Question

Two long parallel 20 -cm-diameter cylinders are located \(30 \mathrm{~cm}\) apart from each other. Both cylinders are black, and are maintained at temperatures \(425 \mathrm{~K}\) and \(275 \mathrm{~K}\). The surroundings can be treated as a blackbody at \(300 \mathrm{~K}\). For a 1 -m-long section of the cylinders, determine the rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings.

Short answer

Answer: The rate of radiation heat transfer between the two cylinders is approximately 529.52 W, and the rate of radiation heat transfer between the hot cylinder and the surroundings is approximately 766.64 W.

Step-by-step solution

Determine the surface area of the 1-meter-long section of each cylinder

First, we need to find the surface area of a 1-meter-long section of each cylinder, since the rate of radiation heat transfer depends on the surface area. Given the diameter of each cylinder is 20 cm, we can find the radius: $$ r = \frac{d}{2} = \frac{20}{2} = 10 \mathrm{~cm} = 0.1 \mathrm{~m} $$ Next, we will calculate the surface area of a 1-meter-long section: $$ A = 2\pi r L = 2\pi (0.1) (1) $$ \(A \approx 0.2\pi \mathrm{~m}^2\)

Calculate the view factor between the cylinders

For parallel cylinders, the "view factor" represents the fraction of the radiative energy leaving one cylinder that reaches the other cylinder. In this case, we can find the view factor using the formula: $$ F = \frac{\sqrt{1-(\frac{2r}{D})^2}}{1-\frac{2r}{D}} $$ where \(D\) is the distance between the cylinders. Plugging in the given values, we have: $$ F = \frac{\sqrt{1-(\frac{2(0.1)}{0.3})^2}}{1-\frac{2(0.1)}{0.3}} $$ \(F \approx 0.462\)

Calculate the rate of radiation heat transfer between the cylinders

Now, we will calculate the rate of radiation heat transfer \(Q\) between the cylinders using the following formula: $$ Q = F \sigma A(T_1^4 - T_2^4) $$ where \(\sigma\) is the Stefan-Boltzmann constant, which has a value of \(5.67\times10^{-8}\mathrm{~W/(m^2K^4)}\), \(A\) is the surface area of each cylinder, \(T_1\) and \(T_2\) are the temperatures of the cylinders, and \(F\) is the view factor between them. Plugging in the values, we get: $$ Q = (0.462)(5.67\times10^{-8})(0.2\pi)(425^4 - 275^4) $$ \(Q \approx 529.52 \mathrm{~W}\)

Calculate the rate of radiation heat transfer between the hot cylinder and the surroundings

Finally, we will calculate the rate of radiation heat transfer between the hot cylinder and the surroundings. To do this, we will use the formula: $$ Q_\text{surr} = \sigma A (T_1^4 - T_\text{surr}^4) $$ where \(T_\text{surr}\) is the temperature of the surroundings. Plugging in the values, we get: $$ Q_\text{surr} = (5.67\times10^{-8})(0.2\pi)(425^4 - 300^4) $$ \(Q_\text{surr} \approx 766.64 \mathrm{~W}\) To summarize: - The rate of radiation heat transfer between the two cylinders is approximately 529.52 W. - The rate of radiation heat transfer between the hot cylinder and the surroundings is approximately 766.64 W.

Key concepts

Understanding the Stefan-Boltzmann Constant

The Stefan-Boltzmann constant, denoted as \(\sigma\), plays a critical role in radiation heat transfer calculations. Its value is approximately \(5.67\times10^{-8} \text{W/m}^2\text{K}^4\), and it helps us understand the amount of heat energy a black body emits per unit area based on the body's temperature. In our example with the cylindrical bodies, knowing the values of their temperatures is crucial as the rate of heat transfer is directly proportional to the fourth power of the absolute temperature difference. This means even small changes in temperature can significantly affect the heat transfer rate.

To illustrate this, we use the Stefan-Boltzmann law, which states that the power emitted per unit area of a black body is proportional to the fourth power of its absolute temperature. Mathematically, this is expressed as \(P = \sigma T^4\), where \(P\) is the emitted power per unit area, \(T\) is the absolute temperature, and \(\sigma\) is the Stefan-Boltzmann constant. This concept allows us to calculate the amount of radiation emitted by objects, such as the cylinders in our exercise, and determine the heat exchange between them and their surroundings.

Surface Area Calculation for Heat Transfer

In the realm of heat transfer, the surface area of an object is crucial because it dictates how much heat energy can be exchanged with its environment. Our example problem involves calculating the surface area of a 1-meter-long cylindrical body. The surface area is the part of the cylinder that is exposed and can radiate heat.

The formula for the lateral surface area of a cylinder is \( A = 2\pi rL \), where \( r \) is the radius and \( L \) is the length of the cylinder. After substituting the provided dimensions into the equation, we found the surface area to be \(0.2\pi \mathrm{m}^2\). With this surface area, we can further compute the radiative heat transfer, emphasizing the indispensable role surface area serves in thermal physics. Understanding surface area is not just about plugging values into a formula; it's also about grasping its influence on heat radiation potential. The larger the area, the more significant the energy exchange, which is particularly important for engineers and scientists designing systems that involve temperature regulation or heat dissipation.

Deciphering the View Factor in Radiative Heat Exchange

When dealing with radiation heat transfer between objects, the view factor, or configuration factor, is a dimensionless quantity that measures how well one body 'sees' another with respect to radiative exchange. In our cylinders' case, the view factor defines the percentage of radiative energy that leaves one cylinder and directly reaches the other.

Calculating the view factor involves both geometry and relative positioning. For parallel cylinders, as in our example, a specific formula is used which takes into account the diameter of each cylinder and the distance between them. By applying this formula, we have determined that the view factor for our system is approximately \(0.462\). This figure implies that about 46.2% of the radiation leaving one cylinder reaches the other. This concept is crucial for designing and analyzing heating systems, manufacturing processes, and even in environmental studies where radiation plays a role in energy budgets. Understanding the geometrical relationship between objects in terms of how they exchange radiative heat can greatly enhance the accuracy and efficiency of thermal systems.