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A 2-in-diameter spherical ball whose surface is maintained at a temperature of \(170^{\circ} \mathrm{F}\) is suspended in the middle of a room at \(70^{\circ} \mathrm{F}\). If the convection heat transfer coefficient is \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) and the emissivity of the surface is \(0.8\), determine the total rate of heat transfer from the ball.

Short Answer

Expert verified
Answer: The total rate of heat transfer from the spherical ball is approximately \(10.2197\pi\) Btu/h.

Step by step solution

01

Convert diameter to feet

First, we should convert the diameter of the spherical ball from inches to feet. There are 12 inches in a foot, so we'll divide the diameter by 12: $$ d_{ft} = \frac{2 \, \text{in}}{12 \, \frac{\text{in}}{\text{ft}}} = \frac{1}{6} \, \text{ft} $$
02

Calculate the surface area of the spherical ball

Next, we need to determine the surface area of the spherical ball. The surface area of a sphere is given by: $$ A = 4\pi r^2 $$ where \(r\) is the radius of the sphere. Since we have the diameter, we can calculate the radius in feet: $$ r = \frac{d_{ft}}{2} = \frac{1/6 \, \text{ft}}{2} = \frac{1}{12} \, \text{ft} $$ Now we can substitute the radius into the surface area formula: $$ A = 4\pi \left(\frac{1}{12} \, \text{ft}\right)^2 = \frac{\pi}{9} \, \text{ft}^2 $$
03

Calculate the convection heat transfer rate

The convection heat transfer rate is given by: $$ q_{conv} = hA(T_s - T_r) $$ where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, \(T_s\) is the surface temperature of the ball, and \(T_r\) is the room temperature. Plugging in the given values, we have: $$ q_{conv} = (15 \, \frac{\text{Btu}}{\text{h} \cdot \text{ft}^2 \cdot ^{\circ}\text{F}})\left(\frac{\pi}{9} \, \text{ft}^2\right)(170^{\circ}\text{F} - 70^{\circ}\text{F}) = 10\pi \, \text{Btu/h} $$
04

Calculate the radiation heat transfer rate

The radiation heat transfer rate is given by: $$ q_{rad} = \epsilon \sigma A (T_s^4 - T_r^4) $$ where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant of 0.1714脳10鈦烩伕 Btu/h路ft虏路掳R鈦, \(A\) is the surface area, and \(T_s^4\) and \(T_r^4\) are the fourth powers of the surface temperature and room temperature in absolute units, respectively (掳R). First, convert the temperature from 掳F to 掳R: $$ T_s = 170^{\circ}\text{F} + 460 = 630\, ^{\circ}\text{R} $$ $$ T_r = 70^{\circ}\text{F} + 460 = 530\, ^{\circ}\text{R} $$ Now substitute the given values into the radiation heat transfer formula: $$ q_{rad} = (0.8)(0.1714 \times 10^{-8} \, \frac{\text{Btu}}{\text{h} \cdot \text{ft}^2 \cdot ^{\circ}\text{R}^4})\left(\frac{\pi}{9} \, \text{ft}^2\right)(630^{4} - 530^{4}) = 0.2197\pi \, \text{Btu/h} $$
05

Calculate the total rate of heat transfer

Now that we have both the convection and radiation heat transfer rates, we can find the total rate of heat transfer by adding them together: $$ q_{total} = q_{conv} + q_{rad} = (10\pi \, \text{Btu/h}) + (0.2197\pi \, \text{Btu/h}) = 10.2197\pi \, \text{Btu/h} $$ The total rate of heat transfer from the spherical ball is approximately \(10.2197\pi\) Btu/h.

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Key Concepts

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

Convection Heat Transfer
Understanding convection heat transfer is crucial for analyzing how heat moves through gases and liquids. Convection occurs when heat is carried away by moving fluid, which comprises both gases and liquids. This process involves the thermal energy being transferred from the hot surface to the cooler fluid moving past it.

In the context of the solved exercise, the 'convection heat transfer coefficient' describes how efficiently heat is transferred from the spherical ball鈥檚 surface to the air around it in the room. A higher coefficient indicates a greater ability to transfer heat. The formula used to calculate the convection heat transfer rate is: \[q_{conv} = hA(T_s - T_r)\], where \(h\) represents the convection heat transfer coefficient, \(A\) is the surface area of the heated object, \(T_s\) stands for the surface temperature of the object, and \(T_r\) is the temperature of the surrounding fluid, in this case, the room temperature.

Generally, convection heating can be influenced by factors such as the surface area of the object, the temperature difference between the object and surroundings, and the properties of the fluid, including its velocity and viscosity.
Radiation Heat Transfer
Radiation heat transfer is another key concept in thermodynamics and involves the emission of energy as electromagnetic waves. Every object emits and absorbs radiant energy and the amount of radiation emitted by a surface depends on its temperature and its ability to emit energy, described by its emissivity coefficient (\(\text{蔚}\)).

The exercise uses the Stefan-Boltzmann law to calculate the rate of heat transfer due to radiation using the formula: \[q_{rad} = \text{蔚蟽}A(T_s^4 - T_r^4)\]. Here, \(蟽\) refers to the Stefan-Boltzmann constant, and the temperatures \(T_s\) and \(T_r\) must be expressed in absolute units to account for the radiation emitted and absorbed by the object and its surroundings. The inclusion of the emissivity coefficient means that a perfect black body, with an emissivity of 1, would emit the maximum amount of radiation possible at a given temperature, while real objects typically have lower emissivities, as seen with the surface emissivity of 0.8 used in our exercise.
Spherical Surface Area
The surface area of a sphere is a fundamental geometric calculation often needed in thermal analysis. It's important because it determines the amount of area available for heat transfer. The surface area of a sphere is given by the formula: \[A = 4\text{蟺}r^2\], where \(r\) is the sphere鈥檚 radius.

In our exercise, we calculate the surface area of a spherical ball to determine the extent of the surface involved in both convection and radiation heat transfer processes. The area plays a pivotal role in both calculations, as it directly relates to the amount of heat that can be transferred per unit time. The larger the surface area, the more area is available for heat transfer.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant (\(蟽\)) is a pivotal proportionality constant in physics, used to describe the power radiated from a black body in terms of its temperature. Specifically, it's utilized in the Stefan-Boltzmann law, which states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time \(j^{\text{*}}\) is directly proportional to the fourth power of the black body's temperature \(T\): \[j^* = \text{蟽}T^4\].

The constant鈥檚 value in the given exercise is 0.1714脳10鈦烩伕 Btu/h路ft虏路掳R鈦, which, when combined with the sphere鈥檚 surface area and the differential of the fourth powers of the surface and room temperatures in Rankine, provides the rate of radiant heat loss. Understanding this constant and the law it pertains to is essential for thermal physics, enabling the calculation of radiative heat transfer for various applications, including our example of a heated spherical ball.

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Most popular questions from this chapter

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) \(2365 \mathrm{~W}\)

The heat generated in the circuitry on the surface of a silicon chip \((k=130 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is conducted to the ceramic substrate to which it is attached. The chip is \(6 \mathrm{~mm} \times 6 \mathrm{~mm}\) in size and \(0.5 \mathrm{~mm}\) thick and dissipates \(5 \mathrm{~W}\) of power. Disregarding any heat transfer through the \(0.5-\mathrm{mm}\) high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation.

Over 90 percent of the energy dissipated by an incandescent light bulb is in the form of heat, not light. What is the temperature of a vacuum-enclosed tungsten filament with an exposed surface area of \(2.03 \mathrm{~cm}^{2}\) in a \(100 \mathrm{~W}\) incandescent light bulb? The emissivity of tungsten at the anticipated high temperatures is about \(0.35\). Note that the light bulb consumes \(100 \mathrm{~W}\) of electrical energy, and dissipates all of it by radiation. (a) \(1870 \mathrm{~K}\) (b) \(2230 \mathrm{~K}\) (c) \(2640 \mathrm{~K}\) (d) \(3120 \mathrm{~K}\) (e) \(2980 \mathrm{~K}\)

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.

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

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