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Chapter 13: Problem 2

How can you determine the view factor \(F_{12}\) when the view factor \(F_{21}\) and the surface areas are available?

Short Answer

Expert verified
Question: Determine the view factor \(F_{12}\) given the view factor \(F_{21}\) and the surface areas of surface 1 and surface 2. Answer: \(F_{12} = \frac{A_2 F_{21}}{A_1}\)

Step by step solution

01

Identifying the given information

In this exercise, we have the following given information: - View factor \(F_{21}\) - Surface area of surface 1, denoted as \(A_1\) - Surface area of surface 2, denoted as \(A_2\)
02

Using the reciprocity relationship

We can now use the reciprocity relationship between the view factors, which states that: \(A_1 F_{12} = A_2 F_{21}\)
03

Solving for \(F_{12}\)

Finally, we need to solve for \(F_{12}\). We can do this by rearranging the equation from Step 2: \(F_{12} = \frac{A_2 F_{21}}{A_1}\) Insert the given values of \(F_{21}\), \(A_1\), and \(A_2\) into the equation and calculate the value of \(F_{12}\).

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Most popular questions from this chapter

Two very long concentric cylinders of diameters \(D_{1}=\) \(0.35 \mathrm{~m}\) and \(D_{2}=0.5 \mathrm{~m}\) are maintained at uniform temperatures of \(T_{1}=950 \mathrm{~K}\) and \(T_{2}=500 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=1\) and \(\varepsilon_{2}=0.55\), respectively. Determine the net rate of radiation heat transfer between the two cylinders per unit length of the cylinders.

Give examples of radiation effects that affect human comfort. 13-84 A thin aluminum sheet with an emissivity of \(0.15\) on both sides is placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=900 \mathrm{~K}\) and \(T_{2}=650 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.8\), respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result with that without the shield.Give examples of radiation effects that affect human comfort.

Two-phase gas-liquid oxygen is stored in a spherical tank of \(1-m\) diameter, where it is maintained at its normal boiling point. The spherical tank is enclosed by a \(1.6-\mathrm{m}\) diameter concentric spherical surface at \(273 \mathrm{~K}\). Both spherical surfaces have an emissivity of \(0.01\), and the gap between the inner sphere and outer sphere is vacuumed. Assuming that the spherical tank surface has the same temperature as the oxygen, determine the heat transfer rate at the spherical tank surface.

Consider a 20-cm-diameter hemispherical enclosure. The dome is maintained at \(600 \mathrm{~K}\) and heat is supplied from the dome at a rate of \(50 \mathrm{~W}\) while the base surface with an emissivity of \(0.55\) is maintained at \(400 \mathrm{~K}\). Determine the emissivity of the dome.

What is operative temperature? How is it related to the mean ambient and radiant temperatures? How does it differ from effective temperature?
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