Question

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

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

Step-by-step solution

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: \(A = 4 \pi (0.125 \mathrm{~m})^2 = 0.196 \cdot 10^{-1} \mathrm{m}^2\)

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: \(q_c = 12 \cdot 0.196 \cdot 10^{-1}(130 - 25) = 247.68 \mathrm{~W}\)

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: \(q_r = 1 \cdot 5.67 \cdot 10^{-8} \cdot 0.196 \cdot 10^{-1}(403.15^4 - 288.15^4) = 217 \mathrm{~W}\)

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).

Key concepts

Convection

Convection is a mode of heat transfer that occurs through fluids, such as gases and liquids. In this process, heat is transferred by the movement of warmer fluid particles to cooler regions within the medium.
For instance, when a hot object, like the black ball in the exercise, is in contact with air, it initiates convection currents.

• The warm air near the ball surface rises.
• Cooler air descends to replace it.

This cycle continues, facilitating the transfer of heat principally through this fluid movement. The convection heat transfer can be quantified using the formula: \( q_c = hA(T_s - T_\infty) \), where:

• \( q_c \) represents the convection heat transfer rate.
• \( h \) denotes the heat transfer coefficient.
• \( A \) stands for the surface area.
• \( T_s \) and \( T_\infty \) are the surface and ambient air temperatures, respectively.

The efficiency of this process depends on factors like fluid velocity, surface area, and temperature difference.

Radiation

Radiation is another form of heat transfer that does not require a medium. It involves electromagnetic waves and can even occur in a vacuum. Every object emits radiation based on its temperature.
In the context of the exercise, the black ball radiates heat energy to its surroundings, which are at a cooler temperature.

• This mode of transfer is pivotal when there is a significant temperature difference between surfaces.
• Unlike convection, which depends on direct contact with fluids, radiation relies on energy emitted as waves.

The emitted energy increases with the fourth power of the object's temperature, making radiation a powerful heat transfer mechanism when significant temperature differences are present. Hence, surfaces emitting or receiving radiation need to be considered in calculating heat loss accurately in thermal analyses.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in physics that defines the power radiated from an object in terms of its temperature. This law applies specifically to perfect black bodies, which are idealized surfaces that absorb all incident radiation without reflecting any.
According to the law, the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature. It is described as:

• \( q_r = e \sigma A(T_s^4 - T_{surr}^4) \)
• \( \sigma \) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^4 \).
• \( e \) is the emissivity of the surface.

This law is instrumental in understanding and calculating the radiant energy exchange between bodies. It's crucial when analyzing heat transfer scenarios where radiation plays a significant role, such as in the exercise above.

Emissivity

Emissivity is a measure of an object's ability to emit thermal radiation compared to a perfect black body, which has an emissivity of 1.0. It ranges from 0 to 1, where:

• A perfect black body has an emissivity (\( e \)) of 1, meaning it emits the maximum amount of heat possible.
• Real-world objects typically have emissivities less than 1, as they emit less heat.

For the black ball described in the exercise, it is treated as a perfect black body for simplicity, with an emissivity of 1. This assumption helps simplify calculations using the Stefan-Boltzmann Law. Understanding emissivity is crucial because it affects the calculation of radiation heat transfer rates. This parameter helps in precisely determining how much heat is radiated by bodies at different temperatures, thereby playing a key role in thermal energy management solutions.