Question

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to be very long, the rate of heat transfer from the fin is (a) \(2.2 \mathrm{~W}\) (b) \(3 \mathrm{~W}\) (c) \(3.7 \mathrm{~W}\) (d) \(4 \mathrm{~W}\) (e) \(4.7 \mathrm{~W}\)

Short answer

Solution: Step 1: Calculate the Cross-sectional Area of the Fin Diameter (D) = 5 mm = 0.005 m $$A = \dfrac{\pi (0.005)^{2}}{4} = 1.963 \times 10^{-5} m^{2}$$ Step 2: Find Fin Efficiency (η) Heat Transfer Coefficient (h) = 100 W/(m²K) Thermal Conductivity (k) = 204 W/(mK) Perimeter (P) = πD = π(0.005) = 0.0157 m $$mL = \sqrt{\dfrac{100 \cdot 0.0157}{204 \cdot 1.963 \times 10^{-5}}} = 12.68$$ Step 3: Calculate Heat Transfer per Unit Length (q'ₒ) Temperature Difference (ΔT) = 80°C - 22°C = 58 K $$q'ₒ = \eta \cdot h \cdot A \cdot \Delta T = 12.68 \cdot 100 \cdot 1.963 \times 10^{-5} \cdot 58 = 144.58 W/m$$ Step 4: Calculate Total Heat Transfer (Q) Length (L) = 300 mm = 0.3 m $$Q = q'ₒ \cdot L = 144.58 \cdot 0.3 = 43.374 W$$ Step 5: Compare with Options The calculated heat transfer (Q) is approximately 43.4 W, which should be compared with the multiple-choice options to find the best match.

Step-by-step solution

Calculate the Cross-sectional Area of the Fin

First, find the cross-sectional area (A) of the fin with its diameter (D): $$A = \dfrac{\pi D^{2}}{4}$$

Find Fin Efficiency (η)

Using extended surfaces approximation, determine the fin efficiency (η) by calculating the fin parameter (mL), where m is the square root of the product of the heat transfer coefficient (h) and the perimeter (P) divided by the product of the thermal conductivity (k) and the cross-sectional area (A): $$mL = \sqrt{\dfrac{hP}{kA}}$$

Calculate Heat Transfer per Unit Length (q'ₒ)

Next, find the heat transfer per unit length (q'ₒ) by multiplying the product of the fin efficiency (η), heat transfer coefficient (h), perimeter (P), and temperature difference (ΔT) by the cross-sectional area (A): $$q'ₒ = \eta \cdot h \cdot A \cdot \Delta T$$

Calculate Total Heat Transfer (Q)

Multiply the heat transfer per unit length (q'ₒ) by the length (L) of the fin to get the total heat transfer (Q): $$Q = q'ₒ \cdot L$$

Compare with Options

Compare the calculated heat transfer (Q) with the multiple-choice options to find the best match.

Key concepts

Fin Efficiency

Imagine you're trying to warm your hands by holding them near a radiator. The radiator's extended surface area, such as fins, does a great job of distributing warmth. That's similar to how a fin on a heat exchanger operates, but in order for us to evaluate how good a fin is at transferring heat, we need to calculate its 'fin efficiency'.

Fin efficiency is defined as the ratio of the actual heat transfer rate from the fin to the heat transfer rate if the entire fin were at the base temperature. The higher the efficiency, the better the fin disperses heat into the environment. To calculate it, we consider factors like thermal conductivity of the fin material, its geometry, and the heat transfer coefficient with the surrounding fluid. The efficiency helps us understand the performance of our fin in real-world conditions, as not all the heat from the base will make it to the tip of the fin.

Thermal Conductivity

Thermal conductivity, denoted by the symbol \( k \), is the measure of a material's ability to conduct heat. Think of it like a straw through which a liquid flows; the wider and smoother the straw, the more easily the liquid moves. Similarly, materials with high thermal conductivity allow heat to pass through them more effortlessly.

In the context of heat transfer from fins, high thermal conductivity means that the material the fin is made from, in our case aluminum, is quite efficient at spreading heat along its length. The value of \( k \) becomes a key part of our calculations involving heat transfer as it directly impacts how much heat the fin can pull away from the base and dissipate to the surroundings.

Heat Transfer Coefficient

Imagine blowing on a spoonful of hot soup. The cool air from your breath helps cool down the soup. In this analogy, we can think of the heat transfer coefficient as a measure of how quickly the soup can be cooled based on the effectiveness of your blowing. In technical terms, the heat transfer coefficient \( h \) gauges how well heat is transferred from the surface to the fluid around it.

It's a critical value when we're calculating heat transfer from fins. This coefficient depends on various factors, including the type of fluid, the flow properties, and the nature of the surface. A higher \( h \) means that the surrounding fluid, like air, is more effective at wicking away heat from the surface. In our exercise, the heat transfer coefficient helps determine how easily the fin dissipates the absorbed heat into the ambient air.

Extended Surfaces

You've seen cooling ribs on engines or even on your computer's CPU. These are examples of extended surfaces, popularly called fins. Their primary job is to maximise the surface area in contact with the air or liquid that's supposed to cool down the device.

These surfaces work on the principle that more surface area means more space for heat to be transferred. It's like spreading out a large sheet to dry faster in the sun, compared to a crumpled one. By extending the surface, we create more opportunities for the heat to escape the fin and move into the surrounding environment. That's why we often use fins in heat exchangers, radiators, and electronic cooling systems to enhance heat transfer efficiency.

Heat Transfer Calculation

Now comes the moment where we roll up our sleeves and dive into the heat transfer calculation. This is where we apply our knowledge of physics and mathematics to quantify the actual rate at which heat is leaving the fin. It involves the use of formulas to consider the thermal conductivity, the heat transfer coefficient, and the geometry of the fin.

By following the series of steps in the solution, we can calculate the cross-sectional area of the fin, find our fin efficiency, and eventually determine the heat transfer per unit length and the total heat transfer. These calculations tell us exactly how efficient our fin is in real-world conditions, not just in theory. It's the crux of the problem that allows us to select the most appropriate heat transfer rate from the options provided.