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Consider a 15-cm-diameter sphere placed within a cubical enclosure with a side length of \(15 \mathrm{~cm}\). The view factor from any of the square-cube surfaces to the sphere is (a) \(0.09\) (b) \(0.26\) (c) \(0.52\) (d) \(0.78\) (e) 1

Short Answer

Expert verified
Answer: The view factor between any of the cube's surfaces and the sphere in this case is approximately 0.26.

Step by step solution

01

Calculate critical radius of the sphere

We'll first need to find the critical radius \(r_c\) of the sphere. The critical radius is half of the diameter, and since the diameter is 15cm, the critical radius is: \(r_c = \frac{15}{2} = 7.5\mathrm{~cm}\).
02

Calculate the string lengths

Now, we will find the length of the strings connecting the cube's surface corners to the sphere's diametrically opposite points. We can call these strings A and B. Since the cube has a side length of 15cm and considering the sphere is entirely inside the cube touching all sides, the lengths of strings A and B will be equal to the cube's side length and the sphere's diameter: \(A = B = 15\mathrm{~cm}\).
03

Apply Hottel's cross string method formula

Hottel's formula for calculating view factor between a sphere and a planar surface (i.e., the cube surface) is given as: \(F_{sphere\rightarrow surface} = \frac{1}{2}\left(1 - \frac{AB}{(AB + r_c^2)^{\frac{1}{2}}}\right)\) Substitute the values of A, B, and \(r_c\) into the formula: \(F_{sphere\rightarrow surface} = \frac{1}{2}\left(1 - \frac{15\times15}{(15\times15 + 7.5^2)^{\frac{1}{2}}}\right)\)
04

Calculate the view factor

Now, we can find the numerical value of the view factor: \(F_{sphere\rightarrow surface} = \frac{1}{2}\left(1 - \frac{15\times15}{(15\times15 + 7.5^2)^{\frac{1}{2}}}\right)\approx 0.26\) Comparing the result with the given options, 0.26 is the correct answer, therefore the view factor from any of the square-cube surfaces to the sphere is (b) \(0.26\).

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Key Concepts

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

Radiative Heat Transfer
Radiative heat transfer is the process of energy transfer in the form of electromagnetic waves, primarily due to thermal emission. This mode of heat transfer does not require a material medium, which makes it unique from conduction and convection. Radiative heat transfer plays a critical role in many engineering applications, such as thermal management in spacecraft and building environments.

In the context of this exercise, radiative heat transfer is studied by analyzing the interaction between the surfaces of two geometries—a sphere and a cube. The view factor is essential here as it determines the fraction of radiation leaving one surface that directly strikes another. Understanding radiative heat transfer involves:
  • Recognizing that all objects emit thermal radiation proportional to their temperature and emissive properties.
  • Using view factors to calculate how much radiation is exchanged between different surfaces.
  • Applying formulas like Stefan-Boltzmann law to determine heat exchange based on temperature and surface properties.
Grasping these concepts helps to analyze and predict temperature distribution and heat flux in complex systems.
Hottel's Cross-String Method
Hottel's Cross-String Method is a technique used to calculate view factors in radiative heat transfer scenarios especially involving complex geometries like spheres and cubes. View factors are fundamental in estimating the radiation heat exchange between surfaces, as they consider both the relative orientation and distance between surfaces.

The cross-string method provides a way to calculate the view factor for non-planar or compound geometries. It involves imaginary strings connecting pairs of points on the surfaces, forming crossing patterns. The method is effective because:
  • It simplifies the geometric complexity by using straightforward mathematical relationships to relate surface interactions.
  • The approach considers the spatial orientations and shapes, which is crucial for non-flat surfaces like a sphere inside a cube.
  • By using formulas specific to the cross-string approach, engineers calculate the interaction extent precisely.
In the given exercise, Hottel's formula calculates the view factor between a cube's surface and an inscribed sphere, providing essential input parameters for heat transfer assessments.
Geometric Configurations in Heat Transfer
Geometric configurations significantly impact how heat is transferred between surfaces, particularly through radiation. Understanding the spatial arrangement and dimensional relationships of surfaces is key to accurately modeling and predicting heat transfer by radiation.

In scenarios involving different shapes, like the sphere within the cube from the exercise, geometric factors affect how much radiation energy is mutually exchanged. When dealing with geometric configurations:
  • Identify the shapes and their positioning relative to each other, as it influences view factors.
  • Recognize symmetry in arrangements, which can simplify calculations.
  • Define and use critical dimensions, such as diameters and corner-to-corner lengths, in equations for precise calculation. In this exercise, the critical radius of the sphere and the cube's side length are pivotal in calculating the view factor.
By comprehensively understanding geometric configurations, one can leverage them efficiently to solve engineering problems involving thermal radiation.

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Most popular questions from this chapter

A solar flux of \(1400 \mathrm{~W} / \mathrm{m}^{2}\) directly strikes a spacevehicle surface which has a solar absortivity of \(0.4\) and thermal emissivity of \(0.6\). The equilibrium temperature of this surface in space at \(0 \mathrm{~K}\) is (a) \(300 \mathrm{~K}\) (b) \(360 \mathrm{~K}\) (c) \(410 \mathrm{~K}\) (d) \(467 \mathrm{~K}\) (e) \(510 \mathrm{~K}\)

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.

Define spectral emissivity of a medium of thickness \(L\) in terms of the spectral absorption coefficient.

Consider a cubical furnace with a side length of \(3 \mathrm{~m}\). The top surface is maintained at \(700 \mathrm{~K}\). The base surface has an emissivity of \(0.90\) and is maintained at \(950 \mathrm{~K}\). The side surface is black and is maintained at \(450 \mathrm{~K}\). Heat is supplied from the base surface at a rate of \(340 \mathrm{~kW}\). Determine the emissivity of the top surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.

What does the view factor represent? When is the view factor from a surface to itself not zero?

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