Question

b) Determine the size of conductor for a two-core cable required to carry the maximum current of 60
Amps: Given: length of cable is 60m, declared supply voltage 240 volts.

Answer 1

Answer:

Electrical load of 80KW, distance between source and load is 200 meters, system voltage 415V three phase, power factor is 0.8, permissible voltage drop is 5%, demand factor is 1.

Answer 2

Complete Question:-

Determine the size of conductor for a two-core Copper cable required to carry the maximum current of 60 Amps: Given: that the length of the cable is 60m, declared supply voltage 240 volts.

The size of the conductor or the radius of the conductor is 2.87* 10-⁴m.

Given:-

Current (i) = 60A

Length of the Cable (l) = 60m

Supply Voltage (v) = 240V

To Find:-

The size of the conductor or the radius of the conductor.

Solution:-

We can simply calculate the size of the conductor or the radius of the conductor by using the following procedure.

As

Current (i) = 60A

Length of the Cable (l) = 60m

Supply Voltage (v) = 240V

The resistivity of Copper = 1.7* 10-⁸

We all know

$v = i \times r$

$r = \frac{v}{i}$

$r = \frac{240}{60}$

$r = \frac{24}{6}$

$r = 4ohm$

Now, according to the formula of Resistance

$R= ¶ \frac{l}{a}$

$4 = 1.7 \times {10 }^{ - 8} \times \frac{60}{a}$

$a = \frac{60 \times {10}^{ - 8} \times 1.7}{4}$

$a = 15 \times 1.7 \times {10}^{ - 8}$

$a = 25.5 \times {10}^{ - 8} {m}^{2}$

And,

$Area = π r²$

$a = \pi {r}^{2}$

${r}^{2} = \frac{a}{\pi}$

${r}^{2} = \frac{25.5 \times {10}^{ - 8} }{3.14}$

${r}^{2} = 8.121 \times {10}^{ - 8}$

$r = \sqrt{8.121 \times {10}^{ - 8} }$

$r = 2.87 \times {10}^{ - 4} m$

Hence, The size of the conductor or the radius of the conductor is 2.87* 10-⁴m.

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