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Chapter 3: Problem 213

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness

Short Answer

Expert verified
The first finned surface has higher efficiency and higher effectiveness compared to the second finned surface.

Step by step solution

01

1. Identify relevant parameters

Let's denote the convection heat transfer coefficients for the first and second finned surfaces as h_1 and h_2, respectively. Given that h_1 = 2 * h_2. Now we need to relate fin efficiency and fin effectiveness to the convection heat transfer coefficient h. To do this, let's use the standard fin efficiency formula for a straight rectangular fin: Fin Efficiency (畏) = tanh(m * L) / (m * L), where m = sqrt(h * P / (k * A_c)), L is the fin length, P is the perimeter, k is the thermal conductivity, and A_c is the cross-sectional area for the fin.
02

2. Efficiency Comparison

Now, we can directly compare the efficiency of the first and second finned surfaces. For example, let the efficiency of the first fin be 畏_1 and that of the second fin be 畏_2. Since h_1 = 2 * h_2, the value of m for the first fin will be sqrt(2) times the value of m for the second fin, as m = sqrt(h * P / (k * A_c)) Thus, m_1 = sqrt(2) * m_2. Fin Efficiency comparison: 畏_1 = tanh(m_1 * L) / (m_1 * L) 畏_2 = tanh(m_2 * L) / (m_2 * L)
03

3. Determine Effectiveness Comparison

For comparing effectiveness, we can use the definition of fin effectiveness: Fin Effectiveness (蠄) = 1 + 畏(m * L) Therefore, 蠄_1 = 1 + 畏_1 * m_1 * L 蠄_2 = 1 + 畏_2 * m_2 * L Now we can compare 蠄_1 and 蠄_2.
04

4. Conclusion

Based on the comparison of efficiency and effectiveness, we can revisit the statement options: (a) Higher efficiency and higher effectiveness: Since 畏_1 > 畏_2 due to higher convection heat transfer coefficient, and the effectiveness formula states 蠄 = 1 + 畏(m * L), it's possible that 蠄_1 > 蠄_2. (b) Higher efficiency but lower effectiveness: This is not possible due to the way the effectiveness formula is formulated (蠄 = 1 + 畏(m * L)). (c) Lower efficiency but higher effectiveness: This is not possible since higher h results in higher fin efficiency. (d) Lower efficiency and lower effectiveness: This option is not possible due to higher convection heat transfer coefficient. (e) Equal efficiency and equal effectiveness: This is not possible since 畏_1 > 畏_2 and 蠄_1 > 蠄_2. Therefore, the correct answer is: (a) Higher efficiency and higher effectiveness.

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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 measurement of how well a fin can transfer heat from the base to the tip. It compares the fin's actual heat transfer with the maximum possible transfer under ideal conditions. The formula used to calculate fin efficiency often involves parameters such as the perimeter of the fin, thermal conductivity, and convection heat transfer coefficient. The fin efficiency (\(\eta\)) can be represented mathematically as:
\[ \eta = \frac{\tanh(m \cdot L)}{m \cdot L} \]where:
  • \(m\) is a term that depends on the convection heat transfer coefficient (\(h\)), perimeter (\(P\)), thermal conductivity (\(k\)), and cross-sectional area of the fin (\(A_c\)).
  • \(L\) is the length of the fin.
The higher the efficiency, the better the fin is at transferring heat over its entire length.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient (\(h\)) is a critical parameter in heat transfer that quantifies how easily heat is transferred from a solid surface to a fluid or vice versa. It is essential in determining how effective a finned surface will be at cooling or heating applications.
A higher convection heat transfer coefficient means that the fin can more efficiently transfer heat to the surrounding environment. In the given exercise, the first surface has a convection heat transfer coefficient twice that of the second, meaning it has a greater potential for heat transfer. This increase in \(h\) also affects other calculations related to fin efficiency and effectiveness.
Thermal Conductivity
Thermal conductivity (\(k\)) is a material property that describes the ability of a material to conduct heat. It appears in the formula for calculating the efficiency and effectiveness of fins, particularly in terms of the variable \(m\):
\[m = \sqrt{\frac{h \cdot P}{k \cdot A_c}}\]Thermal conductivity directly influences the performance of the fin in dissipating heat. A higher thermal conductivity means heat is more easily conducted through the fin, improving its overall efficiency.
In practical applications, materials with high thermal conductivity are often chosen for fins to maximize their heat dissipation capacity.
Fin Effectiveness
Fin effectiveness (\(\psi\)) is another important measure, indicating how effective a fin is at enhancing heat transfer relative to a planar surface without fins. The formula for fin effectiveness takes the form:
\[ \psi = 1 + \eta(m \cdot L) \]This formula shows how fin efficiency impacts fin effectiveness.
  • Increased fin efficiency results in greater fin effectiveness, indicating that fins are not only optimizing heat transfer locally but are also significantly enhancing overall heat transfer capabilities.
  • The value \(m \cdot L\) accounts for the relationship between the convection heat transfer coefficient and fin geometry.
In the exercise, due to a higher convection heat transfer coefficient, fin effectiveness for the first surface is greater, aligning with the given solution conclusion.

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Most popular questions from this chapter

Consider a stainless steel spoon \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) partially immersed in boiling water at \(200^{\circ} \mathrm{F}\) in a kitchen at \(75^{\circ} \mathrm{F}\). The handle of the spoon has a cross section of \(0.08\) in \(\times\) \(0.5\) in, and extends 7 in in the air from the free surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is \(3 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), determine the temperature difference across the exposed surface of the spoon handle. State your assumptions. Answer: \(124.6^{\circ} \mathrm{F}\)

Consider a 25-m-long thick-walled concrete duct \((k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section. The outer dimensions of the duct are \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\), and the thickness of the duct wall is \(2 \mathrm{~cm}\). If the inner and outer surfaces of the duct are at \(100^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer through the walls of the duct. Answer: \(47.1 \mathrm{~kW}\)

A mixture of chemicals is flowing in a pipe \(\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\right.\), and \(L=10 \mathrm{~m}\) ). During the transport, the mixture undergoes an exothermic reaction having an average temperature of \(135^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent any incident of thermal burn, the pipe needs to be insulated. However, due to the vicinity of the pipe, there is only enough room to fit a \(2.5\)-cm-thick layer of insulation over the pipe. The pipe is situated in a plant, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the insulation for the pipe such that the thermal conductivity of the insulation is sufficient to maintain the outside surface temperature at \(45^{\circ} \mathrm{C}\) or lower.

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature of the surface of the wood is \(-10^{\circ} \mathrm{C}\) and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

One wall of a refrigerated warehouse is \(10.0\)-m-high and \(5.0\)-m-wide. The wall is made of three layers: \(1.0\)-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fibreglass \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0-\mathrm{cm}\) thick gypsum board \((k=\) \(0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts \((k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), each \(2.0 \mathrm{~cm}\) in diameter and \(12.0 \mathrm{~cm}\) long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?
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