Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 is approximately 0.25, which is closest to option (b) 0.26.

Step by step solution

01

Write down the given values

We are given the diameter of the sphere, D = 15 cm and side length of the cube, L = 15 cm.
02

Determine the size of the contact area

Since the sphere is housed within the cube, and its diameter is equal to the side length of the cubical enclosure, we infer that the contact area between the sphere and any side of the cube will be a circular region with a diameter equal to the diameter of the sphere (D). Hence, the contact area diameter will also equal 15 cm.
03

Calculate the radius of the contact area

The contact area, being circular, has a radius. To calculate this, we divide the diameter by 2. The radius of the contact circle (r) will be: r = D/2 = 15 cm/2 = 7.5 cm
04

Calculate the radius of the sphere

Since the diameter of the sphere is given as 15 cm, we can calculate the radius of the sphere (R) by dividing the diameter by 2. Hence, R = D/2 = 15 cm/2 = 7.5 cm
05

Apply the Hottel's Crossed-Strings method

We now apply Hottel's Crossed-Strings method to find the view factor from any of the square-cube surfaces to the sphere. According to the method, view factor, F = r^2 / (2 * R * L) Substituting the values, we get F = (7.5^2) / (2 * 7.5 * 15) = 56.25 / 225 = 0.25 Since the given options are not exact and have some rounding errors, we can choose the closest option to our answer.
06

Select the correct option

Our calculated view factor F is approximately 0.25, which is closest to the option (b) 0.26. Therefore, the correct answer is (b) 0.26.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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}\( and 1 atm. The partial pressure of \)\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 long cylindrical fuel rod with a diameter of \(25 \mathrm{~mm}\) is enclosed by a concentric tube with a diameter of \(50 \mathrm{~mm}\). The tube is an ASTM A249 904L stainless steel tube with a maximum use temperature of $260^{\circ} \mathrm{C}$ specified by the ASME Code for Process Piping (ASME B31.32014 , Table A-1M). The gap between the fuel rod and the tube is a vacuum. The fuel rod has a surface temperature of \(550^{\circ} \mathrm{C}\). The emissivities of the fuel rod and the ASTM A249 \(904 \mathrm{~L}\) tube are \(0.97\) and \(0.33\), respectively. Outside of the tube, convection with the ambient air and radiation with the surroundings occur. The convection heat transfer coefficient of the tube with the ambient air is $15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( at \)20^{\circ} \mathrm{C}$. The temperature of the surroundings is the same as the ambient air temperature. Determine whether the temperature of the tube surface is below the maximum use temperature for ASTM A249 904L stainless steel tube.

Two concentric spheres are maintained at uniform temperatures \(T_{1}=45^{\circ} \mathrm{C}\) and \(T_{2}=280^{\circ} \mathrm{C}\) and have emissivities \(\varepsilon_{1}=0.25\) and \(\varepsilon_{2}=0.7\), respectively. If the ratio of the diameters is \(D_{1} / D_{2}=0.30\), the net rate of radiation heat transfer between the two spheres per unit surface area of the inner sphere is (a) \(86 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(1169 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(1181 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(2510 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(3306 \mathrm{~W} / \mathrm{m}^{2}\)

Consider a \(4-\mathrm{m} \times 4-\mathrm{m} \times 4-\mathrm{m}\) cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of \(550 \mathrm{~K}\) and \(1100 \mathrm{~K}\), respectively. Determine the net rate of radiation heat transfer between the floor and the ceiling of the furnace.

A large number of long tubes, each of diameter \(D\), are placed parallel to each other and at a center-to-center distance of s. Since all of the tubes are geometrically similar and at the same temperature, these could be treated collectively as one surface \(\left(A_{j}\right)\) for radiation heat transfer calculations. As shown in Fig. P13-140, the tube bank \(\left(A_{j}\right)\) is placed opposite a large flat wall \(\left(A_{j}\right)\) such that the tube bank is parallel to the wall. (a) Calculate the view factors \(F_{i j}\) and \(F_{j i}\) for $s=3.0 \mathrm{~cm}\( and \)D=1.5 \mathrm{~cm}$. (b) Calculate the net rate of radiation heat transfer between the wall and the tube bank per unit area of the wall when $T_{i}=900^{\circ} \mathrm{C}, T_{j}=60^{\circ} \mathrm{C}, \varepsilon_{i}=0.8\(, and \)\varepsilon_{j}=0.9$. (c) A fluid flows through the tubes at an average temperature of $40^{\circ} \mathrm{C}\(, resulting in a heat transfer coefficient of \)2.0 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Assuming \)T_{i}=900^{\circ} \mathrm{C}, \varepsilon_{i}=0.8\( and \)\varepsilon_{j}=0.9$ (as above) and neglecting the tube wall thickness and convection from the outer surface, calculate the temperature of the tube surface in steady operation.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free