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Chapter 3: Problem 50

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掳C, a bottom temperature of 20掳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掳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掳C, and the bottom temperature is 20掳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掳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路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掳C at one end and 20掳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

To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield with thickness of \(5 \mathrm{~mm}\) and thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outside ambient temperature is \(-10^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature inside the automobile is \(25^{\circ} \mathrm{C}\). Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting.

A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20 -cm-thick concrete walls and ceiling \((k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The kiln is maintained at \(40^{\circ} \mathrm{C}\) by injecting hot steam into it. The two ends of the kiln, \(4 \mathrm{~m} \times 5 \mathrm{~m}\) in size, are made of a 3 -mm-thick sheet metal covered with 2 -cm-thick Styrofoam \((k=0.033 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The convection heat transfer coefficients on the inner and the outer surfaces of the kiln are \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Disregarding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at \(-4^{\circ} \mathrm{C}\).

Consider a power transistor that dissipates \(0.2 \mathrm{~W}\) of power in an environment at \(30^{\circ} \mathrm{C}\). The transistor is \(0.4 \mathrm{~cm}\) long and has a diameter of \(0.5 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a \(24-h\) period, in kWh; \((b)\) the heat flux on the surface of the transistor, in \(\mathrm{W} / \mathrm{m}^{2}\); and \((c)\) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The roof of a house consists of a 15-cm-thick concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The convection heat transfer coefficients on the inner and outer surfaces of the roof are 5 and \(12 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\), respectively. On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\), while the night sky temperature is \(100 \mathrm{~K}\). The house and the interior surfaces of the wall are maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). The emissivity of both surfaces of the concrete roof is \(0.9\). Considering both radiation and convection heat transfers, determine the rate of heat transfer through the roof, and the inner surface temperature of the roof. If the house is heated by a furnace burning natural gas with an efficiency of 80 percent, and the price of natural gas is \(\$ 1.20 /\) therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-h period.

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a \(10-\mathrm{cm}\)-thick layer of insulation \((k=0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)
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