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Chapter 13: Problem 27

To enhance heat rejection from a spacecraft, an engineer proposes to attach an array of rectangular fins to the outer surface of the spacecraft and to coat all surfaces with a material that approximates blackbody behavior. Consider the U-shaped region between adjoining fins and subdivide the surface into components associated with the base (1) and the side (2). Obtain an expression for the rate per unit length at which radiation is transferred from the surfaces to deep space, which may be approximated as a blackbody at absolute zero temperature. The fins and the base may be assumed to be isothermal at a temperature \(T\). Comment on your result. Does the engineer's proposal have merit?

Short Answer

Expert verified
In conclusion, attaching rectangular fins coated with a material that approximates blackbody behavior to a spacecraft can increase the rate of radiation transfer to deep space, with the total rate per unit length being \(Q_\text{total} = \sigma \cdot T^4 \cdot (A_1 + A_2)\). While the proposal has merit, it is important to consider other factors such as materials and manufacturing processes, as well as other heat transfer mechanisms like conduction and convection, for a comprehensive evaluation of its effectiveness.

Step by step solution

01

Understand blackbody radiation and emissive power

A blackbody is an idealized object that absorbs all incident electromagnetic radiation while maintaining thermal equilibrium. It also emits electromagnetic radiation at a rate proportional to the fourth power of its temperature, known as the Stefan-Boltzmann law. The emissive power (E) of a blackbody is given by: \(E = \sigma \cdot T^4\) where \(\sigma\) is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} \mathrm{W/(m^{2}K^{4})}\)) and T is the temperature of the blackbody in Kelvin. Since the fin surfaces are coated with a material that approximates blackbody behavior, we can assume that they radiate with the same emissive power.
02

Calculate radiation transfer from base

To calculate the rate of radiation transfer from the base surface (1), we can use the emissive power formula for a blackbody: \(E_1 = \sigma \cdot T^4\) The rate of radiation transfer per unit length from the base surface is then given by: \(Q_1 = E_1 \cdot A_1\) where \(A_1\) is the area of the base surface per unit length.
03

Calculate radiation transfer from side

Similarly, to calculate the rate of radiation transfer from the side surface (2), we can use the emissive power formula for a blackbody: \(E_2 = \sigma \cdot T^4\) The rate of radiation transfer per unit length from the side surface is then given by: \(Q_2 = E_2 \cdot A_2\) where \(A_2\) is the area of the side surface per unit length.
04

Sum the radiation transfer from all surfaces to find total rate

Now, to find the total radiation transfer per unit length from both surfaces, we can sum the rates found in steps 2 and 3: \(Q_\text{total} = Q_1 + Q_2\) Since the emissive power for both surfaces is the same, we can rewrite the equation as: \(Q_\text{total} = \sigma \cdot T^4 \cdot (A_1 + A_2)\)
05

Analyze the result and discuss the merit of the proposal

The total rate of radiation transfer per unit length (\(Q_\text{total}\)) depends on the sum of the areas of the base and side surfaces. The rate of radiation transfer is proportional to the surface area, so if the fin design can significantly increase the overall surface area, it will improve the rate at which heat is rejected from the spacecraft. However, some other factors need to be considered, such as the materials and manufacturing processes involved in attaching fins to the spacecraft. Also, the radiation transfer to deep space is only one aspect of heat rejection, and other heat transfer mechanisms like conduction and convection might be affected or have more impact in certain scenarios. In conclusion, the engineer's proposal has merit in increasing the rate of radiation transfer to deep space, but other factors need to be considered for a comprehensive evaluation of its effectiveness.

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the energy emitted by a perfect "blackbody" 鈥� an idealized physical entity that absorbs all incoming radiation without reflecting any. Such entities are in thermal equilibrium and the radiation they emit is determined solely by their temperature. This concept is vital for spacecraft, where efficient heat management is crucial.
Blackbodies can serve as a standard in understanding real-world objects. Although no perfect blackbody exists, certain materials can approximate this behavior. When applied to spacecraft surfaces, these materials can maximize heat emission into the cold vastness of space, crucial for maintaining operational temperatures.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law helps quantify the power radiated by a blackbody. It states that the emissive power (E) is proportional to the fourth power of the blackbody's temperature. It can be mathematically expressed as:
\[ E = \sigma T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant (approximately \( 5.67 \times 10^{-8} \ \mathrm{W/(m^{2}K^{4})} \)) and \( T \) is the temperature in Kelvin.
For spacecraft engineers, understanding this relationship is crucial. It implies that as the temperature of a surface increases, the energy it radiates grows dramatically. This principle guides the design of thermal management solutions, ensuring that spacecraft can effectively shed excess heat via radiation.
Radiative Heat Transfer
Radiative heat transfer within a spacecraft involves energy emission due to thermal radiation, primarily following the Stefan-Boltzmann Law. This form of heat transfer is especially pertinent in space, where traditional conduction and convection are limited.
To enhance radiative heat transfer, spacecraft can use surfaces coated with materials resembling blackbodies, which maximizes their emissive power. By improving the emissive surfaces, engineers work to facilitate the efficient transfer of heat from the spacecraft to the cold vacuum of space, an essential factor given the absence of an atmosphere to aid in heat dissipation.
Spacecraft Thermal Management
Spacecraft thermal management entails the regulation of heat within a spacecraft to ensure all components operate within their designated temperature ranges. Given the extreme conditions of space, managing heat through radiation is essential.
Several strategies are employed, including designing surfaces with high emissivity to maximize radiative heat loss. Using fins or other extensions is one method to increase surface area, thus enhancing heat rejection capabilities. Any efficient thermal management system must consider the implications of increased surface areas, material properties, and the specific mission environment.
Successful management requires a balance between retaining necessary heat for operations and preventing overheating, thus these thermal systems are a critical aspect of spacecraft design.

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Most popular questions from this chapter

A double-glazed window consists of two panes of glass, each of thickness \(I=6 \mathrm{~mm}\). The inside room temperature is \(T_{t}=20^{\circ} \mathrm{C}\) with \(h_{i}=7.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the outside temperature is \(T_{o}=-10^{\circ} \mathrm{C}\) with \(h_{a}=25 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The gap between the glass sheets is of thickness \(L=5 \mathrm{~mm}\) and is filled with a gas. The glass surfaces may be treated with a low-emissivity coating to reduce their emissivity from \(\varepsilon=0.95\) to \(\varepsilon=0.05\). Determine the heat flux through the window for case 1: \(\varepsilon_{1}=\varepsilon_{2}=0.95\), case \(2: \varepsilon_{1}=\varepsilon_{2}=0.05\), and case 3: \(\varepsilon_{\mathrm{L}}=0.05, \varepsilon_{2}=0.95\). Consider either air of argon of thermal conductivity \(k_{A R}=17.7 \times 10^{-3}\) W/m. \(\mathrm{K}\) to be within the gap. Radiation heat transfer occurring at the external surfaces of the two glass sheets is negligible, as is free convection between the glass sheets.

A radiant oven for drying newsprint consists of a long duct \((L=20 \mathrm{~m})\) of semicincular cross section. The newsprint moves through the oven on a conveyor belt at a velocity of \(V=0.2 \mathrm{~m} / \mathrm{s}\). The newsprint has a water content of \(0.02 \mathrm{~kg} / \mathrm{m}^{2}\) as it enters the oven and is completely dry as it exits. To assure quality, the newsprint must be maintained at room temperature \((300 \mathrm{~K})\) during drying. To aid in maintaining this condition, all system components and the air flowing through the oven have a temperature of \(300 \mathrm{~K}\). The inner sarface of the semicaircular duct, which is of emissivity \(0.8\) and temperature \(T_{1}\), provides the radiant heat required to accomplish the drying. The wet surface of the newsprint can be considered to be black. Air entering the oven has a temperature of \(300 \mathrm{~K}\) and a relative humidity of \(20 \%\). Since the velocity of the air is large, its temperature and relative humidity can be assumed to be constant over the entire duct length. Calculate the required evaporation rate, air velocity \(u_{m}\), and temperature \(T_{1}\) that will ensure steady-state conditions for the process.

Boiler tubes exposed to the products of coal combustion in a power plant are subject to fouling by the ash (mineral) content of the combustion gas. The ash forms a solid deposit on the tube outer surface, which reduces heat transfer to a pressurized wated'steam mixture flowing through the tubes. Consider a thin-walled boiler tube \(\left(D_{t}=0.05 \mathrm{~m}\right)\) whose surface is maintained at \(T_{t}=600 \mathrm{~K}\) by the boiling process. Combustion gases flowing over the tube at \(T_{-}=1800 \mathrm{~K}\) provide a convection coefficient of \(\bar{h}=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), while radiation from the gas and boiler walls to the tube may be approximated as that originating from large surroundings at \(T_{\text {arr }}=1500 \mathrm{~K}\). (a) If the tube surface is diffuse and gray, with \(\varepsilon_{t}=0.8\), and there is no ash deposit layer, what is the rate of heat transfer per unit length, \(q^{\prime}\), to the boiler tube? (b) If a deposit layer of diameter \(D_{d}=0.06 \mathrm{~m}\) and thermal conductivity \(k=1 \mathrm{~W} / \mathrm{m}\), \(\mathrm{K}\) forms on the tube, what is the deposit surface temperature, \(T_{d} ?\) The depos茂t is diffuse and gray, with \(\varepsilon_{a}=0.9\), and \(T_{m} T_{\ldots}, T_{\text {}}\), and \(T_{\text {sar }}\) remain unchanged. What is the net rate of heat transfer per wnit length, \(y^{\prime}\) ', to the boxler tube? (c) Explore the effect of variations in \(D_{d}\) and \(\bar{h}\) on \(q^{r}\), as well as on relative contributions of convection and radiation to the net heat transfer fate. Represent your results graphically.

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

A long uniform rod of \(50-\mathrm{mm}\) diameter with a thermal conductivity of \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is heated internally by volumetric energy generation of \(20 \mathrm{~kW} / \mathrm{m}^{3}\). The rod is positioned coaxially within a larger circular tube of \(60-\mathrm{mm}\) diameter whose surface is maintained at \(500^{\circ} \mathrm{C}\). The annular region between the rod and the tube is evacuated, and their surfaces are diffuse and gray with an emissivity of \(0.2\). (a) Determine the center and surface temperatures of the rod. (b) Determine the center and surface temperatures of the rod if atmospheric air occupies the annular space. (c) For tube diameters of 60,100 , and \(1000 \mathrm{~mm}\) and for both the evacuated and atmospheric conditions, compute and plot the center and surface temperatures as a function of equivalent surface emissivities in the range from \(0.1\) to \(1.0\).
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