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.

The Grashof number is a dimensionless number that represents the ratio of buoyancy forces (due to density differences caused by temperature) to the viscous forces. It can be calculated using the following equation: \(Gr = \dfrac{g \beta (T_1 - T_2) D^3}{\nu^2}\) Where \(g\) is the acceleration due to gravity, \(\beta\) is the thermal expansion coefficient, \(T_1\) and \(T_2\) are the temperatures of the inner and outer spheres, \(D\) is the diameter of the gap between the two spheres, and \(\nu\) is the kinematic viscosity of air.

The diameter of the gap between the two spheres is given by: \(D = \dfrac{D_2 - D_1}{2} = 5\, cm\). Calculate the film temperature (average temperature): \(T_f = \dfrac{T_1 + T_2}{2} = \dfrac{350 + 275}{2} = 312.5\, K\)

Consult a table or use approximations to find the thermal expansion coefficient of air and its kinematic viscosity at the film temperature: \(\beta \approx \dfrac{1}{T_f} = \dfrac{1}{312.5} = 3.2 \times 10^{-3}\, K^{-1}\) \(\nu \approx 16.5 \times 10^{-6}\, m^2/s\) (at \(312.5\, K\))

Calculate the Grashof number using the values obtained above: \(Gr = \dfrac{9.81 \times 3.2 \times 10^{-3} \times 75 \times (0.05^3)}{(16.5 \times 10^{-6})^2} \approx 2.07 \times 10^8\) Calculate the Nusselt number using an appropriate correlation for concentric spheres (available in references): \(Nu = c \times (Gr \times Pr)^n\) Here, we assume that the values of \(c\) and \(n\) are 1, and \(Pr\) is the Prandtl number of air at the film temperature (about 0.7). Then, \(Nu = 1 \times (2.07 \times 10^8 \times 0.7)^1 \approx 1.45 \times 10^8\)

Now, find the convective heat transfer coefficient: \(h = \dfrac{Nu \cdot k}{D}\) Assuming a thermal conductivity (\(k\)) of air at the film temperature to be \(0.026\, W/m\cdot K\), we get \(h = \dfrac{1.45 \times 10^8 \times 0.026}{0.05} \approx 1.01 \times 10^6\, W/m^2\cdot K\) The natural convection heat transfer rate is given by: \(q_{conv} = h A (T_1 - T_2)\) Where \(A\) is the surface area of the inner sphere. Since \(D_1 = 15\, cm\), we have \(A = 4\pi(0.075^2) = 0.071\, m^2\). So, the heat transfer rate due to natural convection is: \(q_{conv} = 1.01 \times 10^6 \times 0.071 \times (350 - 275) \approx 5.41 \times 10^6\, W\) #b)# Radiation

The radiation heat transfer rate between two concentric spheres is given by the following equation: \(q_{rad} = \dfrac{eA \sigma (T_1^4 - T_2^4)} { (1/e_1 + 1/e_2 - 1)}\) Here, \(e\) is the emissivity of the spheres (both have the same emissivity, given as \(0.75\)), \(A\) is the area of the inner sphere (already calculated above), \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\, W/m^2\cdot K^4\)), and \(T_1\) and \(T_2\) are the temperatures of the inner and outer spheres, respectively. Substitute the given values and calculate the radiation heat transfer rate: \(q_{rad} = \dfrac{0.75 \times 0.071 \times 5.67 \times 10^{-8} \times (350^4 - 275^4)}{(1/0.75 + 1/0.75 - 1)} \approx 2651\, W\) The rate of heat transfer from the inner sphere to the outer sphere by \((a)\) natural convection is approximately \(5.41 \times 10^6\, W\), and by \((b)\) radiation is approximately \(2651\, W\).