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Chapter 13: Problem 164

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) \(141 \mathrm{~kW}\)

Short Answer

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Question: Determine the net rate of radiation heat transfer to the bottom surface of a cubical furnace with side length of 3 m, a bottom surface temperature of 400 K, a top surface radiosity of 7500 W/m², and a side surface radiosity of 3200 W/m². Choose the closest answer. a) 3.21 kW b) 8.27 kW c) 12.55 kW d) 15.48 kW Answer: b) 8.27 kW

Step by step solution

01

Find the surface area of the bottom surface

The bottom surface of the furnace is a square with side length of 3m, so its surface area can be calculated as follows: Area = side × side = 3 m × 3 m = 9 m²
02

Calculate the emissive power of the bottom surface

We have the temperature of the bottom surface, which is 400 K. We can use the Stefan-Boltzmann law to find the emissive power E_bot of the bottom surface: E_bot = σ × T^4 where σ is the Stefan-Boltzmann constant, which is approximately equal to 5.67 × 10^(-8) W/(m²·K^4), and T is the temperature of the surface in kelvin. E_bot = 5.67 × 10^(-8) W/(m²·K^4) × (400 K)^4 = 5670 W/m²
03

Calculate the incident radiation on the bottom surface

The radiosities of the top and side surfaces are given as 7500 W/m² and 3200 W/m², respectively. Since the bottom surface is completely black, it absorbs all the incident radiation from the top and side surfaces. We can calculate the incident radiation on the bottom surface as the average of the top and side surfaces radiosities: Incident_radiation = (7500 W/m² + 3 × 3200 W/m²) / 4 = 5025 W/m²
04

Calculate the net radiation heat transfer to the bottom surface

The net rate of radiation heat transfer to the bottom surface can be calculated as the difference between the incident radiation on the bottom surface and the emissive power of the bottom surface, multiplied by the surface area of the bottom surface: Q_net = Area × (Incident_radiation - E_bot) = 9 m² × (5025 W/m² - 5670 W/m²) = 9 m² × (-645 W/m²) = -5805 W Since we are asked to find the net rate of radiation heat transfer to the bottom surface, we will consider its magnitude (ignoring the negative sign): Q_net = 5805 W, which is approximately equal to 5.81 kW. None of the given options matches the answer exactly, but the closest option is (b) 8.27 kW.

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Most popular questions from this chapter

The number of view factors that need to be evaluated directly for a 10 -surface enclosure is (a) 1 (b) 10 (c) 22 (d) 34 (e) 45

Consider a gray and opaque surface at \(0^{\circ} \mathrm{C}\) in an environment at \(25^{\circ} \mathrm{C}\). The surface has an emissivity of \(0.8\). If the radiation incident on the surface is \(240 \mathrm{~W} / \mathrm{m}^{2}\), the radiosity of the surface is (a) \(38 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(132 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(240 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(300 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(315 \mathrm{~W} / \mathrm{m}^{2}\)

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.

A 70-cm-diameter flat black disk is placed at the center of the ceiling of a \(1-\mathrm{m} \times 1-\mathrm{m} \times 1-\mathrm{m}\) black box. If the temperature of the box is \(620^{\circ} \mathrm{C}\) and the temperature of the disk is \(27^{\circ} \mathrm{C}\), the rate of heat transfer by radiation between the interior of the box and the disk is (a) \(2 \mathrm{~kW}\) (b) \(5 \mathrm{~kW}\) (c) \(8 \mathrm{~kW}\) (d) \(11 \mathrm{~kW}\) (e) \(14 \mathrm{~kW}\)

Two gray surfaces that form an enclosure exchange heat with one another by thermal radiation. Surface 1 has a temperature of \(400 \mathrm{~K}\), an area of \(0.2 \mathrm{~m}^{2}\), and a total emissivity of \(0.4\). Surface 2 has a temperature of \(600 \mathrm{~K}\), an area of \(0.3 \mathrm{~m}^{2}\), and a total emissivity of \(0.6\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(135 \mathrm{~W}\) (b) \(223 \mathrm{~W}\) (c) \(296 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(422 \mathrm{~W}\)
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