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A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will
[2015]
- a)increase heat loss continuously
- b)decrease heat loss continuously
- c)increase heat loss to a maximum and then decrease heat loss
- d)decrease heat loss to a minimum and then increase heat loss
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A 10 mm diameter electrical conductor is covered by an insulation of 2...
rc = 8mm
∴ the heat lost increases to maximum and then decreases.
∴ the heat lost increases to maximum and then decreases.
Most Upvoted Answer
A 10 mm diameter electrical conductor is covered by an insulation of 2...
Explanation:
1. Heat Loss Calculation:
To understand why the heat loss will initially increase and then decrease, we need to calculate the heat loss through the conductor before and after the addition of the insulation.
Heat loss through the conductor:
The heat loss through the conductor can be calculated using the formula:
Q = (k * A * ΔT) / L
Where:
Q = Heat loss (W)
k = Thermal conductivity of the conductor (W/mK)
A = Cross-sectional area of the conductor (m^2)
ΔT = Temperature difference between the inner and outer surfaces of the conductor (K)
L = Length of the conductor (m)
Assuming the temperature difference ΔT remains constant, the heat loss through the conductor can be simplified as:
Q = (k * A) / L
2. Heat Transfer Calculation:
Heat transfer through the insulation:
The heat transfer through the insulation can be calculated using the formula:
Q = (k * A * ΔT) / L
Where:
Q = Heat transfer (W)
k = Thermal conductivity of the insulation (W/mK)
A = Surface area of the insulation (m^2)
ΔT = Temperature difference between the inner and outer surfaces of the insulation (K)
L = Thickness of the insulation (m)
Assuming the temperature difference ΔT remains constant, the heat transfer through the insulation can be simplified as:
Q = (k * A) / L
3. Analysis:
Initially, the heat loss through the conductor is given by Q = (k * A) / L. When the insulation is added, the heat transfer through the insulation can be calculated using Q = (k * A) / L. Since the insulation thickness is 2 mm, the new thickness of the combined insulation is 4 mm.
When additional insulation of the same material is added, the total thickness of the insulation increases. As a result, the heat transfer through the insulation decreases because the insulation thickness in the formula is in the denominator.
However, the heat loss through the conductor remains the same since the conductor's cross-sectional area and length do not change.
Initially, the heat loss is predominantly through the conductor, and the added insulation reduces the heat loss through the insulation. But as more insulation is added, the heat loss through the conductor becomes relatively smaller compared to the heat loss through the insulation. Therefore, the heat loss initially increases as the insulation thickness increases.
Eventually, there will be a point where the heat loss through the insulation becomes significant enough that further adding insulation will not significantly affect the overall heat loss. At this point, the heat loss reaches a maximum and starts decreasing as the additional insulation further reduces the heat transfer through the insulation.
4. Conclusion:
Hence, the correct answer is option C: the heat loss will increase to a maximum and then decrease as further insulation of the same material is added.
1. Heat Loss Calculation:
To understand why the heat loss will initially increase and then decrease, we need to calculate the heat loss through the conductor before and after the addition of the insulation.
Heat loss through the conductor:
The heat loss through the conductor can be calculated using the formula:
Q = (k * A * ΔT) / L
Where:
Q = Heat loss (W)
k = Thermal conductivity of the conductor (W/mK)
A = Cross-sectional area of the conductor (m^2)
ΔT = Temperature difference between the inner and outer surfaces of the conductor (K)
L = Length of the conductor (m)
Assuming the temperature difference ΔT remains constant, the heat loss through the conductor can be simplified as:
Q = (k * A) / L
2. Heat Transfer Calculation:
Heat transfer through the insulation:
The heat transfer through the insulation can be calculated using the formula:
Q = (k * A * ΔT) / L
Where:
Q = Heat transfer (W)
k = Thermal conductivity of the insulation (W/mK)
A = Surface area of the insulation (m^2)
ΔT = Temperature difference between the inner and outer surfaces of the insulation (K)
L = Thickness of the insulation (m)
Assuming the temperature difference ΔT remains constant, the heat transfer through the insulation can be simplified as:
Q = (k * A) / L
3. Analysis:
Initially, the heat loss through the conductor is given by Q = (k * A) / L. When the insulation is added, the heat transfer through the insulation can be calculated using Q = (k * A) / L. Since the insulation thickness is 2 mm, the new thickness of the combined insulation is 4 mm.
When additional insulation of the same material is added, the total thickness of the insulation increases. As a result, the heat transfer through the insulation decreases because the insulation thickness in the formula is in the denominator.
However, the heat loss through the conductor remains the same since the conductor's cross-sectional area and length do not change.
Initially, the heat loss is predominantly through the conductor, and the added insulation reduces the heat loss through the insulation. But as more insulation is added, the heat loss through the conductor becomes relatively smaller compared to the heat loss through the insulation. Therefore, the heat loss initially increases as the insulation thickness increases.
Eventually, there will be a point where the heat loss through the insulation becomes significant enough that further adding insulation will not significantly affect the overall heat loss. At this point, the heat loss reaches a maximum and starts decreasing as the additional insulation further reduces the heat transfer through the insulation.
4. Conclusion:
Hence, the correct answer is option C: the heat loss will increase to a maximum and then decrease as further insulation of the same material is added.
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Community Answer
A 10 mm diameter electrical conductor is covered by an insulation of 2...
rc = 8mm
∴ the heat lost increases to maximum and then decreases.
∴ the heat lost increases to maximum and then decreases.
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A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer?
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A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer?.
A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer?.
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A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness.The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m2K. Addition of further insulation of the same material will[2015]a)increase heat loss continuouslyb)decrease heat loss continuouslyc)increase heat loss to a maximum and then decrease heat lossd)decrease heat loss to a minimum and then increase heat lossCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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