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Post

Rate of Heat Transfer in a Cylindrical Object

Created by @nathanedwards

at October 31st 2023, 9:51:04 pm.

AP_Physics_2-Questions

#thermal-conductivity

#conduction-heat-transfer

#cylindrical-object

Question

A solid cylindrical object with a radius of 0.03 m and a height of 0.1 m is made of a material with a thermal conductivity of 100 W/m·K. The object is initially at a temperature of 200°C and is surrounded by air at a temperature of 25°C.

a) Calculate the rate of heat transfer through the cylindrical object.

b) How long does it take for the cylindrical object to cool down to 75°C?

Assume steady-state conditions, neglect heat loss from the top and bottom surfaces of the cylinder, and use the equation for conduction heat transfer through a cylindrical object:

Q = \frac{{2πkL\Delta T}}{{\ln \left( \frac{{r_2}}{{r_1}} \right)}}

where: Q = rate of heat transfer (in watts), k = thermal conductivity of the material (in W/m·K), L = length of the object (in meters), ΔT = temperature difference between the object and surrounding environment (in Kelvin), r1 = inner radius of the object (in meters), r2 = outer radius of the object (in meters).

c) Based on the calculated rate of heat transfer, explain which factor(s) affect the rate of heat transfer the most and why.

Answer

a) To calculate the rate of heat transfer through the cylindrical object, we can use the equation:

Q = \frac{{2πkL\Delta T}}{{\ln \left( \frac{{r_2}}{{r_1}} \right)}}

Given values: k = 100 W/m·K (thermal conductivity), L = 0.1 m (height of the object), ΔT = (200 + 273) K - (25 + 273) K = 448 K (temperature difference), r1 = 0.03 m (inner radius of the object), r2 = r1 (since it is a solid cylinder).

Substituting these values into the equation, we have:

Q = \frac{{2π(100)(0.1)(448)}}{{\ln \left( \frac{{0.03}}{{0.03}} \right)}}

Simplifying, we get:

Q = 44800π \, \text{W}

Therefore, the rate of heat transfer through the cylindrical object is equal to 44800π W.

b) To find the time taken for the cylindrical object to cool down to 75°C, we can determine the time required for a certain amount of heat to transfer.

Since heat transfer is proportional to the temperature difference (ΔT), we can write:

\frac{{Q_1}}{{\Delta T_1}} = \frac{{Q_2}}{{\Delta T_2}}

where Q1 and ΔT1 represent the initial heat transfer rate and temperature difference, and Q2 and ΔT2 represent the final values.

Given values: Q1 = 44800π W (rate of heat transfer), ΔT1 = (200 + 273) K - (25 + 273) K = 448 K (initial temperature difference), ΔT2 = (75 + 273) K - (25 + 273) K = 248 K (final temperature difference).

Rearranging the formula, we find:

Q_2 = Q_1 \times \frac{{\Delta T_2}}{{\Delta T_1}}

Substituting the values, we get:

Q_2 = 44800π \times \frac{{248}}{{448}} \, \text{W}

Finally, we can find the time taken for the cylindrical object to cool down by dividing the change in heat (Q2) by the rate of heat transfer (Q):

\text{Time} = \frac{{Q_2}}{{Q}}

Substituting the values, we have:

\text{Time} = \frac{{44800π \times \frac{{248}}{{448}}}}{{44800π}} \, \text{s}

Simplifying, we get:

\text{Time} = \frac{{248}}{{448}} \, \text{s}

Therefore, it takes approximately 0.5536 seconds for the cylindrical object to cool down to 75°C.

c) The factors that affect the rate of heat transfer the most are the thermal conductivity of the material (k) and the temperature difference (ΔT).

Thermal conductivity (k): A material with higher thermal conductivity will transfer heat more rapidly. In this case, a higher value of k (100 W/m·K) results in a larger rate of heat transfer.
Temperature difference (ΔT): The greater the temperature difference between the object and the surrounding environment, the higher the rate of heat transfer. In this case, the initial temperature difference (ΔT1 = 448 K) leads to a higher rate of heat transfer compared to the final temperature difference (ΔT2 = 248 K).

It is important to note that the length of the object and the cylindrical dimensions of the object (r1 and r2) also influence the rate of heat transfer, but in this particular scenario, they have negligible impact on the overall rate.