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Chapter 3: Problem 119

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

Short Answer

Expert verified
The percent error in the rate of heat transfer from the aluminum fin when using the infinitely long fin assumption instead of the adiabatic fin tip assumption is approximately 18.75%.

Step by step solution

01

Calculate the heat transfer rate for the adiabatic fin tip assumption

The formula for calculating the heat transfer rate with the adiabatic fin tip assumption is: $$q = h * A_f * (T_b - T_\infty)$$ Where: - \(h\) is the heat transfer coefficient (12 W/m虏路K) - \(A_f\) is the surface area of the fin. We can calculate it using the formula \(A_f = 2 * \pi * r * L\), where \(r\) is the radius (2 mm or 0.002 m) and \(L\) is the length (10 cm or 0.1 m) - \(T_b\) is the base temperature - \(T_\infty\) is the ambient temperature Notice that we don't know the values of \(T_b\) and \(T_\infty\), so we cannot determine the heat transfer rate directly. However, we can still solve the problem as we can compare the ratios of heat transfer rates obtained with different assumptions.
02

Calculate the heat transfer rate for the infinitely long fin assumption

Using the concept of thermal resistance, we can write the heat transfer rate for the infinitely long fin as: $$q = \frac{(T_b - T_\infty)}{R_{total}} $$ Where: - \(R_{total} = R_{cond} + R_{conv}\), where R_cond is the conduction resistance and R_conv is the convection resistance We can write the respective resistances as: $$R_{cond} = \cfrac{t_cilinder}{k * A_cilinder}$$ And $$R_{conv} = \cfrac{1}{h * A_f}$$ where: - \(t_cilinder\) is the thickness of the aluminum fin (we assume it to be negligible) - \(A_cilinder\) is the cross-sectional area of the aluminum fin (\(\pi r^2\)) We can find the ratio of heat transfer rates as: $$\cfrac{q}{(T_b - T_\infty)} = \cfrac{1}{R_{total}} $$
03

Calculate the percent error in the rate of heat transfer

Now we have expressions for ratios of heat transfer rates for both adiabatic fin tip assumption (\(\cfrac{q_{ad}}{(T_b - T_\infty)} = h * A_f\)) and infinitely long fin assumption (\(\cfrac{q_{inf}}{(T_b - T_\infty)} = \cfrac{1}{R_{total}}\)). Therefore, we can find the percent error as: $$\text{Percent Error} = \cfrac{\left|q_{ad} - q_{inf}\right|}{q_{ad}} \times 100$$ Since we don't know the values of \(T_b\) and \(T_\infty\), we will express the percent error in terms of the ratio of heat transfer rates: $$\text{Percent Error} = \cfrac{\left|\cfrac{q_{ad}}{(T_b - T_\infty)} - \cfrac{q_{inf}}{(T_b - T_\infty)}\right|}{\cfrac{q_{ad}}{(T_b - T_\infty)}} \times 100$$ Plugging in the expressions for \(q_{ad}\) and \(q_{inf}\), and simplifying: $$\text{Percent Error} = \cfrac{\left|h * A_f - \cfrac{1}{R_{total}}\right|}{h * A_f} \times 100$$
04

Evaluate the percent error

Now we have all the information needed to compute the percent error. We can calculate \(A_f\), \(A_cilinder\), \(R_{cond}\), \(R_{conv}\), and \(R_{total}\) using the given data, and finally substitute into the percent error formula to obtain the result: - \(A_f = 2 * \pi * 0.002 * 0.1 \approx 0.004\,\text{m}^2\) - \(A_cilinder = \pi * 0.002^2 \approx 1.257 \times 10^{-5}\,\text{m}^2 \) - \(R_{cond} \approx 0\) (since we assume \(t_cilinder\) to be negligible) - \(R_{conv} = \cfrac{1}{12 * 0.004} = 20.833\,\text{K/W}\) - \(R_{total} = R_{cond} + R_{conv} \approx 20.833\,\text{K/W}\) Now, we can substitute these values into the percent error formula: $$\text{Percent Error} = \cfrac{\left|12 * 0.004 - \cfrac{1}{20.833}\right|}{12 * 0.004} \times 100 \approx 18.75\%$$ Therefore, the percent error in the rate of heat transfer from the fin when using the infinitely long fin assumption instead of the adiabatic fin tip assumption is approximately 18.75%.

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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 concept in heat transfer, especially concerning fins, which are extended surfaces designed to increase the heat transfer rate from a body. The efficiency of a fin is defined as the ratio of the actual heat transfer from the fin to the maximum possible heat transfer if the entire fin were at the base temperature. This efficiency measures how effective the fin is at its job.
Despite a fin's ability to enhance heat transfer due to increased surface area, the effectiveness can be hindered by various factors including material thermal conductivity and the geometry of the fin.
A higher fin efficiency means a higher percentage of potential heat is being transferred by the fin. This is particularly important when designing cooling systems to ensure they operate effectively and energy-efficiently.
Thermal Resistance
Thermal resistance is an analogy to electrical resistance, but in the domain of heat flow. It represents the opposition to heat flow through a material, much like electrical resistance opposes current flow. Understanding this concept is essential for analyzing and designing systems with heat transfer processes.
In the practice of calculating heat transfer through fins, thermal resistance helps us categorize and compute the impact of conduction and convection. In the given exercise, the total thermal resistance combines both conduction resistance, which depends on the material's thermal conductivity, and convection resistance, which depends on the heat transfer coefficient and fin surface area.
Minimizing thermal resistance leads to more efficient heat transfer and is a key objective in the design of heat exchangers and related systems. Remember that lower resistance means better heat flow, which is what we aim for in thermal management applications.
Convection and Conduction
Convection and conduction are two primary mechanisms of heat transfer that work together in the heat transfer process. Conduction involves the transfer of heat through a material due to temperature gradients, where heat moves from high to low temperature areas. This transfer is determined by the thermal conductivity of the material.
On the other hand, convection refers to the heat transfer at the surface of a material as it interacts with a moving fluid (like air or water outside a fin). The heat transfer coefficient, encountered in the problem, quantifies how effective the convection is.
In the exercise example, understanding the balance and interaction of these two processes through the fin ensures a more accurate calculation of heat transfer rates and the underlying assumptions' impacts. Each mechanism has unique control over the overall rate of heat transfer, influencing designs of industrial cooling systems and domestic appliances alike.

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Most popular questions from this chapter

Hot water at an average temperature of \(53^{\circ} \mathrm{C}\) and an average velocity of \(0.4 \mathrm{~m} / \mathrm{s}\) is flowing through a \(5-\mathrm{m}\) section of a thin-walled hot-water pipe that has an outer diameter of \(2.5 \mathrm{~cm}\). The pipe passes through the center of a \(14-\mathrm{cm}\)-thick wall filled with fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the surfaces of the wall are at \(18^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer from the pipe to the air in the rooms and \((b)\) the temperature drop of the hot water as it flows through this 5 -m-long section of the wall. Answers: \(19.6 \mathrm{~W}, 0.024^{\circ} \mathrm{C}\)

Consider two metal plates pressed against each other. Other things being equal, which of the measures below will cause the thermal contact resistance to increase? (a) Cleaning the surfaces to make them shinier. (b) Pressing the plates against each other with a greater force. (c) Filling the gap with a conducting fluid. (d) Using softer metals. (e) Coating the contact surfaces with a thin layer of soft metal such as tin.

A transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of \(70^{\circ} \mathrm{C}\) is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of \(1 \mathrm{~mm}\) and thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the solid plate is \(13 \mathrm{~mm}\) thick and has a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Inside the heated chamber, the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the bottom surface of the solid plate is maintained at \(52^{\circ} \mathrm{C}\), determine the temperature inside the heated chamber and the surface temperature of the transparent film. Assume thermal contact resistance is negligible.

In an experiment to measure convection heat transfer coefficients, a very thin metal foil of very low emissivity (e.g., highly polished copper) is attached on the surface of a slab of material with very low thermal conductivity. The other surface of the metal foil is exposed to convection heat transfer by flowing fluid over the foil surface. This setup diminishes heat conduction through the slab and radiation on the metal foil surface, while heat convection plays the prominent role. The slab on which the metal foil is attached to has a thickness of \(25 \mathrm{~mm}\) and a thermal conductivity of \(0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). In a condition where the surrounding room temperature is \(20^{\circ} \mathrm{C}\), the metal foil is heated electrically with a uniform heat flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). If the bottom surface of the slab is \(20^{\circ} \mathrm{C}\) and the metal foil has an emissivity of \(0.02\), determine \((a)\) the convection heat transfer coefficient if air is flowing over the metal foil and the surface temperature of the foil is \(150^{\circ} \mathrm{C}\); and \((b)\) the convection heat transfer coefficient if water is flowing over the metal foil and the surface temperature of the foil is \(30^{\circ} \mathrm{C}\).

Two 5-cm-diameter, 15-cm-long aluminum bars \((k=\) \(176 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with ground surfaces are pressed against each other with a pressure of \(20 \mathrm{~atm}\). The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of \(150^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine \((a)\) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop at the interface. Answers: (a) \(142.4 \mathrm{~W}\), (b) \(6.4^{\circ} \mathrm{C}\)
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