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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.
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.
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 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? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 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? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 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? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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ample number of questions to practice 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? tests, examples and also practice Mechanical Engineering tests.
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