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We often turn the fan on in summer to help us cool. Explain how a fan makes us feel cooler in the summer. Also explain why some people use ceiling fans also in winter.

Short Answer

Expert verified
Answer: A fan makes us feel cooler in the summer by increasing the airflow around our body, which enhances the convection process, and speeding up the evaporation of sweat from our skin. Some people use ceiling fans in the winter to redistribute warm air that has risen to the ceiling, maintaining a more comfortable room temperature and potentially leading to energy savings.

Step by step solution

01

Understanding heat transfer

In order to explain how a fan makes us feel cooler in the summer, we first need to understand the principles of heat transfer. Our body loses heat through convection, conduction, radiation, and evaporation. The heat transfer from our body to the surrounding air mostly happens through convection. When the surrounding air is in contact with our body, it becomes warmer as it absorbs our body heat. Then, this warm air rises and is replaced by cooler air from the surroundings.
02

Evaporation of sweat

Another important aspect of the cooling process is the evaporation of sweat. When we sweat, the liquid on our skin evaporates, turning into vapor. This phase change from liquid to vapor requires energy, which is taken from our skin in the form of heat. As a result, our skin loses heat and helps us feel cooler.
03

The role of the fan in summer

During hot summer days, a fan helps to increase the airflow around our body. As the fan blows air towards us, it increases the rate at which the warm air surrounding our body is replaced with cooler air. This enhances the convection process and helps our body lose heat more effectively. Additionally, the increased airflow also speeds up the evaporation of sweat from our skin, leading to more effective cooling.
04

Using ceiling fans in winter

Some people use ceiling fans during winter as well to help with heating. The reason behind this is that warm air rises to the ceiling, creating a temperature difference between the upper and lower parts of the room. Running the ceiling fan in reverse (clockwise direction) at a low speed will circulate this warm air back down into the room, redistributing the heat more evenly. This helps to maintain a comfortable temperature in the room and may even lead to energy savings, as the heating system does not need to work as hard to compensate for the temperature difference.

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Most popular questions from this chapter

A soldering iron has a cylindrical tip of \(2.5 \mathrm{~mm}\) in diameter and \(20 \mathrm{~mm}\) in length. With age and usage, the tip has oxidized and has an emissivity of \(0.80\). Assuming that the average convection heat transfer coefficient over the soldering iron tip is \(25 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) and the surrounding air temperature is \(20^{\circ} \mathrm{C}\), determine the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

A series of ASME SA-193 carbon steel bolts are bolted to the upper surface of a metal plate. The bottom surface of the plate is subjected to a uniform heat flux of \(5 \mathrm{~kW} / \mathrm{m}^{2}\). The upper surface of the plate is exposed to ambient air with a temperature of \(30^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2}\). K. The ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HF-300) limits the maximum allowable use temperature to \(260^{\circ} \mathrm{C}\) for the SA-193 bolts. Determine whether the use of these SA-193 bolts complies with the ASME code under these conditions. If the temperature of the bolts exceeds the maximum allowable use temperature of the ASME code, discuss a possible solution to lower the temperature of the bolts.

An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is $14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between \(\mathrm{W}\) and \(\mathrm{Btu} / \mathrm{h}, \mathrm{m}\) and \(\mathrm{ft}\), and \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\), express the given convection heat transfer coefficient in Btu/h \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). Answer: \(2.47\) Btu/h $\cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$

A hollow spherical iron container with outer diameter \(20 \mathrm{~cm}\) and thickness \(0.2 \mathrm{~cm}\) is filled with iced water at $0^{\circ} \mathrm{C}\(. If the outer surface temperature is \)5^{\circ} \mathrm{C}$, determine the approximate rate of heat gain by the iced water in \(\mathrm{kW}\) and the rate at which ice melts in the container. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Treat the spherical shell as a plain wall, and use the outer area.

Using the conversion factors between W and Btu/h, \(\mathrm{m}\) and \(\mathrm{ft}\), and \(\mathrm{K}\) and \(\mathrm{R}\), express the Stefan-Boltzmann constant $\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\( in the English unit \)\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot \mathrm{R}^{4}$.

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