Estimation of Capacitance and Inductance for Electrical Transmission Lines


Estimation of Capacitance and Inductance for Electrical Transmission lines



Here's a quick intro to the ideas which have been covered in greater detail in the tutorial document at the end of this page



As, we have learn from the previous tutorials, Electricity is generated at massive power plants which are generally located away from the load points. The generated electricity is transmitted from the plants to load centers through transmission lines. Hence the study of transmission lines is vital in context of understanding the voltage and current levels sent and received. Also, the transmission line losses need to minimized.

Characteristics of a conductor : 1. Magnetic field density

When a current is sent through a conductor of length l, then there is a magnetic field created all around the conductor. The field is given as shown below

The above diagram indicates the direction of the field around a conductor having a current of I flowing in it. The magnitude of the flux is given by

B= μI4πsinθr2

For an infinite conductor the above formula becomes B= μI2Rπ
Here we discuss another important property magnetic field intensity which is defined as H = B /μ .
Problem 1 : Calculate the magenetic flux density of 5A current carrying conductor of the semi infinite length at a distance of 5 m from its center perpendicularly located to it.
(Solved in the tutorial document at the end of this page).


Inductance of  1-Φ Transmission line :

The TL is as shown below. Now we are interested in calculating the inductance effect of forward conductor on return conductor. The converse will be also be same due to symmetry.

The inductance in the above cases is due to two main factors o n forward conductor. First : internal inductance due the current from inside creating a flux on the external conductor. The internal flux linkage which is denoted by λ = flux * no of turns . This expression is valid for all cases.


Problem : 2  Calculate the inductance for unsymmetrical 3 phase transmission line having 3 conductors which are separated by distances a,b, c and having the currents as Ia Ib Ic respectively.
(Solved in the tutorial document at the end of this page).






Transposition of the Transmission lines

Transposition means exchanging of the position of the transmission line after equal distances. Even if the voltage across the conductors are equal during the transmission, there will be slight differences and will result in varied inductances of the conductors. Hence if we have symmetrical conductors running through transmission line in a similar fashion, there may be unbalances in voltage. Also mutual inductances vary for untransposed lines. The topology for a transposed line is shown below. The transposing also reduces the interference caused by power lines in the telephone lines. As the current carrying conductors transpose regularly , the effect of a conductor is not much felt on the other line.

In modern days, the transposing is not being done. It is being t=done only at intermediate sub stations or etc. Also the difference in inductance is very little and it is often taken as the average and added to inductance of each semiconductor. The negligible small nature is due to asymmetrical placing of the semiconductors. Any practical transmission line can be viewed as an example.

Conductors practically used

Some of the conductors which are practically used are stranded and composite as they are made up of two or more materials each. Some of the materials are Aluminum Conductor Steel reinforced (ACSR), hollow copper conductors, stranded  conductors. Some times inner copper followed by outer aluminum strands are used to make conductors of high voltage carrying capability.

ACSR has very good advantages like high tensile strength and high conductance and resistance to rust. These conductors are used now for very wide variety of applications.  ACSR wires are cheaper than their copper counter parts and we can afford them at no compromise on quality and reliability if the system. It has superior strength which can be used to have greater spans of conductor lengths and supporting structures and other equipment becomes cheaper. A reduction is number of supports also decreases the number of outages of the transmission lines.


Inductance calculation of some configurations : problems



( These problems have been solved in the tutorial document at the end of this page).


Problem 3 : Determine the inductance of a 3 phase line operating at 50 Hz and arranged as follows. The conductor diameter is 1 cm.



Problem : 4 :

Determine the inductance of the 1 phase system as show  below. The conductors radius are 2.5mm for a,b,c and 5mm for d,e

R1  corresponding to the conductor with radius 2.5 mm can be obtained as given in the above problem. Self GMD of a,b,c  = 0.001947m.


Problem : 5 :

          A bundled conductor having seven identical strands each having radius of r. Determine the factor by which r should be multiplied to find the self GMD of the conductor. The configuration of the conductor is as shown below. Neglect the spacing between the conductors and assume that all are in perfect contact ie distance between centers of any two conductors is 2r.


Steps to solve problems on finding inductance of transmission line :
  1. Find the distance of each conductor from the other.
  2. Take a conductor and multiply all distances of it from every other conductor.
  3. Square root the obtained result with number of conductors
  4. Now, multiply all such results obtained for each conductor
  5. Again square root the result with total number of conductors after multiplying with a constant 0.7788
  6. For the mutual GMD, we need to do the same procedure but each time take a conductor and multiply with distances from the opposite group of conductors.
  7. Finally, square root with both number of conductors in each group.
  8. Substitute in the formula above to obtain the expression.

Skin effect and Proximity effect in conductors

        When direct current flows in the conductor . the current is uniformly distributed across the section of the conductor whereas the alternating current is not uniform with the outer filaments of the conductor carrying more current than the filaments closed to the center. This results in higher resistance to alternating current  than the direct current and is popularly known as skin effect. The effect is prominently observed in the supplies having higher frequencies ie a conductor having 60 HZ supply exhibits higher skin effect than a conductor having 50 Hz supply. The effect is also dependent on the size of the conductor. The flux linkage/amperes of the inner conductor is more because they link only inner elements and the flux linkage for the outer conductor is low because it links both inner and outer layers. Hence the inner ones have higher inductance and lower current.The alternating magnetic flux in a conductor caused by the current flowing in a neighboring conductor gives rise to the circulating currents which cause and apparent increase in the resistance of a conductor. This phenomenon is called proximity effect.




Capacitance of the Transmission lines


               The above can be shown as a practical representation of Transmission line in the form of a circuit. It is simply two capacitors in shunt with a series inductor. Having understood calculating the inductance in the previous part of tutorial, now we try to understand the capacitance issue. Actually line capacitance in the Transmission line is represented as two capacitance in shunt as C/2 and C/2.

Capacitance is proportional to the length of the line. So for short line we virtually assume no capacitance and this gives the results with high accuracy. Inductance is a phenomenon of Magnetic field where as Capacitance is due to electric field. For a current carrying conductor, the electrical field intensity at a point in space is given by following expression. The capacitor can be defined as a electric device with two conductors separated by a dielectric medium and which stores the electrostatic energy and the capacitance is the fundamental quantity for a capacitor and is defined as the ration of charge on one of the conductors to the potential difference between the conductors. Covered in greater detail in the tutorial document at the end of this page.

                        


( These problems have been solved in the tutorial document at the end of this page).                                                   

Problem : Determine the charging current and capacitance per km when the transmission line given below is operated at 132kv.  


Problem : Determine the capacitance and the charging current for the problem a single phase symmetrical transmission line separated by a h = 0.5m and r = 0.4 cm .Line voltage is to be taken as 220KV


Effect of Earth on Capacitance of a Transmission Line

             While calculating the capacitance , we assume  that the flux from conductors emanate from the positive conductor and terminate at infinity in a imaginary conductor. If earth is take into consideration, the conductor lines will have to cut the equipotential surface of the earth which extend to infinity. Thus this phenomenon should be accounted while calculating the conductors capacitance. Another explanation for this can be inductance is not grounded but the capacitance is grounded in transmission line modeling.To take the effect of earth into consideration , we take the principal of images by Kelvin to understand the behavior of a infinite equipotential surface. The principle states that, we can replicate the situation by removing the equipotential surface  and adding a imaginary conductor at a distance of the conductors distance from the earth at exactly the opposite side of the present conductor. This is illustrated as below.

   We require the capacitance to be calculated  for the conductor given above which is at a h height. For this, we assume the equipotential earth is absent and place a conductor of same but opposite potential charge at a same distance in the opposite side of the earth as stated by Kelvin’s principle. Now it’s a regular transmission line problem


C=2Πεolog2hr


( This problem have been solved in the tutorial document at the end of this page).  

  Problem : In the above case, calculate the capacitance of a conductor which is placed above the earth at a height of 5m and with a radius of 30 cms and compare it with capacitance with earth neglected3 From the results, we can understand that the earth effect will increase the capacitance of a line.


Complete Tutorial Document with Problems and Solutions :