Unit 3

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UNIT-3
UNIT-3 : Mechanical design of OH lines Line Supports - Types of towers - Stress and Sag Calculation – Effects of Wind and Ice loading - Insulators - Types, Voltage distribution - String Efficiency – Testing
Line Supports • Definition: The supporting structures for overhead line conductors are various types of poles and towers called line supports • In general, the line supports should have the following properties : (i) High mechanical strength to withstand the weight of conductors and wind loads etc. (ii) Light in weight without the loss of mechanical strength. (iii) Cheap in cost and economical to maintain. (iv) Longer life. (v) Easy accessibility of conductors for maintenance.
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• The line supports used for transmission and distribution of electric power are of various types including wooden poles, steel poles, R.C.C. poles and lattice steel towers . • The choice of supporting structure for a particular case depends upon the line span, X-sectional area, line voltage, cost and local conditions . 1. Wooden poles. These are made of seasoned wood and are suitable for lines of moderate X-sectional area and of relatively shorter spans, say upto 50 metres . Such supports are cheap, easily available, provide insulating properties and, therefore, are widely used for distribution purposes in rural areas as an economical proposition. The wooden poles generally tend to rot below the ground level, causing foundation failure. In order to prevent this, the portion of the pole below the ground level is impregnated with preservative compounds like creosote oil.
Double pole structures of the ‘A’ or ‘H’ type are often used ( See Figure ) to obtain a higher transverse strength than could be economically provided by means of single poles. • Disadvantage s of wooden supports are : (i) tendency to rot below the ground level (ii) comparatively smaller life (20-25 years) (iii) cannot be used for voltages higher than 20 kV (iv) less mechanical strength and (v) require periodical inspection .
2. Steel poles The steel poles are often used as a substitute for wooden poles . They possess greater mechanical strength , longer life and permit longer spans to be used. Such poles are generally used for distribution purposes in the cities. This type of supports need to be galvanised or painted in order to prolong its life. The steel poles are of three types viz., (i) rail poles (ii) tubular poles (iii) rolled steel joints
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3. RCC poles The reinforced concrete poles have become very popular as line supports in recent years . They have greater mechanical strength, longer life and permit longer spans than steel poles. Moreover, they give good outlook, require little maintenance and have good insulating properties . Figure shows R.C.C. poles for single and double circuit The main difficulty with the use of these poles is the high cost of transport owing to their heavy weight. Disadvantage
4. Steel towers In practice, wooden, steel and reinforced concrete poles are used for distribution purposes at low voltages , say upto 11 kV. However, for long distance transmission at higher voltage , steel towers are invariably employed. Steel towers have greater mechanical strength, longer life, can withstand most severe climatic conditions and permit the use of longer spans. Tower footings are usually grounded by driving rods into the earth. This minimizes the lightning troubles as each tower acts as a lightning conductor o Fig. (i) shows a single circuit tower o Fig. (ii) shows double circuit tower has the advantage that it ensures continuity of supply, when there is breakdown of one circuit, so that the continuity of supply can be maintained by the other circuit. Types :
Stress and Sag Calculation Sag in Overhead Lines The difference in level between points of supports and the lowest point on the conductor is called sag. Fig. (i) shows a conductor suspended between two equilevel supports A and B. The conductor is not fully stretched but is allowed to have a dip. The lowest point on the conductor is O and the sag is S. When the conductor is suspended between two supports at the same level, it takes the shape of catenary. However, if the sag is very small compared with the span, then sag-span curve is like a parabola.
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Stress (Tension): The tension at any point on the conductor acts tangentially. Thus tension T o at the lowest point O acts horizontally as shown in Figure (ii) . The horizontal component of tension is constant throughout the length of the wire. The tension at supports is approximately equal to the horizontal tension acting at any point on the wire. Thus if T is the tension at the support B, then T = T o . • Conductor sag and tension. This is an important consideration in the mechanical design of overhead lines. The conductor sag should be kept to a minimum in order to reduce the conductor material required and to avoid extra pole height for sufficient clearance above ground level. It is also desirable that tension in the conductor should be low to avoid the mechanical failure of conductor and to permit the use of less strong supports. However , low conductor tension and minimum sag are not possible. It is because low sag means a tight wire and high tension, whereas a low tension means a loose wire and increased sag. Therefore, in actual practice, a compromise in made between the two.
Sag
Calculation of Sag • In an overhead line, the sag should be so adjusted that tension in the conductors is within safe limits. • The tension is governed by conductor weight, effects of wind, ice loading and temperature variations. • It is a standard practice to keep conductor tension less than 50% of its ultimate tensile strength i.e., minimum factor of safety in respect of conductor tension should be 2. Calculate sag and tension of a conductor when, (i) supports are at equal levels, (ii). supports are at unequal levels
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(i) When supports are at equal levels . Consider a conductor between two equilevel supports A and B with O as the lowest point as shown in Fig. It can be proved that lowest point will be at the mid-span. Let, l = Length of span w = Weight per unit length of conductor T = Tension in the conductor. Consider a point P on the conductor. Taking the lowest point O as the origin, let the co-ordinates of point P be x and y. Assuming that the curvature is so small that curved length is equal to its horizontal projection (i.e., OP = x), the two forces acting on the portion OP of the conductor are : (a) The weight w.x of conductor acting at a distance x/2 from O. (b) The tension T acting at O.
• Equating the moments of above two forces about point O, we get, • The maximum dip (sag) is represented by the value of y at either of the supports A and B. At support A, Therefore,
(ii) When supports are at unequal levels . In hilly areas , we generally come across conductors suspended between supports at unequal levels. Figure shows a conductor suspended between two supports A and B which are at different levels. The lowest point on the conductor is O. Let, l = Span length h = Difference in levels between two supports x1 = Distance of support at lower level (i.e., A) from O x2 = Distance of support at higher level (i.e. B) from O T = Tension in the conductor If w is the weight per unit length of the conductor, then,
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Having found x 1 and x 2 , values of S 1 and S 2 can be easily calculated.
Problem: A 132 kV transmission line has the following data : Wt. of conductor = 680 kg/km ; Length of span = 260 m Ultimate strength = 3100 kg ; Safety factor = 2. Calculate the height above ground at which the conductor should be supported. Ground clearance required is 10 metres. Solution : Wt. of conductor/metre run, w = 680/1000 = 0·68 kg Span length, l = 260 m ∴ Conductor should be supported at a height of 10 + 3·7 = 13·7 m
Problem : The towers of height 30 m and 90 m respectively support a transmission line conductor at water crossing. The horizontal distance betwen the towers is 500 m. If the tension in the conductor is 1600 kg, find the minimum clearance of the conductor and water, and clearance mid-way between the supports . Weight of conductor is 1·5 kg/m. Bases of the towers can be considered to be at water level. Solution, Figure shows the conductor suspended between two supports A and B at different levels with O as the lowest point on the conductor. Here, l = 500 m ; w = 1·5 kg ; T = 1600 kg. Difference in levels between supports, h = 90 − 30 = 60 m. Let the lowest point O of the conductor be at a distance x1 from the support at lower level (i.e., support A) and at a distance x2 from the support at higher level (i.e., support B). Obviously, x1 + x2 = 500 m
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(i) (ii)
Effect of wind and ice loading. • The sag formula are true only in still air and at normal temperature when the conductor is acted by its weight only. • However, in actual practice, a conductor may have ice coating and simultaneously subjected to wind pressure. • The weight of ice acts vertically downwards i.e., in the same direction as the weight of conductor. • The force due to the wind is assumed to act horizontally i.e., at right angle to the projected surface of the conductor. • Hence, the total force on the conductor is the vector sum of horizontal and vertical forces as shown in Fig.
When the conductor has wind and ice loading also, the following points may be noted : (i) The conductor sets itself in a plane at an angle θ to the vertical where,
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(ii) The sag in the conductor is given by Hence S represents the slant sag in a direction making an angle θ to the vertical. If no specific mention is made in the problem, then slant slag is calculated by using the above formula. (iii) The vertical sag = S cos θ (iv) Safety factor ( f ) = Ultimate strength (Breaking Stress) / Tension on the condutor(T) Note : always multiply cross-sectional area of the condutor with Ultimate strength (Breaking Stress)
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Problem : A transmission line has a span of 214 metres between level supports. The conductors have a cross-sectional area of 3·225 cm^2 . Calculate the factor of safety under the following conditions : Vertical sag = 2·35 m ; Wind pressure = 1·5 kg/m run Breaking stress = 2540 kg/cm^2 ; Wt. of conductor = 1·125 kg/m run. Solution,
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Insulators The insulators provide necessary insulation between line conductors and supports and thus prevent any leakage current from conductors to earth. In general, the insulators should have the following desirable properties : (i) High mechanical strength in order to withstand conductor load, wind load etc. (ii) High electrical resistance of insulator material in order to avoid leakage currents to earth. (iii) High relative permittivity of insulator material in order that dielectric strength is high. (iv) The insulator material should be non-porous, free from impurities and cracks otherwise the permittivity will be lowered. (v) High ratio of puncture strength to flashover . Note : The most commonly used material for insulators of overhead line is porcelain but glass, steatite and special composition materials are also used to a limited extent.
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Types of Insulators There are several types of insulators but the most commonly used are (i). pin type, (ii). suspension type, (iii). strain insulator (iv). shackle insulator . 1. Pin type insulators • The part section of a pin type insulator is shown in Figure (i). • As the name suggests, the pin type insulator is secured to the cross-arm on the pole . There is a groove on the upper end of the insulator for housing the conductor is shown in Figure (ii). The conductor passes through this groove and is bound by the annealed wire of the same material as the conductor. (i) (ii) Pin type insulators are used for transmission and distribution of electric power at voltages upto 33 kV . Beyond operating voltage of 33 kV, the pin type insulators become too bulky and hence uneconomical .
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Causes of insulator failure. o Insulators are required to withstand both mechanical and electrical stresses . o The electrical breakdown of the insulator can occur either by flash-over or puncture . In practice, sufficient thickness of porcelain is provided in the insulator to avoid puncture by the line voltage. The ratio of puncture strength to flashover voltage is known as safety factor i.e., It is desirable that the value of safety factor is high so that flash-over takes place before the insulator gets punctured. For pin type insulators, the value of safety factor is about 10 . In flashover, an arc occurs between the line conductor and insulator pin (i.e., earth) and the discharge jumps across the air gaps, following shortest distance. Figure shows the arcing distance (i.e. a + b + c) for the insulator. In case of flash-over , the insulator will continue to act in its proper capacity unless extreme heat produced by the arc destroys the insulator. In case of puncture , the discharge occurs from conductor to pin through the body of the insulator. When such breakdown is involved, the insulator is permanently destroyed due to excessive heat.
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2. Suspension type insulators For high voltages (>33 kV), it is a usual practice to use suspension type insulators shown in Fig. These insulators consist of a number of porcelain discs connected in series by metal links in the form of a string. The conductor is suspended at the bottom end of this string while the other end of the string is secured to the cross-arm of the tower . Each unit or disc is designed for low voltage, say 11 kV . The number of discs in series would obviously depend upon the working voltage. Example : For instance, if the working voltage is 66 kV , then six discs in series will be provided on the string.
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• Advantages (i) Suspension type insulators are cheaper than pin type insulators for voltages beyond 33 kV. (ii) Each unit or disc of suspension type insulator is designed for low voltage, usually 11 kV . Depending upon the working voltage, the desired number of discs can be connected in series. (iii) if any one disc is damaged , the whole string does not become useless because the damaged disc can be replaced by the sound one. (iv) The suspension arrangement provides greater flexibility to the line. The connection at the cross arm is such that insulator string is free to swing in any direction and can take up the position where mechanical stresses are minimum. (v) In case of increased demand on the transmission line , it is found more satisfactory to supply the greater demand by raising the line voltage than to provide another set of conductors. The additional insulation required for the raised voltage can be easily obtained in the suspension arrangement by adding the desired number of discs . (vi) The suspension type insulators are generally used with steel towers .
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3. Strain insulators o When there is a dead end of the line or there is corner or sharp curve, the line is subjected to greater tension . In order to relieve the line of excessive tension, strain insulators are used . o For low voltage lines (< 11 kV) , shackle insulators are used as strain insulators . o However, for high voltage transmission lines , strain insulator consists of an assembly of suspension insulators as shown in Figure. o The discs of strain insulators are used in the vertical plane . When the tension in lines is exceedingly high, as at long river spans, two or more strings are used in parallel.
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4. Shackle insulators • In early days , the shackle insulators were used as strain insulators. • But now a days , they are frequently used for low voltage distribution lines . • Such insulators can be used either in a horizontal position or in a vertical position . • They can be directly fixed to the pole with a bolt or to the cross arm. Fig. shows a shackle insulator fixed to the pole. The conductor in the groove is fixed with a soft binding wire.
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Potential Distribution over Suspension Insulator String A string of suspension insulators consists of a number of porcelain discs connected in series through metallic links . Fig. (i) shows 3-disc string of suspension insulators. The porcelain portion of each disc is in between two metal links. Therefore, each disc forms a capacitor C as shown in Fig. (ii). This is known as mutual capacitance or self-capacitance . If there were mutual capacitance alone, then charging current would have been the same through all the discs and consequently voltage across each unit would have been the same i.e., V/3 as shown in Fig. (ii). However, in actual practice, capacitance also exists between metal fitting of each disc and tower or earth . This is known as shunt capacitance C 1 . Due to shunt capacitance, charging current is not the same through all the discs of the string [See Fig. (iii)]. Therefore, voltage across each disc will be different. Obviously, the disc nearest to the line conductor will have the maximum* voltage. Thus referring to Fig. (iii), V 3 will be much more than V 2 or V 1 .
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The following points may be noted regarding the potential distribution over a string of suspension insulators : (i) The voltage impressed on a string of suspension insulators does not distribute itself uniformly across the individual discs due to the presence of shunt capacitance . (ii) The disc nearest to the conductor has maximum voltage across it. As we move towards the cross-arm, the voltage across each disc goes on decreasing. (iii) The disc nearest to the conductor is under maximum electrical stress and is likely to be punctured . Therefore, means must be provided to equalize the potential across each unit. (iv) If the voltage impressed across the string were d.c., then voltage across each unit would be the same. It is because insulator capacitances are ineffective for d.c .
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String Efficiency The voltage applied across the string of suspension insulators is not uniformly distributed across various units or discs. The disc nearest to the conductor has much higher potential than the other discs. This unequal potential distribution is undesirable and is usually expressed in terms of string efficiency . The ratio of voltage across the whole string to the product of number of discs and the voltage across the disc nearest to the conductor is known as string efficiency Where, n = number of discs in the string. Note : String efficiency is an important consideration since it decides the potential distribution along the string. The greater the string efficiency, the more uniform is the voltage distribution. Thus 100% string efficiency is an ideal case for which the voltage across each disc will be exactly the same. Although it is impossible to achieve 100% string efficiency, yet efforts should be made to improve it as close to this value as possible.
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Mathematical expression Figure shows the equivalent circuit for a 3-disc string. Let us suppose that self capacitance of each disc is C. Let us further assume that shunt capacitance C 1 is some fraction K of self-capacitance i.e., C 1 = KC . Starting from the cross-arm or tower, the voltage across each unit is V 1 ,V 2 and V 3 respectively as shown. Applying Kirchhoff’s current law to node A, we get , Applying Kirchhoff’s current law to node B, we get ,
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Voltage between conductor and earth (i.e., tower) is
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The following points may be noted from the above mathematical analysis : (i) If K = 0·2 (Say), then from exp. (iv), we get, V 2 = 1·2 V 1 and V 3 = 1·64 V 1 . This clearly shows that disc nearest to the conductor has maximum voltage across it; the voltage across other discs decreasing progressively as the cross-arm in approached. (ii) The greater the value of K (= C 1 /C), the more non-uniform is the potential across the discs and lesser is the string efficiency. (iii) The inequality in voltage distribution increases with the increase of number of discs in the string. Therefore, shorter string has more efficiency than the larger one.
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Methods of Improving String Efficiency It has been seen above that potential distribution in a string of suspension insulators is not uniform . The maximum voltage appears across the insulator nearest to the line conductor and decreases progressively as the cross-arm is approached. If the insulation of the highest stressed insulator (i.e. nearest to conductor) breaks down or flash over takes place, the breakdown of other units will take place in succession. This necessitates to equalize the potential across the various units of the string i.e. to improve the string efficiency . The various methods to improve the string efficiency, (i) By using longer cross-arms (ii) By grading the insulators (iii)By using a guard ring
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(i). By using longer cross-arms o The value of string efficiency depends upon the value of K i.e., ratio of shunt capacitance to mutual capacitance. o The lesser the value of K , the greater is the string efficiency and more uniform is the voltage distribution . o The value of K can be decreased by reducing the shunt capacitance . o In order to reduce shunt capacitance , the distance of conductor from tower must be increased i.e., longer cross-arms should be used . o However, limitations of cost and strength of tower do not allow the use of very long cross-arms. In practice, K = 0·1 is the limit that can be achieved by this method.
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(ii). By grading the insulators In this method, insulators of different dimensions are so chosen that each has a different capacitance . The insulators are capacitance graded i.e. they are assembled in the string in such a way that the top unit has the minimum capacitance, increasing progressively as the bottom unit (i.e., nearest to conductor) is reached. Since voltage is inversely proportional to capacitance, this method tends to equalize the potential distribution across the units in the string. This method has the disadvantage that a large number of different-sized insulators are required . However, good results can be obtained by using standard insulators for most of the string and larger units for that near to the line conductor.
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(iii) By using a guard ring The potential across each unit in a string can be equalised by using a guard ring which is a metal ring electrically connected to the conductor and surrounding the bottom insulator as shown in the Figure. The guard ring introduces capacitance between metal fittings and the line conductor. The guard ring is contoured in such a way that shunt capacitance currents i 1 , i 2 etc. are equal to metal fitting line capacitance currents i′ 1 , i′ 2 etc. The result is that same charging current I flows through each unit of string. Consequently, there will be uniform potential distribution across the units.
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Problem : In a 33 kV overhead line, there are three units in the string of insulators. If the capacitance between each insulator pin and earth is 11% of self-capacitance of each insulator, find (i) the distribution of voltage over 3 insulators and (ii) string efficiency Solution. Fig. 8.14. shows the equivalent circuit of string insulators. Let V 1 , V 2 and V 3 be the voltage across top, middle and bottom unit respectively. If C is the self- capacitance of each unit, then KC will be the shunt capacitance.
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Home work PROBLEM : A 3-phase transmission line is being supported by three disc insulators. The potentials across top unit (i.e., near to the tower) and middle unit are 8 kV and 11 kV respectively. Calculate (i) the ratio of capacitance between pin and earth to the self-capacitance of each unit (ii)the line voltage (iii) string efficiency
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Testing of Insulators The insulators are tested according to the standards . The following tests are performed while testing an insulators, 1. Mechanical insulation tests 2. Electrical insulation tests 3. Environmental tests and Temporary cycle tests 4. Corona and radio interference tests 5. Other important test 1. Mechanical insulation tests The mechanical tests are performed to judge the ability of an insulator to withstand various mechanical stresses . The mechanical tests include : Mechanical strength test Compression test Torsional Test Minimum Bending Test Mechanical Vibration Test
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2. Electrical Insulation Test The main test in this category is the flash over tests. Flash over tests are classified as, (i). 50 cylce dry Flash over test In this test, the voltage applied across the insulators is gradually increased. The voltage at which the surrounding air breaks down is called the flash- over voltage. This flash-over voltage must be greater than the specific limit. Insulator must sustain the minimum voltage for one minutes. (ii). 50 cylce wet Flash over test This test similar to dry test, additionally the water is sprayed over the surface at an angle of 45 degree. The insulator must be capable to withstand the minimum standard voltage for 30 seconds under wet conditions.
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The other electrical tests, (i) Power frequency withstand test: • Normal power frequency voltage is applied to the insulator, this causes dust particle aligned on the surface causing leakage current. • The voltage magnitude is twice the specific rated voltage, and applied for 1 minutes. • There should not be flash-over or puncture. (ii) Impulse voltage withstand test: – In this test, standard impulse voltage surge (by generator) is applied to the insulator. – Surges are caused due to the lightening. – The standard lightning impulse wave is 1.2 micro sec to 50 micro sec. – There should not be flash-over or puncture. (iii) Puncture voltage test: o In this test, the insulator is suspended in the oil and certain minimum voltage is applied. o There should not be flash-over or puncture.
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3. Environment Test In these tests, the insulator is subjected to the alternate temperature cycles, sudden temperature changes, pollution and some environmental stresses. In temperature cycle test, the insulator is heated in water at 70º C for 1 hour and then immediately cooled in water at 7º C for another 1 hour. After the test, the glaze of the insulator should not be damaged. 4. Corona and radio interference test When the voltage stress level increases beyond corona inspection level, corona discharge starts. 5. other important tests – Porosity test – Proof loading test – Galvanising test – Corrosion test
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A cable is attached to a 1.40 kg block A, is looped over a fixed peg at C, and is attached to plate B.(Figure 1) The coefficient of static friction between the plate and the block is u4 = 0.150, the coefficient of static friction between the plate and the inclined plane is uR = 0.300, and the coefficient of static friction between the cable and the peg is uç = 7.00x10-2. If the plane's angle is 30.0 degrees, what is the maximum mass that plate B can have such that it does not slide down the plane? Express your answer numerically in kilograms to three significant figures. • View Available Hint(s) vec ? mB = kg
2. For the linkage shown to the right: a. Determine which rigid body experiences complex motion b. Locate the instant center for that rigid body Use trigonometry to calculate distance from the IC to points B and C C.
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