A single core lead covered cable is to be designed for 66 kV to earth. Its conductor radius is 0.5 cm and its three insulating materials A, B and C have relative permittivities of 4, 2.5 and 4.0 with maximum permissible stresses of 50, 30 and 40 kV/cm respectively. Determine the minimum internal

Question

A single core lead covered cable is to be designed for 66 kV to earth. Its conductor radius is 0.5 cm and its three insulating materials A, B and C have relative permittivities of 4, 2.5 and 4.0 with maximum permissible stresses of 50, 30 and 40 kV/cm respectively. Determine the minimum internal diameter of the lead sheath. Discuss the arrangement of the insulating materials.

Accepted Answer

In order to have minimum internal diameter of the lead sheath and consistency in the electric stresses, it is desired that the material to be placed near the surface of the conductor should be one which has maximum of the product of electric stress and the permittivity i.e., in this case the material A has the product 4 × 50 = 200 as the maximum and will be placed near the surface of the conductor and material C what has the product 4 × 40 = 160 will be placed next and then materials B.

Here r = 0.5 cm

Let q be the charge per unit length of the cable. Then

[latex]\begin{aligned}&g_{1 \max }=\frac{q}{2 \pi \varepsilon_{0} \varepsilon_{1} r} \&g_{2 \max }=\frac{q}{2 \pi \varepsilon_{0} \varepsilon_{2} r_{1}} \&g_{3 \max }=\frac{q}{2 \pi \varepsilon_{0} \varepsilon_{3} r_{2}}\end{aligned}[/latex]

From these equations

[latex]q=2 \pi \varepsilon_{0} \varepsilon_{1} r g_{1 \max }=2 \pi \varepsilon_{0} \varepsilon_{2} r_{1} g_{2 \max }=2 \pi \varepsilon_{0} \varepsilon_{3} r_{2} g_{3 \max }[/latex]

or [latex]\quad \varepsilon_{1} r g_{1 ext { max }}=\varepsilon_{2} r_{1} g_{2 \max }=\varepsilon_{3} r_{2} g_{3 \max }[/latex]

or [latex]\quad \frac{r_{1}}{r} =\frac{\varepsilon_{1} g_{1 \mathrm{max}}}{\varepsilon_{2} g_{2 \max }}[/latex]

or [latex]\quad r_{1} =\frac{200}{160} imes 0.5=\frac{2.5}{4}=0.625 [/latex] cms

Similarly, [latex]\quad \frac{r_{2}}{r_{1}}=\frac{\varepsilon_{2} g_{2 \max }}{\varepsilon_{3} g_{3 \max }}=\frac{4 imes 40}{2.5 imes 30}[/latex]

or [latex]\quad r_{2}=0.625 imes \frac{160}{75} [/latex] = 1.33 cms

[latex]\begin{aligned} ext {Now} \quad V &=66=r g_{1 ext { max }} \ln \frac{r_{1}}{r}+r_{1} g_{2 \max } \ln \frac{r_{2}}{r_{1}}+r_{2} g_{3 \max } \ln \frac{R}{r_{2}} \&=0.5 imes 50 \ln \frac{0.625}{0.5}+0.625 imes 40 \ln \frac{1.33}{0.625}+1.33 imes 30 \ln \frac{R}{1.33}\end{aligned}[/latex]

[latex]\begin{aligned} ext { or } \quad 66 &=25 \ln \frac{0.625}{0.5}+25 \ln \frac{1.33}{0.625}+39.9 \ln \frac{R}{1.33} \&=5.578+18.88+39.9 \ln \frac{R}{1.33} \end{aligned}[/latex]

[latex]39.9 \ln \frac{R}{1.33} =41.54[/latex]

[latex]\ln \frac{R}{1.33} =1.0411635 [/latex]

[latex] ext { or } \quad \frac{R}{1.33} =2.83251 [/latex]

or D = 7.53 cms.

Question 9.2: A single core lead covered cable is to be designed for 66 kV...

Related Answered Questions

Determine the economic overall diameter of a 1-core cable metal sheathed for a working voltage of 85 kV if the dielectric strength of the insulating material is 65 kV/cm. ...

Verified Answer:

If the lead sheath resistance in the previous problem is 2.14 ohm per km and that of the conductor is 0.1625 ohm/km, determine the ratio of sheath loss to core loss of the cable. ...

Verified Answer:

Verified Answer:

Verified Answer:

The capacitance of a 3-core lead sheathed cable measured between any two of the conductors with sheath earthed is 0.19 μF per km. Determine the equivalent star connected capacity and the kVA required to keep 16 kms of the cable charged when connected to 20 kV, 50 Hz supply. ...

Verified Answer:

A 66 kV concentric cable with two inter-sheaths has a core diameter 1.8 cm. Dielectric material 3.5 mm thick constitutes the three zones of insulation. Determine the maximum stress in each of the three layers if 20 kV is maintained across each of the inner two. ...

Verified Answer:

Verified Answer: