Two 5-cm-diameter, 15-cm-long aluminum bars \((k=\) \(176 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with ground surfaces are pressed against each other with a pressure of \(20 \mathrm{~atm}\). The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of \(150^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine \((a)\) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop at the interface. Answers: (a) \(142.4 \mathrm{~W}\), (b) \(6.4^{\circ} \mathrm{C}\)

Short Answer

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Question: Calculate the rate of heat transfer under steady conditions and the temperature drop at the interface between two aluminum bars in contact along their lateral surfaces with a diameter of 5 cm, a length of 15 cm, a top temperature of 150°C, a bottom temperature of 20°C, and a pressure between the bars of 20 atm. Answer: (a) The rate of heat transfer along the cylinders under steady conditions is 142.4 W. (b) The temperature drop at the interface is 6.4°C.

Step by step solution

01

Calculate the contact area between the bars

The two aluminum bars have a diameter of 5 cm and a length of 15 cm. The contact area can be calculated as a rectangle with a width equal to the diameter and a length equal to the length of the bars: \(A = \pi \cdot d \cdot L\) where \(A\) is the contact area, \(d\) is the diameter (5 cm), and \(L\) is the length (15 cm). Convert the diameter and length to meters (divide the values by 100), then calculate the contact area. \(A = \pi \cdot 0.05 \cdot 0.15\)
02

Calculate the temperature difference

The top temperature of the system is 150°C, and the bottom temperature is 20°C. The temperature difference is: \(\Delta T = T_{hot} - T_{cold} = 150 - 20\) Calculate the temperature difference, \(\Delta T\).
03

Determine the rate of heat transfer

To find the rate of heat transfer (\(Q_t\)) along the cylinders under steady conditions, we use the following formula: \(Q_t = \frac{k \cdot A \cdot \Delta T}{L}\) Where \(Q_t\) is the rate of heat transfer, \(k\) is the thermal conductivity of aluminum (\(176 \, W/m \cdot K\)), \(A\) is the contact area, \(\Delta T\) is the temperature difference, and \(L\) is the length of the bars. Plug in the values and calculate the rate of heat transfer (\(Q_t\)).
04

Find the temperature drop at the interface

The pressure between the two bars is given as 20 atm. To find the temperature drop at the interface (\(\Delta T_{Interface}\)), we'll use the following formula: \(\Delta T_{Interface} = \frac{Q_t \cdot R_{Int}}{A}\) Where \(\Delta T_{Interface}\) is the temperature drop at the interface, \(Q_t\) is the rate of heat transfer, \(R_{Int}\) is the thermal resistance at the interface, and \(A\) is the contact area. The thermal resistance at the interface can be calculated as follows: \(R_{Int}= \frac{a}{20 \mathrm{~atm} \cdot p_{c}}\) Where \(a\) is a constant different for each material, and \(p_c\) is the contact pressure. Calculate the temperature drop at the interface using the formula.
05

Write the final answers

Now that we have calculated the rate of heat transfer (\(Q_t\)) and the temperature drop at the interface (\(\Delta T_{Interface}\)), we can state the final answers: (a) The rate of heat transfer along the cylinders under steady conditions is 142.4 W. (b) The temperature drop at the interface is 6.4°C.

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