Sag Tension Calculations MCQ Quiz - Objective Question with Answer for Sag Tension Calculations - Download Free PDF

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% 

Concept:

Sag is defined as the difference in level between points of supports and the lowest point on the conductor.

Sag of conductor’s between two poles can be determined by for Equal level

S = WX2/(8T)

Where, S is the sag of conductors

W is the weight per unit length of the conductor

L is the length of span

T is the tension in the conductor

For unequal level,

S = WX2/(2T)

Observations:

• The SAG is directly proportional to the weight of the conductor.
• The SAG is directly proportional to the span length.
• The SAG is inversely proportional to the working tensile strength of the conductor at constant temperature.

When the temperature rises:

• As the temperature rises, the tension in the transmission line decreases
• As temperature rises, the sag in transmission lines increases
• Tension and sag in transmission lines are complementary to each other

Important Points:

• Stringing chart is useful in knowing the sag and tension at any temperature
• Stringing chart gives the data per sag to be allowed and the tension to be allowed for a particular temperature
• Stringing chart prepared by calculating the sag and tension on the conductor under worst conditions such as maximum wind pressure and minimum temperature by assuming a suitable safety factor

​

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

\(W_{t}=\sqrt{(1.98)^2+(1.5)^2}\) 

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

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).

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,

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

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.

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 = 220 m

Breaking strength (Ultimate strength) = 5758 kg

Weight of conductor (W) = 804 kg/km

= 804 kg/1000m   (1km = 1000m)

W = 0.804 kg/m

working tension, T = ultimate strength/safety factor

\( = \frac{{5758}}{2}\)

T = 2879 kg

Sag \(s = \frac{{W{l^2}}}{{8T}}\)

= \(\frac{{\left( {0.804 \times {{220}^2}} \right)}}{{8 \times 2879}}\)

= 1.69 m

The maximum sag of an overhead transmission line is 1.69 m

• Stringing chart is useful in knowing the sag and tension at any temperature
• Stringing chart gives the data per sag to be allowed and the tension to be allowed for a particular temperature
• Stringing chart prepared by calculating the sag and tension on the conductor under worst conditions such as maximum wind pressure and minimum temperature by assuming a suitable safety factor

​

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

Here in the question length of the span is only increased from 100 m to 120 m

l₁ = 100 m

l₂ = 120m

Sag (S) ∝ l²

_{\({S_2 \over S_1}={l_2^2 \over l_1^2}= {120^2 \over 100^2}=1.44\)}

S₂  = 1.44 × 2 = 2.88 m 

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

Given:

Span length(l) = 150m   ;  Working Tension(T) = 2500 Kg

Wind force/m length of conductor(W_w)= 0.87 Kg

Calculation:

Wt. of conductor/m length(W) = Sp. Gravity \(×\) Volume of 1 m conductor

or, W = 9\(×\)2\(×\)100 = 1800 gm = 1.8 kg

Total weight of 1 m length of conductor(W_t) =  \(\sqrt{(W^2+W_w^2 )}\) = \(\sqrt{(1.8^2+0.87^2 )}\) = 1.99 Kg

Sag(S)  = \(\frac{W_t l^2}{8T}\) = \(\frac{1.99×150^2}{8×2500}\) = 2.25 m

Concept:

Sag is defined as the difference in level between points of supports and the lowest point on the conductor.

Sag of conductor’s between two poles can be determined by

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

Where, S is the sag of conductors

W is the weight per unit length of the conductor

L is the length of span

T is the tension in the conductor

Factors affecting the sag:

Conductor weight: Sag of the conductor is directly proportional to its weight. The weight of the conductors is increased due to ice loading.

Span: Sag is directly proportional to the square of the span length. Longer span gives more sag.

Tension: The sag is inversely proportional to the tension in the conductor. Higher tension increases the stress in the insulators and supporting structures.

Wind: It increases sag in the inclined direction.

Temperature: The sag is reduced at low temperatures and is increases at higher temperatures. In summer due to the increase in average temperature tension decreases and hence sag increases comparison to winter.

The sag of the transmission line is least affected by current passes through the conductor.