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Chapter 2: Problem 29

The steady-state temperature distribution in a onedimensional wall of thermal conductivity \(k\) and thickness \(L\) is of the form \(T=a x^{3}+b x^{2}+c x+d\). Derive expressions for the heat generation nate per unit volume in the wall and the heat fluxes at the two wall faces \((x=0, L)\).

Short Answer

Expert verified
Heat generation rate: \( q''' = -k(6ax + 2b) \). Heat fluxes: \( q(0) = -kc \) and \( q(L) = -k(3aL^{2} + 2bL + c) \).

Step by step solution

01

Identify the Task

We need to find expressions for the heat generation rate per unit volume and the heat fluxes at the two faces of a one-dimensional wall with a given temperature distribution.
02

Understand the Temperature Distribution

The temperature distribution is given by the polynomial form \[ T = a x^{3} + b x^{2} + c x + d. \] This equation describes how temperature changes with position \(x\) within the wall.
03

Calculate the Heat Generation Rate

The heat generation rate per unit volume is given by Fourier's Law for heat conduction and is related to the second derivative of temperature with respect to position, \[ q''' = -k \frac{d^{2}T}{dx^{2}}. \] Calculate the second derivative of the temperature distribution:
04

Differentiate the Temperature Function

First, find the first derivative of the temperature function: \[ \frac{dT}{dx} = 3ax^{2} + 2bx + c. \]Then, find the second derivative:\[ \frac{d^{2}T}{dx^{2}} = 6ax + 2b. \]
05

Substitute into Heat Generation Formula

Substitute the second derivative into the formula for heat generation rate: \[ q''' = -k (6ax + 2b). \] This expression gives the heat generation rate per unit volume at any position \(x\) in the wall.
06

Calculate the Heat Flux at the Faces

The heat flux \( q \) at a position \( x \) is related to the temperature gradient:\[ q = -k \frac{dT}{dx}. \] Use this to find the heat flux at the two faces by evaluating the first derivative at \( x = 0 \) and \( x = L \).
07

Evaluate Heat Flux at \( x = 0 \)

The first derivative of temperature at \( x = 0 \) is \[ \frac{dT}{dx}\bigg|_{x=0} = c. \]Then, the heat flux at \( x = 0 \) is \[ q(0) = -k \cdot c. \]
08

Evaluate Heat Flux at \( x = L \)

The first derivative of temperature at \( x = L \) is \[ \frac{dT}{dx}\bigg|_{x=L} = 3aL^{2} + 2bL + c. \]Thus, the heat flux at \( x = L \) is \[ q(L) = -k (3aL^{2} + 2bL + c). \]

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Key Concepts

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

Temperature Distribution
The concept of temperature distribution is crucial in understanding how heat flows through materials, like our one-dimensional wall. Here, the temperature is described by a polynomial equation, \[ T = ax^3 + bx^2 + cx + d \]. This equation shows how temperature varies with position \( x \) within the wall.
It includes different terms, each representing the influence of various factors on the temperature gradient:
  • The \( ax^3 \) term indicates temperature changes more rapidly with depth in more complex materials.
  • The \( bx^2 \) and \( cx \) terms capture simpler variations in temperature with the wall's position.
  • The constant \( d \) might represent the starting temperature at one side of the wall.
Understanding how these terms affect the distribution helps predict how efficiently a wall can insulate or transfer heat.
Heat Generation Rate
The heat generation rate per unit volume is a measure of how much heat is being produced within a specific area of the wall. Using Fourier's Law, we relate this concept to the second derivative of the temperature distribution, \[ q''' = -k \frac{d^2T}{dx^2} \].
First, determine the second derivative of our temperature equation: \[ \frac{d^2T}{dx^2} = 6ax + 2b \]. Plug this into the heat generation rate formula:\[ q''' = -k(6ax + 2b) \].
This formula suggests several insights:
  • The heat generation rate depends on both the position \( x \) and the temperature gradient.
  • It informs us how much heat is being internally generated across the wall per unit volume.
Knowing this rate is vital for designing materials that must either effectively dispel generated heat or retain it.
Heat Flux
Heat flux is the rate at which heat energy passes through a given surface area. It's essentially about how much heat flows through the wall at any point. To find the heat flux, apply the first derivative from Fourier's Law:\[ q = -k \frac{dT}{dx} \].
Evaluate this at the faces of the wall:
  • At \( x = 0 \), the heat flux is \[ q(0) = -k \cdot c \].
  • At \( x = L \), the heat flux is given by \[ q(L) = -k (3aL^2 + 2bL + c) \].
These results highlight that:
  • The flux at any surface depends directly on the thermal conductivity \( k \) and the temperature gradient \( \frac{dT}{dx} \) at that position.
  • In practical terms, heat flux tells us how efficiently a wall can handle the passage of heat across its surfaces.
With this knowledge, engineers can optimize materials to ensure desired heat transfer characteristics.
Fourier's Law
Fourier鈥檚 Law of heat conduction is the foundational principle in understanding how heat moves through materials. It states that the heat transfer rate through a material is proportional to the negative gradient of the temperature and the area perpendicular to that gradient across which it flows:\[ q = -k \frac{dT}{dx} \].
This law underpins our calculations for both heat flux and heat generation rate.
  • \( k \) represents the thermal conductivity, a key property defining how much heat a material can conduct.
  • The negative sign shows that heat moves from higher to lower temperature regions.
  • The temperature gradient \( \frac{dT}{dx} \) provides the direction and rate of temperature change within the material.
By applying Fourier's Law, we decipher how effectively materials can serve as conductors or insulators, which is crucial for various applications such as thermal management in electronics or construction material design.

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Most popular questions from this chapter

A method for determining the thermal conductivity \(k\) and the specific heat \(c_{p}\) of a material is illustrated in the sketch. Initially the two identical samples of diameter \(D=60 \mathrm{~mm}\) and thickness \(I\), \(=10 \mathrm{~mm}\) and the thin heater are at a uniform temperature of \(T_{i}=23.00^{\circ} \mathrm{C}\). while surrounded by an insulating powder. Suddenly the heater is energized to provide a uniform heat flux \(q_{e}^{\prime \prime}\) on each of the sample interfaces, and the heat fiux is maintained constant for a period of time, \(\Delta t_{t r}\) A short time after sudden heating is initiated, the temperature at this interface \(T_{\omega}\) is related to the heat flux as $$ T_{d}(t)-T_{i}=2 e^{\prime \prime}\left(\frac{t}{\pi \alpha c k}\right)^{i / 2} $$ For a particular test run, the electrical heater dissipates 15.0 W for a period of \(\Delta t_{e}=120 \mathrm{~s}\) and the temperature at the interface is \(T,(30 \mathrm{~s})=24.57^{\circ} \mathrm{C}\) after \(30 \mathrm{~s}\) of heating. A long time after the heiter is deenergized, \(t>\Delta t_{\text {e. }}\) the samples reach the uniform temperature of \(T_{d}(\propto)=\) \(33.50^{\circ} \mathrm{C}\). The density of the sample materials, determined by measurement of volume and mass, is \(\rho=\) \(3965 \mathrm{~kg} / \mathrm{m}^{3}\), Determine the specific heat and thermal conductivity of the test material. By looking at values of the thermophysical properties in Table A.I or A.2, identify the test sample material.

A plane wall of thickness \(2 \mathrm{~L}=40 \mathrm{~mm}\) and themal conductivity \(k=5 \mathrm{~W} / \mathrm{m}\) - \(\mathrm{K}\) experiences uniform volumetric heat generation at a rate \(q\), while convection heat transer occurs at both of its surfaces \(\left(x=-L_{4}+L\right)\). each of which is exposed to a fluid of temperature \(T_{m}=20^{\circ} \mathrm{C}\). Under steady-state conditions, the termperature distribution in the wall is of the form \(T(x)=a+\) \(b x+c x^{2}\), where \(a=82.0^{\circ} \mathrm{C}, b=-210^{\circ} \mathrm{C} / \mathrm{m}, c=\) \(-2 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. The origin of the \(x\)-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation \(\dot{4}\) in the wall? (c) Determine the surface heat fluxes, \(q_{1}^{\prime \prime}(-L)\) and \(q_{:}^{\prime \prime}(+L)\). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the sarfaces at \(x=-L\) and \(x=+L\) ? (e) Obtain an expression for the heas flux distribution, \(q\) " \((x)\). Is the heat flux zero at any location? Explain any significant features of the distribution. (i) If the source of the heat generation is suddenly deactivated \((\dot{q}=0)\), what is the rate of change of energy stored in the wall at this instant? (2) What temperuture will the wall eventually reach with \(\dot{q}=0\) ? How much energy mus be removed by the fluid per unit area of the wall \(\left(1 / \mathrm{m}^{2}\right)\) to reach this state? The density and specific heat of the wall material are \(2600 \mathrm{~kg} / \mathrm{m}^{3}\) and \(800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively,

A large plate of thickness \(2 L\) is at a uniform temperature of \(T_{t}=200^{\circ} \mathrm{C}\), when it is suddenly quenched by dipping it in a liquid buth of temperature \(T_{\mathrm{w}}=20^{\circ} \mathrm{C}\) Heat transfer to the liquid is characterized by the convection ccefficient \(h\). (a) If \(x=0\) corresponds to the midplane of the wall, on \(T-x\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and two intermediate times. (b) On \(q_{r}^{*}-r\) coordinates, sketch the variation with time of the heat flux at \(x=L\) (c) If \(h=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), what is the heat flux at \(x=L\) and \(t=0\) ? If the wall has a thermal conductivity of \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}_{\text {, what is }}\) is the corresponding temperature gradient at \(x=L\) ? (d) Consider a plate of thickness \(2 L=20 \mathrm{~mm}\) with a density of \(\rho=2770 \mathrm{~kg} / \mathrm{m}^{3}\) and a specific heat \(c_{p}=875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). By performing an energy balance on the plate, determine the amount of energy per unit surface area of the plate \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) that is transferred to the bath over the time required to reach steady-state conditions. (c) From other considerations, it is known that, during the quenching process, the heal flux at \(x=+L\) and \(x=-L\) decays exponentially with time according to the relation, \(q^{\prime \prime}=A \exp (-B r)\), where \(t\) is in seconds, \(A=1.80 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2}\), and \(B=4.126 \times\) \(10^{-3} \mathrm{~s}^{-1}\). Use this information to determine the \(\mathrm{en}^{-}\) ergy per unit surface area of the plate that is transferred to the fluid during the quenching process.

A young engineer is asked to design a themal protection barricr for a sensitive clectronic device that might be exposed to irradiation from a high- powered infrarcd laser. Having learmed as a student thut a low thermal conductivity material provides good insulating characteristics, the enginecr specifies use of a nanostructured aerogel, characterized by a thermal conductivity of \(k_{e}=\) 0.0ns W/m - \(\mathrm{K}\), for the protective harrier. The engineer's boss questions the wisdom of selecting the acrogel because it has a low themal conductivity. Consider the sudden laser irradiation of (a) peire aluminum, (b) glass, and (c) aerogel. The laser provides irradiation of \(G=10 \times 10^{6}\) Whm \({ }^{2}\). The absorptivities of the materials are \(a=0.2,0.9\), and \(0.8\) for the aluminum, glass, and acrogel, rerpectively, and the initial ternperature of the barrier is \(T_{i}=300 \mathrm{~K}\). Explain why the boss is concerncd. Hint: All materials experience thetmal expunsion (or contraction), and local stresses that develop within a material are, to a first approximation, proportional io the local temperahure gradicnt.

A plane layer of coal of thickness \(L=1 \mathrm{~m}\) experiences uniform volumetric generation at a rate of \(\hat{q}=20 \mathrm{~W} / \mathrm{m}^{2}\) due to slow exidation of the coal particles. Averuged over a daily period, the top surface of the layer transfers heat by convection to ambient air for which \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\mathrm{m}}=25^{\circ} \mathrm{C}\), while receiving solar irrudiation in the amoent \(G_{5}=400 \mathrm{~W} / \mathrm{m}^{2}\). Irnatiation from the amosphere may be neglected. The solar absorptivity and emissivity of the surface are each \(\alpha_{S}=\varepsilon=0.95\). (a) Write the steady-sate form of the heat diffusion equation for the layer of coul. Verify that this equation is satisfied by a temperiture distribution of the form $$ T(x)=T_{x}+\frac{\dot{q} L^{2}}{2 k}\left(1-\frac{x^{2}}{L^{2}}\right) $$ From this distribution, what can you say about conditions at the bottom surface \((x=0)\) ? Sketch the temperature distribution and label key features. (b) Obtain an expression for the rate of heat transfer by conduction per unit area at \(x=L\). Applying an energy balance to a control surface about the top surface of the layer, obtain an expression for \(T_{r}\) Evaluate \(T_{4}\) and \(T(0)\) for the prescribed conditions. (c) Daily average values of \(G_{5}\) and \(h\) depend on a number of factors such as time of year, cloud cover, and wind conditions. For \(h=5 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\), compute and plot \(T_{\text {t and }} T(0)\) as a function of \(G_{5}\) for \(50 \leq G_{s} \leq 500 \mathrm{~W} / \mathrm{m}^{2}\). For \(G_{5}=400 \mathrm{~W} / \mathrm{m}^{2}\), compute and plot \(T_{\text {, and }} T(0)\) as a function of \(h\) for \(5 \leq h \leq 50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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