Electrical Power Systems

22 Electrical Power Systems
or l int =
or L int = l int
4p´ 10
8p
-7
× I Wb–T/m
I
1 -
= ´ 10
2
7 H/m ...(2.21)
Note that Lint is independent of the radius of the conductor.
2.5 INDUCTANCE DUE TO EXTERNAL LUX LINKAGE
igure 2.2 shows two points P and Q at distances D 1 and
D 2 from a conductor which carries a current of I Amperes.
Since the magnetic lines of flux are concentric circles around
the conductor, whole of the flux between points P and Q
lies within the concentric cylindrical surfaces which pass
through these points. The field intensity at a distance x is
Hx =
I
AT/m
2p x
...(2.22)
and lux density
Bx = m o × I
Wb/m
2px
2 ig. 2.2: lux linkages between two
external points P, Q.
...(2.23)
The flux outside the conductors links the entire current I and hence the flux linkage dlx is
numerically equal to the flux dfx. The flux dfx for a small region of thickness dx and one meter
length of the conductor is given by
dlx = dfx = Bx · dx · 1 = m o × I
px
dx Wb/m length of the conductor
2
...(2.24)
Therefore, the total flux linkages of the conductor due to flux between points P and Q is
m oImo lPQ =
px
p
dx
D2
I D z 2I
= ln Wb–T/m ...(2.25)
2 2 HG D
D1
The inductance between two points external the conductor is then
Lext = lPQ mo
= ln
I 2p
\ L ext = 2 × 10 –7 ln D
HG
HG
2I
1KJ
D
D
D
2
1
I
KJ
1
KJ
H/m
H/m ...(2.26)
2.6 INDUCTANCE O A SINGLE PHASE TWO WIRE LINE
igure 2.3 shows a single phase line consisting of two solid round conductors of radius r 1 and
r 2 spaced distance D apart. The conductor carry equal currents but in the opposite directions.