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Electric Traction: Calculation of Sag and Tension for Trolley Wire
An overhead wire of large cross-section made of copper or a copper alloy and provides the current required for the trolleys of electric vehicle is known as trolley wire.
When a flexible wire is suspended horizontally between two pole supports, it would assume the form of catenary. But in case of tramways, the sag in the trolley wire rarely exceeds 1% to 1.5% of the span length and thus the trolley wire may be considered to hang in the form of a parabola as shown in the figure below.
Let,
$$\mathrm{Span\: length\: of\: trolley\: wire\: (in \: meters)\, \mathrm{\: =\: }\, 2\mathit{l} }$$
$$\mathrm{Weight\: of \: wire\: in\: kg\: per\: meter\, \mathrm{\: =\: }\, \mathit{W}}$$
$$\mathrm{Tension\: in\: wire\: in\: kg\, \mathrm{\: =\: }\, \mathit{T}}$$
Then, by taking the moments about the supports, we have,
$$\mathrm{\mathit{TS\, \mathrm{\: =\: }\,Wl\times \left ( \frac{l}{\mathrm{2}} \right )} }$$
Where, S is the sag in the trolley wire.
$$\mathrm{\therefore Sag,\mathit{S\, \mathrm{\: =\: }\, \frac{Wl^{\mathrm{2}}}{\mathrm{2}T}} \: \: \: \cdot \cdot \cdot \left ( 1 \right )}$$
Now, length of the wire required can be calculated as follows −
As the equation of parabola is given by the following expression,
$$\mathrm{\mathit{y\, \mathrm{\: =\: }\, \frac{Sx^{\mathrm{2}}}{l^{\mathrm{2}}}}}$$
On substituting the value of δ from eqn. (1), we get,
$$\mathrm{\mathit{y\, \mathrm{\: =\: }\, \frac{Wx^{\mathrm{2}}}{\mathrm{2}T}} \: \: \: \cdot \cdot \cdot \left ( 2 \right )}$$
Since,
$$\mathrm{\mathit{\frac{ds}{dx}\mathrm{\: =\: }\sqrt{\mathrm{1}\mathrm{\: +\: }\left ( \frac{dy}{dx} \right )^{\mathrm{2}}}}}$$
$$\mathrm{\left ( \because \mathit{\frac{dy}{dx}\mathrm{\: =\: }\frac{Wx}{T}} \right )}$$
$$\mathrm{\therefore \mathit{\frac{ds}{dx}\mathrm{\: =\: }\sqrt{\mathrm{1\mathrm{\: +\: }}\left ( \frac{Wx}{T} \right )^{\mathrm{2}}}}}$$
$$\mathrm{\Rightarrow \int \mathit{ds\mathrm{\: =\: }\int \sqrt{\mathrm{1}\mathrm{\: +\: }\left ( \frac{Wx}{T} \right )^{\mathrm{2}}}dx\mathrm{\: =\: }\int \left [ \mathrm{1}\mathrm{\: +\: }\frac{W^{\mathrm{2}}x^{\mathrm{2}}}{T^{\mathrm{2}}} \right ]^{\mathrm{1/2}}dx}}$$
$$\mathrm{\Rightarrow\mathit{s\mathrm{\: =\: }\int \left ( \mathrm{1}\mathrm{\: +\: }\frac{W^{\mathrm{2}}x^{\mathrm{2}}}{\mathrm{2}T^{\mathrm{2}}}\mathrm{\: +\: }\frac{\mathrm{3}W^{\mathrm{4}x^{\mathrm{4}}}}{\mathrm{8}T^{\mathrm{4}}}\mathrm{\: +\: }\cdot \cdot \cdot \right )dx}} $$
$$\mathrm{\therefore \mathit{s\mathrm{\: =\: }x\mathrm{\: +\: }\frac{W^{\mathrm{2}}x^{\mathrm{3}}}{\mathrm{6}T^{\mathrm{2}}}}\mathrm{\: +\: }\cdot \cdot \cdot } $$
Now, when 𝑥 = 1, i.e., half of the span length, then the length of the conductor in the half of the span is given by,
$$\mathrm{\mathit{s\mathrm{\: =\: }l\mathrm{\: +\: }\frac{W^{\mathrm{2}}l^{\mathrm{3}}}{\mathrm{6}T^{\mathrm{2}}}\mathrm{\: =\: }l\mathrm{\: +\: }\left ( \frac{\mathrm{2}TS}{l^{\mathrm{2}}} \right )^{\mathrm{2}}\times \left ( \frac{l^{\mathrm{3}}}{\mathrm{6}T^{\mathrm{2}}} \right )}} $$
$$\mathrm{\mathit{\Rightarrow s\mathrm{\: =\: }l\mathrm{\: +\: }\frac{\mathrm{2}S^{\mathrm{2}}}{\mathrm{3}l}}\: \: \cdot \cdot \cdot \left ( 3 \right )} $$
Therefore, the total length of the wire required for the full span length is
$$\mathrm{2\mathit{s\mathrm{\: =\: }\mathrm{2}\left [ l\mathrm{\: +\: }\frac{\mathrm{2}S^{\mathrm{2}}}{\mathrm{3}l} \right ]}\: \: \cdot \cdot \cdot \left ( 4 \right )} $$
Also, the length of the conductor changes with the temperature which will cause the change in sag and tension in a given span. Here, it should be noted that reduction in the temperature reduces the sag and increases the tension.
Numerical Example
A trolley wire of a tramway is suspended from two poles 50 meters apart. If the diameter of the wire is 2 cm and weight per meter is 1 kg, find its sag if the tension applied is 650 kg. Also find the total length of the wire required.
Solution
Given data −
$$\mathrm{Span\: length,\, 2\mathit{l}\mathrm{\: =\: }50\, m} $$
$$\mathrm{Half\: span \: length,\mathit{l}\mathrm{\: =\: }25\, m}$$
$$\mathrm{Weight\: of\: wire\: per\: meter,\mathit{W}\mathrm{\: =\: }1\, kg}$$
$$\mathrm{Tension\: applied,\mathit{T}\mathrm{\: =\: }650\, kg} $$
Therefore, sag in the wire is,
$$\mathrm{Sag, \mathit{S\mathrm{\: =\: }\frac{Wl^{\mathrm{2}}}{\mathrm{2}T}}\mathrm{\: =\: }\frac{1\times 25^{2}}{2\times 650}\mathrm{\: =\: }0.481\, m} $$
Again, the total length of wire required is,
$$\mathrm{2\mathit{s\mathrm{\: =\: }\mathrm{2}\left [ l\mathrm{\: +\: }\frac{\mathrm{2}S^{\mathrm{2}}}{\mathrm{3}l} \right ]}\mathrm{\: =\: }2\times \left [ 25\mathrm{\: +\: }\frac{2}{3}\times \frac{0.481^{2}}{25} \right ]\mathrm{\: =\: }50.0123 \, m} $$
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