Long Transmission Lines Analysis
Progress Report
Contents
1.4.2 Interpretation of the equations
1.4.2 Hyperbolic form of equations
1.6 Hardware / software description
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]
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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/dx
and dV/dx
from 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 A1 and A2 can be evaluated since at x = 0, V=VR and 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 Zc = z/y
and solving for A1 and A2 gives:
A1=VR+IRZC2
and
A2=VR–IRZC2
Then, substituting the values for A1 and A2 in equation (7) and (9) and letting γ=yz
and called propagation constant.
| V=VR+IRZC2eγx+VR–IRZC2e–γx | (10) | ||
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| I=VR/ZC+IR2eγx–VR/ZC–IR2e–γx | (11) | ||
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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, IR and the parameters of the line are known.
1.4.2 Interpretation of the equations
Both γ
and ZC
are complex quantities. The real part of the propagation constant γ
is called attenuation constant α and is measured in nepers per unit length. The quadrature part of γ
is 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) | ||
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| I=VR/ZC+IR2eαxejβx–VR/ZC–IR2e–αxe–jβx | (13) | ||
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The properties eαx
and ejβx
helps 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–jβx
decreases 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) | |||
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| I= VR/ZCeγx–e–γx2+IReγx+e–γx2 | (15) | |||
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Recognising the hyperbolic sin and cos equation (14) and equation (15) yields;
| V= VRcoshγx+IRZCsinhγx | (16) | ||
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| I= VRZCsinhγx+IRcoshγx | (17) | ||
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If we let x=l
to obtain voltages and currents at the sending end, we obtain;
| VS= VRcoshγl+IRZCsinhγl | (18) | ||
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| Is= VRZCsinhγl+IRcoshγl | (19) | ||
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Solving for VR
and IR
from equations (18) and (19) in terms of VS
and IS
yields;
| VR= VScoshγl–ISZCsinhγl | (20) | ||
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| IR= IScoshγl–VSZCsinhγl | (21) | ||
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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 Zc being 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:
- A single-phase stainless-steel conductor.
- The transmission line is 965mm long.
- The spacing between the two conductor is 5mm.
- 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=0
and 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= zy
reduces 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= VRcoshβl
and
IS=VRZCsinhβl
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.
- Vernier calliper – A pair of callipers was used to measure the diameter of transmission line in lab 401.Refer to appendix.
- Tape measure – Tape measure was used to measure the length of the line.
1.6.2 Software
- 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=0.0121log10501
=7.12×10–9F/km
L=0.1+0.92log1050
=1.66mH/km
∴V=17.12×10–9×1.66×10–3 =290874.70Kms–1
Since F= Vλ
By converting velocity V from Kms–1
to ms–1
we 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=0.0121log10501 ×0.00965
CT=6.87×10–11F
LT=0.1+0.92log1050×0.00965
LT=0.016mH
RT= ρlA
Since the line is made of stainless steel its resistivity is found to be 70×10–8Ω/m
RT=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. |
