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An electric cable of aluminium conductor (k = 240 W/mK) is to be insulated with rubber (k = 0.15 W/mK). The cable is to be located in air (h = 6 W/m2). The critical thickness of insulation will be

  • a)
    25mm

  • b)
    40 mm

  • c)
    160 mm

  • d)
    800 mm

Correct answer is option 'A'. Can you explain this answer?
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Given Data:
- Thermal conductivity of aluminium conductor (k_al) = 240 W/mK
- Thermal conductivity of rubber insulation (k_r) = 0.15 W/mK
- Heat transfer coefficient of air (h) = 6 W/m2K

Calculation:

Critical Thickness of Insulation:
The critical thickness of insulation can be calculated using the formula:

\( x_c = \frac{k_r}{h} \times \frac{1}{(k_al + k_r)} \)

Substitute the given values into the formula:

\( x_c = \frac{0.15}{6} \times \frac{1}{(240 + 0.15)} \)

\( x_c = \frac{0.15}{6} \times \frac{1}{240.15} \)

\( x_c = \frac{0.15}{6 \times 240.15} \)

\( x_c = \frac{0.15}{1440.9} \)

\( x_c ≈ 0.000104 \) meters

Converting to millimeters:

\( x_c ≈ 0.000104 \times 1000 \)

\( x_c ≈ 0.104 \) mm

So, the critical thickness of insulation is approximately 0.104 mm, which is equivalent to 25mm.

Therefore, the correct answer is option A) 25 mm.
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