Question

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e. \(20^{\circ} \mathrm{C}\). Its width is \(5.0 \mathrm{~cm}\); thickness is \(1.0 \mathrm{~mm}\); thermal conductivity is \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\); and base temperature is \(40^{\circ} \mathrm{C}\). The heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the fin temperature at a distance of \(5.0 \mathrm{~cm}\) from the base and the rate of heat loss from the entire fin.

Short answer

Answer: The temperature at a distance of 5.0 cm from the base is approximately 24.82°C, and the rate of heat loss from the entire fin is approximately 0.481 W.

Step-by-step solution

Convert units to SI

We are given the measurements in cm and mm, so let's first convert them to meters for consistency. - Width: \(5.0 cm = 0.05 m\) - Thickness: \(1.0 mm = 0.001 m\) - Distance from the base: \(5.0 cm = 0.05 m\)

Calculate the fin area and perimeter

To determine the fin area (A) and perimeter (P), we use the following formulas: \(A = W \cdot t\) \(P = 2(W + t)\) - Area: \(A = 0.05m * 0.001m = 0.00005 m^2\) - Perimeter: \(P = 2(0.05m + 0.001m) = 0.102m\)

Calculate the fin parameter m

We can find the fin parameter (m) using the formula: \(m = \sqrt{\frac{hP}{kA}}\) Where: - \(h = 20 W/m^2K\) (heat transfer coefficient) - \(k = 200 W/mK\) (thermal conductivity) Now, plug in the values and solve for m: \(m = \sqrt{\frac{20W/m^2K \cdot 0.102m}{200W/mK \cdot 0.00005m^2}} = 20.3 m^{-1}\)

Calculate the fin efficiency

The fin efficiency can be found using the formula: \(\eta_f = \frac{\tanh{(mL)}}{mL}\) Where: - \(L = 0.05 m\) (the length of the fin) Plug in the values and solve for \(\eta_f\): \(\eta_f = \frac{\tanh{(20.3m^{-1} \cdot 0.05m)}}{20.3m^{-1} \cdot 0.05m} = 0.481\)

Calculate the fin temperature at a distance of 5 cm from the base

The temperature at a distance x from the base is given by: \(T(x) = T_{\infty} + (T_b - T_{\infty}) e^{-(mx)}\) Where: - \(T(x)\): Temperature at a distance x from the base - \(T_{\infty} = 20^{\circ}C\): Surrounding air temperature - \(T_b = 40^{\circ}C\): Base temperature - \(x = 0.05 m\): Distance from the base Plug in the values and solve for \(T(0.05m)\): \(T(0.05m) = 20^{\circ}C + (40^{\circ}C - 20^{\circ}C) e^{-(20.3m^{-1} \cdot 0.05m)} = 24.82^{\circ}C\)

Calculate the rate of heat loss from the entire fin

The rate of heat loss, \(Q_f\), can be found using the formula: \(Q_f = \eta_f A k \Delta T \) Where: - \(\Delta T = T_b - T_{\infty} = 20K\) Plug in the values and solve for \(Q_f\): \(Q_f = 0.481 \cdot 0.00005 m^2 \cdot 200 W/mK \cdot 20 K = 0.481 W\) The temperature at a distance of 5.0 cm from the base is approximately \(24.82^{\circ}C\), and the rate of heat loss from the entire fin is approximately \(0.481 W\).

Key concepts

Understanding Thermal Conductivity in Fin Heat Transfer

Thermal conductivity is a fundamental property of materials that measures a material's ability to conduct heat. It is represented by the symbol \( k \) and is typically expressed in units of watts per meter per kelvin (\( W/m\cdot K \)). A higher thermal conductivity means that the material can conduct heat more efficiently.

In the context of a fin as described in our exercise, thermal conductivity is crucial because it affects how well the fin transfers heat from the base to the environment. Fins are typically made of materials with high thermal conductivity to maximize the heat dissipation. The formula used in our exercise, \( Q_f = \eta_f A k \Delta T \), shows that the rate of heat loss from the fin is directly proportional to the thermal conductivity. If the thermal conductivity were lower, the fin would not be as effective in transferring heat away from the surface to which it is attached.

It's important for students to remember that thermal conductivity is an intrinsic property of the material and does not change with the dimensions of the fin or the temperature difference across it. However, the actual heat transfer does depend on the geometry of the fin and other factors like the heat transfer coefficient and fin efficiency.

The Role of Heat Transfer Coefficient in Fin Performance

The heat transfer coefficient, represented by \( h \), is a measure of the convective heat transfer between a surface and a fluid flowing over it. In SI units, it is expressed in watts per square meter per kelvin (\( W/m^2\cdot K \)). This coefficient is critical in evaluating how efficiently heat is transferred from the fin to the surrounding environment.

In our exercise, the value of \( h \) is provided and is used to determine the fin parameter \( m \) by the formula \( m = \sqrt{\frac{hP}{kA}} \). The value of \( m \) is then used to calculate the fin efficiency and ultimately, the rate of heat loss from the fin. The heat transfer coefficient depends on various factors, including the type of fluid, fluid velocity, and surface roughness. A higher heat transfer coefficient indicates a more effective convection process and typically leads to higher rates of heat loss from the fin.

Understanding the heat transfer coefficient is essential for tuning the performance of the fin in practical applications. Engineers may alter the conditions around the fin, such as increasing the flow rate of a fluid, to enhance the heat transfer coefficient and, thereby, the efficiency of the heat transfer process.

Calculating Fin Efficiency and Its Impact on Heat Dissipation

Fin efficiency is a crucial measure that tells us how effectively a fin transfers heat compared to an ideal fin with the same dimensions that transfers heat perfectly. It's represented by \( \eta_f \) and is given by the formula \( \eta_f = \frac{\tanh{(mL)}}{mL} \), where \( mL \) is the product of the fin parameter and the length of the fin.

In our exercise, we determined the fin efficiency to be approximately 0.481 or 48.1%. This means that the actual heat transfer rate of the fin is about 48.1% of what it would be if the fin were ideally efficient. The lower the fin efficiency, the less effective the fin is at dissipating heat, often leading to a greater temperature difference between the base and the tip of the fin.

Improving fin efficiency could involve changing the material or altering the fin's dimensions to reduce temperature variations along its length. It's a critical design parameter that can significantly affect the overall performance of systems that rely on fins for heat dissipation, such as in radiators, heatsinks, or air conditioning units. Students should understand that fin efficiency is a direct indicator of the fin's performance, influencing how designers choose fin materials and geometries for specific industrial applications.