/*! 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

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is \(3 \mathrm{~mm}\) thick and whose interior surface is maintained at a temperature of \(36^{\circ} \mathrm{C}\). On a calm day the convection heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), but with \(30 \mathrm{~km} / \mathrm{h}\) winds it reaches \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In both cases the ambient air temperature is \(-15^{\circ} \mathrm{C}\). (a) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at \(-15^{\circ} \mathrm{C}\) on the windy day?

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?

Determine the percentage increase in heat transfer associated with attaching aluminum fins of rectangular profile to a plane wall. The fins are \(50 \mathrm{~mm}\) long, \(0.5 \mathrm{~mm}\) thick, and are equally spaced at a distance of \(4 \mathrm{~mm}\) ( 250 fins \(/ \mathrm{m})\). The convection coefficient associated with the bare wall is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while that resulting from attachment of the fins is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigerant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of \(-3^{\circ} \mathrm{C}\), what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator's defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\). Frost may be assumed to have a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and natural convection maintains a convection coefficient of \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long will it take for the frost to melt? The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.
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