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Chapter 2: Problem 118

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.

Short Answer

Expert verified
For 95% efficiency, thinner fins are needed for materials with higher thermal conductivity. Aluminum requires the thinnest fins, followed by mild steel, and then plastic.

Step by step solution

01

Understanding the Fin Efficiency Formula

To achieve 95% fin efficiency, we use the formula for fin efficiency: \[ \eta_f = \frac{\tanh(m L_f)}{m L_f} \]where \( L_f \) is the fin length, and \( m \) is given by:\[ m = \sqrt{\frac{2 h}{k t}} \]In the problem, \( \eta_f = 0.95 \), and \( L_f = \frac{L}{2} = 2.5 \text{ mm} = 0.0025 \text{ m} \). We need to find the thickness \( t \) for each material.
02

Rearrange the Formula to Solve for t

Rearrange the equation for \( \eta_f \) to find \( t \):\[ 0.95 = \frac{\tanh(m \cdot 0.0025)}{m \cdot 0.0025} \]Substitute \( m \) in terms of \( t \):\[ 0.95 = \frac{\tanh(0.0025 \sqrt{\frac{2 h}{k t}})}{0.0025 \sqrt{\frac{2 h}{k t}}} \]This equation should be solved numerically for \( t \).
03

Solve for Aluminum Alloy

Using \( k = 180 \text{ W/m K} \) and \( h = 160 \text{ W/m}^2\text{K} \) in the equation, we solve numerically to find \( t \). Plug these values into the equation to find the necessary \( t \) so that the fins are 95% efficient.
04

Solve for Mild Steel

Using \( k = 64 \text{ W/m K} \) and \( h = 160 \text{ W/m}^2\text{K} \), solve the rearranged fin efficiency equation numerically to determine the required thickness \( t \) for mild steel.
05

Solve for Plastic

With \( k = 0.33 \text{ W/m K} \) and \( h = 160 \text{ W/m}^2\text{K} \), solve the equation numerically to find the thickness \( t \) required for plastic to achieve 95% efficiency.
06

Discussion of Results

Aluminum, with its higher thermal conductivity, requires a thinner fin than mild steel or plastic to achieve the same efficiency. This affects design choices by implying that materials with higher thermal conductivity can achieve the desired efficiency with less material, potentially reducing weight and cost.

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Key Concepts

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

Fin Efficiency
Fin Efficiency is a crucial factor in the design of heat exchangers. It represents how effectively a fin can transfer heat compared to an ideal fin that is perfectly conductive. The efficiency is determined by the formula:
  • \( \eta_f = \frac{\tanh(m L_f)}{m L_f} \), where \( m = \sqrt{\frac{2 h}{k t}} \)
  • \( L_f \) is the length of the fin and dependent on the spacing between heat exchanger plates.

The goal in the given exercise is 95% fin efficiency, meaning the fins must be optimized for maximum heat transfer. The fin efficiency can be influenced by the material's thermal conductivity, the thickness of the fin, and the convective heat transfer coefficient. Each of these parameters plays a fundamental role in achieving efficient fin design. When modifying any of these parameters, it's critical to consider their interdependent effects to maintain or improve efficiency levels.

Ultimately, materials with higher thermal conductivity will typically require thinner fins to achieve the same level of efficiency. This might reduce material costs and weight, providing a beneficial trade-off in the overall design.
Thermal Conductivity
Thermal Conductivity (\( k \)) is the property of a material to conduct heat. It determines how quickly heat is transferred across a material and plays a pivotal role in the calculations for heat exchanger designs. High thermal conductivity materials transfer heat more effectively.
  • Aluminum Alloy: \( k = 180 \text{ W/m K} \)
  • Mild Steel: \( k = 64 \text{ W/m K} \)
  • Plastic: \( k = 0.33 \text{ W/m K} \)

When designing a heat exchanger, selecting a material with the appropriate thermal conductivity is a strategic decision. A material with a higher thermal conductivity can achieve the desired thermal performance with thinner fins, reducing overall material requirements and costs.

It's important to consider that materials with higher thermal conductivity might be more expensive or less feasible due to other physical properties like weight or structural integrity. Therefore, the choice of material should balance the cost implications, efficiency, and the particular design needs of the heat exchanger.
Convective Heat Transfer Coefficient
The Convective Heat Transfer Coefficient (\( h \)) measures how efficiently heat is transferred between a fluid and a solid surface. In the context of heat exchangers, this coefficient affects how well the heat exchanger can transfer heat from one fluid to another across the finned surfaces.
  • Given in the problem: \( h = 160 \text{ W/m}^2\text{K} \)

This value is a constant in our calculations and impacts the required thickness of the fins for achieving a specified efficiency. Understanding the mechanisms of convection can help in optimizing the heat exchanger design for different fluid flow conditions. The higher the convective coefficient, the more effective the heat transfer between the fluid and solid, which reduces the demand on the fins' performance.

In practical terms, increasing the convective heat transfer coefficient can entail altering fluid dynamics, such as increasing the velocity of the fluid or modifying the surface geometry to enhance turbulence. This would enhance heat transfer capabilities without necessarily changing the material or geometric properties of the fins.

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Most popular questions from this chapter

A test rig for the measurement of interfacial conductance is used to determine the effect of surface anodization treatment on the interfacial conductance for aluminum-aluminum contact. The specimens themselves form the heat flux meters because they are each fitted with a pair of thermocouples as shown. The heat flux is determined from the measured temperature gradient and a known thermal conductivity of the aluminum of \(185 \mathrm{~W} / \mathrm{m} \mathrm{K}\) as $$ q_{1}=\frac{k\left(T_{1}-T_{2}\right)}{L_{1}} ; \quad q_{2}=\frac{k\left(T_{3}-T_{4}\right)}{L_{2}} ; \quad q=\frac{1}{2}\left(q_{1}+q_{2}\right) $$ and the interfacial temperatures obtained by linear extrapolation. $$ \begin{gathered} T_{5}=\frac{1}{2}\left(T_{1}+T_{2}\right)-\left[\frac{L_{3}+\frac{L_{1}}{2}}{L_{1}}\right]\left(T_{1}-T_{2}\right) ; \\ T_{6}=\frac{1}{2}\left(T_{3}+T_{4}\right)+\left[\frac{L_{4}+\frac{L_{2}}{2}}{L_{2}}\right]\left(T_{3}-T_{4}\right) \end{gathered} $$ The interfacial conductance is then obtained as $$ h_{i}=\frac{q}{\left(T_{5}-T_{6}\right)} $$ The main possible sources of error in the values of \(h_{i}\) so determined are due to uncertainty in temperature measurement and thermocouple locations. Since only temperature differences are involved, the absolute uncertainty in the individual temperature measurements is not of concern; rather it is the relative uncertainties. Previous calibrations of similar type and grade thermocouples indicate that the relative uncertainties are \(\pm 0.2^{\circ} \mathrm{C}\). Also, the techinician who drilled the thermocouple holes and installed the thermocouples estimates an uncertainty of \(\pm 0.5 \mathrm{~mm}\) in the thermocouple junction locations. At a particular pressure, the temperatures recorded are \(T_{1}=338.7 \mathrm{~K}, T_{2}=328.7 \mathrm{~K}, T_{3}=305.3 \mathrm{~K}\), \(T_{4}=295.0 \mathrm{~K}\). (i) Assuming that the uncertainties in temperature measurement and thermocouple location can be treated as random errors, estimate the uncertainty in the interfacial conductance. (ii) In reality, the uncertainties in temperature and location are bias errors (for example, the location of a thermocouple does not vary from test to test). Thus it is more appropriate to determine bounds on the possible error in \(h_{i}\), by considering best and worst cases. Determine these bounds.

A \(6 \mathrm{~mm}\)-O.D. tube is to be insulated with an insulation of thermal conductivity \(0.08 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and a very low surface emittance. Heat loss is by natural convection, for which the heat transfer coefficient can be taken as \(\bar{h}_{c}=1.3(\Delta T / D)^{1 / 4} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) for \(\Delta T=T_{s}-T_{e}\) in kelvins and the diameter \(D\) in meters. Determine the critical radius of the insulation and the corresponding heat loss for a tube surface temperature of \(350 \mathrm{~K}\), and an ambient temperature of \(300 \mathrm{~K}\).

Aluminum alloy straight rectangular fins for cooling a semiconductor device are 1 \(\mathrm{cm}\) long and \(1 \mathrm{~mm}\) thick. Investigate the effect of choice of tip boundary condition on heat loss as a function of convective heat transfer coefficient. Use \(k=175 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\) for the alloy and a range of \(h_{c}\) values from 10 to \(200 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

Heat is generated at a rate \(\dot{Q}_{v}^{\prime \prime \prime}\) in a long solid cylinder of radius \(R\). The cylinder has a thin metal sheath and is immersed in a liquid at temperature \(T_{e}\). Heat transfer from the cylinder surface to the liquid can be characterized by an overall heat transfer coefficient \(U\). Obtain the steady-state temperature distributions for the following cases: (i) \(\dot{Q}_{v}^{\prime \prime \prime}\) is constant. (ii) \(\dot{Q}_{v}^{\prime \prime \prime}=\dot{Q}_{v 0}^{\prime \prime \prime}\left[1-(r / R)^{2}\right]\). (iii) \(\dot{Q}_{v}^{\prime \prime \prime \prime}=a+b\left(T-T_{e}\right)\).

A \(5 \mathrm{~kW}\) electric heater using Nichrome wire is to be designed to heat air to \(400 \mathrm{~K}\). The maximum allowable wire temperature is \(1500 \mathrm{~K}\), and a minimum heat transfer coefficient of \(600 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) is expected. A variable voltage power supply up to \(130 \mathrm{~V}\) is available. Determine the length of \(1.0 \mathrm{~mm}\)-diameter wire required. Also check the current and voltage. Take the electrical resistivity of Nichrome wire as \(100 \mu \Omega \mathrm{cm}\) and its thermal conductivity as \(30 \mathrm{~W} / \mathrm{m} \mathrm{K}\).
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