Question

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) that 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 a convection heat transfer coefficient of $130 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after $5 \mathrm{~min}$ of reentry. Assume heat transfer occurs only on the satellite shell.

Short answer

Question: Determine the temperature of a spherical shell satellite after 5 minutes of reentry into the atmosphere, considering the convection heat transfer during reentry. The initial temperature is 10°C, the convection heat transfer coefficient is 130 W/m²K, the outer diameter is 4 meters, the shell thickness is 0.01 meters, and the material is stainless steel with a density of 8238 kg/m³ and a specific heat of 468 J/kgK. Answer: After 5 minutes of reentry, the shell temperature of the satellite is approximately -26.85°C.

Step-by-step solution

Calculate the Initial Parameters

First, we need to calculate the volume, mass, and surface area of the shell. Outer radius \(r_{o} = \frac{4}{2} m = 2 m\) Inner radius \(r_{i} = 2 m - 0.010 m = 1.990 m\) Volume \(V = \frac{4}{3}\pi (r_{o}^3 - r_{i}^3) = \frac{4}{3}\pi [(2^3) - (1.990)^3] = 0.251327 m^3\) Mass \(m = \rho V = 8238 \times 0.251327 = 2070.44 kg\) Surface area \(A_s = 4\pi r_{o}^2 = 4\pi (2^2) = 50.2655 m^2\)

Calculate the Convective Heat Transfer Rate

The heat transfer rate due to convection can be calculated by the equation: \(Q_{conv} = hA_s(T_{\infty} - T_s)\) Where, \(h\) is the convection heat transfer coefficient (\(130 W/m^{2}K\)) \(T_{\infty}\) is the effective atmosphere temperature (\(1250°C = 1523.15 K\)) \(T_s\) is the shell temperature.

Write the energy balance equation

We can write that the change in the internal energy of the shell satellite over the reentry time will be equal to the heat added by convection. We'll assume no heat losses. So, the energy balance equation can be written as: \(m c_{p} \Delta T = Q_{conv} t\) Where, \(c_p\) is the specific heat of stainless steel (\(468 J/kgK\)) \(\Delta T = T_{s} - T_{0}\) where \(T_{0}\) is the initial temperature of the shell (\(10°C = 283.15 K\)) \(t\) is the reentry time (\(5 min = 300 s\))

Substitute the values in energy balance equation and solve for \(T_s\)

Substituting the previously calculated values and data provided, we can rewrite the energy balance equation as: \(2070.44 \times 468 (T_s - 283.15) = (130)(50.2655)(1523.15 - T_s) \times 300\) Solve the equation for \(T_s\): \(967027(T_s - 283.15) = 588344850 - 3991285T_s\) \(10069512T_s - 2876782300 = 588344850 - 3991285T_s\) \(14062797T_s = 3464627150\) \(T_s = \frac{3464627150}{14062797}\) \(T_s = 246.3 K\)

Convert the final temperature to Celsius

Convert the final temperature from Kelvin to Celsius: \(T_s^{\circ}C = T_s^K - 273.15\) \(T_s^{\circ}C = 246.3 - 273.15\) \(T_s^{\circ}C = -26.85 ^{\circ}C\) After 5 minutes of reentry, the shell temperature of the satellite is approximately -26.85°C.