Capacitance of a Single-phase Transmission Line We know that any two conductors which are separated by an insulating medium constitute a
41.16. Capacitance of a Single-phase Transmission Line We know that any two conductors which are separated by an insulating medium constitute a
capacitor. When a potential difference is established across two such conductors, the current flows in at one conductor and out at the other so long as that p.d. is maintained. The conductors of an over- head transmission line fulfil these conditions, hence when an alternating potential difference is applied across a transmission line, it draws a leading current even when it is unloaded. This leading current is in quadrature with the applied voltage and is known as the charging current . Its value depends upon voltage, the capacitance of the line and the frequency of alternating current.
As shown in Art 5.10, the capacitance between conductors of a single-phase line is approxi- mately given by
* It may be noted that standard conductors have a slightly higher inductance.
A.C. Transmission and Distribution
log 10 Dr / Here, D is the distance between conductor centres and r the radius of each conductor, both
log hDr /
2.3 log 10 Dr /
expressed in the same units (Fig. 41.15).
Fig. 41.15
Fig. 41.16
As shown in Fig. 41.16, the capacitance to neutral C n = 2C where point O is the neutral of the system. Obviously, the total capacitance between conductors A and B is given by finding the resultant of two capacitances each of value C n joined in series, the resultant, obviously, being equal to C.
It is important to remember that if capacitance to neutral is used for calculating the charging current, then voltage to neutral must also be used.
line charging current = V =
V =π 2 fC V A/m X C n 1/(2 π fC n )
where V is the voltage to neutral . However, it may be noted that ground effect has been neglected while deriving the above expres-
sion. This amounts to the tacit assumption that height h of the conductors is very large as compared to their spacing ‘d’. In case ground effect is to be taken into account, the expression for capacitance becomes.
Example 41.5. What is the inductance per loop metre of two parallel conductors of a single phase system if each has a diameter of 1 cm and their axes are 5 cm apart when conductors have a relative permeability of (a) unity and (b) 100. The relative permeability of the surrounding medium
is unity in both cases. End effects may be neglected and the current may be assumed uniformly distributed over cross-section of the wires.
Solution. ( a)
log /0.5 h + 1 = 1.02 µµµµµ H/m
log /0.5 h 5 + 100 = 10.9 µµµµµ H/m
Example 41.6. A 20-km single-phase transmission line having 0.823 cm diameter has two line conductors separated by 1.5 metre. The conductor has a resistance of 0.311 ohm per kilometre. Find the loop impedance of this line at 50 Hz.
(Gen. Trans. & Dist. Bombay Univ. 1992)
Solution. 4 Loop length = 20 km = 2 × 10 m
Total loop inductance is 4 L µ
= × 2 10 log e Dr / µ +
henry
Here, –3 D = 1.5 m ; r = 0.823/2 = 0.412 cm = 4.12 × 10 m Assuming µ r = 1 for copper conductors since they are non-magnetic and also taking µ r = 1 for air,
we have
1618 Electrical Technology
=8 –3 × –3 10 (5.89 + 0.25) = 49.12 × 10 H Reactance X = 2 π× 50 × 49.12 × 10 = 15.44 Ω ; Loop resistance = 2 × 20 × 0.311 = 12.44 Ω
2 Loop impedance = 2 12.44 + 15.44 = 19.86 Ω Ω Ω Ω Ω
Example 41.7. The conductors in a single-phase transmission line are 6 m above ground. Each conductor has a diameter of 1.5 cm and the two conductors are spaced 3 m apart. Calculate the capacitance per km of the line (i) excluding ground effect and (ii) including the ground effect.
Solution. The line conductors are shown in Fig. 41.17.
log hdr / log 3/0.75 10 h × − 2
= 9.27 –3 × 10 F/m = 9.27 × × × × × 10 µF/km
( ii) In this case,
d 2 h 1 2 + /4 − 2 12 π× 8.854 10 ×
log h 3 − 12 2 × 2 0.75 10 + 1 3 /4 6 ×
= 9.33 –3 × 10 F/m = 9.33 ××××× 10 mF/km
It is seen that line capacitance when considering ground
Fig. 41.17
effect is more (by about 0.64% in the present case).