Question

A soldering iron has a cylindrical tip of \(2.5 \mathrm{~mm}\) in diameter and \(20 \mathrm{~mm}\) in length. With age and usage, the tip has oxidized and has an emissivity of \(0.80\). Assuming that the average convection heat transfer coefficient over the soldering iron tip is \(25 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) and the surrounding air temperature is \(20^{\circ} \mathrm{C}\), determine the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

Short answer

Answer: To find the total power required to maintain the temperature of the soldering iron tip, follow these steps: 1. Calculate the surface area of the tip, considering the lateral area and the base area: - Convert the diameter to radius: \(r_\text{tip} = 0.00125 \mathrm{m}\) - Find the surface area: \(A_\text{tip} = \pi r_\text{tip}^2 + 2 \pi r_\text{tip} L_\text{tip}\) 2. Calculate the heat transfer due to convection using Newton's Law of Cooling: - \(Q_\text{convection} = hA_\text{tip}(T_\text{tip} - T_\text{air})\) 3. Calculate the heat transfer due to radiation using the Stefan-Boltzmann Law: - \(Q_\text{radiation} = \epsilon A_\text{tip} \sigma (T_\text{tip}^4 - T_\text{air}^4)\) 4. Calculate the total power required by summing the heat transfer due to convection and radiation: - \(Q_\text{total} = Q_\text{convection} + Q_\text{radiation}\) Plug in the given values and convert the temperatures from Celsius to Kelvin, then calculate the total power required.

Step-by-step solution

Calculate the surface area of the soldering iron tip

The shape of the tip is cylindrical, so we can calculate the surface area as the sum of the cylinder lateral area and the circular base area. The lateral area of the cylinder can be calculated using the formula \(A_\text{lateral} = 2 \pi r_\text{tip} L_\text{tip}\), where \(r_\text{tip}\) is the radius of the tip and \(L_\text{tip}\) is the length of the tip. For the base area, we can use the formula \(A_\text{base} = \pi r_\text{tip}^2\). First, let's convert the diameter to radius: $$ r_\text{tip} = \dfrac{2.5\mathrm{~mm}}{2} = 1.25\mathrm{~mm} = 0.00125\mathrm{~m} $$ Now, we can find the surface area of the tip: $$ A_\text{tip} = A_\text{base} + A_\text{lateral} = \pi r_\text{tip}^2 + 2 \pi r_\text{tip} L_\text{tip} $$

Calculate the heat transfer due to convection

To calculate the heat transfer due to convection, we use Newton's Law of Cooling: $$ Q_\text{convection} = hA_\text{tip}(T_\text{tip} - T_\text{air}) $$ Where \(Q_\text{convection}\) is the power (W) required to maintain the tip temperature, \(h\) is the convection heat transfer coefficient, \(A_\text{tip}\) is the tip surface area, \(T_\text{tip}\) is the temperature of the soldering iron tip, and \(T_\text{air}\) is the air temperature.

Calculate the heat transfer due to radiation

To calculate the heat transfer due to radiation, we use the Stefan-Boltzmann Law: $$ Q_\text{radiation} = \epsilon A_\text{tip} \sigma (T_\text{tip}^4 - T_\text{air}^4) $$ Where \(Q_\text{radiation}\) is the power (W) required to maintain the tip temperature, \(\epsilon\) is the emissivity, \(A_\text{tip}\) is the tip surface area, \(\sigma\) is the Stefan-Boltzmann constant \((5.67\times10^{-8} \mathrm{W}/\mathrm{m}^2\mathrm{K}^4)\), and \(T_\text{tip}\) and \(T_\text{air}\) are the temperatures of the soldering iron tip and air, respectively.

Calculate the total power required

To find the total power required to maintain the soldering iron tip at the desired temperature, we sum the heat transfer due to convection and radiation: $$ Q_\text{total} = Q_\text{convection} + Q_\text{radiation} $$ Now, we can plug in the given values and calculate the total power required. Notice that we need to convert the given temperatures from Celsius to Kelvin before performing the calculations: $$ T_\text{tip} = 400 + 273.15 = 673.15\mathrm{K} $$ $$ T_\text{air} = 20 + 273.15 = 293.15\mathrm{K} $$