Question

Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin is the \((a)\) fin effectiveness and \((b)\) fin efficiency higher? Explain.

Short answer

Based on the provided step-by-step solution, determine which fin has a higher fin effectiveness and fin efficiency. Answer: To determine which fin has a higher fin effectiveness and fin efficiency, you will need to compare the values of Fin Effectiveness_1 and Fin Effectiveness_2, as well as Fin Efficiency_1 and Fin Efficiency_2. Unfortunately, numerical values for the required parameters (length L, thermal conductivity k, heat transfer coefficient h, temperature T_base, temperature T_ambient, and temperature gradient dT/dx) were not given, so it's not possible to provide a definitive answer to which fin has higher effectiveness and efficiency without this information.

Step-by-step solution

Define fin effectiveness and fin efficiency

Fin effectiveness is a measure of the actual heat transfer by the fin compared to the maximum possible heat transfer. The formula for fin effectiveness is: \[ \text{Fin Effectiveness} = \frac{\text{Actual Heat Transfer by Fin}}{\text{Max Possible Heat Transfer}} \] Fin efficiency is a measure of how effectively the fin material is used in transferring heat. The formula for fin efficiency is: \[ \text{Fin Efficiency} = \frac{\text{Actual Heat Transfer by Fin}}{\text{Heat Transfer with Perfect Conduction}} \]

Calculate fin effectiveness and fin efficiency for Fin 1

For Fin 1, we are given the thickness as t. Let's assume its length is L and its thermal conductivity is k. We need to find the actual heat transfer by the fin and the heat transfer with perfect conduction. Actual Heat Transfer by Fin for Fin 1: \[ q_\text{actual,1} = h \cdot A_\text{s,1} \cdot (T_\text{base} - T_\text{ambient}) \] Where h is the heat transfer coefficient, A_s,1 is the fin surface area for Fin 1, T_base is the base temperature, and T_ambient is the ambient temperature. Heat Transfer with Perfect Conduction for Fin 1: \[ q_\text{perfect,1} = k \cdot A_\text{c,1} \cdot \frac{dT}{dx} \] Where A_c,1 is the fin cross-sectional area for Fin 1 and dT/dx is the temperature gradient. Now we can calculate the fin effectiveness and efficiency for Fin 1: Fin Effectiveness for Fin 1: \[ \text{Fin Effectiveness}_1 = \frac{q_\text{actual,1}}{q_\text{max,1}} \] Fin Efficiency for Fin 1: \[ \text{Fin Efficiency}_1 = \frac{q_\text{actual,1}}{q_\text{perfect,1}} \]

Calculate fin effectiveness and fin efficiency for Fin 2

For Fin 2, we are given the thickness as 2t. We will use the same length and thermal conductivity as Fin 1, i.e., L and k, respectively. Actual Heat Transfer by Fin for Fin 2: \[ q_\text{actual,2} = h \cdot A_\text{s,2} \cdot (T_\text{base} - T_\text{ambient}) \] Where A_s,2 is the fin surface area for Fin 2. Heat Transfer with Perfect Conduction for Fin 2: \[ q_\text{perfect,2} = k \cdot A_\text{c,2} \cdot \frac{dT}{dx} \] Where A_c,2 is the fin cross-sectional area for Fin 2. Now we can calculate the fin effectiveness and efficiency for Fin 2: Fin Effectiveness for Fin 2: \[ \text{Fin Effectiveness}_2 = \frac{q_\text{actual,2}}{q_\text{max,2}} \] Fin Efficiency for Fin 2: \[ \text{Fin Efficiency}_2 = \frac{q_\text{actual,2}}{q_\text{perfect,2}} \]

Compare fin effectiveness and fin efficiency of both fins

Now that we have the fin effectiveness and efficiency for both fins, we can compare them to determine which fin has higher fin effectiveness and fin efficiency. For fin effectiveness: Compare \(\text{Fin Effectiveness}_1\) and \(\text{Fin Effectiveness}_2\) to find out which fin has higher fin effectiveness. For fin efficiency: Compare \(\text{Fin Efficiency}_1\) and \(\text{Fin Efficiency}_2\) to find out which fin has higher fin efficiency. By comparing these quantities, it will become clear which fin has higher effectiveness and efficiency.

Key concepts

Heat Transfer Coefficient

The heat transfer coefficient (h) is a pivotal factor in thermal management and energy conservation. It quantifies the rate of heat transfer between a physical system's surface and its environment per unit surface area per unit temperature difference. The higher the value of this coefficient, the more effective the system is at transferring heat. For example, in the context of fin performance, a high heat transfer coefficient indicates that the fin is more capable of releasing heat into the surrounding air, which is particularly vital for heat sinks in electronic devices or other applications where cooling is essential.

Consider two fins with different dimensions in a heat transfer scenario. Despite sharing the same material, the one with the better heat transfer coefficient will dissipate heat more efficiently. This concept helps us understand part of the Step 1 calculation, where the actual heat transfer by a fin is computed using this coefficient. By comparing the coefficient values and incorporating other variables, students can predict which fin design would provide better thermal management.

Thermal Conductivity

Thermal conductivity (k) is a property that measures a material's ability to conduct heat. It signifies how easily heat flows through a material due to a temperature difference, generally expressed in watts per meter kelvin (W/mK). Materials with high thermal conductivity, such as copper or aluminum, are excellent conductors of heat, making them ideal for use in fins and other heat dissipation tools.

When evaluating the fin effectiveness and efficiency, as shown in the exercise, we involve thermal conductivity in calculations for both the actual and perfect conduction heat transfer. Understanding thermal conductivity allows students to discern why certain materials are chosen for specific heat transfer applications. It's essential to note that fin effectiveness and efficiency can greatly vary with the thermal conductivity of the fin material; high conductivity means the fin can spread heat away from the base more easily, leading to potentially improved performance.

Fin Surface Area

The fin surface area is a crucial parameter influencing a fin's heat transfer capabilities. This area is the part of the fin exposed to the ambient environment where the heat dissipates. A larger surface area means more space for heat exchange, which typically leads to higher heat dissipation rates and thus improved effectiveness. In the exercise solution, the fin's surface area directly impacts the actual heat transfer rate, being part of the product with the heat transfer coefficient and the temperature difference.

To enhance a student's comprehension of fin design, the relationship between surface area and heat transfer can be emphasized. Visual aids, such as comparisons between fins with varying surface areas, can be particularly illustrative, demonstrating how surface area differences could result in more or less effective heat dissipation. This understanding is vital for grasping why, in the problem provided, a fin's effectiveness could alter due to changes in its dimensions.

Temperature Gradient

A temperature gradient, symbolized as \(dT/dx\), is the rate of temperature change relative to distance in a given direction; it's a measure of how quickly temperatures change from one point to another. In heat transfer problems, such as the one in our exercise, temperature gradients drive the flow of heat from a hot spot towards cooler regions. The greater the temperature gradient, the more rapid the heat transfer, which is significant for both efficiency and effectiveness calculations.

By understanding the temperature gradient, students can delve into why certain fins might perform better under specific conditions. They would recognize that for heat transfer with perfect conduction, as used in the Step 2 and Step 3 solutions, a high temperature gradient would suggest a high thermal flux, meaning more efficient heat conduction through the fin. In real-world applications, managing the temperature gradient is essential for designing fins that can effectively keep devices at optimal operating temperatures.