Question

Consider a 20-cm-diameter hemispherical enclosure. The dome is maintained at \(600 \mathrm{~K}\) and heat is supplied from the dome at a rate of \(50 \mathrm{~W}\) while the base surface with an emissivity of \(0.55\) is maintained at \(400 \mathrm{~K}\). Determine the emissivity of the dome.

Short answer

Answer: The emissivity of the dome is approximately 0.0802.

Step-by-step solution

Write down the given information

Diameter of the hemisphere: D = 20 cm or 0.2 m Dome temperature: T_dome = 600 K Base surface temperature: T_base = 400 K Heat supplied to the dome: Q = 50 W Emissivity of the base surface: e_base = 0.55

Calculate the surface area of the hemisphere and the area of the base surface

A_hemisphere = 2 * pi * r^2 A_base = pi * r^2 where r is the radius of the hemisphere r = D/2 = 0.1 m A_hemisphere = 2 * pi * (0.1)^2 = 0.0628 m^2 A_base = pi * (0.1)^2 = 0.0314 m^2

Use the Stefan-Boltzmann Law to find the total power radiated from the base surface

The Stefan-Boltzmann Law states that the power radiated from a surface is given by: P = emissivity * A * sigma * T^4 For the base surface: P_base = e_base * A_base * sigma * T_base^4 where sigma is the Stefan-Boltzmann constant: sigma = 5.67 × 10^-8 W/(m^2·K^4) P_base = 0.55 * 0.0314 * 5.67 × 10^-8 * (400)^4 = 43.56 W

Calculate the power absorbed by the dome

The power absorbed by the dome should be equal to the difference between the heat supplied to the dome and the power radiated from the base surface. P_absorbed_by_dome = Q - P_base P_absorbed_by_dome = 50 - 43.56 = 6.44 W

Determine the emissivity of the dome

Using the Stefan-Boltzmann Law again for the dome: P_absorbed_by_dome = e_dome * A_hemisphere * sigma * T_dome^4 Solving for e_dome, we get: e_dome = P_absorbed_by_dome / (A_hemisphere * sigma * T_dome^4) e_dome = 6.44 / (0.0628 * 5.67 × 10^-8 * (600)^4) = 0.0802 The emissivity of the dome is approximately 0.0802.

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in the field of thermodynamics, particularly concerning radiative heat transfer. This law relates the power radiated from a black body in terms of temperature to the fourth power. The mathematical expression of this law is given by \( P = \epsilon \cdot A \cdot \sigma \cdot T^4 \), where \( P \) is the total power radiated per unit surface area, \( \epsilon \) is the emissivity of the material, \( A \) is the area of the surface that is emitting or absorbing radiation, \( \sigma \) is the Stefan-Boltzmann constant, approximately equal to \( 5.67 \times 10^{-8} \text{W/(m}^2\text{K}^4\text{)} \), and \( T \) is the absolute temperature in Kelvin.

To understand emissivity, one must grasp that it's a measure of a surface's ability to emit thermal radiation relative to that of an ideal black body. A black body, with an emissivity of 1, is a perfect emitter, absorbing and radiating all energy without reflection. Real-world objects have emissivities less than 1, and this property varies with temperature and the characteristics of the material.

Hemispherical Enclosure

A hemispherical enclosure is a half-sphere shape that is often used in physics and engineering problems related to heat transfer and radiation. In our textbook exercise, this enclosure helps us analyze how heat is being radiated and absorbed due to its shape and properties.

The significance of the hemispherical geometry lies in the uniform distribution of radiated energy on its inside surface. When dealing with such enclosures, the surface area calculation becomes essential. This is because the radiative heat transfer at any point on the inside surface is uniform due to the shape of the hemisphere. To calculate the surface area of a hemispherical enclosure, the following equations are used: \( A_{\text{hemisphere}} = 2 \cdot \pi \cdot r^2 \) and \( A_{\text{base}} = \pi \cdot r^2 \), where \( r \) is the radius. The total surface area is critical since it directly impacts the heat transfer calculations through the Stefan-Boltzmann Law.

Heat Transfer

Heat transfer is a vital concept in various fields, including physics, engineering, and environmental studies. It entails the movement of heat from one area or substance to another and can occur through various mechanisms: conduction, convection, and radiation. In the context of our exercise, we're focused on radiative heat transfer, which is the energy transferred in the form of electromagnetic waves and does not require any medium.

Radiative heat transfer is especially relevant in applications involving vacuum or space where other heat transfer methods are not applicable. In the scenario with the hemispherical enclosure, we calculate the emissivity of the dome by understanding that heat is radiated from the base and absorbed by the dome. The balance of this radiation, as it is applied through the Stefan-Boltzmann Law, helps us determine the heat transfer rate and characterize the dome's material in terms of its ability to emit or absorb heat.