/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 During radiant heat treatment of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solar flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\) is incident on the top side of a plate whose surface has a solar absorptivity of \(0.9\) and an emissivity of \(0.1\). The air and surroundings are at \(17^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the plate and air is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the bottom side of the plate is insulated, determine the steady-state temperature of the plate.

Consider the metallic surface of Example 12.7. Additional measurements of the spectral, hemispherical emissivity yield a spectral distribution which may be approximated as follows: (a) Determine corresponding values of the total, hemispherical emissivity \(\varepsilon\) and the total emissive power \(E\) at \(2000 \mathrm{~K}\). (b) Plot the emissivity as a function of temperature for \(500 \leq T \leq 3000 \mathrm{~K}\). Explain the variation.

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?

A radiator on a proposed satellite solar power station must dissipate heat being generated within the satellite by radiating it into space. The radiator surface has a solar absorptivity of \(0.5\) and an emissivity of \(0.95\). What is the equilibrium surface temperature when the solar irradiation is \(1000 \mathrm{~W} / \mathrm{m}^{2}\) and the required heat dissipation is \(1500 \mathrm{~W} / \mathrm{m}^{2}\) ?

Approximations to Planck's law for the spectral emissive power are the Wien and Rayleigh-Jeans spectral distributions, which are useful for the extreme low and high limits of the product \(\lambda T\), respectively. (a) Show that the Planck distribution will have the form $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{\lambda^{5}} \exp \left(-\frac{C_{2}}{\lambda T}\right) $$ when \(C_{2} / \lambda T\) sor 1 and determine the error (compared to the exact distribution) for the condition \(\lambda T=\) \(2898 \mu \mathrm{m} \cdot \mathrm{K}\). This form is known as Wien's law. (b) Show that the Planck distribution will have the form $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}} $$ when \(C_{2} / \lambda T \& 1\) and determine the error (compared to the exact distribution) for the condition \(\lambda T=100,000 \mu \mathrm{m} \cdot \mathrm{K}\). This form is known as the Rayleigh-Jeans law.
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Engineering Physics

Read Explanation

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Astrophysics

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.