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Since the two spheres are separated by air at 1 atm pressure, we will first need to find the air properties at the film temperature: $$T_f = \frac{T_i + T_o}{2}$$ Substitute the given temperature values. $$T_f = \frac{320 \mathrm{~K} + 280 \mathrm{~K}}{2} = 300 \mathrm{~K}$$ Now, using the air properties at 300 K, we get: Density, \(\rho = 1.177 kg/m^3\) Thermal conductivity, \(k = 0.02624 W/m \cdot K\) Dynamic viscosity, \(\mu = 1.846 \times 10^{-5} kg/m \cdot s\) Coefficient of thermal expansion, \(\beta = 3.410 \times 10^{-3} K^{-1}\) Specific heat, \(c_p = 1005 J/kg \cdot K\) Prandtl number, \(Pr = 0.708\)

The Rayleigh number (\(Ra\)) is a dimensionless number that characterizes a fluid flow's stability, accounting for the balance between buoyancy and viscosity forces. It is used to determine if natural convection dominates the heat transfer process. For concentric spheres, the Rayleigh number can be calculated as: $$Ra = \frac{g \beta (T_i - T_o)(R_o^3 - R_i^3)}{\nu \alpha \cdot R_o R_i}$$ Where: \(g = 9.81 m/s^2\) is the gravitational constant, \(R_i\) and \(R_o\) are the radii of the inner and outer spheres, respectively, \(\nu = \frac{\mu}{\rho}\) is the kinematic viscosity, and \(\alpha = \frac{k}{\rho c_p}\) is the thermal diffusivity. First, we find radii \(R_i\) and \(R_o\): $$R_i = \frac{D_i}{2} = \frac{20 cm}{2} = 10 cm$$ $$R_o = \frac{D_o}{2} = \frac{30 cm}{2} = 15 cm$$ Convert them to meters: $$R_i = 0.1 m$$ $$R_o = 0.15 m$$ Next, calculate kinematic viscosity \(\nu\) and thermal diffusivity \(\alpha\): $$\nu = \frac{1.846 \times 10^{-5} kg/m \cdot s}{1.177 kg/m^3} = 1.568 \times 10^{-5} m^2/s$$ $$\alpha = \frac{0.02624 W/m \cdot K}{(1.177 kg/m^3)(1005 J/kg \cdot K)} = 2.243 \times 10^{-5} m^2/s$$ Finally, calculate the Rayleigh number: $$Ra = \frac{9.81 m/s^2(3.41 \times 10^{-3} K^{-1})(320 K - 280 K)(0.15^3 m^3 - 0.1^3 m^3)}{(1.568 \times 10^{-5} m^2/s)(2.243 \times 10^{-5} m^2/s)(0.15 m)(0.1 m)} = 42216.4$$

The Nusselt number (\(Nu\)) is a dimensionless number representing the ratio of convective heat transfer to conduction heat transfer. For concentric spheres, we can determine the Nusselt number: $$Nu = 0.615 \times Ra^{1/4} \times Pr^{0.074}$$ Substitute the values of \(Ra\) and \(Pr\) to find the Nusselt number: $$Nu = 0.615 \times (42216.4)^{1/4} \times (0.708)^{0.074} = 15.489$$

Now that we have the Nusselt number, we can calculate the heat transfer rate using the following equation: $$q = Nu \times \frac{k}{R_o - R_i} \times (T_i - T_o)$$ Substitute the values of \(Nu\), \(k\), \(R_i\), \(R_o\), \(T_i\), and \(T_o\): $$q = 15.489 \times \frac{0.02624 W/m \cdot K}{0.15 m - 0.1 m} \times (320 \mathrm{~K} - 280 \mathrm{~K}) = 18.657 W$$

The rate of heat transfer from the inner sphere to the outer sphere by natural convection is approximately 18.657 W. This means that the inner sphere is losing heat at this rate to the outer sphere and ultimately to the surroundings.