Pressure Drop - how do you "guess" the value? 3

[ddkm]

We are planning to construct a pipe for overflow (at a given flowrate). As such, we need to determine the minimum diameter for this pipe to cater for this overflow.

Looking at the calculation:

Diameter D = Unknown (to be determined)
Velocity u = Unknown

and Pressure Drop is required input to determine the above using Reynolds number and Friction Factor charts.

My question is: what kind of assumptions to be made to "guess" the value of Pressure Drop?

---engineering your life---

 

[katmar]

I generally start by guessing a velocity rather than a pressure drop. If you know the flowrate and the velocity you can calculate the pipe diameter. Now you have enough information to calculate the pressure drop. From here you iterate until you get a satisfactory answer.

If you want to be completely mechanistic about it (for example if you were putting this into a computer program) you could always start of with a velocity of 1 m/s (or 3 ft/s) for liquids, and 10 m/s (or 30 ft/s) for gases. If you are doing the calculation by hand you could probably improve on these guesses based on actual circumstances.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Latexman]

At the outlet destination of the overflow line, will the overflow fluid enter as a submerged fill line or will it be higher in elevation than the liquid surface, i.e. splash filled?

Good luck,
Latexman

 

[ddkm]

Katmar,

In a way, I also agree with your approach. I start off by guessing the Diameter of the pipe and use that with other flow data to compute the Friction Factor and thereafter, the Reynolds number (using the Charts). With this value of Re, the velocity is determined. And in one big circle, I use the flowrate and this determined velocity to compute back the Diameter. This must be done iteratively until the ASSUMPTION Diameter matches the COMPUTED Diameter.

Hope that makes sense. Is that correct?

My problem with the above is this: for me to even get the Friction Factor, the Pressure Drop is required! How can this value be obtained???


(Note: From the FAQ, it seems we can use the Darcy-Weibach equations to determine the Pressure Drop, but I wanted to verify this with you all here)

---engineering your life---

 

[ddkm]

Latex,

should be higher in elevation than the liquid surface, if I understand yr terms correctly.

---engineering your life---

 

[Latexman]

I will clarify. At the opposite end of the pipe where this overflow fluid comes out (not where it goes into the pipe), does it fall onto a liquid (or solid) surface through an air gap or vapor space, or does it submerge fill into a liquid?

Good luck,
Latexman

 

[Ashereng]

ddkm,

We are currently using a maximum of 35 kPa/100m of pipe for liquid flow.

"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?

 

[katmar]

ddkm,

Whether you guess the velocity or the diameter as the first step, it comes to the same thing in the end. If you are using a computer to do the iterations it does not matter whether it takes 5 loops or 105, so the accuracy of the starting point is usually not too important.

I do not understand why you need the pressure drop to calculate the friction factor. The friction factor is a function of the Reynolds Number and the relative roughness (or Re only for laminar flow). You do not need to know the pressure drop to calculate Re or e/d.

Am I missing something?

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ddkm]

Katmar,

The method I'm using is the correlation between [Φ] Re ² and Re (from a chart)

Since I don't have velocity u and diameter D, Re cannot be computed directly. So I use the above correlation to obtain Re, if I can get the value of [Φ] Re ² .

BUT the formula for [Φ] Re ² is:

= (Pressure Drop)* D^3 * Density / ( 4*L*(visc^2) )


That's why I need to know how Pressure Drop should be determined.

---engineering your life---

 

[ddkm]

Latex,

There is an air gap at the other end.



---engineering your life---

 

[katmar]

ddkm,

but you do have the diameter and the velocity. I proposed guessing the velocity and then calculating the diameter, knowing the flowrate. You proposed the other way around - guess diameter and calculate velocity. Either way you have both diameter and velocity, and with the fluid properties of density and viscosity you can calculate the Reynolds Number. You can get a chart of friction factor vs Re and relative roughness (the Moody diagram) or you can use an approximation formula like Churchills.

You are making an easy calculation unnecessarily difficult.


Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ddkm]

Katmar, by george, I think you're right! I'll try that out and see the outcome. What if the answers are different? <Gasp>

---engineering your life---

 

[rbcoulter]

Maybe I'm missing something here but isn't your pressure drop already known? This is an overflow condition so you must have an idea of how high you want the fluid level above the overflow pipe above the entrance. Correct? Also, the elevation and discharge pressure of the pipe must also be known. So delta h = z1 - z2 + p1 - p2 . If both p1 and p2 are atmospheric then delta h = z1 - z2 where z1 is the height of fluid above the entrance pipe and z2 is your discharge elevation.

Do the iteration on the Darcy and Colebrook equations to get the Diameter you need to get the flow rate you desire.

 

[katmar]

rbcoulter - I think we are all on the same wavelength now. What ddkm and I were proposing was iterating with trial-and-error values for the diameter (or velocity) until the calculated pressure drop was equal to the known pressure drop.

There was an interesting thread (see thread135-149386) discussing the use of explicit friction factor equations instead of the implicit Colebrook-White equation or the Moody diagram. If you have an explicit relation for the friction factor then with a bit of inventive math you can calculate (without trial-and-error) any of diameter, flow or pressure drop if the other two are known. The results can be very close to the "true" answer using the Colebrook-White equation. As someone pointed out in that thread, the uncertainty in the pipe roughness and diameter often exceed the inaccuracy introduced by the approximation of the friction factor. This is even more true if you have valves and fittings where you have to estimate k values.

My own preference (see thread135-149386) is to combine the Colebrook-White and Churchill equations which gives a continuous function that covers the full range of possible flow regimes. This is ideally suited to mechanistic computer solution by iteration, and the answers match the results obtained by hand calculations using the Moody chart.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[rbcoulter]

The Hazen-Williams formula may possibly work here. It is surprisingly applicable in a lot of situations. Certainly a lot easier to use. With it the diameter can be solved for explicitly (no iterations) based on the input of pressure drop and desired flow rate, pipe length, flow coefficient (steel pipe c= 140).

 

[quark]

There is a very good article in Chemical Engineering magazine(Estimate friction factor accurately by TK Serghides) about explicit equations of friction factor and comparision of their accuracies with Colebrook's equation.

One equation close to the Colebrook's equation, but explicit function of 'f' is,

(for Re>2100 and at any e/D)

f = [A-[(B-A)²/(C-2B+A)]]^-2

where A = -2.0log ((e/D/3.7) + (12/Re))
B = -2.0log((e/D/3.7)+(2.51A/Re))
C = -2.0log((e/D/3.7)+(2.51B/Re))

Churchill's equation, for any Re and e/D (as given in the paper)

f = 8((8/Re)¹² + (1/(A+B)^1.5))^1/12

Where, A = (-2.457 ln((7/Re)^0.9 + 0.27e/D))¹⁶
B = (37530/Re)¹⁶

The author concludes that the deviation in friction factor values from Colebrook's equation is maximum for Churchill's equation, with an average deviation of 5.16% and maximum deviation is 55.6%. The first equation has more accuracy with an average deviation of 0.0002% and a maximum deviation of 0.0023%.

An excel spreadsheet for the iteration of Colebrook's equation can be downloaded from

http://me.queensu.ca/courses/mech441/spr/moody.xls

Katmar,

When you say combining Colebrook's and Churchill's equations, do you mean that Churchill's should be used upto transition and Colebrook's for turbulent region? Let me know if you want to have a look at the paper.

 

[katmar]

Quark, I simply substitute the first 5 terms of the Colebrook expansion in place of Churchill's term "A", raised to the correct power of course. This general procedure (i.e. using different expressions for Churchill's "A", "B" and "C") is described by Chandra Verma, Hydrocarbon Processing, Aug 1979, p122-124.

I have not seen the Serghides paper. I am very surprised that he gives such wide deviations for Churchill's equation. This equation has been very widely accepted and I had assumed it was more accurate than that. Perhaps there are wide variations in the transition regime between Re = 2100 and Re = 3000 that skew the averages? I would be very wary of anybody's equation in this zone because the process itself is unstable.

regards
Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Latexman]

ddkm,

I'm curious, what reference are you looking at for this ? Re^2 vs. Re method?

Good luck,
Latexman

 

[Montemayor]

Quark/Katmar:

I have Word for Windows documents of both the 1977 Stuart Churchill Article as well as the 1984 write-up that T. K. Serghides authored. Both appeared in Chemical Engineering Magazine and I hand-typed in the documents myself in order to preserve these important articles with their original formulas.

Unfortunately, as you know, I can’t give you my email address so that I can forward them to you – should you be interested. However, may I suggest I send them to you Harvey at your website (if you are interested)?

I also was impressed by Seghides’ work and claim and have used his version of the explict solution to Colebrook’s famous equation. I’ve found his results to be quite accurate – how accurate I’m unable to prove. However, there is no reason to question the veracity of his equations 22 years after he formally published them --- that is, unless someone has found yet another, better version or has successfully disputed his claim or findings.

I tried most, if not all of the explicits: Churchill, Chen, Serghides, Wood, Moody, Barr, Jain, etc. etc.
Serghides is the one that I’m most content with in obtaining a virtual Colebrook result.

 

[ddkm]

rbcoulter, I think your comments are even closer to what I'm trying to achieve here.

Maybe I'm missing something here but isn't your pressure drop already known? This is an overflow condition so you must have an idea of how high you want the fluid level above the overflow pipe above the entrance. Correct? Also, the elevation and discharge pressure of the pipe must also be known. So delta h = z1 - z2 + p1 - p2 . If both p1 and p2 are atmospheric then delta h = z1 - z2 where z1 is the height of fluid above the entrance pipe and z2 is your discharge elevation.

Perhaps I've been using the wrong approach and wrong equations all this while. RB, is your approach coming from a Bernoulli equation point-of-view?

Example, ½u² + P/rho + gz = constant ??

If so, I've drawn out my "problem" in this link:

http://i76.photobucket.com/albums/j24/dd_km/Overflow.jpg

I now plan to modify my approach based on all the above discussions by you guys, as follows:
- Get velocity based on the Bernoulli equation (where Pressure is atmospheric, i.e. P1-P2 = 0 and where initial velocity = 0)
- From the velocity and the expected INPUT flow (of 60m3/hr), the overflow pipe must now be calculated based on

D = SQRT [ 4 * Vol rate / (u Pi) ]

---engineering your life---