رب اشرح لى صدرى ويسر لى أمر واحلل عقدة من لسانى يفقهوا قولى

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Wilfrid Johnston
6 years ago
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1 بسم هللا الرحمن الرحيم رب اشرح لى صدرى ويسر لى أمر واحلل عقدة من لسانى يفقهوا قولى صدق هللا العظيم

2 Chapter 11 Overhead Line Insulators part 2 pp

3 Potential Distribution over a String of Suspension Insulators p. 376 The voltage across the discs of the string is not uniformly distributed. This is because of the capacitances formed between the metal parts of the insulators and tower structure. These capacitances could be made negligibly small by increasing the distance between the insulators and the tower structure which requires larger lengths bigger size of the towers and hence it becomes uneconomical. 3

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6 In practice the insulators are not very far from the tower structure and hence these capacitances affect the voltage distribution across the string Since the insulator discs are identical, each disc is represented by its mutual capacity C A shunt capacitance C 1 is the capacitance of metal part of the insulator disc to the tower structure (ground). So that C 1 = m C Where m is the ratio between the capacitance to ground to the mutual (unit) capacitance 6

7 Let V be the operating voltage and V 1,V 2,V 3 and V 4 the voltage drops across the units starting from cross towards the power conductor. V = V 1 + V 2 + V 3 + V 4 The objective is to find out the voltage across each disc as a multiple of the operating voltage and to compare the voltage across each unit. 7

8 C : capacitance of each insulator unit. C 1 = m C : capacitance to ground is the capacitance of metal part of the insulator unit to the tower (m<1). V 1, V 2, V 3 the voltage across each unit starting from the cross arm towards the power conductor. V = V1 + V2 +V3 Line voltage 8

9 At point P: I 2 = I 1 +I c1 ωc.v 2 = ωc.v 1 + ωmc.v 1 V 2 = (1+m).V 1 At Point Q: I 3 = I 2 + I c2 ωc.v 3 = ωc.v 2 + ωmc.(v 1 +V 2 ) V 3 =m.v 1 +(1+m).V 2 =(m +(1+m) 2 ).V 1 V 3 = (1+3m +m 2 ).V 1 9

10 The voltage across the units cam also given by: n= 1 V 2 = (1+ m). V 1 n=2 V 3 = (1+ 3 m +m 2 ). V 1 n=3 V 4 = V 3.(1+ 6 m +5 m 2 + m 3 ). V 1 n=4 V 5 = V 4.(1+ 10 m + 15 m 2 + 7m 3 + m 4 ). V 1 10

11 To Find the voltage across each disk For 3 disc string: V = V 1 + V 2 + V 3 V = V 1 + V 1 (1+ m) + V 3 (1+ 3m + m 2 ) V = V 1 (3 + 4 m + m 2 ) Then you can find V 1, V 2 and V 3 Where V: Voltage across the insulator string, (phase Volt) 11

12 For m < 1 V 3 > V 2 > V 1 Insulator (String) Efficiency: %η = (V/n.V max ) x 100 V: Voltage across the insulator string, (phase Volt) n: number of insulator units. V max : Voltage across the insulator unit near to the power line (for n = 3, V max = V 3 ). 12

13 Example (1) Find the voltage distribution of an insulator of 3 units, if the maximum voltage of each unit is 17 kv, and the capacitance to ground is 20% of unit capacitance, also find the insulator efficiency. 13

14 V 3 = 17 kv, m=20% = 0.2 V 2 = (1+m).V 1 = 1.2 V 1 V 3 = (1+ 3 m +m 2 ). V 1 =1.64 V1 V 1 = 17/1.64 = kv V 2 = 1.2x10.36 = kv V= V 1 + V 2 + V 3 = 39.8 kv Insulator efficiency = (39.8/3x17)x100=78.03% 14

15 In general, voltage across the units can given by: V n+1 = V n.(1+m) + (V 1 +V V n-1 ).m n= 1 V 2 = (1+m). V 1 n=2 V 3 = V 2.(1+m) + V 1.m n=3 V 4 = V 3.(1+m) + (V 1 +V 2 ).m n=4 V 5 = V 4.(1+m) + (V 1 + V 2 +V 3 ).m 15

16 Example (2) An insulator string for 66 kv line has 4 units. The capacitance to ground is 10% of the capacitance of each insulator unit. Find the voltage across each insulator unit and string efficiency. 16

17 V 2 = (1+m). V 1 = 1.1 V 1 V 3 = V 2.(1+m) + V 1.m = 1.31 V 1 V 4 = V 3.(1+m) + (V 1 +V 2 ).m =1.651 V 1 V 1 +V 2 +V 3 +V 4 = = 38.1 kv V 1 ( )=38.1 V 1 =7.53 kv, V 2 = 8.28 kv, V 3 =9.86 kv, V 4 = kv String efficiency = (38.1/4x12.43)x100=76.6% 17

18 Improvement of String Efficiency Methods of Equalizing Potential (p.386) 1- Reducing the ground capacitance relative to the capacitance of insulator unit (reduce m where m = c 1 /c ): This can be done by increasing the length of cross arm and hence taller supporting tower which uneconomical. 18

19 Improvement of String Efficiency 2- Grading of insulator units: It can be seen that the unequal distribution of voltage is due to the leakage current from the insulator pin to the tower structure. The solution is to use insulator units with different capacitances. This requires that unit nearest the cross arm should have minimum capacitance (maximum X c ) and the capacitance should increase as we go towards the power line. 19

20 This means that in order to carry out unit grading, units of different types are required. This requires large stocks of different units which is uneconomical and impractical. 20

21 Improvement of String Efficiency (حلقة الحماية ( Ring): 3- Static Shielding (Guard This method uses a large metal ring surrounding the bottom insulator unit and connected to the line. This ring is called a grading or guard ring which gives a capacitance which will cancel the charging current of ground capacitance. 21

22 Guard ring serves two purposes: - Equalizing the voltage drop across each insulator unit. - protects the insulator against flash over. 22

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29 Example (3) A 3-unit insulator string with guard ring. The capacitance to ground and to guard ring are 25 % and 10 % of the capacitance of each unit. Determine the voltage distribution and string efficiency. 29

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31 At point A: I 1 + I y = I x + i 1 0.1ωC.(V 2 +V 3 ) + ωc.v 2 = ωc.v ωc.v V V V 3 =0.0 At point B: I 2 + I z = I y + i 2 0.1ωC.V 3 + ωc.v 3 = ωc.v ωC.(V 1 +V 2 ) 0.25 V V 2-1.1V 3 =0.0 31

32 Also: V 1 + V 2 + V 3 = V Solve, to get: V 1 = 0.295V, V 2 = V, V 3 = 0.406V η = V/(3x0.406V)x100= 82.1 % Find the voltage distribution and insulator efficiency without a guard ring. 32

Prof. Dr. Magdi El-Saadawi

Prof. Dr. Magdi El-Saadawi بسم هللا الرحمن الرحيم رب اشرح لى صدرى ويسر لى أمر واحلل عقدة من لسانى يفقهوا قولى صدق هللا العظيم Prof. Dr. Magdi M. El-Saadawi www.saadawi1.net E-mail : saadawi1@gmail.com www.facebook.com/magdi.saadawi