/*! 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 132 A spherical satellite in near-ea... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 132

A spherical satellite in near-earth orbit is exposed to solar irradiation of \(1368 \mathrm{~W} / \mathrm{m}^{2}\). To maintain a desired operating temperature, the themal control engineer intends to use a checker pattern for which a fraction \(F\) of the satellite surface is coated with an evaporated aluminum film \(\left(\varepsilon=0.03, \alpha_{S}=0.09\right)\), and the fraction \((1-F)\) is coated with a white, zinc-oxide paint \(\left(\varepsilon=0.85, \quad \alpha_{5}=0.22\right)\). Assume the satellite is isothermal and has no internal power dissipation. Determine the fraction \(F\) of the checker pattern required to maintain the satellite at \(300 \mathrm{~K}\).

Short Answer

Expert verified
The fraction of the satellite surface required to be covered with evaporated aluminum film to maintain the satellite at 300 K is approximately 0.318 or 31.8%.

Step by step solution

01

Calculate the incoming solar energy

The solar energy absorbed by the satellite depends on its absorptivity. Let's consider the first part with the aluminum film coating. The absorbed solar energy per unit area can be calculated as follows: \(q_{inA} = \alpha_{SA} * I\) Where: - \(q_{inA}\) is incoming solar energy per unit area for the aluminum film, - \(\alpha_{SA} = 0.09\) is the solar absorptivity of the aluminum film, - I = 1368 W/m虏 is the solar irradiation. Let's now consider the second part with the white zinc-oxide paint coating. The absorbed solar energy per unit area can be calculated as follows: \(q_{inZ} = \alpha_{SZ} * I\) Where: - \(q_{inZ}\) is incoming solar energy per unit area for the white zinc-oxide, - \(\alpha_{SZ} = 0.22\) is the solar absorptivity of the zinc-oxide paint.
02

Calculate the outgoing thermal energy

The satellite conducts heat due to its temperature, which is given as 300 K. The satellite emits thermal energy according to the fractions of coated surfaces and their respective emissivities. For the aluminum film coating, the emitted thermal energy per unit area can be calculated as follows: \(q_{outA} = \varepsilon_A * \sigma * T^4\) Where: - \(q_{outA}\) is the outgoing thermal energy per unit area for the aluminum film, - \(\varepsilon_A = 0.03\) is the emissivity of the aluminum film, - \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\) is the Stefan-Boltzmann constant, - T = 300 K is the temperature of the satellite. For the white zinc-oxide paint coating, the emitted thermal energy per unit area can be calculated as follows: \(q_{outZ} = \varepsilon_Z * \sigma * T^4\) Where: - \(q_{outZ}\) is the outgoing thermal energy per unit area for the zinc-oxide, - \(\varepsilon_Z = 0.85\) is the emissivity of the zinc-oxide paint.
03

Set up the energy balance equation

According to the energy balance, Incoming Solar Energy = Outgoing Thermal Energy. So, we will set up the following equation: \(F * q_{inA} + (1-F) * q_{inZ} = F * q_{outA} + (1-F) * q_{outZ}\) Now, we can plug in the values calculated in Step 1 and 2: \(F * \alpha_{SA} * I + (1-F) * \alpha_{SZ} * I = F * \varepsilon_A * \sigma * T^4 + (1-F) * \varepsilon_Z * \sigma * T^4\)
04

Solve for F

We can now solve for the fraction F: \(F(\alpha_{SA} * I - \varepsilon_A * \sigma * T^4) + \alpha_{SZ} * I - \varepsilon_Z * \sigma * T^4 = 0\) Now, divide the entire equation by I, and isolate F: \(F = \frac{\varepsilon_Z * \sigma * T^4 - \alpha_{SZ} * I}{\alpha_{SA} * I - \varepsilon_A * \sigma * T^4 - \alpha_{SZ} * I + \varepsilon_Z * \sigma * T^4}\) Now, substitute the given values: \(F = \frac{0.85 * 5.67 \times 10^{-8} * 300^4 - 0.22 * 1368}{0.09 * 1368 - 0.03 * 5.67 \times 10^{-8} * 300^4 - 0.22 * 1368 + 0.85 * 5.67 \times 10^{-8} * 300^4}\) After calculating the equation, we get: \(F \approx 0.318\) Hence, the fraction of the satellite surface required to be covered with evaporated aluminum film to maintain the satellite at 300 K is approximately 0.318 or 31.8%.

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Key Concepts

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

Solar Absorptivity
Solar absorptivity is a crucial property in managing the thermal characteristics of a satellite. It indicates how much solar radiation a material can absorb. High absorptivity means the surface will absorb more solar energy, increasing the satellite's temperature.
In our exercise, two coating materials are used, each with different levels of solar absorptivity. The aluminum film has a low absorptivity of 0.09, meaning it absorbs only 9% of the solar radiation, making it a poor heat absorber. On the other hand, the white zinc-oxide paint has a higher absorptivity of 0.22, absorbing 22% of the incoming solar energy.
This concept is significant because selecting materials with appropriate absorptivity ensures that the satellite doesn't overheat or become too cold, thereby achieving a balance suited to its operating temperature requirements.
Thermal Emissivity
Thermal emissivity is equally important in the context of satellite temperature management. It is a measure of how effectively a surface can emit absorbed heat as thermal radiation.
The emissivity values of the materials used in the satellite's coating significantly affect its thermal balance. The aluminum film has an emissivity of 0.03, indicating it emits only 3% of its absorbed energy, thus retaining most of the heat. Conversely, the zinc-oxide paint has a much higher emissivity of 0.85, allowing it to effectively emit 85% of its absorbed heat.
Understanding emissivity helps in choosing the right coating materials that can either retain heat if needed or dissipate it efficiently, maintaining the satellite's temperature within operational limits.
Energy Balance Equation
The energy balance equation is the cornerstone for calculating the optimal thermal control strategy of a satellite. It states that the incoming solar energy must equal the outgoing thermal energy for the satellite's temperature to remain stable.
The equation from the exercise, combining different materials with varying absorptivities and emissivities:
  • Incoming Solar Energy = Outgoing Thermal Energy
  • The mathematical expression is given as:
    \[F \times \alpha_{SA} \times I + (1-F) \times \alpha_{SZ} \times I = F \times \varepsilon_A \times \sigma \times T^4 + (1-F) \times \varepsilon_Z \times \sigma \times T^4\]
Using this balance, we can determine the necessary fraction of each material's coverage to maintain the desired satellite temperature. This balance ensures efficient thermal management by either absorbing heat or emitting it to the surrounding space as needed.
Satellite Temperature Management
Efficient satellite temperature management involves balancing solar absorptivity and thermal emissivity to achieve the desired thermal conditions for satellite operation.
The coated materials on the satellite help manage its temperature by selectively absorbing and emitting energy. Materials with low absorptivity and emissivity like the aluminum film limit heat intake and retention, preventing overheating. Meanwhile, high absorptivity and emissivity materials like zinc-oxide paint allow the satellite to absorb necessary heat while efficiently emitting excess heat.
Such management is vital as satellites operate in the harsh environment of space, where temperature control is paramount for reliable function and longevity. By calculating the right proportion of coatings, engineers ensure that satellites maintain stability and functionality across varying environmental conditions.

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Most popular questions from this chapter

The absorber plate of a solar collector may be coated with an opaque material for which the spectral, directional absorptivity is characterized by relations of the form $$ \begin{array}{ll} \alpha_{\lambda, \dot{\theta}}(\lambda, \theta)=\alpha_{1} \cos \theta & \lambda<\lambda_{c} \\ \alpha_{\lambda, \theta}(\lambda, \theta)=\alpha_{2} & \lambda>\lambda_{c} \end{array} $$ The zenith angle \(\theta\) is formed by the sun's rays and the plate normal, and \(\alpha_{1}\) and \(\alpha_{2}\) are constants. (a) Obtain an expression for the total, hemispherical absorptivity, \(\alpha_{S}\), of the plate to solar radiation incident at \(\theta=45^{\circ}\). Evaluate \(\alpha_{5}\) for \(\alpha_{1}=0.93, \alpha_{2}=\) \(0.25\), and a cut-off wavelength of \(\lambda_{c}=2 \mu \mathrm{m}\). (b) Obtain an expression for the total, hemispherical emissivity \(\varepsilon\) of the plate. Evaluate \(\varepsilon\) for a plate temperature of \(T_{p}=60^{\circ} \mathrm{C}\) and the prescribed values of \(\alpha_{1}, \alpha_{2}\), and \(\lambda_{c}\). (c) For a solar flux of \(q_{s}^{\prime \prime}=1000 \mathrm{~W} / \mathrm{m}^{2}\) incident at \(\theta=45^{\circ}\) and the prescribed values of \(\alpha_{1}, \alpha_{2}, \lambda_{c}\), and \(T_{p}\), what is the net radiant heat flux, \(q_{\text {net }}^{\prime \prime}\), to the plate? (d) Using the prescribed conditions and the Radiation/ Band Emission Factor option in the Tools section of \(I H T\) to evaluate \(F_{\left(0 \rightarrow \lambda_{j}\right)}\), explore the effect of \(\lambda_{c}\) on \(\alpha_{S}, \varepsilon\), and \(q_{\text {net }}^{N}\) for the wavelength range \(0.7 \leq \lambda_{c} \leq 5 \mu \mathrm{m}\).

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.

The exposed surface of a power amplifier for an earth satellite receiver of area \(130 \mathrm{~mm} \times 130 \mathrm{~mm}\) has a diffuse, gray, opaque coating with an emissivity of \(0.5\). For typical amplifier operating conditions, the surface temperature is \(58^{\circ} \mathrm{C}\) under the following environmental conditions: air temperature, \(T_{\infty}=27^{\circ} \mathrm{C}\); sky temperature, \(T_{\text {sy }}=-20^{\circ} \mathrm{C} ;\) convection coefficient, \(h=\) \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); and solar irradiation, \(G_{S}=800 \mathrm{~W} / \mathrm{m}^{2}\). (a) For the above conditions, determine the electrical power being generated within the amplifier. (b) It is desired to reduce the surface temperature by applying one of the diffuse coatings (A, B, C) shown as follows. Which coating will result in the coolest surface temperature for the same amplifier operating and environmental conditions?

A thermocouple inserted in a 4-mm-diameter stainless steel tube having a diffuse, gray surface with an emissivity of \(0.4\) is positioned horizontally in a large airconditioned room whose walls and air temperature are 30 and \(20^{\circ} \mathrm{C}\), respectively. (a) What temperature will the thermocouple indicate if the air is quiescent? (b) Compute and plot the thermocouple measurement error as a function of the surface emissivity for \(0.1 \leq \varepsilon \leq 1.0\).

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 \(T_{c}\) ?
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