Obtain a relation for the fin efficiency for a fin of constant cross-sectional area \(A_{c}\), perimeter \(p\), length \(L\), and thermal conductivity \(k\) exposed to convection to a medium at \(T_{\infty}\) with a heat transfer coefficient \(h\). Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly \(T_{\infty}\). Take the temperature of the fin at the base to be \(T_{b}\) and neglect heat transfer from the fin tips. Simplify the relation for \((a)\) a circular fin of diameter \(D\) and \((b)\) rectangular fins of thickness \(t\).

Short Answer

Expert verified
Answer: (a) For a circular fin with diameter D, the fin efficiency is given by: \(\eta_f = 1 - \frac{1}{e^{\frac{4hL}{kD}} - 1}\). (b) For a rectangular fin with thickness t, the fin efficiency is given by: \(\eta_f = 1 - \frac{1}{e^{\frac{2h(L + t)}{kt}} - 1}\).

Step by step solution

01

1. Define the Fin Efficiency

Fin efficiency is defined as the ratio of the actual heat transfer rate from the fin to the maximum possible heat transfer rate if the entire fin were at the base temperature. Let's denote the fin efficiency as \(\eta_f\).
02

2. Obtain the Governing Equations

The governing equation for heat transfer in a fin with constant cross-sectional area can be obtained from the conduction and convection heat transfer equations. The equation for heat conduction through the fin is: \(q_{cond} = kA_{c}\frac{dT}{dx}\). The equation for heat convection away from the fin is: \(q_{conv} = hp(T - T_{\infty})\), where \(T\) is the temperature of the fin at a distance \(x\) from the base.
03

3. Apply Energy Balance

At a steady state, the rate at which heat is conducted along the fin and the rate at which heat is convected away from the fin must be equal. So, we can equate \(q_{cond}\) and \(q_{conv}\): \(kA_{c}\frac{dT}{dx} = hp(T - T_{\infty})\).
04

4. Non-dimensionalize and Simplify the Equation

Introducing a dimensionless temperature \(\theta = \frac{T - T_{\infty}}{T_b - T_{\infty}}\) and a dimensionless distance \(\xi = \frac{x}{L}\), the equation becomes: \(\frac{d\theta}{d\xi} = \frac{hpL}{kA_c}(1 - \theta)\).
05

5. Integrate the Equation and Find the General Solution

Integrating the equation with respect to \(\xi\), we get: \(\ln{(1 - \theta)} = -\frac{hpL}{kA_c}\xi + C\), where C is the integration constant.
06

6. Apply the Boundary Conditions

We know that at the base of the fin (\(\xi = 0\)), the dimensionless temperature \(\theta = 1\). So, from the general solution, we find the integration constant \(C = 0\). At the tip of the fin (\(\xi = 1\)), \(\theta(\xi = 1) = 0\). Using this boundary condition, we find the fin efficiency: \(\eta_f = 1 - \frac{T_\infty - T_b}{T_\infty - T(1)} = 1 - \frac{1}{e^{\frac{hpL}{kA_c}} - 1}\).
07

7. Simplify the Fin Efficiency for Circular Fins

For circular fins with diameter \(D\), the cross-sectional area is \(A_c = \pi(\frac{D}{2})^2\) and the perimeter is \(p = \pi D\). Plugging these into the fin efficiency equation, we get: \(\eta_f = 1 - \frac{1}{e^{\frac{h\pi DL}{k\pi(\frac{D}{2})^2}} - 1}\). By simplifying, we obtain the fin efficiency for circular fins: \(\eta_f = 1 - \frac{1}{e^{\frac{4hL}{kD}} - 1}\).
08

8. Simplify the Fin Efficiency for Rectangular Fins

For rectangular fins with thickness \(t\), the cross-sectional area is \(A_c = Lt\) and the perimeter is \(p = 2(L + t)\). Plugging these into the fin efficiency equation, we get: \(\eta_f = 1 - \frac{1}{e^{\frac{h(2L + 2t)L}{kLt}} - 1}\). By simplifying, we obtain the fin efficiency for rectangular fins: \(\eta_f = 1 - \frac{1}{e^{\frac{2h(L + t)}{kt}} - 1}\).

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Key Concepts

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

Heat Transfer Coefficient
The heat transfer coefficient, denoted as h, is a measure of the thermal exchange rate between a surface and the surrounding fluid. In the context of fins, which are extended surfaces used to increase the heat transfer area and enhance heat dissipation from a surface to the surrounding medium, the heat transfer coefficient plays a crucial role.

When a fin is in contact with a fluid, such as air or water, the fluid either takes heat away or delivers it to the fin, depending on their respective temperatures. The heat transfer coefficient quantifies how effectively the fluid performs this task per unit area and per degree of temperature difference. It is expressed as watt per square meter per Kelvin (W/m²K).

The higher the value of h, the more efficient the fluid is in transferring heat to or from the surface. It is influenced by various factors, including the type of fluid, fluid velocity, surface roughness, and the temperature gradient between the surface and the fluid.

Practical Implications for Fins

In the case of fins, a high heat transfer coefficient means that the fin efficiently releases heat to the environment, which is crucial for devices that rely on maintaining certain operating temperatures, such as electronic components or engines.
Thermal Conductivity
Thermal conductivity, represented by the symbol k, is a material property that indicates the ability of a material to conduct heat. It is defined as the amount of heat that passes in unit time through a unit area of a substance with a temperature gradient of one degree per unit distance. The SI unit for thermal conductivity is watt per meter Kelvin (W/mK).

Materials with high thermal conductivity are excellent conductors of heat and are often used in applications where efficient heat transfer is required. Conversely, materials with low thermal conductivity are good insulators.

Involvement in Fin Efficiency

The role of thermal conductivity in fin efficiency is pivotal. Since fins are designed to conduct heat from the base to the tip, before it is then convected away by the fluid, a high k value allows the fin material to more effectively transfer heat along its length.

This property directly impacts the performance of fins. The governing equation for a fin's heat conduction involves the thermal conductivity (k) and demonstrates how it influences the temperature distribution along the fin and thus the fin's ability to dissipate heat.
Boundary Conditions in Heat Transfer
Boundary conditions in heat transfer provide information about how heat behaves at the limits or interfaces of different materials or states. They are an essential aspect of solving heat transfer problems as they define limit values for the physical system being analyzed, such as temperatures, heat fluxes, or convection circumstances at a surface.

In the analysis of fin efficiency, it's typical to consider boundary conditions at the fin's base and its tip. For instance, we usually assume that the fin base is at a constant temperature Tb, which is higher than the surrounding fluid temperature T. At the fin tip, the temperature might be considered equal to the fluid temperature if the fin is sufficiently long, as stated in the exercise.

These assumptions allow us to construct and solve the mathematical model for heat transfer in fins.

Application to Simplifying Equations

Applying appropriate boundary conditions simplifies the calculations for the dimensionless temperature and helps in deriving a more manageable form for the fin efficiency relation. As in the given solution, knowing the boundary temperatures allows the integration of the heat transfer equation to solve for fin efficiency, ultimately leading to the simplified relations for circular and rectangular fin geometries.

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Most popular questions from this chapter

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

What is the reason for the widespread use of fins on surfaces?

Consider two identical people each generating \(60 \mathrm{~W}\) of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1 -mm-thick leather \((k=\) \(0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) that covers half of the body while the second one is wearing clothes made of 1 -mm-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that covers the body completely. The ambient air is at \(30^{\circ} \mathrm{C}\), the heat transfer coefficient at the outer surface is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the inner surface temperature of the clothes can be taken to be \(32^{\circ} \mathrm{C}\). Treating the body of each person as a 25 -cm-diameter, \(1.7-\mathrm{m}\)-long cylinder, determine the fractions of heat lost from each person by perspiration.

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=\) \(0.73 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The tube diameter is \(D_{1}=20 \mathrm{~cm}\) and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\), determine the rate of heat transfer per unit length of the tube through the insulation.

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