Consider a \(0.3\)-m-diameter and \(1.8-\mathrm{m}\)-long horizontal cylinder in a room at \(20^{\circ} \mathrm{C}\). If the outer surface temperature of the cylinder is \(40^{\circ} \mathrm{C}\), the natural convection heat transfer coefficient is (a) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(3.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(4.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(5.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

The given values are as follows: - Diameter (d) of the cylinder: \(0.3\,\mathrm{m}\) - Length (L) of the cylinder: \(1.8\,\mathrm{m}\) - Room temperature (Tr): \(20^{\circ} \mathrm{C}\) - Outer surface temperature (Tc) of the cylinder: \(40^{\circ} \mathrm{C}\)

Calculate the temperature difference between the cylinder surface and the room: $$T_{diff} = T_c - T_r = 40 - 20 = 20\,\text{K}$$

For a horizontal cylinder, the characteristic length (Lc) is given by the diameter (d): $$L_c = d = 0.3\,\mathrm{m}$$

The Grashof number (Gr) is a dimensionless number that measures the influence of buoyancy forces relative to viscous forces in natural convection. We assume the given parameters are for air, and use the following properties at the film temperature (\(\frac{T_r + T_c}{2}\)): - Coefficient of thermal expansion (beta): \(3.41 \times 10^{-3}\,\mathrm{K^{-1}}\) - Dynamic viscosity (mu): \(1.97 \times 10^{-5}\,\mathrm{kg \cdot m^{-1} \cdot s^{-1}}\) - Kinematic viscosity (nu): \(1.56 \times 10^{-5}\,\mathrm{m^2 \cdot s^{-1}}\) - Acceleration due to gravity (g): \(9.81\,\mathrm{m \cdot s^{-2}}\) $$Gr = \frac{g \cdot \beta \cdot T_{diff} \cdot L_c^3}{\nu^2} = \frac{9.81 \cdot 3.41 \times 10^{-3} \cdot 20 \cdot (0.3)^3}{(1.56 \times 10^{-5})^2} \approx 5.77 \times 10^8$$

The Prandtl number (Pr) is a dimensionless number that measures the ratio of momentum diffusivity to the thermal diffusivity. For air, at the film temperature, the thermal conductivity (k) is \(0.027\,\mathrm{W \cdot m^{-1} \cdot K^{-1}}\). The Prandtl number is given by the following formula: $$Pr = \frac{\mu \cdot c_p}{k} = \frac{1.97 \times 10^{-5} \cdot 1004}{0.027} \approx 0.72$$

We can use the Churchill-Chu correlation for the Nusselt number (Nu) for a horizontal cylinder in a natural convection situation: $$Nu = 0.60 + \frac{0.387 \cdot Gr^{1/6}}{[1+ (0.559/Pr)^{9/16}]^{8/27}}$$ $$Nu = 0.60 + \frac{0.387 \cdot (5.77 \times 10^8)^{1/6}}{[1+ (0.559/0.72)^{9/16}]^{8/27}} \approx 12.47$$

The heat transfer coefficient (h) can be calculated using the Nusselt number and the thermal conductivity of air: $$h = \frac{Nu \cdot k}{L_c} = \frac{12.47 \cdot 0.027}{0.3} \approx 1.381\,\mathrm{W \cdot m^{-2} \cdot K^{-1}}$$

The calculated heat transfer coefficient value is closest to \(1.4\,\mathrm{W \cdot m^{-2} \cdot K^{-1}}\), but none of the options provided in the problem statement match this result. It seems there is an issue with either the problem statement or the given answer options, as the calculated value does not correspond to any of the options provided.