Design of 3-Phase Transmission Line

Solution:
Figure – 1: Design of 3-Phase Transmission Line
As shown in the figure – 1, the shaded portion is the place which can be taken as
the possible positions of six conductors of 3 phase transmission line. Also, it must
be ensured that no two conductors are closer by less than 5 meters.
With above conditions, following 6 designs are considered for evaluation
1. Design – 1:
a. a1 = (-6, 26, 0), b1 = (-6, 20, 0), c1 = (-6, 14, 0)
b. a2 = (6, 14, 0), b2 = (6, 20, 0), c2 = (6, 26, 0)
2. Design – 2:
a. a1 = (-6, 26, 0), b1 = (0, 18, 0), c1 = (-3, 10, 0)
b. a2 = (3, 10, 0), b2 = (6, 18, 0), c2 = (-1, 26, 0)
3. Design – 3:
a. a1 = (-6, 12, 0), b1 = (-6, 20, 0), c1 = (-6, 25, 0)
b. a2 = (5, 25, 0), b2 = (5, 20, 0), c2 = (5, 12, 0)
4. Design – 4:
a. a1 = (-4, 26, 0), b1 = (-4, 20, 0), c1 = (-4, 14, 0)
b. a2 = (4, 14, 0), b2 = (4, 20, 0), c2 = (4, 26, 0)
5. Design – 5:
a. a1 = (-2.5, 26, 0), b1 = (-4.25, 19, 0), c1 = (-6, 12, 0)
b. a2 = (6, 12, 0), b2 = (4.25, 19, 0), c2 = (6, 12, 0)
6. Design – 5:
a. a1 = (-2.5, 12, 0), b1 = (-4.25, 19, 0), c1 = (-6, 26, 0)
b. a2 = (6, 26, 0), b2 = (4.25, 19, 0), c2 = (2.5, 12, 0)
Let us start with Design – 1
Design – 1
(16,1)(-
16,1)
ROW = 32
meters
(6,12)(-
6,12)
(6,26)(-
6,26)

Vector from a1 to TP – 1 ¿ a11= ( x2 −x1 ) ^i+ ( y2− y1 ) ^j+( z2−z1) ^k
¿ (−16− ( −6 ) ) ^i+ ( 1−26 ) ^j+0 ^k
a11=−10 ^i−25 ^J -------------------------------(1)
|a11|= √(10¿¿ 2+252)= √725 ¿ --------------- (2)
θa11=Phase of a11=tan−1 2.5 -----------(3)
I 1=1∗ ^k (assuming 1 Amp current flowing∈Z−direction)------------(4)
Unit vector∈the direction of a11=−cos θa 11 ^i−sin θa 11 ^j---------------(5)
Unit vector∈the direction of current I1= ^k-----------------------------------(6)
Direction of magnetic field is the cross product of equations (6) and (5)
Direction of magnetic field ¿ ^Ha 1=
| ^i ^j ^k
0 0 1
−cos θa 11 −sin θa 11 0|------------(7)
Expanding equation (7) results
^Ha 1= ( sinθa 11 ) ^i−¿ --------------(8)
Putting value of θa11 from equation (3) in equation (8)
^Ha 1= ( 0.928 ) ^i−(0.371) ^j----------------------(9)
Magnetic field because of current flowing in conductor passing through a1 at TP1 is
defined as
Ha 1= I
2 π |a11| × ^Ha 11= 1
2 π × √ 725 ¿--------(10)
On similar line list of magnetic fields by current flowing through points in design – 1
at TP1 and TP2 is available in Table – 1.
At TP1 At TP2
Ha1 5.49 ×10−3 ^i−2.19 ×10−3 ^j 3.5 ×10−3 ^i+3.1× 10−3 ^j
Hb1 6.56 ×10−3 ^i−3.45 × 10−3 ^j 3.6 ×10−3 ^i+4.1 ×10−3 ^j
Hc1 7.69 ×10−3 ^i−5.91 ×10−3 ^j 3.1 ×10−3 ^i+ 5.3× 10−3 ^j
Ha2 3.2 ×10−3 ^i−5.4 × 10−3 ^j 7.7 ×10−3 ^i+5.9 ×10−3 ^j
Hb2 3.6 ×10−3 ^i−4.1× 10−3 ^j 6.5 ×10−3 ^i+3.4 ×10−3 ^j