/*! 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 2 A horizontal, opaque surface at ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
The absorptivity of the surface is 0.60. The net radiation heat transfer rate is 200 W/m², and heat transfer is to the surface. The combined heat transfer rate is 1600 W/m², and the heat transfer is also to the surface.

Step by step solution

01

Calculate absorptivity

We have the reflectivity, so we can use the formula to find the absorptivity: Absorptivity = 1 - Reflectivity = 1 - 0.40 = 0.60 The absorptivity of the surface is 0.60.
02

Calculate net radiation heat transfer rate

Now we'll calculate the net irradiation on the surface: Net irradiation = (1 - Reflectivity) * Irradiation - Emissive power = (1 - 0.40) * 1380 W/m² - 628 W/m² = 0.60 * 1380 W/m² - 628 W/m² = 828 W/m² - 628 W/m² = 200 W/m² The net radiation heat transfer rate is 200 W/m². Since the value is positive, the heat transfer is to the surface.
03

Calculate convective heat transfer rate

We will use the formula for convective heat transfer rate and the given surface temperature and airflow temperature: Convective heat transfer rate = h * A * (T_surface - T_airflow) = 28 W/m²K * (77℃ - 27℃) = 28 W/m²K * 50K = 1400 W/m² The convective heat transfer rate is 1400 W/m².
04

Calculate combined heat transfer rate

Now we will combine the net radiation heat transfer rate and the convective heat transfer rate: Combined heat transfer rate = Net radiation heat transfer rate + Convective heat transfer rate = 200 W/m² + 1400 W/m² = 1600 W/m² The combined heat transfer rate is 1600 W/m². Since the value is positive, the heat transfer is to the surface.

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 91Ó°ÊÓ!

Key Concepts

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

Convection Heat Transfer Coefficient
Understanding the convection heat transfer coefficient is crucial for evaluating how effectively heat is being transferred from a surface to a fluid moving across it. This coefficient, represented by 'h', is a measure of the convection heat transfer rate per unit area and per degree of temperature difference between the fluid and the surface.

The formula to calculate the heat transfer due to convection is given by
\[ Q_{conv} = h \cdot A \cdot (T_{surface} - T_{fluid}) \]
where \( Q_{conv} \) is the convective heat transfer rate, 'h' is the convective heat transfer coefficient, 'A' is the surface area, \( T_{surface} \) is the temperature of the surface, and \( T_{fluid} \) is the temperature of the fluid. The higher the coefficient, the more efficient the process of heat transfer by convection.
Thermal Radiation
Thermal radiation is a form of heat transfer that occurs through the emission of electromagnetic waves. This process does not require a medium; it can occur in a vacuum, making it distinct from other forms of heat transfer like conduction and convection.

Objects at a temperature above absolute zero emit radiation across a spectrum of wavelengths, with a greater proportion of energy in specific ranges depending on their temperature. The radiative heat transfer from a surface is calculated by using its emissive power, which is dependent on its temperature and emissivity. The emissive power is given by the Stefan-Boltzmann law: \[ E = \epsilon \cdot \sigma \cdot T^4 \]
where \( E \) is the emissive power, \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature of the surface in Kelvin.
Net Radiation Heat Transfer Rate
The net radiation heat transfer rate is essentially the balance of all radiant heat transfer into and out of a surface. Determining this rate is important in understanding whether a surface is gaining or losing heat through radiation.

To calculate this, you need to account for the emissive power of the surface and the irradiation, which is the radiant energy received by the surface. The formula looks like this:\[ Q_{net, rad} = (1 - \rho) \cdot G - E \]
where \( Q_{net, rad} \) represents the net radiation heat transfer rate, \( \rho \) is the reflectivity, \( G \) is the irradiation, and \( E \) is the surface's emissive power. When the result is positive, it indicates that the net effect is an absorption of heat by the surface, whereas a negative value would signify a net loss of heat.
Surface Absorptivity
Surface absorptivity refers to the ability of a material to absorb radiation that falls onto it. This characteristic is crucial in thermal radiation calculations since it influences how much radiation is taken in by the surface versus how much is reflected or transmitted.

Absorptivity, denoted by \( \alpha \), is a number between 0 and 1 and is typically related to the surface's reflectivity (\( \rho \)) and transmissivity through the relation:\[ \alpha + \rho + \tau = 1 \]
where \( \tau \) stands for transmissivity. For opaque surfaces like the one in our exercise, transmissivity is zero, and the relationship simplifies to \( \alpha = 1 - \rho \). It is a key factor in determining the amount of thermal energy absorbed by the surface, which influences the overall heat transfer process.

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

After being cut from a large single-crystal boule and polished, silicon wafers undergo a high-temperature annealing process. One technique for heating the wafer is to irradiate its top surface using highintensity, tungsten-halogen lamps having a spectral distribution approximating that of a blackbody at \(2800 \mathrm{~K}\). To determine the lamp power and the rate at which radiation is absorbed by the wafer, the equipment designer needs to know its absorptivity as a function of temperature. Silicon is a semiconductor material that exhibits a characteristic band edge, and its spectral absorptivity may be idealized as shown schematically. At low and moderate temperatures, silicon is semitransparent at wavelengths larger than that of the band edge, but becomes nearly opaque above \(600^{\circ} \mathrm{C}\). (a) What are the \(1 \%\) limits of the spectral band that includes \(98 \%\) of the blackbody radiation corresponding to the spectral distribution of the lamps? Over what spectral region do you need to know the spectral absorptivity? (b) How do you expect the total absorptivity of silicon to vary as a function of its temperature? Sketch the variation and explain its key features. (c) Calculate the total absorptivity of the silicon wafer for the lamp irradiation and each of the five temperatures shown schematically. From the data, calculate the emissivity of the wafer at 600 and \(900^{\circ} \mathrm{C}\). Explain your results and why the emissivity changes with temperature. Hint: Within IHT, create a look-up table to specify values of the spectral properties and the LOOKUPVAL and INTEGRAL functions to perform the necessary integrations. (d) If the wafer is in a vacuum and radiation exchange only occurs at one face, what is the irradiation needed to maintain a wafer temperature of \(600^{\circ} \mathrm{C}\) ?

A diffuse, opaque surface at \(700 \mathrm{~K}\) has spectral emissivities of \(\varepsilon_{\lambda}=0\) for \(0 \leq \lambda \leq 3 \mu \mathrm{m}, \varepsilon_{\lambda}=0.5\) for \(3 \mu \mathrm{m}<\lambda \leq 10 \mu \mathrm{m}\), and \(\varepsilon_{\lambda}=0.9\) for \(10 \mu \mathrm{m}<\lambda<\infty\). A radiant flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), which is uniformly distributed between 1 and \(6 \mu \mathrm{m}\), is incident on the surface at an angle of \(30^{\circ}\) relative to the surface normal. Calculate the total radiant power from a \(10^{-4} \mathrm{~m}^{2}\) area of the surface that reaches a radiation detector positioned along the normal to the area. The aperture of the detector is \(10^{-5} \mathrm{~m}^{2}\), and its distance from the surface is \(1 \mathrm{~m}\).

Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front side is exposed to ambient air that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), solar irradiation of \(600 \mathrm{~W} / \mathrm{m}^{2}\), and an effective sky temperature of \(-40^{\circ} \mathrm{C}\). What is the electrical power \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) required to maintain the plate surface temperature at \(T_{s}=60^{\circ} \mathrm{C}\) if the plate is diffuse and has the designated spectral, hemispherical reflectivity?

It is not uncommon for the night sky temperature in desert regions to drop to \(-40^{\circ} \mathrm{C}\). If the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefticient for still air conditions is approximately \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), can a shallow pan of water freeze?

The directional total emissivity of nonmetallic materials may be approximated as \(\varepsilon_{\theta}=\varepsilon_{n} \cos \theta\), where \(\varepsilon_{n}\) is the normal emissivity. Show that the total hemispherical emissivity for such materials is \(2 / 3\) of the normal emissivity.

See all solutions

Recommended explanations on Physics Textbooks

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.