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Chapter 3: Problem 20

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 \(W / \mathrm{m}^{2} \cdot \mathbf{K}\). The wall is composed of a 10 -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
Heat loss is approximately 35406 W/m虏. Distribution shows linear temperature drop across layers.

Step by step solution

01

Understand the Problem

We have a composite wall made of two layers with contact resistance in between, separating hot gases and a cooler liquid. We need to find the heat loss per unit surface area through the composite wall.
02

Clarify the System Components

The system involves three main resistances for heat transfer: the convection resistance of the gas side, conduction resistance through the composite layers, and convection resistance of the liquid side. Contact resistance between the solid layers must also be included.
03

Calculate Convection Resistance (Gas Side)

Convection resistance on the gas side, \( R_{conv,gas} \), is given by \( R = \frac{1}{hA} \). Since we focus on unit area, \( A = 1 \ m^2 \). Thus, \( R_{conv,gas} = \frac{1}{50} \ m^2 \cdot K/W \).
04

Calculate Conduction Resistance (Beryllium Oxide)

The conduction resistance \( R_{cond,BeO} \) in a 0.01 m thick BeO layer is \( R = \frac{L}{kA} \). Take the thermal conductivity value for beryllium oxide (\( k \approx 30 \ W/m \cdot K \)), hence, \( R_{cond,BeO} = \frac{0.01}{30} \ m^2 \cdot K/W \).
05

Calculate Contact Resistance

Given contact resistance \( R_{contact} \) between BeO and stainless steel is \( 0.05 m^2 \cdot K/W \).
06

Calculate Conduction Resistance (Stainless Steel)

For stainless steel, \( R_{cond,SS} = \frac{0.02}{k} \ m^2 \cdot K/W \), using \( k = 15 \ W/m \cdot K \) for AISI 304 steel yields \( R_{cond,SS} = \frac{0.02}{15} \ m^2 \cdot K/W \).
07

Calculate Convection Resistance (Liquid Side)

Convection resistance on the liquid side, \( R_{conv,liquid} \), is \( R = \frac{1}{1000} \ m^2 \cdot K/W \).
08

Total Resistance Equation

Total resistance for heat transfer through the composite system is the sum of individual resistances:\[R_{total} = R_{conv,gas} + R_{cond,BeO} + R_{contact} + R_{cond,SS} + R_{conv,liquid}\]
09

Calculate Total Resistance

Substituting all resistance values:\[R_{total} = \frac{1}{50} + \frac{0.01}{30} + 0.05 + \frac{0.02}{15} + \frac{1}{1000} \approx 0.07067 \ m^2 \cdot K/W\]
10

Calculate Heat Loss per Unit Area

The heat loss per unit area \( q \) is given by the temperature difference over total resistance: \[q = \frac{T_{gas} - T_{liquid}}{R_{total}} = \frac{2600 - 100}{0.07067} \approx 35406 \ W/m^2\]
11

Sketch Temperature Distribution

Draw a linear temperature drop across each resistance: steep across small resistances like convection layers, and gradual across thicker layers and the contact resistance.

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Key Concepts

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

Conduction Resistance
Conduction resistance is an important concept in understanding how heat moves through solid materials. It opposes the flow of heat by conduction, much like how electrical resistance hinders the flow of electrons in a circuit.
In our example of a composite wall, conduction resistance is calculated for each layer separately. The formula used is \( R = \frac{L}{kA} \), where \( L \) is the thickness of the material, \( k \) is the thermal conductivity, and \( A \) is the area through which heat is conducted. Because we're considering per unit area, \( A = 1 \, m^2 \) makes calculations straightforward.
For beryllium oxide, with high thermal conductivity (30 \( W/m \cdot K \)), the conduction resistance is relatively low, allowing heat to pass through more easily. In contrast, the stainless steel layer has a lower thermal conductivity (15 \( W/m \cdot K \)), offering more resistance to heat conduction. This impacts the overall heat flow rate through the composite wall.
Convection Resistance
Convection resistance is associated with the transfer of heat between a solid surface and a fluid (such as gas or liquid) moving over the surface. It disciplines how well heat can enter into or leave the solid surface.
Defined by the formula \( R = \frac{1}{hA} \), it depends heavily on the convection heat transfer coefficient \( h \), which varies based on fluid properties and flow characteristics. In our problem, gas-side convection resistance is higher than the liquid side, due to a lower \( h \) coefficient (50 versus 1000 \( W/m^2 \cdot K \), respectively). This means heat transfer is more hindered on the gas side.
This difference is crucial because it dictates how easily heat can move away from or towards the solid wall, impacting the temperature gradient and rate of thermal energy transfer.
Thermal Conductivity
Thermal conductivity \( k \) is a material property that denotes its capability to conduct heat. High thermal conductivity means that the material can conduct heat effectively, while low conductivity suggests that the material is more of an insulator.
In thermal analysis of composite systems, different materials will have varying thermal conductivities. Beryllium oxide, for example, has a relatively high thermal conductivity, meaning it's quite effective in transferring heat. Conversely, the AISI 304 stainless steel's lower thermal conductivity implies more resistance to heat flow. This variance affects the composite wall's overall heat conduction ability.
Understanding thermal conductivity is key for materials selection in engineering applications, especially where efficient heat transfer is required.
Contact Resistance
Contact resistance occurs when two surfaces meet, creating a barrier to heat flow across the interface. It's like a small gap that the heat must "jump" over, causing an additional thermal resistance.
In our composite wall, contact resistance is present between the beryllium oxide and stainless steel layers. This resistance is given as \(0.05 \, m^2 \cdot K/W\), which is notably significant because it features in the total resistance calculation, affecting the rate of heat flow.
Contact resistance must be minimized in applications requiring efficient thermal management, but it can never be entirely eliminated due to material imperfections and surface roughness. Engineers may address this by ensuring smoother contact surfaces or reducing the interface's thermal resistance through design modifications.

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Most popular questions from this chapter

A thin flat plate of length \(L\) thickness \(t\), and width \(W=L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{e}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q^{*}\) - (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2}\) + \(\mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at 5-mm increments along the tube length, what is the heat loss per meter of tube length?

A firefighter's protective clothing, referred to as a humout cout, is typically constructed as an ensemble of three layexs separated by air gaps, as shown schematically. Representative dimensions and thermal conductivities for the layers are as follows. \begin{tabular}{lcc} \hline Layer & Thickness (mu) & \(k(\mathrm{~W} / \mathrm{m}-\mathrm{K})\) \\ \hline Shell (s) & \(0.8\) & \(0.047\) \\ Moisture barricr \((\mathrm{mb})\) & \(0.55\) & \(0.012\) \\ Thermal liner (t) & \(3.5\) & \(0.038\) \\ \hline \end{tabular} The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation ex. change through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rat }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right)=4 o T_{m z}^{3}\), where \(T_{\text {ax }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{n d}^{*}=h_{\text {ad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W})\) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\text {es }}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-flash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the tumout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the tumout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in bum injury?

The wall of a drying oven is constructed by sandwiching an insulation material of thermal conductivity \(k=\) \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) between thin metal shects. The oven air is at \(T_{s j}=300^{\circ} \mathrm{C}\), and the corresponding convection coefficient is \(h_{4}=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner wall surface absorbs a radiant flux of \(q_{\text {iat }}^{\prime \prime}=100 \mathrm{~W} / \mathrm{m}^{2}\) from hotter objects within the oven. The room air is at \(T_{m e}=25^{\circ} \mathrm{C}\), and the overall coefficient for convection and radiation from the outer surface is \(h_{e}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Draw the thermal circuit for the wall and label all temperatures, heat rates, and thermal resistances. (b) What insulation thickness \(L\) is required to maintain the outer wall surface at a sofe-to-touch temperature of \(T_{e}=40^{\circ} \mathrm{C}\) ?

An air heater may be fabricated by coiling Nichfome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(p_{e}=10^{-6} \mathrm{n} \cdot \mathrm{m}\), thernal conductivity \(k=25 \mathrm{~W} / \mathrm{m}=\mathrm{K}\), and emissivity \(\varepsilon=0.2\). The heater is designed to deliver air at a temperatiue of \(T_{m}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {er }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable clextric current \(n\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power raing of the heater? Himt: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this arsumption.
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