1. Ok

this is a bit of a strange question but this is my last place to turn.

Ok High output fans or any fan really is measured by CFM (Cubic feet ,Minute)

Ok now what i am interested in is how to convert a fans CFM to actual WINDSPEED MPH

I am sure some variables would include the following.

Size of Area, Resistance, CFM, and possibly other things.

But is there a way to calculate this or is that something that is just impossible. and i would just have to build a wind tunnel and run a fan with X CFM coming out of it in a specific area and then measure windspeed.

i would hope not, i would like to think that is a way to mathmatically figure it out

thanks for anyone here who can help with this,

And what the basis of this question is for is wind tunnel testing.

X  2.

3. If you take the cubic feet per minute and divide by the cross-sectional area in square feet you will get the average air speed in feet per minute. Divide by 60 for feet per second. Then multiply by 60/88 to get miles per hour. (60 mph=88 fps)  4. i made a really fast (2 minute) php script so that you can review my work to determine if i am on the right track or if i am way off in my numbers
Code:
```<?php
\$cfm = 1900;
\$length = 30;
\$width = 10;
\$height = 10;
\$minute_seconds = 60;
\$hour = 60;
\$mile = 5280;

\$squarefoot = \$length * \$width * \$height;
echo "
total cfm = \$cfm
";
echo "
square foot \$squarefoot
";
\$minute_air_speed = \$squarefoot / \$cfm; // per minute air speed
echo "
minute aire speed: \$minute_air_speed
";

\$feet_per_second = \$minute_seconds / \$minute_air_speed;
\$mph = (\$mile / (\$minute_air_speed * \$hour));

echo "
feet per second: \$feet_per_second
";
echo "
MPH= \$mph ";
?>```
the above script will produce this following output

total cfm = 1900

square foot 3000

minute aire speed: 1.5789473684211

feet per second: 38

MPH= 55.733333333333  5. No, the square feet will be the height of the wind tunnel times the width, so it should be 100 instead of 3000. Length of the wind tunnel should not affect your answer. I get 1900 cfm/100 square feet = 19 feet/minute. Then divide by 60 and that's 0.3 feet per second. Multiply by 60/88 and get 0.2 mph. Doesn't seem like much of a wind tunnel. Is your tunnel really 10 feet by 10 feet? That's huge. Your 1900 cfm fan is not nearly big enough.  6. in the bottom line i am devising an alternative method for generating energy.

i was just using the numbers above to see if i was preforming the equation correctly.

I would have thought that length of tunnel would have been a factor, because i would think there would be some form of resistance. but if that is not a factor, then it would appear as if my overall invention plan will work out exponentially better then i had originally anticipated after your last post.

so i guess my new equation would be something like this because i clearly made a few errors.

Code:
```<?php
\$cfm = 1900;
//\$length = 30;
\$width = 10;
\$height = 10;
\$minute_seconds = 60;
\$hour = 60;
\$mile = 5280;

\$squarefoot = /*\$length * */ \$width * \$height;
echo "
total cfm = \$cfm
";
echo "
square foot \$squarefoot
";
\$minute_air_speed = \$cfm / \$squarefoot; // per minute air speed
echo "
minute aire speed: \$minute_air_speed
";

\$feet_per_second = \$minute_seconds / \$minute_air_speed;
\$mph = (\$mile / (\$minute_air_speed * \$hour));

echo "
feet per second: \$feet_per_second
";
echo "
MPH= \$mph ";
?>```
will output this

total cfm = 1900

square foot 100

minute aire speed: 19

feet per second: 3.1578947368421

MPH= 4.6315789473684

and let med take a second for all your help on this equation,  7. I don't know the php script program you are using so I don't know exactly where you went wrong but if you got 19 feet per minute you could not end up with 3.1 feet per second because there are 60 seconds in a minute, not 6.

If the system resistance is big enough it will have some effect by giving you less cfm than the fan is rated for, and the static pressure also has an effect, but the conversion of actual cfm to air speed will be the same.

I'm not really an expert on fans. I suppose the fan would come with some kind of performance curve that you could use to get the actual cfm, and you would have to calculate the resistance as well. But the method above should get you pretty close.  8. basically it works like this.

i define all the variables and their values etc

essentially this is what happens

square foot calculation = Width * height

then

the Square Footage is divided by the CFM

this will tell me how much air is bieng moved through the fan which in this case is 19 feet minute

then minute air speed is divided by 60 because there is 60 seconds in a minute
this returns the vale of feet moved per second which in this case is 3.15

and then to derive MPH speed of the air

that is essentially what happens step by step

i set the equation 5,280 / (19 * 60)= 4.6 miles per hour

and i could be off i am not that good with math, and i am more of a programmer, and PHP math is not as exact as you would expect, but there should not be very large discrepancies in numbers, no mattter our methods may be different.

i attempted to represent the equation in a manner of programming language, maybe if you just laid down the exact equation with a defination of the variables it might be more clear to me.

but the numbers i am coming up with, are reasonable and i feel you have helped me out emmencly. and possibly the world, with large abundant sources of clean energy in almost limitless supply  9. 19 divided by 60 is still 0.315, not 3.15, no matter how you slice it.

I hope you are not trying to make a perpetual motion machine. If so, you are doomed to failure.

i attempted to represent the equation in a manner of programming language, maybe if you just laid down the exact equation with a defination of the variables it might be more clear to me.
All right. Let's say Q is the volumetric flow rate in cubic feet per minute.
Q=V/t
V=Qt
where V=volume in cubic feet and t=time in minutes.
If you have a duct of uniform cross-sectional area A, which could be cylindrical, rectangular, or any shape, the volume of air in any length of the duct is equal to the cross-sectional area multiplied by the length L.

V=A*L
L=V/A

Now imagine there is an invisible wall or piston in the duct that moves with the air flow in the duct as you force air into it. The velocity v (note lower case v, not the Volume V) will be equal to the change in L per unit time

v=L/t.
Substituting V/A for L,
velocity v=(V/At) but volume V=Qt so velocity v=Qt/At = Q/A
For a rectanglular cross-section Area(A) = width (w) * height (h) and
velocity v = Q/hw
If you use flow in cfm, feet for all your dimensions and minutes for time, this will give your velocity in feet per minute. Then convert to mph by
velocity v(mph) =v(fpm)*60/5280=v(fpm)/88.  Bookmarks
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