Question

This experiment is conducted to determine the emissivity of a certain material. A long cylindrical rod of diameter \(D_{1}=0.01 \mathrm{~m}\) is coated with this new material and is placed in an evacuated long cylindrical enclosure of diameter \(D_{2}=0.1 \mathrm{~m}\) and emissivity \(\varepsilon_{2}=0.95\), which is cooled externally and maintained at a temperature of \(200 \mathrm{~K}\) at all times. The rod is heated by passing electric current through it. When steady operating conditions are reached, it is observed that the rod is dissipating electric power at a rate of \(8 \mathrm{~W}\) per unit of its length and its surface temperature is \(500 \mathrm{~K}\). Based on these measurements, determine the emissivity of the coating on the rod.

Short answer

Question: Determine the emissivity of the coating on the cylindrical rod given the following data: diameter of the rod is 0.01 m, diameter of the enclosure is 0.1 m, emissivity of the enclosure is 0.95, temperature of the enclosure is 200 K, power dissipation rate of the rod per unit length is 8 W/m, and the temperature of the rod is 500 K. Answer: The emissivity of the coating on the cylindrical rod is approximately 0.348.

Step-by-step solution

Understand given information

We have the following information given: - Diameter of the rod, \(D_1 = 0.01 m\) - Diameter of the enclosure, \(D_2 = 0.1 m\) - Emissivity of the enclosure, \(\varepsilon_2 = 0.95\) - Temperature of the enclosure, \(T_2 = 200 K\) - Power dissipation rate of the rod per unit length, \(q_s = 8 W/m\) - Temperature of the rod, \(T_1 = 500 K\)

Determine the radiative heat transfer between the rod and the enclosure

As the rod is in an evacuated enclosure, the only mode of heat transfer will be radiation. The radiative heat transfer between the rod and the enclosure can be determined using the Stefan-Boltzmann law, given by: \(q = \sigma \times A_1 \times (\varepsilon_1(T_1^4 - \varepsilon_2 T_2^4))\) Where, \(q\) is the heat transfer per unit length between the rod and enclosure (in \(W/m\)) \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} W/(m^2K^4)\) \(A_1\) is the surface area of the rod per unit length, \(A_1 = \pi D_1\) \(\varepsilon_1\) is the emissivity of the coating on the rod, which we need to determine.

Relate the power dissipation rate with the radiative heat transfer

The power dissipation rate of the rod per unit length, \(q_s\), is equal to the radiative heat transfer per unit length, \(q\). We can rewrite the equation for radiative heat transfer as: \(q_s = \sigma \times A_1 \times (\varepsilon_1(T_1^4 - \varepsilon_2 T_2^4))\)

Solve for the emissivity of the coating on the rod, \(\varepsilon_1\)

We can now solve the equation from Step 3 for \(\varepsilon_1\): \(q_s = \sigma \times A_1 \times \varepsilon_1 (T_1^4 - \varepsilon_2 T_2^4)\) \(\varepsilon_1 = \frac{q_s}{\sigma \times A_1 \times (T_1^4 - \varepsilon_2 T_2^4)}\) Plugging in the given values and solving for \(\varepsilon_1\): \(\varepsilon_1 = \frac{8}{(5.67 \times 10^{-8})(\pi \times 0.01)(500^4 - 0.95 \times 200^4)}\) After evaluating the expression, we get: \(\varepsilon_1 \approx 0.348 \) So, the emissivity of the coating on the rod is approximately 0.348.

Key concepts

Radiative Heat Transfer

Radiative heat transfer plays a vital role in understanding how energy is exchanged in the form of electromagnetic radiation between bodies at different temperatures. It is especially significant in environments with a vacuum or where conduction and convection are negligible.

In our scenario involving a heated rod within an evacuated enclosure, the transfer of energy occurs solely through radiation. When two surfaces with differing temperatures face one another, the radiative heat transfer can be described quantitatively using the equation:
\[ q = \text{Emissive Power} - \text{Absorbed Power} \]
where q represents the net heat transfer per unit time (in watts), the emissive power is the energy emitted by the rod, and the absorbed power is that which is absorbed from the enclosure. The surface properties, geometry of the bodies, and temperature difference all influence the net radiative heat transfer. This process is beautifully depicted by the Stefan-Boltzmann law in our experiment.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a pillar of understanding radiative heat transfer. It relates the heat emitted by a blackbody to its temperature, and can be expressed as:\[ q = \text{Emissive Power} =\sigma \times A \times (\text{Temperature})^4 \]
where A stands for the emitting surface area, \sigma is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} W/(m^2K^4)\)), and Temperature is the absolute temperature in kelvins raised to the fourth power.

This relation is crucial for our emissivity determination experiment. The law postulates that the power radiated from the rod, which we equate to the electrical power dissipated, is proportional to the fourth power of the rod’s absolute temperature. However, since real materials are not perfect blackbodies, their emissivity, \varepsilon, must be factored into the equation, leading us to the modified form that includes the product \varepsilon \times \sigma \times A \times (\text{Temperature})^4. This adjusted equation allows us to solve for the emissivity of our rod's coating by comparing the power dissipated with the radiative heat transfer.

Thermal Conductivity

Thermal conductivity is a physical property of materials that quantifies the ability to conduct heat. It is denoted by the symbol k and typically measured in units of W/(m⋅K). Materials with high thermal conductivity, like metals, easily transfer heat, whereas those with low thermal conductivity, such as foam insulations, are effective thermal insulators.

In the context of our experiment, thermal conductivity does not have a direct role as the rod is placed in a vacuum and there are no contacting materials to conduct the heat. However, understanding thermal conductivity is essential when considering heat transfer in other non-vacuum contexts or if the rod were in contact with other materials. For comprehensive thermal analyses, the concept of thermal conductivity serves as a foundational component alongside radiative heat transfer, particularly in situations where conduction, convection, and radiation work together to transfer energy.