Mechanical Design of Transmission Lines MCQ Quiz - Objective Question with Answer for Mechanical Design of Transmission Lines - Download Free PDF

Explanation:

Determining the Lowest Point of a Conductor in a Transmission Line with Supports at Different Heights

Introduction: In an electrical transmission system, conductors are suspended between two supports to transmit electricity from one point to another. These conductors are subjected to various forces such as their weight, wind load, and ice load (if applicable). When the supports are at different heights, the conductor does not form a symmetric curve. Instead, the lowest point of the conductor is influenced by the relative heights of the supports and the distribution of these forces.

Such a system can be analyzed using the principles of catenary curves, which describe the shape of a flexible cable or conductor suspended by its ends under its own weight and external forces.

Correct Option Analysis:

The correct option is:

Option 2: Closer to the lower support.

This is correct because, in a transmission line with supports at different heights, the lowest point of the conductor is closer to the lower support. This phenomenon can be explained as follows:

1. Forces Acting on the Conductor:

• The primary force acting on the conductor is its weight, which causes it to sag. This weight acts uniformly along the length of the conductor.
• The tension in the conductor varies along its length, being higher at the higher support and lower at the lower support.

2. Tension Distribution:

• At the higher support, the tension is greater to maintain equilibrium against the downward pull caused by the weight of the conductor.
• At the lower support, the tension is less because the vertical component of the tension at the higher support compensates for the weight of the conductor in between.

3. Sag and the Lowest Point:

• The sag of the conductor is directly related to the tension. Higher tension results in less sag, and lower tension results in more sag.
• Since the tension is lower near the lower support, the conductor sags more in that region, causing the lowest point to shift closer to the lower support.

4. Mathematical Representation:

• The shape of the conductor is approximately a catenary curve, described by the equation:

  y = (T₀/w) × cosh(wx/T₀) - (T₀/w)

  where:
  • y is the vertical displacement (sag) of the conductor.
  • T₀ is the horizontal tension in the conductor.
  • w is the weight of the conductor per unit length.
  • x is the horizontal distance from the lowest point.
• When the supports are at different heights, the horizontal tension component remains constant, but the vertical components vary, leading to an asymmetric catenary curve.

5. Practical Observations:

• In real-world transmission lines, the lowest point is often observed to be closer to the lower support in cases where the supports are at different heights. This has been validated through numerous field observations and engineering analyses.
• The exact position of the lowest point depends on the height difference between the supports, the span length, and the conductor’s mechanical properties.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Exactly at the midpoint of the span.

This is incorrect. The lowest point of a conductor would be at the midpoint only if the supports are at equal heights, resulting in a symmetric catenary curve. When the supports are at different heights, the curve becomes asymmetric, and the lowest point shifts closer to the lower support.

Option 3: Closer to the higher support.

This is incorrect. The lowest point cannot be closer to the higher support because the tension is greater at the higher support, resulting in less sag in that region. The sag is more pronounced near the lower support, causing the lowest point to shift closer to it.

Option 4: At the higher support.

This is incorrect. The conductor cannot have its lowest point at the higher support because the higher support is, by definition, at a greater height than the lower support. The conductor must sag below the level of the higher support, forming a curve with the lowest point closer to the lower support.

Conclusion:

In a transmission line with supports at different heights, the lowest point of the conductor is located closer to the lower support. This is due to the variation in tension along the conductor and the resulting asymmetric sag. Understanding this concept is essential for designing and analyzing transmission lines to ensure their stability, efficiency, and safety.

Explanation:

Length of a Conductor in a Catenary:

Definition: When a transmission line conductor is suspended freely between two towers, it forms a curve known as a catenary. The shape of the catenary is determined by several factors, including the horizontal tension, the weight of the conductor, and the span length. Calculating the length of the conductor requires applying the catenary equation and understanding the parameters involved.

Given Data:

• Constant of the catenary, c = 487.68 m
• Span between the two towers, L = 152 m
• Weight of the conductor per unit length, w = 1160 kg/km = 1.16 kg/m

Formula:

The length of the conductor S in a catenary is calculated using the following equation:

S = 2c sinh(L / 2c)

Where:

• S = Length of the conductor
• c = Constant of the catenary
• L = Span between the two towers
• sinh = Hyperbolic sine function

Solution:

Step 1: Calculate the value of L / 2c

L / 2c = 152 / (2 × 487.68) = 152 / 975.36 = 0.1558

Step 2: Find the hyperbolic sine of L / 2c

sinh(0.1558) can be calculated using the formula for hyperbolic sine:

sinh(x) = (e^x - e^-x) / 2

Substituting x = 0.1558:

sinh(0.1558) = (e^0.1558 - e^-0.1558) / 2

Using exponential values:

• e^0.1558 ≈ 1.1688
• e^-0.1558 ≈ 0.8557

Therefore:

sinh(0.1558) = (1.1688 - 0.8557) / 2 = 0.3131 / 2 = 0.15655

Step 3: Calculate the length of the conductor

Using the formula:

S = 2c sinh(L / 2c)

Substitute the values:

S = 2 × 487.68 × 0.15655

S = 487.68 × 0.3131

S ≈ 487.68 m

Final Answer: The length of the conductor is approximately 487.68 m.

Correct Option: Option 1

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: 152.614 m

This option is incorrect because it represents the horizontal span between the two towers, not the actual length of the conductor. The length of the conductor in a catenary is always greater than the horizontal span due to the sag in the cable.

Option 3: 5.934 m

This option is incorrect because it is significantly smaller than the actual length of the conductor. Such a value does not align with the given data and the catenary equation.

Option 4: 11.9 m

This option is incorrect for similar reasons as Option 3. It is far too small to represent the length of the conductor, given the span and other parameters provided.

Conclusion:

The calculation of the length of a conductor in a catenary is essential for designing and installing transmission lines. The correct answer, as derived, is Option 1: 487.68 m. This value aligns with the given data and the catenary equation. Understanding the principles of catenary curves is crucial for ensuring the structural stability and efficiency of overhead power lines.

Concept

The breaking force is given by:

\(F_b=σ_b\times A\)

The tension experienced by the line is given by:

\(T={F_b\over SF}\)

where, Fb = Breaking force

σb = Breaking stress

A = Cross-sectional area

SF = Safety factor

Calculation

Given, σb = 4000 kg/cm2

A = 2 cm²

\(F_b=4000\times2=8000 \space kg\)

\(T={8000\over 4}\)

T = 2000 kg

Concept

The sag in an overhead transmission line is given by:

\(S={WL^2\over 8T}\)

where, S = Sag

W = Weight per unit length of the conductors

L = Length of span

T = Tension of the conductor

Sag for unequal tower length:

\(h=S_2-S_1\)

\(h={WL_2^2\over 8T}-{WL_1^2\over 8T}\)

\(h={W\over 8T}(L_2^2-L_1^2)\)

\(h={W\over 8T}(x_2^2-x_1^2)\)

\(h={W\over 8T}(x_2-x_1)(x_2+x_1)\)

∴ (S2 - S1) α  (x2 - x1)

Concept:

For overhead transmission lines, sag (S) is directly proportional to the square of the span (L), when weight per unit length and tension are constant:

\( S \propto L^2 \)

Calculation:

Let the initial span be \(L_1\) with sag \(S_1\), and a new span \(L_2\) with sag \(S_2\).

Given that sag is reduced by 25%,

\( \frac{S_2}{S_1} = 0.75 \)

Now using the sag-span relation:

\( \left( \frac{L_2}{L_1} \right)^2 = 0.75 \Rightarrow \frac{L_2}{L_1} = \sqrt{0.75} \approx 0.866 \)

This means the span must be reduced to 86.6% of its original length.

Conclusion:

To decrease the sag by 25%, the span should be reduced by approximately 13.4%. The closest option given is:

Option 3: reduced by 25% 

• Pin insulators are used for holding the line conductors on the straight running of poles. These are commonly used in power networks up to 33 kV system.
• Suspension insulators consist of a number of porcelain discs connected in series by metal links in the form of a string. The conductor is suspended at the bottom end of this string while the other end of the string is secured to the cross- arm of the tower. For high voltage (>33KV), it is a usual practice to use suspension type insulators.
• When there is a dead-end of the line or there is a corner or sharp curve, the line is subjected to greater tension. In order to relieve the line of excessive tension, strain insulators are used.
• For low voltage lines (<11 kV) shackle insulators are used as strain insulators.
• Stay insulators are also known as strain insulators and are generally used up to 33 kV line. These insulators should not be fixed below three meters from the ground level. These insulators are also used where the lines are strained.

Concept

Sag: It is defined as the vertical distance between two points of support of the transmission towers and the lowest peak point of the conductor. 

The sag in an overhead transmission line is given by:

\(S={WL^2\over 8T}\)

where, S = Sag

W = Weight per unit length of the conductors

L = Length of span

T = Tension of the conductor

Explanation: 

From the figure:

The height between the two supports of a transmission and distribution overhead line = Vertical distance between the higher height support point of the conductor and the Lowest point of the conductor - Vertical distance between the lower height support point of the conductor and lowest point of the conductor

\(h=S_2-S_1\)

\(h={Wx_2^2\over 8T}-{Wx_1^2\over 8T}\)

Concept:

Sag (s):

The distance between the highest point of electric poles or towers and the lowest point of a conductor connected between two poles or towers.

Span length:

It is the shortest distance between two towers or poles. 

Sag

 \(s = \frac{{W{l^2}}}{{8T}}\)

Where,

S is the sag of the conductor

W is the weight of the conductor

l is the span length of the conductor

T is the working tension on the conductor

Calculation:

Span length = 200 m

Breaking strength (Ultimate strength) = 6000 kg

Weight of conductor W = 900 kg/km

= 900 kg/1000m   (1 km = 1000 m)

W = 0.9 kg/m

Working tension, T = ultimate strength/safety factor

\( = \frac{{6000}}{2}\)

T = 3000 kg

Sag \(s = \frac{{W{l^2}}}{{8T}}\)

= \(\frac{{\left( {0.9 \times {{200}^2}} \right)}}{{8 \times 3000}}\)

= 1.5 m

Concept

The effective weight per meter of the conductor is calculated by:

Wmaterial is the weight of the conductor material per meter

Wmaterial  =  Specific Gravity × Volume of 1 m conductor

W_wind is the wind pressure acting on the conductor per meter

Calculation

Given, A = 2 cm² 

Specific Gravity = 9.9 gm/cm3 

Wwind = 1.5 kg/m

Wmaterial  = 9.9 × 2 × 100 = 1980 gm = 1.98 kg/m

\(W_{t}=\sqrt{(1.98)^2+(1.5)^2}\) 

\(W_{t}=2.48\space kg/m\)

Concept of internal inductances of cylindrical conductor:

Consider a straight round (cylindrical) conductor, having a radius of r and carries a current I.

According to Ampere law states that the magnetomotive force (MMF) in ampere-turns around a closed path is equal to the net current in ampere enclosed by the path.

It is given by 

\(MMF = \oint H\;ds = I\)

Where H is the magnetic field intensity in At/m, s is the distance along the path in meter and I is the current in ampere.

Let the field intensity at a distance x from the center of the conductor by H_x. It is to be noted that H_x is constant at all points that are at a distance x from the center of the conductor. Therefore H_x is constant over the concentric circular path with a radius of x and is tangent to it. The current enclosed by I_x  then, 

\(\oint {H_x} = dx = {I_x}\)

\({H_x} = \frac{{{I_x}}}{{2\pi x}}\)    ......(1)

Cross-section of a round conductor

Assume that the current density is uniform over the entire conductor then we can write,

\(\frac{I}{{\pi {r^2}}} = \frac{I}{{\pi {x^2}}}\)

\({I_x} = \frac{{\pi {x^2}}}{{\pi {r^2}}}I\)      ----(2)

Put the value of equation (2) in equation (1) then we get,

\(H = \frac{I}{{2\pi {r^2}}}x\)       ----(3)

Assuming a relative permeability of 1, the flux density at a distance of x from the center of the conductor is given by

\({B_x} = {\mu _0}{H_x} = \frac{{{\mu _0}I}}{{2\pi {r^2}}}x\)       ----(4)

where µ₀ is the permeability of the free space and is given by 4π X 10^-7 H/m.

The flux inside (or outside) the conductor is in the circumferential direction. The two directions that are perpendicular to the flux are radial and axial. Let us consider an elementary area that has a dimension of dx m along the radial direction and 1 m along the axial direction. Therefore the area perpendicular to the flux at all angular positions is dx X 1 m². Let the flux along the circular strip be denoted by dφ _x and this is given by,

\(d{\phi _x} = {B_x}dx \times 1 = \frac{{{\mu _0}I}}{{2\pi {r^2}}}x\;dx\)       ----(5)

Note that the entire conductor cross-section does not enclose the above flux. The ratio of the cross-sectional area inside the circle of radius x to the total cross-section of the conductor can be thought of as a fractional turn that links the flux dφ _x. Therefore the flux linkage is

\(d{\lambda _x} = \frac{{\pi {x^2}}}{{\pi {r^2}}}\;d{\phi _x} = \frac{{{\mu _0}I}}{{2\pi {r^4}}}{x^3}dx\)       ----(6)

Integrating equation (6) over the range of x, i.e., from 0 to r, we get the internal flux linkage as,

\({\lambda _{int}} = \mathop \smallint \limits_0^\gamma \frac{{{\mu _0}I}}{{2\pi {r^4}}}{x^3}dx = \frac{{{\mu _0}I}}{{8\pi }}\)     ----(7)

Inductance is given by, 

\(L = \frac{\lambda }{I}\)

we get the internal inductance per unit length as

\({L_{int}} = \frac{1}{2} \times {10^{ - 7}}H/m\)

Given:

Inner circle diameter = d,  Radius = d/2

Let d/2 = R

Thickness of the ice around the conductor = t

Concept:

Area of the shaded region is the difference of the areas of the bigger circle and the smaller circle.

Formula used:

Area of circle = \(\pi r^2\)

Calculation:

∵ Area of the shaded region = \(\pi (R + t)^2\) - \(\pi R^2\)

= \(\pi (R^2 + t^2 + 2Rt)\) - \(\pi R^2\)

= \(\pi (t^2 + 2Rt)\)

= \(\pi t(t + 2R)\)

∵ d/2 = Rd/2 = R

= \(\pi\text{t}(t+2\times\dfrac{d}{2})\)

= πt(d + t)

∴ The expression for volume of ice per unit length is πt(d + t).

Concept of Sag (s):

The distance between the highest point of electric poles or towers and the lowest point of a conductor connected between two poles or towers.

Span length: It is the shortest distance between two towers or poles. 

\(S = \frac{{W{l^2}}}{{8T}}\)

Where,

S is the sag of the conductor

W is the weight of the conductor

l is the span length of the conductor

T is the working tension on the conductor

Stringing chart

• The stringing chart gives the data per sag to be allowed and the tension to be allowed in a transmission line for a particular temperature.
• The stringing chart represents a graph of sag and tension vs temperature.
• The stringing chart is prepared by calculating the sag and tension on the conductor under the worst conditions such as maximum wind pressure and minimum temperature by assuming a suitable safety factor.
• As we want low Tension and minimum sag in our conductor but that is not possible as sag is inversely proportional to tension.
• It is because low sag means a tight wire and high tension whereas low tension means a loose wire and increased sag.
• Therefore, we make a compromise between the two but if the case of temperature is considered and we draw a graph then that graph is called a Stringing chart.

The correct answer is option 2): 6.4

Concept:

Maximum Sag (S)  =Wl² / 8T

Where

W = weight of conductor per meter length in Kg per meter

l = Span length in meter

T = Working tension in Kg

Calculation:

Resultant force per meter length of conductor = Weight of conductor per meter length

W = 2 kg/m,

T = 4000 kg,

l = 320 meters,

\(S = \frac{{2 \times {{320}^2}}}{{8 \times 4000}} = 6.4\)