Question

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.
4 x 10^7 ln âDad. Dln/Dnk. Dtet

where, for example, Dud denotes the distance in meters between conductors a and.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.
b. Find the GO-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A.

Answer 1

Complete question is;

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.

The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m

where, for example, D_ad denotes the distance in meters between conductors a and d.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.

b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A

Answer:

A) M = 1.01 × 10^(-4) H/km

B) v_cd = 5.712 V/km

Explanation:

A) From the distances given in the question, we can deduce that;

D_ac = √(((2.5/2) - (1/2))² + 1.8²)

D_ac = 1.95 m

Also;

D_ad = √(((2.5/2) + (1/2))² + 1.8²)

D_ad = 2.51 m

I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;

φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))

Mutual inductance per km is given as;

M = φ_cd/I_a

Thus;

M = 4 x 10^(-7)( ln (2.51/1.95))

M = 1.01 × 10^(-7) H/m

Per km;

M = 1.01 × 10^(-7) × 1000

M = 1.01 × 10^(-4) H/km

B) voltage per km is gotten by;

v_cd = ωMI

Now, ω = 2πf = 2π × 60 = 377 rad/s

Thus;

v_cd = 377 × 1.01 × 10^(-4) × 150

v_cd = 5.712 V/km