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Chapter 2: Problem 33

A \(2 \mathrm{~mm}\)-diameter electrical wire has a \(1 \mathrm{~mm}\)-thick electrical insulation with a thermal conductivity of \(0.12 \mathrm{~W} / \mathrm{m} \mathrm{K}\). The combined convection and radiation heat transfer coefficient on the outside of the insulation is \(12 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). (i) Would increasing the thickness of the insulation to \(3 \mathrm{~mm}\) increase or decrease the heat transfer? (ii) Would the presence of a contact resistance between the wire and insulation of \(5 \times 10^{-4}\left[\mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\right]^{-1}\) affect your conclusion?

Short Answer

Expert verified
Increasing insulation thickness decreases heat transfer. Contact resistance supports this conclusion.

Step by step solution

01

Analyze Initial Conditions

The electrical wire has a diameter of 2 mm and an insulation thickness of 1 mm. The thermal conductivity of the insulation is 0.12 W/mK, and the convection and radiation heat transfer coefficient outside the insulation is 12 W/m虏K.
02

Determine Heat Transfer for Initial Thickness

The thermal resistance for the conduction through the insulation is calculated using the formula \( R_{ ext{cond}} = \frac{\ln(r_2/r_1)}{2 \pi k L} \), where \( r_1 = 1 \) mm, \( r_2 = 2 \) mm, and \( k = 0.12 \) W/mK. The convective heat transfer resistance \( R_{ ext{conv}} = \frac{1}{hA} \) where \( A = 2 \pi r_2 L \).
03

Change Thickness and Calculate New Resistances

Increasing the thickness to 3 mm changes \( r_2 \) to 4 mm. Recalculate \( R_{\text{cond}} \) with these conditions. For thicker insulation, \( R_{\text{conv}} \) stays the same but \( R_{\text{cond}} \) increases since the natural log term \( \ln(r_2/r_1) \) becomes larger.
04

Compare Total Resistance Before and After

The total thermal resistance is the sum of \( R_{\text{cond}} \) and \( R_{\text{conv}} \). Calculate the total resistance before and after increasing the insulation. An increase in \( R_{\text{total}} \) generally indicates decreased heat transfer.
05

Analyze Impact of Contact Resistance

If a contact resistance exists, add this new resistance \( R_c = 5 \times 10^{-4} [\mathrm{W}/\mathrm{m}^{2}\mathrm{K}]^{-1} \) to \( R_{\text{total}} \). Evaluate how this affects the total resistance and determine if the initial conclusion about heat transfer change holds.
06

Draw Conclusion

With increased insulation, \( R_{\text{cond}} \) increases, raising \( R_{\text{total}} \) and reducing heat transfer, consistent even with additional contact resistance because this added resistance further increases \( R_{\text{total}} \).

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Key Concepts

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

Heat Transfer Coefficient
The heat transfer coefficient is essential in determining how easily heat moves from one area to another. It combines the effects of both convection and radiation. When looking at the outer surface of the insulation, this coefficient is represented as a single value. In our example, it is given as 12 W/m虏K. This value tells us how effective these processes are at the surface level.
  • The higher the heat transfer coefficient, the more heat can be transferred.
  • This factor is crucial in ensuring that the heat generated within the wire is effectively dissipated into the surrounding environment.
Understanding this coefficient helps us analyze how changes in insulation thickness can affect overall heat transfer behavior. As insulation thickness increases, the conductive resistance also increases, affecting the overall heat dissipated by convection and radiation.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is intrinsic to every material and is represented as 'k' in equations. For the insulation in our example, the thermal conductivity is 0.12 W/mK. This relatively low value suggests that the insulation is designed to minimize heat transfer.
  • A material with high thermal conductivity transfers heat efficiently, which isn鈥檛 the goal for insulation.
  • A lower thermal conductivity means the material resists transferring heat, aiding insulation purposes.
In the context of altering the insulation thickness, it's important to realize that the same material is used throughout. This means thermal conductivity remains constant despite changes in thickness. It directly influences the conductive heat resistance, where increasing the thickness results in higher resistance.
Convection and Radiation
Convection and radiation are two primary methods by which heat can escape from the surface of an insulated wire. Convection involves the transfer of heat by the movement of fluids (gases or liquids) over a surface. Radiation, on the other hand, is the transfer of energy through electromagnetic waves.
  • Convection efficiency is influenced by the speed and temperature of the fluid in contact with the surface.
  • Radiation doesn鈥檛 require a medium and can occur in a vacuum.
In our exercise, these two processes are combined into a single heat transfer coefficient. When insulation thickness increases, the conductive resistance increases, but the convective resistance remains the same. This means more insulation makes it harder for heat to reach the outer surface, reducing the overall heat lost through convection and radiation.
Contact Resistance
Contact resistance occurs at the interface between two materials. In this problem, it exists between the wire and its insulation. Even though it is a minor component compared to other thermal resistances, it can still influence heat transfer, especially at high precision levels.
  • Contact resistance is often measured in \(\left[\mathrm{W}/\mathrm{m}^{2}\mathrm{K}\right]^{-1}\), indicating how this resistance affects heat flow through microscopic contact points.
  • This resistance represents the imperfections and gaps at the material boundary, hindering heat flow.
Adding contact resistance to the system increases the total thermal resistance. Consequently, it impacts the heat transfer calculation, reinforcing the prediction that increased insulation leads to decreased heat transfer. The presence of contact resistance further validates that the insulation鈥檚 additional thickness effectively reduces heat transfer, as seen by a unified increase in overall resistance.

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Most popular questions from this chapter

An air-cooled R-22 condenser has 10 mm-O.D., 9 mm-I.D. aluminum tubes with \(50 \mathrm{~mm}-\mathrm{O} . \mathrm{D} .\) annular rectangular aluminum fins of thickness \(0.2 \mathrm{~mm}\) at a pitch of \(2 \mathrm{~mm}\). The heat transfer coefficient inside the tubes is \(800 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and on the fins it is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the \(\mathrm{R}-22\) is at \(320 \mathrm{~K}\) and the air is at \(300 \mathrm{~K}\), calculate the heat transfer per unit length of tube. Take \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum.

An electrical current is passed through a \(1 \mathrm{~mm}\)-diameter, \(20 \mathrm{~cm}\)-long copper wire located in an air flow at \(290 \mathrm{~K}\). If the ends of the wire are maintained at \(300 \mathrm{~K}\), determine the maximum current that can be passed if the midpoint temperature is not to exceed \(400 \mathrm{~K}\). The convective heat transfer coefficient is estimated to be 20 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\). For the copper wire, take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}, \varepsilon=0.8\). and an electrical resistance of \(2.2 \times 10^{-2} \Omega / \mathrm{m}\).

A gas turbine rotor has 54 AISI 302 stainless steel blades of dimensions \(L=6 \mathrm{~cm}\), \(A_{c}=4 \times 10^{-4} \mathrm{~m}^{2}\), and \(\mathscr{P}=0.1 \mathrm{~m}\). When the gas stream is at \(900^{\circ} \mathrm{C}\), the temperature at the root of the blades is measured to be \(500^{\circ} \mathrm{C}\). If the convective heat transfer coefficient is estimated to be \(440 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), calculate the heat load on the rotor internal cooling system.

A thermal conductivity cell consists of concentric thin-walled copper tubes with an electrical heater inside the inner tube and is used to measure the conductivity of granular materials. The inner and outer radii of the annular gap are 2 and \(4 \mathrm{~cm}\). In a particular test the electrical power to the heater was \(10.6 \mathrm{~W}\) per meter length, and the inner and outer tube temperatures were measured to be \(321.4 \mathrm{~K}\) and \(312.7\) \(\mathrm{K}\), respectively. Calculate the thermal conductivity of the sample.

So-called "compact" heat exchanger cores often consist of finned passages between parallel plates. A particularly simple configuration has square passages with the effective fin length equal to half the plate spacing \(L\). In a particular application with \(L=5 \mathrm{~mm}\), a convective heat transfer coefficient of \(160 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) is expected. If \(95 \%\) efficient fins are desired, how thick should they be if the core is constructed from (i) an aluminum alloy with \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) ? (ii) mild steel with \(k=64 \mathrm{~W} / \mathrm{m} \mathrm{K}\) ? (iii) a plastic with \(k=0.33 \mathrm{~W} / \mathrm{m} \mathrm{K}\) ? Discuss the significance of your results to the design of such cores.
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