91Ó°ÊÓ

The radiation heat gage shown in the diagram is made from constantan metal foil, which is coated black and is in the form of a circular disk of radius \(R\) and thickness \(t\). The page is located in an evacuated enclosure. The incident radiation flux absorbed by the foil, \(q_{n}^{*}\) diffuses toward the cuter circumference and into the larger copper ring, which acts as a heat sink at the constant temperature \(T(R)\). Tuo copper lead wires are attached to the center of the foil and to the ring to complete a thermocouple circuit that allows for measurement of the temperature difference between the foil center and the foil edge, \(\Delta T=T(0)-T(R)\). Obtain the differential equation that determines \(T(r)\), the temperature distribution in the foil, under steady- state conditions. Solve this equation to obtuin an expression relating \(\Delta T\) to \(q^{\prime \prime}\). You may neglect radiation exchange between the foil and its surroundings.

Step by step solution

01

Identify Heat Equation for Cylinder

In this problem, the heat transfer occurs radially across a thin circular disk. In cylindrical coordinates, the radial heat conduction equation is used due to symmetry and steady-state conditions. The equation is given by \[\frac{d}{dr}\left(k r \frac{dT}{dr}\right) + q'' = 0\]with \(k\) being the thermal conductivity of constantan and \(q''\) the heat generation per unit volume, here assumed as the flux \(q_n^*\).

02

Simplify the Heat Equation

Since the problem is at a steady-state and the thickness is small, assume that there is no heat generation within the foil itself, i.e. \(q'' = 0\). Therefore, the equation simplifies to:\[\frac{d}{dr}\left(k r \frac{dT}{dr}\right) = 0\]

03

Integrate the Simplified Equation

Integrate the simplified equation once with respect to \(r\):\[k r \frac{dT}{dr} = C_1\]where \(C_1\) is an integration constant. Then solve for \(\frac{dT}{dr}\):\[\frac{dT}{dr} = \frac{C_1}{k r}\]

04

Integrate Again to Find Temperature Distribution

Integrate again to find the temperature \(T(r)\):\[T(r) = \frac{C_1}{k} \ln r + C_2\]where \(C_2\) is another integration constant.

05

Apply Boundary Conditions

Apply the boundary conditions to determine constants \(C_1\) and \(C_2\). At \(r = R\), the temperature \(T(R)\) is constant:\[T(R) = \frac{C_1}{k} \ln R + C_2 = T(R)\]At \(r = 0\) (center), the temperature \(T(0)\):Due to symmetry, \(\frac{dT}{dr}\) is finite which implies \(C_1 = 0\). Hence, \(\Delta T = T(0) - T(R)\). The linear profile equates to gradient \(\frac{dT}{dr} = \frac{T(0) - T(R)}{R}\) with flux causing \(C_1 = q_n^* R / k\).

06

Express Temperature Difference

Substitute back into the simplified gradient equation to express \(\Delta T\) in terms of flux:\[\Delta T = T(0) - T(R) = \frac{q_n^* R^2}{4k}\]This gives the relationship between temperature difference and absorbed heat flux.

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Key Concepts

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

Radiation Heat Transfer

Radiation heat transfer deals with the energy exchange between surfaces at different temperatures, without the need for a medium. In the case of the radiation heat gage exercise, the foil disk absorbs radiative energy. This energy is then transferred towards the larger copper ring. Radiation heat transfer is less intuitive than conduction or convection, as it depends heavily on the temperature and emissivity of the surfaces. Here, the foil absorbs the incident radiation flux, denoted as \(q_{n}^{*}\), which must be accounted for in solving the foil's temperature distribution. This type of heat transfer is crucial in environments such as vacuum, where other modes like convection are negligible.

Cylindrical Coordinates

In the context of the problem, cylindrical coordinates are used because the heat transfer occurs radially in the circular foil. This coordinate system includes radius \(r\), angle \(\theta\), and height \(z\). For radial problems, the height doesn't come into play, simplifying the analysis to two dimensions. The radial heat conduction equation simplifies for this symmetric problem under steady-state conditions. Using cylindrical coordinates, you can focus on the radial dimension, reducing complexity compared to Cartesian coordinates. This approach results in the set of mathematical equations, such as \(\frac{d}{dr}\left(k r \frac{dT}{dr}\right) + q'' = 0\), which help in modeling how the temperature varies from the center to the edge of the disk.

Temperature Distribution

Temperature distribution in an object tells us how temperature varies across it. For the radiation heat gage, finding the temperature distribution \(T(r)\) was key, as it helped relate the temperature difference \(\Delta T = T(0) - T(R)\) to the absorbed heat flux \(q^{\prime\prime}\). Solving the heat conduction equation gives insight into the steady-state temperature profile through methods like integration and applying boundary conditions. The equation integrates to \(T(r) = \frac{C_1}{k} \ln r + C_2\), where \(C_1\) and \(C_2\) are integration constants determined by boundary conditions. By understanding this temperature profile, we gain valuable information about how materials respond to radiative heat and how thermal gradients can influence structural and functional aspects of materials.