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Consider a hemispherical furnace with a flat circular base of diameter \(D\). Determine the view factor from the dome of this furnace to its base. Answer: \(0.5\)

Short Answer

Expert verified
Answer: The view factor from the dome of a hemispherical furnace to its base is 0.5.

Step by step solution

01

Understand the geometry and terminology

In this problem, we have a hemispherical furnace with a flat circular base of diameter D. The view factor we are interested in is from the dome of the furnace to its base. The view factor is a dimensionless quantity that gives the fraction of the radiation leaving one surface that is intercepted by another surface.
02

Recall the view factor reciprocity rule

The view factor reciprocity rule states that: \(A_1F_{12} = A_2F_{21}\) Where: \(A_1\) and \(A_2\) are the areas of the emitting and receiving surfaces, respectively \(F_{12}\) is the view factor from surface 1 to surface 2 \(F_{21}\) is the view factor from surface 2 to surface 1
03

Calculate the areas of the dome and the base

First, let's calculate the area of the dome (\(A_1\)) and the base (\(A_2\)) of the hemisphere. The area of the dome, which corresponds to the curved surface area of a hemisphere, is given by: \(A_1 = 2\pi R^2\) The area of the base, which is a circle, can be obtained using the diameter D: \(A_2 = \pi (\frac{D}{2})^2\) Where: \(R\) is the radius of the hemispherical dome Since \(D = 2R\), we can also express \(A_2\) as: \(A_2 = \pi R^2\)
04

Apply the view factor reciprocity rule

Now that we have the areas of the dome and the base, we can apply the view factor reciprocity rule. \(A_1F_{12} = A_2F_{21}\) Rearrange for \(F_{12}\) (the view factor from the dome to the base): \(F_{12} = \frac{A_2F_{21}}{A_1}\)
05

Understand the hemisphere's view factor from base to dome

Recall that in a closed hemispherical space, the view factor from the base to the dome is 1, since the whole energy emitted from the base will directly be intercepted by the dome: \(F_{21} = 1\)
06

Calculate the view factor from the dome to the base

Now we can find the view factor from the dome to the base using the obtained expression: \(F_{12} = \frac{A_2F_{21}}{A_1} = \frac{\pi R^2 * 1}{2\pi R^2} = \frac{1}{2}\) Therefore, the view factor from the dome of the hemispherical furnace to its base is \(0.5\).

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Most popular questions from this chapter

Two very large parallel plates are maintained at uniform temperatures \(T_{1}=750 \mathrm{~K}\) and \(T_{2}=500 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.85\) and \(\varepsilon_{2}=0.7\), respectively. If a thin aluminum sheet with the same emissivity on both sides is to be placed between the plates in order to reduce the net rate of radiation heat transfer between the plates by 90 percent, the emissivity of the aluminum sheet must be (a) \(0.07\) (b) \(0.10\) (c) \(0.13\) (d) \(0.16\) (e) \(0.19\)

Give examples of radiation effects that affect human comfort.

Two concentric spheres are maintained at uniform temperatures \(T_{1}=45^{\circ} \mathrm{C}\) and \(T_{2}=280^{\circ} \mathrm{C}\) and have emissivities \(\varepsilon_{1}=0.25\) and \(\varepsilon_{2}=0.7\), respectively. If the ratio of the diameters is \(D_{1} / D_{2}=0.30\), the net rate of radiation heat transfer between the two spheres per unit surface area of the inner sphere is (a) \(86 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(1169 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(1181 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(2510 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(3306 \mathrm{~W} / \mathrm{m}^{2}\)

Explain all the different mechanisms of heat transfer from the human body \((a)\) through the skin and \((b)\) through the lungs.

Two gray surfaces that form an enclosure exchange heat with one another by thermal radiation. Surface 1 has a temperature of \(400 \mathrm{~K}\), an area of \(0.2 \mathrm{~m}^{2}\), and a total emissivity of \(0.4\). Surface 2 has a temperature of \(600 \mathrm{~K}\), an area of \(0.3 \mathrm{~m}^{2}\), and a total emissivity of \(0.6\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(135 \mathrm{~W}\) (b) \(223 \mathrm{~W}\) (c) \(296 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(422 \mathrm{~W}\)

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