Long Transmission Lines Analysis

Long Transmission Lines Analysis

Long Transmission Lines Analysis

Progress Report

Contents

1.1 Abstract

1.2 Acknowledgements

1.3 Introduction

1.4 Research description

1.4.1 Equations derivation

1.4.2 Interpretation of the equations

1.4.2 Hyperbolic form of equations

1.5 Preliminary design

1.5.1 System block diagram

1.5.2 Design process

1.5.3 Circuit diagram

1.5.4 System description

1.6 Hardware / software description

1.6.1 Hardware

1.6.2 Software

Conclusion

Appendix

Budget

Time plan

Calculations

Photos

References

1.1 Abstract

This project is concerned with reflection of waves on transmission lines. The project involves using the characteristics of long transmission lines, which means that the parameters are not lumped together but are instead spread through out the line. The projects involve obtaining data from MATLAB software and the physical prototype of the transmission line. Since high voltages are not available in labs and the propagation of the signals tends to remain same at higher frequencies, we opt to use high frequency in the lab to analyse the signal. The test will be carried out on an open circuit, from the basic understanding of the circuit theory, the signal will be fully reflected considering looses along the line.

1.3 Introduction

This project is concerned with mathematical representation, and analysis of travelling waves on power system transmission lines. The transmission lines are mathematically represented by Telegraphers equations.

The exact solution of any transmission line and the one required for a high degree of accuracy in calculating 50-Hz lines more than approximately 200 km long must consider the fact that the parameters of the line are not lumped but distributed uniformly throughout the length of the line. [1]

There are several reasons why such an analysis of transmission lines is of benefit to engineering science. The ability to describe the propagation of signal on power system network is becoming increasingly important as system voltage rises and increase in system complexity. Protection, insulation and other over voltage related problems benefit from the better understand of the travelling waves on the transmission lines. [1]

1.4 Research description

To achieve the exact solution of a transmission line and with a high level of accuracy in calculating 50Hz lines more than 200km long one must consider the fact that parameters of the line are not lumped but distributed uniformly throughout the length of the line [2].

The figure 1 shows one phase and the neutral connection of a three phase line. In figure 1.a consider differential element of length dx in the line at a distance x from the receiving end of the line. Then z dx and y dx are respectively, series impedance and shunt admittance of the elemental section. V and I are phasors which vary with x [2].

Fig 1.a Schematic diagram of a transmission line showing one phase and neutral return.

If we consider a small change in length and calculate the difference in voltage and current at any point in the line. We should let x be distance measured from the receiving end of the line to the small element of the line, if we let length of the element be dx. Then zdx is the series impedance of the elementary line and ydx will be the shunt admittance [2].

1.4.1 Equations derivation

V+dV–V=Izdx

∴dVdx=zI (1)

Similarly,

(

I+dI)–I=Vydx

∴dIdx=yV (2)

Since there are multiple variable, let us differentiate Equations. (1) And (2) with respect to x, and we obtain:

d2Vdx2=zdIdx (3)
and
d2Idx2=ydVdx (4)

If we substitute the value of

dI/dxand

dV/dxfrom equation (1) and equation (2) in equation (3) and equation (4), respectively, we obtain

d2Vdx2=yzV (5)

and

d2Idx2=yzI (6)

Since now we have equation 5 the only variables are V and x and equation 6 in which only variable is I and x. The solution of equation 5 and 6 must yield the original expression time the constant yz when differentiated twice.

Since homogenous equation of second order equation is

Aekx+Bekx  , hence the solution is assumed to be:

V=A1ezyx+ A2e–zyx (7)

If we take second derivative of V with respect to x in equation (7) yields

d2Vdx2=yz(A1ezyx+ A2e–zyx) (8)

Equation (8) is the solution of Equation (5), which is yz times the assumed solution of V.

If we take the solution of equation (7) and plug it into equation (1) we get

I=1zyA1ezyx– 1zyA2e–zyx (9)

The constant Aand A2 can be evaluated since at x = 0, V=Vand I = IR. Substitution of these values in equation (7) and equation (9) we get,

VR=A1+A2

and

IR=1z/y(A1–A2)

If we let Z=

z/yand solving for Aand A2 gives:

A1=VR+IRZC2

and

A2=VR–IRZC2

Then, substituting the values for Aand A2 in equation (7) and (9) and letting

γ=yzand called propagation constant.

V=VR+IRZC2eγx+VR–IRZC2e–γx (10)
I=VR/ZC+IR2eγx–VR/ZC–IR2e–γx (11)

Equation (10) and equation (11) gives RMS value of the voltage and current and their phase angle at any point on the line in terms of distance x from the receiving end of the line, provided VR, Iand the parameters of the line are known.

1.4.2 Interpretation of the equations

Both

γand

ZCare complex quantities. The real part of the propagation constant

γis called attenuation constant α and is measured in nepers per unit length. The quadrature part of

γis called the phase constant β and is measured in radian per unit length. Thus,

γ=α+jβ

And therefore equations (10) and (11) becomes

V=VR+IRZC2eαxejβx+VR–IRZC2e–αxe–jβx (12)
I=VR/ZC+IR2eαxejβx–VR/ZC–IR2e–αxe–jβx (13)

The properties

eαxand

ejβxhelps explain how the magnitude and the phase changes as the signal travel along the transmission line at any instance as a function of distance x from the receiving end.

From the first term of equation (12),

VR+IRZC2eαxejβx  increases in magnitude and advances in phase. However, as the signal advances towards the receiving end is considered from the sending end, the term diminishes in magnitude and is retracted in phase. It is the similar behaviour of a rope tied on one end, which varies in magnitude with time at any point and maximum value diminishes as it approaches the receiving end [2].

The second term

VR–IRZC2e–αxe–jβxdecreases in magnitude and rerated in phase from the receiving end to the sending end. It is called the reflected voltage. This shows that at any point along the line the voltage is the sum of the component incident and reflected voltage sat that point.

Since equation (13) for current is the same, current is incident and reflected.

From the above equations (12) and (13), if the characteristic impedance of the line matches the load impedance there will be no reflected voltages or currents.

1.4.3 Hyperbolic form of equations

The differential equation of the transmission lines is helpful in better understanding the behaviour of the signals on the transmission lines. But however more convenient way of expressing the voltages and currents on transmission lines are using hyperbolic functions.

If we re – arrange equations (10) and (11) it gives:

V= VReγx+e–γx2+IRZCeγx–e–γx2 (14)
I= VR/ZCeγx–e–γx2+IReγx+e–γx2 (15)

Recognising the hyperbolic sin and cos equation (14) and equation (15) yields;

V= VRcosh⁡γx+IRZCsinh⁡γx (16)
I= VRZCsinh⁡γx+IRcosh⁡γx (17)

If we let

x=lto obtain voltages and currents at the sending end, we obtain;

VS= VRcosh⁡γl+IRZCsinh⁡γl (18)
Is= VRZCsinh⁡γl+IRcosh⁡γl (19)

Solving for

VRand

IRfrom equations (18) and (19) in terms of

VSand

ISyields;

VR= VScosh⁡γl–ISZCsinh⁡γl (20)
IR= IScosh⁡γl–VSZCsinh⁡γl (21)

1.5 Preliminary design

1.5.1 System block diagram

Oscilloscope

Open circuit transmission line

Function generator

1.5.2 Design process

1.5.3 Circuit diagram

The diagram below shows one phase of open circuit transmission line with Zbeing the characteristics impedance and V the voltage function generator.

1.5.4 System description

There is an increase usage of electricity in day to day activity, and transmission lines plays an important role in conducting electricity form one point to another. As the system voltages and complexity are rising it is important to understand the behaviour of the signals as the approach several different types load.

The system available for this project in the lab has the following characteristics:

  1. A single-phase stainless-steel conductor.
  2. The transmission line is 965mm long.
  3. The spacing between the two conductor is 5mm.
  4. The diameter of the line is 2mm.

Using the above parameters and the power system knowledge the capacitance, inductance and resistance of the line was calculated. Refer to the appendix below for the calculations.

If there is no load on a line,

IR=0and as determined from the above equations (12) and (13), incident voltage will equal to reflected voltage in terms of magnitude and phase at the receiving end [2].

To achieve a long transmission line effect in lab it is recommended to use higher frequencies. When dealing with higher frequencies losses are often neglected, which means that resistance and conductance are zero and characteristics impedance reduces to

L/C. Also, propagation constant

γ= zyreduces to

jβ=jωLC/l(imaginary number) as attenuation constant α is zero for lossless lines [2]. In this case, from the calculation’s frequency used in this project will be around 15MHz.Refer to appendix.

For this project equations (18) and (19) at the sending end reduces to

VS= VRcosh⁡βl

and

IS=VRZCsinh⁡βl

In exponential form

VS= VReβl+e–βl2

and

IS= VRZCeβl–e–βl2

These voltage equation shows that at any point along the line from sending end, voltage will be the sum of two components i.e. incident and reflected voltages.

1.6 Hardware / software description

1.6.1 Hardware

There were several hardware components used to achieve the deliverables.

  1. Vernier calliper – A pair of callipers was used to measure the diameter of transmission line in lab 401.Refer to appendix.
  2. Tape measure – Tape measure was used to measure the length of the line.

1.6.2 Software

  1. MATLAB software – The use of MATLAB software was important to check how the transmission signal behave when approached to open circuit.

Conclusion

The project is currently proceeding behind schedule for two weeks as some of the activities took longer than planned for it. All the adequate research and calculations has been done for the project to enter the next phase. Initially, MATLAB program was to be completed in this phase of the project, but it’s now delayed for the next phase. The next phase of the project will include writing the MATLAB code and taking practical measures in the lab and the two results will be compared for analysis as per the project deliverables set for the project to be completed.

Appendix

Budget

2018 Project Budget
Project: Long transmission line analysis
Designer: Sanjay Mepani
Supervisor Nigel Shepstone
EXPENDITURE  
A: SALARY  
Design charge rate:   $23.50 <—- enter $/hr
Budget Actual Budget – Actual
Start date Hours Cost Start date Hours Cost Hours Cost
1 Initial negotiation (ie understanding the problem and requirements). 15-Aug-18 10 $235.00 15-Aug-18 8 $188.00 $47.00
2 initial plan, budget & contract 21-Aug-18 10 $235.00 29-Aug-18 12 $282.00 ($47.00)
3 Meeting supervisor 15-Aug-18 25 $587.50 15-Aug-18 6.5 $152.75 $434.75
4 Background history 20-Aug-18 12 $282.00 20-Aug-18 7 $164.50 $117.50
5 Understanding Equations 04-Sep-18 17 $399.50 04-Sep-18 13.5 $317.25 $82.25
6 MATLAB Programming 15-Oct-18 15 $352.50 01-Nov-18 5 $117.50 $235.00
7 Pratical Measures 11-Mar-19 15 $352.50 $0.00 $352.50
10 Preparation of Interim Report 08-Oct-18 10 $235.00 $0.00 $235.00
11 Preparation of Presentation 03-Apr-19 15 $352.50 $0.00 $352.50
12 Preparation of Final Report 09-Apr-19 25 $587.50 $0.00 $587.50
154 $3,619.00 52 $1,222.00 $2,397.00
B: FINAL QUOTE & PROFIT / (LOSS)  
Overhead allowance: 50.00%
Profit 40.00%
Budget   Actual  
Estimated Cost before profit: $5,428.50 $1,833.00
Gross Profit: $2,171.40 Profit: $5,766.90
Quoted Price: $7,599.90 $7,599.90

Time plan

Calculations

CL=0.0121log10⁡501

=7.12×10–9F/km

L=0.1+0.92log10⁡50

=1.66mH/km

∴V=17.12×10–9×1.66×10–3 =290874.70Kms–1

Since

F= VλBy converting velocity V from

Kms–1to

ms–1we get frequency as:

∴F = 290874707.219.3 ≈15MHz

Calculating for beta gives

β=2πλ=2π19.3=0.325 rad m–1

Total capacitance, inductance and the resistance of the line is as follows

CT=0.0121log10⁡501 ×0.00965

CT=6.87×10–11F

LT=0.1+0.92log10⁡50×0.00965

LT=0.016mH

 

RT= ρlA

Since the line is made of stainless steel its resistivity is found to be

70×10–8Ω/mRT=70×10–8×9.65π×0.0012

RT=2.15Ω

Characteristic impedance (Zc) for the line is

LC=0.016×10–36.87×10–11

=482.59Ω

Photos

Vernier calliper used to take readings.

References

[1] P. Barnett, “A thesis presented for the Degree of Doctor of Philosophy in Electrical Engineering in the University of Cantebury,Christchurch ,New Zealand,” University of Cantebury, Christchurch, 1974.
[2] J. John J. Grainger & William D. Stevenson, POWER SYSTEM ANALYSIS, McGraw-hill,Inc., 1994.


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