Mechanical Design of Overhead Transmission Lines

EE301

MECHANICAL DESIGN OF OVERHEAD TRANSMISSION LINE

Instructed By: Ms. P.M.A.U. Karunapala

Name

: G.R. Raban

Index Number

: 070384P

Field

: EE

Group

:8

Date of Performance

: 12/11/2009

Date of Submission

: 09/12/2009

Problem 1

h

d2

d1 18 m 18 m

l1 20 m

y1 x2

x1 300 m

H

=

9000 kg

h

=

20 m

w

=

8 kg/m

x1

=

y

=

d1

=

L 2

–

Hh wL

=

300 2

–

x1

y1

=

=

m

=

2H 8 × x 21 2 × 9000

=

8 × 75 2 2 × 9000

20 300 20 300

× 75 m

=

5m

From the above diagram; l1 + y1 =

18 – d1

l1

18 – 2.5 – 5 m =

=

8 × 300

wx 2

From the slope characteristics; y1

9000 × 20

∴ Minimum clearance

=

10.5 m

10.5 m

m

=

2.5 m

75 m

Problem 2 25 m

d2

d1 l2 l1

h0 20 m

y0 x2

x1 150 m

Safety Factor

T max

=

H

30 × 10 3 H

H

x1

=

L 2

–

Hh wL

=

=

1.5

=

1.5

=

20 kN

150 2

–

20 × 10 3 × 25 2 × 9.8 × 150

From the slope characteristics; y0 x1

=

y

=

d1

=

20 150

--------------------- (1)

wx 2 2H 2 × 9.8 × x 21 2 × 20 × 10 3

--------------------- (2)

From the above diagram; d1 + h0 + y0 h0

=

l1

=

l1 – d1 – y0

= (-) 95.07 m

(1) & (2); h0

=

l1 –

2 × 9.8 × x 21 2 × 20

× 10 3

–

2 × x1 15

Since the minimum clearance allowed between conductor and the ground is 20 m; h0 l1 –

2 × 9.8 × x 21 2 × 20

× 10 3

–

2 × x1 15

l1

≥

20

≥

20

≥

20 –

∴ Minimum height of first tower

l1 + 25 =

l2 + 20

l2

l1 + 5 =

=

∴ Minimum height of towers are

2 × 9.8 × x 21

2 × x1

2 × 20 × 10

15

2 × 9.8 × 95.07 2

=

20 –

=

28.25 m

28.25 + 5 m

=

– 3

2 × 20

=

28.25 m

× 10 3

–

2 ×(−95.07) 15

33.25 m

&

33.25 m

-75 -65 -55 -45 -35 -25 -15 -5 0 5 15 25 35 45 55 65 75 85 95 105 115 125 135 145 155 165 175 185 195 205 215 225

Problem 3 y

y

=

=

wx 2 2H

Problem 1;

x2

2250

Problem 1 - Sag Template

25

20

15

10

5

0

Problem 2; y

=

(4.9 × 10-4) x2

Problem 2 - Sag Template 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 -95

-85

-75

-65

-55

-45

-35

-25

-15

-5

5

15

25

35

45

55