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

Short Answer

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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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Key Concepts

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

Reciprocity Relationship
The concept of view factors is crucial when assessing how much radiation is exchanged between two surfaces. The reciprocity relationship is a key principle in radiation view factors. It helps simplify complex heat transfer calculations, especially when dealing with surfaces of different sizes. This guideline suggests that the product of the area of the first surface and its view factor to the second ( \(A_1 F_{12}\)) is equal to the product of the area of the second surface and its view factor to the first ( \(A_2 F_{21}\)).
  • This relationship is universal for all surface pairs involved in radiation heat transfer.
  • Knowing this allows us to easily derive any unknown view factor once other variables are known.
  • It is particularly helpful for asymmetric surfaces where direct measurement or calculation might be difficult.
Understanding that there is a direct proportional relationship between these factors makes radiation heat transfer calculations more manageable and less time-consuming.
Surface Area in Radiation
Surface area plays a pivotal role in radiation heat transfer. Each surface involved acts as both an emitter and absorber of radiation. The larger a surface area, the more radiation it can potentially emit or receive. This is why accurate knowledge of surface areas is essential in radiation heat transfer problems.
  • In our context, the surface area helps determine the amount of radiant energy exchanged between two surfaces.
  • When calculating view factors using the reciprocity relationship, knowing the precise areas allows us to compute accurate results for unknowns.
  • Errors in measuring the surface area can lead to significant inaccuracies in final heat transfer calculations.
In the specific exercise, the areas of surfaces 1 and 2 are used to substitute values into the equation for finding unknown view factors, highlighting the necessity of having exact surface area measurements in radiation calculations.
Heat Transfer Calculations
Radiation heat transfer calculations can appear daunting due to the depth of physics concepts involved. However, breaking down the problem into smaller steps as done in the exercise makes it simpler.
  • Start by identifying known factors such as view factors and surface areas.
  • Apply known relationships, like the reciprocity relation, to establish crucial equations.
  • Finally, solve the equations by substituting known values to determine unknowns.
Utilizing view factors reduces complex geometry and interactions to mathematical equations. This makes it feasible to predict and calculate the radiant energy exchange between different surfaces. By following systematic steps and applying consistent rules, radiation heat transfer calculations can be well within reach, even without advanced physics knowledge. This method ensures reliable, accurate results for any applied engineering or physics problem where radiation is a key factor.

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Most popular questions from this chapter

What is a reradiating surface? What simplifications does a reradiating surface offer in the radiation analysis?

Two parallel black disks are positioned coaxially with a distance of \(0.25 \mathrm{~m}\) apart in a surrounding with a constant temperature of \(300 \mathrm{~K}\). The lower disk is \(0.2 \mathrm{~m}\) in diameter and the upper disk is \(0.4 \mathrm{~m}\) in diameter. If the lower disk is heated electrically at \(100 \mathrm{~W}\) to maintain a uniform temperature of \(500 \mathrm{~K}\), determine the temperature of the upper disk.

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface of the furnace is black and has a temperature of \(400 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be \(7500 \mathrm{~W} / \mathrm{m}^{2}\) and \(3200 \mathrm{~W} / \mathrm{m}^{2}\), respectively. The net rate of radiation heat transfer to the bottom surface is (a) \(2.61 \mathrm{~kW}\) (b) \(8.27 \mathrm{~kW}\) (c) \(14.7 \mathrm{~kW}\) (d) \(23.5 \mathrm{~kW}\) (e) \(141 \mathrm{~kW}\)

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 from the bottom surface is (a) \(-93.6 \mathrm{~kW}\) (b) \(-86.1 \mathrm{~kW}\) (c) \(0 \mathrm{~kW}\) (d) \(86.1 \mathrm{~kW}\) (e) \(93.6 \mathrm{~kW}\)

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.

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