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A copper wire of radius 0.5 mm is insulated with a sheathing ofthickness 1 mm having a thermal conductivity of 0.5 W/m – K. Theoutside surface convective heat transfer coefficient is 10 W/m2 – K. Ifthe thickness of insulation sheathing is raised by 10 mm, then theelectrical current-carrying capacity of the wire will: 
  • a)
    Increase
  • b)
    Decrease
  • c)
    Remain the same
  • d)
    Vary depending upon theelectrical conductivity of the wire
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A copper wire of radius 0.5 mm is insulated with a sheathing ofthickne...
Given that:
ri = 0.5 mm
ro = 0.5 + 1 + 10
= 11.5 mm
κ = 0.5 W/mK
h = 10 W/m2K
Critical radius of insulation for wire is
rc = κ/ho = 0.5/10
= 0.05 m
= 5 mm
Since rc < ro, there will be decrease in rate of heat
transfer.
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Most Upvoted Answer
A copper wire of radius 0.5 mm is insulated with a sheathing ofthickne...
Explanation:

When the thickness of insulation sheathing is raised by 10 mm, the current-carrying capacity of the wire will increase. This can be explained by the following points:

Effect of insulation thickness on heat transfer:

- Heat transfer through insulation occurs by conduction.
- The rate of heat transfer through insulation is given by Fourier's law of heat conduction: q = -kA(dT/dx), where q is the heat transfer rate, k is the thermal conductivity, A is the area of heat transfer, and (dT/dx) is the temperature gradient.
- The temperature gradient (dT/dx) decreases with an increase in insulation thickness. This is because the temperature drop across the insulation is distributed over a larger thickness.
- Therefore, the rate of heat transfer through insulation decreases with an increase in insulation thickness.

Effect of heat transfer on current-carrying capacity:

- When a current flows through a wire, it generates heat due to the resistance of the wire.
- The rate of heat generation is given by Joule's law: q = I^2R, where q is the heat generation rate, I is the current, and R is the resistance of the wire.
- The wire dissipates this heat to the surroundings by convection and radiation.
- The rate of heat dissipation is proportional to the surface area of the wire and the convective heat transfer coefficient: q = hA(Ts - T∞), where h is the convective heat transfer coefficient, A is the surface area of the wire, Ts is the surface temperature of the wire, and T∞ is the surrounding temperature.
- Therefore, the rate of heat dissipation increases with an increase in surface area and convective heat transfer coefficient.
- This means that the current-carrying capacity of the wire increases with an increase in surface area and convective heat transfer coefficient.

Conclusion:

- When the thickness of insulation sheathing is raised by 10 mm, the rate of heat transfer through insulation decreases.
- This reduces the temperature gradient across the insulation and increases the surface temperature of the wire.
- Therefore, the rate of heat dissipation from the wire increases due to an increase in surface area and convective heat transfer coefficient.
- This increases the current-carrying capacity of the wire.
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A copper wire of radius 0.5 mm is insulated with a sheathing ofthickne...
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A copper wire of radius 0.5 mm is insulated with a sheathing ofthickness 1 mm having a thermal conductivity of 0.5 W/m – K. Theoutside surface convective heat transfer coefficient is 10 W/m2 – K. Ifthe thickness of insulation sheathing is raised by 10 mm, then theelectrical current-carrying capacity of the wire will:a)Increaseb)Decreasec)Remain the samed)Vary depending upon theelectrical conductivity of the wireCorrect answer is option 'A'. Can you explain this answer?
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