POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. Chapter four Over head insulator The overhead line conductors should be supported on the poles or towers in such a way that currents from conductors do not flow to earth through supports i.e., line conductors must be properly insulated from supports. This is achieved by securing line conductors to supports with the help of insulators. The insulators provide necessary insulation between line conductors and supports and thus prevent any leakage current from conductors to earth. *The function of the insulators are :. 1- Insulate the conductors from each other and from the towers under highest voltage and under bad air estimate circumstance 2-Carry the conductors under the bad estimate mechanical stresses *The materials that used for made the insulators of overhead line are 1- porcelain 2-Tougheneed glass. Although the two above materials are brittle and inelastic but it are the best materials that used to insulate the conductors. Types of insulator: 1-Pin type insulators 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. POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. . 2-Suspension- type insulator. The cost of pin type insulator increases rapidly as the working voltage is increased. Therefore, this type of insulator is not economical beyond 33 kV. For high voltages (>33 kV), it is a usual practice to use suspension type 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. For instance, if the working volt-age is 66 kV, then six discs in series will be provided on the string. POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. 3. Strain insulators. 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. For low voltage lines (< 11 kV), shackle insulators are used as strain insulators. However, for high voltage transmission lines, strain insulator consists of an assembly of suspension insulators as shown in Fig. 8.8. 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. 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. The conductor in the groove is fixed with a soft binding wire. POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. *The eclectic behavior of each Insulator is as like a capacitance Appeared in the string 1- C ( self capacitance ) Between the cap of insulator and the pin in the bottom of insulator and its value of about 30 pf 2- Ce ( air capacitance ) Where Ce is another capacitance between the tower and the connection point between the pin of each insulator and the cap of the next insulator and it is have a value of about ( K where k (0.1 The insulator make the distribution of voltage through the insulator of the string is different (where the insulator that nearest to the conductor has a higher voltage then the next and up) *There is another capacitance ( air capacitance ) between the conductor and the connection point between the pin of each insulator and the cap of the next insulator and is very little and it can be neglected. POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. (1) At point B (2) POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. At point C (3) At point D (4) If we have n insulator (5) (6) POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. Where V is the voltage of line (string voltage ƪ is the efficiency of the string. Also there is another method (general method ) to solve this problem, where (7) (8) Where X is the number of insulator that you need to get the voltage across it. *Methods of increasing the string efficiency The string efficiency increase when there is a uniformly distribution voltage across the insulator, that means when if V (phase votage of the conductor) Is equal to n ( n : number of insulator and insulator ) We can get ƪ = 100 % ( maximum efficiency ) is the voltage of the last The voltage across the capacitors ( insulator) are equally ( uniformly). But this case is a theoretical case because of the air capacitance (ce) that make the voltage across the insulator are not uniformly. Also the string efficiency decrease if the number of insulator (n) increase because of the not equally . Distribution voltage across the insulators. Many method are used to increase the string efficiency such as :. POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. 1-Reducing the value of k :. When We can reduce the value of k by reducing the value of by increasing the length of cross- arm for the tower but this method leads to reduce rigidity for the cross-arm and increase the cost 2-Grading of units :. This method depend on the change Of the self capacitance where The capacitance that near to The conductor is greater than The others due to the location Of insulator in the string and The capacitance (self capacitance) That near to the cross-arm is Less than the others If we have four insulator (n =4) and the self capacitance that nearest to the cross- arm is C then we make a grading for self capacitance in order to get an equal voltage across the insulators (maximum efficiency) ( POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. At point A (9) At point B (10) At point C Ƶ= y+3kc Ƶ= c(1+3k)+3kc (11) And so on This method is difficult in practical application because of different value of the capacitance . POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. 3) Guard Ring :. It is possible to make the voltage distribution across the units of string of insulator uniformly by a guard ring which is a ring of conductor put or placed across or around the nearest insulator to the conductor . The main reason of different the Voltage across the insulator of The string is the air capacitance (Ce) And so that the guarding reduce Ce by adding a number of Air capacitance between the Conductor or guard ring and Metal point of insulator and these air capacitances are in opposite by the effect with Ce and make the voltage across the insulators more uniformly . POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. Ex1) Over string of five insulator and the ratio of Ce/C=0.1. If the string is connected to line of voltage 33kv. 1- What is the voltage distribution over the unit of string and what will be the string efficiency. 2- When the string is supplied by a guard Ring and this lead to add a two air capacitance of value 0.2c,0.1c respectively to the nearest to the conductor. Find the new voltage distribution and the new efficiency. So1) 0.1c A 0.1c cc B 0.1c (1) C 0.1c c D (2) c Conductor (3) (4) POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. } *Also we can solve it by POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. 2) After the guard Ring 0.1c c A At point C c 0.1c B c 0.1c C 0.1c c 0.1c D c 0.2 c POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. At point D 1.299 Now After guard Ring Before guard Ring POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. Ex2) Each of the 4- insulator have after guard ring 0.6C 3C=5Ce Calculator cd, cf, cg in 0.6C C A cf C cg B 0.6C C C C So1) at point A At point B Conductor cd POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. At point C Ex3) over string of 3 insulators and the ratio of if the string is connected to line voltage of 33kv. Also if the break down voltage of each insulator is 7kv. 1- Find the voltage distribution over the unit of the string and what is the string efficiency. 2- When the string is supplied by a guard ring and this load to add at air capacitance of 0.15 c,0.15c respectively to the nearest to the conductor . find the new voltage distribution and the new efficiency So1) = 19050 voltage 0.15c C 0.15c C V *1 C *2 conductor POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. We see that the break down voltage is 7000 volt Then the case will be C 0.15C V C conductor (Break down voltage) Then The string failed to insulator the conductor from the tower 2) at point A 0.15C 0.15C C A C 0.15C B 0.15C C At point B conductor POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. } This string successes in insulation the conductor POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. Ex4) Over string of 3- Transmission line has four insulator and Ce=0.12C. if the voltage between the metol cap of the last insulator and the earth is 33kv. 1- Find the voltage distribution across the insulators and what is the string efficiency 2- If the string is supplied by a guard ring and this lead to add a two Air capacitance of o.2C, 0.1 C respectively to the nearest to the conductor. Find the new voltage distribution and the new string efficiency So1) C 0.12c 0.12c 0.12c C V C C K= 0.12 POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. ( KV At point B (1) POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING /ELECTRICAL DEP. C 0.12C A C 0.12C 0. 1C B C 0.12C 0. 2C C C At point C Also as F before