Question

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) is reentering the atmosphere. The shell satellite is made of stainless steel with properties of \(\rho=8238 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=468 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=13.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). During the reentry, the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after 5 minutes of reentry. Assume heat transfer occurs only on the satellite shell.

Short answer

Answer: After 5 minutes of reentry, the spherical shell satellite's temperature is approximately \(826.3^{\circ} \mathrm{C}\).

Step-by-step solution

Calculate the shell surface area and volume

We'll start by finding the surface area and volume of the spherical shell, which will be crucial in later steps to determine the heat generated and the temperature change. Given that, the satellite's outer diameter is \(4 \mathrm{~m}\), and its shell thickness is \(10 \mathrm{~mm}\). The surface area of a sphere can be calculated as: \(A = 4 \pi R^2\) The volume of a sphere can be calculated as: \(V = \frac{4}{3} \pi R^3\) Since we have a shell with a thickness, we will find the outer surface area and the volume of the shell: Surface area of the shell: \(A = 4 \pi (2)^2 = 16 \pi \mathrm{~m}^2\) Volume of the shell: \(V = \frac{4}{3} \pi [(2)^3 - (1.99)^3] = 0.1256 \pi \mathrm{~m}^3\)

Calculate the conduction heat flux and its rate

In this step, we will calculate the heat flux due to conduction across the shell by using Fourier's law of conduction. And then we'll determine the rate at which heat is transferred from the outer surface of the satellite to the inner surface. Fourier's law states that the heat flux across any layer of thickness (shell thickness in this case) is proportional to the temperature difference (ΔT) and inversely proportional to the layer thickness (l): \(q = -k \frac{ΔT}{l}\) Given the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(130 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\). Let \(T_s\) be the shell temperature, and the initial temperature is \(10^{\circ} \mathrm{C}\). We can calculate the heat flux as follows: \(q = -k \frac{(T_s - 10) - 1250}{0.01}\) Now, we need to find the rate of heat transfer from the outer surface to the inner surface. The heat transfer rate (Q) can be calculated as: \(Q = hA(T_{atm} - T_s)\) Where h is the convection heat transfer coefficient and A is the shell surface area.

Apply energy conservation to find the shell temperature

As per the energy conservation principle, the heat generated during the reentry and the heat transferred across the shell should be equal, so we can write: \(Q_{gen} = Q_{trans}\) We will use the heat transfer equation \(Q = mcΔT\), where m is the mass of the shell, c is the specific heat capacity, and ΔT is the temperature change. The mass can be calculated as the product of volume and density, \(m = Vρ\). Calculating mass: \(m = 0.1256 \pi (8238) = 3275.27 \mathrm{~kg}\) Substituting mass, specific heat capacity, and temperatures into the equation, we have: \(3275.27(468)(T_s - 10) = 130(16 \pi)(T_{atm} - T_s)\) Now we can plug in the given values for \(T_{atm}\) which is \(1250^{\circ} \mathrm{C}\) and solve for \(T_s\). After solving the equation, we get: \(T_s \approx 816.3^{\circ} \mathrm{C}\)

Determine the shell temperature after 5 minutes of reentry

We have determined the temperature change in the shell due to the heat generated during reentry. As we assume that the heat transfer occurs only on the satellite shell, we can now add this temperature change to the initial temperature to find the shell temperature after 5 minutes of reentry. \(T_{final} = T_{initial} + \Delta T\) \(T_{final} = 10^{\circ} \mathrm{C} + 816.3^{\circ} \mathrm{C} \approx 826.3^{\circ} \mathrm{C}\) After 5 minutes of reentry, the spherical shell satellite's temperature is approximately \(826.3^{\circ} \mathrm{C}\).

Key concepts

Fourier's Law of Conduction

Fourier's law of conduction is pivotal in understanding heat transfer through materials, and it's particularly relevant when analyzing systems like a reentering spherical shell satellite. According to this law, the heat flux, denoted by \( q \), is the amount of heat transferred per unit area per unit time, and it is directly proportional to the temperature gradient. In its simplest form, the law is written as:
\[ q = -k \frac{\Delta T}{l} \]
Here, \( k \) is the thermal conductivity of the material, which is a measure of its ability to conduct heat; \( \Delta T \) is the temperature difference across the material, and \( l \) is the thickness of the material through which heat is being transferred.

For our satellite problem, the steel shell has a specific thermal conductivity that enables us to calculate the heat flux during reentry. The negative sign in Fourier's law indicates that heat flows from the region of higher temperature to the region of lower temperature, which aligns with the second law of thermodynamics. Applying Fourier's law allows us to determine how much heat is conducted through the satellite's shell and is essential for predicting the temperature changes that the satellite will experience.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, often denoted as \( h \), is a crucial parameter representing the convection heat transfer between a solid surface and a fluid in motion. This coefficient quantifies the rate of heat transfer per unit surface area per unit temperature difference between the surface and the fluid. The related formula is:
\[ Q = hA(T_{fluid} - T_{surface}) \]
In the context of the satellite, the fluid is the atmospheric gases that the satellite contacts during reentry, and the surface is the outer layer of the satellite's shell. The convection heat transfer coefficient depends on various factors, such as the velocity of the fluid, its properties, and the nature of the surface.

For our satellite during reentry, the convection heat transfer coefficient is a given value that helps us calculate the rate at which the satellite's surface loses or gains heat due to the atmosphere's temperature. Understanding this concept is vital for engineers to design thermal protection systems that can withstand the extreme temperatures encountered during reentry.

Energy Conservation in Heat Transfer

Energy conservation in heat transfer is a fundamental principle that ensures the energy balance within a system. It states that energy cannot be created or destroyed, only transferred or converted from one form to another. When heat is introduced to a system, such as our spherical shell satellite, the heat must go somewhere—either it's stored in the system by increasing the temperature or it's lost through transfer mechanisms.

In mathematical terms, for a closed system we can express it as:
\[ Q_{generated} = Q_{stored} + Q_{transferred} \]
During reentry, our satellite generates heat due to friction and aerodynamic heating. According to the principle of energy conservation, this heat will cause the temperature of the satellite shell to increase, unless it's transferred away. We calculate the heat generated using the specific heat capacity of the materials and their mass, and the transferred heat using the convection heat transfer coefficient. By setting the generated heat equal to the transferred heat, we can solve for the unknown temperature, assuming a steady-state condition where internal energy generation equals heat loss. This approach is essential in assessing the thermal response of an object subjected to extreme heating, like a satellite reentering Earth's atmosphere.