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Chapter 2: Problem 93

The end of a soldering iron consists of a \(4 \mathrm{~mm}\)-diameter copper rod, \(5 \mathrm{~cm}\) long. If the tip must operate at \(350^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\), determine the base temperature and heat flow. The heat transfer coefficient from the rod to the air is estimated to be about \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the copper.

Short Answer

Expert verified
The base temperature is 350掳C and the heat flow is 0.0415 W.

Step by step solution

01

Determine Cross Sectional Area

To calculate the cross-sectional area of the copper rod, we use the formula: \[A = \pi \left( \frac{d}{2} \right)^2\]where \(d = 4 \, \text{mm} = 0.004 \, \text{m}\).\[A = \pi \left( \frac{0.004}{2} \right)^2 = \pi \times (0.002)^2 = \pi \times 0.000004 = 1.2566 \times 10^{-5}\, \text{m}^2\]
02

Calculate the Thermal Resistance of the Fin

The thermal resistance due to conduction is given by:\[R_{cond} = \frac{L}{kA}\]where \(L = 0.05 \, \text{m}\) and \(k = 386 \, \text{W/mK}\).\[R_{cond} = \frac{0.05}{386 \times 1.2566 \times 10^{-5}} = 0.0104 \, \text{K/W}\]
03

Calculate the Thermal Resistance to Convection

The thermal resistance due to convection is:\[R_{conv} = \frac{1}{hA}\]where \(h = 10 \, \text{W/m}^2\text{K}\).\[R_{conv} = \frac{1}{10 \times 1.2566 \times 10^{-5}} = 7954.3 \, \text{K/W}\]
04

Total Thermal Resistance

The total thermal resistance of the system is the sum of the two resistances because they are in series:\[R_{total} = R_{cond} + R_{conv} = 0.0104 + 7954.3 \approx 7954.31 \, \text{K/W}\]
05

Calculate Heat Transfer Rate

Using the formula for heat transfer rate:\[Q = \frac{T_{base} - T_{ambient}}{R_{total}}\]where \(T_{ambient} = 20^{\circ}\, \text{C}\), \(T_{base} = 350^{\circ}\, \text{C}\).\[Q = \frac{350 - 20}{7954.31} = 0.0415 \, \text{W}\]
06

Calculate the Base Temperature, Assuming Ideal Conditions

To find the base temperature when the heat flow is known, rearrange the heat transfer equation to isolate \(T_{base}\):\[T_{base} = R_{total} \cdot Q + T_{ambient}\]Since we already calculated \(Q\), we use previous values.\[T_{base} = 7954.31 \times 0.0415 + 20 = 350^{\circ}\, \text{C}\]

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Key Concepts

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

Heat Transfer Rate
The heat transfer rate is a key concept that determines how quickly heat is transferred from one region to another. In the context of the exercise, we calculate the heat transfer rate to understand how much heat flows from the base of the copper rod to its tip.
Understanding this helps us gauge how efficiently the soldering iron operates. The heat transfer rate (\( Q \)) is calculated using the temperature difference between the base and the ambient air divided by the total thermal resistance:\[ Q = \frac{T_{base} - T_{ambient}}{R_{total}} \]where \( T_{base} \) is the temperature at the base, \( T_{ambient} \) is the ambient air temperature, and \( R_{total} \) is the total thermal resistance of the system.
  • The higher the temperature difference, the greater the heat transfer rate.
  • A lower thermal resistance implies a more efficient system, allowing more heat to flow.
Breaking this down into understandable parts makes it clear that improving the heat transfer rate is all about managing resistance and temperature differences effectively.
Conduction
Conduction is the process through which heat is directly transferred through a material without any movement of the material itself. In our exercise, conduction occurs along the length of the copper rod.
It is vital to calculate the thermal resistance due to conduction to accurately assess total heat transfer efficiency. The thermal resistance to conduction (\( R_{cond} \)) is calculated as:\[ R_{cond} = \frac{L}{kA} \]where \( L \) is the length of the rod, \( k \) is the thermal conductivity of the material (copper in this case), and \( A \) is the cross-sectional area.Points to remember about conduction:
  • Copper, with its high thermal conductivity, effectively transfers heat along its length.
  • The rod's dimensions play a major role in determining the resistance due to conduction.
Effective conduction is crucial for the soldering iron tip to reach its working temperature efficiently.
Convection
Convection is the process of heat transfer through the movement of fluids (which can be gases or liquids). In the soldering iron exercise, heat is transferred from the rod to the surrounding air by convection.
This process is pivotal because it represents how heat is lost to the environment. To understand convection in this context, we compute the thermal resistance to convection (\( R_{conv} \)), determined by:\[ R_{conv} = \frac{1}{hA} \]where \( h \) is the heat transfer coefficient between the rod and air, and \( A \) is the surface area.
  • A larger area or a higher heat transfer coefficient decreases \( R_{conv} \).
  • Efficient convection is key to maintaining a consistent and reliable operational temperature of the soldering iron.
By managing these parameters, one can vastly improve the soldering process' efficiency.
Copper Rod
The copper rod is a central component in the soldering iron, crucial for conducting heat efficiently. Copper is favored due to its excellent thermal conductivity, allowing quick and effective heat transfer throughout the rod's length.
The copper rod's properties, such as diameter and length, directly influence how heat is transferred (both by conduction and convection) and consequently, the performance of the soldering iron.Important characteristics of the copper rod include:
  • Thermal Conductivity: Copper's high thermal conductivity means it can easily transfer heat from its base to tip, crucial for soldering applications.
  • Material Dimensions: The rod's dimensions, such as its 4 mm diameter and 5 cm length, not only determine the area for conduction (\( A \)) but also the surface area for convection.
  • Resistance Calculation: These physical dimensions are incorporated into formulas that calculate both conduction and convection resistances, affecting the total thermal resistance calculation.
Overall, the copper rod's involvement in the heat transfer processes underscores its critical role in the soldering iron's functionality.

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Most popular questions from this chapter

The convective heat transfer coefficient around a cylinder held perpendicular to a flow varies in a complicated manner. A test cylinder to investigate this behavior consists of a \(0.001\) in-thick, \(12.7 \mathrm{~mm}\)-wide stainless steel heater ribbon (cut from shim stock) wound around a \(2 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness Teflon tube. A single thermocouple is located just underneath the ribbon and measures the local ribbon temperature \(T_{s}(\theta)\). The cylinder is installed in a wind tunnel, and a second thermocouple is used to measure the ambient air temperature \(T_{e}\). The power input to the heater is metered, from which the electrical heat generation per unit area \(\dot{Q} / A\) can be calculated ( \(A\) is the surface area of one side of the ribbon). As a first approximation, the local heat transfer coefficient \(h_{c}(\theta)\) can be obtained from $$ h_{c}(\theta)=\frac{\dot{Q} / A}{T_{s}(\theta)-T_{e}} $$ Hence, by rotating the cylinder with the power held constant, the variation of \(h_{c}\) can be obtained from the variation of \(T_{s}\) : where \(T_{s}(\theta)\) is low, \(h_{c}(\theta)\) is high, and vice versa. A typical variation of \(T_{s}(\theta)\) is shown in the graph. A problem with this technique is that conduction around the circumference of the tube causes the local heat flux \(q_{s}(\theta)\) to not exactly equal \(\dot{Q} / A\). (i) Derive a formula for \(h_{c}(\theta)\) that approximately accounts for circumferential conduction. (ii) The following table gives values of \(T_{s}(\theta)\) in a sector where circumferential conduction effects are expected to be large. Use these values together with \(\dot{Q} / A=5900 \mathrm{~W} / \mathrm{m}^{2}\) and \(T_{e}=25^{\circ} \mathrm{C}\) to estimate the conduction effect at \(\theta=110^{\circ}\). \begin{tabular}{cc} Angle (degrees) & \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline 100 & \(65.9\) \\ 110 & \(65.7\) \\ 120 & \(64.4\) \end{tabular} (iii) Comment on the design of the cylinder. Would a \(3 \mathrm{~mm}\)-thick brass tube, directly heated by an electric current, be a suitable alternative? Use \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel and \(0.38 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for Teflon.

(i) Derive an expression for the relation between heat loss and temperature difference across the inner and outer surfaces of a hollow sphere, the conductivity of which varies with temperature in manner given by \(k=k_{0}\left[1+a\left(T-T_{0}\right)\right]\), where \(T_{0}\) is a reference temperature. (ii) Find the corresponding result for a hollow cylinder. (iii) Compare the expressions derived for parts (i) and (ii) for the special case of the outside radius becoming infinite. Explain the different values obtained.

A \(2 \mathrm{~mm}\)-diameter resistor, for an electronic component on a space station, is to have a sheath of thermal conductivity \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). It is cooled by forced convection with \(\bar{h}_{c} \simeq 1.1 D^{-1 / 2} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), for diameter \(D\) in meters, and by radiation with \(q_{\mathrm{rad}}=\sigma \varepsilon\left(T_{s}^{4}-T_{e}^{4}\right)\). Determine the radius of the sheath that maximizes the heat loss when the resistor is at \(400 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). Take the value of the surface emittance \(\varepsilon\) as (i) \(0.9\). (ii) \(0.5\).

An electric heater consists of a thin ribbon of metal and is used to boil a dielectric liquid. The liquid temperature \(T_{e}\) is uniform at its boiling point, and the heat transfer coefficient on the ribbon can be assumed to be uniform as well. A resistance measurement allows the average temperature of the ribbon \(\bar{T}\) to be determined. Obtain an expression for \(\bar{T}-T_{e}\) in terms of the ribbon dimensions (width \(W\), thickness \(2 t\) ), the ribbon thermal conductivity \(k\), the heat transfer coefficient \(h_{c}\), and the ribbon electrical conductivity \(\sigma\left[\Omega^{-1} \mathrm{~m}^{-1}\right]\).

A steel heat exchanger tube of \(2 \mathrm{~cm}\) outer diameter is fitted with a steel spiral annular fin of \(4 \mathrm{~cm}\) outer diameter, thickness \(0.4 \mathrm{~mm}\), and wound at a pitch of 3 \(\mathrm{mm}\). If the outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the original bare tube and \(15 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the finned tube, determine the reduction in the outside thermal resistance achieved by adding the fins. Take \(k_{\text {steel }}=42 \mathrm{~W} / \mathrm{m} \mathrm{K}\).
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