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Chapter 1: Problem 141

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 \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), 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

Expert verified
Answer: 1. Calculate the surface area of the tip: \(A = 2\pi(0.00125)(0.02) \approx 0.000157 \mathrm{~m}^{2}\) 2. Calculate the heat transfer by convection: \(Q_{conv} = (25)(0.000157)(400-20) \approx 1.491 \mathrm{~W}\) 3. Calculate the heat transfer by radiation: \(Q_{rad} = (0.80)(5.67\times 10^{-8})(0.000157)((400+273)^{4}-(20+273)^{4}) \approx 1.164 \mathrm{~W}\) 4. Calculate the total heat transfer and power required: \(P_{required} = Q_{total} = Q_{conv} + Q_{rad} = 1.491 + 1.164 \approx 2.655 \mathrm{~W}\) The power required to maintain the temperature of the soldering iron tip at \(400^{\circ} \mathrm{C}\) is approximately \(2.655 \mathrm{~W}\).

Step by step solution

01

Calculate the surface area of the tip

The first step is to find the surface area of the cylindrical tip. The surface area of a cylinder can be calculated using the formula \(A=2 \pi r h\) where \(r\) is the radius of the base and \(h\) is the height. The diameter is given as \(2.5 \mathrm{~mm}\), so the radius is \(1.25 \mathrm{~mm}\). Convert the radius and height to meters: - radius: \(1.25 \mathrm{~mm} = 0.00125 \mathrm{~m}\) - length: \(20 \mathrm{~mm} = 0.02 \mathrm{~m}\) Now, calculate the surface area of the tip: \(A = 2\pi(0.00125)(0.02)\)
02

Calculate the heat transfer by convection

The heat transfer by convection can be calculated using the formula \(Q_{conv} = hA(T_{tip} - T_{air})\), where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, \(T_{tip}\) is the temperature of the tip, and \(T_{air}\) is the air temperature. Plug in the values given: \(Q_{conv} = (25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K})(2\pi(0.00125)(0.02))(400-20)\)
03

Calculate the heat transfer by radiation

The heat transfer by radiation can be calculated using the formula \(Q_{rad} = \epsilon \sigma A(T_{tip}^{4} - T_{air}^{4})\), where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67\times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4})\), and \(A\) is the surface area. Plug in the values given: \(Q_{rad} = (0.80)(5.67\times 10^{-8})(2\pi(0.00125)(0.02))((400+273)^{4}-(20+273)^{4})\)
04

Calculate the total heat transfer and power required

To determine the power required to maintain the temperature, we need to add the heat transfer by convection and radiation: \(Q_{total} = Q_{conv} + Q_{rad}\) Finally, the power required is equal to the total heat transfer, so: \(P_{required} = Q_{total}\) Calculate the values using the above equations and obtain the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convection
Convection is a method of heat transfer that involves the movement of fluid, which can be a gas or a liquid. In the context of a soldering iron, convection occurs as the heat travels from the hot surface of the soldering iron tip to the cooler surrounding air.

When examining how convection works with a soldering iron, it's important to understand a few key points:
  • Convection Heat Transfer Coefficient: This is a measure of how effectively heat is transferred from a solid surface to a moving fluid or vice versa. In this exercise, the coefficient is given as 25 W/m虏路K, which means for every square meter of surface area, 25 watts of thermal energy can be transferred for every degree of temperature difference.
  • Temperature Difference: The heat transfer by convection depends significantly on the temperature difference between the soldering iron tip and the surrounding air. The greater this difference, the more heat will be transferred.
Convection helps in dissipating the heat from the tip, which is crucial for regulating the temperature and preventing overheating.
Radiation
Radiation is another form of heat transfer where energy is emitted by a heated surface and travels through space without needing a medium. This principle is at play with a soldering iron as it disperses heat through its oxidized tip.
  • Emissivity: In the exercise, the soldering iron's tip has aged and oxidized, affecting its emissivity, which is the efficiency at which a surface emits thermal radiation. Here, the emissivity is 0.80, indicating that 80% of the potential radiation energy is emitted.
  • Stefan-Boltzmann Law: This principle is used to calculate the power radiated and involves the temperature raised to the fourth power. This non-linear relationship means that even a minor increase in temperature can lead to a much larger increase in radiated heat.
  • Temperature in Kelvin: When using the Stefan-Boltzmann Law, it's essential to convert temperatures to Kelvin, which is the absolute temperature scale. This ensures accuracy in the calculations of energy emitted.
Radiation plays a vital role in the total energy balance of the soldering iron, especially when dealing with high temperatures like 400掳C.
Soldering Iron
A soldering iron is a hand tool that heats up to melt solder, which is used in joining components in electronics. The tip of the soldering iron is crucial as it directly influences how well the soldering is performed due to its heat transmission capabilities.

There are several aspects of behavior in soldering iron usage:
  • Tip Material and Shape: The tip is typically metallic, often using materials such as iron or copper, known for their thermal conductivity. A cylindrical shape is common, providing consistent heat distribution.
  • Oxidation of Tip: Over time, use can cause the tip of a soldering iron to oxidize. This affects its efficiency in transferring heat but also increases emissivity, which can affect the overall power needed to maintain temperature.
  • Power Requirements: Keeping the soldering iron at a stable high temperature, such as 400掳C, requires considering both convective and radiative heat losses. These factors dictate how much power is essential to ensure the soldering iron functions effectively without overheating or cooling excessively during use.
Understanding the principles of heat transfer through convection and radiation is essential in determining the power needs and efficiency of a soldering iron, especially when maintained over prolonged periods.

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Most popular questions from this chapter

Determine a positive real root of this equation using \(E E S\) : $$ 3.5 x^{3}-10 x^{0.5}-3 x=-4 $$

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(120 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

A 2.1-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at \(20^{\circ} \mathrm{C}\). Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be \(180^{\circ} \mathrm{C}\) in steady operation. Also, the voltage drop and electric current through the wire are measured to be \(110 \mathrm{~V}\) and \(3 \mathrm{~A}\), respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: \(156 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Solve this system of three equations with three unknowns using EES: $$ \begin{aligned} 2 x-y+z &=5 \\ 3 x^{2}+2 y &=z+2 \\ x y+2 z &=8 \end{aligned} $$
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