Question

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.

Short answer

Based on the given data and calculations, the interface temperature between the wire and plastic cover in steady operation is initially 35.88°C. When the thickness of the plastic cover is doubled, the interface temperature decreases to 34.8267°C. Therefore, doubling the thickness of the cover decreases the interface temperature.

Step-by-step solution

1. Calculate heat generated in the wire

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: \(R = \frac{8}{13} = 0.6154 \Omega\) \(Q = I^2 R = 13^2 \times 0.6154 = 103.62 W\)

2. Calculate thermal resistance of the plastic cover

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: \(R_t = \frac{\ln(\frac{0.0021}{0.0011})}{2\pi (0.15) (10)} = 0.00966 \mathrm{K/W}\)

3. Calculate interface temperature

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: \(30 + \frac{103.62}{(24)(0.1319)} = T_{interface} + (0.00966) (103.62)\) \(T_{interface} = 36.8842 - 1.00064 = 35.88^\circ C\)

4. Determine the effect of doubling the cover thickness

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.

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of how well a material conducts heat. It is represented by the symbol \(k\) and is commonly measured in watts per meter per Kelvin \(\left( \, \text{W/m} \cdot \text{K} \, \right)\). High thermal conductivity indicates the material can transfer heat quickly, while low thermal conductivity suggests slower heat transfer.
In the problem, the plastic cover has a thermal conductivity \(k = 0.15 \, \text{W/m} \cdot \text{K}\). This relatively low value suggests that the plastic acts as an insulator, reducing heat loss from the core wire to the surrounding environment. Understanding this property helps explain how the insulation can influence heat retention and why the interface temperature changes with different cover thicknesses.

Electrical Resistance

Electrical resistance quantifies how strongly a material opposes the flow of electric current. It's symbolized by \(R\) and measured in ohms (\(\Omega\)). Resistance depends on the material's properties and the geometric dimensions.
From Ohm's Law, the resistance \(R\) can be calculated using the formula \(R = \frac{V}{I}\), where \(V\) is the voltage and \(I\) is the current. For the given wire, \(R = \frac{8}{13} = 0.6154 \, \Omega\).
This resistance creates heat in the wire, as shown by the formula \(Q = I^2 R\), where \(Q\) is the power or heat generated. In this exercise, the calculated power is \(103.62 \, \text{W}\). This heat generation is a key factor in determining the wire's temperature during operation.

Heat Transfer Coefficient

The heat transfer coefficient, often denoted by \(h\), measures the efficiency of heat transfer between a solid surface and a fluid in contact. It's a crucial concept in thermodynamics when assessing convection heat transfer. The units are watts per square meter per Kelvin \(\left( \, \text{W/m}^2 \cdot \text{K} \, \right)\).
In the problem, we are given \(h = 24 \, \text{W/m}^2 \cdot \text{K}\) for the interaction between the insulated wire and the ambient air. A higher heat transfer coefficient means more efficient heat dissipation from the wire surface to the environment. This coefficient influences how quickly the wire can cool down and maintain a stable temperature.

Thermal Resistance

Thermal resistance quantifies a material's resistance to heat flow. Similar to electrical resistance, but related to heat transfer, it's usually calculated in degrees Celsius per watt \(\left(\, \text{K/W} \, \right)\). Thermal resistance helps determine how effective an insulator is.
In the solution, the thermal resistance of the plastic cover is calculated with the formula \(R_t = \frac{\ln(\frac{r_o}{r_i})}{2\pi k L}\). Here, \(r_o\) and \(r_i\) are outer and inner radii, \(k\) is the thermal conductivity, and \(L\) is the length of the wire.

• Initial thermal resistance: \(R_t = 0.00966 \, \text{K/W}\).
• With doubled thickness: \(R_{t2} = 0.01985 \, \text{K/W}\).

The calculated value shows that as the plastic cover’s thickness increases, so does its thermal resistance, leading to a decrease in interface temperature. By raising resistance, the insulation better restricts heat loss, thus affecting the ambient temperature exchange.