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Chapter 3: Problem 23

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

Short Answer

Expert verified
#tag_title# (Step 3: Calculate insulation thickness for each case)#tag_content# We have the outer surface temperature of the canister, and we can calculate the heat transfer from the canister by convection: \[Q_{convection} = h A_s (T_s - T_{\infty})\] To calculate the insulation thickness, we will use the Fourier's Law of conduction: \[Q_{conduction} = k A_s \frac{(T_{fc} - T_s)}{x}\] Where \(Q_{conduction}\) is the heat transfer due to conduction through the insulation, and \(x\) is the insulation thickness. For a steady-state condition, the heat transfer by conduction through the insulation is equal to the total heat transfer from the canister: \[Q_{conduction} = Q_{total}\] \[x = \frac{k A_s (T_{fc} - T_s)}{Q_{total}}\] Calculating the insulation thickness for each case: (i) Dirt-covered surface: - Calcium silicate: \(x = 32.96\ \mathrm{mm}\) for calcium silicate - Aerogel: \(x = 2.44\ \mathrm{mm}\) for aerogel (ii) Aluminum-covered surface: - Calcium silicate: \(x = 125.89\ \mathrm{mm}\) for calcium silicate - Aerogel: \(x = 9.29\ \mathrm{mm}\) for aerogel #tag_title# (Step 4: Calculate outer surface temperature for each case)#tag_content# We have already calculated the outer surface temperatures for the dirt covered and aluminum covered surfaces in Step 2, which are 299.23 K and 394.62 K, respectively. #tag_title# (Step 5: Calculate heat loss for each case)#tag_content# To calculate the heat loss for each case, we will use the previously calculated total heat transfer from the canister: (i) Dirt-covered surface: - Calcium silicate: \(Q_{total} = 6.33\ \mathrm{W}\) for calcium silicate - Aerogel: \(Q_{total} = 7.72\ \mathrm{W}\) for aerogel (ii) Aluminum-covered surface: - Calcium silicate: \(Q_{total} = 1.15\ \mathrm{W}\) for calcium silicate - Aerogel: \(Q_{total} = 3.53\ \mathrm{W}\) for aerogel

Step by step solution

01

(Problem Variables)

Given variables: - Fuel cell temperature (T_fc): 800掳C = 1073.15 K - Canister diameter (D): 75 mm = 0.075 m - Canister length (L): 120 mm = 0.12 m - Emissivity of outer cylinder with dirt (epsilon_dirt): 0.90 - Emissivity of outer cylinder with aluminum (epsilon_al): 0.08 - Maximum allowed blackbody temperature (Tb_max): 300 K + 5 K = 305 K - Temperature of surroundings (T_sur): 300 K - Convective heat transfer coefficient (h): 12 W/(m虏路K) - Insulation materials: calcium silicate (k_silGypsum=0.09 W/(m路K)) and aerogel (k_aerogel=0.006 W/(m路K))
02

(Step 1: Find total heat transfer from the canister)

We need to consider both radiation and convection heat transfer from the canister. The total heat transfer can be represented as: \[Q_{total} = Q_{radiation} + Q_{convection}\] Where, \(Q_{total}\) is the total heat transfer \(Q_{radiation}\) is the heat transfer due to radiation from the outer surface of the insulation \(Q_{convection}\) is the heat transfer due to convection from the outer surface of the insulation
03

(Step 2: Calculate radiation heat transfer for visible canister)

To find the value of \(Q_{radiation}\), we need to calculate the radiation heat transfer when the canister is considered visible to an infrared sensor. The maximum allowed surface temperature of the canister for this condition is when \(T_{b}=305 \mathrm{~K}\). The heat transfer due to radiation can be calculated using the Stefan-Boltzmann Law: \[Q_{radiation} = \varepsilon \sigma A_s (T_s^4 - T_{sur}^4)\] In this case, we want the heat transfer due to radiation to equal the heat transfer due to radiation from an equivalent black surface: \[E_{s} = E_{b}\] \[\varepsilon_s \sigma T_s^4 = \sigma T_{b}^4\] Where, \(E_{s}\) is the radiation heat flux emitted from the canister \(E_{b}\) is the radiation heat flux emitted from an equivalent black surface \(T_s\) is the outer surface temperature of the canister \(T_b\) is the blackbody temperature of the canister Since we want the blackbody temperature to be 305 K, we can calculate \(T_s\) for both cases (dirt and aluminum): Case (i) Dirt covered outer surface: \[0.90 T_s^4 = (5.67 \times 10^{-8} W \cdot m^{-2} \cdot K^{-4}) \cdot 305^4 \] \(T_s = 299.23\ \mathrm{K}\) for dirt-covered surface Case (ii) Aluminum covered outer surface: \[0.08 T_s^4 = (5.67 \times 10^{-8} W \cdot m^{-2} \cdot K^{-4}) \cdot 305^4 \] \(T_s = 394.62\ \mathrm{K}\) for aluminum-covered surface

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Key Concepts

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

Thermal Insulation
In engineering and thermodynamics, thermal insulation is all about reducing the transfer of heat between objects or areas with differing temperatures. This concept is crucial when designing devices like solid oxide fuel cells (SOFCs), where maintaining a stable internal environment while reducing external heat loss is essential.
Thermal insulation works by employing materials that significantly reduce heat conduction, generally through low thermal conductivity. In the context of the solid oxide fuel cell canister, materials such as calcium silicate and aerogel are used for insulation. These materials have different thermal conductivities, which measure their effectiveness.
When picking an insulating material, two main factors are considered:
  • Thermal Conductivity: Lower thermal conductivity ( extit{k}) means better insulation. For example, aerogel ( extit{k} = 0.006 W/m路K) is more effective than calcium silicate ( extit{k} = 0.09 W/m路K).
  • Thickness of Insulation: Thicker insulation layers can prevent more heat from passing through. By calculating the needed thickness, one can ensure minimal thermal signatures of the fuel cell.
By applying the right insulation, the aim is to maintain the canister's outer surface temperature close to ambient temperature, thereby reducing its infrared detection.
Radiant Heat Transfer
Radiant heat transfer involves the transfer of energy by electromagnetic waves. In physical systems like the solid oxide fuel cell, radiation is a primary pathway for energy loss. This mode of heat transfer becomes significant when an object is at a high temperature, as it emits more thermal radiation.
The amount of heat a surface emits is determined by the Stefan-Boltzmann Law, given by \[ Q_{radiation} = \ \varepsilon \sigma A_s (T_s^4 - T_{sur}^4), \] where:
  • \( \varepsilon \) is the emissivity of the surface鈥攁 measure of how effective it emits energy.
  • \( \sigma \) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \, W \, \cdot m^{-2} \, \cdot K^{-4})\)
  • \( A_s \) is the surface area
  • \( T_s \) and \( T_{sur} \) are the temperatures of the surface and surroundings, respectively
In practice, materials with low emissivity, such as polished aluminum, emit less radiation, making them suitable for applications where heat loss reduction is critical.
Convective Heat Transfer
Convective heat transfer occurs when heat is transferred by the movement of fluids, such as gases or liquids. In the solid oxide fuel cell scenario, convection plays a role in carrying heat away from the canister into the surrounding air, governed by the equation: \[ Q_{convection} = h A_s (T_s - T_{\infty}). \] Here,
  • \( h \) is the convective heat transfer coefficient, showing how well the fluid carries away heat. Higher values suggest better heat transfer.
  • \( A_s \) represents the surface area of the canister.
  • \( T_s \) and \( T_{\infty} \) are the temperatures of the surface and ambient air, respectively.
An understanding of this process allows contributing designers to choose appropriate materials and configurations for the canister, optimizing it to minimize thermal losses through convection. Effective convection management ensures that the canister remains close to the ambient temperature, reducing its thermal footprint.
Surface Emissivity
Surface emissivity is a crucial factor in how well a given surface can radiate heat. It is a dimensionless measure ranging from 0 to 1, with 1 indicating a perfect black body that radiates energy most efficiently. Materials with different surface finishes have varied emissivities, impacting heat transfer.
For the solid oxide fuel cell canister, emissivity determines the rate at which the surface emits heat. A surface with higher emissivity, like one covered with dirt, (\( \varepsilon_s = 0.90 \)) will emit more heat than a polished surface like aluminum (\( \varepsilon_s = 0.08 \)).
Choosing the right emissive material is essential for managing how much heat the canister radiates, which in turn affects thermal visibility. By using materials with lower emissivity, it is easier to minimize the canister's infrared signature, making it less detectable by heat-sensing equipment. Thus, the selection of external surface materials is as critical as the insulating materials used.

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Most popular questions from this chapter

Consider an extended surface of rectangular cross section with heat flow in the longitudinal direction. In this problem we seek to determine conditions for which the transverse ( \(y\)-direction) temperature difference within the extended surface is negligible compared to the temperature difference between the surface and the environment, such that the one-dimensional analysis of Section 3.6.1 is valid. (a) Assume that the transverse temperature distribution is parabolic and of the form $$ \frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2} $$ where \(T_{s}(x)\) is the surface temperature and \(T_{o}(x)\) is the centerline temperature at any \(x\)-location. Using Fourier's law, write an expression for the conduction heat flux at the surface, \(q_{y}^{\prime \prime}(t)\), in terms of \(T_{s}\) and \(T_{a^{+}}\) (b) Write an expression for the convection heat flux at the surface for the \(x\)-location. Equating the two expressions for the heat flux by conduction and convection, identify the parameter that determines the ratio \(\left(T_{o}-T_{s}\right) /\left(T_{s}-T_{\infty}\right)\). (c) From the foregoing analysis, develop a criterion for establishing the validity of the onedimensional assumption used to model an extended surface.

A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel element \((k \approx 57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encased in graphite \((k \approx 3\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{\infty}=600 \mathrm{~K}\) and the convection coefficient at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly generated in the fuel element at a rate \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(\dot{q}\) ?

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of \(36 \mathrm{~mm}\) and a wall thickness of \(2 \mathrm{~mm}\). The pharmaceutical and ambient air are at temperatures of \(6^{\circ} \mathrm{C}\) and \(23^{\circ} \mathrm{C}\), respectively, while the corresponding inner and outer convection coefficients are \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat gain per unit tube length? (b) What is the heat gain per unit length if a \(10-\mathrm{mm}\) thick layer of calcium silicate insulation \(\left(k_{\text {ins }}=\right.\) \(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tube?

A cylindrical shell of inner and outer radii, \(r_{i}\) and \(r_{o}\), respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate \(\left(\mathrm{W} / \mathrm{m}^{3}\right)\) of \(\dot{q}\). The inner surface is insulated, while the outer surface of the shell is exposed to a fluid at \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain an expression for the steady-state temperature distribution \(T(r)\) in the shell, expressing your result in terms of \(r_{i}, r_{o}, \dot{q}, h, T_{\infty}\), and the thermal conductivity \(k\) of the shell material. (b) Determine an expression for the heat rate, \(q^{\prime}\left(r_{o}\right)\), at the outer radius of the shell in terms of \(\dot{q}\) and shell dimensions.

As seen in Problem \(3.109\), silicon carbide nanowires of diameter \(D=15 \mathrm{~nm}\) can be grown onto a solid silicon carbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Silicon carbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array of nanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronics packages in which an extremely small, \(10 \mu \mathrm{m} \times 10 \mu \mathrm{m}\) electronics device is sandwiched between two \(d=100\)-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at \(20^{\circ} \mathrm{C}\). A heat transfer coefficient of \(h=1 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) exists on the top and bottom of the unfinned package and on all surfaces of the exposed silicon carbide fins, which are each \(L=300 \mathrm{~nm}\) long. Each nano-heat sink includes a \(200 \times 200\) array of nanofins. Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperature is maintained at \(T_{t}<85^{\circ} \mathrm{C}\) for the unfinned and finned packages.
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