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An electronic package with a surface area of \(1 \mathrm{~m}^{2}\) placed in an orbiting space station is exposed to space. The electronics in this package dissipate all \(1 \mathrm{~kW}\) of its power to the space through its exposed surface. The exposed surface has an emissivity of \(1.0\) and an absorptivity of \(0.25\). Determine the steady state exposed surface temperature of the electronic package \((a)\) if the surface is exposed to a solar flux of \(750 \mathrm{~W} /\) \(\mathrm{m}^{2}\), and \((b)\) if the surface is not exposed to the sun.

Short Answer

Expert verified
(a) The steady-state surface temperature of the electronic package when exposed to a solar flux is approximately 335.97 K. (b) The steady-state surface temperature of the electronic package when not exposed to a solar flux is approximately 314.63 K.

Step by step solution

01

Energy Balance Equation

We need to find the energy balance equation for power conservation between the energy going in and the energy going out through the package's surface. The energy going in will be the power \(\mathrm{1 \ kW}\) from the electronics and the absorbed fraction of solar flux \((750 \mathrm{~W} / \mathrm{m}^{2} \times 0.25)\) when exposed to the Sun. In both cases, the surface will be radiating energy out according to the Stefan-Boltzmann law \((e \sigma T^4 \times A)\), where \(\sigma = 5.67 \times 10^{-8} \mathrm{W} / (\mathrm{m}^{2} \ \mathrm{K}^{4})\) is the Stephan-Boltzmann constant, \(e\) is the emissivity, and \(A\) is the surface area. For case (a), we have the following energy balance equation: $$P_{in} + P_{solar} = P_{out}$$ For case (b), the energy balance equation is: $$P_{in} = P_{out}$$
02

Solve for Steady State Temperature (a)

Since we have the energy balance equation for the case when the package is exposed to solar flux, we can now substitute the given values and solve for the steady-state temperature \(T_a\). $$P_{in} + P_{solar} = e \sigma T_a^4 A$$ $$1000 + (750 \times 0.25) = 1.0 \times 5.67 \times 10^{-8} \times T_a^4 \times 1$$ $$1000 + 187.5 = 5.67 \times 10^{-8} \times T_a^4$$ $$T_a^4 = \frac{1187.5}{5.67 \times 10^{-8}}$$ $$T_a = \sqrt[4]{\frac{1187.5}{5.67 \times 10^{-8}}}$$ $$T_a = 335.97 \ \mathrm{K}$$ The steady-state temperature when the surface is exposed to solar flux is \(335.97 \ \mathrm{K}\).
03

Solve for Steady State Temperature (b)

Similarly, using the energy balance equation for the case when the package is not exposed to the Sun, we can solve for the steady-state temperature \(T_b\). $$P_{in} = e \sigma T_b^4 A$$ $$1000 = 1.0 \times 5.67 \times 10^{-8} \times T_b^4 \times 1$$ $$T_b^4 = \frac{1000}{5.67 \times 10^{-8}}$$ $$T_b = \sqrt[4]{\frac{1000}{5.67 \times 10^{-8}}}$$ $$T_b = 314.63 \ \mathrm{K}$$ The steady-state temperature when the surface is not exposed to the Sun is \(314.63 \ \mathrm{K}\). So we have the steady-state temperatures for (a) and (b) as: (a) \(T_a = 335.97 \ \mathrm{K}\) and (b) \(T_b = 314.63 \ \mathrm{K}\).

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Key Concepts

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

Energy Balance Equation
Heat transfer involves understanding how energy enters and exits a system. The Energy Balance Equation helps us to account for this energy flow. In this context, we analyze an electronic package on a space station. The package dissipates 1 kW of energy into space or adjusts for additional solar energy when exposed to sunlight.

The energy balance for such systems is formulated as:
  • For a sunny case: \[P_{in} + P_{solar} = P_{out}\] where \(P_{in}\) represents the electronic power and \(P_{solar}\) is the absorbed solar power.
  • For a non-sunny case: \[P_{in} = P_{out}\]
This indicates that in steady-state scenarios, the energy entering a system equals the energy leaving it, thereby allowing us to solve for unknowns like temperature.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is key in determining how much energy a body will radiate. This law dictates that the power radiated by a surface is proportional to the fourth power of its temperature:\[P_{out} = e \, \sigma \, T^4 \, A\]where:
  • \(e\) is the emissivity of the surface.
  • \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \, \text{W/m}^2 \, \text{K}^4\).
  • \(T\) is the absolute temperature in Kelvin.
  • \(A\) is the surface area in square meters.
This formula forms the basis for calculating temperatures in varying environments, as it quantifies how temperature affects radiative energy loss.
Emissivity and Absorptivity
Emissivity and absorptivity are characteristics of materials that affect heat transfer. Emissivity (\(e\)) describes how efficiently a surface emits thermal radiation compared to a perfect emitter. A perfect emitter, or blackbody, has an emissivity of 1.

Absorptivity is a measure of how much radiation a surface absorbs. Here, the package has:
  • Emissivity \(e = 1.0\), making it a perfect emitter.
  • Absorptivity of 0.25, meaning it absorbs 25% of incoming radiation.
Understanding these concepts allows the accurate determination of steady-state temperatures for the electronic package.
Steady-State Temperature
In thermal physics, the steady-state temperature implies a condition where temperatures stabilize, and energy inflows are balanced by energy outflows.

To find it, substitute known values into the Stefan-Boltzmann equation. For example, when exposed to solar radiation:
  • \(T_a\) is calculated using: \[T_a = \sqrt[4]{\frac{P_{in} + P_{solar}}{e \, \sigma \, A}}\]
  • Without solar exposure, \(T_b\) is determined by: \[T_b = \sqrt[4]{\frac{P_{in}}{e \, \sigma \, A}}\]
This will yield distinct temperatures depending on solar exposure, exemplifying thermal equilibrium in space environments.
Orbital Environment
A space station's environment greatly influences heat transfer processes. In orbit, the absence of atmospheric convection necessitates reliance purely on radiative methods.

Key factors in such environments include:
  • Direct solar radiation exposure, varying due to orbital path.
  • Significant thermal gradients due to alternating shadow and direct sunlight phases.
  • Essential design strategies ensuring thermal balance for instruments.
Adapting to orbital environments is pivotal for maintaining the functionality of space-installed electronics and equipment.

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Most popular questions from this chapter

A \(0.3\)-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.06 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board. Answer: \(0.042^{\circ} \mathrm{C}\)

A soldering iron has a cylindrical tip of \(2.5 \mathrm{~mm}\) in diameter and \(20 \mathrm{~mm}\) in length. With age and usage, the tip has oxidized and has an emissivity of \(0.80\). Assuming that the average convection heat transfer coefficient over the soldering iron tip is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surrounding air temperature is \(20^{\circ} \mathrm{C}\), determine the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

Heat treatment is common in processing of semiconductor material. A 200-mm- diameter silicon wafer with thickness of \(725 \mu \mathrm{m}\) is being heat treated in a vacuum chamber by infrared heater. The surrounding walls of the chamber have a uniform temperature of \(310 \mathrm{~K}\). The infrared heater provides an incident radiation flux of \(200 \mathrm{~kW} / \mathrm{m}^{2}\) on the upper surface of the wafer, and the emissivity and absorptivity of the wafer surface are \(0.70\). Using a pyrometer, the lower surface temperature of the wafer is measured to be \(1000 \mathrm{~K}\). Assuming there is no radiation exchange between the lower surface of the wafer and the surroundings, determine the upper surface temperature of the wafer. (Note: A pyrometer is a non-contacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)

Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. If the walls of the brick house are twice as thick, which house do you think will be more energy efficient?

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