Influence of restrictions/underpressure on RPM & CFM

[SpyderCat]

First of all: I don't have a problem, just being curious.

I'm trying to evaluate my current setup with measurements where possible, and educated guesses for those I can't take measurements on.

Current setup:

Room temperature: 20*C
Case temperature: 25*C
The case temperaure stems from HD-cooling, graphic-card cooling, Mobo cooling.
Both PSU and CPU exhaust air is kept isolated from the air in the case.

PSU:
Nexus 3000 with it's stamped grill cut out, original fan (@12volts: 3000RPM/38CFM/29 DbA)
Air temp at intake of the PSU: 25*C (= case temperature).
Air temp at exhaust of the PSU: 34*C. --> Delta(T)= 9*C

CPU:
AMD 2400+ with modded AX-7 heatsink in sucking mode, with duct, NO fan on the heatsink, folding like mad under 100% load.
Panaflo A12L (@12volts: 1900RPM/24CFM/21 DbA) @9 volts sucks air through the heatsink, and dumps it at the rear, outside of the case.
Air temp at case exhaust (= CPU exhaust): 41*C. --> Delta(T)= 16*C


That's all. Only two fans, both undervolted.

From Mike's review of the Nexus PSU: The PSU-fan gets 4.6 volts @ 20*C ambient, and 5 volts @ 38*C ambient. I provide 25*C ambient, so I guess the fan gets 4.7 volts.

Now, how much RPM is this PSU-fan making, and how much CFM is it moving out of the case?

All quoted RPM/CFM numbers from manufacturers are 'free flow' numbers.
Hardly anyone here uses his fans in a 'free flow' situation.
When there were no restrictions to inhibit airflow, there would be no pressure difference between intake/exhaust of the fan; a true 'free flow' situation. Because we have restrictions, and because we have multiple fans we have under-pressure, and this MUST influence both RPM and CFM of the fan.

Suppose we have temporally shut off the power to the PSU fan. The case fan is pumping air, and some of it flows in through the PSU.
Now we supply some power to the PSU-fan, just enough to make the airflow through the PSU come to a standstill. Here we have a situation where we are supplying power to the fan, it has a certain RPM, and a CFM of 0 (ZERO).
Now we increase the PSU-fan voltage to 4.7 volts (see the example above)
What is the CFM of this fan now?
Just by holding my hand at the back of the rig I can tell it's clearly less than the exhaust of the Panaflo case-fan @ 9 volts.

As the PSU-fan is doing work, it's RPM must be lower than in a 'free-flow' situation.
How much lower? 90% of 'free-flow', 80% ????

Does anyone have 'educated guesses', knowledge regarding these questions?

Regards, Han

[dukla2000]

Hi Han

folding like mad under 100% load.

Well sorry, you dont appear to be on the SPCR team so unfortunately we cant help

Here we have a situation where we are supplying power to the fan, it has a certain RPM, and a CFM of 0 (ZERO).

Well, OK, just this once! The key to this (I think, no expert) is the static pressure curves of the fan. The 2 fundamental gotchas are first you have lowered the voltage of the fan so the manufacturers curves no longer apply (although Panaflo at least show curves for 4 voltages), and second we have no way of measuring our static pressures so even if we were at 12V, we haven't a clue where we are on the curve.

Now we increase the PSU-fan voltage to 4.7 volts (see the example above) What is the CFM of this fan now?

If I understand fans correctly the rpm will be proportionate to the voltage (a little later in your post you query the rpm), and (from the Panaflo curves) at a constant voltage but increasing pressure the rpm are constant while the cfm drop.

Overall this is the part of fan theory I dont really understand, but you need to plot your "System Impedance Curve" against the fan's volume/static pressure curve. To complicate things even more, in this case the voltage of your psu fan is changing so you need a series of fan curves. Hopefully Dorothy or someone else who knows what they are talking about will chip in.

[SpyderCat]

Hi Dukla,

A reasonable estimate of CFM might be obtained by using the PSU's &#916(T).
More CFM --> smaller &#916(T) Right?

At the moment I can't easily get a "PSU's &#916(T)" in "free-flow" condition.
Still &#916(T) seems hold promise for getting CFM-numbers in real life situations.

I retreat to think some more

Edit:
Mike,
Did you ever take delta(T) measuments when reviewing PSU's?

[SpyderCat]

Hi Dukla,

Done some thinking...

In an other post you wrote:

Say you CPU is 80W (peak), then the basic aircooling theory says your temp rise (with 20cfm) is
dt = (80 * 1.76)/20 = 7C

Using the same formula on my case/CPU exhaust fan:
Delta(T)=16
Heating power=64 Watts

Formula: CFM=(power*1.76)/(delta(T))

CFM = (64*1.76)/16 = 7.04 CFM

From feeling with my hand at the exhaust I estimate the PSU exhaust CFM at half the volume of the case exhaust. --> PSU exhaust = 3.5 CFM !

Ambient temp=20*C
PSU delta(T) with regards to the room=14*C
Case/CPU-fan delta(T) with regards to the room=21*C

Formula: (CFM*delta(T))/1.76 =power

Power expelled by PSU-fan: (3.5*14)/1.76 = 28 watts

Power expelled by case-fan: (7*21)/1.76 = 83 watts

Total watts for the system= 28+83 = 111 watts.

This could be close !

Amazingly low CFM-values, don't you think?

Hey, hey, hey... we're actually getting somewhere!

Regards, Han.


Edit:

Using the graphs in Panasonic's fba08a.pdf (Panaflo specs)
I arrive at pressure of 0.6 mm H2O for 7 CFM at 9 volts
Using a pressure of 0.6 mm H2O in the graph of the Panaflo A12H @ 4.7 volts then 3.5 cfm doesn't seem unlikely.

There is one thing bugging me:
Mike's review of the Nexus 3000 gave an efficiency of about 70% at power consumption around 110 watts.
So, if the PSU's CFM=3.5 and delta(T)PSU=9*C than using the formula again:
Formula: (CFM*delta(T))/1.76 =power
(3.5*9)/1.76 = 18 watts
This value suggests I my estimate of CFM(PSU)=.5*CFM(case) is likely to low.

Also:
When the total power consumption = 111 watts and the PSU's efficiency=70% -->
Net output PSU= 70%*111 = 78 watts

CPU 64 watts ???
HD 7 watts ???
Graphic 12 watts ???
Mobo 20 watts ???
----------------------
Total 103 watts

The numbers aren't exactly fiiting.
What numbers should be corrected ?

[dukla2000]

Yeah - was about to post that formula again. Except you got in your next long one which I need to digest.

The other thought I had (while doing some gardening) is that the voltage at the point where your cfm=0 gives you a point on your System Impedance Curve (I think), except I am not sure which point!

ps - the SPCR folding team # is 31574

[Athlon Powers]

CPU:
AMD 2400+ with modded AX-7 heatsink in sucking mode, with duct, NO fan on the heatsink, folding like mad under 100% load.
Panaflo A12L (@12volts: 1900RPM/24CFM/21 DbA) @9 volts sucks air through the heatsink, and dumps it at the rear, outside of the case.
Air temp at case exhaust (= CPU exhaust): 41*C. --> Delta(T)= 16*C

Cool! I have that exact same configuration! Will you tell me what mods you did and if it cools any better? Thanks!

[SpyderCat]

CPU:
AMD 2400+ with modded AX-7 heatsink in sucking mode, with duct, NO fan on the heatsink, folding like mad under 100% load.
Panaflo A12L (@12volts: 1900RPM/24CFM/21 DbA) @9 volts sucks air through the heatsink, and dumps it at the rear, outside of the case.
Air temp at case exhaust (= CPU exhaust): 41*C. --> Delta(T)= 16*C

Cool! I have that exact same configuration! Will you tell me what mods you did and if it cools any better? Thanks!

My temps are rather high as I've used any gains in cooling to reduce the noise.

Case temp: 5*C over ambient
Mobo sensor: 10*C over ambient
PSU intake: 5*C over ambient
PSU exhaust: 14*C over ambient
Case/CPU exhaust: 41*C @ 20*C(ambient)

CPU-fan (=case exhaust fan) is powered by the MoBo, and voltage (8 to 12 volts) is regulated by ASUS-Q-fan.
ASUS-Q-fan gives 8 volts when the CPU <= 50*C
ASUS-Q-fan gives 12 volts when the CPU >= 62*C

At 20*C ambient, after 30 minutes CPUBurn, and the fan fed with 12 volts, the CPU has 53*C.
To me this is an indication I can run CPUBurn for half an hour during a heatwave. But who would do THAT
The highest temps we have over here in summer are around 32*C.

The fact that ASUS-Q-fan starts reducing the voltage to the CPU-fan at 62*C (or lower) is an indication to
me that ASUS thinks 62*C for a CPU is just fine.

In a few days/weeks I expect to put up some pictures.

Regards, Han.

[dukla2000]

There is one thing bugging me:
Mike's review of the Nexus 3000 gave an efficiency of about 70% at power consumption around 110 watts.
...
When the total power consumption = 111 watts and the PSU's efficiency=70% -->
Net output PSU= 70%*111 = 78 watts
...
What numbers should be corrected ?

OK, after 2 days came up with 1 correction. The net/output is 110W, so the actual input is 110/0.7 = 157W. (70% or 110W goes out the cables, 30% or 47W is keeping the psu warm.)

Now, coming back to my rig (and the chain of thought you PM'd)

My updated (to 1 decimal) measurements are:
Air intake 19.8
Case exhaust 25.5
PSU exhaust 37.5
Motherboard 25.0 (and assume for PSU input)

Power use assume 110W
My PSU efficiency assume 65%, so PSU generating additional 59W heat.

To see how realistic these are we can reverse calculate the power from the temps and airflows:

With 7V L1A on PSU pushing 14cfm (theoretical), assume 60% Verneuker's constant because of PSU static pressure, then fan is actually pushing 8.4cfm.
power = cfm * deltaT / 1.76 = 8.4 * 12.5 / 1.76 = 60W - spot on

Similarly NMB at 7V is 34.2 (theoretical), assume 97% Verneuker's constant because case has several exhaust opportunities, so fan is actually pushing 33.2cfm.
power = 33.2 * 5.7 / 1.76 = 108W - pretty close

Now need to try work this on your case: I think something missing from your working above is the mobo/peripheral power/usage: you only calculate for the CPU and PSU.

Footnote for those not familiar with Verneuker's constant: as above it is actually variable, and is defined as the ratio between the correct answer and the answer you calculate.

[SpyderCat]

Verneuker's constant

Footnote for those not familiar with Verneuker's constant: as above it is actually variable, and is defined as the ratio between the correct answer and the answer you calculate.

Dammit Chris,

You made me splash half a glass of wine over my keyboard and screen !

For those of you who don't have a decent amount of Dutch blood running through their veins:

"Verneuker" translates to something like "f**king quack" or "f**king fraud"
Never expected to read something like this over here

Now it's my time to digest your post ! (Have to get up early in the morning )

Regards, Han.

[dukla2000]

Hi Han, looking at your system you have 3 enclosures:

1 - the mobo: theoretically generating 48W of heat (VCore regulator inefficiency, RAM, Graphics & other). Intake temp 20, exhaust 25, cfm 17.93 (total of the cfm in 2 and 3 below).
So calculated power = 17.93 * 5 / 1.76 = 51W - pretty close

2 - the CPU: theoretically 62W. Intake temp 25, exhaust 41, cfm = 'flo L1A @ 9V = 18. Assume Verneuker's constant 50% then actual cfm = 9.
So calculated power = 9 * 16 /1.76 = 82W - way out

3 - the psu: theoretically 47W (assuming 70% efficient at 110W output). Intake temp 25, exhaust 34, cfm from 12V/38cfm fan at 4.7V = 14.88. Verneuker's constant 60% (same as my psu) means actually 8.93cfm.
So calculated power = 8.93 * 9 / 1.76 = 46W - pretty close

The real problem is with the CPU: Boundary element analysis gives a Verneuker's constant of 38%, rather than the 50% I assumed. This may be correct: perhaps the static pressure sucking through the AX-7 at 9V really is this high? Or the intake temp could be higher? Need to balance your model a bit better but anything we do with the CPU affects the case calculations as well.

[SpyderCat]

There is one thing bugging me:
Mike's review of the Nexus 3000 gave an efficiency of about 70% at power consumption around 110 watts.
...
When the total power consumption = 111 watts and the PSU's efficiency=70% -->
Net output PSU= 70%*111 = 78 watts
...
What numbers should be corrected ?

OK, after 2 days came up with 1 correction. The net/output is 110W, so the actual input is 110/0.7 = 157W. (70% or 110W goes out the cables, 30% or 47W is keeping the psu warm.)

The 111 Watts already included the PSU-losses.
Please check, and come back with your findings.

Now, coming back to my rig (and the chain of thought you PM'd)

My updated (to 1 decimal) measurements are:
Air intake 19.8
Case exhaust 25.5
PSU exhaust 37.5
Motherboard 25.0 (and assume for PSU input)

I use the semi-professional Dynatek 120 thermometer for most of my temp. readings. It rounds its readings to 1*C (no decimals), but I trust the readings.
On my MoBo I found that the MoBo-sensor reads consistantly higher than the case-temps taken with the Dynatek. About 3 to 4 *C higher.
Depending on where your MoBo has its sensor this might apply to you too, or not.

Power use assume 110W
My PSU efficiency assume 65%, so PSU generating additional 59W heat.

To see how realistic these are we can reverse calculate the power from the temps and airflows:

With 7V L1A on PSU pushing 14cfm (theoretical), assume 60% Verneuker's constant because of PSU static pressure, then fan is actually pushing 8.4cfm.
power = cfm * deltaT / 1.76 = 8.4 * 12.5 / 1.76 = 60W - spot on

Nonsense ! Dukla's rig can be seen here
As I have argued before, your case has positive pressure so I assume your theoretical CFM for the Panaflo to be 16, at least higher than 14 --> To make your arithmatic work out you need to lower your "Verneuker's constant" to 52%.

Can we please replace your phrase "Verneuker's constant" by the word "efficiency" ? It would make it easier for others to understand this thread.

Similarly NMB at 7V is 34.2 (theoretical), assume 97% Verneuker's constant because case has several exhaust opportunities, so fan is actually pushing 33.2cfm.
power = 33.2 * 5.7 / 1.76 = 108W - pretty close

Come on Chris, here you are really joking. 97% efficiency for a 120mm fan that has to push it's air through a few tiny holes?
Get real ! When you do enough "assuming" you can always make the numbers fit, but this is ridicules.

Now need to try work this on your case: I think something missing from your working above is the mobo/peripheral power/usage: you only calculate for the CPU and PSU.

I'm not aware (even now) of any omissions.
It IS very difficult to estimate a difference in airflow between two blowholes when both blowholes don't have the same temperature.

I can really get along with you Chris, but in discussions like these I will attack you like I would my worse enemy!
You're a sloppy debater.

Now I'll try to take apart your other post.

Regards, Han

[SpyderCat]

Hi Han, looking at your system you have 3 enclosures:

1 - the mobo: theoretically generating 48W of heat (VCore regulator inefficiency, RAM, Graphics & other). Intake temp 20, exhaust 25, cfm 17.93 (total of the cfm in 2 and 3 below).
So calculated power = 17.93 * 5 / 1.76 = 51W - pretty close

2 - the CPU: theoretically 62W. Intake temp 25, exhaust 41, cfm = 'flo L1A @ 9V = 18. Assume Verneuker's constant 50% then actual cfm = 9.
So calculated power = 9 * 16 /1.76 = 82W - way out

3 - the psu: theoretically 47W (assuming 70% efficient at 110W output). Intake temp 25, exhaust 34, cfm from 12V/38cfm fan at 4.7V = 14.88. Verneuker's constant 60% (same as my psu) means actually 8.93cfm.
So calculated power = 8.93 * 9 / 1.76 = 46W - pretty close

The real problem is with the CPU: Boundary element analysis gives a Verneuker's constant of 38%, rather than the 50% I assumed. This may be correct: perhaps the static pressure sucking through the AX-7 at 9V really is this high? Or the intake temp could be higher? Need to balance your model a bit better but anything we do with the CPU affects the case calculations as well.

As already mentioned to you before in a PM:
I found that running CPU-Burn gives considerable higher thermal loads folding!
Today I ran a few more tests, both running Folding@Home and CPU-Burn



Clearly you can see that the CPU runs much hotter with CPU-Burn than with Folding@Home. As both programs ran with the Case/CPU fan @ 12 volts we can try to calculate the power drawn by an AMD Thoroughbred while Folding.
AMD states max power for Thoroughbred(B) 2400+ =68.3 Watts
Using the formula: CFM=(Watts*1.76)/Delta(T) , and assuming that running CPU-Burn equals the max. thermal output for a AMD CPU:
CFM = (68.3*1.76) / 12.5 = 9.6 CFM --->
In my setup the Panaflo runs at a efficiency of 40% @ 12 volts
You might expect this efficiency to rise a little bit at lower voltages as air resistance climbs exponentially with rising speed.

Using the same airflow to cool the CPU while running Folding@Home:
Watts = (CFM/1.76)*Delta(T) = (9.6/1.76)*10 = 54.6 Watts.

Please prove to me that this CPU uses more than 54.6 Watts while Folding.

Okay, now we use this 54.6 Watts to calculate the CFM of the Panaflo @ 9 volts.
Delta(T) @ 9 volts = 16*C
CFM = (54.6*1.76) / 16 = 6.0 CFM

It's getting time to catch some sleep.
I'll continue tomorrow.

Regards, Han.

[dukla2000]

Get real ! When you do enough "assuming" you can always make the numbers fit, but this is ridicules.

Sure, but as per my first post on this thread the actual science requires, from what I have read, knowledge of your System Impedance Curve, which in turn requires either a Wind Tunnel or a double chamber pressure test environment. In the absence of this test equipment we are guessing.

The 111 Watts already included the PSU-losses.
Please check, and come back with your findings.

I referred to the first table on this page under test results: 90W load at 130W AC power = 68.5% efficiency, 150W load at 207W AC power = 72.5% efficiency. So figured as you were using a 70% efficiency you were talking about a (111W) load between 90W and 150W.

97% efficiency for a 120mm fan that has to push it's air through a few tiny holes?

Again depends on the Impedance. However in the absence of that knowledge I compare the intake and exhaust areas. The intake is radius 58mm, for exhaust I have (as PM'd) 2 rows of 'stamped holes' on each side panel: 27 holes per row approx 3mm*8mm per hole in addition to the 2 stamped grills on the back panel. This has to exhaust whatever does not go through the psu. In total 10,568mm2 intake area, 7,010mm2 (assuming the rear stamped grills are 50% free) + psu for exhaust. IF the fan could pump against the static pressure without dropping cfm, the result is the exhaust air velocity would be slightly higher than intake velocity (depends where we settle on flow through the psu of course). Indeed there is static pressure, the fan efficiency is reduced, pick a number?

The bottom line is I am prepared to accept the equation
cfm = (1.76 * P) / delta T (P in Watts, T in centigrade)
as applicable for heat dissipation. I believe it originates in thermodynamics (not my field) and is certainly used by Sunon, Papst and others in their literature. The limitation, IMHO, is we have no idea of our Impedance/Static pressure and thus our actual cfm, but we absolutely have to assume efficiencies that conform to the formula.

If we can agree on the power consumption, then perhaps it would be more valuable to use the formula to calculate actual cfm and hence efficiency. Methinks the bottom line would be my psu fan is 52 or 60% efficient, my case fan is 97% efficient?

[dukla2000]

Hi Han

We were writing our posts above at the same time, but I think similar angles. Basically it is clearer to calculate cfm efficiency using that equation and present it as the result, rather than (as I first did) 'assuming' cfm efficiency to 'prove' suitable results .

Your Panaflo efficiency of 40% at 12V is the valid way to present. (BTW the 62W I assumed was the AMD quoted 'typical'.) And the 6.0 cfm at 9V appears valid to me => 33.3% efficient. There are 2 immediate weaknesses, is CPUBurn running the CPU at maximum load and also the A7V333 seems to run VCore 0.05V higher than it is set (according to MBM). However these 2 unknowns will tend to balance out: if the VCore really is 1.70V instead of 1.65 then the maximum power is 72.4W and so there is some room for CPUBurn not to disipate the maximum sustained power and still assume 68.3W.

The other issue is the overall PC power consumption. I used 110W (net from the psu) as you mentioned 111W earlier, I have seen 100 or 110W mentioned elsewhere on SPCR, and my own guess is 134W
12W mobo (somewhere in the AMD Builders Guide)
62W CPU
16W VCore regulation inefficiency while regulating 62W net
10W Vid card
10W RAM
13W hdd (I measured my 12V & 5V lines)
6W optical drives & floppy at idle
5W other PCI cards at idle & fans
A while back (not my exact current config) I measured the voltage across a 1 ohm resistor in the AC input (I think Mike posted this technique, and it's limitations, somewhere) and have noted (at that time) I was running 122W.

[SpyderCat]

Dammit !
Chris, I worked on a long post for at least 2 hours, only to get a "invalid session" message when I hit the "submit" button.

I'll try to recreate the message, this time in notepad !

To all: worked on a long post? Copy/Paste a backup before sending your message !

[dukla2000]

Hey - if it was as critical as your first post last night then take it as a message from above

[jojo4u]

Dammit !
Chris, I worked on a long post for at least 2 hours, only to get a "invalid session" message when I hit the "submit" button.
!

Backbutton did not help?

[jojo4u]

I already thought about the impact of impendance to fan performance some time ago:

I was interested in the fan impedance in "fire and ice"(http://www.intel.com/idf/us/fall2002/pr ... I130PS.pdf) page 4-6. As my reference, I took the slowest Ys- Tech 80x80x25 12V ball bearing.
http://www.yensun.com.tw/FanHtml/FD8125.htm
It's maximum airflow is 27cfm. That makes ~46m³/h.
Now the hole venting area of a psu is 80x80mm. That makes 0,0064m². This is the usable area for the airflow.
Now we have to calculate the velocity. Imagine a cylindre with 1m length and 0,0064m² foot size. In 1 second there fits 0,0064m³ in. If we exchange this amount every second, there are fitting 0,0064m³ * 60s * 60m = 23m³ per hour. Result: with a velocity of 1m/s we have a airflow of 23m³/h. With a velocity of 2m/ s we have our wanted 46m³/h.
Now with a velocity of 2m/s and 56% FAR there is a pressure drop of estimated 0,03 inH20 in the Intel paper, which corresponds to 0,75mmH2O.
When this Ys-Tech fan encounters a pressure of 0,75mmH2O, it's airflow is around 7cfm. Quite a large drop.
Am I right with my calculations? When i tried a smaller airflow area, the airflow dropped under zero...

Sorry for grammar mistakes, Englisch is not my native language.

[dukla2000]

Hi jojo4u - that Intel link wont work for me. Browsing round trying to get to the Fall 2002 papers I get to the stage where I can sign up for $199 for the full set! Does the link work for you? (Are you an IDF member?)

[SpyderCat]

Okay Chris,
Here a second try:

... Basically it is clearer to calculate cfm efficiency using that equation and present it as the result, rather than (as I first did) 'assuming' cfm efficiency to 'prove' suitable results .

Your Panaflo efficiency of 40% at 12V is the valid way to present. (BTW the 62W I assumed was the AMD quoted 'typical'.) And the 6.0 cfm at 9V appears valid to me => 33.3% efficient. There are 2 immediate weaknesses, is CPUBurn running the CPU at maximum load and also the A7V333 seems to run VCore 0.05V higher than it is set (according to MBM). However these 2 unknowns will tend to balance out: if the VCore really is 1.70V instead of 1.65 then the maximum power is 72.4W and so there is some room for CPUBurn not to disipate the maximum sustained power and still assume 68.3W.

I have the same problem with my MoBo you have: VCore @ 1.70 volts, and no way to lower it.
Today I found at benchtest.com :
#1, a power calculator, and
#2, according to the developer of CBUBurn, running CPUBurn constitutes a thermal load equal to 88% of the max. thermal load AMD provides for this CPU.
For a Tbred(B) @ 2000 MHz with 1.7 v VCore: Max Wattage= 72.5 Watts; Wattage running CPUBurn = 63.8 Watts.
I take this info to be true, so have to redo yesterdays calcs, which part follows here:

Clearly you can see that the CPU runs much hotter with CPU-Burn than with Folding@Home. As both programs ran with the Case/CPU fan @ 12 volts we can try to calculate the power drawn by an AMD Thoroughbred while Folding.
From benchtest.com we learn that running CPU-Burn represents a load of 63.8 Watts on my CPU.
Using the formula: CFM=(Watts*1.76)/Δ(T):
CFM = (63.8*1.76) / 12.5 = 9.0 CFM --->
In my setup the Panaflo runs at a efficiency of 9/24 = 37.5% @ 12 volts
You might expect this efficiency to rise a little bit at lower voltages as air resistance climbs exponentially with rising speed. Only, here the case-fan has to compete with another fan (in the PSU) for the same air. It is working against a pressure difference, so you can't expect it produce as much CFM at reduced RPM, compared to a single fan at reduced RPM.

Using the same airflow to cool the CPU while running Folding@Home:
Watts = (CFM/1.76)*Δ(T) = (9.0/1.76)*10 = 51.1 Watts.

Okay, now we use this 51.1 Watts to calculate the CFM of the Panaflo @ 9 volts.
Δ(T) @ 9 volts = 16*C
CFM = (51.1*1.76) / 16 = 5.6 CFM
Theoretical CFM @ 9 volts = 18 CFM
Efficiency = ( 5.6 / 18 ) = 31.1%

The other issue is the overall PC power consumption. I used 110W (net from the psu) as you mentioned 111W earlier, I have seen 100 or 110W mentioned elsewhere on SPCR, and my own guess is 134W

The 111 Watts I mentioned was AC-power !

12W mobo (somewhere in the AMD Builders Guide)
62W CPU
16W VCore regulation inefficiency while regulating 62W net
10W Vid card
10W RAM
13W hdd (I measured my 12V & 5V lines)
6W optical drives & floppy at idle
5W other PCI cards at idle & fans
A while back (not my exact current config) I measured the voltage across a 1 ohm resistor in the AC input (I think Mike posted this technique, and it's limitations, somewhere) and have noted (at that time) I was running 122W.

12 Watts mobo (somewhere in the AMD Builders Guide)
51.1 Watts CPU (as calculated above)
13 Watts VCore regulation inefficiency while regulating 51.1W net
15 Watts Vid card (still no data, but my heatsink = 63*C)
6 Watts RAM
6.2W hdd (source: IBM in a document on the 120 GB - GXP120 )
6W optical drives & floppy at idle
5W other PCI cards at idle & fans
--------------
114.3 Watts Total

Of this 114 Watts, the CPU's heat doesn't contribute to the case temps, it's heat is dumped outside the case. This leaves the other components to be accountable for the rise in temperature from "ambient" to "case temperature".
In the list above these components amount to 63.2 watts
Previously I measured a Δ(T)(case) of 5*C.
As my thermometer has a resolution of 1*C, in reality it might have been 5.5*C
Now let's calculate (again) how much CFM is needed to account for a Δ(T)(case) of 5*C, 5.5*C and 6*C.
Using the formula: CFM=(Watts*1.76)/Δ(T):
CFM= (63.2*1.76) / 5.0 = 22.25 CFM
CFM= (63.2*1.76) / 5.5 = 20.24 CFM
CFM= (63.2*1.76) / 6.0 = 18.54 CFM

I find it very difficult to "feel" how much air is blown from PSU when compared to the blowhole in the case, as the temperatures of the air from both holes is different. Even when I take the most favourable outcome of 18.54 CFM to do the calcs, it just doesn't make sense!

18.54 CFM minus ( 5.6 CFM from the case/cpu fan ) = 12.94 CFM for the PSU, or more than two times the air blown from the case.
Also, when 12.94 CFM is flowing through the PSU, then why is its Δ(T) 9*C ? This would indicate losses in the PSU of some 66.5 watts, and this again would indicate a DC power used by the system of 154 Watts.

I would estimate it at about half the air blown from the case.
I can't come to another conclusion than that the sum of:

mobo
VCore regulation inefficiency while regulating 51.1W net
Vid card (still no data, but my heatsink = 63*C)
RAM
6.2 Watts(source: IBM in a document on the 120 GB - GXP120 )
optical drives & floppy at idle
other PCI cards at idle & fans

Is much smaller than 63.2 watts.


I keep thinking and try to solve this puzzle.



Regards, Han.

[dukla2000]

it just doesn't make sense!

Agreed, but it was all looking logical & correct up till then!

3 thoughts:
- your 'the power must be less' and my '97% efficient fan' are symptoms of the same problem
- typically heat radiated from the case is ignored. Makes sense especially when/if we have then lined with acoustic damping, but I can dig out some formulae I have for this and see how large this could be (assuming something, hopefully metal, I have a thermal value for)
- I have hatched a cunning plan to measure the AC power 100% accurately (without any test equipment!) - will take me a few days to execute, watch this space!

[SpyderCat]

Hi jojo4u,

I can't discus a whole lot with you, as I couldn't get to the Intel PDF you pointed to. If you happen to have a copy on your disk, mail it to me please?

... Now we have to calculate the velocity. Imagine a cylindre with 1m length and 0,0064m² foot size. In 1 second there fits 0,0064m³ in. If we exchange this amount every second, there are fitting 0,0064m³ * 60s * 60m = 23m³ per hour. Result: with a velocity of 1m/s we have a airflow of 23m³/h. With a velocity of 2m/ s we have our wanted 46m³/h.
Now with a velocity of 2m/s and 56% FAR there is a pressure drop of estimated 0,03 inH20 in the Intel paper, which corresponds to 0,75mmH2O.
When this Ys-Tech fan encounters a pressure of 0,75mmH2O, it's airflow is around 7cfm. Quite a large drop.
Am I right with my calculations? When i tried a smaller airflow area, the airflow dropped under zero...

You probably try to get an airspeed because the Intel-paper is set up that way. Until you start talking about pressure-drops all seems correct.
The pressure drop you do talk about ( 7.5 mm H2O ) is almost the same as we arrived at using the manufacturers graphs. I had numbers more like 6 mm H2O.
An airflow dropping under zero ? Do you mean it reverses direction?

Regards, Han.

[SpyderCat]

Hi Chris,

I completely missed this part of your post yesterday.
I have some doubts about 122 Watts number, but Hey, I didn't bring it up and it fits me !!

....
A while back (not my exact current config) I measured the voltage across a 1 ohm resistor in the AC

input (I think Mike posted this technique, and it's limitations, somewhere) and have noted (at that

time) I was running 122W.

Let's assume 140 Watts AC-power. (Fits me even better, and less extreme that your number. )

Based on Mike's review of the Nexus PSU, I assume an efficiency of 66% for the PSU. --->
DC power from PSU = 140 * 66% = 92.4 Watts, and
Losses in the PSU = 140 * 34% = 47.6 Watts.

CFM calculation for the whole case:
Total DC-power minus CPU-power = 92.4 Watts - 51.1 Watts = 41.3 Watts
This 41.3 Watts is the power accountable for the rise in temperature from "ambient" to "case temperature".

Previously I measured a &#916(T)(case) of 5*C.
As my thermometer has a resolution of 1*C, in reality it might have been 5.5*C
Now let's calculate (again) how much CFM is needed to account for a &#916(T)(case) of 5*C, 5.5*C and 6*C.

Using the formula: CFM=(Watts*1.76)/&#916(T):
CFM= (41.3*1.76) / 5.0 = 14.53 CFM
CFM= (41.3*1.76) / 5.5 = 13.22 CFM
CFM= (41.3*1.76) / 6.0 = 12.11 CFM

14.53 CFM minus ( 5.6 CFM from the case/cpu fan ) = 8.93 CFM for the PSU

Now a check with this 8.93 CFM to see if it fits the PSU losses:
Using power = (CFM*&#916(T))/1.76 and &#916(T)(PSU)= 9 degrees C
power = (8.93*9)/1.76 = 45.66 Watts PRETTY CLOSE

Now we have to redistribute these 41.3 Watts (=power available for other components) over the components:

6.2 Watts hdd (source: IBM in a document on the 120 GB - GXP120 )
12 Watts mobo (somewhere in the AMD Builders Guide)
0 Watts VCore regulation (included in the MoBo power, I guess)
12 Watts Vid card (still no data, but my heatsink = 63*C)
5.1 Watts RAM
1W optical drives & floppy at idle
5W other PCI cards at idle & fans
--------------
41.3 Watts Total


Let's see if this part fits the puzzle:

The 8012H fan in the Nexus PSU is 38 CFM @ 12 volts.
Lets assume a 60% efficiency for the PSU fan. --->
CFM @ 12 volts = 38 * 60% = 22.8 CFM
Previously we found 8.93 CFM going through the PSU.
Volts needed to produce 8.93 CFM = (8.93 / 22.8 ) * 12 = 4.7 volts (Now, were did I see this number before .... )

By the looks of it the puzzle is almost done.
We need to find confirmation for a whole lot assumptions we made, but hey, I think this is a reasonable representation of my rig, and I feel I've acquired a good intuition for estimating CFM's in real life situations.

Comments ?

Perhaps now it's time to take your rig apart

Regards, Han.

[jojo4u]

Sorry for the late posting. But i was out for the whole day. I found the intel paper on my disk. Should be available at www.computer-networking.de/~kluge/DPI130PS.pdf I can't comment any more. Its 1h after midnight here.

[SpyderCat]

Sorry for the late posting. But i was out for the whole day. I found the intel paper on my disk. Should be available at www.computer-networking.de/~kluge/DPI130PS.pdf. I can't comment any more. Its 1h after midnight here.

Hi Jojo4u,

Thanks for the PDF.
Please remove the last dot from your link, it works better that way
Time to study the pdf.

Regards, Han.

[dukla2000]

I completely missed this part of your post yesterday.
I have some doubts about 122 Watts number, but Hey, I didn't bring it up and it fits me !!

....
A while back (not my exact current config) I measured the voltage across a 1 ohm resistor in the AC

input (I think Mike posted this technique, and it's limitations, somewhere) and have noted (at that

time) I was running 122W.

Let's assume 140 Watts AC-power. (Fits me even better, and less extreme that your number. )

OK - the cunning plan was that my house has an electricity meter that reads KWh. Actually it reads to the nearest W. So managed to get some time when I could power off everything in the house except 2 PC towers and the smoke detectors (run trickle chargers which I choose to ignore). The PCs are pretty much similar in my paper power calculations, both are dominated by Athlon CPUs - a (maximum) 72W Thunderbird 1.4GHz and a (maximum) 68W XP2700. Both were folding for the duration, both are Via mobos, single hard drives, 'old tech' vid cards, NIC and not a lot else active. They also each have another PCI card (TV in 1, ADSL modem the other), keyboards & mice (both optical) were of course powered through the test. The Thunderbird box could draw slightly more power: the CPU itself has a higher rating and it has 2 sticks of SDRAM as opposed to 1 DDR with the XP2700. The monitors were powered off, so pretty much a measure of the gross tower power only.

The answer is - 120W.

Ran for 4 hours, total power consumed 957Wh. I measured every hour, the hourly readings were 240W, 235W, 245W and 237W, so pretty constant and (as I only discovered how to read the unit Watts after the start) consistent with the average.

So if my typical tower runs 120W (gross AC power), then the net must be a lot lower. It seems to be general practice that ATX psu are less efficient at the low end of their capacity, so assuming 60% efficiency means the DC usage is 72W, and I am sure it is probably lower (efficiency and DC usage). Which completely blows away the paper calculation of 130W DC, but is also consistent with my old measurement on a resistor!

But also explains how my fan was so efficient (mainly as it wasn't shifting so much heat!).

Now, as per my current sig I have also revamped my case since we started this thread: I no longer have a fan in the psu, the front fan has changed and both remaining fans are running PWM. But may take some time out to reconsider the thread and what we are left with. (Mainly much lower power than I expected, and also pretty low fan efficiencies I think.)

[SpyderCat]

OK - the cunning plan was that my house has an electricity meter that reads KWh.
......
The answer is - 120W.
.....
So if my typical tower runs 120W (gross AC power), then the net must be a lot lower. It seems to be general practice that ATX

psu are less efficient at the low end of their capacity, so assuming 60% efficiency means the DC usage is 72W, and I am sure it is probably lower (efficiency and DC usage). Which completely blows away the paper calculation of 130W DC, but is also consistent with my old measurement on a resistor!

But also explains how my fan was so efficient (mainly as it wasn't shifting so much heat!).

Now, as per my current sig I have also revamped my case since we started this thread: I no longer have a fan in the psu, the

front fan has changed and both remaining fans are running PWM. But may take some time out to reconsider the thread and what we are left with. (Mainly much lower power than I expected, and also pretty low fan efficiencies I think.)

Hi Chris,

In the meantime I've undertaken some action too in order to find some confirmation of our assumptions.
From my job I could borrow a fairly professional multimeter, with the ability to read Amperes AC. Hooked up the multimeter and took readings in two configurations, running a selection of programs.
One set running the CPU overclocked @ 2158 MHz, and one set running at stock speed = 2000 MHz.

CPU= AMD 2400+ @ 13*166 = 2158 MHz

Code: Select all

Task                        Volts    Amps         Watts AC

Machine switched OFF         227    .04              9
Idle                         227    .40             91
3DMark 2000 (Demo-mode)      227    .44-.46        102
3DMark 2000 (Benchmarking)   227    .53-.55        123
DOS-prog                     227    .51            116
F@H                          227    .54-.55        124
CPU-Burn                     227    .63-.64        144
 

CPU= AMD 2400+ @ 15*133 = 2000 MHz

Code: Select all

Task                        Volts    Amps         Watts AC

Machine switched OFF         228    .04              9
Idle                         228    .38-.39         88
3DMark 2000 (Demo-mode)      228    .43-.45        100
3DMark 2000 (Benchmarking)   228    .50-.52        117
DOS-prog                     228    .49-.50        113
F@H                          228    .51-.52        117
CPU-Burn                     228    .60-.61        138

Did you run you rigs OC-ed or at stock speed while performing your test ?
I assume you ran Folding@Home with your CPU @ 2158 MHz.
In de same circumstances I arrived at an AC-power of 124 Watts, so, fortified by your findings, I take this reading to be precise and true for my rig.
The whole previous discussion was based on findings with my rig running at stock speed, and in order not to confuse the discussion I will only use the AC-power readings of my rig at stock speed.

CPU = AMD 2400+ @ 2000 MHz.
AC power = 117 Watts while folding.

I made an Excel spreadsheet with the breakdown of measurements, and conclusions drawn from it.
Chris, I’ll send you this spreadsheet by E-mail, and I will send it to others requesting it.

Conclusions from the spreadsheet:
While running Folding@Home, giving a 100% CPU load, it is likely that:

Total airflow through the case = 9 CFM
Airflow through the PSU = 2.3 CFM
Airflow through the CPU-heatsink = 6.7 CFM
Efficiency of the Nexus 300 Watts PSU @ 117 Watts AC load = 71.5%
Efficiency of the 80mm Panaflo A12L, @ 9 volts, sucking air through a shrouded Thermalright AX-7 heatsink = 37%

Before I continue I’ll let you attack the spreadsheet.

A question for the people with more knowledge regarding PSUs ( AnthonyK ? ):
Is it possible that the Nexus 300 Watts PSU works at an efficiency of 71.5% while taking 117 Watts AC @ 228 volts ?

EDIT: May 20 2003
I just found this posting in another thread:

I just got the efficiency figures for the Seasonic SS-350FS (active PFC) power supply. Figures are efficiency at full load as offered to me by the manufacturer:

70~71% for 115v 60 Hz
74~75% for 230v 50Hz
.....
Halcyon

By the looks of it there might be an efficiency gain of 4% by operating your PSU @ 230 volts.
This makes my efficiency number of 71.5% for the Nexus a lot more likely.

Regards, Han.

[dukla2000]

Did you run you rigs OC-ed or at stock speed while performing your test ?

My ASUS/XP is always 13*166. My Thunderbird would never overclock (or at least I guess AMD did that at the factory )

That's interesting that you get differences in power between 2000 & 2158: according to AMD the XP2400, XP2600 (266 & 333) and XP2700 are all the same thermal power. Which always struck me as dubious (a bit of something for nothing!). Some may be memory, but there sure seems to be some that is CPU. Your readings make more sense to me.

[jafb2000]

Generally PCs create less heat that people realise.
To find the actual power consumption you can use a Clamp Meter,
which does RMS, to ensure you get correct current measurement.

Obviously that will not help dynamic variation, you can buy
refurbished "credit meters" but obvious safety risks here.
A clamp meter is a partly non-invasive measurement device,
I say partly because you must clamp the live only wire. That
is best done with a similar attention to safety re mains voltages.


As a general guide 1500W needs 300cfm of cooling.
A PC is around 1/10th of that so you need 30cfm of cooling.

That is realised cfm, and it is usual to de-rate your fans by 50%.
I should add the same derating is already in place - eg, that 37cfm
fan you are replacing is performing as a 18cfm fan in-situ.


To improve cooling, my advice would be via ducting:
o Fan to pull hot air thro a duct direct to outside
o Fan to push cool air thro a duct from outside to a specific area

The benefit can be seen by considering simple CPU coolers.
Impingement CPU coolers are the most common form, they involve
a fan blowing at heatsink fins & plate - plate at 90-degrees to airflow.
This involves an air direction change, creating static pressure and
also due to close proximity of plate to fan blades, turbulence. In
combination, one can say impingement coolers are usually noisy
and present a high static resistance reducing net cfm airflow.

Blow-thro CPU coolers by comparison blow air thro fins, usually
with the fan vertical at one end, a duct on top to make a windtunnel.
These are commonly seen with the Intel Xeon "windtunnel" coolers.
Inherently these are quieter, less static resistance so less cfm needed.

Despite this, both types of CPU cooler suffer the same basic problem:
o They recirculate the same air over-&-over within the case
o Recirculation can be as high as 70%
o So your proudly boasting 37cfm fan is actually 11cfm at best
o At best, because static resistance probably reduces it another 30%
o Thus your heatsink has an effective ~8cfm of cool air

What this means is, if you can get 100% cool air to the CPU cooler,
you can reduce the cfm / noise of the CPU cooler fan correspondingly.

The QuietPC FCPGA2 cooler handles ~40W at <21dB(A) fitted, with
an 8cfm 50x50x10mm fan - itself capable of little static pressure. It
does this as a blow thro design, yet still suffers air recirculation.

Ducting resolves the noise / cfm trap of hot CPUs, and thermal
solutions going forward will require this to achieve any quiet spec.

When you replace a PSU fan with a quieter fan, you require the
case fans to handle it instead. The ATX Spec requires ~37cfm to cool
envisaged processors, without case fan, and use that hot heated case
air to cool the PSU additionally. Far less than 37cfm is required to cool
a PSU, yet 37cfm from an 80mm fan in a restrictive PSU is ~35dB(A).
Fitting an 18cfm case fan & 18cfm PSU fan gets you similar cfm in total
(probably slightly more re PSU resistance), but a combined ~20dB(A).

The cfm numbers aren't so critical.
Use of that air is.

It should be remembered that the formuli listed are only a simple
approximation - they do not take account of point heat sources. They
assume a large-body source, and not hot smaller objects. Even so,
unless you plan on extremely low cfm this is not really material.

In a packed rack of 48 dual CPU PCs it does get important, and the
individual case thermal profile (like an infra-red picture) does matter.

So don't go OTT on the watts/cfm numbers, once you have the right
ballpark of cfm/noise - try to improve on the routing of that cfm inside.

Ducting is the most boring part of silent product design, but
it can have 2 huge benefits in terms of production systems:
o Doesn't cost much - relative to fans in terms of beans
o Can act as an auditory baffle as well as cfm-directive device

Every 180-degree you turn noise, you knock 6dB(A) off the noise.
You also cost cfm, so as always there is a balance. Fortunately PCs
have both motherboard & CPU temperature sensors - and charting.

Soundproofing only works between your ear & the noise source.
Soundproofing inside a PC with fans on the case structure puts the
soundproofing after the noise source as far as the ear is concerned.

For AMD processors, the best cooling benefit you can do is get cool
air direct to the intake of the CPU fan - from case-rear or side. It only
needs a tiny amount of cool air to have a huge benefit, just 8cfm of
cool air has a considerable impact in offsetting air-recirculation.

Often worth cardboarding case sides before swiss-cheesing them!
Try a few places, and minimal cfm fans before brute air-force.
--
Dorothy Bradbury
www.stores.ebay.co.uk/panaflofan (Ebay)
http://homepage.ntlworld.com/dorothy.br ... anaflo.htm (Direct Prices)