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A spherical interplanetary probe of \(0.5-\mathrm{m}\) diameter contains electronics that dissipate \(150 \mathrm{~W}\). If the probe surface has an emissivity of \(0.8\) and the probe does not receive radiation from other surfaces, as, for example, from the sun, what is its surface temperature?

Short Answer

Expert verified
The surface temperature of the interplanetary probe is approximately 288 K.

Step by step solution

01

Determine the total radiated power

According to Stefan-Boltzmann law, the total power radiated by an object is given by: \[P = \sigma \epsilon A T^4\] Where: - \(P\) is the total power radiated - \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\) - \(\epsilon\) is the emissivity of the surface - \(A\) is the surface area - \(T\) is the temperature in Kelvins The power radiated by the probe should equal the power being dissipated (150W) since it does not receive any radiation from other sources.
02

Calculate surface area

The surface area of a sphere can be calculated using the formula: \[A = 4\pi r^2\] Where: - \(r\) is the radius of the sphere The diameter of the probe is given as \(0.5\mathrm{m}\). Thus, the radius is \(r = 0.25\mathrm{m}\). Calculating the surface area, we obtain: \[A = 4\pi (0.25)^2 = 0.785 \mathrm{m^2}\]
03

Solve for the surface temperature

Now we have everything we need to solve the Stefan-Boltzmann equation for the temperature. From the known power dissipated in the probe P=150W: \[150 \mathrm{W} = \sigma \epsilon A T^4\] Plug in the values for \(\sigma\), \(\epsilon\), and \(A\): \[150 \mathrm{W} = (5.67 \times 10^{-8} \mathrm{W/m^2 K^4})(0.8)(0.785 \mathrm{m^2}) T^4\] Solving for temperature T: \[T^4 = \frac{150 \mathrm{W}}{(5.67 \times 10^{-8} \mathrm{W/m^2 K^4})(0.8)(0.785 \mathrm{m^2})} = 4.452 \times 10^6\mathrm{K^4}\] Now we can take the fourth root of the result to get the temperature in Kelvins: \[T = \sqrt[4]{4.452 \times 10^6\mathrm{K^4}} = 288 \mathrm{K}\] Thus, the surface temperature of the interplanetary probe is approximately 288 K.

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Key Concepts

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

Radiative Heat Transfer
Radiative heat transfer is the process by which heat energy is emitted from a surface in the form of electromagnetic radiation. Unlike conduction or convection, radiative heat transfer does not require a medium, which makes it particularly significant in space environments, like our interplanetary probe. The Stefan-Boltzmann Law is key here, as it explains how the power radiated by a surface area is related to its temperature raised to the fourth power. This law is crucial for understanding how objects like the probe can maintain energy balance by dissipating heat effectively.
In the absence of any external radiation like from the sun, the probe must rely solely on radiating its own heat to prevent overheating. The equation used to calculate this is precisely employed to equate the amount of power being wasted as heat ( 150 W in our case) with the power that the probe can radiate into space.
Emissivity
Emissivity is a measure of a material's ability to emit infrared radiation at a given temperature. It is expressed as a value between 0 and 1, where 1 means a perfect black body that radiates energy most efficiently. For our probe, the emissivity value is given as 0.8, indicating that its surface is quite effective at radiating heat, but not perfect.
  • A material with high emissivity will radiate more energy at a given temperature compared to one with low emissivity.
  • This property helps in calculating the total power radiated by linking it with other variables such as surface area and temperature through the Stefan-Boltzmann equation.
The high emissivity value of the probe ensures that it can release the significant 150 W of internal energy consistently, without absorbing additional energy from its surroundings.
Spherical Geometry
Spherical geometry refers to the three-dimensional shape used in this scenario, which is a sphere. The relevant property of spherical objects here is their symmetrical surface area, which can be calculated using the formula \(A = 4\pi r^2\).
In our problem, the probe has a diameter of 0.5 m, giving it a radius of 0.25 m. Plugging this into the formula shows us the total surface area available for dissipating heat, essential in calculating radiative heat transfer. This requires breaking down the geometry of the object to understand how its shape affects its thermal properties. The larger the surface, the more heat it can radiate, which explains why geometry is a core consideration in thermal regulation for spacecraft.
Thermal Equilibrium
Achieving thermal equilibrium means that an object's heat input and output are balanced. For the probe, this is necessary to prevent overheating or supercooling in space. At equilibrium, the power being dissipated by the electronics (150 W) is equal to the power radiated by the probe’s surface. This equilibrium state ensures that the surface temperature remains constant, provided external conditions also remain unchanged. In our example, the calculated surface temperature of 288 K is the equilibrium point where this balance is maintained, allowing the probe to operate efficiently without temperature swings that could negatively affect its operation.
Heat Dissipation
Heat dissipation is the process of losing or transferring heat from one object to another or into its environment. In the context of the interplanetary probe, heat dissipation occurs through radiation, ensuring that the electronics remain within operational temperature limits. The capability of the probe surface to dissipate heat effectively is a function of several factors:
  • The power generated by electronics (150 W).
  • The probe’s surface area involved in radiation.
  • Its emissivity value.
Efficient heat dissipation allows for safe and consistent energy release in the vacuum of space, preventing damage that could be caused by excessive heat buildup inside the probe.

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Most popular questions from this chapter

Single fuel cells such as the one of Example \(1.5\) can be scaled up by arranging them into a fuel cell stack. A stack consists of multiple electrolytic membranes that are sandwiched between electrically conducting bipolar plates. Air and hydrogen are fed to each membrane through fiw channels within each bipolar plate, as shown in the sketch. With this stack arrangement, the individual fuel cells are connected in series, electrically, producing a stack voltage of \(E_{\text {stack }}=N \times E_{c}\), where \(E_{c}\) is the voltage produced across each membrane and \(N\) is the number of membranes in the stack. The electrical current is the same for each membrane. The cell voltage, \(E_{c}\), as well as the cell efficiency, increases with temperature (the air and hydrogen fed to the stack are humidified to allow operation at temperatures greater than in Example 1.5), but the membranes will fail at temperatures exceeding \(T \approx 85^{\circ} \mathrm{C}\). Consider \(L \times w\) membranes, where \(L=w=100 \mathrm{~mm}\), of thickness \(t_{m}=0.43 \mathrm{~mm}\), that each produce \(E_{c}=0.6 \mathrm{~V}\) at \(I=60 \mathrm{~A}\), and \(\dot{E}_{c g}=45 \mathrm{~W}\) of thermal energy when operating at \(T=80^{\circ} \mathrm{C}\). The external surfaces of the stack are exposed to air at \(T_{\infty}=25^{\circ} \mathrm{C}\) and surroundings at \(T_{\text {sur }}=30^{\circ} \mathrm{C}\), with \(\varepsilon=0.88\) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the electrical power produced by a stack that is \(L_{\text {stack }}=200 \mathrm{~mm}\) long, for bipolar plate thickness in the range \(1 \mathrm{~mm}

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Liquid oxygen, which has a boiling point of \(90 \mathrm{~K}\) and a latent heat of vaporization of \(214 \mathrm{~kJ} / \mathrm{kg}\), is stored in a spherical container whose outer surface is of \(500-\mathrm{mm}\) diameter and at a temperature of \(-10^{\circ} \mathrm{C}\). The container is housed in a laboratory whose air and walls are at \(25^{\circ} \mathrm{C}\). (a) If the surface emissivity is \(0.20\) and the heat transfer coefficient associated with free convection at the outer surface of the container is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the rate, in \(\mathrm{kg} / \mathrm{s}\), at which oxygen vapor must be vented from the system? (b) Moisture in the ambient air will result in frost formation on the container, causing the surface emissivity to increase. Assuming the surface temperature and convection coefficient to remain at \(-10^{\circ} \mathrm{C}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, compute the oxygen evaporation rate \((\mathrm{kg} / \mathrm{s})\) as a function of surface emissivity over the range \(0.2 \leq \varepsilon \leq 0.94\).

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

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