# Heating element + DC fan

Hi, I need to blow hot air to different liquids and measure the time needed to cool down each one of them. Currently, I’m using DS18B20 sensors to measure the temperature of the air and the liquids, but the hot air is being produced by a hair dryer (60°C), so I can’t set a different temperature. I thought I could replace the hair dryer with some power resistors and a DC fan like this one:
www.amazon.com/GDSTIME-80mm-15mm-Brushless-Cooling/dp/B00N1Y4U8U/ref=sr_1_4?keywords=80x80+fan&qid=1570564829&sr=8-4

I would like to heat the air from 20°C to 30, 40 and 60°C. For example, to heat the air from 20 to 60°C I would need:

Q[W] = mairCpair (T2 - T1)

I get the air flow from the fan specs. 33 CFM = 0.016 m3/s = 0.018 kg/s @40°C
Cp(@40°C) = 1007 [J/kg/K]
T2-T1 = 40 K

Q= 0.018100740 = 725 W. I understand that this is the heat that the power resistors would dissipate but I don’t know how to supply that power (at 24V It requires 30A ). Any ideas on how to do it or am I doing the math wrong?. Thanks.

Why not continue using the hair drier. The amount of heat input to the liquid will be determined by how close (or far away) the heater is from the liquid. Sometimes we tend to over-engineer solutions to problems

I think if I change the distance, the air velocity changes too. The idea is to test changes in cold liquids (8°C) with a set of air temperatures and velocities. For example for a liquid A:
test 1 = Heating liquid A with no forced convection. (fan off, room temperature = 20°C)
test 2 = Heating liquid A with 10 CFM of air at 20°C
test 3 = Heating liquid A with 10 CFM of air at 60°C
test 4 = Heating liquid A with 20 CFM of air at 60°C

It will indeed but you omitted to specify the fixed flow requirement in your original enquiry.
Have you considered the problems of creating laminar air flow with resistors tending to create turbulence

jackrae:
It will indeed but you omitted to specify the fixed flow requirement in your original enquiry.
Have you considered the problems of creating laminar air flow with resistors tending to create turbulence

Yes sorry I forgot. If I enclose the resistors and pass air through them, turbulence will increase the heat transfer so I should test the location of the resistors first. Do I need to use the socket to power the resistors?

jackrae:
Why not continue using the hair drier. The amount of heat input to the liquid will be determined by how close (or far away) the heater is from the liquid. Sometimes we tend to over-engineer solutions to problems

You could separate out the heater supply in the hair dryer and pass it through a SSR, which you duty cycle modulate at a rate 1Hz or so to control the heater element. This way the airflow will be largely constant
and the air temperature varied.

MarkT:
You could separate out the heater supply in the hair dryer and pass it through a SSR, which you duty cycle modulate at a rate 1Hz or so to control the heater element. This way the airflow will be largely constant
and the air temperature varied.

That’s a good idea. Thank you!

Depending upon the quality of the hair drier, most of the small cheap versions use a small DC motor in series with the heating element so preventing separation.

If you are measuring time for liquids to cool, why does it matter how said liquids are heated?

raymw:
If you are measuring time for liquids to cool, why does it matter how said liquids are heated?

It shouldn't matter. I just wanted to have control on different air conditions and the hair dryer just blows at a fixed temperature and velocity. I though it could be easier but for now I'm going see if someone with more experience can give me a hand on this (maybe modifying the hair dryers , I'm not experienced with electronics).

Quite often, the hair dryer heating element is in series with the motor, and it may be tricky to adjust each separately. If you want to make comparisons between different liquids, that is not the same as getting specific values for each liquid, and perhaps a fixed rig, with the air in a duct, aimed over the top of the pan of liquid, take measurements, then replace pan of liquid with next test sample. Use a longer duct would reduce the temperature of the air at exit, but the airflow should be nominally the same. (I'm guessing a bit here, but you could measure it). For a radiant bod, iirc something varies as the square of the distance, for convection in a duct, it'll be different...