/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 A spherical Pyrex glass shell ha... [FREE SOLUTION] | 91Ó°ÊÓ

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A spherical Pyrex glass shell has inside and outside diameters of \(D_{1}=0.1 \mathrm{~m}\) and \(D_{2}=0.2 \mathrm{~m}\), respectively. The inner surface is at \(T_{s, 1}=100^{\circ} \mathrm{C}\) while the outer surface is at \(T_{s, 2}=45^{\circ} \mathrm{C}\). (a) Determine the temperature at the midpoint of the shell thickness, \(T\left(r_{m}=0.075 \mathrm{~m}\right)\). (b) For the same surface temperatures and dimensions as in part (a), show how the midpoint temperature would change if the shell material were aluminum.

Short Answer

Expert verified
The temperature at the midpoint of the shell thickness (\(T\left(r_{m}=0.075 \mathrm{~m}\right)\)) for both Pyrex glass and Aluminum shells is approximately \(100^{\circ} C\).

Step by step solution

01

Determine Pyrex glass thermal conductivity

To solve this problem, we first need the thermal conductivity of Pyrex glass. It can be found in a standard material properties table. The thermal conductivity of Pyrex glass is approximately \(k_g=1.1\,W/(m\cdot K)\).
02

Determine Aluminum thermal conductivity

Similarly, we also need the thermal conductivity of Aluminum for part (b). From the material properties table, the thermal conductivity of Aluminum is approximately \(k_a=237\,W/(m\cdot K)\).
03

Find the radii of the shell

Now we compute the radii of the shell with the given diameters. The inside radius will be \(r_1 = D_1/2 = 0.1~m/2 = 0.05~m\), and the outside radius will be \(r_2 = D_2/2 = 0.2~m/2 = 0.1~m\).
04

Determine the conduction heat rate

We first find the conduction heat rate, using the formula: \[\dot{Q}_{conduction} = k_g \times A \times \frac{T_{s1}-T_{s2}}{R}\] Where: - \(k_g\) is the thermal conductivity of Pyrex glass; - \(A = 4\pi r^2\) is the surface area of the sphere; - \(T_{s1}\) and \(T_{s2}\) are the temperatures of the shell's inner and outer surfaces; and - \(R = r_2 - r_1\) is the thermal resistance. Before solving for \(\dot{Q}_{conduction}\), let's first find the values of A and R. We will use the midpoint radius \(r_m = 0.075~m\): \[A = 4\pi (0.075)^2 = 0.070685\,m^2\] \[R = 0.1 - 0.05 = 0.05~m\] Now we can calculate the conduction heat rate: \[\dot{Q}_{conduction} = 1.1 \times 0.070685 \times \frac{100 - 45}{0.05} = 8.968\,W\]
05

Determine the temperature at the midpoint for Pyrex glass

Using the conduction heat rate, we can determine the temperature at the midpoint of the Pyrex glass shell: \[T_m = T_{s1} - \frac{\dot{Q}_{conduction} \cdot (R - r_1)}{k_g \cdot A}\] \[T_m = 100 - \frac{8.968 \times (0.05 - 0.05)}{1.1 \times 0.070685} \approx T_{s1} = 100^{\circ} C\] Hence, for the Pyrex glass shell, the temperature at the midpoint (\(T\left(r_{m}=0.075 \mathrm{~m}\right)\)) is approximately \(100^{\circ} C\).
06

Determine the temperature at the midpoint for Aluminum

Using the formula provided above, we should change the thermal conductivity value to that of Aluminum to find the midpoint temperature for the shell: \[T_m = T_{s1} - \frac{\dot{Q}_{conduction} \cdot (R - r_1)}{k_a \cdot A}\] \[T_m = 100 - \frac{8.968 \times (0.05 - 0.05)}{237 \times 0.070685} \approx T_{s1} \approx 100^{\circ} C\] For the Aluminum shell, the temperature at the midpoint (\(T\left(r_{m}=0.075 \mathrm{~m}\right)\)) is also approximately \(100^{\circ} C\). In conclusion, for both the Pyrex glass and Aluminum shells with given dimensions and surface temperatures, the temperature at the midpoint of the shell thickness is approximately \(100^{\circ} C\).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property that indicates how well a material can conduct heat. It is usually denoted by the symbol 'k' and is expressed in units of watts per meter-kelvin (W/m·K). Materials with high thermal conductivity, like metals (for example, aluminum), are able to transfer heat quickly, which makes them great conductors. In contrast, materials with low thermal conductivity, such as glass or plastic, are not as efficient at transferring heat and are considered insulators.

The value of thermal conductivity plays a pivotal role in solving heat transfer problems. For instance, if you were to wrap a hot object in a material with high thermal conductivity, the heat would be lost rapidly, whereas wrapping it in an insulative material with a low 'k' value would help retain the heat longer. In the exercise given, the difference between thermal conductivity of Pyrex glass (\(k_g = 1.1\text{ W/m·K}\)) and aluminum (\(k_a = 237\text{ W/m·K}\)) starkly illustrates why metals feel colder to the touch in cold environments and hotter in warm environments compared to glass; they transfer the heat to or from our skin much more efficiently.
Conduction Heat Rate
The conduction heat rate, often expressed as \(\dot{Q}_{conduction}\), quantifies the amount of heat transferring through a material by conduction per unit time. It's calculated using the formula \(\dot{Q}_{conduction} = k \times A \times \frac{T_1 - T_2}{R}\) where 'k' is the thermal conductivity, 'A' is the area through which heat is being transferred, \(T_1\) and \(T_2\) are temperatures at different sides of the material, and 'R' is the thermal resistance or thickness of the material.

Understanding this concept is instrumental when designing systems where temperature control is crucial, such as insulation for buildings or cooling systems for electronic devices. In our example with the spherical shell, we see that despite aluminum's high thermal conductivity, the conduction heat rate calculation also depends on the surface area and the temperature gradient, leading to a situation where the midpoint temperature remains unchanged for both Pyrex glass and aluminum.
Temperature Distribution
Temperature distribution refers to how temperature varies within a material or system. In the context of solid materials, temperature distribution is generally not uniform due to the presence of thermal gradients — differences in temperature that lead to heat flow from the hotter areas to the colder ones. Analyzing temperature distribution is key to many engineering applications, from preventing thermal stresses in construction materials to optimizing the performance of heat exchangers.

In the exercise on the spherical shell, the temperature distribution can be visualized as a gradient from the hot inner surface to the cooler outer surface. The temperature at any point within the material depends on the thermal properties of the material and the boundary conditions (in this case, the surface temperatures). The exercise demonstrates that, interestingly, even when a material with vastly different thermal conductivity is chosen, the midpoint temperature can remain the same if the heat rate and the geometry are not altered. This counterintuitive result highlights the intricacies of how temperature distribution is influenced by a combination of factors, not just thermal conductivity alone.

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Most popular questions from this chapter

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

Two stainless steel plates \(10 \mathrm{~mm}\) thick are subjected to a contact pressure of 1 bar under vacuum conditions for which there is an overall temperature drop of \(100^{\circ} \mathrm{C}\) across the plates. What is the heat flux through the plates? What is the temperature drop across the contact plane?

Determine the parallel plate separation distance \(L\), above which the thermal resistance associated with molecule-surface collisions \(R_{t, m-s}\) is less than \(1 \%\) of the resistance associated with molecule-molecule collisions, \(R_{t, m-m}\) for (i) air between steel plates with \(\alpha_{t}=0.92\) and (ii) helium between clean aluminum plates with \(\alpha_{t}=0.02\). The gases are at atmospheric pressure, and the temperature is \(T=300 \mathrm{~K}\).

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

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