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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

Expert verified
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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steady State Heat Transfer
In steady state heat transfer, the process reaches an equilibrium where the temperature field inside the object does not change with time. This means the amount of heat entering a system is equal to the amount of heat leaving. For the aluminum bars in question, the heat transfer takes place along the length of the bars from the hotter top surface to the cooler bottom surface. No heat is lost through the sides due to the insulation sleeve, which makes the calculation straightforward. This kind of analysis is crucial for systems where maintaining a constant temperature gradient across certain points is needed. Understanding the steady state conditions helps in predicting the behavior of materials and structures under thermal stress, ensuring safety and efficiency in applications like engineering and construction.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. In the context of our aluminum bars, the thermal conductivity value of 176 W/m·K indicates how easily heat is transferred through the metal. A high thermal conductivity means that the material is an excellent conductor of heat, which is why metals like aluminum are often used in heat transfer applications. In the given exercise, knowing the thermal conductivity helps calculate the rate of heat transfer using the formula:
  • \(Q_t = \frac{k \cdot A \cdot \Delta T}{L}\)
Where \(k\) is the thermal conductivity. The concept is essential for designing devices and systems that rely on effective heat dissipation, like heat sinks and radiators.
Thermal Resistance
Thermal resistance is the opposite of thermal conductivity. It measures a material's ability to resist heat flow. In a practical sense, it's like the 'insulating effect' a material provides. In the case of the exercises with aluminum bars, the thermal resistance at the interface affects how much total heat flows through them.
  • Calculated as \( R_{Int} = \frac{a}{p \cdot p_{c}} \)
The effect of pressure (20 atm in the exercise) on the interface resistance also illustrates the impact of contact quality. Higher contact pressure generally decreases thermal resistance, allowing heat to transfer more readily between surfaces.
Temperature Gradient
The temperature gradient is the rate at which temperature changes in space within an object. It's calculated by the difference in temperature across a given distance. For the aluminum bars, the temperature gradient is set by maintaining 150°C at one end and 20°C at the other.
  • Calculated as \( \Delta T = T_{hot} - T_{cold} \)
This difference drives the heat transfer across the bars. The concept is crucial for determining how quickly or slowly heat moves through materials. The steeper the temperature gradient, the faster the heat transfer, due to a larger driving potential. The temperature gradient helps engineers configure materials and shapes to manage thermal conditions efficiently in various practical applications.

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Most popular questions from this chapter

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain.

Consider a stainless steel spoon \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) partially immersed in boiling water at \(200^{\circ} \mathrm{F}\) in a kitchen at \(75^{\circ} \mathrm{F}\). The handle of the spoon has a cross section of \(0.08\) in \(\times\) \(0.5\) in, and extends 7 in in the air from the free surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is \(3 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), determine the temperature difference across the exposed surface of the spoon handle. State your assumptions. Answer: \(124.6^{\circ} \mathrm{F}\)

A 0.6-m-diameter, 1.9-m-long cylindrical tank containing liquefied natural gas (LNG) at \(-160^{\circ} \mathrm{C}\) is placed at the center of a 1.9-m-long \(1.4-\mathrm{m} \times 1.4-\mathrm{m}\) square solid bar made of an insulating material with \(k=0.0002 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the outer surface temperature of the bar is \(12^{\circ} \mathrm{C}\), determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be \(425 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3.475 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively.

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

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