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Chapter 12: Problem 12

During radiant heat treatment of a thin-film material, its shape, which may be hemispherical (a) or spherical (b), is maintained by a relatively low air pressure (as in the case of a rubber balloon). Irradiation on the film is due to emission from a radiant heater of area \(A_{\text {h }}=0.0052 \mathrm{~m}^{2}\), which emits diffusely with an intensity of \(I_{e, \mathrm{~h}}=169,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\). (a) Obtain an expression for the irradiation on the film as a function of the zenith angle \(\theta\). (b) Based on the expressions derived in part (a), which shape provides the more uniform irradiation \(G\) and hence provides better quality control for the treatment process?

Short Answer

Expert verified
We derived the expression for irradiation as a function of the zenith angle 胃 to be \(G(\theta) = A_{h} \cdot \frac{I_{e, h} \cdot \cos(\theta)}{\pi}\). Comparing the uniformity for hemispherical and spherical shapes, we found that the hemispherical shape provides more uniform irradiation, leading to better quality control for the treatment process.

Step by step solution

01

Consider the geometry of the problem

The geometry of the problem consists of a radiant heater that is emitting diffusely on the thin-film material, which can have a hemispherical or spherical shape. The heater has an area of \(A_{h} = 0.0052 \,\text{m}^2\) and emits with an intensity of \(I_{e, h} = 169,000 \, \text{W/m}^2\text{sr}\). The zenith angle 胃 is the angle between the normal to the surface and the line connecting the center of the heater to the point where the irradiation is being calculated.
02

Find the irradiance on the film

The irradiance on the film can be found using the formula: \[E(\theta) = \frac{I_{e, h} \cdot \cos(\theta)}{\pi}\] Where \(E(\theta)\) is the irradiance on the film as a function of the zenith angle 胃, and the \(\cos(\theta)\) term takes into account the angle between the normal to the surface and the line connecting the center of the heater to the point where the irradiation is being calculated.
03

Calculate the irradiation on the film

The irradiation on the film can be calculated as the product of the heater area, \(A_{h}\), and the irradiance on the film, \(E(\theta)\): \[G(\theta) = A_{h} \cdot E(\theta)\] Substituting the expression for \(E(\theta)\) from Step 2, we get: \[G(\theta) = A_{h} \cdot \frac{I_{e, h} \cdot \cos(\theta)}{\pi}\]
04

Analyze the uniformity of irradiation for each shape

Now that we have the expression for the irradiation on the film, we need to determine which shape, hemispherical or spherical, provides more uniform irradiation. For a hemispherical shape, the zenith angle 胃 varies between 0 and \(\frac{\pi}{2}\). For a spherical shape, 胃 varies between 0 and \(\pi\). Since the irradiation \(G(\theta)\) depends on the cosine of the zenith angle, we can observe that the irradiation will be relatively uniform for small values of 胃 and will decrease as the zenith angle increases. For the hemispherical shape, the irradiation will be relatively uniform because the zenith angle varies within a smaller range, while for the spherical shape, the irradiation will be less uniform due to larger variations of the zenith angle. Therefore, the hemispherical shape provides more uniform irradiation and hence better quality control for the treatment process.

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Key Concepts

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

Thin-film material
When we talk about a thin-film material in the context of radiant heat transfer, we're considering a material that has one dimension significantly smaller compared to the other two. This thinness allows for certain advantageous properties, like flexibility and quick heat response.
Thin-film materials can be quite versatile and are often used in applications where lightweight and conformability are crucial. In radiant heat treatment, these materials absorb energy from a heat source and distribute it across their surface.
Because of their structure, different shapes, such as hemispherical or spherical, can play a significant role in how evenly the material receives and spreads this energy.
Irradiation
Irradiation is the process of exposing a material to radiation, like light or heat. In our context, it refers specifically to the energy received by the thin-film from the radiant heater. This energy falls onto the material's surface and is typically measured in Watts per square meter (W/m虏).
Understanding irradiation is key to ensuring efficient heating and uniformity across the material. The precise amount of energy and the way it's distributed affect the material's treatment quality. It's important for manufacturing processes that the amount of irradiation aligns with desired outcomes, such as uniform hardening or drying.
The formula to determine irradiation involves the zenith angle, which tells us how direct the radiation hits the surface.
Zenith angle
The zenith angle is an important parameter in calculating how much irradiation a surface receives. It鈥檚 the angle between a line perpendicular to the surface and the line from the radiant source to a point on the surface.
In mathematical terms, it influences the calculation as it appears in the cosine function: \[ E(\theta) = \frac{I_{e, h} \cdot \cos(\theta)}{\pi} \]The zenith angle determines how directly the irradiation hits the surface: a smaller angle means more direct irradiation, resulting in greater intensity.
Understanding this concept is vital for optimizing the heating process, as different angles can lead to variations in energy distribution, impacting the uniformity of the treatment.
Hemispherical shape
The shape of the thin-film material plays a large role in heat distribution. A hemispherical shape suggests only half of the sphere is interacting with the radiation.
This shape influences the zenith angle and, consequently, the irradiation distribution across its surface. Compared to a complete sphere, a hemisphere has a smaller range of zenith angles鈥攆rom 0 to \(\frac{\pi}{2}\)鈥攚hich helps in maintaining more uniform irradiation.
  • The cosine of smaller zenith angles tends to be larger, improving uniformity.
  • This shape offers better quality control as it minimizes extreme variations in irradiation.
This makes the hemispherical shape preferable in processes requiring precise control, resulting in consistent treatment outcomes.

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Most popular questions from this chapter

A horizontal, opaque surface at a steady-state temperature of \(77^{\circ} \mathrm{C}\) is exposed to an airflow having a free stream temperature of \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissive power of the surface is \(628 \mathrm{~W} / \mathrm{m}^{2}\), the irradiation is \(1380 \mathrm{~W} / \mathrm{m}^{2}\), and the reflectivity is \(0.40\). Determine the absorptivity of the surface. Determine the net radiation heat transfer rate for this surface. Is this heat transfer to the surface or from the surface? Determine the combined heat transfer rate for the surface. Is this heat transfer to the surface or from the surface?

A horizontal semitransparent plate is uniformly irradiated from above and below, while air at \(T_{c}=300 \mathrm{~K}\) flows over the top and bottom surfaces, providing a uniform convection heat transfer coefficient of \(h=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The absorptivity of the plate to the irradiation is \(0.40\). Under steady-state conditions measurements made with a radiation detector above the top surface indicate a radiosity (which includes transmission, as well as reflection and emission) of \(J=5000 \mathrm{~W} / \mathrm{m}^{2}\), while the plate is at a uniform temperature of \(T=350 \mathrm{~K}\). Determine the irradiation \(G\) and the emissivity of the plate. Is the plate gray \((\varepsilon=\alpha)\) for the prescribed conditions?

The extremely high temperatures needed to trigger nuclear fusion are proposed to be generated by laserirradiating a spherical pellet of deuterium and tritium fuel of diameter \(D_{p}=1.8 \mathrm{~mm}\). (a) Determine the maximum fuel temperature that can be achieved by irradiating the pellet with 200 lasers, each producing a power of \(P=500 \mathrm{~W}\). The pellet has an absorptivity \(\alpha=0.3\) and emissivity \(\varepsilon=0.8\). (b) The pellet is placed inside a cylindrical enclosure. Two laser entrance holes are located at either end of the enclosure and have a diameter of \(D_{\mathrm{LEH}}=2 \mathrm{~mm}\). Determine the maximum temperature that can be generated within the enclosure.

Growers use giant fans to prevent grapes from freezing when the effective sky temperature is low. The grape, which may be viewed as a thin skin of negligible thermal resistance enclosing a volume of sugar water, is exposed to ambient air and is irradiated from the sky above and ground below. Assume the grape to be an isothermal sphere of \(15-\mathrm{mm}\) diameter, and assume uniform blackbody irradiation over its top and bottom hemispheres due to emission from the sky and the earth, respectively. (a) Derive an expression for the rate of change of the grape temperature. Express your result in terms of a convection coefficient and appropriate temperatures and radiative quantities. (b) Under conditions for which \(T_{\text {sky }}=235 \mathrm{~K}, T_{\mathrm{s}}=\) \(273 \mathrm{~K}\), and the fan is off \((V=0)\), determine whether the grapes will freeze. To a good approximation, the skin emissivity is 1 and the grape thermophysical properties are those of sugarless water. However, because of the sugar content, the grape freezes at \(-5^{\circ} \mathrm{C}\). (c) With all conditions remaining the same, except that the fans are now operating with \(V=1 \mathrm{~m} / \mathrm{s}\), will the grapes freeze?

A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.
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