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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.

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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.

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Most popular questions from this chapter

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.

An opaque surface, \(2 \mathrm{~m} \times 2 \mathrm{~m}\), is maintained at \(400 \mathrm{~K}\) and is simultaneously exposed to solar irradiation with \(G_{S}=1200 \mathrm{~W} / \mathrm{m}^{2}\). The surface is diffuse and its spectral absorptivity is \(\alpha_{\lambda}=0,0.8,0\), and \(0.9\) for \(0 \leq \lambda \leq\) \(0.5 \mu \mathrm{m}, 0.5 \mu \mathrm{m}<\lambda \leq 1 \mu \mathrm{m}, 1 \mu \mathrm{m}<\lambda \leq 2 \mu \mathrm{m}\), and \(\lambda>2 \mu \mathrm{m}\), respectively. Determine the absorbed irradiation, emissive power, radiosity, and net radiation heat transfer from the surface.

A radiation thermometer is a device that responds to a radiant flux within a prescribed spectral interval and is calibrated to indicate the temperature of a blackbody that produces the same flux. (a) When viewing a surface at an elevated temperature \(T_{s}\) and emissivity less than unity, the thermometer will indicate an apparent temperature referred to as the brightness or spectral radiance temperature \(T_{\lambda}\). Will \(T_{\lambda}\) be greater than, less than, or equal to \(T_{s}\) ? (b) Write an expression for the spectral emissive power of the surface in terms of Wien's spectral distribution (see Problem 12.27) and the spectral emissivity of the surface. Write the equivalent expression using the spectral radiance temperature of the surface and show that $$ \frac{1}{T_{x}}=\frac{1}{T_{\lambda}}+\frac{\lambda}{C_{2}} \ln \varepsilon_{\lambda} $$ where \(\lambda\) represents the wavelength at which the thermometer operates. (c) Consider a radiation thermometer that responds to a spectral flux centered about the wavelength \(0.65 \mu \mathrm{m}\). What temperature will the thermometer indicate when viewing a surface with \(\varepsilon_{\lambda}(0.65 \mu \mathrm{m})=0.9\) and \(T_{x}=1000 \mathrm{~K}\) ? Verify that Wien's spectral distribution is a reasonable approximation to Planck's law for this situation.

Four diffuse surfaces having the spectral characteristics shown are at \(300 \mathrm{~K}\) and are exposed to solar radiation. Which of the surfaces may be approximated as being gray?

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at \(2500 \mathrm{~K}\), a heated metal at \(1500 \mathrm{~K}\), human skin at \(305 \mathrm{~K}\), and a cryogenically cooled metal surface at \(60 \mathrm{~K}\). Estimate the fraction of the solar emission that is in the following spectral regions: the ultraviolet, the visible, and the infrared.

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