Question

Why is the radiation analysis of enclosures that consist of black surfaces relatively easy? How is the rate of radiation heat transfer between two surfaces expressed in this case?

Short answer

Answer: Radiation analysis for enclosures with black surfaces is relatively easy because black surfaces absorb all incident radiation and have an emissivity of 1, meaning they are perfect emitters. This property simplifies the analysis since there is no need to account for reflection or transmission. The rate of radiation heat transfer between two black surfaces can be expressed using the simplified equation: \(q_{1 \rightarrow 2} = A_1 F_{1-2} (\sigma T_1^4 - \sigma T_2^4)\), where \(q_{1 \rightarrow 2}\) is the heat transfer rate, \(A_1\) is the area of surface 1, \(F_{1-2}\) is the view factor, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_1\) and \(T_2\) are the temperatures of the surfaces.

Step-by-step solution

Black Surfaces

A black surface is a theoretical surface that absorbs all incident radiation and has an emissivity (e) of 1, which means it is a perfect emitter. In other words, a black surface absorbs all the radiation falling on it without reflecting or transmitting any portion. This property simplifies the radiation analysis since we don't need to account for reflections or transmission.

Radiation Heat Transfer

Radiation heat transfer is the energy transfer as electromagnetic waves. Radiosity, which is the total radiation leaving a surface, can be written as \(J_i = G_i + \frac{E_i}{1 - \epsilon_i}\), where \(J_i\) is the radiosity, \(G_i\) is the irradiation, \(E_i\) is the blackbody emittance, and \(\epsilon_i\) is the emissivity of a surface i. For black surfaces, this relationship is very simple: since \(\epsilon_i = 1\), then \(J_i = E_i\).

Rate of Radiation Heat Transfer between Two Surfaces

The rate of radiation heat transfer between two surfaces can be expressed using the radiosity concepts and Kirchhoff's laws. For black surfaces, the equation can be simplified, as they have an emissivity of 1. The simplified equation for the rate of radiation heat transfer between two black surfaces, A1 and A2, is given by: \(q_{1 \rightarrow 2} = A_1 F_{1-2} (\sigma T_1^4 - \sigma T_2^4)\) where \(q_{1 \rightarrow 2}\) is the heat transfer rate from surface 1 to 2, \(A_1\) is the area of surface 1, \(F_{1-2}\) is the view factor from surface 1 to 2, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_1\) and \(T_2\) are the temperatures of the surfaces 1 and 2. This equation for radiation heat transfer is much simpler for black surfaces, since their radiosity simplifies the analysis and there is no need to account for reflection or transmission.