Question

Two parallel disks of diameter \(D=0.6 \mathrm{~m}\) separated by \(L=0.4 \mathrm{~m}\) are located directly on top of each other. Both disks are black and are maintained at a temperature of \(450 \mathrm{~K}\). The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at \(300 \mathrm{~K}\). Determine the net rate of radiation heat transfer from the disks to the environment.

Short answer

According to the information provided and the solutions outlined in the steps above, please follow these calculations: Step 2: Calculate the radiative heat flux from the disks \(q_1 = (1)(5.67 \times 10^{-8}\ \mathrm{W/m^2K^4})(450 \mathrm{~K})^4 = 4010.1\ \mathrm{W/m^2}\) Step 3: Calculate the radiative heat flux returning to the disks from the environment \(q_2 = (1)(5.67 \times 10^{-8}\ \mathrm{W/m^2K^4})(300 \mathrm{~K})^4 = 459.3\ \mathrm{W/m^2}\) Step 4: Calculate the net rate of radiation heat transfer from the disks to the environment First, calculate the area of one disk: \(A = \pi (\frac{0.6}{2})^2 = \pi(0.3)^2 = 0.2827\ \mathrm{m^2}\) Now, calculate the net rate: \(q_{net} = 2A(q_1 - q_2) = 2(0.2827\ \mathrm{m^2}) (4010.1\ \mathrm{W/m^2} - 459.3\ \mathrm{W/m^2}) = 2 \times 0.2827 \times 3550.8\ \mathrm{W/m^2} = 2006.6\ \mathrm{W}\) Thus, the net rate of radiation heat transfer from the disks to the environment is 2006.6 W.

Step-by-step solution

Define the variables

The diameter of the disks is \(D=0.6 \mathrm{~m}\), the separation between the disks is \(L=0.4 \mathrm{~m}\), the temperature of the disks is \(T_1 = 450 \mathrm{~K}\), and the temperature of the environment is \(T_2 = 300 \mathrm{~K}\). We also know that both disks are black, and thus have the emissivity value \(\epsilon = 1\).

Calculate the radiative heat flux from the disks

Since the disks are black, we can use the Stefan-Boltzmann Law to determine the radiative heat flux emitted by the disks. The formula to find the radiative heat flux is: \(q = \epsilon \sigma T^4\) where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, which is equal to \(5.67 \times 10^{-8}\ \mathrm{W/m^2K^4}\), and \(T\) is the temperature in Kelvin. For each disk, the radiative heat flux emitted is: \(q_1 = \epsilon_1 \sigma T_1^4\) \(q_1 = (1)(5.67 \times 10^{-8}\ \mathrm{W/m^2K^4})(450 \mathrm{~K})^4\) Calculate \(q_1\).

Calculate the radiative heat flux returning to the disks from the environment

We also need to determine the radiative heat flux that comes back to the disks from the environment. The radiative heat flux from the environment returning to the disks is governed by its temperature \(T_2\). Using the same formula, we have: \(q_2 = \epsilon_2 \sigma T_2^4\) \(q_2 = (1)(5.67 \times 10^{-8}\ \mathrm{W/m^2K^4})(300 \mathrm{~K})^4\) Calculate \(q_2\).

Calculate the net rate of radiation heat transfer from the disks to the environment

To find the net rate of radiation heat transfer from the disks to the environment, we can subtract the radiative heat flux returning to the disks from the total radiative heat flux emitted by the disks. We also need to account for both disks, so we multiply the net flux by 2 and by the area of one disk, \(A\). The formula for the area of a disk is: \(A = \pi (\frac{D}{2})^2\) Calculate \(A\). Now, we can calculate the net rate like this: \(q_{net} = 2A(q_1 - q_2)\) Plug the values of \(A\), \(q_1\), and \(q_2\) into the formula and calculate the net rate of radiation heat transfer from the disks to the environment.

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is fundamental in understanding radiation heat transfer. It relates the amount of energy radiated from a black body in terms of its temperature. Specifically, it states that the energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature.

Mathematically, the law is expressed as:
\[ q = \.\epsilon \sigma T^4 \]
where \( q \) is the total radiated energy per unit area, \( \.\epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \mathrm{W/m^2K^4} \)), and \( T \) is the absolute temperature in kelvin.

This law is crucial for calculating the radiative heat transfer between objects and their surroundings, and for the design of systems where such transfer is significant, such as thermal insulation materials, radiators, and solar panels.

Emissivity

Emissivity is a measure of an object's ability to emit infrared energy compared to that of an ideal emitter, known as a black body. It is dimensionless and ranges from 0 to 1, where 1 corresponds to a perfect black body that emits the maximum amount of radiation possible at a given temperature, and 0 corresponds to a perfect reflector that emits no thermal radiation.

Factors Affecting Emissivity

The emissivity of a material can depend on several factors, such as:

• Temperature: Some materials change their emissivity with temperature.
• Surface roughness: Rougher surfaces tend to have higher emissivity.
• Material composition: Different materials inherently have different emissivity values.
• Surface coatings: Coatings can be applied to alter a material's emissivity.

In practical applications, engineers must consider the emissivity of materials when designing systems and predicting heat transfer rates, as it directly affects the Stefan-Boltzmann Law calculations.

Blackbody Radiation

Blackbody radiation is the thermal electromagnetic radiation emitted by an object that is an ideal emitter at all wavelengths. The term 'black body' might seem counterintuitive, but it signifies that the object absorbs all incident radiation, regardless of frequency or angle of incidence, which means it also emits the maximum amount of radiation possible.

A perfect black body with an emissivity of 1 is a theoretical concept, as all real objects have emissivities less than 1. Blackbody radiation forms the basis of the field of thermal radiation and is characterized by a spectrum dependent solely on the body's temperature, as described by the Planck's law.

Importance in Heat Transfer

Understanding blackbody radiation is fundamental for thermography, the development of infrared sensors, and in astrophysics, for studying the thermal properties of stars and planets. Moreover, calculating the net radiative heat transfer, as given in the original exercise, requires considering both the emission from the object and the absorption of blackbody radiation from its environment.