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Chapter 1: Problem 160

A 25 -cm-diameter black ball at \(130^{\circ} \mathrm{C}\) is suspended in air, and is losing heat to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat transfer from the black ball is (a) \(217 \mathrm{~W}\) (b) \(247 \mathrm{~W}\) (c) \(251 \mathrm{~W}\) (d) \(465 \mathrm{~W}\) (e) \(2365 \mathrm{~W}\)

Short Answer

Expert verified
Answer: The total rate of heat transfer from the black ball is approximately 465 W.

Step by step solution

01

Find the surface area of the black ball

The surface area of a sphere (the black ball) can be calculated using the formula: \(A = 4 \pi r^2\) Given the diameter, we can find the radius: \(r = \frac{d}{2} = \frac{25 \mathrm{~cm}}{2} = 12.5 \mathrm{~cm} = 0.125 \mathrm{~m}\) Now, we can find the surface area: \(A = 4 \pi (0.125 \mathrm{~m})^2 = 0.196 \cdot 10^{-1} \mathrm{m}^2\)
02

Calculate the convection heat transfer rate

To find the convection heat transfer rate, we can use the formula: \(q_c = hA(T_s - T_\infty)\) where \(q_c\) is the convection heat transfer rate, \(h = 12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the heat transfer coefficient, \(A\) is the surface area, \(T_s = 130^{\circ} \mathrm{C}\) is the surface temperature of the ball, and \(T_\infty = 25^{\circ} \mathrm{C}\) is the surrounding air temperature. Plugging in the values, we get: \(q_c = 12 \cdot 0.196 \cdot 10^{-1}(130 - 25) = 247.68 \mathrm{~W}\)
03

Calculate the radiation heat transfer rate

To find the radiation heat transfer rate, we can use the formula: \(q_r = e \sigma A(T_s^4 - T_surr^4)\) where \(q_r\) is the radiation heat transfer rate, \(e = 1\) is the emissivity of the black ball (assuming it's a black body), \(\sigma = 5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^4\) is the Stefan-Boltzmann constant, and \(T_surr = 15^{\circ} \mathrm{C}\) is the surrounding surface temperature. First, convert the temperatures to Kelvin: \(T_{s,K} = 130 + 273.15 = 403.15 \mathrm{~K}\) \(T_{surr,K} = 15 + 273.15 = 288.15 \mathrm{~K}\) Now we can plug in the values: \(q_r = 1 \cdot 5.67 \cdot 10^{-8} \cdot 0.196 \cdot 10^{-1}(403.15^4 - 288.15^4) = 217 \mathrm{~W}\)
04

Calculate the total heat transfer rate

Now we can find the total heat transfer rate by adding the rates calculated in steps 2 and 3: \(q_{total} = q_c + q_r = 247.68+ 217 = 464.68 \mathrm{~W}\) Since all the answers are given in whole numbers, we can safely round the result, and the total rate of heat transfer from the black ball is approximately: \(q_{total} \approx 465 \mathrm{~W}\) The correct option is (d).

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Most popular questions from this chapter

Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?

On a still clear night, the sky appears to be a blackbody with an equivalent temperature of \(250 \mathrm{~K}\). What is the air temperature when a strawberry field cools to \(0^{\circ} \mathrm{C}\) and freezes if the heat transfer coefficient between the plants and air is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) because of a light breeze and the plants have an emissivity of \(0.9\) ? (a) \(14^{\circ} \mathrm{C}\) (b) \(7^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-3^{\circ} \mathrm{C}\)

Define emissivity and absorptivity. What is Kirchhoff's law of radiation?

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.

Consider a person standing in a room maintained at \(20^{\circ} \mathrm{C}\) at all times. The inner surfaces of the walls, floors, and ceiling of the house are observed to be at an average temperature of \(12^{\circ} \mathrm{C}\) in winter and \(23^{\circ} \mathrm{C}\) in summer. Determine the rates of radiation heat transfer between this person and the surrounding surfaces in both summer and winter if the exposed surface area, emissivity, and the average outer surface temperature of the person are \(1.6 \mathrm{~m}^{2}, 0.95\), and \(32^{\circ} \mathrm{C}\), respectively.
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