A long aluminum wire of diameter \(3 \mathrm{~mm}\) is (t)s extruded at a temperature of \(280^{\circ} \mathrm{C}\). The wire is subjected to cross air flow at \(20^{\circ} \mathrm{C}\) at a velocity of \(6 \mathrm{~m} / \mathrm{s}\). Determine the rate of heat transfer from the wire to the air per meter length when it is first exposed to the air.

To determine the rate of heat transfer, we first need to calculate the surface area (A) of the wire per meter length. We can find the surface area of a cylinder (ignoring the ends) using the formula: $$ A = 2 \pi r L $$ where A is the surface area, r is the radius of the wire, and L is the length of the wire. Since we want the surface area per meter length, we can assume L = 1 meter. Given the diameter of the wire is 3 mm, we can calculate the radius: $$ r = \frac{d}{2} = \frac{3}{2} \mathrm{~mm} = \frac{3}{2} \cdot 10^{-3} \mathrm{~m} $$ Now, we can calculate the surface area of the wire per meter length: $$ A = 2 \pi \cdot \frac{3}{2} \cdot 10^{-3} \mathrm{~m} $$

To find the convective heat transfer coefficient, we first need to calculate the Reynolds number for the air flow (Re). The Reynolds number is given by the formula: $$ Re = \frac{\rho v d}{\mu} $$ where ρ is the density of the air, v is the velocity of the air, d is the diameter of the wire, and μ is the dynamic viscosity of the air. At 20°C, we can assume the air to have properties: - Density (ρ) = 1.2 kg/m³ - Dynamic viscosity (μ) = 1.8 × 10⁻⁵ Pa s We are given the air velocity (v) as 6 m/s. Now, we can calculate the Reynolds number for the air flow: $$ Re=\frac{(1.2)(6)(3 \times 10^{-3})}{1.8\times10^{-5}} $$

With the Reynolds number calculated, let's now find the convective heat transfer coefficient (h). For this problem, we will use the Dittus-Boelter equation for turbulent flow to calculate h: $$ Nu = h \frac{d}{k} = C Re^m Pr^n $$ where Nu is the Nusselt number, k is the thermal conductivity of the air, C, m, and n are constants based on the flow situation. As we have a cooling situation, we choose C=0.023, m = 0.8, and n = 0.4. Now, we need to find the Prandtl number, Pr, which is the ratio of momentum diffusivity to thermal diffusivity. The Prandtl number is given by the formula: $$ Pr = \frac{c_p \mu}{k} $$ At 20°C, we can assume the air properties: - Specific heat capacity (c_p) = 1005 J/kg·K - Thermal conductivity (k) = 0.026 W/m·K Using the given dynamic viscosity (μ) from Step 2, we can calculate the Prandtl number (Pr): $$ Pr = \frac{(1005)(1.8 \times 10^{-5})}{0.026} $$ With the Reynolds number, Prandtl number, and the Dittus-Boelter equation, we can now find the convective heat transfer coefficient (h): $$ h = \frac{C Re^m Pr^n k}{d} $$

Now that we have found the convective heat transfer coefficient (h), we can calculate the rate of heat transfer from the wire to the air. The convective heat transfer equation is given as: $$ Q = h A \Delta T $$ where Q is the rate of heat transfer, A is the surface area of the wire per meter length calculated in Step 1, and ΔT is the temperature difference between the wire and the air. We are given that the wire's initial temperature is 280°C and the air temperature is 20°C, so we can calculate the temperature difference: $$ \Delta T = T_{wire} - T_{air} = 280 - 20 = 260 \mathrm{^{\circ}C} $$ Now, we can find the rate of heat transfer from the wire to the air per meter length (Q): $$ Q = h A \Delta T $$