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Transmission and Distribution 10EE53

Solution to Question Bank

Chapter 1. Basics of Transmission and Distribution

1. Draw the line diagram of a typical power supply scheme indicating the standard voltages Jan/Feb- 2013

Corona on transmission lines causes power loss, radio and television interference, and

audible noise (in terms of buzzing, hissing, or frying sounds) in the vicinity of the line.A

transmission line should he designed to operate just below the disruptive critical voltage

in fair wether so that corona only takes place during adverse atmospheric conditions.

Therefore, the calculated disruptive critical voltage is an indicator of corona performance

of the line. However, a high value of the disruptive critical voltage is not the only

criterion of satisfactory corona performance. The sensitivity of the conductor to foul

weather should also be considered (e.g., corona increases more slowly on stranded

conductors than on smooth conductors). Due to the numerous factors involved, the

precise calculation of the peak value of corona loss is extremely difficult, if not

impossible. The minimum voltage at which the ionization occurs in fair weather is called

the disruptive critical voltage and can be determined from

where E0= value of electric stress (or critical gradient) at which disruption startsin kilovolts per centimeters

V0= disruptive critical voltage to neutral in kilovolts (rms)r = radius of conductor in centimetersD = spacing between two conductors in centimeters

Since, in fair weather, the E0 of air is 21.1 kV/cm rms,


which is correct for normal atmospheric pressure and temperature (76cm Hg at 25°C). For other atmospheric pressures and temperatures,

where δ is the air density factor. Further, after making allowance for the surface condition of the conductor by using the irregularity factor, the disruptive critical voltage can be expressed as

where m = irregularity factor (0<m0 1)=1 for smooth, polished, solids, cylindrical conductors

= 0.93—0.98 for weathered, solid, cylindrical conductors =0.87—0.90 for weathered conductors with more than evenstrands

2. What are the effects of corona? Derive expression for the visual critical voltage. Explain the following terms with reference to corona: Disruptive critical voltage; ii) Visual critical voltage; iii) Corona power less.(06) Dec2015, Jan/Feb-2013

Corona on transmission lines causes power loss, radio and television interference, and

audible noise (in terms of buzzing, hissing, or frying sounds) in the vicinity of the line. At

extra high-voltage levels (i.e., at 345 kV and higher), the conductor itself is the major source

of audible noise, radio interference, television interference, and corona loss. The audible

noise is a relatively new environmental concern and is becoming more important with

increasing voltage level. For example, for transmission lines up to 800 kV, audible noise and

electric field effects have become major design factors and have received considerable

testing and study. It had been observed that the audible noise from the corona process mainly

takes place in foul weather. In (Icy conditions, the conductors normally operate below the

corona detection level, and therefore, very few corona sources exist. In wet conditions,

however, water drops on the conductors cause large number of corona discharges and a

resulting burst of noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise

is the limiting environmental design factor.


A transmission line should he designed to operate just below the disruptive critical

voltage in fair wether so that corona only takes place during adverse atmospheric conditions.

Therefore, the calculated disruptive critical voltage is an indicator of corona performance of

the line. However, a high value of the disruptive critical voltage is not the only criterion of

satisfactory corona performance. The sensitivity of the conductor to foul weather should also

be considered (e.g., corona increases more slowly on stranded conductors than on smooth

conductors). Due to the numerous factors involved, the precise calculation of the peak value

of corona loss is extremely difficult, if not impossible. The minimum voltage at which the

ionization occurs in fair weather is called the disruptive critical voltage and can be

determined from

where E0= value of electric stress (or critical gradient) at which disruption starts inkilovolts per centimetersV0= disruptive critical voltage to neutral in kilovolts (rms)

r = radius of conductor in centimetersD = spacing between two conductors in centimetersSince, in fair weather, the E0 of air is 21.1 kV/cm rms,




where m = irregularity factor (0<m0 1)=1 for smooth, polished, solids, cylindrical conductors

= 0.93—0.98 for weathered, solid, cylindrical conductors= 0.87—0.90 for weathered conductors with more than seven strands

=0.80—0.87 for weathered conductors with up to seven strandsNote that at the disruptive critical voltage V,, there is no visible corona. In the event that the potential difference (Or critical gradient) is further increased, a second point is reached at which a weak luminous glow of violet color can be seen to surround each conductor. The voltage value at this point is called the visual critical voltage and is given by

where Vv=visual critical voltage in kilovolts (rms) mv= irregularity factor for visible

corona (0<m1, 1) =1 for smooth, polished. solid, cylindrical conductors

visual corona on weathered stranded conductors

Note that the voltage equations given in this section are for fair weather. For wet weather

voltage values, multiply the resulting fair weather voltage values, multiply the resulting fair

weather voltage values by 0.80. For a three-phase horizontal conductor configuration, the

calculated disruptive critical voltage should be multiplied by 0.96 and 1.06 for the middle

conductor and for the two outer conductors, respectively.

3.Write short note on disruptive critical voltage. July/Aug-2014

A transmission line should he designed to operate just below the disruptive critical

voltage in fair wether so that corona only takes place during adverse atmospheric conditions.

Therefore, the calculated disruptive critical voltage is an indicator of corona performance of

the line. However, a high value of the disruptive critical voltage is not the only criterion of

satisfactory corona performance. The sensitivity of the conductor to foul weather should also

be considered (e.g., corona increases more slowly on stranded conductors than on smooth

conductors). Due to the numerous factors involved, the precise calculation of the peak value

of corona loss is extremely difficult, if not impossible. The minimum voltage at which the

ionization occurs in fair weather is called the disruptive critical voltage and can be


determined from

4.Explain the terms with reference to corona.i)disruptive critical voltage ii) Power loss due to corona Jul/Aug-2013

The minimum voltage at which the ionization occurs in fair weather is called the disruptive critical voltage and can be determined from

where E0= value of electric stress (or critical gradient) at which disruption starts in kilovolts per centimeters V0= disruptive critical voltage to neutral in kilovolts (rms)r = radius of conductor in centimeters D = spacing between two conductors in centimeters

Since, in fair weather, the E0 of air is 21.1 kV/cm rms,




Where m =irregularity factor (0<m0 1)=1 for smooth, polished, solids, cylindrical conductors

Where, f =frequency in hertzV = line-to-neutral operating voltage in kilovoltsV0 = disruptive critical voltage in kilovolts

The wet weather corona can be calculated from the above equations by multiplying V0by 0.80. Peek’s equation gives a correct result if

(1) the frequency is between 25 and 150 Hz, (2) the conductor radius is greater than 0.25 cm. and (3) the ratio of V to V0 is greater than 1.8.

The power LOSS IS proportional to the square root of the size of the conductor.Therefore, the larger the radius of the conductor, the larger the power loss. Also, the larger the spacing between conductors, the smaller the power loss. Similarly,

that is, for a given voltage level, the larger the conductor site, the larger the disruptive critical voltage and therefore the smaller the power loss. According to Peterson (11], the fair weather corona loss per phase or conductor can

where d = conductor diameterD = spacing between conductorsf = frequency in hertzV= line-to-neutral operating voltage in kilovoltsF= corona factor determined by test and is a function of ratio of V to V0

In general, the corona losses due to fair weather conditions arc not significantly large at extra-high-voltage range. Therefore, their effects are not significant from technical andlor economic points of view. Whereas the corona losses due to foul weather conditions are very significant.


5.Write short note on corona in transmission lines. Jul/Aug-2015

Transmission lines causes power loss, radio and television interference, and audible noise

(in terms of buzzing, hissing, or frying sounds) in the vicinity of the line. At extra high-

voltage levels (i.e., at 345 kV and higher), the conductor itself is the major source of audible

noise, radio interference, television interference, and corona loss. The audible noise is a

relatively new environmental concern and is becoming more important with increasing

voltage level. For example, for transmission lines up to 800 kV, audible noise and electric

field effects have become major design factors and have received considerable testing and

study. It had been observed that the audible noise from the corona process mainly takes place

in foul weather. In (Icy conditions, the conductors normally operate below the corona

detection level, and therefore, very few corona sources exist. In wet conditions, however,

water drops on the conductors cause large number of corona discharges and a resulting burst

of noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise is the limiting

environmental design factor.

6.Write short note on factors affecting corona and methods of reducing corona effect. Jan/Feb-15


Causes power loss, radio and television interference, and audible noise (in terms of

buzzing, hissing, or frying sounds) in the vicinity of the line. At extra high-voltage levels

(i.e., at 345 kV and higher), the conductor itself is the major source of audible noise, radio

interference, television interference, and corona loss. The audible noise is a relatively new

environmental concern and is becoming more important with increasing voltage level. For

example, for transmission lines up to 800 kV, audible noise and electric field effects have

become major design factors and have received considerable testing and study. It had been

observed that the audible noise from the corona process mainly takes place in foul weather.

In (Icy conditions, the conductors normally operate below the corona detection level, and

therefore, very few corona sources exist. In wet conditions, however, water drops on the

conductors cause large number of corona discharges and a resulting burst of noise. At

ultrahigh-voltage levels (1000 kV and higher), such audible noise is the limiting


7. Explain the terms with reference to corona.i)visual critical voltage ii)Power loss due to corona Jul/Aug-2015

Assume that a three-phase overhead transmission line is made up of three equilaterally

spaced conductors, each with overall diameter of 3 cm. The equilateral spacing between

conductors is 5.5 m. The atmosphere pressure is 74 cm Hg and the temperature is 10°C. If

the irregularity factor of the conductors is 0.90 in each case, determine the following:

(a) Disruptive critical rms line voltage.


(b) Visual critical rms line voltage.

9. Assume that a three-phase overhead transmission line is made up of three equilaterally spaced conductors, each with overall diameter of 3 cm. The equilateral spacing between conductors is 5.5 m. The atmosphere pressure is 74 cm Hg and the temperature is 10°C.Consider Example 8.1 and assume that the line operates at 345 kV at 60 Hz and the Peek’s formula. Jan/Feb-13,Jul/Aug13,2015


10. Draw the line diagram of a typical transmission and distribution scheme indicating the standard voltages and also explain feeder, distributor and services main of a distribution scheme.

December 2015

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11. Why it is necessary to use high voltage for power transmission? And state the effect of high voltage used in transmission on: i) Volume of copper required ii) Line efficiency iii) Line voltage drop.

JUL-2013• A process in the delivery of electricity to consumers, • Is the bulk transfer of electrical power. • Power transmission is between the power plant and a substation near a populated area. • Electric power transmission allows distant energy sources (such as hydroelectric power

plants) to be connected to consumers in population centers. Line voltage drop. • Line efficiency electricity distribution is the delivery from the substation to the

consumers.• Electricity is usually transmitted over long distance through overhead power transmission

lines. • Underground power transmission is used only in densely populated areas due to its high

cost of installation and maintenance, • A power transmission system is referred to as a "grid";

A transmission grid is a network of power stations, transmission circuits, and substations. Energy is usually transmitted within the grid with three-phaseac.

Volume of copper required• A substation is a high-voltage electric system facility. • It is used to switch generators, equipment, and circuits or lines in and out of a system. • It also is used to change ac voltages from one level to another, and/or change alternating

current to direct current or direct current to alternating current.

Chapter 2. Overhead Transmission Lines

1. Show that a transmission line conductor suspended between level support assumes the shape of a catenory. Derive the expression for sag.

December 2015,Jul/Aug 2013, Jul/Aug 2015CatenaryMethodFigure 1 shows a span of conductor with two supports at the same level and separated by a horizontal distance L.

Let ‘0’ be the lowest point on the catenary curve and ‘l’ be the length of the conductor between two supports.Let ‘w’ be the weight of the conductor per unit length, ‘T’ be the tension of the conductor at any point ‘P’ in the direction of the curve, and H be the tension at origin 0.Further, let ‘s’ be the length of the curve between points 0 and P, so that the weight of the portion s is ws. Tension T can be resolved into two components, T the horizontal component and T,, thc vertical component. Then, for equilibrium,

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Thus, the portion OP of the conductor is in equilibrium under the tension T at P. the weight wsacting vertically downward, and the horizontal tension H. In the triangle shown in Figure 2, ds represents a very short portion of the conductor, in the region of point P When s is increased by ds the corresponding x and y are increased by dxand dy, respectively.Hence,

Figure 1. Conductor suspended between supports at same elevation.

Integrating both sides gives

Therefore,

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Where, K is the constant of integration. When x = 0, $ = 0, and K = 0,

If the lowest point of the curve is taken as the origin, when x=0, y = 0,

since, by the series, cosh0=1. Therefore,

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2. Explain what is sag and why it is inevitable in over head transmission lines? What are the factors influencing it?

Dec/Jan 2015,Jul/Aug 2013,Dec/Jan 2013Conductor sag and tension analysis is an important consideration in overhead distribution

line design as well as in overhead transmission line design. The quality and continuity of

electric service supplied over a line (regardless of whether it is a distribution, a sub-

transmission, or a transmission line) depend largely on whether the conductors have been

properly installed. Thus, the designing engineer must determine in advance the amount of sag

and tension to be given the wires or cables of a particular line at a given temperature. In order

to specify the tension to be used in stringing the line conductors, the values of sag and

tension for winter and summer conditions must be known. Tension in the conductors

contributes to the mechanical load on structures at angles in the line and at dead ends.

3. With usual notations derive an expression for maximum sag of a tranmn. Line where the supports are at different levels? Jan2013/2014

Consider a span L between two supports, as shown in Figure 10.5, whose elevations differ by a distance h. Let the horizontal distance from the lowest point of the curve to the lower and the higher supports be x1 and x2, respectively. By using equation (10.46), that is,

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Figure 4. Supports at different levels.

By adding the above two equations,

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Figure 5. Case of negative x1

On Subtracting the same two equations,

In this equation,

Obtain the expression for sag in a freely suspended conductor when the supports are at equal levels.(8) – refer Q.No-1

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4. With usual notations derive an expression for maximum sag of a tranmn. Line where the supports are at different levels?

Jul2015/Jan13Sag and tension analysis is an important consideration in overhead distribution line

design as well as in overhead transmission line design. The quality and continuity of

electric service supplied over a line (regardless of whether it is a distribution, a sub-

transmission, or a transmission line) depend largely on whether the conductors have been

properly installed. Thus, the designing engineer must determine in advance the amount of

sag and tension to be given the wires or cables of a particular line at a given temperature.

In order to specify the tension to be used in stringing the line conductors, the values of

sag and tension for winter and summer conditions must be known. Tension in the

conductors contributes to the mechanical load on structures at angles in the line and at

dead ends. Excessive tension may cause mechanical failure of the conductor itself.

5. Obtain the expression for sag in a power conductor when the supports are at equal levels, taking into the effect of wind and ice loading

Dec/Jan 2015,Jul/Aug 2015,Dec/Jan 2013,Dec/Jan2014

In mountainous geographic areas, the thickness of ice formed on the conductor becomes

very significant. Depending on the circumstances, it might be as much as several times

the diameter of the conductor. Ice accumulations on the conductor affect the design of the

line

(1) by increasing the dead weight per foot of the line and

(2) by increasing the projected surface of the line subject to wind pressure. Mostly used for distribution lines.

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Figure 6. Probable configuration of ice covered conductor cross-sectional

Even though the more likely configuration of a conductor with a coating of ice is as

shown in Figure 6, for the sake of simplicity, it can be assumed that the ice coating, of

thickness t, inches, is uniform over the surface of a conductor, as shown in Figure 7.

Then the cross-sectional area of the ice is

6. Explain the effects of sag in overhead trasmn. Line.Dec/Jan 2015,Jul/Aug 2013,Dec/Jan 2013

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If itie ice load is assumed to be uniform the

volume of ice per foot isThe weight of the ice is 57 lb/It3, so that the weight of ice per foot is

or approximately

Therefore, the total vertical load on the conductor per unit length is

where wT= total vertical load on conductor per unit length w = weight of conductor per unit lengthw1= weight of ice per unit length

It is customary to assume that the wind blows uniformly and horizontally across the projected area of the conductor covered with no ice and ice, respectively. The projected area per unit length of the conductor with no ice is

where Sni= projected area of conductor covered with no ice in square feet per Unit length Ani = cross-sectional area of conductor covered with no icc in square feetl= length of conductor in unit lengthfor a 1-ft length of conductor with no ice,

whereas with ice, it is

where Swi= projected area of conductor covered with icc in square feet per unit length

Figure 8. Force of wind on conductor covered with no ice.

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7. Obtain the expression for sag in a power conductor when the supports are at equal levels, taking into the effect of wind and ice loading

Dec/Jan2015

Awi= cross-sectional area of conductor covered with ice in square feet1 = length of conductor in unit lengthfor a 1-ft length of conductor,

Therefore, the horizontal force exerted on the line as a result of the wind pressure with noice (Figure 8) is

for a 1-ft length of conductor,

where P= horizontal wind force (i.e., load) exerted on line in pounds per feetp =wind pressure in pounds per square feetwhereas with ice (Figure 9), it is

Figure 10


Therefore, the effective load acting on the conductor is

acting at an angle θto the vertical, as shown in Figure 10.

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By replacing w by wein the previously derived equations for tension and sag of the line in still air, these equations can he applied to a wind- and ice-loaded line. For example, the sag equation becomes

8. Write short note on effect of ice load and wind effect on sag of transmission Line. Dec/Jan 2015,Jul/Aug 2013,Dec/Jan 2013,Dec/Jan2014

The span design consists in determining the sag at which the line is constructed so that

heavy winds, accumulations of ice or snow, and excessive temperature changes will not

stress the conductor beyond its elastic limit, cause a serious permanent stretch, or result in

fatigue failures from continued vibrations, in other words, the lines will be erected under

warmer and nearly still-air conditions and yet must comply with the worst conditions.

In mountainous geographic areas, the thickness of ice formed on the conductor becomes very

significant. Depending on the circumstances, it might be as much as several times the

diameter of the conductor. Ice accumulations on the conductor affect the design of the line

(1) by increasing the dead weight per foot of the line and

(2) by increasing the projected surface of the line subject to wind pressure. Mostly used for

distribution lines.It is customary to assume that the wind blows uniformly and horizontally

across the projected area of the conductor covered with no ice and ice, respectively.

9. From the first principles derive the expression for sag in a freely suspended conductor when the supports are at unequal levels

Dec/Jan 2013,Jul/Aug 2015,Dec/Jan 2013

Conductor tension pulls the conductor up and decreases its sag. At the same time, tension

elongates the conductor, from elastic stretching, which tends to relieve tension and

increase sag. The elastic property of metallic wire is measured by its modulus of

elasticity. The modulus of elasticity of a material equals the stress per unit of area divided

by the deformation per unit of length. That is, since

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10. Give the list of supporting structures and the main components used in overhead transmission line. December 2015

In order to determine the conductor load properly, the factors that need to be taken into account are:

1. Weight of conductor itself.

2. Weight of ice or snow clinging to wire.

3. Wind blowing against wire.

The maximum effective weight of the conductor is the vector sum of the vertical weight

and the horizontal wind pressure. It is very important to include the most adverse

condition. The wind is considered to be blowing at right angles to the line and to act

against the projected area of the conductor, including the projected area of ice or snow

that may be clinging to it. Economic design dictates that conductor sag should be

minimum to refrain from extra pole height, so provide sufficient clearance above ground

level, and to avoid providing excessive horizontal spacing between conductors to prevent

them swinging together in mid-span.

Conductor tension pulls the conductor up and decreases its sag. At the same time, tension

elongates the conductor, from elastic stretching, which tends to relieve tension and

increase sag. The elastic property of metallic wire is measured by its modulus of

elasticity. The modulus of elasticity of a material equals the stress per unit of area divided

by the deformation per unit of length. That is, since

where σ= stress per unit area in pounds per square inchesT = conductor tension in poundsA = actual metal cross section of conductor in square inchesThe resultant elongation e of the conductor due to the tension is

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Of course, if the modulus of elasticity is low, the elongation is high, and vice versa. Thus,

a small change in conductor length has a comparatively large effect on conductor sag and

tension. Sags and stresses in conductors are dependent on the initial tension put on them

when they are clamped in place and are due to the weight of the conductors themselves,

to ice or sleet clinging to them, and to wind pressure. The stress in the conductor is the

parameter on which everything else is based. But the stress itself is determined by the sag

in the conductor as it hangs between adjacent poles or towers. Since the stress depends on

sag, any span can be used provided the poles or towers are high enough and strong

enough. The matter is merely one of extending the catenary in both loading. Thus, the

problem becomes the balancing of a larger number of lighter and shorter poles or towers

against a smaller number of heavier and taller ones.

11. What is sag in a conductor? Derive the expression for the sag when the supports are at equal heights. Jan 2013/December 2015

A conductor suspended freely from two supports, which are at the same level and

spaced L unit length apart, as shown in Figure 2 takes the form of a catenary curve

providing the conductor is perfectly flexible and its weight is uniformly distributed along

its length. if the conductor is tightly stretched (i.e., when sag d is very small in

comparison to span L), the resultant curve can be considered a parabola. If the

conductor’s sag is less than 6 percent of its span length, the error in sag computed by the

parabolic equations is less than 0.5 percent. If the conductor’s sag is less than 10 percent

of the span, the error is about 2 percent. In distribution systems, determining accurate

values of sag is not so important as it is in transmission systems. Nevertheless, even in

the distribution lines, if the conductor is strung with too low tension, the resultant sag will

be excessive, with the likelihood of wires swinging together and short-circuited. The

usual tendency, however, is to pull the

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conductor too tight, which causes the conductor to be overstressed and stretched when the

heaviest loading takes place and the normal sag after this loading becomes excessive.

Then the excessive sag needs to be pulled out of the conductor, a process that also causes

the conductor to be overstressed on heaviest loading. This process of overstressing and

pulling up may cause the conductors, especially the smaller ones, tohe broken. This can

be eliminated by measuring the line tension more accurately.

12. A transmission line conductor at a river crossing is supported from two towers of height of 40m and 30m respectively above water level. The horizontal distance between the towers is 300m. If the tension in the conductor is 1500 kg, find the clearance of the conductor at a point midway between the supports. Weight of the conductor is 0.8 kg/m. Assume bases of the tower to be at the water level.

December 2013

catenaryMethod

A span of conductor with two supports at the same level and separated by a horizontal distance L.

Let ‘0’ be the lowest point on the catenary curve and ‘l’ be the length of the conductor between two supports.

Let ‘w’ be the weight of the conductor per unit length, ‘T’ be the tension of the conductor

at any point ‘P’ in the direction of the curve, and H be the tension at origin 0. Further, let

‘s’ be the length of the curve between points 0 and P, so that the weight of the portion s is

ws.

Tension T can be resolved into two components, T the horizontal component and T,, thc vertical component. Then, for equilibrium,

Thus, the portion OP of the conductor is in equilibrium under the tension T at P. the weight wsacting vertically downward, and the horizontal tension H.

In the triangle shown in Figure 2, ds represents a very short portion of the conductor, in

the region of point P When s is increased by ds the corresponding x and y are increased

by dxand dy, respectively.

The factors affecting the sag of a conductor strung between supports arc: 1.Conductorloadperunitlength.

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2.Span,thatis,distancebetweensupports.

3.Temperature.

4.Conductor

Chapter 4. Insulators

1. Explain the various tests conducted on insulators. (7) Jul-2015

In mountainous geographic areas, the thickness of ice formed on the conductor becomes very significant. Depending on the circumstances, it might be as much as several times the diameter of the conductor. Ice accumulations on the conductor affect the design of the line

(1) by increasing the dead weight per foot of the line and (2)by increasing the projected surface of the line subject to wind

pressure. Mostly used for distribution lines.

Even though the more likely configuration of a conductor with a coating of ice is as shown in Figure 6, for the sake of simplicity, it can be assumed that the ice coating, of thickness t, inches, is uniform over the surface of a conductor, as shown in Figure 7. Then the cross-sectional area of the ice .

2. With usual notations, derive the general expression for the metal link of string to line capacitance, when guard ring is used for the string of insulators. Dec/Jan 2015

Page 25


If it ice load is assumed to be uniform throughout the length of the conductor, the volume

of ice per foot isThe weight of the ice is 57 lb/It3, so that the weight of ice per foot is

or approximately

Therefore, the total vertical load on the conductor per unit length is

where wT= total vertical load on conductor per unit lengthw = weight of conductor per unit length1= weight of ice per unit lengthEffect of WindIt is customary to assume that the wind blows uniformly and horizontally across theprojected area of the conductor covered with no ice and ice, respectively.The projected area per unit length of the conductor with no ice is

where Sni= projected area of conductor covered with no ice in square feet per Unit length Ani = oss-sectional area of conductor covered with no icc in square feetl= length of conductor in unit lengthfor a 1-ft length of conductor with no ice,

3. What is string efficiency in the context of suspension insulators? Explain themethods of improving the same.(8) Jul-2013

where Swi= projected area of conductor covered with icc in square feet per unit length

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Figure 8. Force of wind on conductor covered with no ice.

Awi= cross-sectional area of conductor covered with ice in square feet1 = length of conductor in unit lengthfor a 1-ft length of conductor,

Therefore, the horizontal force exerted on the line as a result of the wind pressure with noice (Figure 8) is


where P= horizontal wind force (i.e., load) exerted on line in pounds per feetp =wind pressure in pounds per square feet whereas with ice it is

4. Write short note on different types of O H line insulators.Dec/Jan 2013

A stress-crossing overhead sub-transmission line has a span of 500 ft over the stream.

The line is located in a heavy-loading district in which the horizontal wind pressure is 4

lb/ft2 and the radial thickness of the ice is 0.50 in. Use an ACSR conductor of 195 kcmil

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having an outside diameter of 1.093 in., a weight of 5399 lb/mi. and an ultimate strength

of 28,500 lb. Also use a safety factor of 2 and 57 lb/ft3 for the weight of ice. Using the

parabolic method, (a) Weight of ice in pounds per feet.

(h) Total vertical load on conductor in pounds per feet.

(c) Horizontal wind force exerted on line in pounds per feet.

(d) Effective load acting on conductor in pounds per feet.

(e) Sag in feet

4 & 5. Define string efficiency and hence calculate the mathematical expression for it. Explain the methods of improving the string efficiency.(6)

Jul-2015As the conductor becomes energized on each half cycle of the ac voltage wave,

the electrons in the air near its surface are accelerated toward the conductor on its

positive half cycle and away from the conductor on its negative half cycle. The velocity

attained by a free electron is dependent on the Intensity of the electric field. If the

intensity of the electric field exceeds a certain critical value, any free electron in this field

will acquire a sufficient velocity and energy to knock one of the outer orbit electrons

clear out of one of the two atoms of the air molecule. This process is called ionization,

and the molecule with the missing electron is called a positive ion. The initial electron,

which lost most of its velocity in the collision, and the electron According to Nasser ,

“coronas have various industrial applications

Page 28


6. What are the insulators with O H Lines? Discuss the desirable properties of insulators and name the types of insulators.

Jul2013Audible noise (in terms of buzzing, hissing, or frying sounds) in the vicinity of

the line. At extra high-voltage levels (i.e., at 345 kV and higher), the conductor itself

is the major source of audible noise, radio interference, television interference, and

corona loss. The audible noise is a relatively new environmental concern and is

becoming more important with increasing voltage level. For example, for

transmission lines up to 800 kV, audible noise and electric field effects have become

major design factors and have received considerable testing and study. It had been

observed that the audible noise from the corona process mainly takes place in foul

weather. In (Icy conditions, the conductors normally operate below the corona

detection level, and therefore, very few corona sources exist. In wet conditions,

however, water drops on the conductors cause large number of corona discharges and

Page 29


a resulting burst of noise. the limiting environmental design factor.

7. State the various properties of an insulator. Give the list of important insulators and make the comparision of pin and suspension type insulators.

Dec2015

Succinctly put, corona is a luminous partial discharge due to ionization of the air

surrounding a conductor caused by electrical overstress. Many tests show that dry air

at normal atmospheric pressure and temperature (25 °C and 76 cm barometric

pressure) breaks down at 29.8 kV/cm (maximum, or peak, value) or 21.1 kV/cm (rms,

or effective, value). There are always a few free electrons in the air due to ultraviolet

radiation from the sun, cosmic rays from outer space, radioactivity of the earth, etc.

As the conductor becomes energized on each half cycle of the ac voltage wave, the

electrons in the air near its surface are accelerated toward the conductor on its

positive half cycle and away from the conductor on its negative half cycle. The

velocity attained by a free electron is dependent on the Intensity of the electric field.

If the intensity of the electric field exceeds a certain critical value, any free electron in

this field will acquire a sufficient velocity and energy to knock one of the outer orbit

Page 30


electrons clear out of one of the two atoms of the air molecule. This process is called

ionization, and the molecule with the missing electron is called a positive ion. The

initial electron, which lost most of its velocity in the collision, and the electron

According to Nasser , “coronas have various industrial applications, such as in high-

speed printout devices, in air purification devices by electronic precipitators, in dry-

ore separation systems, as chemical catalysts, in radiation detectors and counters, and

in discharging undesirable electric charges from airplanes and plastics. Coronas are

used as efficient means of discharging other statically electrified surfaces of wool and

paper in the manufacturing industry.

8. Define string efficiency. Explain the method of calculating the string efficiencyfor a given three insulator string. Dec2015

As a rule of thumb, if the ratio of spacing between conductors w the radius

conductors before corona phenomenon occurs. Since for overhead lines this ratio is

much greater than 15, the flashover can be considered as impossible under normal

circumstances. At a given voltage level, the factors affecting corona include line

configuration, conductor type. condition of conductor surface, and weather. In a

horizontal configuration, the field near the middle conductor is larger than the field

near the outer conductors. Therefore, the disruptive critical voltage is lower for the

middle conductor, causing larger corona loss than the ones for the two other

conductors. If the conductors are not spaced cquilaterally, the surface gradients of the

conductors and therefore the corona losses are not equal. Also, the conductor height

affects the corona loss, that is, the greater the height, the smaller the corona loss. The

corona loss is proportional to the frequency of the voltage. Therefore, the higher the

frequency, the higher the corona losses. Thus, the corona loss at 60 Hz is greater than

the one at 50 Hz. Of course, the corona loss at zero frequency, that is, direct current,

is far less than the one for ac current. The irregularity of the conductor surface in

terms of scratches, raised strands, die burrs, die grease, and the particles of dust and

dirt that clog the conductor can significantly increase the corona loss.

Page 31


9. An insulator string consists of 3 units each having a safe working voltage of 15 KV. The ratio of self capacitance to shunt capacitance of each unit is 8:1. Find the maximum safe working voltage of the string. Also find the string efficiency. Dec2013, Jul-2013

Therefore, the calculated disruptive critical voltage is an indicator of corona

performance of the line. However, a high value of the disruptive critical voltage is not the

only criterion of satisfactory corona performance. The sensitivity of the conductor to foul





the disruptive critical voltage and can be determined from

where E0= value of electric stress (or critical gradient) at which disruption startsinkilovoltspercentimeters

V0= disruptive critical voltage to neutral in kilovolts (rms)

r = radius of conductor in centimeters

D = spacing between two conductors in centimeters Since, in fair weather, the E0 of air is 21.1 kV/cm rms,


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where m = irregularity factor (0<m0 1)

=1 for smooth, polished, solids, cylindrical conductors

= 0.93—0.98 for weathered, solid, cylindrical conductors

Chapter 5. Under ground Cables

Page 33


1. Draw the cross sectional view of a single core cable and explain the construction. Dec/jan2015

2. Compare the dielectric stress of a homogeneous cable with that of a capacitancegraded cable. Jul/Aug 2015

For transmission lines up to 800 kV, audible noise and electric field effects have become

major design factors and have received considerable testing and study. It had been

observed that the audible noise from the corona process mainly takes place in foul

weather. In (Icy conditions, the conductors normally operate below the corona detection

level, and therefore, very few corona sources exist. In wet conditions, however, water

drops on the conductors cause large number of corona discharges and a resulting burst of

noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise is the limiting


3. Write short notes on testing of cables. Dec/jan2013

Underground cables may have one or more conductors within a protective sheath.

The protective sheath is an impervious covering over insulation, and it usually is lead.

The conductors are separated from each other and from the sheath by insulating

Page 34


materials. The insulation materials used are Rubber is used in cables rated 600 V—35

kV, whereas polyethylene (FE), propylene (PP) and polyvinyl chloride (PVC) arc used in

cables rated 600 V—l38kV. The high-moisture resistance of rubber makes it ideal for

submarine cables. Varnished cambric is used in cables rated 600 V—28 kV. Oil-

impregnated paper is used in solid-type cables up to 69 kV and in pressurized cables up

to 345 kV. In the solid-type cables, the pressure within the oil-impregnated cable is not

raised above atmospheric pressure. In the pressurized cables, the pressure is kept above

atmospheric pressure either by gas in gas pressure cables or by oil in oil-filled cables.

Impregnated paper is used for higher voltages because of its low dielectric losses and

lower cost. Cables used for 59 kV and below are either (1) low pressure, not over 15 psi,

or (2) medium pressure, not over 45 psi. High-pressure cables, up to 200 psi, installed

pipes arc not economical for voltages of 69 kV and below. Voids or cavities can appear

as the result of faulty product or during the operation of the cable under varying load.

Bending the cable in handling and on installation, and also the different thermal

expansion coefficient of the insulating paper, the impregnating material and the lead

sheath result in voids in the insulation of cable not under pressure. The presence of higher

electrical field strength ionization that appears in the voids in the dielectric leads to

destruction of the insulation. The presence of ionization can he detected by means of the

power factor change as a test voltage is applied. The formation of voids is avoided in the

case of the oil-filled cable. With the gas-filled cable, the pressure in the insulation is

increased to such a value that existing voids or cavities are ionization free. Ionization

increases with temperature and decreases with increasing pressure.

4. Derive an expression for insulation resistance of a cable. Jul/Aug 2013

Assume that three 35-kV, 350-kcmil, single-conductor belted cables are located in

touching equilateral formation with respect to each other and the sheaths are bounded to

ground a several points. The cables are operated at 34.5kV and 60Hz. The cable has a

conductor diameter of 0.681 in., insulation thickness of 345 cmil, lead sheath thickness of

105 cmil, and a length of l0mi. Conductor ac resistance is 0.l9P/Wmi per phase at 50°C.

Calculatethefollowing:

(a) Mutual reactance between conductors and sheath.

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(b)Sheathresistanceofcable.

(c)increasein onductorresistance due to sheath currents.

(d)Total resistance of conductor including sheath loss.

5. Explain capacitance grading of cables with appropriate derivation. Dec/jan2014

The conductors used in underground cables can be copper or aluminum.

Aluminum dictates larger conductor sizes to carry the same current as copper. The need

for mechanical flexibility requires stranded conductors to be used. The equivalent

aluminum cable is lighter in weight and larger in diameter in comparison to copper cable.

Stranded conductors can be in various configurations, for example, concentric,

compressed, compact,andrope. Cables are classified in numerous ways. For example,

they can be classified as (1) underground, (2) submarine, and (3) aerial, depending on

location. They can be classified according to the type of insulation, such as (1) rubber and

rubberlike compounds, (2) varnished cambric, and (3) oil-impregnated paper. They can

be classified as (1) single conductor, (2) two conductor duplex, three conductor, etc.,

depending on the number of conductors in a given cable. They can be classified as

shielded or nonshielded (belted), depending on the presence or absence of metallic

shields over the insulation. Shielded cables can be solid, oil filled, or gas filled. They can

he classified by their protective finish such as (1) metallic (e.g., a lead sheath) or (2)

nonmetallic (e.g., plastic). insulation shields help to (1) confine the electric field within

the cable; (2) protect cable better from induced potentials; (3) limit electromagnetic or

electrostatic interference; (4) equalize voltage stress within the insulation, minimizing

surface discharges; and (5) reduce shock hazard (when properly grounded). In general,

shielding should be considered for nonmetallic covered cables operating at a circuit

voltage over 2 kV and where any of the following conditions exist

6. What is meant by grading of cables? Briefly explain the various methods of grading. Dec/jan2014

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(f) Total sheath losses of feeder in watts if current in conductor is 400 A.

7. Write short note on thermal rating of cables.(5) Jul2015

There arc various methods for Locating faults in underground cables. The method

used for locating any particular fault depends on the nature of the fault and the extent of

the experience of the testing engineer. Cable faults In general, conductor failures are

located by comparing the capacity of the insulated conductors. On the other hand,

insulation failures are located by fault tests that compare the resistance of the conductors.

In short cables, the fault is usually located by inspection, that is, looking for smoking

manholes or listening for cracking sound when the kenetron is applied to the faulty cable.

The location of ground faults on cables of known length can be determined by means of

the balanced-bridge principle. Kenetron is a two-electrode high-vacuum tube. They arc

used as power rectifiers tor

Page 37


applications requiring low currents at high dc voltages, such as for electronic dust

precipitation and high-voltage test equipment

8. Derive expressions for the maximum and minimum dielectric stress in a single core cable and ontain the criterion for keeping the diectric stress to a minimum value. (6) Dec/jan2015

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9. State five advantages of using underground cables for power distribution.10. Write short note on laying of UG Cables. Jul 2015/jan2013

Newly installed cables should be subjected to a nondestructive test at higher than normal

use values. Megger testing is a common practice The word

Meggeris the trade name of a line of ohmmeters manufactured by the James G. Diddle

Company. Certain important information regarding the quality condition of insulation

can be determined from regular Megger readings that is a form of preventive

maintenance.

For example, Figure 13 shows a portable high-resistance bridge for cable-fault-locating

work. Faults can be between two conductors or between a conductor and its conducting

sheath, concentric neutral, or ground. Figure 14 shows a heavy-duty cable test and fault-

locating system, which can be used for either grounded or ungrounded neutral 15-kV

cables. The full 100 mA output current allows rapid reduction of high-resistance faults on

cables rated 35 kV ac or higher to the lcvcl of 25 kV or lower for fault- locating purposes.

Figure 15 shows a lightweight battery-operated cable route trace that can be used to

locate, trace, and measure the depth of buried energized power cables.

11. Derive an expression for inductance of a single phase, two-wire transmission line. DEC 2015

Page 39


12. Explain the terms self GMD and mutual GMD. Jan 2014

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13. A three phase transmission line has conductor diameter of 1.8cm each, the conductors being spaced as shown in the Fig.Q.5(c). The loads are balanced and the

Page 41


line is transposed. Find the inductance per phase of 50 km long transmission line. Dec/jan2013

It is the simplest of the bridge methods for locating cable failures between

conductors and ground in any cable where there is a second conductor of the same size as

the one with the fault. It is one of the best methods of locating high-resistance faults in

low-conductor-resistance circuits. Figure 13 shows a Murray loop. The faulty conductor

is looped to an un faulted conductor of the same cross-sectional area, and a slide-wire

resistance box with two sets of coils is connected across the open ends of the loop.

Obviously, the Murray loop cannot be established if the faulty conductor is broken at any

point. Therefore, the continuity of the loop should he tested before applying the bridge

principle. In order to avoid the effects of earth currents, the galvanometer is connected as

shown in the figure. A battery energizes the bridge between the sliding contact or

resistance box center and the point at which the faulty line is grounded. Balance is

obtained by adjustment of the sliding contact or resistance. If the non-grounded (un-

faulted) line and the grounded (faulted) line have the same resistance per unit length and

if the slide wire is of uniform cross-sectional area,

Chapter 6. Transmission Line Parameters

1. Explain the terms self GMD and mutual GMD and prove that the inductance of a group of parallel wires carrying current can be represented in terms of theirgeometric distances. Dec/jan2015

Page 42


According to Peek, the fair weather corona loss per phase or conductor can be calculated from

Where, f =frequency in hertzV = line-to-neutral operating voltage in kilovoltsV0 = disruptive critical voltage in kilovolts

The wet weather corona can be calculated from the above equations by multiplying V0by 0.80. Peek’s equation gives a correct result if(1) the frequency is between 25 and 150 Hz, (2) the conductor radius is greater than 0.25 cm. and (3) the ratio of V to V0 is greater than 1.8.

The power LOSS IS proportional to the square root of the size of the conductor. Therefore, the larger the radius of the conductor, the larger the power loss. Also, the larger the spacing between conductors, the smaller the power loss. Similarly,

that is, for a given voltage level, the larger the conductor site, the larger the disruptivecritical voltage and therefore the smaller the power loss.According to Peterson (11], the fair weather corona loss per phase or conductor can be calculated from

where d = conductor diameterD = spacing between conductorsf = frequency in hertzV= line-to-neutral operating voltage in kilovoltsF= corona factor determined by test and is a function of ratio of V to V0

Page 43


In general, the corona losses due to fair weather conditions arc not significantly large at

extra-high-voltage range. Therefore, their effects are not significant from technical andlor

economic points of view. Whereas the corona losses due to foul weather conditions are very

significant.

2. Derive an expression for the capacitance of a unsymmetrically spaced but regularlytransposed line. (10) Jul/Aug 2015

3. Derive the expression for the inductance of a 3 phase transmission line with unsymmetrcal spacing without transposition. Use the Flux linkage concept. (10)

Knocked out of the air molecule, which also has a low velocity, are both accelerated

by the electric field, and therefore, each electron is capable of ionizing an air molecule at the

next collision. Of course, after the second collision, there are now four electrons to repeat the

process, and so on, the number of electrons doubling after each collision. All this time, the

electrons arc advancing toward the positive electrode, and after many collisions, their number

has grown enormously. Therefore, this process is called the avalanche process. Note that

each so-called electron avalanche is initiated by a single free electron that finds itself in an

Page 44


intense electrostatic field. Also note that the intensity of the electrostatic field around a

conductor is nonuniform. Therefore, it has its maximum strength at the surface of the

conductor and its intensity diminishes inversely as the distance increases from the center of

the conductor. Thus, as the voltage level in the conductor is increased, the critical field

strength is approached, and the initial discharges take place only at or near the conductor

surface. For the positive half cycle, the electron avalanches move toward the conductor and

continue to grow until they hit the surface. For the negative half cycle, the electron

avalanches move way from the conductor surface toward a weaker field and ceases to

advance when the field becomes too weak to accelerate

4. Write short notes on transposition of transmn. Lines. Dec/jan2013

Electric field, and therefore, each electron is capable of ionizing an air molecule

at the next collision. Of course, after the second collision, there are now four electrons to

repeat the process, and so on, the number of electrons doubling after each collision. All

this time, the electrons arc advancing toward the positive electrode, and after many

collisions, their number has grown enormously. Therefore, this process is called the

avalanche process. Note that each so-called electron avalanche is initiated by a single

free electron that finds itself in an intense electrostatic field. Also note that the intensity

of the electrostatic field around a conductor is nonuniform. Therefore, it has its maximum

strength at the surface of the conductor and its intensity diminishes inversely as the

distance increases from the center of the conductor. Thus, as the voltage level in the

conductor is increased, the critical field strength is approached, and the initial discharges

take place only at or near the conductor surface. For the positive half cycle, the electron

avalanches move toward the conductor and continue to grow until they hit the surface.

For the negative half cycle, the electron avalanches move way from the conductor

surface. a weaker field and ceases to advance when the field becomes too weak to

accelerate

5. Derive the expression for the inductance of a 3 phase unsymmetrically spaced buttransmission line/km.

6. Derive the expression for the capacitance of a 3 phase single ckt. Line withequilateral spacing Jul/Aug 2013 Dec/jan2014

Page 45


The flashover can be considered as impossible under normal circumstances. At a

given voltage level, the factors affecting corona include line configuration, conductor

type. condition of conductor surface, and weather. In a horizontal configuration, the field

near the middle conductor is larger than the field near the outer conductors. Therefore,

the disruptive critical voltage is lower for the middle conductor, causing larger corona

loss than the ones for the two other conductors. If the conductors are not spaced

cquilaterally, the surface gradients of the conductors and therefore the corona losses are

not equal. Also, the conductor height affects the corona loss, that is, the greater the

height, the smaller the corona loss. The corona loss is proportional to the frequency of the

voltage. Therefore, the higher the frequency, the higher the corona losses. Thus, the

corona loss at 60 Hz is greater than the one at 50 Hz. Of course, the corona loss at zero

frequency, that is, direct current, is far less than the one for ac current.

7. Show how the inductanceof 3 phase trasmn. Line with equilateral and symmetricalspacing between conductors can be calculated.

Dec/jan2015

For the same diameter, a stranded conductor is usually satisfactory for about 80—

85 percent of the voltage of a smooth conductor. As said before, the size of the

conductors and their spacings also have considerable effect on corona loss. The larger the

diameter, the less likelihood of corona. Therefore, the use of conductors with large

diameters or the use of hollow conductors or the use of bundled conductors increases the

effective diameter by reducing the electric stress at the conductor surfaces. The

breakdown strength of air varies with atmospheric conditions. Therefore, the breakdown

strength of air is directly proportional to the density of the air. The air— density factor is

defined as

where p = barometric pressure in centimeters of mercury t = ambient temperature in degrees Celsius

Rain affects corona loss usually more than any other factor.

Page 46


For example. it may cause the corona loss to be produced on a conductor at

voltages as low as 65 percent of the voltage at which the same loss takes place during fair

weather. Heavy winds have no effect on the disruptive critical voltage or on the loss, but

presence of smoke lowers the critical voltage and increases the loss. Corona in fair

weather may be negligible up to a voltage close to the disruptive critical voltage for a

particular conductor. Above this voltage, the impacts of corona increase very quickly. A

transmission line should he designed to operate just below the disruptive critical voltage

in fair wether so that corona only takes place during adverse atmospheric conditions.

Therefore, the calculated disruptive critical voltage is an indicator of corona performance

of the line. However, a high value of the disruptive critical voltage is not the only

criterion of satisfactory corona performance. The sensitivity of the conductor to foul





the disruptive critical voltage and can be determined.

8. Derive the expression for the capacitance of a 3 phase line withUn symmetrical spacing. Jul/Aug 2015

Page 47


9. What is skin effect? Which are the factors influencing skin effect?Dec/jan2013

The frequency of the ACvoltage producesase condeffecton the conductor

resistance due to the nonuniformdistribution ofthe current.

Thisphenomenonisknownasskineffect.Asthecurrenttendstogotowardthesurfaceofthecond

uctorandthecurrentdensity decreasesatthecenter.Skineffectreducestheeffectivecross-

sectionareausedbythecurrent,andthus,theeffectiveresistanceincreases.Also,althoughinsma

llamount,afurtherresistanceincreaseoccurswhenothercurrent-

arryingconductorsarepresentintheimmediatevicinity.Askincorrectionfactork,obtainedbydif

ferentialequationsandBesselfunctions,isconsideredtoreevaluatetheACresistance.For60Hz,

kisestimatedaround1.02

Othervariationsinresistancearecausedby

. Temperature

. Spiralingofstrandedconductors

Page 48


. Bundleconductorsarrangement

5. Derive from first principles, an expression for the inductance per phase per km of a 3 phase regularly transposed trn. Line. The conductors are of diamt. d mt. and placed at the corner of a triangle of sides a, b, c. Dec/jan2015 6. Calculate the

inductance of single phase two wire line starting from fundamentals.Jul/Aug 2013

reactanceareconsidered.Assumingbalancedconditions,thelinecanberepresentedbyt heequivalentcircuitofasinglephasewithresistanceR,andinductivereactanceXL inseries (seriesimpedance),asshowninFig.13.1.Ifthe ransmissionlinehasalengthbetween80km(50miles)and240km(150miles),the lineisconsideredamedium-length lineand itssingle-phase equivalentcircuitcanberepresentedinanominalpcircuitconfiguration[1].Theshuntcapacita nceofthelineisdividedintotwoequalparts,eachplaced atthesendingandreceivingendsoftheline.Figure13.2showstheequivalent circuitforamedium-lengthline.Bothshort- andmedium-length transmissionlinesuseapproximated lumped-parametermodels.However,ifthelineislargerthan240km,themodelmustconsiderparameters uniformlydistributedalongtheline.Theappropriate seriesimpedanceand shunt capacitance arefoundbysolvingthecorrespondingdifferentialequations,wherevoltagesandcurrentsarede scribedasafunctionofdistanceandtime.Figure13.3showstheequivalentcircuitforalongline.

Page 49


Chapter 7. Performance of Transmission Line

1. Derive expressions for generalized ABCD constants for a long transmission line using rigorous method of analysis. Dec/jan2015

1. MagneticfieldintensityH 2. MagneticfielddensityB 3. Fluxlinkagel

InductanceofaSolid, Round,InfinitelyLongConductor

Consideraninfinitelylong,solidcylindrical conductor withradiusr,carryingcurrentIasshowninFig.13.6.Iftheconductorismadeofanonmagneticma terial,andthecurrentisassumeduniformlydistributed(noskineffect),thenthegeneratedinterna landexternalmagneticfieldlinesareconcentriccirclesaroundtheconductorwithdirectiondefi nedbytheright-handrule.

2. Derive expressions for ABCD constants for a medium transmission line usingnominalT model. Hence prove AD-BC = 1. Jul/Aug 2015

Page 50


3. Write short note on classification of transmission lines. Dec/jan2013

Ampere’slawdetermines the magneticfieldintensity Hx, constant at anypoint alongthecirclecontouras

4. What are ABCD constants of a trasmn. Line? Determine the same for a a medium transmission line using nominal TI model. Hence prove AD-BC = 1.(10)

Dec/jan2015

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5. Derive the expressions for sending end voltage and current of a medium trasmn. Line(nominal T method) interms of Y, Z, Vr and Ir. (8) Dec- 2014

For the same diameter, a stranded conductor is usually satisfactory for about 80—85 percent of

the voltage of a smooth conductor. As said before, the size of the conductors and their spacings also have considerable effect on corona loss. The larger the diameter, the less likelihood of corona.

Therefore, the use of conductors with large diameters or the use of hollow conductors or the use of

bundled conductors increases the effective diameter by reducing the electric stress at the conductor

surfaces. The breakdown strength of air varies with atmospheric conditions. Therefore, the

breakdown strength of air is directly proportional to the density of the air. The air—density factor is

defined as

Page 52


Page 53


6. Write short note on surge impedance loading. (5) Dec/jan2015

7. Derive expressions for generalized ABCD constants for equivalent T representation of long transmission line. Jul/Aug 2015

If the ratio of spacing between conductors w the radius conductors before corona phenomenon

occurs. Since for overhead lines this ratio is much greater than 15, the flashover can be considered as

impossible under normal circumstances. At a given voltage level, the factors affecting corona include

line configuration, conductor type. condition of conductor surface, and weather. In a horizontal

configuration, the field near the middle conductor is larger than the field near the outer conductors.

Therefore, the disruptive critical voltage is lower for the middle conductor, causing larger corona loss

than the ones for the two other conductors. If the conductors are not spaced quilaterally, the surface

gradients of the conductors and therefore the corona losses are not equal. Also, the conductor height

affects the corona loss, that is, the greater the height, the smaller the corona loss. The corona loss is

Page 54


proportional to the frequency of the voltage. Therefore, the higher the frequency, the higher the

corona losses. Thus,

8. Discuss the terms voltage regulation and transmission efficiency as applied totransmission line. Dec/jan2013

Page 55


9. Write and explain the classification of overhead trasmn. Lines. Jul/Aug 2013

15. Calculate the effcy. And voltage regulation for medium trn. Line assuming nominal Tmethod. Dec- 2014

13.Derive an expression for the effcy. And voltage regulation for short trn. Line givingthe vector diagm. Dec/jan201514.Explain how the transmission lines are classified. (04 Marks Jul/Aug 2015

Page 56


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Chapter 8. Distribution Systems

1. Write short note on radial and ring main distributors. Dec/jan2015

A current-carrying conductor produces concentric magnetic flux lines around the

conductor .If the current varies with the time, the magnetic flux changes and

avoltageisinduced.Therefore,aninductanceispresent,definedastheratioofthemagneticfluxli

nkageandthecurrent.Themagneticfluxproduced by the current in transmission line

totalinductancewhosemagnitudedependsonthelineconfiguration.Todeterminetheinductan

ceoftheline,itisnecessarytocalculate,asinanymagneticcircuitwithpermeabilitym,thefollowi

ngfactors:

The sensitivity of the conductor to foul weather should also be considered (e.g.,

corona increases more slowly on stranded conductors than on smooth conductors). Due to

the numerous factors involved, the precise calculation of the peak value of corona loss is

extremely difficult, if not impossible. The minimum voltage at which the ionization

occurs in fair weather is called the disruptive critical voltage and can be determined.

2. What is meant by DC distribution? Explain with diagram different types of DC distribution and discuss their merits and demerits. Jul/Aug 2015

Page 58


Ampere’slawdetermines the magneticfieldintensity Hx, constant at anypoint

3. Write short note on radial distribution system. (5) Dec/jan2013

4. Show different types of distribution systems with single line diagrams and state the merits and demerits of ring main and radial distribn. Systems.

Jul/Aug 2013

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5. Calculate the total voltage drop in uniformly loaded distributor, when it is fed atone end. DEC 2014

6. Write short note on ring main distributors. (5) Dec/jan20157. Write short note on feeders, distributors and service mains. (5) Jul/Aug 2015

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8. Explain the radial and ring main distribution system, bring out their merits and demerits. (08 Marks) Dec/jan2013

Rain affects corona loss usually more than any other factor.

For example. it may cause the corona loss to be produced on a conductor at voltages as low as 65

percent of the voltage at which the same loss takes place during fair weather. Heavy winds have

no effect on the disruptive critical voltage or on the loss, but presence of smoke lowers the

critical voltage and increases the loss. Corona in fair weather may be negligible up to a voltage

close to the disruptive critical voltage for a particular conductor. Above this voltage, the impacts

of corona increase very quickly. A transmission line should he designed to operate just below the

disruptive critical voltage in fair wether so that corona only takes place during adverse

atmospheric conditions. Therefore, the calculated disruptive critical voltage is an indicator of

corona performance of the line. However, a high value of the disruptive critical voltage is not the

only criterion of satisfactory corona performance. The sensitivity of the conductor to foul weather

should also be considered (e.g., corona increases more slowly on stranded conductors than on

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smooth conductors). Due to the numerous factors involved, the precise calculation of the peak

value of corona loss is extremely difficult, if not impossible. The minimum voltage at which the

ionization occurs in fair weather is called the disruptive critical voltage and can be determined

from

9. A two wire D.C. distributor AB, 600m long, is loaded as under. Distance from A (meters) 150, 300, 350, 450. Loads in amperes 100, 200, 250, 300. The feeding point A is maintained at 440V and that of B at 430V. If each conductor has a resistance of 0.01 per 100 meters. Calculate: i) The current supplied from A and B. ii) The

power dissipated in the distributor. Jul/Aug 2013

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