/*! 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 30 A spherical interplanetary probe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A computer consists of an array of five printed circuit boards (PCBs), each dissipating \(P_{b}=20 \mathrm{~W}\) of power. Cooling of the electronic components on a board is provided by the forced flow of air, equally distributed in passages formed by adjoining boards, and the convection coefficient associated with heat transfer from the components to the air is approximately \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Air enters the computer console at a temperature of \(T_{i}=20^{\circ} \mathrm{C}\), and flow is driven by a fan whose power consumption is \(P_{f}=25 \mathrm{~W}\). (a) If the temperature rise of the airflow, \(\left(T_{o}-T_{i}\right)\), is not to exceed \(15^{\circ} \mathrm{C}\), what is the minimum allowable volumetric flow rate \(\dot{\forall}\) of the air? The density and specific heat of the air may be approximated as \(\rho=1.161\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) The component that is most susceptible to thermal failure dissipates \(1 \mathrm{~W} / \mathrm{cm}^{2}\) of surface area. To minimize the potential for thermal failure, where should the component be installed on a PCB? What is its surface temperature at this location?

A square silicon chip \((k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is of width \(w=5 \mathrm{~mm}\) on a side and of thickness \(t=1 \mathrm{~mm}\). The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If \(4 \mathrm{~W}\) are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

The diameter and surface emissivity of an electrically heated plate are \(D=300 \mathrm{~mm}\) and \(\varepsilon=0.80\), respectively. (a) Estimate the power needed to maintain a surface temperature of \(200^{\circ} \mathrm{C}\) in a room for which the air and the walls are at \(25^{\circ} \mathrm{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), may be approximated by an expression of the form \(h=0.80\left(T_{s}-T_{\infty}\right)^{1 / 3}\). (b) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.