/*! 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

A nuclear reactor fuel element consists of a solid cylindrical pin of radius \(r_{1}\) and thermal conductivity \(k_{f}\). The fuel pin is in good contact with a cladding material of outer radius \(r_{2}\) and thermal conductivity \(k_{c^{*}}\). Consider steady-state conditions for which uniform heat generation occurs within the fuel at a volumetric rate \(\dot{q}\) and the outer surface of the cladding is exposed to a coolant that is characterized by a temperature \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain equations for the temperature distributions \(T_{f}(r)\) and \(T_{c}(r)\) in the fuel and cladding, respectively. Express your results exclusively in terms of the foregoing variables. (b) Consider a uranium oxide fuel pin for which \(k_{f}=2\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{1}=6 \mathrm{~mm}\) and cladding for which \(k_{c}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{2}=9 \mathrm{~mm}\). If \(\dot{q}=2 \times 10^{8}\) \(\mathrm{W} / \mathrm{m}^{3}, h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty}=300 \mathrm{~K}\), what is the maximum temperature in the fuel element? (c) Compute and plot the temperature distribution, \(T(r)\), for values of \(h=2000,5000\), and 10,000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the operator wishes to maintain the centerline temperature of the fuel element below \(1000 \mathrm{~K}\), can she do so by adjusting the coolant flow and hence the value of \(h\) ?

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.

A cylindrical shell of inner and outer radii, \(r_{i}\) and \(r_{o}\), respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate \(\left(\mathrm{W} / \mathrm{m}^{3}\right)\) of \(\dot{q}\). The inner surface is insulated, while the outer surface of the shell is exposed to a fluid at \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain an expression for the steady-state temperature distribution \(T(r)\) in the shell, expressing your result in terms of \(r_{i}, r_{o}, \dot{q}, h, T_{\infty}\), and the thermal conductivity \(k\) of the shell material. (b) Determine an expression for the heat rate, \(q^{\prime}\left(r_{o}\right)\), at the outer radius of the shell in terms of \(\dot{q}\) and shell dimensions.

Consider a plane composite wall that is composed of two materials of thermal conductivities \(k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{\mathrm{A}}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and material \(\mathrm{B}\) adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?
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