Capacitance of 3-core Belted Cables The capacitance of a cable is much more important than of the overhead wires of the same length

41.33. Capacitance of 3-core Belted Cables The capacitance of a cable is much more important than of the overhead wires of the same length

because in cables ( i) conductors are nearer to each other and the sheath and ( ii) they are separated by

a dielectric medium of higher permittivity as compared to air. Fig. 41.47 shows a system of capacitances

Fig. 41.47

in a belted 3-core cable used for 3-phase system. It can be regarded as equivalent to three-phase cables having a common sheath. Since there is a p.d. between pairs of conductors and between each conductor and sheath, there exist electrostatic fields as shown in Fig. 41.47 (a) which shows average distribution of electrostatic flux, though, actually, the distribution would be continually changing because of changing potential difference between conductors themselves and between conductors and sheath. Because of the existence of this electrostatic coupling, there exist six capacitances as shown in Fig. 41.47 (b). The three capacitances between three cores are delta-connected whereas the other three between each core and the sheath are star-connected, the sheath forming the star-point [Fig. 41.47 (c)].

The three delta-connected capaci- tances each of value C 1 can be converted

Fig. 41.48

into equivalent star-capacitance C 2 which will be three times the delta-capacitances C 1 as shown in Fig. 41.48.

The two star capacitances can now be combined as shown in Fig. 41.49 (a). In this way, the whole cable is equivalent to three star-connected capacitors each of capacitance C n = 3C 1 +C S as shown in Fig. 41.49 (b).

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Fig. 41.49

If Vp is the phase voltage, then the charging current is given by

I C =V P (C S + 3C 1 ) ω =V P ω C n

41.34. Tests for Three-phase Cable Capacitance The following three tests may be used for the measurement of C 1 and C S : ( i) In the first test, all the three cores are bunched together and then capacitance is measured by

usual methods between these bunched cores and the earthed sheath. This gives 3 C S because the three capacitances are in parallel.

( ii) In the second method, two cores are bunched with the sheath and capacitance is measured between these and the third core. It gives 2C 1 +C S . From this value, C 1 and C S can be found.

( iii) In the third method, the capacitance C L between two cores or lines is measured with the third core free or shorted to earth. This gives

1 (3 C 1 + C n

Hence, C n is twice the measured value i.e. C n = 2C L

Therefore, charging current I C = ω V P C n =

2. ω VC L L

where V L is the line voltage (not phase voltage). Example 41.37. A single-core lead-covered cable is to be designed for 66-kV to earth. Its conductor radius is 1.0 cm and its three insulating materials A, B, C have permittivities of 5.4 and 3

respectively with corresponding maximum safe working stress of 38 kV per cm (r.m.s. value), 26-kV per cm and 20-kV per cm respectively. Find the minimum diameter of the lead sheath.

g 2max =

g 3max =

From ( i) and ( ii) , we get, 38/26 = 4r 1 /5, r 1 = 1.83 cm

Similarly, from ( i) and ( ii) ,

A.C. Transmission and Distribution

38/20 = 3r 2 /5 ; r 2 = 3.17 cm

V 1 =g 1max × r ×

V 2 =g 2max × r × 2.3 log 1 2 10

V 3 =g 3max × r 2 × 2.3 log 3 10

r = 145.82 log 3

∴ r 3 = 4.15 cm ∴

diameter of the sheath = 2r 3 =2 ×

8.3 cm

Example 41.38. The capacitances per kilometer of a 3-phase cable are 0.63 µF between the three cores bunched and the sheath and 0.37 µF between one core and the other two connected to sheath. Calculate the charging current taken by eight kilometres of this cable when connected to a 3-phase, 50-Hz, 6,600-V supply.

Solution. As shown in Art. 40.33, 0.63 = 3C S ∴ C S = 0.21 µ F/km From the second test,

0.37 = 2C 1 +C 2 = 2C 1 + 0.21 ∴ C 1 = 0.08 µ F/km ∴

C S for 8 km = 0.21 × 8 = 1.68 µ F;C 1 for 8 km = 0.08 × 8 = 0.64 µ F

0.64) = 3.6 µ F Now,

C n =C S + 3C 1 = 1.68 + (3 ×

V p = (6,600/ 3 ); ω = 314 rad/s

I –6

C = ω V p C n = (6,600/ √ 3) × 3.6 × 10 × 314 =

4.31 ampere

Example 41.39. A 3-core, 3-phase belted cable tested for capacitance between a pair of cores on single phase with the third core earthed, gave a capacitance of 0.4 mF per km. Calculate the charging current for 1.5 km length of this cable when connected to 22 kV, 3-phase, 50-Hz supply.

Solution.

C L = 0.4 µ

L F, V = 22,000 V,

ω = 314 rad/s 2 2 − 6

× 22,000 0.4 10 × × × 314 3.2 = A

Charging current for 15 km

F between shorted conductors and sheath and (ii) capacitance between two conductors shorted with the sheath and the third conductor 0.6 µ

Example 41.40. A 3-core, 3-phase metal-sheathed cable has (i) capacitance of 1 µ

F. Find the capacitance (a) between any two conductors (b) between any two shorted conductors and the third conductor.

(Power Systems-I, AMIE, Sec. B, 1993)

Solution. ( a) The capacitance between two cores or lines when the third core is free or shorted

to earth is given by 1/2 (3C 1 +C S )

Now, ( i)

we have 3C S =1 µ

F or C s = 1/3 µ

F = 0.333 µ F

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1 F, 2C = 0.6 − 0.333 = 0.267 µ F, ∴

From ( ii) we get, 2C 1 +C S = 0.6 µ

C 1 = 0.133 µ F

( b) The capacitance between two shorted conductors and the other is given by

1 + 3 C S =2 × 0.133 + 3 × 0.333 = 3 0.488 mF C

1 S = 2C