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Consider steady conduction across a cylindrical or spherical shell. It is sometimes convenient to express the heat flow as $$ \dot{Q}=\frac{k A_{\mathrm{eff}}\left(T_{1}-T_{2}\right)}{r_{2}-r_{1}} $$ that is, in the same form as for a plane slab where \(A_{\text {eff }}\) is an effective crosssectional area. (i) Determine \(A_{\text {eff }}\) for a cylindrical shell. (ii) Repeat for a spherical shell. (iii) If the arithmetic mean area \(A_{m}=\frac{1}{2}\left(A_{1}+A_{2}\right)\) is used instead of \(A_{\text {eff }}\), determine the error in heat flow for \(r_{2} / r_{1}=1.5,3\), and 5 .

Short Answer

Expert verified
(i) For a cylinder: \( A_{\mathrm{eff}} = \frac{2 \pi L (r_2 - r_1)}{\ln(\frac{r_2}{r_1})} \). (ii) For a sphere: \( A_{\mathrm{eff}} = \frac{4 \pi (r_2 - r_1)^2}{\frac{1}{r_1} - \frac{1}{r_2}} \). (iii) Errors for \( r_2/r_1 = 1.5, 3, 5 \) depend on the discrepancy between these effective areas and \( A_m \).

Step by step solution

01

Understand the Geometry of a Cylindrical Shell

A cylindrical shell has radii \( r_1 \) and \( r_2 \) (inner and outer radii). The effective area \( A_{\mathrm{eff}} \) for a cylindrical shell is usually obtained by integrating over the cylinder's lateral surface between these radii.
02

Determine the Effective Area for a Cylindrical Shell

For the cylindrical shell, the heat flow is best expressed using logarithmic mean area:\[A_{\mathrm{eff}} = \frac{2 \pi L (r_2 - r_1)}{\ln{\left( \frac{r_2}{r_1} \right)}}\]where \( L \) is the length of the cylinder. This accounts for the radial nature of heat conduction.
03

Understand the Geometry of a Spherical Shell

In a spherical shell, the radii still denote inner \( r_1 \) and outer \( r_2 \) boundaries, but, unlike the cylinder, its surface area increases more rapidly due to the square of the radius term.
04

Determine the Effective Area for a Spherical Shell

For a spherical shell, the effective area \( A_{\mathrm{eff}} \) is calculated using:\[A_{\mathrm{eff}} = \frac{4 \pi (r_2 - r_1)^2}{\frac{1}{r_1} - \frac{1}{r_2}}\]This result arises from solving the integration over a spherical surface.
05

Calculate Error Using Arithmetic Mean Area for Cylindrical Shell

Given the arithmetic mean area \( A_m = \frac{1}{2} (A_1 + A_2) \), calculate the percentage error using:\[\text{Percentage Error} = \left|\frac{A_{\mathrm{eff}} - A_m}{A_{\mathrm{eff}}} \right| \times 100\]Compute this for \( r_2 / r_1 = 1.5, 3, \) and \( 5 \) using the effective area formula derived in Step 2.
06

Calculate Error Using Arithmetic Mean Area for Spherical Shell

Using the same arithmetic mean area \( A_m \), calculate the error for a spherical shell using the formula:\[\text{Percentage Error} = \left|\frac{A_{\mathrm{eff}} - A_m}{A_{\mathrm{eff}}} \right| \times 100\]Apply this to \( r_2 / r_1 = 1.5, 3, \) and \( 5 \) with the effective area from Step 4.

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Key Concepts

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

Cylindrical Shell
A cylindrical shell can be thought of as a hollow cylinder with two radii, namely, the inner radius \( r_1 \) and the outer radius \( r_2 \). Understanding the concept of a cylindrical shell in heat conduction involves considering how heat flows radially through this shell.

In heat conduction, the geometric shape of the material affects how heat transfers through it. For cylindrical shells, heat flows outward from the inner radius toward the outer radius. This means that the lateral surface area, which is the curved surface between the two radii, plays a crucial role.

The determination of the effective area \( A_{\mathrm{eff}} \) for heat flow in cylindrical systems not only helps in understanding heat transfer in pipes but is essential in various engineering applications like insulation of cables and fluid containers.
Spherical Shell
A spherical shell differs from a cylindrical shell in that it is shaped like a hollow sphere, encompassing an inner and outer boundary defined by radii \( r_1 \) and \( r_2 \) respectively.

When considering heat conduction across a spherical shell, the heat transfer mimics the growing nature of a sphere where the surface area increases with the square of the radius. This increase in surface area influences the rate at which heat is conducted across the shell.

The calculated effective area for a spherical shell takes into account this geometric complexity, leading to distinct expressions that address the spherical nature of the geometry. Recognizing this geometric effect is vital in applications such as thermal shields for reactors and large-scale storage tanks.

By understanding spherical shells, engineers and scientists can better design materials and structures that involve thermal processes in round geometries.
Effective Area
The concept of effective area \( A_{\mathrm{eff}} \) is crucial when analyzing heat conduction through shells.

The effective area is an equivalent area that simplifies the effects of geometry in heat conduction calculations. In simple terms, \( A_{\mathrm{eff}} \) allows us to treat cylindrical or spherical surfaces like flat planes, making the math way simpler.

Heat transfer equations usually become more complicated when shells are involved, but the effective area helps in expressing them in much simpler forms. For a cylindrical shell, we derive this through the expression \( A_{\mathrm{eff}} = \frac{2 \pi L (r_2 - r_1)}{\ln{\left( \frac{r_2}{r_1} \right)}} \). For spherical shells, it is given as \( A_{\mathrm{eff}} = \frac{4 \pi (r_2 - r_1)^2}{\frac{1}{r_1} - \frac{1}{r_2}} \).

This concept allows for more straightforward calculations, aiding engineers and designers to efficiently determine heat flow across complex geometries.
Logarithmic Mean Area
While effective area simplifies the conduction calculations, one crucial mathematical concept intertwined with it is the logarithmic mean area.

The logarithmic mean area provides a means to incorporate the variable surface area of cylindrical and spherical shells effectively, which isn't uniform like a flat plane.

This logarithmic nature ensures greater accuracy in computations and reflects how surface changes affect heat flow.

In cylindrical shells, the effective area involves a logarithmic mean stemming from the radii ratio, represented as \( \ln{\left( \frac{r_2}{r_1} \right)} \).

The importance of using logarithmic mean areas is emphasized when comparing with simple arithmetic mean areas, where inaccuracies are noticeably high especially when the ratio of \( r_2/r_1 \) increases. As we apply the equation for heat conduction in different geometries, the logarithmic mean deftly balances the variable nature of curved areas, ensuring that engineers achieve precise and reliable results.

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Most popular questions from this chapter

The convective heat transfer coefficient around a cylinder held perpendicular to a flow varies in a complicated manner. A test cylinder to investigate this behavior consists of a \(0.001\) in-thick, \(12.7 \mathrm{~mm}\)-wide stainless steel heater ribbon (cut from shim stock) wound around a \(2 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness Teflon tube. A single thermocouple is located just underneath the ribbon and measures the local ribbon temperature \(T_{s}(\theta)\). The cylinder is installed in a wind tunnel, and a second thermocouple is used to measure the ambient air temperature \(T_{e}\). The power input to the heater is metered, from which the electrical heat generation per unit area \(\dot{Q} / A\) can be calculated ( \(A\) is the surface area of one side of the ribbon). As a first approximation, the local heat transfer coefficient \(h_{c}(\theta)\) can be obtained from $$ h_{c}(\theta)=\frac{\dot{Q} / A}{T_{s}(\theta)-T_{e}} $$ Hence, by rotating the cylinder with the power held constant, the variation of \(h_{c}\) can be obtained from the variation of \(T_{s}\) : where \(T_{s}(\theta)\) is low, \(h_{c}(\theta)\) is high, and vice versa. A typical variation of \(T_{s}(\theta)\) is shown in the graph. A problem with this technique is that conduction around the circumference of the tube causes the local heat flux \(q_{s}(\theta)\) to not exactly equal \(\dot{Q} / A\). (i) Derive a formula for \(h_{c}(\theta)\) that approximately accounts for circumferential conduction. (ii) The following table gives values of \(T_{s}(\theta)\) in a sector where circumferential conduction effects are expected to be large. Use these values together with \(\dot{Q} / A=5900 \mathrm{~W} / \mathrm{m}^{2}\) and \(T_{e}=25^{\circ} \mathrm{C}\) to estimate the conduction effect at \(\theta=110^{\circ}\). \begin{tabular}{cc} Angle (degrees) & \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline 100 & \(65.9\) \\ 110 & \(65.7\) \\ 120 & \(64.4\) \end{tabular} (iii) Comment on the design of the cylinder. Would a \(3 \mathrm{~mm}\)-thick brass tube, directly heated by an electric current, be a suitable alternative? Use \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel and \(0.38 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for Teflon.

A thin-wall, spherical stainless steel vessel has an outside diameter of \(40 \mathrm{~cm}\) and contains biochemical reactants maintained at \(160^{\circ} \mathrm{C}\). The tank is located in a laboratory where the air is maintained at \(20^{\circ} \mathrm{C}\). A layer of insulation is to be added to prevent workers from being burned by accidental contact with the vessel. The insulation chosen has a thermal conductivity of \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). How thick should the layer be if the threshold for a skin burn on a nonmetallic surface can be taken as \(55^{\circ} \mathrm{C}\) ? Also calculate the heat loss. The combined convective and radiative heat transfer coefficients on the outside of the insulation is estimated to be \(9 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A space radiator is made of \(0.3 \mathrm{~mm}\)-thick aluminum plate with heatpipes at a pitch of \(8 \mathrm{~cm}\). The heatpipes reject heat at \(330 \mathrm{~K}\). The back of the radiator is insulated, and the front sees outer space at \(0 \mathrm{~K}\). If the aluminum surface is hard-anodized to give an emittance of \(0.8\), determine the fin effectiveness of the radiator and the rate of heat rejection per unit area. Take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum. (Hint: We do not expect the plate temperature to vary more than a few kelvins: assume \(q_{\mathrm{rad}}\) is constant at an average value to obtain an approximate analytical solution.)

Aluminum alloy straight rectangular fins for cooling a semiconductor device are 1 \(\mathrm{cm}\) long and \(1 \mathrm{~mm}\) thick. Investigate the effect of choice of tip boundary condition on heat loss as a function of convective heat transfer coefficient. Use \(k=175 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\) for the alloy and a range of \(h_{c}\) values from 10 to \(200 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A steel heat exchanger tube of \(2 \mathrm{~cm}\) outer diameter is fitted with a steel spiral annular fin of \(4 \mathrm{~cm}\) outer diameter, thickness \(0.4 \mathrm{~mm}\), and wound at a pitch of 3 \(\mathrm{mm}\). If the outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the original bare tube and \(15 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the finned tube, determine the reduction in the outside thermal resistance achieved by adding the fins. Take \(k_{\text {steel }}=42 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

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