Question

A total of 10 rectangular aluminum fins $(k=$ $203\mathrm{W}/\mathrm{m}\cdot \mathrm{K})$ are placed on the outside flat surface of an electronic device. Each fin is $100\mathrm{mm}$ wide, $20\mathrm{mm}$ high and $4\mathrm{mm}$ thick. The fins are located parallel to each other at a center- tocenter distance of $8\mathrm{mm}$. The temperature at the outside surface of the electronic device is ${60}^{\circ }\mathrm{C}$. The air is at ${20}^{\circ }\mathrm{C}$, and the heat transfer coefficient is $100\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$. Determine $(a)$ the rate of heat loss from the electronic device to the surrounding air and $(b)$ the fin effectiveness.

Short answer

Question: Determine (a) the rate of heat loss from the device to the surrounding air, and (b) the fin effectiveness. a) The rate of heat loss from the device to the surrounding air (total heat transfer rate) can be found with the calculated values: $Q=(100(100\times {10}^{-3})(20\times {10}^{-3})(60-20))(10\times 8\times {10}^{-3})W$ b) The fin effectiveness can be found using the formula: $\epsilon =\frac{Q}{q_u^{″}{L}_{fins}}$ Insert the calculated values for $Q$ and $q_u^{″}{L}_{fins}$ to find the fin effectiveness.

Step-by-step solution

1. Calculate fin thickness and area

First, we need to calculate the fin thickness ($t_f$) and fin area ($A_f$) using the given values of fin width (100 mm), height (20 mm), and thickness (4 mm). Fin thickness, $t_f=4\times {10}^{-3}m$ (converting from mm to m) Fin area, $A_f=(100\times {10}^{-3})(20\times {10}^{-3})m^2$ (converting from mm to m)

2. Calculate heat transfer rate per unit length

Now we will calculate the heat transfer rate per unit length ($q^{″}$) using the heat transfer coefficient ($h$) and the temperature difference between the electronic device ($T_s$) and air ($T_{\mathrm{\infty }}$). Equation for heat transfer rate per unit length is: $q^{″}=h{A}_f(T_s-T_{\mathrm{\infty }})$. Plug in the known values: $h=100W/m^{2}K$ $A_f=(100\times {10}^{-3})(20\times {10}^{-3})m^2$ $T_s={60}^{\circ }C$ $T_{\mathrm{\infty }}={20}^{\circ }C$ $q^{″}=100(100\times {10}^{-3})(20\times {10}^{-3})(60-20)W/m$

3. Calculate the total heat transfer rate

To find the total heat transfer rate ($Q$) for all the fins, we have to multiply the heat transfer rate per unit length ($q^{″}$) by the total length of the fins ($L_{fins}$). Since there are 10 fins and each has a center-to-center distance of 8 mm, the total length can be calculated as follows: $L_{fins}=10\times 8\times {10}^{-3}m$ Now, we can calculate the total heat transfer rate: $Q=q^{″}{L}_{fins}=(100(100\times {10}^{-3})(20\times {10}^{-3})(60-20))(10\times 8\times {10}^{-3})W$

4. Calculate the fin effectiveness

Finally, we will calculate the fin effectiveness ($\epsilon$), which is defined as the ratio of the actual heat transfer rate of the finned surface to the heat transfer rate of an equivalent unfinned surface. First, we'll find the heat transfer rate for an unfinned surface. The unfinned surface area ($A_u$) is the same as a single fin area as per the given width and height. $A_u=(100\times {10}^{-3})(20\times {10}^{-3})m^2$ Now, calculate the heat transfer rate for the unfinned surface ($q_u^{″}$) with the same heat transfer coefficient and temperature difference: $q_u^{″}=h{A}_u(T_s-T_{\mathrm{\infty }})=100(100\times {10}^{-3})(20\times {10}^{-3})(60-20)W/m$ Finally, calculate the fin effectiveness: $\epsilon =\frac{Q}{q_u^{″}{L}_{fins}}$ Plug in the calculated values for $Q$ and $q_u^{″}{L}_{fins}$.

Key concepts

Fins

Fins are extended surfaces designed to increase the heat transfer rate by increasing the surface area available for heat exchange. They are commonly used in various applications where efficient heating or cooling is required. Fins are made from materials with high thermal conductivity, like aluminum, to effectively conduct heat from the base to the tip. In the exercise we are considering, fins are attached to an electronic device to dissipate heat generated during operation. The dimensions of the fins and their arrangement play a crucial role in determining the amount of heat they can remove. By converting the measurements into meters, as done in the original solution, calculations are easier and consistent with standard units used in the equations. This allows calculating the precise area of the fins and consequently the heat they can help dissipate.

Electronic Cooling

Electronic cooling is a crucial aspect of maintaining the performance and longevity of electronic devices. As devices perform their functions, they generate heat. If this heat is not removed efficiently, the devices can overheat, leading to potential damage or failure. In this regard, fins are an effective solution because they increase the surface area for heat dissipation. By conducting heat away from the device to the surrounding air, fins help maintain an optimal operating temperature. This exercise specifically models the use of aluminum fins due to their high thermal conductivity, which ensures efficient heat transfer. Electronic cooling needs to be calculated carefully to keep devices running smoothly, making it essential to consider various factors like fins' material, placement, and environmental conditions. The problem also provides insight into practical thermal management by illustrating how a device loses heat to its surroundings through air convection.

Air Convection

Air convection is the process through which heat is transferred from the surface of the fins to the surrounding air. This mechanism is crucial in cooling systems as it helps remove heat from the fins and, consequently, from the electronic device itself. Convection depends on the temperature difference between the fin and the air, as well as the heat transfer coefficient, which quantifies how effectively heat is being transferred. In the provided exercise, the heat transfer coefficient is given as 100 W/m²K, indicating a fairly efficient transfer of heat from the fins to the air. Calculating the heat transfer rate per unit length helps in understanding the total capacity of the system to cool the device. When air flows over the fins, it carries away heat, thus reducing the fin temperature and preventing the electronic device from overheating. Efficient air convection ensures that a stable temperature is maintained during the operation of the device.

Fin Effectiveness

Fin effectiveness is a measure of how well a fin performs in enhancing the heat transfer compared to a similar surface without fins. It is calculated as the ratio of heat transfer by the fins to the heat transfer from an equivalent area of unfinned surface. In essence, it indicates the multiplier effect of adding fins to the system. The higher the fin effectiveness, the more efficient the fins are at conducting heat away from the device. The exercise uses the heat transfer rate of an unfinned surface as a benchmark to determine the ratio. By comparing the actual heat transfer of the finned surface to this benchmark, we can assess how beneficial the fins are in cooling the device. A fin effectiveness greater than one indicates that fins provide significant benefits in managing thermal performance, confirming their importance in applications like electronic cooling where compact and efficient thermal solutions are needed.