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Chapter 13: Problem 154

A solar flux of \(1400 \mathrm{~W} / \mathrm{m}^{2}\) directly strikes a spacevehicle surface which has a solar absortivity of \(0.4\) and thermal emissivity of \(0.6\). The equilibrium temperature of this surface in space at \(0 \mathrm{~K}\) is (a) \(300 \mathrm{~K}\) (b) \(360 \mathrm{~K}\) (c) \(410 \mathrm{~K}\) (d) \(467 \mathrm{~K}\) (e) \(510 \mathrm{~K}\)

Short Answer

Expert verified
Answer: The equilibrium temperature of the space vehicle surface is approximately 410 K.

Step by step solution

01

Understand the solar properties

We are given the solar flux (\(F_{sun}\)) as \(1400 \mathrm{~W} / \mathrm{m}^{2}\), solar absorptivity (\(\alpha\)) as \(0.4\), and thermal emissivity (\(\epsilon\)) as \(0.6\). \(F_{sun}\) represents the intensity of sunlight on the surface of the space vehicle.
02

Calculate the absorbed power

When sunlight strikes the surface of the space vehicle, a part of the sunlight is absorbed by the surface, which is determined by the solar absorptivity. The absorbed power (\(P_{absorbed}\)) can be calculated using the formula $$P_{absorbed} = \alpha \times F_{sun}.$$Plug in the values provided:$$P_{absorbed} = 0.4 \times 1400 \mathrm{~W} / \mathrm{m}^{2} = 560 \mathrm{~W} / \mathrm{m}^{2}.$$
03

Recall the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power emitted by an object (due to radiation) is $$P_{emitted} = \epsilon \cdot \sigma \cdot A \cdot T_{eq}^4,$$where \(\epsilon\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4\) is the Stefan-Boltzmann constant, \(A\) is the area, and \(T_{eq}\) is the equilibrium temperature in Kelvin.
04

Equate the absorbed and emitted powers

At equilibrium, the absorbed power should be equal to the emitted power. Therefore, we have $$560 \mathrm{~W} / \mathrm{m}^{2} = 0.6 \cdot 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4 \cdot A \cdot T_{eq}^4.$$Since we are dealing with the surface, the area \(A\) will cancel out on both sides. Thus, we have:$$T_{eq}^4 = \frac{560 \mathrm{~W} / \mathrm{m}^{2}}{(0.6)(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4)}.$$
05

Solve for equilibrium temperature

Finally, solve for the equilibrium temperature (\(T_{eq}\)) by taking the fourth root on both sides of the equation:$$T_{eq} = \sqrt[4]{\frac{560 \mathrm{~W} / \mathrm{m}^{2}}{(0.6)(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4)}} \approx 410 \mathrm{~K}.$$The correct answer is (c) \(410 \mathrm{~K}\).

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Most popular questions from this chapter

A furnace is shaped like a long equilateral-triangular duct where the width of each side is \(2 \mathrm{~m}\). Heat is supplied from the base surface, whose emissivity is \(\varepsilon_{1}=0.8\), at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\) while the side surfaces, whose emissivities are \(0.5\), are maintained at \(500 \mathrm{~K}\). Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

A 70-cm-diameter flat black disk is placed in the center of the top surface of a \(1-\mathrm{m} \times 1-\mathrm{m} \times 1-\mathrm{m}\) black box. The view factor from the entire interior surface of the box to the interior surface of the disk is (a) \(0.077\) (b) \(0.144\) (c) \(0.356\) (d) \(0.220\) (e) \(1.0\)

Consider two rectangular surfaces perpendicular to each other with a common edge which is \(1.6 \mathrm{~m}\) long. The horizontal surface is \(0.8 \mathrm{~m}\) wide and the vertical surface is \(1.2 \mathrm{~m}\) high. The horizontal surface has an emissivity of \(0.75\) and is maintained at \(400 \mathrm{~K}\). The vertical surface is black and is maintained at \(550 \mathrm{~K}\). The back sides of the surfaces are insulated. The surrounding surfaces are at \(290 \mathrm{~K}\), and can be considered to have an emissivity of \(0.85\). Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings.

Consider a circular grill whose diameter is \(0.3 \mathrm{~m}\). The bottom of the grill is covered with hot coal bricks at \(950 \mathrm{~K}\), while the wire mesh on top of the grill is covered with steaks initially at \(5^{\circ} \mathrm{C}\). The distance between the coal bricks and the steaks is \(0.20 \mathrm{~m}\). Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.

How is the insulating effect of clothing expressed? How does clothing affect heat loss from the body by convection, radiation, and evaporation? How does clothing affect heat gain from the sun?
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