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I recently had a discussion with a friend of mine regarding the effectiveness of handheld fans. My friend argued that you would always get warmer. I’m sure that this needn’t be the case, as the system (fan with human) is not closed. However, I’m not too sure how effective it really is. I tried to set up some equations comparing cooling vs. heating, but finding the right numbers to fill in seems difficult. Does anyone have an idea on the true effectiveness of handheld fans?

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    $\begingroup$ What equations have you set up? What numbers are you having difficulty with? $\endgroup$ Jul 25 '18 at 11:24
  • $\begingroup$ You might want to see: physics.stackexchange.com/questions/385811/…, since it seems like that's what your friend is thinking of. $\endgroup$
    – Allure
    Jul 26 '18 at 3:44
  • $\begingroup$ @sammygerbil for heating I made an estimate on the amount of air moved by the fan with each fan-swing plus its velocity. Then, by using E=1/2 m v^2, I estimated the energy from each swing. Multiply this by the swinging frequency to get a power. And finally, we need an efficiency to get the amount of power consumed by the human (I have no idea what the efficiency should be). With cooling I got stuck pretty quick, as I could not find good methods for evaporative cooling. Hopefully this clarifies it a bit. $\endgroup$ Jul 26 '18 at 13:57
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In many situations, the fan will help.

If two things are at different temperatures, but in thermal contact with each other, the heat will flow from hot to cold. When the temperature difference isn't too large, an effective way to transfer the heat is by convection. Convection occurs when at least one of the substances in thermal contact is a moving fluid; in the case of a fan blowing on a person, the air is the moving fluid.

If the temperature outside is below the human body temperature; then a fan should absolutely help (assuming the fan motor isn't hot enough to raise the temperature of the air it is blowing above body temperature, which is probably rare).

An even bigger reason to use the fan is to assist with evaporation. People sweat so that the sweat on their skin can evaporate. This evaporation takes thermal energy from you when it occurs. If the air isn't too humid, flowing air across the moist surface will speed up this process. This is why fans can cool you down so much when you're sweating.

So a fan is likely to help, but there are 3 factors to consider when deciding. How hot is it? How humid is it? How hot does the fan get when it's running?

Unless you're in an extremely hot and humid environment, or using really inefficient handheld fans, they are likely to help cool you.

Also, when you say "handheld fans" I'm picturing a handheld electric fan, which is where this heating talk comes in. If you mean a fan that's hand operated by waving it in your face, you might have to do a bit of math on the energy released/used by the movements; but you should still benefit. Your point about open systems has a lot to do with this. The fan will help you evaporate sweat, and as long as less humid air can in to take it's place, you can keep evaporating sweat from your body to cool it down. If the room was closed, you would reach equilibrium with the humidity in the room and be unable to evaporate away the heat effectively.

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  • $\begingroup$ I do indeed mean old traditional fans (so the ones you move by hand). Thanks for the different cooling mechanisms, do you see any reason why the heating should be negligible against the cooling? I understand that these are highly dependent on the situation, but is an order-of-magnitude analysis possible? $\endgroup$ Jul 26 '18 at 13:51
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Movement of air by a fan transports heat at an increased rate by convection and assists in the evaporation of perspiration which transports heat by conduction.

From Wikipedia's - Evaporative Cooler, section: Physical Principles - (Other Sources: #2, #3):

The amount of heat transfer depends on the evaporation rate, however for each kilogram of water vaporized 2,257 kJ of energy (about 890 BTU per pound of pure water, at 95 °F (35 °C)) are transferred. The evaporation rate depends on the temperature and humidity of the air, which is why sweat accumulates more on humid days, as it does not evaporate fast enough.

From Wikipedia's - Wind Chill Factor, section North American and United Kingdom Wind Chill Index Calculations:

Many formulas exist for wind chill because, unlike temperature, wind chill has no universally agreed upon standard definition or measurement. All the formulas attempt to qualitatively predict the effect of wind on the temperature humans perceive.

November 2001, Canada, the United States, and the United Kingdom implemented a new wind chill index developed by scientists and medical experts on the Joint Action Group for Temperature Indices (JAG/TI). It is determined by iterating a model of skin temperature under various wind speeds and temperatures using standard engineering correlations of wind speed and heat transfer rate. Heat transfer was calculated for a bare face in wind, facing the wind, while walking into it at 1.4 metres per second (5.0 km/h; 3.1 mph). The model corrects the officially measured wind speed to the wind speed at face height, assuming the person is in an open field. The results of this model may be approximated, to within one degree, from the following formula:

The standard Wind Chill formula for Environment Canada is:

$${ T_{\mathrm {wc} }=13.12+0.6215T_{\mathrm {a} }-11.37v^{+0.16}+0.3965T_{\mathrm {a} }v^{+0.16}}$$

where $T_{wc}$ is the wind chill index, based on the Celsius temperature scale; $Ta$ is the air temperature in degrees Celsius; and $v$ is the wind speed at 10 m (33 ft) standard anemometer height, in kilometres per hour.

The equivalent formula in US customary units is:

$${T_{\mathrm {wc} }=35.74+0.6215T_{\mathrm {a} }-35.75v^{+0.16}+0.4275T_{\mathrm {a} }v^{+0.16}\,\!}$$

where $T_{wc}$ is the wind chill index, based on the Fahrenheit scale; $T_a$ is the air temperature in degrees Fahrenheit, and $v$ is the wind speed in miles per hour.

Windchill temperature is defined only for temperatures at or below 10 °C (50 °F) and wind speeds above 4.8 kilometres per hour (3.0 mph).

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    $\begingroup$ For the uneducated: this means it's effective? $\endgroup$ Jul 25 '18 at 9:57
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    $\begingroup$ It is commendable that you have provided this information, but your answer would be better if you applied it to the question. It is not obvious that it shows that a hand-held fan will be effective. $\endgroup$ Jul 25 '18 at 11:22
  • $\begingroup$ Where does the first quote come from? Also, this really doesn't seem like a good answer to the question IMO. You've explained why someone might choose to use a fan; but there isn't anything here addressing the real question - the effectiveness of a fan. $\endgroup$
    – JMac
    Jul 25 '18 at 20:12
  • $\begingroup$ @LievenKeersmaekers - I've updated the answer, I hope it is now more detailed. $\endgroup$
    – Rob
    Jul 25 '18 at 20:18
  • $\begingroup$ Where does " The amount of heat transfer depends on the evaporation rate, however for each kilogram of water vaporized 2,257 kJ of energy (about 890 BTU per pound of pure water, at 95 °F (35 °C)) are transferred. The evaporation rate depends on the temperature and humidity of the air, which is why sweat accumulates more on humid days, as it does not evaporate fast enough." come from? $\endgroup$
    – JMac
    Jul 26 '18 at 1:35

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