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I was just trying to lift a toy motor off the ground so mounted a toy propeller and tested it, i think it would have lifted(without battery weight) but i didn't bothered balancing it.

Final result was that it became a very nice little fan.

Coming to the point, winter is coming so i decided to make a heater out of it. i got 8x 2500mAh 3.7v lithium ion batteries (similar to those found in laptops). i use them as a powerbank. that makes about 20Amp. As a heat sources, i decided to use nichrome wire out of soldering iron(rated 25W) and tested it using about 8" piece of it. It worked fine. This is not my field so please bear with my calculations.

  • Power source : 3.7V lithium ion batteries 8x2500=20000mAh(20Amp), 74 Watt

  • Motor consumption : 11 Watt (1.3 Ohm)

  • 4 pieces of nichrome wire each with 3.5 Ohm resistence = 4 Watt

My questions are

  • can i use nichrome wire for this purpose?

  • should i use resistor to control heat?

  • can i produce enough heat using nichrome wire at around 16 Watt (4x 4 Watt pieces) for this purpose?

  • will my batteries last 2 hours of time?

And finally, is it a good idea?

Edit: i am not trying to raise room temperature, just trying to make it portable to make my hands or mostly feet warm using hot air.

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  • \$\begingroup\$ Please properly capitalize your units. It's (capital) "V" for Volts, (capital) "A" for Ampere (and not "amp"), (capital) "W" for Watt, and the unit milliamperehours is thus spelt exactly "mAh", not "mah". \$\endgroup\$ – Marcus Müller Nov 4 '16 at 20:32
  • \$\begingroup\$ Would you consider, say, a 25 W incandescent light bulb to be a good room heater? \$\endgroup\$ – Andrew Morton Nov 4 '16 at 20:33
  • \$\begingroup\$ do u mean i should use one to produce heat? \$\endgroup\$ – user6657161 Nov 4 '16 at 20:37
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    \$\begingroup\$ @user6657161 You might find that the cooling effect of the moving air over your skin (due to evaporation of natural skin moisture) is greater than the heat supplied. Perhaps you could use heating tape inside gloves or socks to more useful effect. \$\endgroup\$ – Andrew Morton Nov 4 '16 at 20:47
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    \$\begingroup\$ A typical hair dryer on high is 1500W. Your solution (16W) produces about 1% of the heat of the hair dryer. The continuous heat output of your construction is approximately the same as running the hair dryer for about half a second once every minute. Will that warm you? \$\endgroup\$ – marcelm Nov 4 '16 at 20:57
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74 W is just maybe enough to heat your hands and your feet, with forced air. but would only last <1hr

. a 100W bulb is even better. I would suggest a 750W interior winter car heater with a thermostat built in. quiet. or a ceramic heater....

Although you could make NiChrome wound socks with 2 Watts each;)

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Let's roll this up from the bottom:

And finally, is it a good idea?

Firstly: at "around 16 W", your heating produces as much energy as let's say half a tealight. In other words, as long as the room you're heating isn't very small, you probably won't easily notice you're heating it.

So, and this is intuitively not very surprising, with a bit of wire fed from a 20 Ah battery running at 3.7 V, you don't get your room really warm.

Let math persuade you:

Air has a specific heat capacity of pretty much \$1 \frac{\text{J}}{\text{gK}}\$, i.e. it takes one Joule (\$=1\,\text{Ws}\$) to heat up a gram of air by one degree Kelvin/Celsius. Now, together with the density of air at sea level at roughly room temperature, that means that you get a volumetric heat capacity of air of roughly \$1.3\cdot10^{3}\frac{\text{J}}{\text{m}^3\text{K}}\$, or: to heat up a cubic meter of air by one Kelvin, you need \$1.3\,\text{kJ}=1.3\,\text{kWs}\$.

Let's say your fan gives you (and this is a wild guess based on a 11W fan I found online) \$80\frac{\text{m}^3}{\text{h}} =\frac{80}{3600}\frac{\text{m}^3}{\text{s}} =0.022\frac{\text m^3}{\text s}\$.

You have 16 Ws = 16 J that you put into this per second. So the air coming out of your heater is warmer by

$$\frac{16\,\text{Ws}}{2.2\cdot 10^{-2}\text m^3 \,\cdot\,1.3\cdot10^{3}\frac{\text{J}}{\text{m}^3\text{K}}} = \frac{16}{22\cdot 1.3}\text{K}\approx 0.6\,\text{K} $$

i.e. barely, if at all, noticable.

how much power is good enough for that purpose so batteries can last for some time and dont blast on my face?

Seriously, there's no magic. Either you drain your batteries fast, or you won't get warm air. You can't have both – conservation of Energy is among the things that this universe (magroscopically) depends on.

@marcelm made an excellent comment:

Your hair dryer has 1500 W. Your heating has 16 W. That is but 1% of that. How warm do you think 1% of the heat of a hair dryer is? Not that warm. The wind will probably make your face feel colder, not warmer.

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  • \$\begingroup\$ its not the room m trying to heat, i wanna use small dc motor to get some hot air. \$\endgroup\$ – user6657161 Nov 4 '16 at 20:40
  • \$\begingroup\$ yes, but you're only using 16 W to heat air. That air is not going to be very hot. And if you use a fan to use the same power to heat more air, then that air would get even less hot. \$\endgroup\$ – Marcus Müller Nov 4 '16 at 20:42
  • \$\begingroup\$ thats why i am here. how much power is good enough for that purpose so batteries can last for some time and dont blast on my face? \$\endgroup\$ – user6657161 Nov 4 '16 at 20:47
  • \$\begingroup\$ see my updated answer. TLDR: no magic. Nothing to keep your face warm. \$\endgroup\$ – Marcus Müller Nov 4 '16 at 21:01

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