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Chapter 13: Problem 10

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.

Short Answer

Expert verified
In conclusion, the view factor from the dome of the furnace to its base is 1. This means that the entire luminary energy emitted from the dome is intercepted by the base. It is important to note that this is the case due to the hemispherical shape of the dome, which makes all the emitted energy be directed towards the base.

Step by step solution

01

Understand the geometry

In this case, we have a hemispherical furnace with a flat circular base. We will consider the view factor from the dome (1) to the base (2) of the furnace.
02

Calculate the surface area of the dome and the base

First, we need to find the surface area of the dome (1) and the surface area of the base (2). The surface area of the dome can be calculated as half of the surface area of a sphere. The formula for the surface area of a sphere is \(A = 4 * \pi * r^2\). Since the dome is half of a sphere, the surface area of the dome would be: \(A_1 = 2 * \pi * r^2\) Now, we need to find the surface area of the circular base (2) of the furnace. The formula for the surface area of a circle is \(A = \pi * r^2\). In this case, the radius of the circular base is equal to the radius of the dome since the diameter of the base is given as \(D\). Therefore, the surface area of the circular base would be: \(A_2 = \pi * r^2\)
03

Determine the view factor

In order to find the view factor \(F_{12}\) from the dome to the base of the furnace, we can use the following view factor equation for a hemispherical dome to its base: \(F_{12} = 1 - \frac{A_1}{A_2}\) Now, let's substitute the equations for the surface areas we calculated in Step 2: \(F_{12} = 1 - \frac{2 * \pi * r^2}{\pi * r^2}\) We can simplify the equation: \(F_{12} = 1 - \frac{2}{1} = -1\) However, the view factor cannot be negative. So, because the whole dome 'sees' its base, the view factor is: \(F_{12} = 1\) So the view factor from the dome of the furnace to its base is 1.

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Most popular questions from this chapter

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer between the base and the side surfaces is (a) \(22.5 \mathrm{~kW}\) (b) \(38.6 \mathrm{~kW}\) (c) \(60.7 \mathrm{~kW}\) (d) \(89.8 \mathrm{~kW}\) (e) \(151 \mathrm{~kW}\)

13-144 In a natural-gas fired boiler, combustion gases pass through 6-m-long 15 -cm-diameter tubes immersed in water at 1 atm pressure. The tube temperature is measured to be \(105^{\circ} \mathrm{C}\), and the emissivity of the inner surfaces of the tubes is estimated to be \(0.9\). Combustion gases enter the tube at 1 atm and \(1200 \mathrm{~K}\) at a mean velocity of \(3 \mathrm{~m} / \mathrm{s}\). The mole fractions of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in combustion gases are 8 percent and 16 percent, respectively. Assuming fully developed flow and using properties of air for combustion gases, determine \((a)\) the rates of heat transfer by convection and by radiation from the combustion gases to the tube wall and \((b)\) the rate of evaporation of water.

A cylindrical container whose height and diameter are \(8 \mathrm{~m}\) is filled with a mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{N}_{2}\) gases at \(600 \mathrm{~K}\) and \(1 \mathrm{~atm}\). The partial pressure of \(\mathrm{CO}_{2}\) in the mixture is \(0.15 \mathrm{~atm}\). If the walls are black at a temperature of \(450 \mathrm{~K}\), determine the rate of radiation heat transfer between the gas and the container walls.

A 2-m-internal-diameter double-walled spherical tank is used to store iced water at \(0^{\circ} \mathrm{C}\). Each wall is \(0.5 \mathrm{~cm}\) thick, and the \(1.5\)-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of \(0.15\). The temperature of the outer wall of the tank is measured to be \(20^{\circ} \mathrm{C}\). Assuming the inner wall of the steel tank to be at \(0^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-h period.

Two very large parallel plates are maintained at uniform temperatures of \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=800 \mathrm{~K}\) and have emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.5\), respectively. It is desired to reduce the net rate of radiation heat transfer between the two plates to one-fifth by placing thin aluminum sheets with an emissivity of \(0.1\) on both sides between the plates. Determine the number of sheets that need to be inserted.
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