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Using the thermal resistance relations developed in Chapter 3, determine shape factor expressions for the following geometries: (a) Plane wall, cylindrical shell, and spherical shell. (b) Isothermal sphere of diameter \(D\) buried in an infinite medium.

Short Answer

Expert verified
The shape factor expressions for the given geometries are as follows: (a) Plane Wall: \(S = \frac{kA}{L}\) Cylindrical Shell: \(S = \frac{2 \pi kL}{\log{(r_2/r_1)}}\) Spherical Shell: \(S = \frac{4 \pi k}{(\frac{1}{r_1} - \frac{1}{r_2})}\) (b) Isothermal Sphere Buried in an Infinite Medium: \(S = \frac{4 \pi kD}{2}\)

Step by step solution

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1. Understanding Thermal Resistance and Shape factors

Thermal resistance is a property that describes the difficulty of heat transfer through a material or across an interface between two materials. It is the ratio of the temperature difference between the two sides of the material to the heat transfer rate, and it is usually expressed in units of \(K/W\). There are three main types of thermal resistances: conduction, convection, and radiation. Shape factor is a non-dimensional quantity used to characterize the geometry of the heat flow path. It plays a significant role in calculating the heat transfer between two surfaces separated by a medium, mainly involving radiation and conduction heat transfer methods. The shape factor, also known as the geometric or configuration factor, is used to predict the total heat transfer between two isothermal surfaces. The shape factors can be related to the thermal resistances using its equations.
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2. Shape Factor for Plane Wall

For a plane wall of thickness L and area A, the heat transfer takes place through conduction, and the governing thermal resistance relation is given by: \(R_{cond}= \frac{L}{kA} \) Where \(R_{cond}\) is the conduction thermal resistance, L is the thickness of the wall, k is the thermal conductivity of the wall material, and A is the area of the wall. The shape factor S (dimensionless) for the plane wall is defined as: \(S = \frac{1}{R_{cond}} = \frac{kA}{L} \)
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3. Shape Factor for Cylindrical Shell

For a cylindrical shell with inner radius \(r_1\), outer radius \(r_2\), and length L, the conduction heat transfer occurs radially, and the governing thermal resistance relation is given by: \(R_{cond} = \frac{\log{(r_2/r_1)}}{2 \pi kL} \) The shape factor S for the cylindrical shell is the inverse of this thermal resistance: \( S = \frac{1}{R_{cond}} = \frac{2 \pi kL}{\log{(r_2/r_1)}} \)
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4. Shape Factor for Spherical Shell

For a spherical shell with inner radius \(r_1\) and outer radius \(r_2\), the conduction heat transfer occurs radially. The governing thermal resistance relation for a spherical shell is given by: \(R_{cond} = \frac{1}{4 \pi k} (\frac{1}{r_1} - \frac{1}{r_2}) \) The shape factor S for the spherical shell is the inverse of this thermal resistance: \( S = \frac{1}{R_{cond}} = \frac{4 \pi k}{(\frac{1}{r_1} - \frac{1}{r_2})} \)
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5. Shape Factor for Isothermal Sphere buried in an Infinite Medium

For an isothermal sphere of diameter D buried in an infinite medium, the heat transfer can be modeled as the sum of an infinite number of concentric spherical shells' conductive heat transfers. The shape factor for this case can be determined by taking the limit as the outer radius approaches infinity and using the shape factor formula for a spherical shell: \( S = \lim_{r_2 \to \infty} \frac{4 \pi k}{(\frac{1}{r_1} - \frac{1}{r_2})} \) Since \(r_1 = \frac{D}{2}\), \( S = \lim_{r_2 \to \infty} \frac{4 \pi k}{(\frac{2}{D} - \frac{1}{r_2})} = \frac{4 \pi kD}{2} \) Note that the result doesn't have a \(r_2\) component, which is expected since the medium is infinite, and the shape factor should only depend on the sphere's size and material properties. In conclusion, we have determined the shape factor expressions for different geometries using the thermal resistance relations as requested in the exercise.

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Key Concepts

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

Understanding Thermal Resistance and Its Importance in Heat Transfer
Thermal resistance is a crucial concept when studying heat transfer, particularly through conduction. Imagine you're trying to keep your coffee warm in a thermal mug. The effectiveness of the mug in preventing heat loss from your coffee to the outside air is essentially an example of thermal resistance at work. In technical terms, thermal resistance represents how much a particular material resists the flow of heat. It's often likened to electrical resistance, which measures how much a material resists the flow of electricity. The higher the thermal resistance of a material, the slower the rate of heat transfer through it.

The mathematical expression for thermal resistance in conduction heat transfer is given by the ratio of temperature difference to heat transfer rate, written as \( R_{thermal} = \frac{\Delta T}{Q} \), where \(\Delta T\) is the temperature difference across the material, and \(Q\) is the heat transfer rate. This property plays a pivotal role in designing and analyzing systems that involve heat transfer such as insulation materials, electronic cooling systems, and building construction.

By understanding thermal resistance, one can select appropriate materials and design configurations that optimally control the temperature profile within a particular application. It essentially guides engineers and product designers towards achieving energy efficiency and maintaining desired thermal conditions.
The Role of Shape Factor in Conduction Heat Transfer
The shape factor is a dimensionless quantity that significantly simplifies the process of analyzing heat transfer for complex geometries. It enables us to compare the heat transfer capabilities between different shapes without the need for extensive calculations. By knowing the shape factor, you can understand how the geometry of an object affects its ability to conduct heat.

In a more technical sense, the shape factor relates the geometry of the heat flow path to the area through which heat is being transferred. For example, in the case of a plane wall, the shape factor is calculated by taking the ratio of the thermal conductivity of the material times the area over the thickness. Essentially, the shape factor provides us with insights into how the physical dimensions and characteristics of an object can influence thermal resistance and, consequently, the rate of heat transfer.

Understanding the shape factor is incredibly beneficial when dealing with heat transfer in practical situations, such as in building designs where certain walls or materials may need to have specific properties to maintain an internal temperature. As you can imagine, it's not just about the material's thermal conductivity but also its shape and size, which ultimately determine the overall heat transfer efficiency.
How Geometries Influence Heat Transfer in Different Situations
Different geometries in heat transfer present various challenges and solutions in the field of thermodynamics. The geometry of an object directly affects how heat is distributed throughout it and how it interacts with surrounding environments. A plane wall has different heat transfer characteristics compared to a cylindrical or spherical shell primarily because of the variations in their surface areas and directions of heat flow.

For example, a cylindrical shell may have heat flowing radially, which means that the surface area through which heat is transferred changes along the length of the cylinder. This is different from the plane wall where the area is constant. Similarly, a spherical shell, like that around the Earth, experiences heat transfer that is omnidirectional, posing distinct challenges in calculation and material choices to maintain specific temperature limits.

In practical applications, understanding geometries in heat transfer allows for the design of more efficient systems, such as radiators with fins that increase the surface area for heat dissipation, or the insulation of pipes where the cylindrical geometry must be taken into account. In essence, recognizing how different shapes transfer heat helps in the creation of better, more energy-efficient structures and components, leading to advancements in technologies ranging from aerospace to household appliances.

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Most popular questions from this chapter

Conduction within relatively complex geometries can sometimes be evaluated using the finite-difference methods of this text that are applied to subdomains and patched together. Consider the two-dimensional domain formed by rectangular and cylindrical subdomains patched at the common, dashed control surface. Note that, along the dashed control surface, temperatures in the two subdomains are identical and local conduction heat fluxes to the cylindrical subdomain are identical to local conduction heat fluxes from the rectangular subdomain. Calculate the heat transfer per unit depth into the page, \(q^{\prime}\), using \(\Delta x=\Delta y=\Delta r=10 \mathrm{~mm}\) and \(\Delta \phi=\pi / 8\). The base of the rectangular subdomain is held at \(T_{h}=20^{\circ} \mathrm{C}\), while the vertical surface of the cylindrical subdomain and the surface at outer radius \(r_{o}\) are at \(T_{c}=0^{\circ} \mathrm{C}\). The remaining surfaces are adiabatic, and the thermal conductivity is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A tube of diameter \(50 \mathrm{~mm}\) having a surface temperature of \(85^{\circ} \mathrm{C}\) is embedded in the center plane of a concrete slab \(0.1 \mathrm{~m}\) thick with upper and lower surfaces at \(20^{\circ} \mathrm{C}\). Using the appropriate tabulated relation for this configuration, find the shape factor. Determine the heat transfer rate per unit length of the tube.

An igloo is built in the shape of a hemisphere, with an inner radius of \(1.8 \mathrm{~m}\) and walls of compacted snow that are \(0.5 \mathrm{~m}\) thick. On the inside of the igloo, the surface heat transfer coefficient is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); on the outside, under normal wind conditions, it is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal conductivity of compacted snow is \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature of the ice cap on which the igloo sits is \(-20^{\circ} \mathrm{C}\) and has the same thermal conductivity as the compacted snow. (a) Assuming that the occupants' body heat provides a continuous source of \(320 \mathrm{~W}\) within the igloo, calculate the inside air temperature when the outside air temperature is \(T_{\infty}=-40^{\circ} \mathrm{C}\). Be sure to consider heat losses through the floor of the igloo. (b) Using the thermal circuit of part (a), perform a parameter sensitivity analysis to determine which variables have a significant effect on the inside air temperature. For instance, for very high wind conditions, the outside convection coefficient could double or even triple. Does it make sense to construct the igloo with walls half or twice as thick?

In Chapter 3 we assumed that, whenever fins are attached to a base material, the base temperature is unchanged. What in fact happens is that, if the temperature of the base material exceeds the fluid temperature, attachment of a fin depresses the junction temperature \(T_{j}\) below the original temperature of the base, and heat flow from the base material to the fin is two-dimensional. Consider conditions for which a long aluminum pin fin of diameter \(D=5 \mathrm{~mm}\) is attached to a base material whose temperature far from the junction is maintained at \(T_{b}=100^{\circ} \mathrm{C}\). Fin convection conditions correspond to \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) What are the fin heat rate and junction temperature when the base material is (i) aluminum ( \(k=\) \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and (ii) stainless steel \((k=15\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ? (b) Repeat the foregoing calculations if a thermal contact resistance of \(R_{t, j}^{\prime \prime}=3 \times 10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) is associated with the method of joining the pin fin to the base material. (c) Considering the thermal contact resistance, plot the heat rate as a function of the convection coefficient over the range \(10 \leq h \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for each of the two materials.

For a small heat source attached to a large substrate, the spreading resistance associated with multidimensional conduction in the substrate may be approximated by the expression [Yovanovich, M. M., and V. W. Antonetti, in Adv. Thermal Modeling Elec. Comp. and Systems, Vol. 1, A. Bar-Cohen and A. D. Kraus, Eds., Hemisphere, \(\mathrm{NY}, 79-128,1988]\) \(R_{r(\mathrm{sp})}=\frac{1-1.410 A_{r}+0.344 A_{r}^{3}+0.043 A_{r}^{5}+0.034 A_{r}^{7}}{4 k_{\mathrm{sub}} A_{s, h}^{1 / 2}}\) where \(A_{r}=A_{s, h} / A_{s, \text { sub }}\) is the ratio of the heat source area to the substrate area. Consider application of the expression to an in- line array of square chips of width \(L_{h}=\) \(5 \mathrm{~mm}\) on a side and pitch \(S_{h}=10 \mathrm{~mm}\). The interface between the chips and a large substrate of thermal conductivity \(k_{\text {sub }}=80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=0.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If a convection heat transfer coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is associated with airflow \(\left(T_{\infty}=15^{\circ} \mathrm{C}\right)\) over the chips and substrate, what is the maximum allowable chip power dissipation if the chip temperature is not to exceed \(T_{h}=85^{\circ} \mathrm{C}\) ?

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