Question

Steam in a heating system flows through tubes whose outer diameter is \(5 \mathrm{~cm}\) and whose walls are maintained at a temperature of \(180^{\circ} \mathrm{C}\). Circular aluminum alloy 2024-T6 fins \((k=186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of outer diameter \(6 \mathrm{~cm}\) and constant thickness \(1 \mathrm{~mm}\) are attached to the tube. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.

Short answer

Answer: The increase in heat transfer from the tube per meter of its length as a result of adding fins is 4145.298 W.

Step-by-step solution

Calculate the fin's thermal resistance

First, we need to calculate the thermal resistance of the fins. Fins increase the heat transfer rate by increasing the exposed surface area, but they also introduce a resistance to heat transfer. We can calculate the fin's thermal resistance using: \(R_{fin} = \frac{t}{k A_{fin}}\) where \(t = 1 \mathrm{~mm}\) (the fin thickness) \(k = 186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (the thermal conductivity of the aluminum alloy 2024-T6) \(A_{fin} = \pi D_{fin} L_{fin}\) (the fin's surface area) \(D_{fin} = 6 \mathrm{~cm}\) (the fin's outer diameter) \(L_{fin} = 3 \mathrm{~mm}\) (the space between fins) Let's calculate \(A_{fin}\): \(A_{fin} = \pi (0.06 \mathrm{~m})(0.003 \mathrm{~m}) = 0.0005655 \mathrm{~m}^2\) Now we can calculate \(R_{fin}\): \(R_{fin} = \frac{0.001 \mathrm{~m}}{186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 0.0005655 \mathrm{~m}^2} = 9.3451 \times 10^{-3} \mathrm{~K} / \mathrm{W}\)

Calculate the heat transfer rate per meter of tube without fins

Now, we will calculate the heat transfer rate per meter of the tube without fins using: \(q_{nf} = h A_{nf} (T_{walls} - T_{\infty})\) where \(h = 40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (the heat transfer coefficient) \(A_{nf} = \pi D_{tube} L_{tube}\) (the exposed surface area of the tube without fins) \(D_{tube} = 5 \mathrm{~cm}\) (the tube's outer diameter) \(L_{tube} = 1 \mathrm{~m}\) (length of the tube) \(T_{walls} = 180^{\circ} \mathrm{C}\) (the tube wall temperature) \(T_{\infty} = 25^{\circ} \mathrm{C}\) (the surrounding air temperature) Let's calculate \(A_{nf}\): \(A_{nf} = \pi (0.05 \mathrm{~m})(1 \mathrm{~m}) = 0.1571 \mathrm{~m}^2\) Now we can calculate \(q_{nf}\): \(q_{nf} = 40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 0.1571 \mathrm{~m}^2 \cdot (180 - 25) \mathrm{~K} = 988.231 \mathrm{~W}\)

Calculate the heat transfer rate per meter of tube with fins

Next, we calculate the heat transfer rate per meter of tube with fins using: \(q_{f} = q_{nf} + \frac{N_f}{R_{fin}L_{tube}}(T_{walls}-T_{\infty})\) where \(N_f = 250\) (number of fins per meter length of the tube) Let's calculate the increase in heat transfer rate due to the fins: \(q_{increase} = \frac{250}{9.3451 \times 10^{-3} \mathrm{~K} / \mathrm{W} \cdot 1 \mathrm{~m}}(180-25) \mathrm{~K} = 4145.298~\mathrm{W}\) Now we can calculate \(q_{f}\): \(q_{f} = 988.231 \mathrm{~W} + 4145.298 \mathrm{~W} = 5133.529 \mathrm{~W}\)

Determine the increase in heat transfer as a result of adding fins

Finally, we can determine the increase in heat transfer from the tube per meter of its length as a result of adding fins by subtracting the heat transfer rate without fins from the heat transfer rate with fins: \(Increase = q_{f} - q_{nf} = 5133.529 \mathrm{~W} - 988.231 \mathrm{~W} = 4145.298\mathrm{~W}\) Hence, the increase in heat transfer from the tube per meter of its length as a result of adding fins is \(4145.298\mathrm{~W}\).

Key concepts

Fin Thermal Resistance Calculation

When designing heat exchangers or enhancing heat transfer, fins play an integral role by increasing the surface area available for heat exchange. However, fins are not perfect conductors of heat; they have an inherent thermal resistance. The calculation of this resistance is a crucial step in determining the overall efficacy of the fin.

Thermal resistance of a fin is calculated using the formula:
\(R_{fin} = \frac{t}{k A_{fin}}\)
where \(t\) represents the thickness of the fin, \(k\) is the thermal conductivity of the fin material, and \(A_{fin}\) is the surface area of the fin. To grasp this concept fully, consider a thin rectangular fin with a constant cross-sectional area along its length. The temperature gradient along the fin causes heat to flow from the base (where it is hotter) to the tip (where it is cooler). The thermal resistance is key to determining how effectively the fin transfers heat from the base to the surrounding environment.

In the exercise, the fin material is aluminum alloy 2024-T6, known for its relatively high thermal conductivity. This is important because the efficiency of the fin in dissipating heat is directly related to the material's ability to conduct heat.

Heat Transfer Rate

Within the realm of thermal systems, understanding the rate at which heat is transferred is fundamental. The heat transfer rate, denoted \(q\), quantifies the amount of heat being transferred per unit time.

In situations without fins, the heat transfer rate is direct and can be found using the equation:
\(q_{nf} = h A_{nf} (T_{walls} - T_{\text{\html{\infty}}})\)
This expression relies on the heat transfer coefficient \(h\), the surface area available for heat transfer \(A_{nf}\), and the temperature difference between the wall and the ambient air \((T_{walls} - T_{\text{\html{\infty}}})\).

In the provided exercise, fins are introduced to enhance heat transfer. They work by increasing the effective surface area and thus the overall heat transfer rate. However, while fins increase the surface area, they also come with their own thermal resistance, so the new heat transfer rate with fins has to account for this additional resistance. The effective heat transfer rate with fins \(q_{f}\) is, therefore, a combination of the natural rate along with the contribution from the fins, factoring in their respective thermal resistance.

Fin Efficiency

Fin efficiency is a measure of the actual heat transfer rate of a fin compared to the heat transfer rate if the entire fin were at the base temperature. It is a dimensionless number that gives an indication of how well a fin performs its function of transferring heat from the base to the surrounding environment.

The efficiency of a fin is impacted by factors such as material thermal conductivity, the surroundings into which the fin is dissipating heat, the geometry of the fin, and the temperature gradient along the fin. In practice, not all parts of the fin will be at the base temperature; the tip of the fin is usually far cooler. This temperature variation means that not all the fin's surface area is equally effective in transferring heat.
Fin efficiency can be improved by using materials with higher thermal conductivity, increasing the fin size to dissipate heat over a larger area, or optimizing fin spacing and thickness to balance material usage and thermal resistance.

Material Thermal Conductivity

The materials used in heat transfer applications are chosen based on their thermal conductivity, \(k\), which is a measure of a material's ability to conduct heat. High thermal conductivity is desirable for heat exchanger components because it enables faster heat dissipation from hot to cold regions.

In the context of the exercise, aluminum alloy 2024-T6 is used for the fins. This material is known for its good balance of strength and high thermal conductivity \(186 \mathrm{~W} / \mathrm{m} \bullet \mathrm{K}\), which makes it an excellent choice for fin applications. When selecting a material for heat transfer, it is essential to not only consider the maximum temperature it can withstand but also how quickly and efficiently it can transfer heat. The higher the thermal conductivity, the more effective the material will be in channeling heat away from the source.