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Chapter 3: Problem 5

A dormitory at a large university, built 50 years ago, has exterior walls constructed of \(L_{s}=25\)-mm-thick sheathing with a thermal conductivity of \(k_{s}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). To reduce heat losses in the winter, the university decides to encapsulate the entire dormitory by applying an \(L_{i}=25\)-mm-thick layer of extruded insulation characterized by \(k_{i}=0.029 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to the exterior of the original sheathing. The extruded insulation is, in turn, covered with an \(L_{g}=5\)-mm-thick architectural glass with \(k_{g}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the heat flux through the original and retrofitted walls when the interior and exterior air temperatures are \(T_{\infty 0, i}=22^{\circ} \mathrm{C}\) and \(T_{\infty, a}=\) \(-20^{\circ} \mathrm{C}\), respectively. The inner and outer convection heat transfer coefficients are \(h_{i}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=\) \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively.

Short Answer

Expert verified
The heat flux through the original wall is found to be \(q_s = \frac{\Delta T}{R_s} = \frac{42 \, \text{K}}{\frac{0.025 \, \text{m}}{0.1 \, \text{W/m路K}}} = 168 \, \text{W/m}^2\). For the retrofitted wall, the total thermal resistance is \(R_{total} = R_s + R_i + R_g = \frac{0.025 \, \text{m}}{0.1 \, \text{W/m路K}} + \frac{0.025 \, \text{m}}{0.029 \, \text{W/m路K}} + \frac{0.005 \, \text{m}}{1.4 \, \text{W/m路K}}\), yielding a heat flux of \(q_r = \frac{\Delta T}{R_{total}} = \frac{42 \, \text{K}}{R_{total}} \approx 102.67 \, \text{W/m}^2\). The heat flux is reduced from 168 W/m虏 in the original wall to 102.67 W/m虏 in the retrofitted wall.

Step by step solution

01

Find the thermal resistances of each layer

For each layer, we can find the thermal resistance \(R\) using the formula: \[R = \frac{L}{kA}\] where \(L\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the area. For the original wall (sheathing), we have the following: \[R_s = \frac{L_s}{k_sA}\] For the retrofitted wall (sheathing + insulation + glass), we have the following resistances: \[R_i = \frac{L_i}{k_iA}\] \[R_g = \frac{L_g}{k_gA}\]
02

Find the total thermal resistance for original and retrofitted walls

For the original wall (sheathing), the total resistance is simply \(R_s\). For the retrofitted wall, the total resistance is the sum of all layers' resistances: \[R_{total} = R_s + R_i + R_g\]
03

Find the heat flux through the original wall

To find the heat flux, first, let's find the temperature difference between the inner and outer surfaces: \[\Delta T = T_{\infty 0, i} - T_{\infty, a}\] For the original wall, the heat flux, \(q_s\), can be expressed as: \[q_s = h_iA(T_{\infty 0, i} - T_s) = h_oA(T_s - T_{\infty, a})\] We can substitute the temperature difference \(\Delta T\) using the total resistance of the wall: \[q_s = \frac{\Delta T}{R_s}\]
04

Find the heat flux through the retrofitted wall

Similarly, for the retrofitted wall, the heat flux, \(q_r\), can be expressed as: \[q_r = h_iA(T_{\infty 0, i} - T_r) = h_oA(T_r - T_{\infty, a})\] Substituting the temperature difference \(\Delta T\) using the total resistance of the retrofitted wall: \[q_r = \frac{\Delta T}{R_{total}}\]
05

Calculate the final results

Now we have all the formulas necessary to find the heat flux through the original and retrofitted walls. After plugging in the values, we will get the final results for \(q_s\) and \(q_r\).

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Key Concepts

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

Thermal Resistance
Thermal resistance is a measure of a material鈥檚 ability to resist the flow of heat. It is often likened to electrical resistance, but instead of impeding electricity, it slows down the transfer of thermal energy. This concept is crucial in understanding how heat moves through different materials, such as walls in a building.

For any layer in a construction arrangement, such as a wall, the thermal resistance, symbolized as R, can be calculated using the equation: \[R = \frac{L}{kA}\], where L is the material's thickness, k is the thermal conductivity, and A represents the area through which heat is being transferred. The thermal resistance is directly proportional to the thickness of the material and inversely proportional to its thermal conductivity.

In practical terms, adding insulation with high thermal resistance to a wall can significantly reduce heat flow, thus improving the building's energy efficiency. The example given in the exercise shows a retrofit with additional layers, each with its thermal resistance. By summing these resistances, we can determine the overall effectiveness of the wall against heat transfer.
Thermal Conductivity
Thermal conductivity, denoted by k, is an intrinsic property of materials that describes how well they can conduct heat. It is defined as the amount of heat, in watts, passing through a material one meter thick, over an area of one square meter, for every degree of temperature difference across the material.

Materials with high thermal conductivity, such as metals, are excellent heat conductors and have a lower thermal resistance, meaning they allow more heat to pass through. Conversely, materials like the extruded insulation mentioned in the exercise possess low thermal conductivity, making them good insulators due to higher thermal resistance.

The understanding of thermal conductivity is essential when selecting materials for insulation purposes. In the exercise, you observed how the addition of an insulation layer with lower conductivity reduced the overall heat flux through the wall. This demonstrates the effectiveness of using materials with appropriate thermal conductivity values to enhance energy conservation in buildings.
Heat Transfer Coefficients
Heat transfer coefficients quantify the rate of heat transfer between a surface and a fluid (liquid or gas) adjacent to it. They are denoted by h and typically measured in watts per square meter per degree Kelvin (W/m虏路K). These coefficients play a pivotal role in calculating the heat transfer in situations where convection is involved.

For example, the exercise discusses an interior convection heat transfer coefficient, hi, and an outer convection heat transfer coefficient, ho. Higher values of h indicate more efficient heat transfer from the surface to the fluid or vice versa. This property is crucial when considering the entire heat transfer system, including not just the solid layers of a wall but also the air layers in contact with it.

The balance of heat transfer across each boundary is maintained, as seen in the exercise, by setting up an equation for the heat transfer through the wall, balanced by the convection on each side. This approach helps provide a comprehensive view of heat flux in real-world scenarios, accounting for both conduction through materials and convection at surfaces.

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Most popular questions from this chapter

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

Consider two long, slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocouple, it was observed that the temperatures of the rods were equal at the positions \(x_{\mathrm{A}}=0.15 \mathrm{~m}\) and \(x_{\mathrm{B}}=0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(\mathrm{A}\) is known to be \(k_{\mathrm{A}}=70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{B}}\) for rod B.

A bonding operation utilizes a laser to provide a constant heat flux, \(q_{o}^{\prime \prime}\), across the top surface of a thin adhesivebacked, plastic film to be affixed to a metal strip as shown in the sketch. The metal strip has a thickness \(d=1.25 \mathrm{~mm}\), and its width is large relative to that of the film. The thermophysical properties of the strip are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=435 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal resistance of the plastic film of width \(w_{1}=40 \mathrm{~mm}\) is negligible. The upper and lower surfaces of the strip (including the plastic film) experience convection with air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The strip and film are very long in the direction normal to the page. Assume the edges of the metal strip are at the air temperature \(\left(T_{\infty}\right)\). (a) Derive an expression for the temperature distribution in the portion of the steel strip with the plastic film \(\left(-w_{1} / 2 \leq x \leq+w_{1} / 2\right)\). (b) If the heat flux provided by the laser is 10,000 \(\mathrm{W} / \mathrm{m}^{2}\), determine the temperature of the plastic film at the center \((x=0)\) and its edges \(\left(x=\pm w_{1} / 2\right)\). (c) Plot the temperature distribution for the entire strip and point out its special features.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

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