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A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electric current. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will
- a)increase
- b)remain the same
- c)decrease
- d)be zero
Correct answer is option 'A'. Can you explain this answer?
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A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0...
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A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0...
Explanation:
The heat transfer from the wire to the ambient is determined by conduction through the plastic sleeve and convection at the outer surface of the sleeve exposed to air. Let's analyze each of these processes separately.
Conduction through the plastic sleeve:
The rate of heat transfer through conduction can be determined using Fourier's law of heat conduction, which states that the rate of heat transfer is proportional to the temperature gradient and the cross-sectional area.
q = -k * A * (dT/dr)
where,
q = rate of heat transfer
k = thermal conductivity of the plastic
A = cross-sectional area
dT/dr = temperature gradient
Since the wire and the plastic sleeve are in contact, the temperature at the interface is the same. Therefore, the temperature gradient across the plastic sleeve is constant.
As the outer radius of the plastic sleeve is larger than the wire, the cross-sectional area of the plastic sleeve is larger than the wire. Hence, the rate of heat transfer through conduction will increase due to the addition of the plastic cover.
Convection at the outer surface of the sleeve exposed to air:
The rate of heat transfer through convection can be determined using Newton's law of cooling, which states that the rate of heat transfer is proportional to the temperature difference between the surface and the ambient and the heat transfer coefficient.
q = h * A * (Ts - Ta)
where,
q = rate of heat transfer
h = heat transfer coefficient
A = surface area
Ts = surface temperature
Ta = ambient temperature
As the heat transfer coefficient on the outer surface of the sleeve exposed to air is a constant value, the rate of heat transfer through convection will remain the same.
Therefore, the overall rate of heat transfer from the wire to the ambient will increase due to the addition of the plastic cover. Hence, the correct answer is option 'A' - the heat transfer will increase.
The heat transfer from the wire to the ambient is determined by conduction through the plastic sleeve and convection at the outer surface of the sleeve exposed to air. Let's analyze each of these processes separately.
Conduction through the plastic sleeve:
The rate of heat transfer through conduction can be determined using Fourier's law of heat conduction, which states that the rate of heat transfer is proportional to the temperature gradient and the cross-sectional area.
q = -k * A * (dT/dr)
where,
q = rate of heat transfer
k = thermal conductivity of the plastic
A = cross-sectional area
dT/dr = temperature gradient
Since the wire and the plastic sleeve are in contact, the temperature at the interface is the same. Therefore, the temperature gradient across the plastic sleeve is constant.
As the outer radius of the plastic sleeve is larger than the wire, the cross-sectional area of the plastic sleeve is larger than the wire. Hence, the rate of heat transfer through conduction will increase due to the addition of the plastic cover.
Convection at the outer surface of the sleeve exposed to air:
The rate of heat transfer through convection can be determined using Newton's law of cooling, which states that the rate of heat transfer is proportional to the temperature difference between the surface and the ambient and the heat transfer coefficient.
q = h * A * (Ts - Ta)
where,
q = rate of heat transfer
h = heat transfer coefficient
A = surface area
Ts = surface temperature
Ta = ambient temperature
As the heat transfer coefficient on the outer surface of the sleeve exposed to air is a constant value, the rate of heat transfer through convection will remain the same.
Therefore, the overall rate of heat transfer from the wire to the ambient will increase due to the addition of the plastic cover. Hence, the correct answer is option 'A' - the heat transfer will increase.
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A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer?
Question Description
A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer?.
A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A plastic sleeve of outer radius ro = 1 mm covers a wire (radius r = 0.5 mm) carrying electriccurrent. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on theouter surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover,the heat transfer from the wire to the ambient willa)increaseb)remain the samec)decreased)be zeroCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice GATE tests.
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