Question

A 1-m-diameter spherical cavity is maintained at a uniform temperature of \(600 \mathrm{~K}\). Now a 5 -mm-diameter hole is drilled. Determine the maximum rate of radiation energy streaming through the hole. What would your answer be if the diameter of the cavity were \(3 \mathrm{~m}\) ?

Short answer

#tag_title#Short Answer#tag_content# To determine the maximum rate of radiation energy streaming through the hole for the two cases, we can follow these steps: 1. Calculate the emitted energy by the walls of the cavity using the Stefan-Boltzmann law and the given cavity diameter, temperature, and constant. 2. Compute the fraction of energy passing through the hole using the solid angle and complete solid angle. 3. Calculate the maximum rate of radiation energy streaming through the hole by multiplying the emitted energy with the fraction of energy passing through the hole. For the case with a 1-meter diameter cavity, the maximum rate of radiation energy streaming through the hole is represented by \(P_1\). For the case with a 3-meter diameter cavity, the maximum rate of radiation energy streaming through the hole is represented by \(P_2\).

Step-by-step solution

Identify the variables given

We are given the following variables: - The diameter of the spherical cavity: \(D_1 = 1~m\) - The diameter of the hole drilled in the cavity: \(d = 5~mm = 0.005~m\) - The temperature of the cavity: \(T = 600~K\) (uniform temperature) - Stefan-Boltzmann constant: \(\sigma = 5.67 \times 10^{-8} \frac{W}{m^2 K^4}\) We need to answer the rate of radiation energy streaming through the hole for two cases: 1. The cavity's diameter is 1 meter. 2. The cavity's diameter is 3 meters.

Calculate the radiation energy emitted by the walls of the cavity

First, we must calculate the radiation energy emitted by the surface of the cavity using the Stefan-Boltzmann law as follows: \(E = A\sigma T^4\) Where \(E\) is the emitted energy, \(A\) is the surface area, and \(\sigma\) is the Stefan-Boltzmann constant. For the spherical cavity, the surface area can be calculated as: \(A = 4\pi R^2\) Where \(R\) is the radius of the cavity. We can compute the surface area for the given diameter \(D_1\): \(A_1 = 4\pi \left(\frac{D_1}{2}\right)^2\) Now calculate the emitted energy for the first case when the diameter of the cavity is \(1~m\): \(E_1 = A_1\sigma T^4\)

Calculate the fraction of energy passing through the hole

Now we need to compute the fraction of energy passing through the hole. We can do it by calculating the solid angle \(\Omega\) created by the hole in the cavity as follows: \(\Omega = \frac{A_{hole}}{R^2}\) Where \(A_{hole}\) is the area of the hole, which can be calculated as: \(A_{hole} = \pi \left(\frac{d}{2}\right)^2\) Now calculate the solid angle: \(\Omega = \frac{A_{hole}}{\left(\frac{D_1}{2}\right)^2}\) Next, we need to find the fraction of the energy that passes through the hole, which is given by the solid angle divided by the complete solid angle \(4\pi\): \(f = \frac{\Omega}{4\pi}\)

Calculate the maximum rate of radiation energy streaming through the hole for the first case

Now we can find the maximum rate of radiation energy streaming through the hole for the first case, when the diameter of the cavity is \(1~m\). The maximum rate can be calculated by multiplying the emitted energy with the fraction of energy passing through the hole: \(P_1 = E_1 \times f\)

Calculate the maximum rate of radiation energy streaming through the hole for the second case

Similarly, we can calculate the maximum rate of radiation energy streaming through the hole for the second case, when the diameter of the cavity is \(3~m\). First, we need to compute the surface area for the given diameter \(D_2 = 3~m\): \(A_2 = 4\pi \left(\frac{D_2}{2}\right)^2\) Now calculate the emitted energy for the second case: \(E_2 = A_2\sigma T^4\) Since the diameter of the hole is the same, the fraction of energy passing through the hole \(f\) remains unchanged. Thus, we can directly calculate the maximum rate of radiation energy streaming through the hole for the second case: \(P_2 = E_2 \times f\) Now, we have the maximum rates for both cases, \(P_1\) and \(P_2\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a cornerstone of thermal physics that relates the energy radiated by a black body in equilibrium per unit area to the fourth power of the temperature. This relationship is expressed mathematically as:
\( E = \text{A}\text{σ}T^4 \)
Where \( E \) is the total emitted radiation energy per unit time, \( A \) is the surface area of the black body, \( \text{σ} \) is the Stefan-Boltzmann constant, which is approximately \( 5.67 \times 10^{-8} \text{W/m}^2\text{K}^4 \), and \( T \) is the absolute temperature in Kelvin.
When considering a spherical cavity as a black body, we use this law to estimate the rate at which radiation energy is emitted. This estimation is especially useful for theoretical and practical applications, where understanding heat transfer rates is vital. By knowing the surface area of the spherical cavity and its temperature, we can calculate the energy radiated from its interior walls, which forms the foundation for determining how much energy streams through any openings, such as a drilled hole.

Spherical Cavity Radiation

Radiation within a spherical cavity is an interesting application of the Stefan-Boltzmann law. The uniform temperature of the cavity's walls ensures a steady emission of thermal radiation towards the center. The emitted energy is based on the interior surface area of the sphere.
For a cavity with radius \( R \), the surface area \( A \) is calculated using the formula:
\( A = 4\textπR^2 \)
From here, the total energy emitted by the spherical cavity can be found using the Stefan-Boltzmann law. If a hole is introduced into the cavity, some of the radiation inside will escape through this opening, which ties directly into understanding concepts such as solid angle and energy streaming outlined in the problem posed. The role of the cavity's size in relation to the hole drilled is essential for determining how much energy is lost; a larger cavity with the same-sized hole would inherently lose less energy through that hole due to geometric considerations.

Solid Angle in Radiation

In the context of radiation and cavities, the solid angle is a measure of how large the 'cone' of vision is from a point inside the cavity towards the hole through which radiation can escape. The larger the solid angle, the more energy can pass through the hole.
Mathematically, a solid angle \( \text{Ω} \) is measured in steradians (sr) and can be found using the area of the opening \( A_{\text{hole}} \) and the radius of the sphere \( R \) as follows:
\( \text{Ω} = \frac{A_{\text{hole}}}{R^2} \)
Once the solid angle is calculated, the fraction of energy \( f \) that escapes through the hole is the solid angle over the full solid angle of a sphere, which is \( 4\textπ \) steradians:
\( f = \frac{\text{Ω}}{4\textπ} \)
The concept of solid angle helps in quantifying the directivity of radiation and is crucial when calculating the maximum rate at which radiation can escape through an opening. The interplay between the hole's size, the sphere's radius, and the resultant solid angle culminates in understanding how changes in these parameters impact the rate of energy streaming, an important factor in thermal system design and analysis.