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The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?

Step by step solution

01

Analyze the problem and define variables

: Let's start by defining the given parameters and variables: - \(T_{i} = 50^{\circ}C\): initial temperature inside the chamber - \(h_{i} = 20\frac{W}{m^{2}K}\): heat transfer coefficient from air to the wall - \(L = 0.2m\): thickness of the wall - \(k = 4\frac{W}{mK}\): thermal conductivity of the wall - \(Q_{gen} = 1000\frac{W}{m^{3}}\): uniform heat generation within the wall - \(T_{\infty, o} = 25^{\circ}C\): temperature outside of the chamber - \(h_{o} = 5\frac{W}{m^2K}\): heat transfer coefficient outside the chamber - \(q_{o}''\): heat flux supplied by the strip heater

02

Temperature distribution in the wall

: To find the temperature distribution in the wall, we need to first determine the heat transfer rate and then use the Fourier's law of conduction. The relation for the temperature distribution is: \[- k \frac{dT}{dx} = h_{i}(T-T_{i})\] Now, integrate the above equation to find the temperature distribution: \[T(x) = T_{i} + \frac{h_{i}}{k}x\]

03

Find Temperature at the wall boundaries \(T(0)\) and \(T(L)\)

: Now, plug in the boundary conditions to find the temperatures at the wall boundaries: a) At \(x=0\), \(T(0) = T_{i} = 50^{\circ}C\) b) At \(x=L\), \(T(L) = T_{i} + \frac{h_{i}}{k}L = 50 + \frac{20}{4}(0.2) = 50 + 10 = 60^{\circ}C\)

04

Find the heat flux supplied by the strip heater \(q_{o}''\)

: To find the value of \(q_{o}''\), we need to first determine the total heat generated in the wall and the heat transfer rate to the inside of the chamber: \[Q_{total} = Q_{gen}(L)A = 1000\times0.2A\] where \(A\) is the area of the wall. Now, the heat transfer rate to the inside of the chamber can be calculated as: \[Q_{transfer} = h_{i}A(T-T_{i}) = 20A(60-50) = 20A(10)\] To maintain the condition where no heat is lost to the outside of the chamber, the heat supplied by the strip heater should be equal to the difference between the total heat generated and the heat transferred to the inside of the chamber: \[q_{o}''A = Q_{total} - Q_{transfer} = 1000\times0.2A - 20A(10)\] Hence, \[q_{o}'' = 1000\times0.2 - 20(10) = 200 - 200 = 0\frac{W}{m^2}\]

05

Steady-state temperature of the outer wall surface \(T(0)\) when the heat generation is switched off

: In the case where the heat generation in the wall is switched off, the heat flux to the strip heater remains constant: \[0 = -k\frac{dT}{dx} + h_{o}(T_{\infty, o}-T)\] Now, let's rearrange the above equation to solve for the steady-state temperature: \[\frac{dT}{dx} = \frac{-h_{o}}{k}(T_{\infty, o}-T)\] Integrate the above equation to find the steady-state temperature: \[T(x) = T_{\infty, o} - \frac{h_{o}}{k}x\] At the outer wall surface \(x=0\), the steady-state temperature can be calculated as: \[T(0) = 25 - \frac{5}{4}(0) = 25^{\circ}C\] So, the steady-state temperature of the outer wall surface is \(25^{\circ}C\) when the heat generation in the wall is switched off.

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Key Concepts

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

Convective Heat Transfer

Heat transfer within a material, and between materials that are in direct contact, is conduction. On the other hand, convective heat transfer occurs when a fluid (gas or liquid) is involved, moving heat from one place to another. This process is governed by the convective heat transfer coefficient, denoted by 'h', measured in units of Watts per square meter Kelvin \(W/m^2K\).

For instance, in the textbook exercise, air inside a chamber transfers heat to the wall through convection with a given coefficient \( h_i \). This coefficient indicates how efficiently heat is being moved from the warm air to the wall's surface. The higher the value of 'h', the more efficient the process of heat transfer. If we place our hand over a heating vent, we can feel the convective heat transfer as warm air moves over our skin.

Understanding convective heat transfer is crucial, especially for designing heating and cooling systems, as it helps in predicting how quickly a space will reach a desired temperature. Additionally, this knowledge is vital to engineers who work on enhancing energy efficiency in buildings and industrial processes.

Thermal Conductivity

The rate at which heat passes through a material is determined by its thermal conductivity, denoted by 'k', with units of Watts per meter Kelvin \(W/mK\). This property quantifies how well a material can conduct heat. In the context of the exercise, the wall has a given thermal conductivity that allows heat to move from one side to the other.

The formula for heat conduction, based on Fourier’s law, is:\[ q = -k \frac{dT}{dx} \], where 'q' is the heat flux, \( \frac{dT}{dx} \) is the temperature gradient, and the negative sign indicates that heat flows from hot to cold regions.

When selecting materials for construction or manufacturing, knowing the thermal conductivity helps engineers ensure that components can handle the thermal loads they will encounter. High thermal conductivity materials are sought for heat sinks and other applications where efficient heat dissipation is crucial, while low thermal conductivity materials make great insulators.

Uniform Heat Generation

Sometimes, a material may generate heat evenly throughout its volume, a phenomenon described as uniform heat generation, denoted usually by \( Q_{gen} \) with the unit \( W/m^3 \). Electrical components like resistors, and mechanical processes such as friction in brakes, can cause uniform heat generation within a solid.

In our exercise, the wall has a specified uniform heat generation. This means that every unit volume of the wall generates the same amount of heat due to some internal mechanism. The presence of uniform heat generation significantly influences the temperature profile within the material, resulting in a non-linear temperature distribution, unless the material is perfectly insulated or in thermal equilibrium with its surroundings.

Managing uniform heat generation within materials is vital in electronic design, material processing, and in understanding geothermal gradients within the earth. Solutions to such problems require integrating concepts of both thermal conductivity and convective heat transfer to predict and control temperature distributions effectively.