pre-processor geometry should look as

pictured in Figure 2.

Figure 2: Completed coil model, ready to be analyzed.


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2.10 Generate Mesh and Run FEA

Now save the file and click on the toolbar button with yellow mesh. This action generates atriangular mesh for your problem. If the mesh spacing seems to fine or too coarse you can select

block labels or line segments and adjust the mesh size defined in the properties of each object.

Once the mesh has been generated, click on the turn the crank button to analyze your model.

Processing status information will be displayed. If the progress bars do not seem to be moving

then you should probably cancel the calculation. This can occur if insufficient boundary

conditions have been specified. For this particular problem, the calculations should be completed

within a second. There is no confirmation for when the calculations are complete, the status

window just disappears when the processing is finished.

3. Analysis Results

Click on the glasses icon to view the analysis results. A post-processor window will appear. The

post-processor window will allow you to extract many different sorts of information from the

solution.

3.1 Point values

Just like the pre-processor, the post-processor window has a set of different editing modes: Point,

Contour, and Area. The choice of mode is specified by the mode toolbar buttons, i.e. where the

first button corresponds to Point mode, the second to Contour mode, and the third to Area mode.

By default, when the program is first installed, the post-processor starts out in Point mode. By

clicking on any point with the left mouse button, the various field properties associated with that

point are displayed in the floating FEMM Output window. Similar to drawing points in the pre-

processor, the location of a point can be precisely specified by pressing the button and

entering the coordinates of the desired point in the dialog that pops up. For example, if the point

(0, 0) is specified in the pop-up dialog, the resulting properties displayed in the output window

are as pictured in Figure 3.


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Figure 3: Display of field values at the point (0,0).

3.2 Coil terminal properties

With FEMM, it is straightforward to determine the inductance and resistance of the coil as seen

from the coil's terminals. Press the button to display the resulting attributes of each Circuit

Property that has been defined. For the Coil property defined in this example, the resulting dialog

is pictured in Figure 4.

Figure 4: Circuit Property results dialog.


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Since the problem is linear and there is only one current, the Flux/Current result can be

unambiguously interpreted as the coil's inductance (i.e. 22.9 mH). The resistance of the coil is

the Voltage/Current result (i.e. 3.34 ).

3.3 Plotting field values along a contour

FEMM can also plot values of the field along a user-defined contour. Here, we will plot the flux

density along the centerline of the coil. Switch to Contour mode by pressing the Contour Mode

toolbar button. You can now define a contour along which flux will be plotted. There are three

ways to add points to a contour:

1. Left Mouse Button Click adds the nearest input node to the contour;

2. Right Mouse Button Click adds the current mouse pointer position to the contour;

3. Key displays a point entry dialog that allows you to enter in the

coordinates of a point to be added to the contour.

Here, method 1 can be used. Click near the node points at (0, 4), (0, 0), and (0,-4) with the left

mouse button, adding the points in the above order. Then, press the Plot toolbar button . Hit OK

in the X-Y Plot of Field Values pop-up dialogthe default selection is magnitude of flux

density. If desired, different types of plot can be selected from the drop list on this dialog.

NOTE: It is often the case in the solution to magnetic problems that the field values are

discontinuous across a boundary. In this case, FEMM determines which side of the boundary

will be plotted based on the order in which points are added. For example, if points are added

around a closed contour in a counterclockwise order, the plotted points will lie just to the inside

of the contour. If the points are added in a clockwise order, the plotted points will lie just to the

outside of the contour. The implication to our example problem is that the contour along the r=0

should be defined in order of decreasing z (i.e. counterclockwise so that the plotted points will lie

inside the solution domain instead of outside it, where the field values are not defined).


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3.4 Plotting Flux Density

By default, when the program is first installed, only a black-and-white graph of flux lines isdisplayed. Flux density can be plotted as a color density plot, if you so desire. To make a color

density plot of flux, click on the purple shaded toolbar button to generate a color flux density

plot. When the dialog box comes up, select the Flux density plot radio button and accept the

other default values. Click on OK. The resulting solution view will look similar to that pictured

in Figure 5.

Figure 5: Color flux density plot of solution.


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Figure 6: Flux density (Tesla) versus length (Inches)

4. Conclusions

You have now completed your first model of a magnetic problem with FEMM. From this basic

introduction, you have been exposed to the following concepts:

How to draw a model using nodes, segments, arc, and block labels;

How to add material to your model and how to assign them to regions;

How to specify the finite element mesh size;

How to define boundary for your model;

How to define and apply boundary conditions;

How to analyze a problem;

How to inspect local field values;

How to plot field values along a line;


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How to compute inductance and resistance;

How to display color flux density plots.

Hopefully, this tutorial has presented you with enough of the basics of FEMM so that you can

explore more complicated problems without getting sidetracked by the mechanics of how a

problem is drawn and analyzed.


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PRACTICAL NO. 2

Aim: Calculation of Leakage reactance of transformer winding.

Problem Statement: 31.5 MVA, 132 KV/33KV, Y/, Transformer, Ampere-turns: 137780, the

dimensions of the transformer are given in Figure 1.

All dimensions are in mm.

Figure 1: Transformer dimensions

Dimensions:

Core Diameter : 560 mm

Core-LV gap : 23 mm

LV radial depth : 70 mm


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LV-HV gap : 50 mm

HV radial depth : 100 mm

rof core = 3000

N1 turns =1000

N2 turns =433

Calculate % reactance of transformer using FEM software (attach flux plot) and comment on

result. Also reduce HV winding height by 2% at one end and do the same.

Simulation:

Calculation:

Current density = mmf/area A/mm2

Find on LV as well as HV side. Here we have taken Short circuit on HV side.


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Using FEM plot, Magnetic field energy W=..Joules

Find current I=AT/(N) Ampere-turn/turns on both side of transformer.

W=(1/2) LI2

From above equation find L and on both side.

XL=2fL

Zb =V2/MVA

% Leakage Reactance %X = XL/ Zb * 100 %

Do same calculation for reduce HV height.

CONCLUSION:


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PRACTICAL NO.- 3

Aim: Find the Force on Plunger of an actuator.

Problem Statement: Obtain the magnetic flux density plot in a solenoid and force applied to the

plunger for d= 0.02.

Find the force on the plunger for various values of d and plot same.

All dimensions are in meter.

Figure 1: Problem sketch

Relative Permeability of air and coil 1

Current Density in coil 1e6 Amp/m


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The B-H curve for the core and plunger

H (A/m) 460 640 720 890 1280 1900 3400 6000

B(T) 0.8 0.95 1 1.10 1.25 1.40 1.55 1.65

SIMULATION:

Figure: Axysymmetric simulation of Plunger of an actuator


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Figure: B-H curve of core and plunger

Find force on plunger using FEMM.

Plot the graph for force on the plunger V/S gap length d

CONCLUSION:


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PRACTICAL NO.- 4

Aim: Simulation on Basic Electrical Engineering Application.

Problem Statement:

(1)Construct a simulink model in order to obtain Sine, Square and Sawtooth waves.(2)Construct a simulink model for average and RMS values(3)Construct a simulink model for (i) CCCS (ii) CCVS (iii) VCCS (iv) VCVS(4)Construct a simulink model for (i) Series Resonance (ii) Parallel Resonance

Simulation and Result:

(1)Construct a simulink model in order to obtain Sine, Square and Sawtooth waves.

Figure 4.1 Model for Sine, Square and Sawtooth waves


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Figure 4.2 Plot for sine, square and sawtooth waves

(2)Construct a simulink model for average and RMS values

Figure 4.3 Circuit model for average and RMS values


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Figure 4.4 Plot showing the peak values of the current and voltage

(3)Construct a simulink model for (i) CCCS (ii) CCVS (iii) VCCS (iv) VCVS

(i) CCCS (Current Controlled current source)

Figure 4.5 Model of current controlled current source


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Figure 4.6 Input and Output currents of CCCS

(ii) CCVS (Current Controlled voltage source)

Figure 4.7 Model of current controlled voltage source


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Figure 4.8 Input voltage and Output current of CCVS

(iii) VCCS (Voltage Controlled Current source)

Figure 4.9 Model for voltage controlled current source


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Figure 4.10 Input voltage and output current of VCCS

(iv) VCVS (Voltage controlled voltage source)

Figure 4.11 Model of voltage controlled voltage source


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Figure 4.12 Input and Output voltages of VCVS

(4) Construct a simulink model for (i) Series Resonance (ii) Parallel Resonance

(i) Series Resonance

Figure 4.13 Series Resonance Circuit


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Figure 4.14 Supply voltage and current waveforms for series resonance

(ii) Parallel Resonance

Figure 4.15 Parallel Resonance


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Figure 4.16 Supply voltage and current for parallel resonance

CONCLUSION:


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PRACTICAL NO.- 5

Aim: Simulation on (I) Measurement of Apparent, Active and Reactive Powers (II) ThreePhase source and load simulation.

Problem Statement:

Construct Matlab Simulink Model for Measurement of Apparent, Active and Reactive Powers:

An AC voltage source of 500 V, 50 Hz in series with a resistance of 10 and an inductance of

10mH in one branch. This branch is in series with two parallel branches; one branch with a

resistance of 35 an capacitance of 40 F, and another having a resistance of 20 and an

inductance of 10 mH and a load resistance of 35 .

Three Phase source and load simulation: The peak voltage amplitude of each phase is 315 V and

frequency 50 Hz. The resistance and inductance of the transmission lines are assumed to be zero.

All the three voltage sources are star connected, i.e., line currents and phase currents are equal.

The phase voltages of this three-phase voltage source are shown in figure. The transmission line

feeds a three-phase star connected balanced load resistance 50 and inductance 40 mH.


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SIMULATION AND RESULT:

(I) Measurement of Apparent, Active and Reactive Powers

Figure 5.1 Measurement of active and reactive powers

Figure 5.2 Waveforms of active and reactive powers


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(II) Three Phase source and load simulation.

Figure 5.3 Star Connected three phase source

Figure 5.4 Three phase active and reactive powers


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Figure 5.5 Three phase symmetrical supply

CONCLUSION:


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PRACTICAL NO.- 6

Aim: Simulation of Transformers- (I) Single Phase (II) Three Phase.

Problem Statement:

Construct Matlab Simulink Model for

Single phase transformer: The primary winding parameters of the transformer are V1(rms) =110

KV, R1(pu)= 0.02 , and L1(pu) =0.08 H and secondary winding parameters are V2(rms) = 220 KV,

R2(pu)= 0.02 , and L2(pu) =0.08 H. The primary winding is supplied with a voltage source of 200

V, 50 Hz. A resistance of 10 is connected in series with primary winding ad of 120 in series

with the secondary winding.

Three Phase transformer: A circuit containing a three phase transformer having two star

connected winding is shown in figure. The primary winding is fed by a three phase supply of 300

V, 50 Hz. The primary winding parameters are V1(rms) =300 V, R1(pu)= 0.002 , and L1(pu) =0.08H and secondary winding parameters are V2(rms) = 600 KV, R2(pu)= 0.02 , and L2(pu) =0.08 H.

The nominal power and frequency of transformer are 250 KW, 50 Hz. The resistances connected

to the primary windings are of 5 each and connected to the secondary windings are of 10

each.


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Simulation of Single phase transformer:

Figure 6.1 Single phase transformer

Figure 6.2 Primary (V1) and Secondary (V2) winding voltages


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Figure 6.3 Primary (I1) and Secondary (I2) winding currents

Simulation of Three phase transformer

Figure 6.4 Three Phase Transformer


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Figure 6.5 Three phase primary and secondary winding voltages

Figure 6.6 Three phase primary and secondary winding currents

CONCLUSION:


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PRACTICAL NO.- 7

Aim: Simulation of Transmission line model.

Problem Statement:

Medium Transmission Line: A 345 KV, three phase transmission line is 130 km long. The

resistance per phase is 0.036 per km and the inductance per phase is 0.8 mH per km. The shunt

capacitance is 0.0112 F per km. The receiving end load is 270 MVA with 0.8 power factor

lagging at 325 kV. Use the medium transmission line model to find the voltage and power at the

sending end and the voltage regulation.

Medium Transmission Line: A 345 KV, three phase transmission line is 130 km long. The

series impedance is z = 0.036 + j0.3 per phase per km, and the shunt admittance is y =

j4.22*10e-6 siemens per phase per km. The sending end voltage is 345 kV, and the sending end

current is 400 A at 0.95 power factor lagging. Use the medium line model to find the voltage,

current and power at the receiving end and the voltage regulation.

Long Transmission Line: A 500 kV, three phase transmission line is 250 km long. The series

impedance is z = 0.045 + j0.4 per phase per km and the shunt admittance is y = j4*10e -6

siemens per phase per km. Evaluate the equivalent model and the transmission matrix.


Medium transmission line:

m-file:

r = .036; g = 0; f = 60;

L = 0.8; % milli-Henry

C = 0.0112; % micro-Farad

Length = 130; VR3ph = 325;


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VR = VR3ph/sqrt(3) + j*0; % kV (receiving end phase voltage)

[Z, Y, ABCD] = rlc2abcd(r, L, C, g, f, Length);

AR = acos (0.8);

SR = 270*(cos (AR) + j*sin (AR)); % MVA (receiving end power)

IR = conj (SR)/ (3*conj (VR)); % kA (receiving end current)

VsIs = ABCD* [VR; IR]; % column vector [Vs; Is]

Vs = VsIs (1);

Vs3ph = sqrt (3)*abs (Vs); % kV (sending end L-L voltage)

Is = VsIs (2); Ism = 1000*abs (Is); % A (sending end current)

pfs= cos (angle (Vs) - angle (Is)); % (sending end power factor)

Ss = 3*Vs*conj (Is); % MVA (sending end power)

REG = (Vs3ph/abs (ABCD (1, 1)) - VR3ph)/VR3ph *100;

fprintf (' Is = %g A', Ism), fprintf(' pf = %g\n', pfs)

fprintf(' Vs = %g L-L kV\n', Vs3ph)

fprintf(' Ps = %g MW', real(Ss)),

fprintf(' Qs = %g Mvar\n', imag(Ss))

fprintf(' Percent voltage Reg. = %g\n', REG)


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Result:

Enter 1 for Medium line or 2 for long line --> 1

Nominal pi model

----------------

Z = 4.68 + j 39.2071 ohms

Y = 0 + j 0.000548899 Siemens

ABCD =

Is = 421.132 A pf = 0.869657

Vs = 345.002 L-L kV

Ps = 218.851 MW Qs = 124.23 Mvar

Percent voltage Reg. = 7.30913

Medium transmission line:

m-file:

z = .036 + j* 0.3; y = j*4.22/1000000; Length = 130;

Vs3ph = 345; Ism = 0.4; %KA;

As = -acos (0.95);

Vs = Vs3ph/sqrt (3) + j*0; % kV (sending end phase voltage)

0.98924 + j 0.0012844 4.68 + j 39.207

-3.5251e-007 + j 0.00054595 0.98924 + j 0.0012844


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Is = Ism*(cos (As) + j*sin (As));

[Z,Y, ABCD] = zy2abcd (z, y, Length);

VrIr = inv (ABCD)* [Vs; Is]; % column vector [Vr; Ir]

Vr = VrIr (1);

Vr3ph = sqrt (3)*abs (Vr); % kV (receiving end L-L voltage)

Ir = VrIr (2); Irm = 1000*abs (Ir); % A (receiving end current)

pfr= cos (angle (Vr) - angle (Ir)); % (receiving end power factor)

Sr = 3*Vr*conj (Ir); % MVA (receiving end power)

REG = (Vs3ph/abs (ABCD (1, 1)) - Vr3ph)/Vr3ph *100;

fprintf(' Ir = %g A', Irm), fprintf(' pf = %g\n', pfr)

fprintf(' Vr = %g L-L kV\n', Vr3ph)

fprintf(' Pr = %g MW', real (Sr))

fprintf(' Qr = %g Mvar\n', imag(Sr))

fprintf(' Percent voltage Reg. = %g\n', REG)


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Result:


Nominal pi model

----------------

Z = 4.68 + j 39 ohms

Y = 0 + j 0.0005486 Siemens

ABCD =

Ir = 441.832 A pf = 0.887501

Vr = 330.68 L-L kV

Pr = 224.592 MW Qr = 116.612 Mvar

Percent voltage Reg. = 5.45863

0.9893 + j 0.0012837 4.68 + j 39

-3.5213e-007 + j 0.00054567 0.9893 + j 0.0012837


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Long transmission line:

m-file:

z = 0.045 + j*.4; y = j*4.0/1000000; Length = 250;

gamma = sqrt (z*y); Zc = sqrt (z/y);

A = cosh (gamma*Length); B = Zc*sinh (gamma*Length);

C = 1/Zc * sinh (gamma*Length); D = A;

ABCD = [A B; C D]

Z = Zc * sinh (gamma*Length)

Y = 2/Zc * tanh (gamma*Length/2)

Result:

ABCD =

Z =

10.8778 +98.3624i

Y =

0.0000 + 0.0010i

CONCLUSION:

0.9504 + 0.0055i 10.8778 +98.3624i

-0.0000 + 0.0010i 0.9504 + 0.0055i


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PRACTICAL NO.- 8

Aim: Simulation of power system for different fault condition.

Problem Statement: Construct a simulink model of power system for different fault condition.


Figure 8.1 Power system model for different fault conditions

Following program is used to set the values of various parameters of the power system model.

% This program sets the parameters for three phase power system model under different fault

% conditions.


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clear all;

clc;

% Three phase power system parameters

f= 50; % frequency of the power system in Hz

T =1/f; % Time period in seconds

R1 = 1; % Resistance of the transmission line in ohms

L1 = 10e-6; % Inductance of the transmission line in H

R2 =50; % Load resistance in ohms

Ton = 0.02; % On time of the power supply in seconds

% Three voltage source

Pia = 0; % Phase angle of the voltage source Va in degrees

Pib = 120; % Phase angle of the voltage source Vb in degrees

Pic = 240; % Phase angle of the voltage source Vc in degrees

V =230; % Peak voltage per phase

% Under voltage, continuous Reduction parameters

k1 = 20.0; % Under voltage, continuous reduction of voltage in

V1 = (k1/100) * V; % Voltage to be reduced

Va = VV1; % Actual reduced peak voltage


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% Voltage Unbalance Parameters

UF = 0.5; % Voltage Unbalance factor

Vneg = Va * UF; % Negative sequence voltage

% Voltage Sag/Swell parameters

N = 3.0; % Duration of sag/swell for number of fundamental cycles

T1 = 0.05; % Time in sec at which sag will be initiated

k = 50; % Reduction in magnitude in % for + values sag

% will be there & forive values will be there

Vu = Va * (k/100); % 50% reduction in phase voltage

T3 = T * N; % end of sag after N cycles

T = T1 + T3;

% Parameters for Harmonics Distortion

h5 = 0.05; % Harmonic Factor of 5th harmonic

h7 = 0.03; % Harmonic Factor of 7th harmonic

f5 = f * 5; % 5th harmonic frequency

f7 = f * 7; % 7th harmonic frequency

Vh5 = V * h5; % 5th harmonic, 5% of fundamental component

Vh7 = V * h7; % 7th harmonic, 3% of fundamental component

% End of the program


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Figure 8.2 Source, Negative sequence and fifth harmonic voltage

Figure 8.3 The 7th harmonic, sag voltages, and line (Ra, La) current


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Figure 8.4 Power line (Rb, Lb, Rc, Lc) currents, load voltage, and rectifier output voltage

CONCLUSION:


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PRACTICAL NO.- 9

Aim: Simulation of Separately excited DC motor.

Problem Statement: Construct a simulink model of Separately excited DC motor.


In case of separately excited DC motors, V = Ea + I*Ra. The speed of a DC motor can be given

by = (V-I*Ra)/K, where K is the machine constant and is the flux per pole. So, the speed

of the DC motor can be varied by varying the armature resistance (which varies Ia) or by varying

the field resistance (which varies the flux per pole). The armature resistance control method

consists of a variable resistance connected in series with the armature as shown in figure 9.1. The

speed of the motor can be reduced to any desired value depending on the armature resistance.

The field current remains unaffected as it is connected to a separately excited DC motor is shown

in figure 9.2. As the circuit breakers connected in parallel are turned ON, the resistances

connected in parallel to them gets short circuited. Initially, all the CBs are OFF and are turned

ON one by one as can be seen from figure 9.2. The supply given to the armature of the DC motor

is 180 V and to the field winding is100 V. The initial speed of the motor is taken as 1 rad/s and is

fed back to the motor torque (input terminal TL) by multiplying it by a gain of 0.405. When the

ideal switch is turned ON at 0.01 s, the 180 V DC supply gets connected with the armature of

the motor through the series resistance. Initially, all the CBs are OFF so the voltage supplied to

the armature is low (20 V) as can be seen in figure 9.3. After 6 s when all the CBs are turned

ON, the voltage at the armature terminal become 180 V. The speed of the motor in rad/s,

armature current in A and electromagnetic torque developed in N.m are shown in figure 9.4. The

graphs of armature current (Ia) versus speed (), and speed () versus electromagnetic torque

(Te) are shown in figure 9.5 and 9.6 respectively. If the mechanical torque input, i.e., TL, to the

motor is negative it acts as a generator.


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In the field control method of speed control, a variable resistance is connected in series with the

field winding of the motor as shown in 9.7. As the field resistance increases, the field current

decreases with a consequent reduction in flux which is inversely proportional to the motor speed.

Thus, as the field resistance increases, the speed of the motor also increases. This method of

speed control does not depend on the motor load and therefore permits the remote control of the

motor speed. The different parameters of the model in figure 9.7 are same as that of figure 9.1,

except that the series resistances are now connected in series with the field winding. Also the

subsystem1 of this model is same as shown in figure 9.2, expect that all the CBs are initially

ON and are turned OFF one by one. Thus, initially the field winding resistance is low and is

increased in steps of 1 s. The voltage at the field winding terminal is shown in figure 9.8. The

motor speed in rad/s, armature current in A, field current in A, and electromagnetic torque in

N*m are shown in figure 9.9. The graphs of armature current (Ia) versus speed (), and speed

() versus electromagnetic torque are shown in figure 9.10 and 9.11 respectively.

Figure 9.1 Separately excited DC motor armature control


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Figure 9.2 Subsystem for varying the armature resistance

Figure 9.3 Armature Voltage


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Figure 9.4 Motor Speed (rad/s), armature current (A) and electromagnetic torque (N m ) of the

motor

Figure 9.5 Graph of armature current versus speed


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Figure 9.6 Graph of speed versus electromagnetic torque

Figure 9.7 Field control of Separately excited DC motor


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Figure 9.8 Field winding voltage

Figure 9.9 Speed (rad/s), armature current (A), field current (A), and

electromagnetic torque (N m)


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Figure 9.10 Armature current versus speed

Figure 9.11 Speed versus electromagnetic torque

CONCLUSION


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PRACTICAL NO.- 10

Aim: Simulation of transmission line for load compensation.

Problem Statement: Simulation of transmission line for load compensation.

A three phase, 60 Hz, 550 kV transmission line is 300 km long. The line parameters per phase

per unit length are found to be

r = 0.016 /km L = 0.97mH/km C =0.0115 F/km


(A) Determine the line performance when load at the receiving end is 800 MW, 0.8 power factor

lagging at 500 KV.

(Note:Use lineperf and run in command window)

lineperf

TRANSMISSION LINE MODEL

Type of parameters for input Select

Parameters per unit length

r(ohms), g(siemens) L(mH) & C (micro F) 1

Complex z and y per unit length

r+j*x (ohms/length), g+j*b (siemens/length) 2

Nominal pi or Eq. pi model 3

A, B, C, D constants 4


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Conductor configuration and dimension 5

To quit 0

Select number of menu --> 1

Enter Line length = 300

Enter Frequency in Hz = 60

Enter line resistance/phase in ohms per unit length r = 0.016

Enter line inductance/phase in millihenry per unit length L = 0.97

Enter line capacitance/phase in micro F per unit length C = 0.0115

Enter line conductance/phase in siemens per unit length g = 0


Equivalent pi model

-------------------

Z' = 4.57414 + j 107.119 ohms

Y' = 6.9638e-007 + j 0.00131631 siemens

Zc = 290.496 + j -6.35214 ohms

alpha l = 0.00826172 neper beta l = 0.377825 radian = 21.6478

0.9295 + j 0.0030478 4.5741 + j 107.12

ABCD =

-1.3341e-006 + j 0.0012699 0.9295 + j 0.0030478


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Hit return to continue ENTER

TRANSMISSION LINE PERFORMANCE

----------Analysis---------- Select

To calculate sending end quantities

for specified receiving end MW, Mvar 1

To calculate receiving end quantities

for specified sending end MW, Mvar 2


when load impedance is specified 3

Open-end line & inductive compensation 4

Short-circuited line 5

Capacitive compensation 6

Receiving end circle diagram 7

Loadability curve and voltage profile 8

To quit 0

Select number of menu -->1

Enter receiving end line-line voltage kV = 500

Enter receiving end voltage phase angle (for Ref. enter 0 ) = 0

Enter receiving end 3-phase real power MW = 800


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Enter receiving end 3-phase reactive power(+ for lagging & - for leading power factor) Mvar =

600

Line performance for specified receiving end quantities

-------------------------------------------------------

Vr = 500 kV (L-L) at 0

Pr = 800 MW Qr = 600 Mvar

Ir = 1154.7 A at -36.8699 PFr = 0.8 lagging

Vs = 623.511 kV (L-L) at 15.5762

Is = 903.113 A at -17.6996 PFs = 0.836039 lagging

Ps = 815.404 MW Qs = 535.129 Mvar

PL = 15.404 MW QL = -64.871 Mvar

Percent Voltage Regulation = 34.1597

Transmission line efficiency = 98.1108



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(b) Determine the receiving end quantities and the line performance when 600 MW and 400

Mvar are being transmitted at 525 KV from the sending end.














To quit 0


Enter sending end line-line voltage kV = 525

Enter sending end voltage phase angle (for Ref. enter 0 ) = 0


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Enter sending end 3-phase real power MW = 600

Enter sending end 3-phase reactive power(+ for lagging & - for leading power factor) Mvar =400

Line performance for specified sending end quantities

-----------------------------------------------------

Vs = 525 kV (L-L) at 0

Ps = 600 MW Qs = 400 Mvar


Vr = 417.954 kV (L-L) at -16.3044


Pr = 588.261 MW Qr = 425.136 Mvar

PL = 11.739 MW QL = -25.136 Mvar





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(C) Determine the sending end quantities and the line performance when the receiving end load

impedance is 290 at 500 kV.














To quit 0





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Enter receiving end complex load impedance Rload + j*Xload in /phase Zload = 290 + j*0

Line performance for specified load impedance

---------------------------------------------

Vr = 500 kV (L-L) at 0

Ir = 995.431 A at 0 PFr = 1

Pr = 862.069 MW Qr = 0 Mvar

Vs = 507.996 kV (L-L) at 21.5037

Is = 995.995 A at 21.7842 PFs = 0.999988 leading

Ps = 876.341 MW Qs = -4.290 Mvar

PL = 14.272 MW QL = -4.290 Mvar





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(d) Find the receiving end voltage when the line is terminated in an open circuit and is energized

with 500 kV at the sending end. Also, determine the reactance and the Mvar of a three-phase

shunt reactor to be installed at the receiving end in order to limit the no-load receiving end

voltage to 500 kV.














To quit 0



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Open line and shunt reactor compensation

----------------------------------------

Vs = 500 kV (L-L) at 0

Vr = 537.92 kV (L-L) at -0.00327893


Desired no load receiving end voltage with shunt reactor compensation kV (L-L) =500

Desired no load receiving end voltage = 500 kV

Shunt reactor reactance = 1519.4 ohm

Shunt reactor rating = 164.538 Mvar


The voltage profile for the uncompensated and the compensated line is also found as shown in

figure 10.1.


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Figure 10.1 Compensated and uncompensated voltage profile of open ended line.

(e) Find the receiving end and the sending end currents when line is terminated in a short circuit.










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To quit 0



Enter sending end voltage phase angle (for Ref. enter 0 ) = 0

Line short-circuited at the receiving end

-----------------------------------------

Vs = 500 kV (L-L) at 0

Ir = 2692.45 A at -87.5549

Is = 2502.65 A at -87.367



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(f) The line loading in part (a) resulted in a voltage regulation of 34.16 percent, which is

unacceptably high. To improve the line performance, the line is compensated with series and

shunt capacitors. For the loading condition in (a):

(1) Determine the Mvar and the capacitance of the shunt capacitors to be installed at the

receiving end to keep the receiving end voltage at 500 kV when the line is energized with 500

kV at the sending end.














To quit 0



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CAPACITIVE COMPENSATION

Analysis Select

-------- ------

Shunt capacitive compensation 1

Series capacitive compensation 2

Series & Shunt capacitive compensation 3

To quit 0

Selectr number of menu--> 1


Enter desired receiving end line-line voltage kV = 500



Enter receiving end 3-phase reactive power (+ for lagging & - for leading power factor) Mvar =

600

Shunt capacitive compensation

-----------------------------

Vs = 500 kV (L-L) at 20.2479


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Vr = 500 kV (L-L) at 0

Pload = 800 MW Qload = 600 Mvar

Load current = 1154.7 A at -36.8699 PFl = 0.8 lagging

Required shunt capcitor: 407.267 ohm, 6.51314 micro F, 613.849 Mvar

Shunt capacitor current = 708.811 A at 90

Pr = 800.000 MW Qr = -13.849 Mvar

Ir = 923.899 A at 0.991732 PFr = 0.99985 leading


Ps = 812.469 MW Qs = -55.006 Mvar

PL = 12.469 MW QL = -41.158 Mvar





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(2) Determine the line performance when the line is compensated by series capacitors for 40

percent compensation with the load condition in (a) at 500 kV.


Analysis Select

-------- ------




To quit 0






600

Enter percent compensation for series capacitor(recommnded range 25 to 75% of the line

reactance) = 40


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Series capacitor compensation

-----------------------------

Vr = 500 kV (L-L) at 0

Pr = 800 MW Qr = 600 Mvar

Required series capacitor: 42.8476 ohm, 61.9074 micro F, 47.4047 Mvar

Subsynchronous resonant frequency = 37.9473 Hz


Vs = 571.904 kV (L-L) at 9.95438


Ps = 815.383 MW Qs = 433.517 Mvar

PL = 15.383 MW QL = -166.483 Mvar




(3)

(3) The line has 40 percent series capacitor compensation and supplies the load in (a). Determine

the Mvar and the capacitance of shunt capacitors to be installed at the receiving end to keep the

receiving end voltage at 500 kV when line is energized with 500 kV at the sending end.


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Analysis Select

-------- ------




To quit 0



Enter desired receiving end line-line voltage kV = 500




600

Enter percent compensation for series capacitor(recommended range 25 to 75% of the line

reactance) = 40

Series and shunt capacitor compensation

---------------------------------------

Vs = 500 kV (L-L) at 12.0224

Vr = 500 kV (L-L) at 0


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Pload = 800 MW Qload = 600 Mvar

Load current = 1154.7 A at -36.8699 PFl = 0.8 lagging

Required shunt capcitor: 432.736 ohm, 6.1298 micro F, 577.72 Mvar

Shunt capacitor current = 667.093 A at 90

Required series capacitor: 42.8476 ohm, 61.9074 micro F, 37.7274 Mvar

Subsynchronous resonant frequency = 37.9473 Hz

Pr = 800 MW Qr = 22.2804 Mvar



Ps = 812.257 MW Qs = -137.023 Mvar

PL = 12.257 MW QL = -159.304 Mvar




(g) Construct the receiving end circle diagram.


Analysis Select

-------- ------




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To quit 0















To quit 0



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Receiving end power circle diagram

----------------------------------


Figure 10.2 Receiving End circle diagram

A plot of the receiving end circle diagram is obtained as shown in figure 10.2.


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(h) Determine the line voltage profile for the following cases: no-load, rated load, line terminated

in the SLL, and short-circuited line.














To quit 0



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Voltage profile and line loadability

Analysis Select

--------- ----------

Voltage profile curves 1

Line loadability curves 2

To quit 0


Voltage profile for line length up to 1/8 wavelength

----------------------------------------------------


Enter rated sending end power, MVA = 1000

Enter power factor = 0.8


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A plot for the voltage profile is shown in figure 10.3.

Figure 10.3 Voltage profile for length up to 1/8 wavelength

Obtain the line loadability curves.

Voltage profile and line loadability

Analysis Select

--------- ----------

Voltage profile curves 1

Line loadability curves 2

To quit 0



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Loadability curve of a lossless line (up to 1/4 wavelength

----------------------------------------------------------



Enter rated line-line voltage kV = 500

Enter line current-carrying capacity, Amp/phase = 3500

The line loadability curve is shown in figure 10.4.

Figure 10.4 Line loadability curve for length up to wavelength.

PSAS Lab Manual

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