A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-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 \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), 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.

To calculate the heat generated in the wire, we can use the formula \(Q=I^2 R\), where \(Q\) is the heat generated, \(I\) is the current passing through the wire, and \(R\) is the electrical resistance of the wire. Using Ohm's law, we can find the resistance from the given voltage drop: \(R = \frac{V}{I}\). Given values are \(I=13 A\) and \(V=8 V\). Let's calculate the resistance and heat generated: \(R = \frac{8}{13} = 0.6154 \Omega\) \(Q = I^2 R = 13^2 \times 0.6154 = 103.62 W\)

We can calculate the thermal resistance of the plastic cover using the formula: \(R_t=\frac{\ln(\frac{r_o}{r_i})}{2\pi k L}\), where \(R_t\) is the thermal resistance, \(r_o\) and \(r_i\) are the outer and inner radius of the plastic cover, \(k\) is the thermal conductivity, and \(L\) is the length. Given values are \(k=0.15 W/(m\cdot K)\), \(L=10 m\), \(r_i=1.1\ mm\) (half of the wire diameter), and \(r_o=1.1+1=2.1\ mm\). First, we need to convert \(r_i\) and \(r_o\) to meters: \(r_i = 0.0011\ m\), \(r_o = 0.0021\ m\). Now let's calculate the thermal resistance of the cover: \(R_t = \frac{\ln(\frac{0.0021}{0.0011})}{2\pi (0.15) (10)} = 0.00966 \mathrm{K/W}\)

Now that we know the heat generated by the wire and the thermal resistance of the cover, we can determine the interface temperature using the formula: \(T_{\infty} + \frac{Q}{hA} = T_{interface} + R_t Q\), where \(h\) is the heat transfer coefficient, \(A\) is the outer surface area of the cover, and \(T_{\infty}\) is the ambient temperature. Given values are \(h=24 W/(m^2 K)\), \(T_{\infty}=30^\circ C\). We need to calculate the outer surface area of the cover: \(A = 2\pi r_o L = 2\pi (0.0021)(10) = 0.1319\ m^2\). Now let's calculate the interface temperature: \(30 + \frac{103.62}{(24)(0.1319)} = T_{interface} + (0.00966) (103.62)\) \(T_{interface} = 36.8842 - 1.00064 = 35.88^\circ C\)

Now we need to calculate the new thermal resistance with the doubled cover thickness and determine the new interface temperature. The new outer radius (\(r_{o2}\)) with doubled cover thickness is \(r_{o2} = r_i + 2(1) = 3.1\ mm\) or \(0.0031\ m\). Now we calculate the new thermal resistance (\(R_{t2}\)): \(R_{t2} = \frac{\ln(\frac{0.0031}{0.0011})}{2\pi (0.15) (10)} = 0.01985 \mathrm{K/W}\) Let's calculate the new interface temperature, \(T_{interface2}\): \(30 + \frac{103.62}{(24)(0.1319)} = T_{interface2} + (0.01985) (103.62)\) \(T_{interface2} = 36.8842 - 2.05747 = 34.8267^\circ C\) Since the new interface temperature (\(34.8267^\circ C\)) is lower than the initial interface temperature (\(35.88^\circ C\)), doubling the thickness of the plastic cover will decrease the interface temperature.