The Fan Story

by John Beale  Aug. 2001

photos
 

Back in the summer of 1993, I was acting director of the ballroom dance club ("acting" only as long as it took to find some new victim "volunteer" for the job). The most critical task of the director was to find a space for the club to hold its dance practices, but all the good spots on campus (e.g. with air conditioning, or even a floor to speak of) had long since been reserved for high-dollar business seminars. As a mere student club we counted ourselves lucky to find anything. What we finally managed to secure was a modular unit on a hill behind FloMo hall. This dusty, disused classroom with a decrepit tile floor was in construction much like a mobile home, and in temperature very like a furnace. The indoor atmosphere was not improved by forty sweating students practising dance moves (when not fainting).

The situation clearly called for serious air-handling technology. I had been impressed by a floor fan I had seen at a local dance hall, and purchased one for the club's use.  This fan did improve matters somewhat in our dance/sauna room but not as much as I expected. I investigated the matter further, and after a few measurements I sent the following letter to the fan's manufacturer (names have been removed to protect the guilty).


XXX Inc.
New Haven, IN
Sept. 12 1993

Dear Sir or Madam:

I purchased one of your floor fans this Summer to ventilate a dance hall. I had estimated the volume of air circulation required, and selected your 12,500 cfm model (Whole House fan, Model TG-1887W 120 V 60 Hz 270 W 2.6 A serial number 93-TG4-002713) on that basis. The fan was helpful in cooling the room, but it did not provide as much cooling as I had estimated. This prompted me to attempt to measure the airflow generated by the fan directly (all the comments which follow refer to the "high" speed setting.)

I have measured air velocity about 5 cm from the face of the fan and this velocity is plotted as a function of radial distance from the fan center on the enclosed graph.  From this data a number of values may be calculated, including the air flow volume (1.0 cubic meters per second), reaction force or thrust (8.6 N) and air kinetic energy/sec (40 watts). This measurement is disturbing since the measured air flow is considerably less than the advertised value of 12,500 cfm (5.9 cubic meters/second).  However, I have independently measured the fan thrust at 9.5 N which is in agreement within 10% of the value calculated from my velocity data.  I also measured the electrical current consumed by the fan to be 2.6 A as specified.  For the stated 270 W electric power, the overall efficiency [air kinetic/electric] is then 15% which is what might be expected for a typical consumer motor and fan blade design. I conclude that my measurements are approximately correct and the actual fan performance is 1/6 of the advertised value.

It might also be pointed out that, given only the diameter of the blades on this fan (18" = 46 cm) and the power used (270 W), it is fundamentally impossible to achieve the stated air flow of 5.9 cubic meters per second.  If the air velocity is uniform across the cross-section (unrealistic, but the minimum-power requirement case), a velocity of 36 m/s (80 mph) is required to achieve 5.9 cubic meters/sec. This becomes absurd when the implications are considered. The power in such an air flow is 4.6 kW, which is 17 times as much (!) as the fan uses.  At the existing 15% efficiency of your fan design, 30 kW of electrical energy would be required. This hypothetical fan would also be quite dangerous as it hops around the room, since the reaction force of 255 N (57 pounds-thrust) is roughly four times the fan's weight.

If we work in the other direction and assume the fan motor and blade are both 100% efficient, we obtain 14 m/s flow, which corresponds to 2.3 cubic meters per second, still less than half the quoted value. With the existing 270 W motor and fan blade design with 15% overall efficiency, a fan blade diameter of 1.5 meters (4' 11") would be required for 5.9 cubic meters/s (12,500 cfm).

I would appreciate hearing from you in detail how you measured the air movement capacity of this fan. Thank you very much.

Sincerely,
(signed)
John Beale

[enclosed graph]

I mailed the letter and awaited details of how the fan company had arrived at their numbers. Perhaps my measurements were in error. Several weeks passed without any reply-- maybe the letter had been misplaced. I sent another copy, this time adding a new cover letter.


XXX Inc.
YYY, director of engineering

Dear YYY,

Enclosed please find a copy of the letter I sent to your company three weeks ago. I have not received any reply so I suspect the letter may have been misdirected or misplaced. Accordingly, I am taking the precaution of sending this letter by registered mail, return receipt requested.

Per the enclosed letter, I have had difficulty in finding agreement between the stated capacity of your floor fan, model TG-1887W, and my own measurements detailed therein. As the letter sets forth, the three independent measurements I have made of the fan output agree on a value which is approximately 1/6 (that is, 17%) of the 12,500 cfm capacity advertised on side of the fan box. Further, based on theoretical grounds I have concluded that the model TG-1887W fan cannot simultaneously achieve an output of 12,500 cfm while also obeying the laws of physics that I am familiar with.

Together with several members of my research group, I have considered the problem of reconciling the measured fan output with the advertised performance. We have arrived at a few possible scenarios:

1) You measured the fan output at some extreme elevation where the pressure is less than 1/16 sea level and the fan would have enough energy (at least in theory) to impart 36 m/s velocity to this low-density air. However, the efficiency of your fan blade design at very low pressures poses an interesting problem, which we have not yet examined, and your product packaging does not mention or suggest use of the product in near vacuum.

2) You might have used a toroidal test chamber, in which the air was accelerated in several stages through the fan blades to achieve a peak velocity greater than any single pass. I imagine the chamber walls would have to be quite smooth to avoid severe losses, and even so I would expect significant boundary layer drag. In any case I have not previously heard of such a test, or found it described in any ASHRAE standards documents to which I have access.

3) You are employing some nonstandard definition of CFM, interpreted not as cubic feet per minute but rather as some novel unit. For example, I calculate that the fan moves a quantity of air equivalent to 3874 cubic furlongs per millennium, and perhaps some other combination of units would agree with 12,500 to a reasonable approximation.

I remain in some doubt that any of these speculations correctly describe the measurements you have performed on the fan. I am curious how you did obtain your fan rating number and I would appreciate the favour of a reply.

Regards,
(signed)
John Beale.

Shortly after I sent the second letter, I received a call from the director of engineering at the company in question. He did not have a great deal of information to convey. He seemed relieved when, in response to his first question, I said that I did not represent a law firm.  He said he had asked around the company, and no one could remember how the advertised number had been determined. I volunteered that I was surprised that they even advertised a specific CFM value, as the average consumer probably wasn't familiar with these measurements. He replied that yes, in fact the packaging had recently been redesigned and the CFM number had been replaced with information felt to be of more use to the customer.

What was this information, I wondered. 

He replied that the package now specifies the color of the fan. It comes in white, teal or black.

I thanked him for his time.

Addendum: this story is completely true. From what I can determine online now in 2001, the company in question is no longer in business.

A few physics notes

To convert CFM to thrust, you need to know the velocity and cross-sectional area of the airstream.

F = M * a

F: force or thrust (newtons)
M: mass (kilograms)
a: acceleration = dv/dt
v: units of meters per second
a: units of meters per second per second

Metric and english unit conversion:
1 m/s = 2.24 mph
1 cubic meter = 35.31 cubic feet
1 cubic meter/sec = 2119 CFM (cubic feet / minute)

At atmospheric pressure and room temp, the density of air is approximately 1 kg/cubic meter, a handy number.
On earth, the acceleration of gravity is 9.8 m/s2, so 1 kg weighs 9.8 newtons. F (newtons) = m (kg) * a (m/s2).

If a fan accelerates 1 cubic meter of air (mass=1 kg) to a uniform velocity of 10 meters per second (22 mph) in 1 second,
it must have a thrust of F = m * a = (1 kg) * (10 m/s per second) = 10 newtons, which is roughly 1 kg or 2.2 pounds.
If the velocity is uniform across the flow, that volume and velocity implies a cross-sectional area of 0.1 square meter,
for example a circle 36 cm in diameter.   Note that the fan described in the letters above did not quite reach a 10 m/sec
air velocity.
Rather than F = M*a, a physicist might prefer to write: F = dp/dt
where p (should be Greek 'rho') is momentum, and dp/dt is the rate of change of momentum with time.

p = m*v so dp/dt = d/dt (m*v) = m(dv/dt) + v(dm/dt).

If the mass does not change with time (dm/dt=0) this becomes just F=ma.

On Jan. 11 2005 Ray wrote:
--------------------------
I was searching the internet for measuring fan output and found your article called Fan Story Aug, 2001.  It was very interesting. I was wondering if you could share how you arrived at your various measurements like air flow.

My daughter in grade 5 would like to do a science project to test various propellers to see how much wind is needed to get them going. So if we use a room fan to try to make the propellers turn then we need to know how many miles per hour or meters per second the wind from the fan is blowing. Then we can try and measure the rotation speed of her various prop designs and also try to measure the lifting capacity of each propeller and use the propeller to drive a generator and measure the electrical output possible.  This is all for a project to do with wind power.

Thank you,
Ray V.R.
Canada
Hi Ray,

I'm glad you enjoyed the fan story. I used several different methods to measure the air flow. The first thing I did was borrow a wind-speed gauge from a friend who is a windsurfer. You can build something simple, but it would need to be calibrated.

You can also buy digital wind-speed gauges with little propellers, but they are expensive. My friend's gauge was a simple mechanical one with an initially vertical hanging bar, that tilts upwards as the wind pushes the free bottom end, and it swings up against a scale with wind speed indications.

The second method I used was to drop small bits of paper in front of the fan, and use a video camcorder from one side to record the image of the paper bits being blown in the wind. I also put a ruler in front of the fan. I knew the video camera takes 30 frames per second so I looked at the individual frames on the computer and measured the distance (based on the ruler) that the paper bits moved in 1/30 of a second.

You could also do this with a still camera, for example use a 1/10 second shutter speed and measure the length of a streak of white paper fragment over 1/10 of a second. (That's just an example, you'd have to experiment to find the best settings on your camera.)

The third method involved some more calculations; I measured thrust by hanging the fan from my back porch roof using two strings, turning it on and measuring how far it pushed itself back from the force of its output. Knowing the weight of the fan from my bathroom scale, and the angle the strings made from the vertical, a little trigonometry gave me the ratio of the fan's thrust to its weight. From that, knowing the density of air (1 kg per cubic meter) and force = mass * acceleration (F=m*a) and the cross-sectional area of the airflow, I could calculate the air velocity.

The windspeed gauge and the paper bits measure the airspeed in just one spot. The thrust measurement is an average over the whole width of the fan. It turns out that usually there is quite a bit of variation in airspeed right in front of the fan, which makes sense- after all, in the very middle of the fan there is no push from the propeller, but there is a lot near the outer edge of the blade, and just beyond the very edge you get a "tip vortex" and the air actually moves backwards (unless you have a "ducted fan").  Once you move several fan-diameters away from the fan, the air velocity evens out, but there is still some turbulence. You can drop a bunch of paper bits in front of the fan, and get a sense for how much there is.

Good luck with the project, I hope this is useful!

regards,
John
9/2020: Tejas asked, why did I first suspect the fan output claim on the box was wrong?

In a word, math. I knew the actual temperature rise in the room, I checked that first because people were complaining about it. To plan which fan to buy, I did some rough estimates of how much heat all the dancers were generating, and I knew the outside air temperature, the room volume, and heat capacity of air. From this I could calculate what the temperature rise inside should be with no air exchange, or with any given amount of air flow, and how far a 12500 CFM fan should be able to drop the inside temperature towards what it was outside. With the fan running, I found the actual drop was a small fraction of my estimate. The error was much larger than what I believed it reasonably could be. Therefore I suspected the fan was not moving nearly as much air as the box claimed, and I decided to measure it directly to confirm that guess, which is where the story above starts.
Back to measurements page.