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A plane wall of thickness \(L=0.1 \mathrm{~m}\) experiences uniform volumetric heating at a rate \(\dot{q}\). One surface of the wall \((x=0)\) is insulated, and the other surface is exposed to \(\mathrm{a}\) fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\), with convection heat transfer characterized by \(h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Initially, the temperature distribution in the wall is \(T(x, 0)=a+b x^{2}\), where \(a=300^{\circ} \mathrm{C}, b=-1.0 \times 10^{40} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. Suddenly, the volumetric heat generation is deactivated ( \(\dot{q}=0\) for \(t \geq 0\) ), while convection heat transfer continues to occur at \(x=L\). The properties of the wall are \(\rho=7000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=90 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Determine the magnitude of the volumetric energy generation rate \(\dot{q}\) associated with the initial condition \((t<0)\). (b) On \(T-x\) coordinates, sketch the temperature distribution for the following conditions: initial condition \((t<0)\), steady-state condition \((t \rightarrow \infty)\), and two intermediate conditions. (c) On \(q_{x}^{\prime \prime}-t\) coordinates, sketch the variation with time of the heat flux at the boundary exposed to the convection process, \(q_{x}^{\prime \prime}(L, t)\). Calculate the corresponding value of the heat flux at \(t=0, q_{x}^{\prime \prime}(L, 0)\). (d) Calculate the amount of energy removed from the wall per unit area \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) by the fluid stream as the wall cools from its initial to steady-state condition.

Step by step solution

01

Apply the heat diffusion equation to the problem

In this problem, heat generation occurs before t=0, and the heat diffuses through the wall. The governing equation for heat diffusion with heat generation is: \( \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} + \frac{\dot{q}}{\rho c_p} \) Considering \( t < 0 \), the temperature distribution \( T(x, 0)=a+b x^{2} \) is constant over time, which means, \( \frac{\partial T}{\partial t} = 0 \)

02

Calculate the temperature gradient

Now to calculate the volumetric heat generation rate \(\dot{q}\), we first need to find \(\frac{\partial^2 T}{\partial x^2}\) with the given temperature distribution, \( T(x, 0)=a+b x^{2} \) The second derivative is: \( \frac{\partial^2 T}{\partial x^2} = 2b \)

03

Calculate the volumetric heat generation rate

Now, substituting the values in the equation from Step 1, we have: \( 0 = \alpha \frac{\partial^2 T}{\partial x^2} + \frac{\dot{q}}{\rho c_p} \) Rearranging the terms for \(\dot{q}\), \( \dot{q} = - \rho c_p \alpha (2b) \) Now we have to substitute the given values in the equation: \( \rho = 7000 \mathrm{~kg} / \mathrm{m}^{3}, c_p = 450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k = 90 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \) and \( \alpha = k / (\rho c_p) \). We get: \( \dot{q} = -7000 \times 450 \times \frac{90}{7000 \times 450} \times (2 \times -1.0 \times 10^{4}) \) \( \dot{q} = 18 \times 10^{4} \mathrm{~W} / \mathrm{m}^{3} \) So, the magnitude of the volumetric energy generation rate for the initial condition is \( \dot{q} = 18 \times 10^{4} \mathrm{~W} / \mathrm{m}^{3} \). (b) and (c) involve sketching graphs and can be done on paper. (d) Calculate the amount of energy removed from the wall per unit area \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) by the fluid stream as the wall cools from its initial to steady-state condition:

04

Calculate initial and steady-state temperatures

For this, we first need to find the initial and steady-state temperature at x=L, At \(t<0\): \( T_0 = a + b \times L^2 \) At steady-state (t→∞): \(T_{\infty} = T_{\infty} + \frac{q_s' \times L}{h} \), where \(q_s'\) is the steady-state heat flux.

05

Calculate heat flux

Substituting Fourier's law i.e., \( q_s' = - k \frac{dT}{dx} \) at x=L, \( q_s' = - k \times 2bL \) \( q_s' = - 90 \times 2 \times (-1.0 \times 10^{4}) \times 0.1 \) \( q_s' = 18 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \)

06

Calculate steady-state temperature at x=L

Substitute the heat flux value in the steady-state temperature equation: \( T_{\infty} = 20 + \frac{(18 \times 10^{3}) \times 0.1}{1000} \) \( T_{\infty} = 20 + 180 = 200^{\circ} \mathrm{C} \)

07

Calculate the heat conduction energy

Now we have to calculate the energy removed from the wall per unit area due to heat conduction as the wall cools from its initial to steady-state condition: \( E_{\text{cond}} = \rho c_p (T_0 - T_{\infty}) \times L \) Substituting the values, \( E_{\text{cond}} = 7000 \times 450 \times (300 - 200) \times 0.1 \) \( E_{\text{cond}} = 3.15 \times 10^6 \mathrm{~J} / \mathrm{m}^{2} \) So, the amount of energy removed from the wall per unit area by the fluid stream as the wall cools from its initial to steady-state condition is \( 3.15 \times 10^6 \mathrm{~J} / \mathrm{m}^{2} \).

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Key Concepts

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

Convection Heat Transfer

Convection heat transfer is a fundamental mechanism by which heat energy is transferred between a solid surface and an adjacent fluid (such as air or water) in motion. Here, the situation involves a fluid with a temperature of \( T_{\infty} = 20^{\circ} \mathrm{C} \) moving over the surface of a plane wall. The convection is quantified using the heat transfer coefficient \( h \), which in this case is \( 1000 \, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \). This coefficient acts as a measure of the thermal interaction between the fluid and the wall.

In practical terms, convection is responsible for either heating or cooling the wall, depending on the temperature difference between the wall and the fluid. A higher \( h \) value indicates more efficient heat transfer, which prompts rapid temperature changes at the surface. This efficient energy exchange keeps processes in engineering and everyday applications running smoothly and helps manage the thermal dynamics efficiently.

Volumetric Heat Generation

Volumetric heat generation refers to the internal generation of heat within a material, which can vary depending on various sources like chemical reactions, electricity, or nuclear reactions. Before the heat generation is turned off in the exercise (for \( t < 0 \)), the wall experiences uniform volumetric heating at a rate \( \dot{q} \). This internal heat generation creates a temperature gradient within the wall, which influences how heat spreads out.

It is essentially calculated using the principle of heat diffusion, where the second derivative of the temperature with respect to the spatial coordinate \( x \) gives a measure of the heat distribution rate. In the exercise, the rate \( \dot{q} \) was calculated to be \( 18 \times 10^{4} \, \mathrm{W} / \mathrm{m}^{3} \), demonstrating a high rate of energy production inside the wall. Understanding \( \dot{q} \) is crucial in designing materials and systems to manage and dissipate excess heat effectively, ensuring safety and efficiency.

Temperature Distribution

Temperature distribution is the variation of temperature within a body, which in this scenario is the plane wall. It is initially expressed as a quadratic function of position \( x \), given by \( T(x, 0)=a+bx^2 \), where \( a=300^{\circ} \mathrm{C} \) and \( b=-1.0 \times 10^{4} \mathrm{C} / \mathrm{m}^{2} \). This mathematical expression shows how temperature differs from one part of the wall to another initially (before \( t=0 \)).

Sketches of the temperature distribution at different moments (initial, intermediate, and steady-state) help visualize how the wall cools over time. Intermediate conditions depict gradual transitions as internal heat dissipates towards the surface and interact with the convection process. Ultimately, the analysis of this distribution provides insight into the thermal behavior and assists in anticipating system responses to changes like heat load variations.

Steady-State Condition

A steady-state condition is achieved when the temperature in the wall becomes uniform and does not change over time. In this context, it occurs when the wall, after having its internal heat generation turned off (for \( t \geq 0 \)), reaches thermal equilibrium with its environment. This is characterized by constant temperature distribution across the wall and consistent heat flux.

For this problem, the final temperature at \( x = L \) approaches \( T_{\infty} + \frac{q_s' \times L}{h} \), leading to a final surface temperature of \( 200^{\circ} \mathrm{C} \). At steady state, there is no net accumulation of heat within the wall, and the convection process fully controls the heat balance. Engineers often design systems to reach steady-state to ensure predictable and stable operating conditions. Understanding when and how systems achieve this condition is essential, particularly for long-term operation and thermal management.