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The outer surface of a long cylinder is maintained at constant temperature. The cylinder does not have any heat source.
The temperature in the cylinder will:
The temperature in the cylinder will:
- a)Increase linearly with radius
- b)Decrease linearly with radius
- c)Be independent of radius
- d)Vary logarithmically with radius
Correct answer is option 'D'. Can you explain this answer?
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The outer surface of a long cylinder is maintained at constant tempera...
Ans. (d)
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Explanation:
The temperature in the cylinder will vary logarithmically with radius. This is because the heat transfer equation for a long cylinder with constant surface temperature is given by:
q = 2πkL(Ts - T)/ln(r2/r1)
where,
q = heat transfer rate
k = thermal conductivity
L = length of cylinder
Ts = surface temperature
T = temperature at any point inside cylinder
r1 and r2 = inner and outer radii of cylinder
From this equation, it can be observed that the temperature inside the cylinder varies logarithmically with radius. This means that the temperature gradient is higher near the surface of the cylinder and decreases as we move towards the center.
Reason for logarithmic variation:
This logarithmic variation occurs due to the nature of heat transfer in a long cylinder. The heat transfer occurs primarily through conduction, and as we move away from the surface of the cylinder, the area available for heat transfer decreases. This results in a decrease in heat transfer rate and hence a decrease in temperature gradient.
Additionally, the logarithmic variation also occurs due to the geometry of the cylinder. The area of each annular section of the cylinder is proportional to its radius, and hence the heat transfer rate is also proportional to the radius. This results in a logarithmic variation in temperature.
Conclusion:
In conclusion, the temperature inside a long cylinder with constant surface temperature varies logarithmically with radius. This variation occurs due to the nature of heat transfer through conduction and the geometry of the cylinder.
The temperature in the cylinder will vary logarithmically with radius. This is because the heat transfer equation for a long cylinder with constant surface temperature is given by:
q = 2πkL(Ts - T)/ln(r2/r1)
where,
q = heat transfer rate
k = thermal conductivity
L = length of cylinder
Ts = surface temperature
T = temperature at any point inside cylinder
r1 and r2 = inner and outer radii of cylinder
From this equation, it can be observed that the temperature inside the cylinder varies logarithmically with radius. This means that the temperature gradient is higher near the surface of the cylinder and decreases as we move towards the center.
Reason for logarithmic variation:
This logarithmic variation occurs due to the nature of heat transfer in a long cylinder. The heat transfer occurs primarily through conduction, and as we move away from the surface of the cylinder, the area available for heat transfer decreases. This results in a decrease in heat transfer rate and hence a decrease in temperature gradient.
Additionally, the logarithmic variation also occurs due to the geometry of the cylinder. The area of each annular section of the cylinder is proportional to its radius, and hence the heat transfer rate is also proportional to the radius. This results in a logarithmic variation in temperature.
Conclusion:
In conclusion, the temperature inside a long cylinder with constant surface temperature varies logarithmically with radius. This variation occurs due to the nature of heat transfer through conduction and the geometry of the cylinder.
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The outer surface of a long cylinder is maintained at constant temperature. The cylinder does not have any heat source.The temperature in the cylinder will:a)Increase linearly with radiusb)Decrease linearly with radiusc)Be independent of radiusd)Vary logarithmically with radiusCorrect answer is option 'D'. Can you explain this answer?
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