Question

A single core cable has to be designed for capacity of 66 kV to ground. The radius of conductor is 10 mm and insulating material A, B and C have relative permittivities of 5, 4, 3 and corresponding maximum stresses of 3.8, 3.6 and 2 kV/mm respectively. The minimum diameter of the sheath will be _________ mm.

• A. 78.86

Answer 1

The correct option is A 78.86

We know, $g_{Amax}=\frac{Q}{2\pi {ϵ}_0{ϵ}_{1}r}$

$g_{Bmax}=\frac{Q}{2\pi {ϵ}_0{ϵ}_2{r}_1}$

$g_{Cmax}=\frac{Q}{2\pi {ϵ}_0{ϵ}_3{r}_2}$

$\frac{g_{Amax}}{g_{Bmax}}=\frac{{ϵ}_2{r}_1}{{ϵ}_{1}r}$

$\frac{3.8}{3.6}=\frac{4}{5}\times \frac{r_1}{10}$

$\Rightarrow r_1=\frac{3.8\times 50}{3.6\times 4}=13.19mm$

$\frac{g_{Bmax}}{g_{Cmax}}=\frac{{ϵ}_3{r}_2}{{ϵ}_2{r}_1}$

$\Rightarrow \frac{3.6}{2}=\frac{3\times r_2}{4\times 13.19}$

$r_2=31.66mm$

$V=g_{max1}rln\left(\frac{r_1}{r}\right)+g_{max2}{r}_{1}ln\left(\frac{r_2}{r_1}\right)+g_{max3}{r}_{2}ln\left(\frac{R}{r_2}\right)$

$66=3.8\times 10\times ln\left(\frac{13.19}{10}\right)+3.6\times 13.19ln\left(\frac{31.66}{13.19}\right)+2\times 31.66ln\left(\frac{R}{31.66}\right)$

$66=(3.8\times 10\times 0.2768)+(3.6\times 13.19\times 0.876)+2\times 31.66ln\left(\frac{R}{31.66}\right)$

$66=\left(10.5184\right)+\left(41.59\right)+63.32ln\left(\frac{R}{31.66}\right)$

$13.89=63.32ln\left(\frac{R}{31.66}\right)$

$ln\left(\frac{R}{31.66}\right)=0.2194$

$\frac{R}{31.66}=1.9515,R=39.43mm$

So, minimum diameter of sheath = 2R = 78.86 mm