Question

A 2.2-mm-diameter and 14-m-long electric wire is tightly wrapped with a \(1-\mathrm{mm}\)-thick plastic cover whose thermal conductivity is $k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire, and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2}\), \(\mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Short answer

Answer: The temperature at the interface of the wire and the plastic cover in steady operation is 131.9°C. Doubling the thickness of the plastic cover will increase the interface temperature to 191.8°C.

Step-by-step solution

Calculate the power generated

Using the given voltage (V = 8V) and current (I = 13A), we can calculate the power generated in the wire using the equation P = V * I. P = 8V * 13A = 104 W

Calculate the resistance of the wire

To compute the resistance, we will use the power equation again, but this time with the given current. R = V/I = 8V/13A = 0.615Ω

Calculate the heat generation rate in the wire

We will use the resistance heating equation Q_generation = I^2 * R to calculate the heat generation rate in the wire. Q_generation = (13A)^2 * 0.615Ω = 109.045 W

Calculate the heat dissipation rate through plastic cover

We will use the heat conduction equation to calculate the heat dissipation rate: q = k*A*(T_interface - T_infinity)/L Where k is the thermal conductivity (0.15 W/mK), A is the surface area of the wire (A = π * D * L, with D = 0.0022m and L = 14m), T_infinity = 30°C, and L is the thickness of the plastic cover (0.001m). Plugging in the values: q = 0.15 * (π * 0.0022 * 14) * (T_interface - 30)/0.001

Equate heat generation and heat dissipation rates

Assuming steady operation, the heat generated by the wire equals the heat dissipated through the plastic cover: 109.045 = 0.15 * (π * 0.0022 * 14) * (T_interface - 30)/0.001 Now, we can solve for T_interface: T_interface = 131.9°C

Determine the effect of doubling the plastic cover's thickness

We can redo the analysis using a thickness (L) of 0.002m instead of 0.001m: q = 0.15 * (π * 0.0022 * 14) * (T_interface_double - 30)/0.002 Since Q_generation remains the same, we equate Q_generation and the new heat dissipation rate: 109.045 = 0.15 * (π * 0.0022 * 14) * (T_interface_double - 30)/0.002 Solving for T_interface_double: T_interface_double = 191.8°C Since T_interface_double > T_interface, doubling the thickness of the plastic cover will increase the interface temperature.