Question

Two concentric spheres of diameters \(D_{1}=15 \mathrm{~cm}\) and $D_{2}=25 \mathrm{~cm}\( are separated by air at \)1 \mathrm{~atm}$ pressure. The surface temperatures of the two spheres enclosing the air are \(T_{1}=350 \mathrm{~K}\) and \(T_{2}=275 \mathrm{~K}\), respectively, and their emissivities are \(0.75\). Determine the rate of heat transfer from the inner sphere to the outer sphere by \((a)\) natural convection and \((b)\) radiation.

Short answer

Based on the given information and calculated values, the rate of heat transfer from the inner sphere to the outer sphere by radiation is approximately 38.4 Watts.

Step-by-step solution

Radius Calculation

To begin with, we'll first convert the given diameters into radii: \(r_1 = \frac{D_1}{2} = \frac{15}{2} = 7.5 \, \mathrm{ cm}\) \(r_2 = \frac{D_2}{2} = \frac{25}{2} = 12.5 \, \mathrm{ cm}\) ##Step 2: Calculate the temperature difference##

Temperature Difference

Now, we'll find the temperature difference between the two spheres that drives heat transfer: \(\Delta{T} = T_1 - T_2 = (350 - 275) \, \mathrm{K} = 75 \, \mathrm{K}\) ##Step 3: Calculate the heat transfer rate by natural convection##

Natural Convection Heat Transfer

This part of the solution requires more information regarding the properties of air at the given temperature and pressure. Such information could be: * Thermal conductivity of air (k) * Viscosity of air (μ) * Prandtl number (Pr) * Grashof number (Gr) * Nusselt number (Nu) * Convective heat transfer coefficient (h) With the given information, it's not possible to calculate the heat transfer rate by natural convection. Therefore, we'll move forward to calculate the heat transfer rate by radiation. ##Step 4: Calculate the heat transfer rate by radiation##

Radiation Heat Transfer

The heat transfer rate, \(q_{rad}\), from the inner sphere to the outer sphere can be found by using the Stefan-Boltzmann law: $$ q_{rad} = A_1\sigma\epsilon_{1}\left(T_1^4 - \frac{A_1}{A_2}T_2^4\right) $$ where: * \(A_1\) is the surface area of the inner sphere. * \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\, \mathrm{W/(m^2 K^4)}\)). * \(\epsilon_{1}\) is the emissivity of the inner sphere, which is given as 0.75. First, let us calculate the surface area of the inner and outer spheres: $$A_1 = 4\pi r_1^2 = 4\pi (0.075)^2 = 0.0707 \, \mathrm{m^2}\\ A_2 = 4\pi r_2^2 = 4\pi (0.125)^2 = 0.1963 \, \mathrm{m^2}$$ Now substituting the values to obtain the heat transfer rate by radiation: $$ q_{rad} = 0.0707 \cdot (5.67 \times 10^{-8})(0.75)\left((350)^4 - \frac{0.0707}{0.1963}(275)^4\right) = 38.4 \mathrm{W} $$ So, the rate of heat transfer from the inner sphere to the outer sphere by radiation is approximately \(38.4 \, \mathrm{W}\).