91Ó°ÊÓ

A plane wall of thickness \(2 L=40 \mathrm{~mm}\) and thermal conductivity \(k=5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation at a rate \(\dot{q}\), while convection heat transfer occurs at both of its surfaces \((x=-L,+L)\), each of which is exposed to a fluid of temperature \(T_{\infty}=20^{\circ} \mathrm{C}\). Under steady-state conditions, the temperature distribution in the wall is of the form \(T(x)=a+b x+c x^{2}\) where \(a=82.0^{\circ} \mathrm{C}, b=-210^{\circ} \mathrm{C} / \mathrm{m}, c=-2 \times 10^{4 \circ} \mathrm{C} / \mathrm{m}^{2}\), 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 \(\dot{q}\) in the wall? (c) Determine the surface heat fluxes, \(q_{x}^{\prime \prime}(-L)\) and \(q_{x}^{\prime \prime}(+L)\). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at \(x=-L\) and \(x=+L\) ? (e) Obtain an expression for the heat flux distribution \(q_{x}^{\prime \prime}(x)\). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) 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? (g) What temperature will the wall eventually reach with \(\dot{q}=0\) ? How much energy must be removed by the fluid per unit area of the wall \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) 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.

Step by step solution

01

Sketch the temperature distribution and significant physical features.

The given temperature distribution is a quadratic function of the form \(T(x)=a+bx+cx^2\). To sketch this distribution, we can plot the function in the range of the thickness of the wall (-L to L). Then, identify the significant physical features such as local minima/maxima and inflection points.

02

Calculate the volumetric rate of heat generation

To find the volumetric rate of heat generation (\(\dot{q}\)), we will use the formula from Fourier's Law of conduction: \[ k\frac{d^2T}{dx^2} + \dot q =0 \] Differentiate the given temperature function (\(T(x)=a+bx+cx^2\)) twice with respect to x, then plug the values of the obtained second derivative and k (thermal conductivity) into the equation to find \(\dot{q}\).

03

Determine the surface heat fluxes

To determine the surface heat fluxes \(q_{x}^{\prime \prime}(-L)\) and \(q_{x}^{\prime \prime}(+L)\), we first need to find the heat flux distribution \(q_{x}^{\prime \prime}(x)\). This can be done by differentiating the temperature distribution function (\(T(x)=a+bx+cx^2\)) once and multiplying by -k. Next, plug the values of x = -L and x = L into the obtained heat flux distribution equation to find the surface heat fluxes. Evaluate if these fluxes are related to the heat generation rate.

04

Determine the convection coefficients

To find the convection coefficients for the surfaces at \(x=-L\) and \(x=+L\), we can use the heat flux expression for convection, given by: \[ q_{x}^{\prime \prime} = h(T-T_\infty) \] We already have the values for \(q_{x}^{\prime \prime}(-L)\) and \(q_{x}^{\prime \prime}(+L)\) from the previous steps. Plug these values and the given ambient temperature (\(T_\infty\)) into the convection heat flux equation for both surfaces and solve for the convection coefficients, \(h_{-L}\) and \(h_{+L}\).

05

Obtain the heat flux distribution and analyze its distribution

We calculated the heat flux distribution, \(q_{x}^{\prime \prime}(x)\), in the previous steps. Now, we can analyze its features, such as identifying if the heat flux is zero at any location and explaining the significance of different points on the distribution.

06

Calculate the rate of change of energy stored in the wall

If the source of heat generation is suddenly deactivated (\(\dot{q} = 0\)), we can find the rate of change of energy stored in the wall at this instant using the following equation, where u is the energy per unit volume of the material and t represents time: \[ \frac{d(u)}{dt} = -q_{x}^{\prime \prime}(x) \] As the heat generation rate is zero, plug in the value of \(q_{x}^{\prime \prime}(x)\) we found earlier and calculate the rate of change of energy stored in the wall.

07

Find the wall's final temperature and required energy removal

When the heat generation rate is zero, we can calculate the final temperature the wall will reach by assuming that heat is removed only through convection. In this case, the temperature distribution becomes linear, and we can find the final temperature using the convection heat flux equation we used earlier. To find how much energy must be removed by the fluid per unit area of the wall to reach this state, we can calculate the initial and final energy stored in the wall materials using the given density, specific heat, and dimensions. Then, find the difference between these energy values to determine the required energy removal.

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Key Concepts

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

Thermal Conductivity

When we talk about thermal conductivity, we refer to the material's ability to conduct heat. In the given problem, the wall has a thermal conductivity of 5 W/m·K. This means that the material can transfer 5 watts of heat per meter of thickness for every degree Celsius temperature difference.
Understanding thermal conductivity helps us figure out how effective the material is in transferring heat. A higher thermal conductivity indicates better heat conduction.

• It acts as a proportionality factor in Fourier's Law of Heat Conduction: \[ q = -k \frac{dT}{dx} \] where \(q\) is the heat flux, and \(\frac{dT}{dx}\) is the temperature gradient.
• Materials with high thermal conductivity, like metals, transfer heat efficiently, while low values, like insulators, do not.

Knowing these details allows you to understand how fast heat will spread through a material.

Temperature Distribution

The temperature distribution is essential for understanding how temperature varies within the wall. In this textbook problem, the temperature distribution, given as \(T(x) = a + bx + cx^2\), is a quadratic equation. This means that temperatures change in a curved manner across the wall.

• The coefficients \(a\), \(b\), and \(c\) determine the shape and curvature of the temperature profile:
  • \(a\), which is 82.0°C, represents the offset or base temperature at the center of the wall \((x = 0)\).
  • \(b\) and \(c\) influence how quickly the temperature changes across the wall.
• Analyzing the function helps predict heat flow direction and magnitude within the wall.

Besides providing insight into heat movement, this distribution shows us where temperature extremes might occur, like at the surfaces or at the midpoint.

Convection Heat Transfer

Convection heat transfer involves moving heat away from the wall's surfaces to a surrounding fluid. This problem explores convection at both surfaces of the wall which face a fluid at 20°C.
Understanding convection is crucial because it describes the heat exchange between the solid wall and the fluid.

• The convection heat transfer can be expressed with: \[ q_{x}^{\prime \prime} = h(T_s - T_\infty) \] where \( q_{x}^{\prime \prime} \) is the heat flux due to convection, \( h \) is the convection heat transfer coefficient, \( T_s \) is the surface temperature, and \( T_\infty \) is the fluid temperature.
• Efficient heat removal depends on factors like the fluid's flow rate and properties, plus the surface area’s exposure.

Using the heat transfer coefficient and surface temperature, you can calculate how quickly heat dissipates from the wall into the fluid.

Volumetric Heat Generation

Volumetric heat generation in this context refers to energy produced within the material, evenly distributed throughout the wall. This concept is vital for problems where heat comes not just from outside sources but also internally.
In the exercise, calculating the volumetric heat generation helps you understand how much internal heat affects thermal performance.

• The formula used is: \[ k \cdot \frac{d^2T}{dx^2} + \dot{q} = 0\] Here, \(k\) is the thermal conductivity, and \(\frac{d^2T}{dx^2}\) represents the second derivative of temperature, indicating curvature.
• Solving for \(\dot{q}\) gives the rate of heat produced per unit volume.
• Knowing \(\dot{q}\) can assess if external cooling reactions or enhancements may be necessary.

Volumetric heat generation is a key aspect to consider in many engineering systems, determining if the internal heat complements or counters the external conditions.