Question

Circular fins of uniform cross section, with diameter of \(10 \mathrm{~mm}\) and length of \(50 \mathrm{~mm}\), are attached to a wall with surface temperature of \(350^{\circ} \mathrm{C}\). The fins are made of material with thermal conductivity of \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and they are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heat transfer rate and plot the temperature variation of a single fin for the following boundary conditions: (a) Infinitely long fin (b) Adiabatic fin tip (c) Fin with tip temperature of \(250^{\circ} \mathrm{C}\) (d) Convection from the fin tip

Short answer

#tag_title#Step 6: Plot the temperature distribution for an infinitely long fin#tag_content#To plot the temperature distribution, we can convert the dimensionless temperature back to the actual temperature using the relationship: $$T(x) = T_\infty + (T_w - T_\infty) \Theta(\zeta)$$ For the infinitely long fin, we have \(\Theta(\zeta) = e^{-\zeta}\), so: $$T(x) = T_\infty + (T_w - T_\infty) e^{-\zeta}$$ Using the given values, we can convert the distances and find the temperatures for different points along the fin: $$T(x) = 23 + (400 - 23) e^{-\zeta}$$ The temperature distribution for an infinitely long fin can be plotted as a function of the distance from the wall or \(\zeta\). In this graph, the temperature decreases exponentially from the wall temperature to the ambient temperature as we move along the fin. In summary, for the infinitely long fin, the heat transfer rate is \(7.086 \mathrm{~W}\) and the temperature distribution can be plotted as a decaying exponential function of the distance from the wall.

Step-by-step solution

Write the fin equation

For a 1D steady-state radial fin, the fin equation is given by: $$\frac{d}{dx}\left( k A_{c} \frac{dT}{dx} \right) - h P (T - T_{\infty}) = 0$$ where \(x\) is the distance from the wall, \(k\) is the thermal conductivity, \(A_c\) is the cross-sectional area, \(T\) is the temperature, \(h\) is the heat transfer coefficient, \(P\) is the perimeter, and \(T_\infty\) is the ambient temperature.

Simplify the equation for a circular fin and use dimensionless variables

For a circular fin with uniform cross-section, the cross-sectional area \(A_c = \pi (d/2)^2\) and the perimeter \(P = \pi d\). We can simplify the equation by introducing dimensionless variables: $$\Theta(x) = \frac{T(x) - T_\infty}{T_{w} - T_\infty}$$ $$\zeta = \sqrt{\frac{h P}{k A_c}} x$$ Substituting the variables into the fin equation, we obtain: $$\frac{d^2 \Theta}{d \zeta^2} - \Theta = 0$$

Solve the general differential equation

The solution for \(\Theta(\zeta)\) can be found using the method of separation of variables. The solution will have the form: $$\Theta(\zeta) = C_1 e^{\sqrt{-1} \zeta} + C_2 e^{-\sqrt{-1} \zeta}$$ Now, we need to apply the specific boundary conditions for each case and find the constants \(C_1\) and \(C_2\). #a. Infinitely long fin

Apply boundary condition for infinitely long fin

For an infinitely long fin, as \(\zeta \to \infty\), \(\Theta(\zeta) \to 0\). Therefore: $$C_1 = 0$$ The remaining boundary condition at the wall (\(\zeta = 0\)) is \(\Theta(0) = 1\): $$\Theta(\zeta) = C_2 e^{-\zeta}$$ $$1 = C_2 e^{0} \Rightarrow C_2 = 1$$ The temperature distribution for an infinitely long fin is: $$\Theta(\zeta) = e^{-\zeta}$$

Calculate heat transfer rate for infinitely long fin

The heat transfer rate from the fin, \(Q_f\), can be found using the temperature distribution: $$Q_f = k A_c \frac{dT}{dz}\Bigg|_{\zeta=0}$$ $$= k A_c \frac{T_w - T_\infty}{\sqrt{k A_c / hP}} \cdot \frac{d\Theta}{d\zeta}\Bigg|_{\zeta=0}$$ $$= k A_c \sqrt{\frac{h P}{k A_c}} \cdot \frac{d\Theta}{d\zeta}\Bigg|_{\zeta=0}$$ Using \(\Theta (\zeta) = e^{-\zeta}\), we can find the derivative at \(\zeta=0\): $$Q_f = k A_c \sqrt{\frac{h P}{k A_c}} \cdot e^{-0} = k A_c \sqrt{\frac{h P}{k A_c}}$$ Now, substitute the given values: $$Q_f = 240 \left(\frac{\pi}{4} \cdot 0.01^2 \right) \sqrt{\frac{250 \cdot \pi \cdot 0.01}{240 \cdot \frac{\pi}{4} \cdot 0.01^2}}$$ $$Q_f = 7.086 \mathrm{~W}$$ The heat transfer rate for an infinitely long fin is \(7.086 \mathrm{~W}\). Now let's plot the temperature distribution:

Key concepts

Conduction and Convection in Fins

The concepts of conduction and convection are fundamental to understanding heat transfer in fins. Conduction is the process by which heat energy is transmitted through a material due to a temperature gradient. In the context of a fin, conduction occurs along the length of the fin from the base, which is at a higher temperature, to the tip, which is cooler. The rate of this heat transfer through the fin depends on the thermal conductivity of the material.

Convection, on the other hand, is the heat transfer that occurs between a solid surface and a fluid, like air or water, that is in motion. The moving fluid carries away heat from the surface. In fin heat transfer analysis, convection plays a vital role in removing the heat conducted through the fin to the surrounding air. Convection is characterized by the convection heat transfer coefficient; a higher coefficient indicates more efficient heat transfer from the fin to the air.

Relation to the Exercise

For the cylindrical fins in the exercise, conduction heat transfer within the fin directs heat toward the fin tip, while convection from the surface of the fin dissipates this heat to the surrounding air at ambient temperature. In the step-by-step solution, the resulting fin equation accounts for both these phenomena and relates the temperature along the fin to these two modes of heat transfer.

Fin Efficiency

Fin efficiency is a measure of how well a fin conducts heat from its base to its surroundings compared to an ideal fin that has no temperature variation along its length. It quantifies the actual heat transfer rate in relation to the maximum possible heat transfer rate. The efficiency of a fin is influenced by factors such as its geometry, the thermal conductivity of the material, convection conditions, and environmental temperature.

An infinitely long fin is often assumed to have the highest efficiency because it offers a large area for heat transfer and is presumed to maintain a constant base temperature throughout. However, in real applications, fins are not infinite, and various tip conditions like adiabatic (insulated), constant temperature, or convection from the fin tip require different approaches to calculation, affecting the fin's efficiency.

Application to the Given Problem

In the provided exercise, the efficiency of the fin under several boundary conditions is analyzed, demonstrating how the actual heat transfer rate is calculated.

Thermal Conductivity

Thermal conductivity is a material-specific property that indicates how well a material can conduct heat. It represents the amount of heat, in watts, that can pass through a meter thickness of the material when there is a temperature difference of one Kelvin across the material. Materials with high thermal conductivity are excellent conductors of heat, such as metals, whereas those with low thermal conductivity are insulators, like foam.

In the context of fins, a higher thermal conductivity allows for more effective heat transfer from the base to the tip of the fin. This effectiveness directly influences the fin's ability to dissipate heat into the surrounding environment.

Thermal Conductivity in the Example

The exercise mentions a thermal conductivity value of the fin material (\(240 \frac{W}{m \times K}\)), which is fundamental in determining the temperature distribution and overall heat transfer rate. In the solution steps, this thermal conductivity plays a critical role in solving the fin equation and assessing the fin's performance.