/*! 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 52 An opaque surface, \(2 \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The absorbed irradiation is 8160 W, the emissive power is 72490.88 W, the radiosity is 73930.88 W, and the net radiation heat transfer from the surface is 65770.88 W.

Step by step solution

01

Calculate absorbed solar irradiation for each wavelength range

To calculate the absorbed solar irradiation for each wavelength range, we need to take the product of the spectral absorptivity and the given solar irradiance value. We then multiply the result by the surface area. For \(0 \leq \lambda \leq 0.5 \mu \mathrm{m}\), \(\alpha_{\lambda} = 0\). Absorbed irradiation, \(Q_{a_1} = \alpha_{\lambda} \times G_{S} \times A = 0 \times 1200 \times 4 = 0\, \mathrm{W}\). For \(0.5 \mu \mathrm{m} < \lambda \leq 1 \mu \mathrm{m}\), \(\alpha_{\lambda} = 0.8\). Absorbed irradiation, \(Q_{a_2} = \alpha_{\lambda} \times G_{S} \times A = 0.8 \times 1200 \times 4 = 3840\, \mathrm{W}\). For \(1 \mu \mathrm{m} < \lambda \leq 2 \mu \mathrm{m}\), \(\alpha_{\lambda} = 0\). Absorbed irradiation, \(Q_{a_3} = \alpha_{\lambda} \times G_{S} \times A = 0 \times 1200 \times 4 = 0\, \mathrm{W}\). For \(\lambda > 2 \mu \mathrm{m}\), \(\alpha_{\lambda} = 0.9\). Absorbed irradiation, \(Q_{a_4} = \alpha_{\lambda} \times G_{S} \times A = 0.9 \times 1200 \times 4 = 4320\, \mathrm{W}\).
02

Calculate total absorbed irradiation from the sun

To find the total absorbed irradiation from the sun, we add up all the absorbed irradiation values calculated in Step 1. Total absorbed irradiation, \(Q_{a_{total}} = Q_{a_1} + Q_{a_2} + Q_{a_3} + Q_{a_4} = 0 + 3840 + 0 + 4320 = 8160\, \mathrm{W}\).
03

Calculate the emissive power

Using the Stefan-Boltzmann law, we can calculate the surface's emissive power (\(E_b\)). The law states that \(E_b = \sigma T^4\), where \(\sigma\) is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m²K^4), and \(T\) is the temperature of the surface in Kelvin. Emissive power, \(E_b = \sigma T^4 = (5.67 \times 10^{-8}) \times (400)^4 = 18122.72\, \mathrm{W/m^2}\). To find the total emissive power, we multiply the emissive power by the surface area. Total emissive power, \(E_{b_{total}} = E_b \times A = 18122.72 \times 4 = 72490.88\, \mathrm{W}\).
04

Calculate radiosity

Radiosity (\(J\)) is the sum of the emissive power and the reflected irradiation. In this problem, the surface is considered diffuse, and so its reflectivity is equal to its spectral absorptivity at each wavelength range. We can calculate the average reflectivity (\(\rho_{avg}\)) by adding the reflectivities at each range and dividing by 4. Average reflectivity, \(\rho_{avg} = \frac{0 + 0.2 + 0 + 0.1}{4} = 0.075\). Reflected irradiation (at each range) is equal to \(\rho_{avg} \times G_{S} \times A\). Summing those, we get total reflected irradiation: Total reflected irradiation, \(Q_{r_{total}} = 4 \rho_{avg} G_{S} A = 4 \times 0.075 \times 1200 \times 4 = 1440\, \mathrm{W}\). Finally, we can calculate the radiosity (\(J_{total}\)): Total radiosity, \(J_{total} = E_{b_{total}} + Q_{r_{total}} = 72490.88 + 1440 = 73930.88\, \mathrm{W}\).
05

Calculate net radiation heat transfer from the surface

To find the net radiation heat transfer from the surface, we can subtract the total absorbed irradiation from the total radiosity: Net radiation heat transfer, \(Q_{net} = J_{total} - Q_{a_{total}} = 73930.88 - 8160 = 65770.88\, \mathrm{W}\). In summary, we found the following values: - Total absorbed irradiation: 8160 W - Total emissive power: 72490.88 W - Total radiosity: 73930.88 W - Net radiation heat transfer from the surface: 65770.88 W

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Key Concepts

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

Spectral Absorptivity
The concept of spectral absorptivity is essential in understanding how materials absorb radiation at different wavelengths. Spectral absorptivity, denoted as \(\alpha_{\lambda}\), quantifies the fraction of incident radiation absorbed by a surface at a particular wavelength. It's a crucial parameter to determine the energy absorbed by the surface when exposed to solar radiation.
In the provided example, spectral absorptivity varies across different ranges of wavelengths. For instance, in the range of \(0 \leq \lambda \leq 0.5 \mu\text{m}\), the surface does not absorb any solar radiation because \(\alpha_{\lambda}\) equals zero. However, in the range from \(0.5 \mu\text{m} < \lambda \leq 1 \mu\text{m}\), \(\alpha_{\lambda} = 0.8\), meaning that 80% of the incident radiation is absorbed. Understanding these variations helps us determine the total absorbed irradiation by summing contributions from all spectral ranges.
In practical terms, engineers use this information to design materials and coatings optimized for specific thermal applications, enabling better control of heat flows in systems.
Emissive Power
Emissive power is a measure of the energy emitted by a surface due to its temperature. It's described by the Stefan-Boltzmann law, which states that the emissive power \(E_b\) of a black body is proportional to the fourth power of its absolute temperature, i.e., \(E_b = \sigma T^4\), where \(\sigma\) is the Stefan-Boltzmann constant.
For the example surface, which is at a temperature of 400 K, we calculate the emissive power to be approximately 18122.72 W/m². Multiplying this by the surface area gives the total emissive power. This output is significant because it represents the thermal radiation energy emitted by the surface due to its temperature.
Understanding emissive power is crucial in systems where heat management is important, such as in radiators, solar panels, and building materials. Knowing the emissive power helps in determining whether the surface is likely to be a source or sink of thermal energy, influencing design and operational decisions.
Radiosity
Radiosity is the total radiation leaving a surface, combining both emitted and reflected energy. In heat transfer problems, it's denoted as \(J\), and accounts for both the emissive power and any reflected ambient radiation.
To calculate radiosity, we consider both the emissive power and the reflected component from the incoming radiation. Reflectivity, the fraction of incident radiation reflected, is used to determine the latter. Since the surface is diffuse, its average reflectivity was calculated over its spectral range and then applied to quantify total reflected irradiation.
In our scenario, radiosity not only includes the emissive power calculated earlier but also adds the contribution of reflected radiation, leading to a substantial value of 73930.88 W. This concept helps in understanding how much energy is leaving the surface, which is vital for thermal regulation, energy balance calculations, and simulation of environmental conditions.
Net Radiation Heat Transfer
Net radiation heat transfer describes the amount of thermal energy exchanged between a surface and its surroundings, defined as the difference between the radiosity and absorbed irradiation. It reflects the net gain or loss of energy, influencing temperature changes and thermal management strategies.
In our example, the net radiation heat transfer from the surface is determined by subtracting the total absorbed irradiation from radiosity. This calculation helps illustrate whether the surface is net radiating or absorbing energy, which is essential for thermal design and efficiency evaluations.
Understanding net radiation heat transfer has practical implications in industries such as aerospace, where spacecraft surfaces' thermal control is crucial, or in building design, affecting heating and cooling loads. Efficient management of net radiation heat transfer can lead to significant energy conservation and improved system performance.

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Most popular questions from this chapter

A procedure for measuring the thermal conductivity of solids at elevated temperatures involves placement of a sample at the bottom of a large furnace. The sample is of thickness \(L\) and is placed in a square container of width \(W\) on a side. The sides are well insulated. The walls of the cavity are maintained at \(T_{w}\), while the bottom surface of the sample is maintained at a much lower temperature \(T_{e}\) by circulating coolant through the sample container. The sample surface is diffuse and gray with an emissivity \(\varepsilon_{s}\). Its temperature \(T_{s}\) is measured optically. (a) Neglecting convection effects, obtain an expression from which the sample thermal conductivity may be evaluated in terms of measured and known quantities \(\left(T_{w}, T_{s}, T_{c}, \varepsilon_{s}, L\right)\). The measurements are made under steady-state conditions. If \(T_{w}=1400 \mathrm{~K}, T_{s}=1000 \mathrm{~K}, \varepsilon_{s}=0.85, L=\) \(0.015 \mathrm{~m}\), and \(T_{c}=300 \mathrm{~K}\), what is the sample thermal conductivity? (b) If \(W=0.10 \mathrm{~m}\) and the coolant is water with a flow rate of \(\dot{m}_{c}=0.1 \mathrm{~kg} / \mathrm{s}\), is it reasonable to assume a uniform bottom surface temperature \(T_{c}\) ?

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.

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.

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.

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?

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