02.07.2013 Views

Electrical Power Systems

Electrical Power Systems

Electrical Power Systems

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

Resistance and Inductance of Transmission Lines 41<br />

\ X L = Lw = 0.904 × 10 –3 × 2p × 50 W/km = 0.284 W/km<br />

Example 2.8: Determine the geometric mean radius of a conductor in terms of the radius r of<br />

an individual strand for (a) three equal strands as shown in ig. 2.19(a) and (b) four equal<br />

strands as shown in ig. 2.19(b).<br />

Solution:<br />

ig. 2.19 (a) ig. 2.19 (b)<br />

(a) D s = (r¢ · 2r · 2r) 1/3 = r(0.7788 × 4) 1/3 = 1.46 r<br />

1/ 4<br />

e j = 1.722 r<br />

(b) Ds = (r¢ · 2r ·2r · 2 2r) 1/4 = r 0. 7788 ´ 8 2<br />

Example 2.9: A three-phase untransposed transmission line and a telephone line are supported<br />

on the same towers as shown in ig. 2.20. The power line carries a 50 Hz balanced current of 150<br />

Amp per phase. The telephone line is located directly below phase C. ind the voltage per km<br />

induced in the telephone line.<br />

Solution:<br />

D<br />

D<br />

b2<br />

b1<br />

2 2<br />

= (.) 4 6 + 4 = 6. 096 m<br />

2 2<br />

= (.) 34 + 4 = 525 .<br />

m<br />

U<br />

V|<br />

W|<br />

ig. 2.20<br />

D<br />

D<br />

a2<br />

a1<br />

2 2<br />

= (.) 8 6 + 4 = 9. 484 m<br />

2 2<br />

= (.) 74 + 4 = 841 .<br />

rom ig. 2.20, the flux linkage between conductors T 1 and T 2 due to current I a is<br />

<br />

HG<br />

l12(Ia) = 0.4605 Ia log Da2<br />

D<br />

a1<br />

I<br />

KJ mWb–T/km<br />

The flux linkage between conductors T 1 and T 2 due to current I b is<br />

m<br />

U<br />

V|<br />

W|

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!