Question

Determine the GMR of each of the unconventional stranded conductors shown in Figure 4.30. All strands have the same radius r.

Short answer

(a) The GMR for unconventional stranded conductor is 1.46r

(b) The GMR for unconventional stranded conductor is 2.155r

(c) The GMR for unconventional stranded conductor is 2.6374r .

Step-by-step solution

Determine the formula for the GMR.

Write the formula for GMR.

$\mathbf{GMR}\mathbf{=}\sqrt[\mathbf{9}]{{\left[\left(\mathbf{0}\mathbf{.}\mathbf{7788}\mathbf{r}\right)\left(\mathbf{2}\mathbf{r}\right)\left(\mathbf{2}\mathbf{r}\right)\right]}^{\mathbf{3}}}$

Determine the GMR for unconventional stranded conductor in figure 4.30(a). 

(a)

All the conductor has the same radius r .

Calculate the GMR for the figure 4.30(a).

$$\begin{gathered} GMR=\sqrt[9]{{\left[\left(0.7788r\right)\left(2r\right)\left(2r\right)\right]}^3} \\ GMR={\left[0.7788\left(4\right){\left(r\right)}^3\right]}^{\frac{1}{3}} \\ GMR={\left[3.115{\left(r\right)}^3\right]}^{\frac{1}{3}} \\ GMR=1.46r \end{gathered}$$

Therefore, the GMR is 1.46r.

Determine the GMR for unconventional stranded conductor in figure 4.30(b).

All the conductor has the same radius r.

Consider the schematic diagram given below.

Calculate the GMR for the figure 4.30(b).

$$\begin{gathered} GMR=\sqrt[16]{{\left[\left(0.7788r\right)\left(2r\right)\left(4r\right)\left(6r\right)\right]}^2{\left[\left(0.7788r\right)\left(2r\right)\left(2r\right)\left(4r\right)\right]}^2} \\ GMR=\sqrt[16]{\left[{\left(0.7788\right)}^4{\left(r\right)}^{16}{\left(2\right)}^6{\left(4\right)}^4{\left(6\right)}^2\right]} \\ GMR=\left[{\left(0.7788\right)}^{\frac{1}{4}}{\left(589824\right)}^{\frac{1}{16}}r\right] \\ GMR=2.155r \end{gathered}$$

Therefore, the GMR is 2.155r

Determine the GMR for unconventional stranded conductor in figure 4.30(c).

(c)

All the conductor has the same radius r.

Consider the schematic diagram given below.

The distance between a and d can be calculated as follows.

$$\begin{gathered} ad=\sqrt{{\left(4r\right)}^2+{\left(2r\right)}^2} \\ ad=\sqrt{16r^2+4r^2} \\ ad=\sqrt{20r^2} \\ ad=\sqrt{20}r \end{gathered}$$

Calculate the GMR for the figure 4.30(c).

localid="1656755891718" $$\begin{gathered} GMR=\sqrt[81]{{\left[\left(0.7788r\right){\left(2r\right)}^2{\left(4r\right)}^2{\left(\sqrt{20}r\right)}^2\left(\sqrt{8}r\right)\left(\sqrt{32}r\right)\right]}^4{\left[\left(0.7788r\right)\left(2r^3\right){\left(\sqrt{8}r\right)}^2{\left(\sqrt{20}r\right)}^2\left(4r\right)\right]}^4\times \left[\left(0.7788r\right){\left(2r\right)}^4{\left(\sqrt{8}r\right)}^4\right]} \\ GMR=\sqrt[81]{\left[{\left(0.7788\right)}^9{\left(r\right)}^{81}{\left(2\right)}^{24}{\left(4\right)}^{12}{\left(\sqrt{20}\right)}^{16}{\left(\sqrt{8}\right)}^{16}{\left(\sqrt{32}\right)}^4\right]} \\ GMR=r\sqrt[81]{\left[\left(0.10539\right)\times 16777216\times 16777216\times 2.56\times {10}^{10}\times 16777216\times 1024\right]} \\ GMR=r{\left[1.304\right]}^{\frac{1}{81}} \end{gathered}$$

Solve further as

$$\begin{gathered} GMR=r\times 2.6374 \\ GMR=2.6374r \end{gathered}$$

Therefore, the GMR is 2.6374r