Question

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

Short answer

The percent error in the rate of heat transfer from the aluminum fin when using the infinitely long fin assumption instead of the adiabatic fin tip assumption is approximately 18.75%.

Step-by-step solution

Calculate the heat transfer rate for the adiabatic fin tip assumption

The formula for calculating the heat transfer rate with the adiabatic fin tip assumption is: $$q = h * A_f * (T_b - T_\infty)$$ Where: - \(h\) is the heat transfer coefficient (12 W/m²·K) - \(A_f\) is the surface area of the fin. We can calculate it using the formula \(A_f = 2 * \pi * r * L\), where \(r\) is the radius (2 mm or 0.002 m) and \(L\) is the length (10 cm or 0.1 m) - \(T_b\) is the base temperature - \(T_\infty\) is the ambient temperature Notice that we don't know the values of \(T_b\) and \(T_\infty\), so we cannot determine the heat transfer rate directly. However, we can still solve the problem as we can compare the ratios of heat transfer rates obtained with different assumptions.

Calculate the heat transfer rate for the infinitely long fin assumption

Using the concept of thermal resistance, we can write the heat transfer rate for the infinitely long fin as: $$q = \frac{(T_b - T_\infty)}{R_{total}} $$ Where: - \(R_{total} = R_{cond} + R_{conv}\), where R_cond is the conduction resistance and R_conv is the convection resistance We can write the respective resistances as: $$R_{cond} = \cfrac{t_cilinder}{k * A_cilinder}$$ And $$R_{conv} = \cfrac{1}{h * A_f}$$ where: - \(t_cilinder\) is the thickness of the aluminum fin (we assume it to be negligible) - \(A_cilinder\) is the cross-sectional area of the aluminum fin (\(\pi r^2\)) We can find the ratio of heat transfer rates as: $$\cfrac{q}{(T_b - T_\infty)} = \cfrac{1}{R_{total}} $$

Calculate the percent error in the rate of heat transfer

Now we have expressions for ratios of heat transfer rates for both adiabatic fin tip assumption (\(\cfrac{q_{ad}}{(T_b - T_\infty)} = h * A_f\)) and infinitely long fin assumption (\(\cfrac{q_{inf}}{(T_b - T_\infty)} = \cfrac{1}{R_{total}}\)). Therefore, we can find the percent error as: $$\text{Percent Error} = \cfrac{\left|q_{ad} - q_{inf}\right|}{q_{ad}} \times 100$$ Since we don't know the values of \(T_b\) and \(T_\infty\), we will express the percent error in terms of the ratio of heat transfer rates: $$\text{Percent Error} = \cfrac{\left|\cfrac{q_{ad}}{(T_b - T_\infty)} - \cfrac{q_{inf}}{(T_b - T_\infty)}\right|}{\cfrac{q_{ad}}{(T_b - T_\infty)}} \times 100$$ Plugging in the expressions for \(q_{ad}\) and \(q_{inf}\), and simplifying: $$\text{Percent Error} = \cfrac{\left|h * A_f - \cfrac{1}{R_{total}}\right|}{h * A_f} \times 100$$

Evaluate the percent error

Now we have all the information needed to compute the percent error. We can calculate \(A_f\), \(A_cilinder\), \(R_{cond}\), \(R_{conv}\), and \(R_{total}\) using the given data, and finally substitute into the percent error formula to obtain the result: - \(A_f = 2 * \pi * 0.002 * 0.1 \approx 0.004\,\text{m}^2\) - \(A_cilinder = \pi * 0.002^2 \approx 1.257 \times 10^{-5}\,\text{m}^2 \) - \(R_{cond} \approx 0\) (since we assume \(t_cilinder\) to be negligible) - \(R_{conv} = \cfrac{1}{12 * 0.004} = 20.833\,\text{K/W}\) - \(R_{total} = R_{cond} + R_{conv} \approx 20.833\,\text{K/W}\) Now, we can substitute these values into the percent error formula: $$\text{Percent Error} = \cfrac{\left|12 * 0.004 - \cfrac{1}{20.833}\right|}{12 * 0.004} \times 100 \approx 18.75\%$$ Therefore, the percent error in the rate of heat transfer from the fin when using the infinitely long fin assumption instead of the adiabatic fin tip assumption is approximately 18.75%.

Key concepts

Fin Efficiency

Fin efficiency is a crucial concept in heat transfer, especially concerning fins, which are extended surfaces designed to increase the heat transfer rate from a body. The efficiency of a fin is defined as the ratio of the actual heat transfer from the fin to the maximum possible heat transfer if the entire fin were at the base temperature. This efficiency measures how effective the fin is at its job.
Despite a fin's ability to enhance heat transfer due to increased surface area, the effectiveness can be hindered by various factors including material thermal conductivity and the geometry of the fin.
A higher fin efficiency means a higher percentage of potential heat is being transferred by the fin. This is particularly important when designing cooling systems to ensure they operate effectively and energy-efficiently.

Thermal Resistance

Thermal resistance is an analogy to electrical resistance, but in the domain of heat flow. It represents the opposition to heat flow through a material, much like electrical resistance opposes current flow. Understanding this concept is essential for analyzing and designing systems with heat transfer processes.
In the practice of calculating heat transfer through fins, thermal resistance helps us categorize and compute the impact of conduction and convection. In the given exercise, the total thermal resistance combines both conduction resistance, which depends on the material's thermal conductivity, and convection resistance, which depends on the heat transfer coefficient and fin surface area.
Minimizing thermal resistance leads to more efficient heat transfer and is a key objective in the design of heat exchangers and related systems. Remember that lower resistance means better heat flow, which is what we aim for in thermal management applications.

Convection and Conduction

Convection and conduction are two primary mechanisms of heat transfer that work together in the heat transfer process. Conduction involves the transfer of heat through a material due to temperature gradients, where heat moves from high to low temperature areas. This transfer is determined by the thermal conductivity of the material.
On the other hand, convection refers to the heat transfer at the surface of a material as it interacts with a moving fluid (like air or water outside a fin). The heat transfer coefficient, encountered in the problem, quantifies how effective the convection is.
In the exercise example, understanding the balance and interaction of these two processes through the fin ensures a more accurate calculation of heat transfer rates and the underlying assumptions' impacts. Each mechanism has unique control over the overall rate of heat transfer, influencing designs of industrial cooling systems and domestic appliances alike.