"250mm" fans vs 120mm: the physics, math, tables &

Felger Carbon · Post by **Felger Carbon** » Sun Dec 31, 2006 11:58 pm

The physics:

The energy of a moving bit of air is proportional to M*V^2. If we double the amount of moving air, there are twice as many moving bits. So the energy of a moving column of air is equal to M*V^3, where M is the mass of air and V is its velocity. Acoustic noise is proportional to the energy of the air column. This is the only information needed to mathematically compare the CFM and noise of different-sized fans, to a first-order approximation (meaning for perfect fans that have, for example, no excess motor noise).

So let's do some comparisons, adopting 120mm as our reference fan size:

Delta noise, other size fans vs 120mm, same CFM:

60mm = +12.04 dBA
70mm = + 9.36
80mm = + 7.04
92mm = + 4.62
100mm = + 3.17
120mm = 0 dBA (reference)
140mm = - 2.68
200mm = - 8.87
220mm = -10.53

The above sez a 60mm fan is over 12dBA noisier than a 120mm fan if both are pushing the same CFM. A 220mm fan is over 10dBA quieter than a 120mm fan at the same CFM.

Delta CFM, other size fans vs 120mm, same noise:

60mm = 0.397
70mm = 0.487
80mm = 0.582
92mm = 0.838
100mm = 0.886
120mm = 1.0 (reference)
140mm = 1.108
200mm = 1.406
220mm = 2.244

At the same noise level a 60mm fan will only push ~four tenths the CFM as a 120mm. A 220mm fan will push ~2 & 1/4 times the CFM of a 120mm fan at the same noise level.

Quickie facts about noise and CFM:

The Bel is equal to the log to the base 10 of the noise divided by a reference noise, which is the accepted limit of human hearing. But we use the deci-Bel, which is a tenth of a Bel. So dBA (the abbreviation for deciBel) = 10Log10(noise power/limit of hearing). The A in dBA once represented a weighting curve vs frequency to compensate for the Fletcher-Munson effect. It is now a historical standard weighting curve. Noise is proportional to M*V^3.

Holding CFM constant as the fan diameter increases: CFM = M*V. If the fan diameter doubles, M increases by 4. If the CFM is constant, then V must decrease by 4 (and RPM by 8).

Holding noise constant as the fan diameter increases: Noise = M*V^3. If the fan diameter doubles, M increases by 4. If noise is constant, then V^3 must decrease by 4, or V by the cube root of 4.

The math:

DR (diameter ratio) = Dia/RefDia; AR (area ratio) = DR^2

For constant CFM, noise increase in dBA = 10*Log10(AR/AR^3)
or, dBA = -10*Log10(AR^2)
or, dBA = -40*Log10(DR) <-- simplest form

For constant noise, CFM ratio = AR * e^(-ln(AR)/3)

A quick reality check:

Panasonic makes the FBA line of fans. The 25mm-thick variety are made in 60, 80, and 92mm sizes: The FBA08L is rated at 24CFM and 21dBA, the FBA08U at 46.9CFM and 38.2dBA. So a 1.954 increase in CFM causes a 17.2dBA increase in noise. Now, those are actual measurements, not theory. But it sounds pretty close to 18.06dBA per 2-1 CFM ratio, doesn't it? Just as if the noise increased by the cube of the air velocity, in fact.

Real-World fans are not ideal devices:

Real fans make more noise than the base theory predicts. There's motor noise. And there's the turbulence at the outside edge of the fan blades. Neither of these two factors is constant as the diameter of the fan increases. Motor noise is largely bearing noise, and every fan has a bearing, no matter its size. It seems to me that very small fans are at a serious disadvantage here. Now, about the turbulence: it's proportional to the circumference of the fan, while the mass of air being pushed is proportional to the square of the fan's diameter. Once again, small fans are at a disadvantage here; this "edge effect" halves with every 2-1 increase in fan diameter.

Opinion: I believe if very careful measurements are made, the smaller fans will perform worse than the tables at the start of this posting indicate, and the bigger fans will perform slightly better.

Summary:

This writeup is about single fans of varying diameter but all 25mm thick. The equations are derived from the two short paragraphs at the end of the "Quickie facts".

edit: disable emoticons to prevent a formula from becoming a smiley.

edit2: the difference between M*V^2 and M*V^3 explained.

pcy · Post by **pcy** » Mon Jan 01, 2007 2:47 am

Hi,

1. The energy of a moving column of air is equal to M*V^3, where M is the mass of air and V is its velocity.

2. Acoustic noise is proportional to the energy.

1. I'd be prepared to beleive that the energy was SxV*3 (where S is the specific graviity of air) as the mass of air passing through the fan is also proportional to V, but if M is the mass of the moving column of air then the energy is MxV*2, surely.

2. I see no obvious reason why noise is proprotional to energy. I'd have thought that noise is a funtion of the turbulence in the air - in a purely lamina flow of classical aerodynamics (which you might achieve, or at least approach at sufficiently low air velocity) there would be no air noise. Again, I'd be prepared to believe that once you had turbulent flow noise would be proportional to total energy, but a proof would be nice.

Thanks

Peter

Felger Carbon · Post by **Felger Carbon** » Mon Jan 01, 2007 4:23 am

The energy of moving air, namely M*V^3, is absolutely fundamental to the "wind energy" industry, now billions of dollars worldwide. The on-line explanations will turn handsprings to avoid confusing the general public with actual facts, so they try to avoid the subject. I found this on-line, however (google):

"The kinetic energy in wind increases exponentially in proportion to its speed, so a small increase in wind speed is in fact a large increase in power potential. The general rule of thumb is that with a doubling a wind speed comes an eight-fold increase in power potential. So theoretically, a turbine in an area with average wind speeds of 26 mph will actually generate eight times more electricity than one set up where wind speeds average 13 mph."

I'm an engineer, not a physicist. It's simply a fact that noise is also proportional to the cube of the air velocity, which is why the noise level increases 18dBA when CFM doubles. This is supported by the Panaflo published info I cited as a reality check. The "why" will have to come from a physicist.

jaganath · Post by **jaganath** » Mon Jan 01, 2007 5:16 am

It's simply a fact that noise is also proportional to the cube of the air velocity

While googling I came across the startling fact that noise from a jet exhaust is proportional to the sixth power of the mean jet exhaust velocity(!). Maybe the noise is so much more because of the hot gases expanding etc?

I see no obvious reason why noise is proportional to energy. I'd have thought that noise is a funtion of the turbulence in the air

Wouldn't turbulence be a function of velocity (and hence energy)?

pcy · Post by **pcy** » Mon Jan 01, 2007 3:24 pm

Hi,

I'm simply trying to understand. There is no way that energy is proportional to mass times velocity cubed because energy it is in fact proportional to mass times velocity squared.

The mass of the air colum may itself be proportional to the velocity, and I'm intirely happy with the idea that total energy is proportional to the velicity cubed, because teh effective mass of teh colums is also increasing with velocity, provided M is the mass of a given volume of the gas (i.e cross sectional area of fan x unit length) we are talking about.

Noise is clearly a function of energy, but not necessaruily proportional to it. I finds the idea that noise might be proportional to energy (at least in some range of flow conditions) enetirely plausible; but I can find neither any experimental results nor a clear reason why it should be so; and actually, I suspect it in not always true.

Clearly noise will increase rapidly with if anything a reduction in air velocity as the fan blades stall and I'd expect more noise at the same velocity with increasing backpressure; there is probably an equally sharp transition when the flow goes form laminar to turbulent, and it seems that at sufficiently high velocity the noise becomes proportional to the 6th power of the velocity.

Peter

spookmineer · Post by **spookmineer** » Mon Jan 01, 2007 4:08 pm

pcy is correct, 1 joule = 1 Nm = 1 kg m2 s-2 = 1 Ws.

Energy in J, mass in kg, speed v in m/s.

1 J = 1 kg m2 s-2 = 1 kg v2.

The lists are helpful anyway because real measurements seem to back them up, but the derivation is not valid.

Felger Carbon · Post by **Felger Carbon** » Mon Jan 01, 2007 4:38 pm

spookmineer wrote:The lists are helpful anyway because real measurements seem to back them up, but the derivation is not valid.

I'm (between football games) in contact with a PHd physicist who consults for EPRI, the electrical power consortium. The partial info I have from him sez that it's V^3 within the fan frame but V^2 elsewhere. Since I'm just writing about fans, I'm comfortable with that.

There's still the quotation I found about wind and windmill power... but there's the football games, too...

So the measurements agree with fan = M*V^3, and my contact agrees within the fan frame. More when I get this sorted out. (I got the V^3 from this guy in the first place, but failed to understand that it only applied in a fan context.)

Felger Carbon · Post by **Felger Carbon** » Mon Jan 01, 2007 4:58 pm

OK, I think I've sorted it out: the energy of a given cubic cm of air as it moves along is indeed M*V^2.

But a fan is stationary, and air flows thru it. The mass of air passing thru the fan in one second is proportional to the airflow velocity. This is where the extra V comes from. For fans, the energy rate of the airflow is indeed M*V^3, and my equations and tables are accurate.

But I did express the physics incorrectly; I should have restricted it to the air movement thru the fan frame.

From Wikipedia:

http://en.wikipedia.org/wiki/Wind_power ... _potential

The power P available in the wind is given by:

The mass flow </wiki/Mass_flow_rate> of air that travels through the swept area of a wind turbine varies with the wind speed and air density. As an example, on a cool 15Â°C (59Â°F) day at sea level, air density is 1.225 kilograms per cubic metre. An 8 m/s breeze blowing through a 100 meter diameter rotor would move almost 77,000 kilograms of air per second through the swept area.

and:

The kinetic energy </wiki/Kinetic_energy> of a given mass varies with the square of its velocity. Because the mass flow increases linearly with the wind speed, the wind energy available to a wind turbine increases as the cube of the wind speed. The power of the example breeze above through the example rotor would be about 2.5 megawatts.

pcy · Post by **pcy** » Tue Jan 02, 2007 5:12 am

Hi,

Excellent, thanks. So that's energy proportional to specific gravity of air times velocity cubed.

Seems intuitive now, and if experimental evidence confirm it as well...

I'm still concernrd that the noise ~ power bit is only true within a limited envelope of conditions - something like subsonic flow with buondary layer attached, or a range of Renalds Numbers or...

Peter

Felger Carbon · Post by **Felger Carbon** » Tue Jan 02, 2007 5:38 am

pcy wrote:I'm still concernrd that the noise ~ power bit is only true within a limited envelope of conditions - something like subsonic flow with buondary layer attached...

I think we can agree that all of the fans of interest here at SPCR will produce subsonic flow.

Now, about that boundary layer, and possibly laminar flow... Consider a 120mm fan pushing a 120mm diameter cylinder of air thru a still-air environment. There will never ever be any disturbance at the edge where the still air meets the moving column??

Once the air has left the fan, and I mean immediately, the moving column of air will start vigorously interacting with the still air. I think this is called turbulence. Since I'm not a physicist I'll express no opinion as to whether this mechanism accounts for the noise being proportional to the energy.

But I don't believe in laminar flow or boundary layers in this scenario.

BTW: thanks to your complaints (and others) I made a small edit to the first posting to explain the squared vs cubed issue succinctly. Keep up the good work!

jaganath · Post by **jaganath** » Tue Jan 02, 2007 6:46 am

But I don't believe in laminar flow or boundary layers in this scenario.

I don't think there can ever be laminar flow through a fan, and I'll explain what makes me say that: the blades on an impeller are basically airfoils, which are designed to produce lift, and one inevitable consequence of generating lift is trailing vortices being shed by the airfoil trailing edge; no vortices, no lift (no CFM).

pcy · Post by **pcy** » Tue Jan 02, 2007 6:57 am

Hi,

I completely agree that thre must be tip vortices and that the flow cannot ne entirely laminar.

But by "laminar flow" I meant tht the flow over the baldes was largly laminar, with turbulence confined to the boundary layer and teh tip vortices.

That's certaily what you get on the sail of a yacht, which is another example of a slow moving. low noise, aerofoil.

Totally laminar flow is only a theoritical possibility, requiring a fluid with zero viscosity. The boundary layer must exsit in any real situation, and as far as I know it is always turbulent. I think that whne the main flow becomes turbulent rather than lamina, that is the point when the foil stalls; and I also think there is a significant increase in turbulence (and noise) at the point where the boundary layer detaches from the foil.

We need a proper aerodynamicist...

Peter

jaganath · Post by **jaganath** » Tue Jan 02, 2007 8:41 am

The boundary layer must exist in any real situation, and as far as I know it is always turbulent.

Agree with the first part, disagree with the second:

http://www.grc.nasa.gov/WWW/K-12/airplane/boundlay.html

Boundary layers may be either laminar (layered), or turbulent (disordered) depending on the value of the Reynolds number. For lower Reynolds numbers, the boundary layer is laminar and the streamwise velocity changes uniformly as one moves away from the wall, as shown on the left side of the figure. For higher Reynolds numbers, the boundary layer is turbulent and the streamwise velocity is characterized by unsteady (changing with time) swirling flows inside the boundary layer.

For most streamlined shapes, the flow starts off laminar and becomes turbulent when it encounters a sufficiently strong adverse pressure gradient.

MC FLMJIG · Post by **MC FLMJIG** » Tue Jan 02, 2007 10:26 am

I prefer a 120mm fan. I had one of the cases with those huge fans and I didn't think it was so silent. Besides it will drag in that much more dust inside the PC. I used a fan controller and then the motor made a slight sound.

It could also be the XCLIO case/fan that I used.