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Consider a 5 -mm-square, diffuse surface \(\Delta \mathrm{A}_{o}\) having a total emissive power of \(E_{o}=4000 \mathrm{~W} / \mathrm{m}^{2}\). The radiation field due to emission into the hemispherical space above the surface is diffuse, thereby providing a uniform intensity \(I(\theta, \phi)\). Moreover, if the space is a nonparticipating medium (nonabsorbing, nonscattering, and nonemitting), the intensity is independent of radius for any \((\theta, \phi)\) direction. Hence intensities at any points \(P_{1}\) and \(P_{2}\) would be equal. (a) What is the rate at which radiant energy is emitted by \(\Delta A_{o n} q_{\text {emit }}\) ? (b) What is the intensity \(I_{o, e}\) of the radiation field emitted from the surface \(\Delta A_{o}\) ? (c) Beginning with Equation \(12.13\) and presuming knowledge of the intensity \(I_{o, x}\), obtain an expression for \(q_{\text {emit }}\) (d) Consider the hemispherical surface located at \(r=R_{1}=0.5 \mathrm{~m}\). Using the conservation of energy requirement, determine the rate at which radiant energy is incident on this surface due to emission from \(\Delta A_{v}\). (e) Using Equation 12.10, determine the rate at which radiant energy leaving \(\Delta A_{o}\) is intercepted by the small area \(\Delta A_{2}\) located in the direction \(\left(45^{\circ}, \phi\right)\) on the hemispherical surface. What is the irradiation on \(\Delta A_{2}\) ? (f) Repeat part (e) for the location \(\left(0^{\circ}, \phi\right)\). Are the irradiations at the two locations equal? (g) Using Equation \(12.18\), determine the irradiation \(G_{1}\) on the hemispherical surface at \(r=R_{1^{-}}\)

Step by step solution

01

a) Finding the rate at which radiant energy is emitted

Given, total emissive power, \(E_o = 4000 W/m^2\) and area \(\Delta A_o = 5 \ mm^2 = 5 \times 10^{-6} \ m^2\). To find the rate at which radiant energy is emitted, we can use the formula: \[q_{emit} = E_o \times \Delta A_o\] \[q_{emit} = 4000 \times 5 \times 10^{-6}\] \(q_{emit} = 0.02 W\)

02

b) Finding the intensity of the radiation field emitted

Since the radiation field is diffuse, we can use the formula for intensity as follows: \[I_{o,e} = \frac{E_o}{\pi}\] \[I_{o,e} = \frac{4000}{\pi}\] \(I_{o,e} = 1273.24 W/m^2\)

03

c) Obtaining an expression for \(q_{\text {emit }}\)

To find an expression for \(q_{\text{emit}}\), we need to perform integration over the surface A and solid angle, which is given by the following equation (12.13): \[q_{\text{emit}} = \int_{A}^{}\int_{2\pi}^{}\int_{\pi /2}^{0} I_{o,x}(\theta, \phi)r^2 \sin \theta d\theta d\phi dA_1\] Since the intensity is independent of radius for any \((\theta, \phi)\), the expression becomes: \[q_{\text{emit}} = I_{o, x} \int_{A}^{}\int_{2\pi}^{}\int_{\pi /2}^{0} r^2 \sin \theta d\theta d\phi dA_1\]

04

d) Finding the rate at which radiant energy is incident on the surface

The conservation of energy requirement tells us that the rate at which energy is emitted by the surface must be equal to the rate atwhich it is received by the hemisphere: \[q_{\text{incident}} = q_{\text{emit}}\] Therefore, the rate at which radiant energy is incident on the hemispherical surface is also \(0.02 W\).

05

e) Finding the radiant energy intercepted by \(\Delta A_2\) and irradiation

We can use Equation 12.10 to determine the rate at which radiant energy leaving \(\Delta A_o\) is intercepted by the small area \(\Delta A_2\) and find irradiation on \(\Delta A_2\): \[q_{12} = I_{o,e} \cos (45^\circ) \cdot \frac{\Delta A_o \Delta A_2}{r^2}\] To find the irradiation on \(\Delta A_2\), we can use the following relation: \[G_2 = \frac{q_{12}}{\Delta A_2}\]

06

f) Finding radiant energy intercepted and irradiation at location \((0^\circ, \phi)\)

Repeating part e for location \((0^\circ, \phi)\), we can determine the radiant energy intercepted by \(\Delta A_2\) and the irradiation on \(\Delta A_2\): \[q_{12} = I_{o,e} \cos (0^\circ) \cdot \frac{\Delta A_o \Delta A_2}{r^2}\] The irradiation on \(\Delta A_2\) will be: \[G_2 = \frac{q_{12}}{\Delta A_2}\]

07

g) Determine the irradiation \(G_{1}\) on the hemispherical surface

To calculate the irradiation \(G_1\) on the hemispherical surface at \(r = R_1\), we can use Equation 12.18: \[G_1 = \int_{0}^{2\pi}\int_{0}^{\pi/2}I_{o,e}(\theta, \phi) \sin \theta \cos \theta d\theta d\phi\] By calculating the integrals and multiplying by the found intensity \(I_{o,e}\), we can find the irradiation \(G_1\) on the hemispherical surface.

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Key Concepts

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

Diffuse Surface Emission

When dealing with radiative heat transfer, the concept of diffuse surface emission is crucial. A diffuse emitter means that the radiation is scattered uniformly in all directions. This simplifies the problem significantly because it allows us to model the radiative emission as being equal across the hemispherical space above the surface. In the context of our problem, the surface \(\Delta A_o\) emits a total power density of \(E_o = 4000 \, \text{W/m}^2\). This parameter is used to calculate the radiant energy emitted by the surface using the formula \(q_{\text{emit}} = E_o \times \Delta A_o\). Breaking down the units and performing the multiplication gives us a straightforward answer, which is particularly helpful when verifying results with complex systems.

Intensity of Radiation Field

The intensity of a radiation field is a measure of how much radiant energy is traveling in a particular direction per unit area and per unit solid angle. For a diffuse surface, the intensity is uniformly distributed across the hemisphere above, making it independent of any particular direction \((\theta, \phi)\). In our example, the intensity \(I_{o,e}\) is calculated by dividing the emissive power \(E_o\) by \(\pi\), leading to the expression \(I_{o,e} = \frac{E_o}{\pi}\). This result shows how the power density from the surface spreads across the hemisphere, providing a consistent mechanism to predict energy radiating in specific directions. Understanding this uniformity is key to keeping consistency when applying the concept in physical simulations or theoretical investigations.

Conservation of Energy in Radiation

In radiative heat transfer, the principle of energy conservation is pivotal. Simply put, the energy emitted by a surface must equal the energy received by any surroundings. In our problem, the conservation principle affirms that the total energy emitted by the surface \(\Delta A_o\) is precisely the energy intercepted by the surrounding hemispherical surface. \ The exercise demonstrates this by equating the emitted and incident energies: \(q_{\text{incident}} = q_{\text{emit}}\). This equality holds as long as the radiation travels through a nonparticipating medium, meaning it isn't absorbed, scattered, or re-emitted by the medium. Applying this concept helps ensure accurate calculations within many engineering and environmental contexts, displaying radiation dynamics in an understandable manner, even when the underlying systems are quite complex.