# Which fan / blower type should I chose?

Goal

I am creating a project where I will need to move small beans around in turbulent air flow. I will have a heat source to which I do not want the heat to leave easily (I want the beans to be allowed to heat up).

I initially tried to purchase a fan that simply had high CFM (about 400 cfm or so) but that seemed to be a poor choice as it just sucks all of my heat away and does not move my beans very effectively. This fan was simply a 120mm computer case fan. I believe my problem was that it pushes air with high velocity but with little momentum behind it.

My thought process

So I feel that this article gives a pretty great definition of the different types of fans and motors. I am not product searching but rather just trying to pick the appropriate blower / fan type that would best help my application.

I think that I want low CFM and high static pressure if I am understanding correctly? Based on this I think that the squirrel cage type blowers may be good enough. Please do correct me if someone has more experience with this though.

Once a type is established, what is the best method to actually size up the right sized blower?

• Please can you provide a diagram showing how the air moves from the fan/blower to the beans etc? Are you trying to make the beans 'jiggle', or move along a path? How does the heat source interact? Is it heating a plate the beans sit on? Heating the air? N.B. "Momentum" is defined as mass*velocity, and so your statement about high velocity, low momentum, doesn't make sense. – Jonathan R Swift Jan 14 '18 at 14:18
• @JonathanRSwift Diagram added. I am trying to keep the beans in a turbulent state. Diagram shows how heater interacts. There is a mesh layer in between to allow air flow without them falling through. Heating the air yes. I know what momentum is. That is how a fan works though for pushing air. It uses the mass and shape of the fin to push air. The air speed is defined as flow rate (CFMs) and "how hard it pushes" is the pressure which is derived from the momentum of the fins. Either way the goal of my question is the type of fan / blower for high obstruction? – Eric F Jan 14 '18 at 19:49
• Thanks for the diagram - looks a lot like a hairdryer (would a hairdryer work?)! Are you able to run an experiment to find out the required pressure/flow characteristics? The two variables that you need to find out are the flowrate of air above the beans (can be measured by timing how long it takes to fill a large bag of known volume), and the pressure below the beans (can be measured using a pressure gauge, or a manometer setup if you don't have one). You will need an over-powered fan that can easily create the turbulence that you need, then, simply restrict the flow until you... [continued] – Jonathan R Swift Jan 14 '18 at 22:33
• [continuation] ...reach the minimum flow rate required to keep the beans jumping. Once you know those two key values, you can look at the pressure/flow curves supplied on fan/blower datasheets to help you to choose your fan/blower. N.B. there's no need for the heater to be switched on for this experiment, but it must be present, as it will present an obstruction to the flow. – Jonathan R Swift Jan 14 '18 at 22:34
• @JonathanRSwift I agree with the hair dryer comment and actually that is what I am using for initial tests. I am trying to think of the most efficient fan type though, regardless of if the hair dryer works or not. I do know some blowers can be much cheaper than blow dryers as some blow dryers are $100-$150 or so. So you seem to be agreeing with my conclusions. I need to determine the pressure needed to raise my beans. As far as the heat component though wouldn't I want to keep my flow rate pretty low? I need my beans to reach a temperature of 400-450F or so . – Eric F Jan 15 '18 at 1:17

I assume you want to keep the beans suspended in the air, like they do with persons at skydive simulators.

In order to do that, you do need a high speed airflow (or CFM). The (static) pressure that the fan can maintain doesn't matter here. The high speed airflow creates dynamic pressure that acts on the beans and keeps them suspended, which is what your looking for if I understand correctly. At terminal velocity, the force that acts on the object due to drag, is equal to the force that pulls it down, which we call gravity.

Consider the following formula for the terminal velocity of an object:

$v=\sqrt(2mg/\rho*C_d*A)$

Where $v$ is the velocity on an object, relative to the medium/stream in metres per second.
$m$ is the mass of the object(beans) in kilogram.
$g$ is the gravitational acceleration, which is about $9.81m/s^2$ everywhere on Earth.
$\rho$ is the density of the medium, which is 1.2kg per cubic metre for air at standard conditions.
$C_d$ is the drag coefficient of the object. This is a factor usually between near 0 and 1. It's defined by the shape and skin texture of the object. A sphere has a $C_d$ of about 0.9, so i'd expect a bean to have a $C_d$ of about 0.85. It depends which orientation the bean has relative to the airflow.
$A$ is the frontal area of the object. This also depends on the orientation of the bean.

Since we can't change the properties of the beans, nor can we significantly change the density of the air, all we're left with is the velocity as tunable parameter.

I can't find the properties of a bean in my physics book, but let's suppose it's as dense as average wood: about 700kg/m3, and that the bean has the shape of a sphere with a diametre of 8mm. (about 1/3rd of an inch) Then the frontal area of the bean will be:

$A=1/4*\pi*D^2=1/4*\pi*8^2=50.3mm^2$ or 5.03e-5 m2.

The volume of the spherical bean will be:

$V=1/6*\pi*D^3=268mm^3$ or 2.68e-7 m3.

Which at a density of 700kg/m3, gives us a mass of the bean of:

$m=\rho*V=700*2.68e-7=0.188grams$ or 1.88e-4 kg.

If we fill in these values in the first formula, it gives us a velocity of:

$\sqrt((2*1.88e-4*9.81)/(1.2*0.85*5.03e-5))=8.5m/s$

That is the needed velocity of the air to keep them suspended. If you multiply it by the surface of your fan, you get the needed airflow. This is assumed the surface of the airduct above the fan is the same surface as the fan. Your 120mm fan has a surface of:

$A=1/4*\pi*D^2=1/4*\pi*0.12^2=0.0113m^2$

Which means that the volume air flow must be:

$Volumeflow=v*A=8.5*0.0113=0.096m^3/s$ or 96L/s or 203CFM.

I'd choose a fan that has a higher CFM rating(300CFM or so), to create a margin to make up for any errors made by assumptions in these calculations. You can always easily throttle down the fan. And also, your mesh, the heater, the beans, and any other construction will restrict the airflow, which makes it impossible for the fan to provide advertised CFM rating. That rating is probably based on an unrestricted fan. That's why you need to aim higher for CFM.

If you want to warm up the beans, i'd recommend to create a construction that makes the fan recirculate the air. So create your airduct like a donut or a square donut. Not only does it allow you to gradually heat up the air and the beans without losing your heat, it also keeps the load of the fan at a minimum.

• N.B. in the 'airduct like a donut' concept, your fan must be able to take the high temperatures involved. You'd certainly melt a CPU fan at $230\unicode{x2103}$! There may be some Blowers from the world of ovens/gas boilers that could manage this, however. – Jonathan R Swift Jan 17 '18 at 9:25
• Very helpful answer and definitely answers the air flow portion of the question. In a donut method, the calculations get to be quite complex I know but do you have a way to calculate the required wattage for a heater at the air flow calculated above, to achieve say 450F on a given diameter donut? – Eric F Jan 17 '18 at 13:15
• @JonathanRSwift I didn't see yet that he meant to heat up the beans to that temperature, but that would indeed create a problem. However, heating 200-300CFM from ambient to 230 degrees C would cost a lot of power. Fans from ovens may not be able to produce the high airflow though. – Bart Jan 17 '18 at 17:44
• @EricF specific heat of air is about 287J/kgK, specific weight at STP is about 1.2kg/m3. So at ambient conditions, about 1.2*287=345J is needed for a temperature rise of 1C per m3. Let's say ambient temp is 20C, which makes 230C a rise of 210C. If we take our results we get 0.096*345*210=6955W. That's a lot, nearly 7kW of power, if you don't recirculate the air. If you do recirculate the air, you can mix fresh air so that the incoming air is about 100C, a temperature that the fan may still able to handle. Needed power then is 0.096*345*110=3.6kW. – Bart Jan 17 '18 at 17:52
• @Bart I don't think that is correct based on a real life test. In my test, it was set up identical to my diagram above with an opened top with a roughly 1500 W heater and no recirculation and I could achieve around 350-400F (176 -204 C) after some time of heating up. The problem is that this is where it tapered off and never went higher. My thought is that the beans also might be insulating the system as well to help it increase in temperature – Eric F Jan 17 '18 at 18:22

To add onto Bart's Excellent Answer (above), we have to step out of theoretical and into applications.

1. Redo the math with air density at the higher temperature! - This changes ${\rho}$ to only 0.55 gm/m^3 at 300C. Terminal velocity shifts to 12.5 m/s Redo the flow rate with new terminal velocity! You need 300 CFM Add onto the safety factor - since beans won't suspend if they go on the non-flat face, we get rid of the safety factor and see we need 450 CFM, or 420 gm/hr. 420 gm/hr of cool air at 1.2 gm/m^3 is only 206 CFM.
2. Now we develop the flow profile. Through the heater, the losses can be expressed as some sort of mesh. There are a few ways known to express this, but this needs to be reviewed. At the end, the fan needs to be able to counter the static pressure loss of flowing 450 CFM through the heater.
3. Similar through the mesh holding up the beans. 450 CFM of air at 0.55 gm/m^3 flows through the mesh, which would be experienced as a pressure loss.
4. The beans themselves will float on velocity pressure alone, not static pressure. As a result, the final static pressure to push through the mesh and heater can be calculated.
5. The fan operates on a curve. This has been discussed elsewhere, but the fan will need to hit the static pressure loss for the heater as well as the mesh, while simultaneously providing 450 CFM.