/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 159 Annular aluminum fins of rectang... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 159

Annular aluminum fins of rectangular profile are attached to a circular tube having an outside diameter of \(50 \mathrm{~mm}\) and an outer surface temperature of \(200^{\circ} \mathrm{C}\). The fins are \(4 \mathrm{~mm}\) thick and \(15 \mathrm{~mm}\) long. The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) If there are 125 such fins per meter of tube length, what is the rate of heat transfer per unit length of tube?

Short Answer

Expert verified
(a) The fin efficiency is calculated using \(\eta_{f}=\frac{\tanh (M L)}{M L}\), where M is the fin parameter and L is the fin length. The fin effectiveness is calculated using \(\epsilon_{f}=\eta_{f} \frac{A_{f}}{A_{b}}\), where 畏f is the fin efficiency, Af is the fin area, and Ab is the base area. (b) The heat transfer rate per unit length (q) can be calculated using the formula \(q = h A_{eff} (T_s - T_\infty)\), where h is the convection coefficient, Aeff is the effective fin area, Ts is the outer surface temperature of the tube, and T鈭� is the surrounding air temperature.

Step by step solution

01

First, convert all the given dimensions (fin thickness, fin length, and tube diameter) from millimeters to meters. This will simplify the calculations and ensure that all units are consistent. #Step 2. Calculate the fin area (Af)#

Next, calculate the fin area (Af) by multiplying the fin thickness by its length. This area will be used to determine the heat transfer from the fin surface to the surrounding air. #Step 3. Calculate the fin parameter (M)#
02

The fin parameter (M) is an important value related to fin efficiency. It can be calculated using the formula: \(M = \sqrt{\frac{2hP}{k_t t_f}}\), where h is the convection coefficient, P is the perimeter of the fin, k_t is the thermal conductivity of aluminum, t_f is fin thickness. #Step 4. Calculate the fin efficiency (畏f)#

Fin efficiency (畏f) can be calculated using the formula: \(\eta_{f}=\frac{\tanh (M L)}{M L}\), where L is the fin length, and M is the fin parameter calculated in step 3. #Step 5. Calculate the fin effectiveness#
03

Fin effectiveness can be calculated using the formula: \(\epsilon_{f}=\eta_{f} \frac{A_{f}}{A_{b}}\) , where 畏f is the fin efficiency, Af is the fin area, and Ab is the base area, which can be calculated as the product of the fin thickness and tube circumference: \(A_b=t_f(2\pi r)\). #Phase 2: Calculating heat transfer rate per unit length of tube# #Step 6. Calculate total area of fins on one meter of tube (Atotal)#

Now, we'll find the total area of fins on one meter of tube (Atotal) by multiplying the fin area (Af) by the number of fins per meter (125). #Step 7. Apply effectiveness factor to calculate effective fin area (Aeff)#
04

The heat transfer per unit length can be obtained by calculating the effective fin area (Aeff) by multiplying the total area of fins (Atotal) by the fin effectiveness calculated in step 5. #Step 8. Calculate the heat transfer rate per unit length (q)#

Finally, we will calculate the heat transfer rate per unit length (q) using the formula: \(q = h A_{eff} (T_s - T_\infty)\), where h is the convection coefficient, Aeff is the effective fin area, Ts is the outer surface temperature of the tube, and T鈭� is the surrounding air temperature. The calculated heat transfer rate is the final answer for part (b) of this exercise.

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Key Concepts

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

Fin Effectiveness
Fin effectiveness (\(\epsilon_{f}\)) is a useful measure to determine how well a fin enhances heat transfer from a surface compared to a surface without it. Essentially, it's about how beneficial the fin is at transferring heat away. This effectiveness is particularly important when designing cooling surfaces because it provides insight into the efficiency of material usage and heat dissipation.To calculate fin effectiveness, we use the relation:\[\epsilon_{f} = \eta_{f} \frac{A_{f}}{A_{b}}\]
  • Where \(\eta_{f}\) is the fin efficiency, calculated as the ratio of actual heat removal to the maximum possible heat removal if the entire fin were at the base temperature.
  • \(A_{f}\) represents the area of the fin.
  • \(A_{b}\) is the base area, typically the face of the surface area which the fin is attached to.
The value of fin effectiveness assists in deciding whether adding fins is worthwhile. If the effectiveness is greater than one, it means the fins significantly aid in increasing the heat transfer rate. It's an indication that enhancements such as added mass and complexity in the system actually lead to better heat dissipation.
Convection Coefficient
The convection coefficient (\(h\)) is a crucial variable in heat transfer as it quantifies the heat transfer rate between a surface and surrounding fluid. Essentially, it bridges the temperature difference between the object surface and the fluid, signifying how rapidly heat can be exchanged by convection.### Importance of Convection CoefficientThe convection coefficient depends on many factors, such as the fluid velocity, fluid properties like viscosity and thermal conductivity, and the nature of the flow, whether turbulent or laminar.
  • The higher the convection coefficient, the more effective the heat transfer rate, indicating a better thermal interaction between the fin and the ambient medium.
  • It plays a role in defining the temperature gradient at the interface, which directly influences the overall heat transfer across the surface.
In our exercise, a convection coefficient of \(40\, \mathrm{W/m}^2 \cdot \mathrm{K}\) is given, depicting the environmental condition and the interaction intensity between the aluminum fin surface and the ambient air at room temperature.
Heat Transfer Rate
The heat transfer rate (\(q\)) indicates the amount of heat energy transferred per unit time from the fins to the surrounding air. It is a fundamental concept in thermal management, revealing the efficiency of cooling or heating systems.### Calculating Heat Transfer RateIn the context of the finned tube system, the formula for determining the heat transfer rate is given by:\[q = h A_{eff} (T_s - T_\infty)\]
  • Where \(h\) denotes the convection coefficient.
  • \(A_{eff}\) is the effective fin area, impacted by the fin's effectiveness.
  • \(T_s\) is the surface temperature of the tube.
  • \(T_\infty\) signifies the ambient air temperature.
The difference \((T_s - T_\infty)\) represents the driving force for heat transfer, with larger differences resulting in higher rates of energy transfer. For our system, knowing the heat transfer rate helps optimize the design and position of fins to manage temperatures effectively in various environmental conditions.

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Most popular questions from this chapter

Determine the parallel plate separation distance \(L\), above which the thermal resistance associated with molecule-surface collisions \(R_{t, m-s}\) is less than \(1 \%\) of the resistance associated with molecule-molecule collisions, \(R_{t, m-m}\) for (i) air between steel plates with \(\alpha_{t}=0.92\) and (ii) helium between clean aluminum plates with \(\alpha_{t}=0.02\). The gases are at atmospheric pressure, and the temperature is \(T=300 \mathrm{~K}\).

Consider an extended surface of rectangular cross section with heat flow in the longitudinal direction. In this problem we seek to determine conditions for which the transverse ( \(y\)-direction) temperature difference within the extended surface is negligible compared to the temperature difference between the surface and the environment, such that the one-dimensional analysis of Section 3.6.1 is valid. (a) Assume that the transverse temperature distribution is parabolic and of the form $$ \frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2} $$ where \(T_{s}(x)\) is the surface temperature and \(T_{o}(x)\) is the centerline temperature at any \(x\)-location. Using Fourier's law, write an expression for the conduction heat flux at the surface, \(q_{y}^{\prime \prime}(t)\), in terms of \(T_{s}\) and \(T_{a^{+}}\) (b) Write an expression for the convection heat flux at the surface for the \(x\)-location. Equating the two expressions for the heat flux by conduction and convection, identify the parameter that determines the ratio \(\left(T_{o}-T_{s}\right) /\left(T_{s}-T_{\infty}\right)\). (c) From the foregoing analysis, develop a criterion for establishing the validity of the onedimensional assumption used to model an extended surface.

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and measurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?

As seen in Problem \(3.109\), silicon carbide nanowires of diameter \(D=15 \mathrm{~nm}\) can be grown onto a solid silicon carbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Silicon carbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array of nanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronics packages in which an extremely small, \(10 \mu \mathrm{m} \times 10 \mu \mathrm{m}\) electronics device is sandwiched between two \(d=100\)-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at \(20^{\circ} \mathrm{C}\). A heat transfer coefficient of \(h=1 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) exists on the top and bottom of the unfinned package and on all surfaces of the exposed silicon carbide fins, which are each \(L=300 \mathrm{~nm}\) long. Each nano-heat sink includes a \(200 \times 200\) array of nanofins. Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperature is maintained at \(T_{t}<85^{\circ} \mathrm{C}\) for the unfinned and finned packages.
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