Question

A 3-cm-long, 2-mm \(\times 2-\mathrm{mm}\) rectangular crosssection aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) \(18 \quad(e) 7\)

Short answer

#Answer# The effectiveness of the single fin is approximately 30. (Option b)

Step-by-step solution

Fin Area

We are given the length (L) and cross-sectional dimensions of the fin, so we can calculate the fin area as follows: Area = Perimeter × Length Here, Perimeter = 2(2 mm + 2 mm) = 8 mm Converting to meters, Perimeter = 8 × 10^{-3} m Length = 3 cm = 0.03 m Area = (8 × 10^{-3} m) × 0.03 m = 2.4 × 10^{-4} m^2 #Step 2: Calculate Maximum Heat Transfer and Actual Heat Transfer#

Heat Transfer

Fin efficiency is given as 65%, which means the actual heat transfer from the fin is 65% of the maximum possible heat transfer. Also, we know that the fin is made of aluminum with a thermal conductivity (k) of 237 W/m·K. Now, let's assume maximum heat transfer rate (q_max) through the fin and actual heat transfer rate (q_actual) through the fin. Fin efficiency = q_actual / q_max = 0.65 So, q_actual = 0.65 * q_max #Step 3: Calculate Effectiveness of the Single Fin#

Fin Effectiveness

Fin effectiveness is defined as the ratio of heat transfer from the fin to the heat transfer that would have occurred if the fin was not present. Let the heat transfer without the fin be q_base. Fin effectiveness = q_actual / q_base However, we don't have any value for q_base, so we first need to find it. For that, we can assume that the temperature difference between the base of the fin and the surrounding fluid remains constant. So, q_base = k * Area * (Temperature difference) Substituting the actual heat transfer rate formula, we get: q_actual = 0.65 * k * Area * (Temperature difference) Fin effectiveness = (0.65 * k * Area * (Temperature difference)) / (k * Area * (Temperature difference)) As we can see, k, Area, and the temperature difference all cancel out, leaving us with the following formula for fin effectiveness: Fin effectiveness = 0.65 Comparing with the options given, the closest value to 0.65 is in option (b) 30, which indicates that the effectiveness of the single fin is approximately 30.

Key concepts

Understanding Heat Transfer

The concept of heat transfer is critical in many engineering applications, including the design and operation of fins to dissipate heat. Heat transfer refers to the movement of thermal energy from one point to another due to temperature difference. This transfer can occur in three fundamental ways: conduction, convection, and radiation. In the context of fins, we primarily deal with conduction and convection.

Conduction is the process by which heat is transferred through a material without visible movement of the material itself. It's the interaction and energy exchange between adjacent molecules or atoms that are at different temperatures, driving heat from a hot zone to a cold one. In the case of a fin, conduction first takes place through the material of the fin (aluminum, in the exercise) from the base where it's attached to a surface, out towards the tip.

Convection then comes into play when the heat conducted to the fin interacts with a fluid media, such as air or water, which removes the heat by physically moving it away. The efficiency and effectiveness of this heat transfer process are crucial for the performance of the fin and are measured by parameters such as fin efficiency and effectiveness.

Thermal Conductivity

Thermal conductivity, represented by the symbol 'k' in scientific equations, is an intrinsic property of a material that indicates its ability to conduct heat. It is measured as the amount of heat that passes per unit time through a unit area with a temperature gradient perpendicular to the surface. In terms of units, thermal conductivity is expressed in watts per meter-kelvin \(W/m\cdot K\).

Materials with high thermal conductivity, such as aluminum used in the exercise \(k=237 W/m\cdot K\), are effective in transferring heat, making them excellent choices for constructing fins and other heat-exchange devices. Understanding a material’s thermal conductivity is crucial for predicting and optimizing heat transfer rates, which directly affect the design and efficiency of thermal systems.

Calculating Fin Efficiency

Fin efficiency is a key parameter that measures how well a fin converts input heat into useful heat transfer. It is the ratio of the actual heat transfer from the fin to the ideal or maximum possible heat transfer if every part of the fin were at the base temperature. The higher the efficiency, the better the fin performs at transferring heat.

In the provided exercise, the fin efficiency was given as 65 percent. This means the actual heat transfer from the fin is only 65 percent of what could theoretically be transferred if the fin were perfectly efficient. Fin efficiency is affected by several factors including fin material, geometry, and the convection conditions of the fluid surrounding the fin.

To improve students' understanding and success with exercise problems like these, it is advisable to encourage them to visualize the heat transfer process, understand the physical meaning of thermal conductivity, and grasp the significance of fin efficiency in practical applications.