Effect of sag parts :- 01
The difference in level
between points of supports and the lowest point on the conductor is called sag
Pole or tower span places of between the two . Span, sag
and height are dependent on the poll. Span will be higher, lower poll numbers
will be greater. But sag increase. Will increase the length of the portion of
the poll. Overhead line design is very important when calculating sag.
If you take too much sag materials conductor. If the weight is too high will put pressure on the support line. The need for stronger support. The wintry wind and air pressure are likely to go off tear in his swing naturally. If the amount is too low while the flag, then reduce the temperature (in winter), cable crunches, or shaking because of the additional strain on the cable. This is likely to go off tear. Therefore, the amount of soot is less or more, both are equally harmful. This means that both sides of the line is placed at a certain inclination. This line increases the security and confidentiality.
Factor affecting the sag
Sag Calculation approximate formula for the -
D = Wm ² / 8T
Where D = sag (meters), W = length of conductor unit weight (kg), m = same letter heights of the span is the distance between the two poll (meters), T = Transporter tensile (kg)
Therefore, all of which depend on the overhead line is sag -
1. Weight of the conductor.
2. Length of the span
3. Working tensile strength or Working tensile strength = ultimate stress / factor of safety.
If you take too much sag materials conductor. If the weight is too high will put pressure on the support line. The need for stronger support. The wintry wind and air pressure are likely to go off tear in his swing naturally. If the amount is too low while the flag, then reduce the temperature (in winter), cable crunches, or shaking because of the additional strain on the cable. This is likely to go off tear. Therefore, the amount of soot is less or more, both are equally harmful. This means that both sides of the line is placed at a certain inclination. This line increases the security and confidentiality.
Factor affecting the sag
Sag Calculation approximate formula for the -
D = Wm ² / 8T
Where D = sag (meters), W = length of conductor unit weight (kg), m = same letter heights of the span is the distance between the two poll (meters), T = Transporter tensile (kg)
Therefore, all of which depend on the overhead line is sag -
1. Weight of the conductor.
2. Length of the span
3. Working tensile strength or Working tensile strength = ultimate stress / factor of safety.
4. Temperature.
5. Wind load and ice.
5. Wind load and ice.
Required formula
- Working stress in kg / cm² =
Ultimate breaking stress (or strength) in kg / cm² / Factor of safety
- Working
tension in kg = Ultimate tensile stress (or strength) in kg /
cm² /
Factor of safety
=
Working stress in kg / cm² × area of cross section in cm²
- maximum
working strength in kg = Ultimate
tensile stress (or strength) in kg / cm² / Factor of safety
(solved problems on sag of transmission line)
Tips: -1. Horizontal distance between the two tower 30 meters in height, with a same letter heights of 250 meters. Conductor cross section 1. 24 cm², weight of 1170 kg / km and Breaking strength of 4218 kg/cm². If safety is a factor of 5, how to land on top of conductive meddle point?
Answer: - Because there is no air pressure, so,
Wc = W
= 1170 Kg / km = 1.17 kg / m
Wc / m = 1.17 kg
Breaking Strength = 4218 kg / cm ²
h = 30 m
Transverse m = 250 m
A = 1.24 cm ²
Ground clearance = ?
Effective tensile , T = transverse × baking stress / safety factor
= 4218 × 1.24 / 5
= 1046 kg.
Sag d = Wm ² / 8 T
= 1.17 × (250) ² / 8 × 1046 = 12.84 = 8.73 m
Ground clearance = h - d = 30 - 8.73 = 21. 27 m
Meddle point land transport to the 21.27 m. On top.
Tips:- 2. Plane between the two tower 200 m and conductor cross section
1.5 cm ². maximum strength 5000 kg / cm ², air pressure of 1.5 kg per meter of cable, but you sag determined ?
Answer: - here,
M = 200 m.
A = 1.5 cm ²
Wc / m = 1.2 kg Ww / = 1.5 kg
Maximum strength = 5000 kg / cm ²
D =?
Meaningful tension T = meaningful stress × area = 5000 × 1.5 = 7500 kg
Total weight Ww / m = √ Ww ² + √ Wc ² = √ (1.5) ² + (1.2) ² = 1.95 kg.
D = Wm ² / 8T
= 1.95 × (200) ² / 8 × 7500 = 1.3 m (Ans.)
Answer: - here,
M = 200 m.
A = 1.5 cm ²
Wc / m = 1.2 kg Ww / = 1.5 kg
Maximum strength = 5000 kg / cm ²
D =?
Meaningful tension T = meaningful stress × area = 5000 × 1.5 = 7500 kg
Total weight Ww / m = √ Ww ² + √ Wc ² = √ (1.5) ² + (1.2) ² = 1.95 kg.
D = Wm ² / 8T
= 1.95 × (200) ² / 8 × 7500 = 1.3 m (Ans.)
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