Question

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

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

Step-by-step solution

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: $$ d_{ft} = \frac{2 \, \text{in}}{12 \, \frac{\text{in}}{\text{ft}}} = \frac{1}{6} \, \text{ft} $$

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: $$ A = 4\pi \left(\frac{1}{12} \, \text{ft}\right)^2 = \frac{\pi}{9} \, \text{ft}^2 $$

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: $$ q_{conv} = (15 \, \frac{\text{Btu}}{\text{h} \cdot \text{ft}^2 \cdot ^{\circ}\text{F}})\left(\frac{\pi}{9} \, \text{ft}^2\right)(170^{\circ}\text{F} - 70^{\circ}\text{F}) = 10\pi \, \text{Btu/h} $$

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 0.1714×10⁻⁸ Btu/h·ft²·°R⁴, \(A\) 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 (°R). First, convert the temperature from °F to °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: $$ q_{rad} = (0.8)(0.1714 \times 10^{-8} \, \frac{\text{Btu}}{\text{h} \cdot \text{ft}^2 \cdot ^{\circ}\text{R}^4})\left(\frac{\pi}{9} \, \text{ft}^2\right)(630^{4} - 530^{4}) = 0.2197\pi \, \text{Btu/h} $$

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.

Key concepts

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’s 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 (\(\text{ε}\)).

The exercise uses the Stefan-Boltzmann law to calculate the rate of heat transfer due to radiation using the formula: \[q_{rad} = \text{εσ}A(T_s^4 - T_r^4)\]. Here, \(σ\) 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: \[A = 4\text{π}r^2\], where \(r\) is the sphere’s 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 (\(σ\)) 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 \(T\): \[j^* = \text{σ}T^4\].

The constant’s value in the given exercise is 0.1714×10⁻⁸ Btu/h·ft²·°R⁴, which, when combined with the sphere’s 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.