Three single core lead sheathed cables carry three-phase current of 400 A. The nominal conductor area of the cores is 1.25 sq. cm, the sheath thickness is 0.152 cm and the diameter over the sheath is 2.28 cms. They are supported in equilateral formation with a distance between cable centres of 5.08

Question

Three single core lead sheathed cables carry three-phase current of 400 A. The nominal conductor area of the cores is 1.25 sq. cm, the sheath thickness is 0.152 cm and the diameter over the sheath is 2.28 cms. They are supported in equilateral formation with a distance between cable centres of 5.08 cms. Calculate the induced e.m.f. in each sheath when the cable is 2 km long and supply is 50 Hz.

Accepted Answer

From the figure, the mean radius of the sheath

[latex]\begin{aligned}&=\frac{2.28}{2}-\frac{0.152}{2}=\frac{2.28-0.152}{2} \&=1.06 \mathrm{~cm}.\end{aligned}[/latex]

[latex]\begin{array}{ll} herefore \quad ext { The mutual inductance } & =2 imes 10^{-7} \ln \frac{d}{r_{m}} \mathrm{H} / ext { metre } \& =2 imes 10^{-7} \ln \frac{5.08}{1.06} \& =3.134 imes 10^{-7} \mathrm{H} / ext { metre } \end{array}[/latex]

[latex]\begin{array}{ll} herefore \quad ext { For } 2 \mathrm{~km} ext { length it will be } & =2 imes 10^{3} imes 3.134 imes 10^{-7} \mathrm{H} \& =6.268 imes 10^{-4} \mathrm{H} \end{array}[/latex]

[latex]\begin{array}{ll} herefore \quad ext { Voltage induced } \omega I M & =314 imes 6.268 imes 10^{-4} imes 400 ext { volts } \& =78.72 ext { volts. }\end{array}[/latex]

Question 9.7: Three single core lead sheathed cables carry three-phase cur...

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