91Ó°ÊÓ

Since the cylinder is placed horizontally, its surface area can be calculated using the formula for the lateral surface area of the cylinder: \(A = 2 \pi rL\), where r is the radius of the cylinder, and L is its length. Given the diameter of the cylinder as 10 cm, the radius is 5 cm or 0.05 m. The length of the cylinder is 10 m. Plugging these values into the formula: \(A = 2 \pi(0.05 \mathrm{~m})(10 \mathrm{~m}) = \pi \mathrm{~m}^2\) Now, let's calculate the heat transfer rate for both cases.

For this case, we will use the Dittus-Boelter equation to find the convective heat transfer coefficient, \(h\). The formula for the Dittus-Boelter equation is: \(h=0.023Re^{0.8}Pr^n\), where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number (assuming air), and \(n\) is 0.3 for heating and 0.4 for cooling. We will first find the Reynolds number using the formula: \(Re = \frac{VD}{\nu}\), where \(V\) is the air velocity, \(D\) is the diameter and \(\nu\) is the kinematic viscosity of air. Given the air velocity of 10 m/s and the diameter of 0.1 m, let's assume the kinematic viscosity (\(\nu\)) of air to be \(1.5 \times 10^{-5} \mathrm{m}^2/\mathrm{s}\). So, we can find the Reynolds number: \(Re = \frac{10 \mathrm{~m}/\mathrm{s} \times 0.1 \mathrm{~m}}{1.5 \times 10^{-5} \mathrm{m}^2/\mathrm{s}} = 66666\) Now, we can calculate the convective heat transfer coefficient using the Dittus-Boelter equation. Assuming Prandtl number (Pr) of air to be 0.7 and since the cylinder is cooling down, we will use n = 0.4. \(h = 0.023(66666)^{0.8} \times 0.7^{0.4} \approx 150.16 \mathrm{~W}/\mathrm{m}^2 \mathrm{K}\) Finally, we can calculate the heat transfer rate using the formula: \(q = hA(T_s - T_\infty) =150.16 \mathrm{~W}/\mathrm{m}^2\mathrm{K} \times \pi\mathrm{m}^2 \times (10^{\circ}\mathrm{C} - 40^{\circ}\mathrm{C}) = -14100.78 \mathrm{~W}\)

For this case, we will use the Churchill-Chu correlation to find the convective heat transfer coefficient, \(h\). The formula for the Churchill-Chu correlation is: \(h = k \left(\frac{0.6Pr\frac{d}{L}(GrPr)^{1/4}}{[(0.5+0.5Pr)^{9/16}]^{-16/9}}\right)\), where \(k\) is the thermal conductivity of air, \(d\) is the diameter, and \(L\) is the length. We will first calculate the Grashof number (\(Gr\)) using the formula: \(Gr = \frac{g \beta \Delta T D^3}{\nu^2}\), where \(g\) is the gravitational acceleration (9.81 m/s²), \(\beta\) is the coefficient of thermal expansion (assumed to be 1/T, T is in Kelvin), and \(\Delta T\) is the temperature difference. We know T as 40°C, which is equivalent to 313.15 K. So, calculating the Grashof number: \(Gr = \frac{9.81 \mathrm{~m}/\mathrm{s}^2 \times \frac{1}{313.15 \mathrm{~K}} \times (40^{\circ}\mathrm{C} - 10^{\circ}\mathrm{C}) \times (0.1 \mathrm{~m})^3}{(1.5 \times 10^{-5} \mathrm{m}^2/\mathrm{s})^2} \approx {1.34 \times 10^{8}}\) Now, we can calculate the convective heat transfer coefficient using the Churchill-Chu correlation. Assume the thermal conductivity (k) of air to be \(0.028 \mathrm{~W}/\mathrm{m}\mathrm{K}\). \(h = 0.028 \mathrm{~W}/\mathrm{m}\mathrm{K} \left(\frac{0.6 \times 0.7 \times \frac{0.1\mathrm{~m}}{10\mathrm{~m}} \times (1.34 \times 10^{8})^{1/4}}{[(0.5+0.5 \times 0.7)^{9/16}]^{-16/9}}\right) \approx 3.68 \mathrm{~W}/\mathrm{m}^2 \mathrm{K}\) Finally, we can calculate the heat transfer rate using the formula: \(q = hA(T_s - T_\infty) = 3.68 \mathrm{~W}/\mathrm{m}^2\mathrm{K} \times \pi\mathrm{m}^2 \times (10^{\circ}\mathrm{C} - 40^{\circ}\mathrm{C}) = -346.19 \mathrm{~W}\) So, the steady-state heat transfer rates for the two cases are: (a) Normal winds: -14,100.78 W (b) No wind: -346.19 W