low Re air in pipe: duct chart or Darcy formula?

[NoHoCa]

I'm checking head losses on an air header that delivers about 1400 cfm at 10.5 psig and 200F (average in pipe) from a blower. The main header is 20" steel pipe, which I think is plenty oversized for minimal friction losses. So I expect the losses to be perhaps only a fraction of an inch. The Reynolds number is less than 100, so apparently quite laminar.

I've been told that the "Friction Loss in Standard Duct" table (MERM, Lindeburg, Fans & Duct chapter) is the way to go. But I'm suspicious because it's intended for building ducts, not pressurized process piping.

So I tried the Darcy equation. But the problem is that at very low Reynolds numbers, the friction factor f=64/Re. Instead of a fractional factor (and low head loss), I get an "f" of about 1 (with Re=63). That translates into a head loss of about 250 feet over 76' of 20" piping. Which is wrong because we know the existing system should only drop a few inches in air pressure, if that.

What am I missing here? Is the Darcy eqn really worthless for laminar flow? Did I miss a conversion (don't think so, my units checked)? Thanks for any advice on Darcy versus the old Duct table...

 

[JStephen]

I get Re= 11.54 ft/s x 1.6 ft /2.4 x10-4 ft^2/s = 77,000.

I was thinking that practically all air flow problems were turbulent.

 

[25362]

From a graph (fig. 2.15) in the Fan Engineering handbook by the Buffalo Forge Company, I found with my weak eyes a [Δ]p ~ 0.035 in. wg per 100 ft of pipe.

The next fig 2.16 gives the equivalent diameter of a round duct knowing the short and long sides of a rectangular duct.

BTW, I get for the Reynolds number a value not far apart from that quoted by JStephen.

 

[Montemayor]

My experience concurs with JStephen’s comment about practically all air pip flow being turbulent.

However, I calculate the following:

Actual average pipe velocity = (1,400 cfm)(4/pi*D^2)(min/60 sec) = 10.69 ft/sec

Where D = 20/12 = pipe diameter in feet.

d = 20 inches, pipe diameter in inches;
v = average pipe velocity = 10.69 ft/sec;
rho = air density, lb/ft3 @ 10 psig & 200 ^oF = 0.101 (source: Engineeringtoolbox);
mu = air absolute viscosity, cP @ 200 ^oF = 0.0215 (source: Crane’s TP 410)

From Crane TP 410, page 3-2:

Reynolds Number = (123.9) (d*v*rho/mu) = (123.9 * 20 * 10.69 *0.101/0.0215) = 124,448

I find that it is, indeed, turbulent. The Darcy Equation holds.

 

[iainuts]

Just to throw one more check into the mix. Note I'm assuming 1400 SCFM, not ACFM... (1.75 lbm/s)
Re: ~ 95,000
Velocity = 8.2 ft/s
For Sched 10 (19.5" ID), dP for 76' = .018" water, dP for 100' = .024" water

 

[NoHoCa]

Yesterday I ordered TP410, but since it won't arrive for a few days, I'll need to use a textbook approach. I think I found my errors:

First, I jacked up the calc by blowing the viscosity units. Had to convert from cP to lbf-sec/ft^2.

Second, I was given air density in lb/ft^3. Then I realized that was shorthand for lbm/ft^3. Once I divided by gravity all the units cancelled out. So Re ~ 121,000. My number may not exactly match yours since density and viscosity vary if referenced or calculated.

As an aside, I used the LMNO "Gas Viscosity Calculator" on their website, which references Crane's.

This sure is good prep for the PE exam. I really hate english units but at least now I have one more reference on how to untangle them. Thanks again for all the pointers!

My undergrad (BS) was ME. Gotta figure out how to change the signature next to my name in these forums...

 

[JStephen]

I used the kinematic viscosity for atmospheric pressure in my Re number up there- didn't figure I'd quickly find values versus pressure and that was good for a quick check.

20" Steel pipe = 20" OD normally, by the way, with ID varying by wall thickness.

 

[NoHoCa]

For the record, I gave the original units as 'actual' cfm. If it was 'standard', I would have said scfm. But I've noticed that not every reference uses the same standard. Sorry if it confused some. Hope no one gets too bent out of shape here.

---> Now I'm seeing another nuance. The Darcy equation is only valid for turbulent gases when the pressure drop is less than 10% of the "starting pressure" (Lindeburg, MERM). So the 10.5 psig discharge on the blower equals 25.2 psia, and ten percent of that is about 2.5 psi. So if Darcy shows that the pressure in my aeration header drops by more than 2 pounds, then apparently I can't use that as a reliable number any more. Does that put me back at the Duct tables instead of Darcy?

Can't wait to get TP410 so I've got a more precise reference...

 

[katmar]

The standard Darcy-Weisbach equation is for incompressible fluids. If the gas pressure drop is less than 10% of the supply pressure then its density will change by less than 10% and you can regard it as incompressible for practical purposes. For higher pressure drops the expansion and acceleration of the gas become significant, but there are ways of incorporating the effects into D-W. When you receive your TP410 it will all be made clear.

I cannot imagine Duct Tables ever being set up for pressure drops greater than 10% of the supply pressure. Remember that it is the absolute supply pressure that is important. In my opinion you will always be better off with D-W than with Duct Tables. Duct Tables were for the days before computers and spreadsheets.

In your particular example I estimate your pressure drop to be less than 0.002 PSI, making your pressure drop 0.008% of the supply pressure. Definitely no problem using D-W here!

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Montemayor]

NoHoCa:

You are showing a lot of savy and smart engineering by paying careful attention to the basics and the fundamental premises that units and equations are based on. This speaks highly of you and I once again compliment you on stating a gas flow condition correctly the first time – unlike a lot of older engineers that frequent our forums. It is a real treat to be able to read a flow specification that means exactly what it states. That’s why it is a pleasure to try to help out on threads such as this one.

Your attention to – or knowledge of – the limitations of the Darcy Equation is commendable. You are exactly correct in your statement that you are limited to a nominal 10% pressure drop in the inlet pressure when you deal with applying it to compressible flow. In fact, when you get your copy of the Crane TP 410, look in page 1-7:

“When dealing with compressible fluids, such as air, steam, etc., the following restriction should be observed in applying the Darcy formula:

1. If the calculated pressure drop is less than about 10% of the inlet pressure P1, reasonable accuracy will be obtained if the specific volume used in the formula is based upon either the upstream or downstream conditions, whichever are known.

2. If the calculated pressure drop is greater than about 10%, but less than about 40% of inlet pressure P1, the Darcy equation may be used with reasonable accuracy by using a specific volume based upon the average of upstream and downstream conditions; otherwise, the method given on page 1-9 may be used.

3. For greater pressure drops, such as are often encountered in long pipe lines, the methods given on the next two pages should be used.”

As you've deduced, you have up to 2.5 psi (or more) of pressure drop to calculate for with “reasonable” accuracy. What I’ve often done is to sub-divide my total pipe length into smaller segments and ensure that each segment contributes less than 10% of its inlet pressure. The more segments you have, the more accurate your calculation. I set this up in a spreadsheet and save a lot of recalculations that way.

I hope this helps out. Keep up the good Basic Data reporting; it not only is refreshing and uplifting, it will greatly contribute to your future career successes. Good Luck.

 

[katmar]

NoHoCa, I am sure that if you write to the site management they will be able to switch you from being "Civil/Environmental" to "Mechanical". See the "Contact Us" link at the top right hand corner of this page.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[NoHoCa]

katmar,

You mentioned a result of 0.002 psi for my given conditions, which roughly corresponds to 0.0046 ft of water column. I can't reproduce that. Here's my attempt:

1400 cfm, 10.5 psig, 200F, 20" diam steel x 76' long:
Re ~ 1.1x10^5
f ~ 0.0183, Darcy frict factor (turb), MERM(11) App. 17.B

hf = (f * L * v^2) / (2 * D * g)
hf = 1.48 ft, which is about 0.65 psi

Where am I going wrong? The minor differences in density or volume conversion at T&P don't explain how far off I am...

 

[Montemayor]

My compressible pressure drop spreadsheet based on Churchill, S.W. Chem. Engineering, Nov 7/77, p 91 yields the following answer:

Flow = 1.080216 lb/s/sqft
Re = 124,609
e/D = 9E-05
f = 0.00221
outlet density = 0.100998 lb/ft3
average density = 0.100999 lb/ft3
p2 calc = 10.4995 psig
Delta P = 0.000502 psi
Inches H2O = 0.014

A very small pressure drop as Harvey predicted. My spreadsheet reiterates down to the converging answer. I hope this helps.

 

[katmar]

NoHoCa, the error in your calculation is that the form of the Darcy-Weisbach equation that you have used gives the pressure drop in feet of flowing fluid and not feet of water.

I have not tried to do an exact calculation, but as you correctly stated the minor differences in properties are not the reason for the large difference in calculated pressure drop. Anyway, for comparison my numbers are

Density = 0.1023 lb/ft³ (from Uconeer)
Length = 76 ft
D = 18.81/12 = 1.567 ft (20" Sched 40)
V = 12.09 ft/s
g = 32.17 ft/s²
Re = 115400
f = 0.0181 (Moody or Darcy-Weisbach form)

[Δ]P = ( 0.018 x 76 x 12.09² )/( 2 x 1.567 x 32.17 )
= 1.98 ft

To convert this head to equivalent head of water column, multiply by the ratio of the densities
i.e. = 1.98 x 0.1023 / 62.43
= 0.00325 ft water
= 0.0014 psi

Note that Montemayor has reported the friction factor in the Stanton form, which was used by Churchill. The conversion of the various friction factors is included in Uconeer which can be downloaded for free by clicking on my signature below.

regards
Harvey (aka katmar)

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[katmar]

PS
I had guessed the viscosity at 0.025 cP, which may be a bit different from what you and the others used.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com