Question

Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length \(L=25.4 \mathrm{~mm}\), made of copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\), and the convection coefficient is \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by 10 -cm section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

Short answer

Question: Express the temperature function θ(x) along a fin and calculate the temperature at x = L/2. Determine the rate of heat transferred through each fin and the fin effectiveness. Calculate the total rate of heat transfer from a 10 × 10 cm section of the wall with 625 fins. a) The temperature function along a fin is given by: θ(x) = 132 - e^((625/2)⋅x +ln(132)) The temperature at x = L/2 (12.7 mm) is approximately: θ(L/2) ≈ 66.4°C b) The rate of heat transfer through each fin is approximately: q_f ≈ 32.7 W The fin effectiveness is approximately: η_f ≈ 0.327 c) The total rate of heat transfer from the 10 × 10 cm section of the wall with 625 fins is approximately: q_total ≈ 21694.7 W

Step-by-step solution

(a) Expressing the Temperature Function along a Fin

First, let's determine the temperature distribution along the fin (\(x\) direction) using the general heat conduction equation: $$\frac{d}{dx}\left(kA\frac{dT}{dx}\right) = q_{conv}$$ where \(A\) is the fin cross-sectional area, and \(q_{conv}\) is the heat transfer due to convection. In the steady state, the temperature distribution in the fin is linear, and we can write: $$kA\frac{dT}{dx} = hP(T - T_\infty)$$ where \(P\) is the perimeter of the fin. Substituting the given data, we obtain: $$400 * \pi (0.0005^2) * \frac{d\theta}{dx} = 100 * \pi(0.001) * (132 - \theta)$$ Solinding for \(\frac{d\theta}{dx}\): $$\frac{d\theta}{dx} = \frac{-625}{2}(132 - \theta)$$ Integrating to find the temperature function: $$\int_{132}^{\theta} \frac{1}{132 - y} dy = \int_0^x -\frac{625}{2} dx$$ $$-\ln(132 - \theta) = -\frac{625}{2}x + C_1$$ Now, we need to apply the boundary condition: at \(x = L\), \(T = T_{s2} = 0^{\circ} \mathrm{C}\): $$C_1 = \ln(132)$$ So, the temperature function \(\theta(x)\) can be expressed as: $$\ln(132 - \theta) = \frac{625}{2}x + \ln(132)$$ Now, let's find the temperature at \(x = \frac{L}{2} = 12.7 \mathrm{mm}\): $$\theta\left(\frac{L}{2}\right) = 132 - e^{\frac{625}{2} \cdot 0.0127+\ln(132)}$$ Calculating: $$\theta\left(\frac{L}{2}\right) \approx 66.4^{\circ} \mathrm{C}$$

(b) Determining Rate of Heat Transfer and Fin Effectiveness

The rate of heat transfer from the hot surface through each fin, \(q_f\), can be calculated by: $$q_f = kA \frac{dT}{dx} |_{x = 0}$$ $$q_f = 400 * \pi (0.0005^2)(-\frac{625}{2}(132 - \theta(0^{\circ} \mathrm{C})))$$ Calculating: $$q_f \approx 32.7 \mathrm{W}$$ Now, let's determine fin effectiveness. Fin effectiveness, \(\eta_f\), is the ratio of the actual heat transfer rate through the fin to the maximum possible heat transfer rate if the entire fin were at the base temperature: $$\eta_f = \frac{q_f}{hPL(T_{s1} - T_\infty)}$$ $$\eta_f = \frac{32.7}{100*\pi(0.001)*25.4*(132 - 0)}$$ Calculating: $$\eta_f \approx 0.327$$ The use of fins is justified if the fin effectiveness is greater than 0.1. In this case, η_f = 0.327, so the use of fins is indeed justified.

(c) Total Rate of Heat Transfer from the Wall Section

To calculate the total rate of heat transfer from a \(10 \times 10\) cm section of the wall with 625 uniformly distributed fins, we first determine the total heat transfer rate from the fins, and then add it to the heat transfer rate from the unfinned wall surface. Heat transfer rate from the fins: $$q_{f, total} = 625 \times q_f = 625 \times 32.7 = 20437.5 \mathrm{W}$$ Next, we need to calculate the heat transfer rate from the unfinned wall surface. The total area of the fins can be found by multiplying the number of fins by their base surface area: $$A_{fins} = 625 * \pi (0.0005^2) = 0.0491 \mathrm{m^2}$$ So, the unfinned wall surface area is: $$A_{wall} = 0.1^2 - A_{fins} = 0.0100 - 0.0491 = 0.095091 \mathrm{m^2}$$ Now, we can calculate the rate of heat transfer from the unfinned wall surface: $$q_{wall} = hA_{wall}(T_{s1} - T_\infty) = 100 * 0.095091 * (132 - 0) = 1257.2 \mathrm{W}$$ Finally, the total rate of heat transfer from the wall section is: $$q_{total} = q_{f, total} + q_{wall} = 20437.5 + 1257.2 = 21694.7 \mathrm{W}$$

Key concepts

Cooling Fins

Cooling fins are devices designed to increase the surface area of heat dissipation. These extensions are attached to surfaces that need enhanced cooling. Enhancing cooling involves the transfer of thermal energy away from hot surfaces into cooler surroundings. Fins accomplish this primarily by increasing the surface area, which facilitates a more efficient heat exchange.

Consider the following advantages of utilizing cooling fins:

• Greater surface area for heat transfer
• Reduced overheating risk in machinery and electronics
• Enhanced thermal management in compact systems

Fins come in various shapes and sizes, with some being circular, as seen in the original exercise. The fins' effectiveness relies on their material, geometry, and the surrounding conditions. Metal fins made of copper or aluminum are common due to their excellent heat conduction properties. The geometry, such as the diameter and length of circular fins, is crucial in defining the effectiveness in dissipating heat.

Convection Heat Transfer

Convection heat transfer plays a pivotal role in the cooling process when using fins. It occurs when thermal energy is transferred between the fin surface and the surrounding air. As the air flows over the fins, it carries away heat from the surface, thereby cooling it.

The rate of convection is often characterized by the convection heat transfer coefficient, denoted as \(h\), which measures how efficiently heat is carried away by the surrounding fluid. A higher \(h\) value indicates better cooling performance.

• Convection improves with increased air movement.
• Fins enhance the surface area for convection, optimizing heat loss.
• Temperature gradients between the fin surface and the air significantly affect convection.

In the exercise scenario, the convection coefficient was provided as \(100 \, \text{W/m}^2 \cdot \text{K}\), indicating the level of air-flow effectiveness in removing heat from the copper cooling fins.

Thermal Conductivity

Thermal conductivity is a material property that describes how well a substance can conduct heat. It is denoted by \(k\), where a higher value signifies better thermal conduction. In the exercise given, the fins are composed of copper, which has a thermal conductivity \(k = 400 \, \text{W/m} \cdot \text{K}\).

Key aspects of thermal conductivity:

• Materials with high thermal conductivity efficiently transfer heat.
• Copper is favored for fins due to its high thermal conductivity.
• Thermal conductivity directly impacts the temperature distribution of the fins.

With cooling fins, understanding the material's thermal conductivity helps in predicting how rapidly and uniformly the heat can be dissipated from the base to the tip of the fin. High thermal conductivity allows the material to quickly absorb and transport heat away from the hot surface, which is essential for the fin's overall cooling effectiveness.

Fin Effectiveness

Fin effectiveness is a measure used to evaluate how well the fins enhance heat transfer as compared to a baseline case without fins. It is a crucial metric used to justify the addition of fins to a system. The fin effectiveness, denoted \(\eta_f\), is defined as the ratio of the actual heat transfer rate of the fin to the maximum possible heat transfer rate if the fin were at the base temperature throughout.

The exercise showed a fin effectiveness value of 0.327, signaling that the fins have a substantial impact on the heat dissipation process, even though this value might seem less than 1.

Key points about fin effectiveness:

• An effectiveness greater than 0.1 often makes fins worthwhile.
• High-performance fins yield \(\eta_f\) values closer to 1.
• It assists in assessing the cost-benefit of implementing fins in a cooling system.

This calculation helps engineers and designers determine if the added complexity and material cost of including fins translate to sufficient improvements in thermal management, justifying their use in the system.