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'[EE]:: Calculation of pneumatic resistance'
2007\05\19@055744 by Russell McMahon

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I'm seeking to find a formula which will allow me to very
approximately establish the pneumatic resistance of smooth walled
ducts at modest flow rates. I know Gargoyle knows but I can't convince
it to tell me. I'm sure more than one of the people receiving this
also know.

This is essentially a HVAC application.
Flow rates are in the few litres per second range.
Temperatures are in the 0 - 30 C range.
Wall material will ultimately be something smooth which may be a metal
on some or all faces.

If replying to this on a list I'd be obliged if you'd also copy any
useful answer to me directly as list servers sometimes swallow
responses for a while.



regards

       Russell McMahon
       Use whatever email address you already have for me, or else
use.
      CHANGE Q'S TO R'S  -->  spam_OUTquvalTakeThisOuTspampaqadise.net.nz   <--- CHANGE Q'S
TO R'S




2007\05\19@072151 by Russell McMahon

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Aaaagh !!!
Definitely more than I wanted to know I didn't know.
An excellent start though.



       Russell

_________________________
       
{Quote hidden}

2007\05\19@084259 by Gerhard Fiedler

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Russell McMahon wrote:

> I'm seeking to find a formula which will allow me to very approximately
> establish the pneumatic resistance of smooth walled ducts at modest flow
> rates. I know Gargoyle knows but I can't convince it to tell me. I'm
> sure more than one of the people receiving this also know.
>
> This is essentially a HVAC application. Flow rates are in the few litres
> per second range. Temperatures are in the 0 - 30 C range. Wall material
> will ultimately be something smooth which may be a metal on some or all
> faces.

I have a paper source (the in Germany well-known Recknagel/Sprenger
"Taschenbuch für Heizung+Klimatechnik"). It contains a few empirical
diagrams, so for certain defined situations, I could look up the values.
This may then enable you to extend it further, as long as only parameters
change that have a known relationship (length, for example).

Gerhard

2007\05\19@092010 by Russell McMahon

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Thanks for the offer - it may be a bit hard to model well and there
are some variable variables :-) - but to give you an idea.

There are two identical ducts in series with an essentially lossless
section in between. .

Per duct:

Duct is a very wide and shallow rectangle (a slot in cross section )

- Duct length is (only) 400mm
- Duct width is 350mm.
- Air flow is about 10l / second

Desired pressure is such that a fan rated in the 5 to 10 Watt range
will maintain the given airflow with the smallest possible duct height
(and therefore cross section).
I'd guesstimate that 5000 to 1000 Pa range would be OK. (about 1/3 to
2/3 psi)

In fact just about everything is potentially variable but that's a
typical configuration. I think that a bit of playing is going to teach
me much. An added factor to make it all more fun is that the air
temperature drops or rises by around 20C across the length of each
short duct :-).

Odds are that's far too variable a set of variables to expect any
useful results but you are most welcome to comment if appropriate.

My friend Ken Mardle provided this reference

       http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html

which looks like it will be highly useful once I take the time to come
to grips with its nuances.



       Russell



I have a paper source (the in Germany well-known Recknagel/Sprenger
"Taschenbuch für Heizung+Klimatechnik"). It contains a few empirical
diagrams, so for certain defined situations, I could look up the
values.
This may then enable you to extend it further, as long as only
parameters
change that have a known relationship (length, for example).

Gerhard

2007\05\19@094741 by Goflo

picon face
Dunno if Grainger catalogs exist in NZ but it contains useful
discussion of just such in the appendices.

regards, Jack

---- Russell McMahon <.....apptechKILLspamspam@spam@paradise.net.nz> wrote:
{Quote hidden}

> --

2007\05\19@103823 by Sean Breheny

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Hi Russell,

I can't find something for air (in a quick search) but the equivalent
plumbing term is "head loss per foot" in pipe. (Head is pressure in
units of (vertical) feet of water, I think) This page has tables for
copper and PVC tubing:

http://www.plumbingsupply.com/pluminfo.html

Sean


On 5/19/07, Russell McMahon <.....apptechKILLspamspam.....paradise.net.nz> wrote:
{Quote hidden}

> -

2007\05\19@150743 by Gerhard Fiedler

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Russell McMahon wrote:

> There are two identical ducts in series with an essentially lossless
> section in between. .
>
> Per duct:
>
> Duct is a very wide and shallow rectangle (a slot in cross section )
>
> - Duct length is (only) 400mm
> - Duct width is 350mm.

I think most formulas for duct resistance will fail here. If I understand
fluid dynamics correctly, this is a length/width ratio that creates a flow
situation that's quite different from the one that the normal resistance
formulas assume (they assume much longer than wide). Normal formulas mostly
assume a laminar flow profile, which can only be expected to start after a
length of about 10 diameters.

> - Air flow is about 10l / second

Here is a formula for round pipes:

(1) delta p = lambda * (l/d) * (rho/2) * w^2

 delta p: pressure difference [N/m^2]
 lambda: friction coefficient [1]
 l, d: length, diameter [m]
 rho: specific mass [kg/m^3]
 w: flow speed [m/s]

For rectangular ducts, an equivalent diameter can be calculated:

(2) d equ = 2 * a * b / (a + b)

 d equ: equivalent diameter [m]
 a, b: sides of the rectangle [m]

You now can express d in (1) in terms of a and b from (2), where a is
constant and b is what you want. You have delta p and l given. rho is known
(at least the range, depending on your temperature range). w can be
expressed as a function of the volume flow (given) and cross section
(composed of the given a and the result b). That leaves us lambda as only
unknown.

lambda (probably something like "friction coefficient" in English) depends
on the flow profile (laminar, turbulent -- in your case the latter, given
your short length), the smoothness of the inner surface and the Reynolds
number.

For turbulent flow and ideally smooth surfaces is

(3) lambda = 0.3164 * Re^(-0.25)

(If you want to go into more detail here and possibly take the surface
smoothness into account, I have other formulas for this. But the principle
remains the same; you'd only have one more variable in here and much more
complex formulas.)

The Reynolds number is

(4) Re = w * d / nu

 w: flow speed [m/s]
 d: (equivalent) diameter [m]
 nu: "kinematische Zãhigkeit" (don't know the English name) [m^2/s]

Now we can insert lambda from (3) in (1) and substitute Re with (4). In
(4), we can substitute w and d like before (with a, b and the given volume
flow), and nu is ~13 at 0°C, ~15 at 20°C, ~17 at 40°C (for air at 1 bar, in
10^-6 m^2/s).

I didn't do all the substitutions, but you'll end up with a formula that
you can solve for b (height of the duct), dependent on only known values.
You'll get even a notion how much the various parameters influence the
result; this could potentially be more interesting than the actual result
itself.

Gerhard

2007\05\19@151454 by Gerhard Fiedler

picon face
Russell McMahon wrote:

> My friend Ken Mardle provided this reference
>
>         www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html
>
> which looks like it will be highly useful once I take the time to come
> to grips with its nuances.

This talks about the same calculations I just wrote about. The page took a
while to load here, so I only saw this now, after I already sent the other
message. In this page you'll find some more detailed explanations (and the
correct terms in English :) of the formulas I used.

Gerhard

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