A 3 -mm-diameter and 12-m-long electric wire is tightly wrapped with a \(1.5-\mathrm{mm}\)-thick plastic cover whose thermal conductivity and emissivity are \(k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\varepsilon=0.9\). Electrical measurements indicate that a current of \(10 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(7 \mathrm{~V}\) along the wire. If the insulated wire is exposed to calm atmospheric air at \(T_{\infty}=30^{\circ} \mathrm{C}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Take the surrounding surfaces to be at about the same temperature as the air. Evaluate air properties at a film temperature of \(40^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

First, we need to determine the power dissipated in the wire due to the passage of electric current. For that, we'll use the formula: P = V * I where P is the power, V is the voltage drop, and I is the current. Given V = 7 V and I = 10 A, we can calculate the power as: P = 7 V * 10 A = 70 W

Now, we need to find the heat transfer rate through the plastic cover due to conduction. The formula for the conduction heat transfer rate q is: q = k * A * (T_wire - T_air) / L where k is the thermal conductivity of the plastic, A is the cross-sectional area of the plastic cover, T_wire and T_air are the wire temperature and air temperature, respectively, and L is the thickness of the plastic cover. We are given that k = 0.20 W/(m·K) and L = 1.5 mm = 0.0015 m. We also know the radius of the wire r_wire = 1.5 mm = 0.0015 m (since it is tightly wrapped by the cover). Hence, we can calculate the area A as: A = 2 * π * r_wire * L_wire = 2 * π * 0.0015 m * 12 m ≈ 0.113 m^2 Plugging all the values back into the conduction heat transfer rate formula gives us the following equation for q: 70 W = 0.20 W/(m·K) * 0.113 m^2 * (T_wire - 30°C) / 0.0015 m Now we can solve this equation for T_wire to find the temperature of the wire.

Solving the equation from step 2: T_wire - 30°C = (70 W * 0.0015 m) / (0.20 W/(m·K) * 0.113 m^2) T_wire - 30°C ≈ 18.35°C T_wire ≈ 48.35°C

In this step, we need to calculate the natural convection heat transfer coefficient h for the surrounding air at a film temperature of 40°C and 1 atm pressure. Since it's not given, we will be using the average Nusselt number for a vertical cylinder (which is 0.59) and the Rayleigh number. We have T_film = 40°C and air properties at this temperature are given by: k_air = 0.027 W/(m·K), α_air (thermal diffusivity) = 22.56 × 10^{-6} m^2/s, ν_air (kinematic viscosity) = 15.89 × 10^{-6} m^2/s, and β_air (thermal expansion coefficient) = 1 / T_film (in K). To find the Rayleigh number: Ra = g * β_air * (T_wire - T_air) * L_wire^3 / (ν_air * α_air) We can calculate Rayleigh number as: Ra ≈ 9.81 m/s^2 * (1/313.15) * (18.35°C) * (12 m)^3 / (15.89 x 10^{-6} m^2/s * 22.56 x 10^{-6} m^2/s) Ra ≈ 6.85 × 10^8 Since the Rayleigh number is less than 10^9, the average Nusselt number is 0.59 for a vertical cylinder. Now we can calculate the heat transfer coefficient h: h = (Nu * k_air) / L_wire h ≈ (0.59 * 0.027 W/(m·K)) / 12 m h ≈ 0.00132 W/(m²·K)

Finally, we can calculate the interface temperature using the convection heat transfer equation, q = h * A * (T_interface - T_air). We know q (70 W) from step 1 and h (0.00132 W/(m²·K)) from step 4. The area A is the same as in step 2. Solving for T_interface: T_interface - T_air = (70 W) / (0.00132 W/(m²·K) * 0.113 m²) T_interface - 30°C ≈ 17.81°C T_interface ≈ 47.81°C This assumption seems reasonable, as the interface temperature is close to the wire temperature of 48.35°C.