Question

A plane wall surface at \(200^{\circ} \mathrm{C}\) is to be cooled with aluminum pin fins of parabolic profile with blunt tips. Each fin has a length of \(25 \mathrm{~mm}\) and a base diameter of \(4 \mathrm{~mm}\). The fins are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\) and the heat transfer coefficient is \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the thermal conductivity of the fins is \(230 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the heat transfer rate from a single fin and the increase in the rate of heat transfer per \(\mathrm{m}^{2}\) surface area as a result of attaching fins. Assume there are 100 fins per \(\mathrm{m}^{2}\) surface area.

Short answer

The heat transfer rate from a single fin is 2.49 W, and the increase in heat transfer rate per m² surface area due to the addition of fins is 249 W/m².

Step-by-step solution

Calculate fin area and perimeter

First, we need to find the area and perimeter of a fin. For a single fin, we can use the given base diameter (\(D = 4 \mathrm{~mm}\)) and length (\(L = 25 \mathrm{~mm}\)) to calculate the fin area (\(A_\mathrm{f}\)) and perimeter (\(P_\mathrm{f}\)): Fin area: \(A_\mathrm{f} = \pi \times (D/2)^2 = \pi \times (4/2)^2 = 12.57\ \mathrm{mm}^2\) Fin perimeter: \(P_\mathrm{f} = \pi \times D = \pi \times 4 = 12.57\ \mathrm{mm}\)

Calculate fin heat transfer area and fin width

To find the heat transfer area of a single fin (\(A_\mathrm{c}\)), we can use the fin length (\(L\)) and the fin perimeter (\(P_\mathrm{f}\)): Fin heat transfer area: \(A_\mathrm{c} = L \times P_\mathrm{f} = 25 \times 12.57 = 313.92 \mathrm{~mm}^2\) Also, we can determine the fin width (\(B\)) using the fin area (\(A_\mathrm{f}\)) and fin length (\(L\)): Fin width: \(B = A_\mathrm{f} / L = 12.57 / 25 = 0.5028\ \mathrm{mm}\)

Calculate the fin efficiency

We can now use the thermal conductivity of the fins (\(k\)), heat transfer coefficient (\(h\)), fin width (\(B\)), fin length (\(L\)) and fin heat transfer area (\(A_\mathrm{c}\)) to find the fin efficiency (\(\eta_\mathrm{f}\)). Fin efficiency: \(\eta_\mathrm{f} = \dfrac{tanh(mL)}{mL}\), where \(m = \sqrt{2h / (k \cdot B)}\) \(m = \sqrt{2 \times 45 / (230 \times 0.5028 \times 10^{-3})} = 2.189\) \(\eta_\mathrm{f} = \dfrac{tanh(2.189 \times 25 \times 10^{-3})}{2.189 \times 25 \times 10^{-3}} = 0.972\)

Calculate the heat transfer rate from a single fin

Now we can use the fin efficiency (\(\eta_\mathrm{f}\)), heat transfer coefficient (\(h\)), fin heat transfer area (\(A_\mathrm{c}\)), and temperature difference (\(\Delta T\)) to compute the heat transfer rate from one fin (\(q_\mathrm{f}\)): Heat transfer rate: \(q_\mathrm{f} = \eta_\mathrm{f} \times h \times A_\mathrm{c} \times \Delta T\) \(\Delta T = 200 - 25 = 175\ \mathrm{K}\) \(q_\mathrm{f} = 0.972 \times 45 \times 313.92 \times 10^{-6} \times 175 = 2.49\ \mathrm{W}\)

Calculate the increase in the rate of heat transfer per m²

Finally, we can determine the increase in the rate of heat transfer per m² surface area (\(q'_\mathrm{increase}\)) by finding the heat transfer rate of 100 fins per m² (\(q_\mathrm{total}\)) and dividing it by the surface area (1 m²): Increase in the rate of heat transfer: \(q'_\mathrm{increase} = 100 \times q_\mathrm{f}\) \(q'_\mathrm{increase} = 100 \times 2.49 = 249\ \mathrm{W/m}^2\) Thus, the heat transfer rate from a single fin is \(2.49\ \mathrm{W}\), and the increase in the rate of heat transfer per m² surface area due to the addition of fins is \(249\ \mathrm{W/m}^2\).

Key concepts

Thermal Conductivity

Thermal conductivity is a fundamental property of materials that describes how well heat is conducted through them. It's represented by the symbol \( k \) and is measured in watts per meter-kelvin (\( W/m \times K \)). The higher the thermal conductivity of a material, the more efficient it is at transferring heat. For instance, metals typically have high thermal conductivity, which is why they feel cold to the touch; they are rapidly conducting heat away from your warm hand.

In the context of fins, high thermal conductivity is desirable as it means the fin is capable of efficiently transferring heat from the base (where it is attached to the heated surface) to the tip (where the heat is dissipated). The aluminum pin fins described in the exercise have a thermal conductivity of \( 230 W/m \times K \), enabling them to effectively remove heat from the plane wall surface.

Heat Transfer Coefficient

The heat transfer coefficient, symbolized by \( h \), measures how effectively heat is transferred from a solid surface to a fluid or from a fluid to a solid surface. The unit is watts per square meter-kelvin (\( W/m^2 \times K \)). A higher heat transfer coefficient means that heat can be exchanged more quickly between the surface and the fluid.

In our fin scenario, the heat transfer coefficient of \( 45 W/m^2 \times K \) signifies that the surrounding air is somewhat effectively absorbing the heat being dissipated by the fins. This value plays a critical role when calculating the fin efficiency and overall heat transfer rate because it indicates how well the fin surface interacts with the air environment.

Fin Efficiency

Fin efficiency, represented as \( \eta_f \), is a measure of the fin's effectiveness in transferring heat compared to an ideal fin with the same dimensions but with infinite thermal conductivity. It is a ratio of the actual heat transfer rate from the fin to the rate that would occur if the entire fin were at the base temperature.

To calculate fin efficiency, we must consider the shape and size of the fin, the material's thermal conductivity, and the heat transfer coefficient. The equation involves hyperbolic functions, as seen in the exercise. The given efficiency of \( 0.972 \) indicates that our aluminum pin fins are performing at nearly 97.2% of their maximum potential, which is quite effective. If the efficiency were lower, it would suggest that the design or the fin material could possibly be optimized for better performance.

Understanding the correlation between these concepts helps us to evaluate how well the fins will perform in real applications, such as cooling electronic devices or enhancing the heat dissipation from an engine component.