Is it possible to calculate cfm, and dba by volt modding?

[anvil82]

I want to get some powerful 120mm fans, but I would also like to use a rheobus to control voltage.

Is there any type of formula to claculate the dba, and cfm difference from 12v to 7v for a particular fan?

 

[emericanchaos]

you could divide the CFM by 12 volts to find CFM per 1 volt than multiply that by 7volts.

 

[CrashOveride]

but even that wouldnt be to preices... you would have to make sure that the cfm doesnt go down at a faster rate with each decreas in voltage (i would guess that it might), but im sure that would work alright for basice use

 

[dukla2000]

The fan rpm are directly proportional to the voltage for a DC fan, and in free air the cfm are directly proportional to the rpm. So for a free air model emericanchaos is right.

dB2 = dB1 - 50 * log (rpm1/rpm2) (log base 10)

 

[anvil82]

The fan rpm are directly proportional to the voltage for a DC fan, and in free air the cfm are directly proportional to the rpm. So for a free air model emericanchaos is right.

dB2 = dB1 - 50 * log (rpm1/rpm2) (log base 10)

Where does the 50 come from?

In my example with a 4050rpm fan at 53dba, and 130cfm I would get

4050/12 = 337.5

337.5 * 7 = 2362.5


db2 = 53.5 - 50 * log(4050/2362)

db2 = 3.5 * .23401082

db2 = .819

wow that's one quiet fan, heh. hmpf

 

[start]

peesh, I get stuck with the exponential dB rating..Ive no idea how on earth you worked that out!

 

[TerroH8er]

Where does the 50 come from?

In my example with a 4050rpm fan at 53dba, and 130cfm I would get

4050/12 = 337.5

337.5 * 7 = 2362.5


db2 = 53.5 - 50 * log(4050/2362)

db2 = 3.5 * .23401082

db2 = .819

wow that's one quiet fan, heh. hmpf

remember your order of operations.. multiplication and division first, then addition and subtraction.

so it would be

db2 = 53.5 - 50 * 1.71464
db2 = 53.5 - 85.732
db2 = -23.232

ok.. nevermind, but still remember your order of operations lol

could someone explain what we are doing wrong? lolol

edit: hmm.. is it actually rpm2/rpm1? Cause that would work out better.

 

[anvil82]

so it would be

db2 = 53.5 - 50 * 1.71464
db2 = 53.5 - 85.732
db2 = -23.232

What does the 50 represent though?

 

[anvil82]

Hmm this is an interesting read

http://www.amdmb.com/article-display.php?ArticleID=132&PageID=8

http://www.amdmb.com/article-display.php?ArticleID=132&PageID=9

 

[TerroH8er]

What does the 50 represent though?

i dunno lol (i wasn't the one who posted the forumla if that's what you're thinking, it was dukla2000)

I'd like to know what log base 10 is..

 

[dukla2000]

db2 = 53.5 - 50 * log(4050/2362)

You were doing fine up till here, then

db2 = 53.5 - 50 * log (1.7146)
= 53.5 - 50 * 0.2341
= 53.5 - 11.7087
= 41.79

Log base 10 (TerroH8er) as opposed to log base e (commonly ln) or potentially any other base for the scale. You can create a logartithmic scale using any number for the base (OK, I forgot most of my maths but think 1 is meaningless, as are probably negative numbers). So base 2, base e, base 69 etc are all feasible, e and 10 being the most commonly used.

 

[TerroH8er]

ah thanks! I now understand.

This helped me a lot as well: http://photography.cicada.com/tips/logbase10.html