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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.

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=4πr2. Since the dome is half of a sphere, the surface area of the dome would be: A1=2πr2 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=πr2. 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: A2=πr2
03

Determine the view factor

In order to find the view factor F12 from the dome to the base of the furnace, we can use the following view factor equation for a hemispherical dome to its base: F12=1A1A2 Now, let's substitute the equations for the surface areas we calculated in Step 2: F12=12πr2πr2 We can simplify the equation: F12=121=1 However, the view factor cannot be negative. So, because the whole dome 'sees' its base, the view factor is: F12=1 So the view factor from the dome of the furnace to its base is 1.

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Key Concepts

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

Hemispherical Furnace
A hemispherical furnace is essentially a furnace that has a dome-shaped top and a flat circular base. Think of it as half of a sphere sitting on a flat surface. This unique shape is often used in various industrial applications where even heat distribution is critical. The design helps maximize the heat transfer by allowing the radiator, or the heat source, to effectively reflect off the curved inner surface towards the center.
  • The dome shape influences how heat is distributed within the furnace.
  • The flat base serves as the primary heat transfer surface.
  • This configuration often accommodates a central heat source or load for processing.
The geometry of a hemispherical furnace is an important aspect when discussing view factors, which are crucial for analyzing how much radiation energy leaves one surface and arrives at another. View factor calculations allow us to account for the geometry in thermal transfer analyses.
Surface Area Calculation
Calculating the surface area is essential in understanding how radiation heat transfer is analyzed within structures like the hemispherical furnace. For a hemisphere, the surface area of the dome is half of that of a full sphere. A full sphere surface area is given by the formula:A=4πr2Since a dome is half of a sphere, we calculate its surface area as:A1=2πr2The flat base is simply a circle, so we use the formula:A2=πr2Knowing these areas helps to compute view factors by showing the proportion of radiation that each part of the furnace exchanges.
  • Calculate the dome area as half of a sphere for simplicity.
  • Base area corresponds to the circle with the same radius.
  • These surfaces determine how heat exchanges between the parts.
In this exercise, the surface areas help confirm that the entire dome sees the base, according to the problem's geometry, ensuring the view factor is considered appropriately.
Heat Transfer
Heat transfer, especially in radiation, is a core concept in thermal physics. In the context of the hemispherical furnace, heat transfer involves the movement of thermal energy between the dome and the base. Radiation heat transfer is significant because it involves the transfer of heat in the form of electromagnetic waves, which in this scenario mostly occur within the furnace's confined space.
To determine how effectively heat is transferred, the view factor is used. For a hemisperical furnace, since the dome encapsulates the base, the view factor from the dome to the base is maximized. In other words, all the radiation leaving the dome impacts the base, meaning:F12=1This indicates that the geometry is perfectly efficient for heat transfer in terms of radiation:
  • All radiative heat leaving the dome reaches the base.
  • The efficiency of heat distribution is high because of the dome shape.
Understanding these principles can help effectively design furnaces and similar devices for optimal energy efficiency and operation. The perfect view factor of 1, as seen here, highlights the seamless transition of energy between the surfaces in question.

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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 m. Heat is supplied from the base surface, whose emissivity is ε1=0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

Two parallel disks of diameter D=0.6 m separated by L=0.4 m are located directly on top of each other. Both disks are black and are maintained at a temperature of 450 K. The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at 300 K. Determine the net rate of radiation heat transfer from the disks to the environment.

How does radiation transfer through a participating medium differ from that through a nonparticipating medium?

In a cogeneration plant, combustion gases at 1 atm and 800 K are used to preheat water by passing them through 6-m-long, 10-cm-diameter tubes. The inner surface of the tube is black, and the partial pressures of CO2 and H2O in combustion gases are 0.12 atm and 0.18 atm, respectively. If the tube temperature is 500 K, determine the rate of radiation heat transfer from the gases to the tube.

Two concentric spheres of diameters D1=0.3 m and D2=0.4 m are maintained at uniform temperatures T1=700 K and T2=500 K and have emissivities ε1=0.5 and ε2=0.7, respectively. Determine the net rate of radiation heat transfer between the two spheres. Also, determine the convection heat transfer coefficient at the outer surface if both the surrounding medium and the surrounding surfaces are at 30C. Assume the emissivity of the outer surface is 0.35.

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