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Chapter 3: Problem 26

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

Short Answer

Expert verified
The heat loss per unit surface area of the composite is \(34474.1\, W/m^2\). The temperature distribution can be sketched by finding the temperature drop across each resistance. The temperature drops are \(\Delta T_{gconv} = 689.5^{\circ}C\), \(\Delta T_{BeO} = 11.4^{\circ}C\), \(\Delta T_{c} = 1723.7^{\circ}C\), \(\Delta T_{AISI} = 45.9^{\circ}C\), and \(\Delta T_{lconv} = 34.5^{\circ}C\). Plot these values on the y-axis and the corresponding layer on the x-axis.

Step by step solution

01

Identify Relevant Parameters

For this exercise, the relevant parameters given are: 1. Temperatures: \(T_g = 2600^{\circ}C\) (Combustion gas) and \(T_l = 100^{\circ}C\) (Liquid coolant) 2. Convection coefficients: \(h_g = 50 W/m^2\cdot K\) (Gas side) and \(h_l = 1000 W/m^2\cdot K\) (Liquid side) 3. Wall thicknesses: \(L_1 = 10 mm\) (Beryllium oxide) and \(L_2 = 20 mm\) (Stainless steel) 4. Contact resistance: \(R_{c} = 0.05 m^2\cdot K/W\) (Between oxide and steel)
02

Calculate Thermal Resistances

Next, we need to calculate the thermal resistance for each section of the system. These include convection resistances at the gas and liquid side, and the conduction resistances for the beryllium oxide, stainless steel, and the contact resistance between them. For convection resistances: \[R_{gconv} = \frac{1}{h_g A} ,\, R_{lconv} = \frac{1}{h_l A}\] For conduction resistances: \[R_{BeO} = \frac{L_1}{k_1 A} ,\, R_{AISI} = \frac{L_2}{k_2 A}\] The material properties of the beryllium oxide and AISI 304 stainless steel are needed to determine the conduction resistances. The corresponding thermal conductivities are \(k_1 = 30 \, W/m\cdot K\) for the beryllium oxide and \(k_2 = 15 \, W/m\cdot K\) for the stainless steel. Calculating the resistances: - \(R_{gconv} = \frac{1}{(50)(1)} = 0.02\, m^2\cdot K/W\) - \(R_{lconv} = \frac{1}{(1000)(1)} = 0.001\, m^2\cdot K/W\) - \(R_{BeO} = \frac{0.01}{(30)(1)} = 0.00033\, m^2\cdot K/W\) - \(R_{AISI} = \frac{0.02}{(15)(1)} = 0.00133\, m^2\cdot K/W\)
03

Calculate Overall Resistance and Heat Transfer Rate

Find the overall resistance by summing all individual resistances: \[R_{total} = R_{gconv} + R_{BeO} + R_{c} + R_{AISI} + R_{lconv}\] \[R_{total} = 0.02 + 0.00033 + 0.05 + 0.00133 + 0.001 = 0.07266\, m^2\cdot K/W\] Now that the total resistance has been determined, we can calculate the heat transfer rate per unit surface area using the heat transfer equation: \[q = \frac{T_g - T_l}{R_{total}}\] \[q = \frac{2600 - 100}{0.07266} = 34474.1\, W/m^2\]
04

Sketch Temperature Distribution

To sketch the temperature distribution graphically, we need to find the temperature drop across each resistance. This can be done using the equation: \[\Delta T_i = q \cdot R_i\] - \(\Delta T_{gconv} = 34474.1 \cdot 0.02 = 689.5^{\circ}C\) - \(\Delta T_{BeO} = 34474.1 \cdot 0.00033 = 11.4^{\circ}C\) - \(\Delta T_{c} = 34474.1 \cdot 0.05 = 1723.7^{\circ}C\) - \(\Delta T_{AISI} = 34474.1 \cdot 0.00133 = 45.9^{\circ}C\) - \(\Delta T_{lconv} = 34474.1 \cdot 0.001 = 34.5^{\circ}C\) Plot the temperature distribution by marking the temperature drop across each section on the y-axis and layer (conduction or convection part) on the x-axis.

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Key Concepts

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

Thermal Resistance Calculation
Understanding thermal resistance is crucial when it comes to heat transfer in composite walls. Think of thermal resistance as an insulator's effectiveness to combat heat flow; the higher the resistance, the slower the heat transfer. The total thermal resistance, often denoted as 'R', is the sum of individual resistances in series.

In practice, for a wall with multiple layers, we calculate resistance for each layer individually. Convection resistance, associated with the fluid motion on either side of the wall, is given by \( R_{conv} = \frac{1}{hA} \), where 'h' is the convection heat transfer coefficient and 'A' is the area through which heat is being transferred. Conduction resistance, on the other hand, is derived from \( R_{cond} = \frac{L}{kA} \), with 'L' being the thickness of the material, and 'k' its thermal conductivity.

When walls are in contact, an additional thermal resistance known as contact resistance may be present. It accounts for the imperfect contact between materials.

To find the total thermal resistance, we simply add up all the resistances: \[ R_{total} = R_{conv1} + R_{cond1} + R_{contact} + R_{cond2} + R_{conv2} \]With the total resistance known, we can then determine the heat loss using the formula \( q = \frac{\Delta T}{R_{total}} \) where \( \Delta T \) is the temperature difference across the composite wall.
Conduction and Convection in Heat Transfer
When heat traverses a solid material, it does so primarily by conduction. The ease with which heat moves through a material is measured by its thermal conductivity, 'k'. Materials like copper have a high 'k', indicating efficient heat transfer. Conversely, insulators like beryllium oxide have a low 'k', reflecting their resistance to heat flow.

Conduction can be mathematically described by Fourier's law, and the rate at which heat transfers through a material depends on its thermal conductivity, the temperature gradient, and the cross-sectional area perpendicular to the heat flow.

Convection is the transfer of heat through a fluid (gas or liquid), which occurs when a fluid moves from a warm location to a cooler one. The convection heat transfer coefficient 'h' quantifies how well a fluid can carry away heat from a surface. High 'h' values, like those for liquids, typically signify efficient heat removal.

Convection can be natural, driven by buoyancy forces that arise from temperature differences, or forced, where a pump or a fan propels the fluid. In calculations, the distinction between conduction and convection is important as it affects the computation of thermal resistances and therefore the overall heat transfer rate.
Temperature Distribution Sketching
A temperature distribution sketch visually represents how temperature changes across different layers of a composite wall. It depicts the temperature drop in each section, essential for understanding the heat transfer process.

To create this sketch, you start by plotting temperature on the y-axis against the wall layers on the x-axis. Mark the initial temperature on the hot side of the wall and begin to subtract the temperature drops, computed by multiplying the heat transfer rate by the corresponding thermal resistance (\( \Delta T_i = q \cdot R_i \) ), from this value at each layer or interface.

For example, the largest temperature drop usually occurs where the thermal resistance is highest鈥攖ypically at a contact resistance or on the side with the lowest convection coefficient. The slope of the line between points in the sketch indicates how quickly temperature changes: a steeper slope means a faster temperature change, linked to lower resistance. The sketch provides a clear visual cue for where the most significant insulation is needed or where heat is most readily lost, aiding in the design of efficient thermal systems.

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Most popular questions from this chapter

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

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.

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a composite wall consisting of an exterior sheet of \(6.35-\mathrm{mm}\)-thick plain carbon steel, an intermediate layer of \(100-\mathrm{mm}\)-thick cork insulation, and an inner sheet of \(6.35\)-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are \(22^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\), respectively?

A nuclear reactor fuel element consists of a solid cylindrical pin of radius \(r_{1}\) and thermal conductivity \(k_{f}\). The fuel pin is in good contact with a cladding material of outer radius \(r_{2}\) and thermal conductivity \(k_{c^{*}}\). Consider steady-state conditions for which uniform heat generation occurs within the fuel at a volumetric rate \(\dot{q}\) and the outer surface of the cladding is exposed to a coolant that is characterized by a temperature \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain equations for the temperature distributions \(T_{f}(r)\) and \(T_{c}(r)\) in the fuel and cladding, respectively. Express your results exclusively in terms of the foregoing variables. (b) Consider a uranium oxide fuel pin for which \(k_{f}=2\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{1}=6 \mathrm{~mm}\) and cladding for which \(k_{c}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{2}=9 \mathrm{~mm}\). If \(\dot{q}=2 \times 10^{8}\) \(\mathrm{W} / \mathrm{m}^{3}, h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty}=300 \mathrm{~K}\), what is the maximum temperature in the fuel element? (c) Compute and plot the temperature distribution, \(T(r)\), for values of \(h=2000,5000\), and 10,000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the operator wishes to maintain the centerline temperature of the fuel element below \(1000 \mathrm{~K}\), can she do so by adjusting the coolant flow and hence the value of \(h\) ?

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.
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