Hot water is to be cooled as it flows through the tubes exposed to atmospheric air. Fins are to be attached in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why?

Short Answer

Expert verified
Answer: The fins should be attached on the outside of the tubes to increase the surface area exposed to the atmospheric air, enhancing the heat transfer process through convection and radiation and cooling the hot water more effectively.

Step by step solution

01

Understand the heat transfer process

The main objective is to cool the hot water flowing through the tubes by exposing them to atmospheric air. Heat transfer occurs mainly through convection from the hot water to the cooler tubes' surfaces, followed by radiation and convection to the surrounding air.
02

Analyze fins' purpose

Fins are attached to increase the effective surface area exposed to the surrounding air, which enhances the heat transfer process. They provide a larger area for convection and radiation to occur, thus cooling the hot water faster.
03

Compare heat transfer inside vs. outside tubes

When analyzing the heat transfer process, we need to consider the materials involved, their thermal conductivity, the atmosphere (air), and the surfaces through which heat is being transferred. Since the air's thermal conductivity is lower than the tubes' material (usually metals), attaching the fins inside the tubes would not be significantly helpful because the fins would be surrounded by hot water which has higher thermal conductivity than air. This results in less heat dissipation by convection from the inner tube surface.
04

Recommend where to attach fins

Based on the analysis of heat transfer processes and the impact of adding fins to the tubes, it is recommended to attach the fins on the outside of the tubes. This placement will increase the surface area exposed to the atmospheric air, enhancing the heat transfer process through convection and radiation and cooling the hot water more effectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convective Heat Transfer
Imagine a pot of water boiling on a stove. The hot water at the bottom gets less dense and rises, while the cooler water sinks down to replace it, forming a circular pattern known as convection currents. These currents are a prime example of convective heat transfer, a process where heat is carried away by the movement of fluids—such as water or air.

When you're trying to cool hot water in tubes exposed to air, the goal is to maximize this type of heat transfer. The cooler air absorbs the heat from the hot water, creating air currents that naturally carry the heat away. Fins enhance this process by increasing the surface area that comes into contact with the air, thus allowing more heat to be transferred from the water to the air. The main takeaway here is that convective heat transfer is most efficient when there's a large temperature difference between the fluid within the tubes (hot water) and the surrounding fluid (atmospheric air), and when there is plenty of surface area available for heat to be exchanged.
Thermal Conductivity
In everyday terms, thermal conductivity is a measure of how well a material can conduct heat. It's like comparing a metal spoon and a wooden spoon when both are placed in a pot of hot soup. The metal spoon gets hot quickly because metals typically have high thermal conductivity, meaning heat moves through them easily. Wooden spoons, on the other hand, stay cool longer because wood has low thermal conductivity.

In a heat exchange scenario, understanding thermal conductivity is vital when deciding where to place fins. High thermal conductivity materials—like metals used in tube construction—transfer heat efficiently, making them excellent for removing heat from the water inside. By contrast, the environment outside the tubes, which is primarily air, has low thermal conductivity. It doesn't transfer heat away as effectively as metals do. This is why using fins outside the tubes boosts heat dissipation into the air by expanding the contact area, allowing heat to escape more readily into the less conductive air.
Fins in Heat Exchange
Now, let's talk about fins in heat exchange. Picture a motorcycle engine with its ridged cylinders or a computer's CPU with a finned heatsink clamped on top; these fins are not there for aesthetics but for a very functional purpose: to boost heat dissipation. The idea is to take advantage of their increased surface area to spread out heat faster and more efficiently into the surrounding air.

In the context of cooling hot water in tubes, attaching fins outside the tubes is akin to giving the tubes a larger 'wingspan' to release heat into the air. Fins work marvelously in such setups because they create extra surface room for heat to be emitted via radiation and picked up by air currents in convection. By improving the efficiency of the system, you can ensure the hot water loses its heat rapidly and the system maintains an optimal temperature with improved effectiveness.

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Most popular questions from this chapter

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

What is the difference between the fin effectiveness and the fin efficiency?

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

Consider a house whose attic space is ventilated effectively so that the air temperature in the attic is the same as the ambient air temperature at all times. Will the roof still have any effect on heat transfer through the ceiling? Explain.

Ice slurry is being transported in a pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\), and \(L=\) \(5 \mathrm{~m}\) ) with an inner surface temperature of \(0^{\circ} \mathrm{C}\). The ambient condition surrounding the pipe has a temperature of \(20^{\circ} \mathrm{C}\), a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a dew point of \(10^{\circ} \mathrm{C}\). If the outer surface temperature of the pipe drops below the dew point, condensation can occur on the surface. Since this pipe is located in a vicinity of high voltage devices, water droplets from the condensation can cause electrical hazard. To prevent such incident, the pipe surface needs to be insulated. Determine the insulation thickness for the pipe using a material with \(k=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to prevent the outer surface temperature from dropping below the dew point.

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