Question

A bundle configuration for UHV lines (above) has identical conductors equally spaced around a circle, as shown in Figure 4.29. N_bis the number of conductors in the bundle, A is the circle radius, and D_sis the conductor GMR. Using the distance D_1nbetween conductors 1 and n given by ${\mathit{D}}_{\mathbf{1}\mathbf{n}}\mathbf{=}\mathbf{2}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left[}\left(n-1\right)\frac{\mathbf{\pi }}{{\mathbf{N}}_{\mathbf{b}}}\mathbf{\right]}$for n=1,2.....N_b and the following trigonometric identity:

$\mathbf{\left[}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\left[\frac{\pi }{N_b}\right]\mathbf{\right]}\mathbf{\left[}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\left[\frac{2\pi }{N_b}\right]\mathbf{\right]}\mathbf{\left[}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\left[\frac{3\pi }{N_b}\right]\mathbf{\right]}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left[}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\left[\left(N_b-1\right)\frac{\pi n}{N_b}\right]\mathbf{\right]}\mathbf{=}{\mathit{N}}_{\mathbf{b}}$

show that the bundle GMR, denoted D_SL, is,

${\mathit{D}}_{\mathbf{S}\mathbf{L}}\mathbf{=}{\mathbf{\left[}{\mathbf{N}}_{\mathbf{b}}{\mathbf{D}}_{\mathbf{S}}{\mathbf{A}}^{\left(N_b-1\right)}\mathbf{\right]}}^{\frac{\mathbf{1}}{{\mathbf{N}}_{\mathbf{b}}}}$

Also show that the above formula agrees with (4.6.19) through (4.6.21)
for EHV lines with N_b=2.3 and 4.

Short answer

Answer

Therefore, the bundle GMR for N_bnumber of conductors is$D_{SL}={\left(D_S{A}^{N_b-1}{N}_b\right)}^{\frac{1}{N_b}}$

The bundle GMR for N_b=2,3 and 4 are$\sqrt{d{D}_s},\sqrt[3]{D_s{d}^2}and\sqrt[4]{D_s{d}^3}\sqrt{\frac{4}{2\sqrt{2}}}$

Step-by-step solution

Write the given data from question.

The distance between the conductor 1 and n.

$D_{1n}=2A\left[\sin \left(n-1\right)\frac{\pi }{N_b}\right]$

Here is the radius of the circle, N_bis the number of the conductor.

The trigonometric identity.

$\left[2sin\left[\frac{\pi }{N_b}\right]\right]\left[2sin\left[\frac{2\pi }{N_b}\right]\right]\left[2sin\left[\frac{3\pi }{N_b}\right]\right].......\left[2sin\left[\left(N_b-1\right)\frac{\pi n}{N_b}\right]\right]=N_b$

Write the formula for the GMR.

The bundle GMR can be calculated as follows:

$$\begin{gathered} {\mathbf{D}}_{\mathbf{SL}}\mathbf{=}\sqrt[{\mathbf{N}}_{\mathbf{b}}^{\mathbf{2}}]{{\left({\mathbf{D}}_{\mathbf{11}}{\mathbf{D}}_{\mathbf{12}}......{\mathbf{D}}_{\mathbf{1}{\mathbf{N}}_{\mathbf{b}}}\right)}^{{\mathbf{N}}_{\mathbf{b}}}} \\ {\mathbf{D}}_{\mathbf{SL}}\mathbf{=}{\left({\mathbf{D}}_{\mathbf{11}}{\mathbf{D}}_{\mathbf{12}}......{\mathbf{D}}_{\mathbf{1}{\mathbf{N}}_{\mathbf{b}}}\right)}^{\frac{\mathbf{1}}{{\mathbf{N}}_{\mathbf{b}}}} \end{gathered}$$

Show that DSL=[NbDsANb-1]1Nb

Consider the diagram given below,

Substitute D_{11 for Ds}$D_{1n}=2A\sin \left[\frac{\left(n-1\right)\pi }{N_b}\right]$where n=2,3.....N_b into above equation.

$D_{SL}=D_s{\left(\left[2sin\left[\frac{\pi }{N_b}\right]\right]\left[2sin\left[\frac{2\pi }{N_b}\right]\right]\left[2sin\left[\frac{3\pi }{N_b}\right]\right].......\left[2sin\left[\left(N_b-1\right)\frac{\pi n}{N_b}\right]\right]\right)}^{\frac{1}{N_b}}$

By using trigonometric identity here.

$D_{SL}={\left(D_S{A}^{N_b-1}{N}_b\right)}^{\frac{1}{N_b}}$

Therefore, the bundle GMR is$D_{SL}={\left(D_S{A}^{N_b-1}{N}_b\right)}^{\frac{1}{N_b}}$

In case of the two conductor the schematic diagram is shown below.

The radius of the circle,$A=\frac{d}{2}$

Here dis the distance between the conductors.

Substitutelocalid="1656753009678" $\frac{d}{2}$for A in equation (1).

localid="1656753014271" $$\begin{gathered} D_{SL}={\left[D_s\left(\frac{d}{2}\right)\left(2\right)\right]}^{\frac{1}{2}} \\ {D}_{SL}={\left(d{D}_s\right)}^{\frac{1}{2}} \\ {D}_{SL}=\sqrt{d{D}_s} \end{gathered}$$

Therefore, the bundle GMR islocalid="1656753019613" $\sqrt{d{D}_s}$

In case of three conductor, the schematic diagram is shown below.

The radius of the circle:$A=\frac{d}{\sqrt{3}}$

Substitutelocalid="1656753026742" $\frac{d}{\sqrt{3}}$for A into equation (1).

localid="1656753031815" $$\begin{gathered} D_{SL}={\left[D_s{\left(\frac{d}{\sqrt{3}}\right)}^2\left(3\right)\right]}^{\frac{1}{3}} \\ {D}_{SL}={\left[D_s\left(\frac{d^2}{3}\right)\left(3\right)\right]}^{\frac{1}{3}} \\ {D}_{SL}={\left(D_s{d}^2\right)}^{\frac{1}{3}} \\ {D}_{SL}=\sqrt[3]{D_s{d}^2} \end{gathered}$$

Therefore, the bundle GMR is localid="1656753039119" $\sqrt[3]{D_s{d}^2}$.

In case of four conductor, the schematic diagram is shown below.

The radius of the circle:$A=\frac{d}{\sqrt{2}}$

Substitute$\frac{d}{\sqrt{2}}$for A into equation (1).

$$\begin{gathered} D_{SL}={\left[D_s{\left(\frac{d}{\sqrt{2}}\right)}^3\left(4\right)\right]}^{\frac{1}{4}} \\ {D}_{SL}={\left[D_s\left(\frac{d^3}{{\left(\sqrt{2}\right)}^3}\right)\left(4\right)\right]}^{\frac{1}{4}} \\ {D}_{SL}=\sqrt[4]{D_s{d}^3\left(\frac{4}{{\left(\sqrt{2}\right)}^3}\right)} \\ {D}_{SL}=\sqrt[4]{D_s{d}^3}\sqrt{\frac{4}{2\sqrt{2}}} \end{gathered}$$

Therefore, the bundle GMR is $\sqrt[4]{D_s{d}^3}\sqrt{\frac{4}{2\sqrt{2}}}$.