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A thick walled cop per cylinder has an inside radius of 1 cm and an outside radius of 2 cm. T he inner and
outside surfaces are held at 310°C and 290°C respectively. Assume 'k' varie with temperature as k = 371.9
[1 – 9.25 × 10–5(T – 150)] W/mK. Determine the heat lost per unit length
outside surfaces are held at 310°C and 290°C respectively. Assume 'k' varie with temperature as k = 371.9
[1 – 9.25 × 10–5(T – 150)] W/mK. Determine the heat lost per unit length
_______________ kW/m
Correct answer is between '63,70'. Can you explain this answer?
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To calculate the temperature gradient in the thick-walled copper cylinder, we can use the formula for thermal conductivity:
Q = k * A * ΔT / L
Where:
- Q is the heat flow
- k is the thermal conductivity of copper (assumed to be constant)
- A is the cross-sectional area of the cylinder
- ΔT is the temperature difference across the cylinder
- L is the thickness of the cylinder
In this case, we are interested in the temperature gradient, which is the change in temperature per unit length. Therefore, we can rearrange the formula as follows:
ΔT / L = Q / (k * A)
To calculate the temperature gradient, we need to find the heat flow (Q). The heat flow can be calculated using Fourier's law of heat conduction:
Q = -k * A * dT / dx
Where:
- dT is the temperature difference across an infinitesimally small thickness of the cylinder
- dx is the infinitesimally small thickness of the cylinder
Since the temperature difference across the cylinder is constant (310 K), we can simplify the equation to:
Q = -k * A * ΔT / L
Now, we can substitute the expression for Q into the temperature gradient formula:
ΔT / L = (-k * A * ΔT / L) / (k * A)
Simplifying further:
ΔT / L = -ΔT / L
The negative sign indicates that the temperature gradient is negative, meaning the temperature decreases as we move from the inside to the outside of the cylinder.
Therefore, the temperature gradient in the thick-walled copper cylinder is a decrease of 310 K per unit length.
Q = k * A * ΔT / L
Where:
- Q is the heat flow
- k is the thermal conductivity of copper (assumed to be constant)
- A is the cross-sectional area of the cylinder
- ΔT is the temperature difference across the cylinder
- L is the thickness of the cylinder
In this case, we are interested in the temperature gradient, which is the change in temperature per unit length. Therefore, we can rearrange the formula as follows:
ΔT / L = Q / (k * A)
To calculate the temperature gradient, we need to find the heat flow (Q). The heat flow can be calculated using Fourier's law of heat conduction:
Q = -k * A * dT / dx
Where:
- dT is the temperature difference across an infinitesimally small thickness of the cylinder
- dx is the infinitesimally small thickness of the cylinder
Since the temperature difference across the cylinder is constant (310 K), we can simplify the equation to:
Q = -k * A * ΔT / L
Now, we can substitute the expression for Q into the temperature gradient formula:
ΔT / L = (-k * A * ΔT / L) / (k * A)
Simplifying further:
ΔT / L = -ΔT / L
The negative sign indicates that the temperature gradient is negative, meaning the temperature decreases as we move from the inside to the outside of the cylinder.
Therefore, the temperature gradient in the thick-walled copper cylinder is a decrease of 310 K per unit length.
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A thick walled cop per cylinder has an inside radius of 1 cm and an outside radius of 2 cm. T he inner andoutside surfaces are held at 310°C and 290°C respectively. Assume 'k' varie with temperature as k = 371.9[1 – 9.25 × 10–5(T – 150)] W/mK. Determine the heat lost per unit length_______________ kW/mCorrect answer is between '63,70'. Can you explain this answer?
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A thick walled cop per cylinder has an inside radius of 1 cm and an outside radius of 2 cm. T he inner andoutside surfaces are held at 310°C and 290°C respectively. Assume 'k' varie with temperature as k = 371.9[1 – 9.25 × 10–5(T – 150)] W/mK. Determine the heat lost per unit length_______________ kW/mCorrect answer is between '63,70'. 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 thick walled cop per cylinder has an inside radius of 1 cm and an outside radius of 2 cm. T he inner andoutside surfaces are held at 310°C and 290°C respectively. Assume 'k' varie with temperature as k = 371.9[1 – 9.25 × 10–5(T – 150)] W/mK. Determine the heat lost per unit length_______________ kW/mCorrect answer is between '63,70'. 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 thick walled cop per cylinder has an inside radius of 1 cm and an outside radius of 2 cm. T he inner andoutside surfaces are held at 310°C and 290°C respectively. Assume 'k' varie with temperature as k = 371.9[1 – 9.25 × 10–5(T – 150)] W/mK. Determine the heat lost per unit length_______________ kW/mCorrect answer is between '63,70'. Can you explain this answer?.
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