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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 = 6W/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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An electric cable of aluminium conductor (k = 240 W/mK) is to be insul...
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An electric cable of aluminium conductor (k = 240 W/mK) is to be insul...
Solution:

Given data:

Conductivity of aluminium, k1 = 240 W/mK

Conductivity of rubber, k2 = 0.15 W/mK

Heat transfer coefficient of air, h = 6W/m2K

Critical thickness of insulation, x = ?

Formula used:

Heat transfer rate through the insulation,

q = (T1 - T2) / [(1/h) + (x/k2) + (x/k1)]

where,

T1 = temperature of cable

T2 = temperature of air

By assuming T1 = 80°C and T2 = 20°C, the above equation can be written as,

q = (80 - 20) / [(1/6) + (x/0.15) + (x/240)]

q = 60 / [(1/6) + (x/0.15) + (x/240)]

For critical thickness of insulation, q = 0

0 = 60 / [(1/6) + (x/0.15) + (x/240)]

0 = [(1/6) + (x/0.15) + (x/240)]

0 = (1/6) + (x/0.15) + (x/240)

Solving the above equation, we get,

x = 0.025 m or 25 mm

Therefore, the critical thickness of insulation will be 25 mm.
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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 = 6W/m2). The critical thickness of insulation will be:a)25mmb)40 mmc)160 mmd)800 mmCorrect answer is option 'A'. Can you explain this answer?
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