Pressure Drop - how do you "guess" the value?
Pressure Drop - how do you "guess" the value?
(OP)
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?
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---





RE: Pressure Drop - how do you "guess" the value?
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
RE: Pressure Drop - how do you "guess" the value?
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
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---
RE: Pressure Drop - how do you "guess" the value?
should be higher in elevation than the liquid surface, if I understand yr terms correctly.
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
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?
RE: Pressure Drop - how do you "guess" the value?
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
RE: Pressure Drop - how do you "guess" the value?
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---
RE: Pressure Drop - how do you "guess" the value?
There is an air gap at the other end.
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
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
RE: Pressure Drop - how do you "guess" the value?
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
Do the iteration on the Darcy and Colebrook equations to get the Diameter you need to get the flow rate you desire.
RE: Pressure Drop - how do you "guess" the value?
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
RE: Pressure Drop - how do you "guess" the value?
RE: Pressure Drop - how do you "guess" the value?
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.
RE: Pressure Drop - how do you "guess" the value?
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
RE: Pressure Drop - how do you "guess" the value?
I'm curious, what reference are you looking at for this ? Re^2 vs. Re method?
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
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.
RE: Pressure Drop - how do you "guess" the value?
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:/
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---
RE: Pressure Drop - how do you "guess" the value?
But basically, the numbers are:
Height 10000mm (or 10m) of the tank
Height 300mm of the tank roof above the overflow line
Both tank and collection pot are at atmospheric
There is an INPUT material from the plant flowing at 60m3/hr
My Concept
The size of the overflow pipe should be determined based on the fact that we DON'T WANT the contents in the Tank to reach the Roof. This means the overflow pipe should be large enough to cater for the INCOMING 60m3/hr while up to an ELEVATION of Z1-Z2.
As such, in tandem with RB's earlier post, it seems that Z1-Z2 should be equal to the height or elevation of the Tank Roof ABOVE the Overflow Line. i.e. about 300mm.
Any comments?
(Please note that I have used arbitrary figures for the above problem, instead of my plant's actual flow data. Hopefully, there's no unrealistic numbers used)
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
Don't forget to include a term in the Bernouilli's equation for frictional losses. What you have so far is for frictionless flow.
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
This is precisely what I'm trying to clarify, as it was not so clear from RB's post. Like you, he seems to suggest to use the overall elevation as the dH - in my example, this would be 10m. HOWEVER, I feel that the dH should actually be the small height ABOVE the overflow line, because this is the only height that is PUSHING the fluid across the Overflow line. Everything else BELOW the overflow line can't contribute to the oveflow, right??
Appreciate your comments.
Note: Your second point on catering for frictional losses would be my next step in this problem. I'm trying to solve the uncertainties one by one. Thanks.
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
Try this mental experiment. Imagine a vertical piece of 2 inch pipe, say 5m long, where you can somehow control the amount of water being fed into the upper end and the water simple falls out the bottom end onto the floor. Now start feeding it at 1 litre per minute. I'm sure we would all imagine that this small amount of water would simply fall through the pipe without filling it. Now gradually increase the feed rate. There will come a time where the water perfectly fills the pipe and it runs out the bottom at the same rate that it comes in. If you increase the feed rate any more the top end will simply overflow because no more can go down the pipe.
This equilibrium point is when the driving force (which is the head of water in the pipe) exactly matches the resistance (which would be the friction in the pipe, plus any entrance or exit effects you want to take into account). Note that although I said a 5m length of pipe, the length is actually irrelevant. If you double the length you double both the driving force and the resistance (neglecting end effects). For a 2" sched 40 pipe with water this equilibrium flowrate is about 55 m3/h.
But you have a complicating factor that has only come to light in your sketch and was not mentioned before. The horizontal section of pipe between the tank and the collection pot will also have a resistance, and the driving force of the head in the vertical pipe has to overcome this as well. What is this length?
In your case the driving force would be due to the height of 9700 mm of vertical pipe. I would not include the 300 mm ullage here - that is your safety margin. Let's guess that the horizontal distance is 20 m and there are just the two bends at the top and bottom of the vertical section. Your driving force is 9700 mm of water (=95 kPa) and the resistance is the two bends plus entrance and exit effects plus friction in 29.7 m of pipe with a flow of 60 m3/h. To get the driving force and resistance to match you would need a 70 mm ID pipe. (This is the trial and error calculation we discussed before.)
It is unlikely that you will find a pipe of exactly the diameter you calculate, so you have to use your engineering judgement of what size you will actually use. How likely are upsets where you get more than 60 m3/h flowing? What is the impact of the tank overflowing? Etc, etc. You may want to go one size bigger and use a 3" pipe, or maybe a 4" pipe - perhaps even bigger if the risks are great. If your pipe ID is larger than the calculated value it simply means that the level in the vertical section will settle out somewhere less than 9700 mm.
In practice I would make one other adaptation. Whatever size pipe I decided on, I would make the actual nozzle in the tank one size bigger, with the first bend that same size. At the top of the vertical section I would put a reducer to get to the pipe size I had decided on. Why would I do this? I just like it that way.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Pressure Drop - how do you "guess" the value?
I'm getting close to see what you mean, but still need some time to digest all the info. Thanks.
By the way, do you think that the linear velocity could be determined using the Bernoulli equation? Like what I've described earlier?
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
I bet the Hazen-Williams formula will give you the answer within 5 %.
RE: Pressure Drop - how do you "guess" the value?
Theoretically you can calculate the velocity from the Bernoulli acceleration term. The kinetic energy due to the velocity is usually termed the "exit" loss.
In practice (with liquids) the energy lost to friction is usually much more than the acceleration term, so you would have to factor this into your calculation of the velocity.
I suspect that rbcoulter is correct in saying that Hazen-Williams will come very close to the right answer, but I find that when I am trying to get my mind around the concepts of all the elements that go into a problem applying Bernoulli is a good way to ensure that you have covered all the bases and interpreted the problem correctly.
Once you have understood it - then you just wack it with the software!
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Pressure Drop - how do you "guess" the value?
My problem now is how to include this term inside the calculations. In the Bernoulli equation itself?? Or in the Darcy-Weisbach equation for pressure loss determination (=f*rho*L*u^2/(2d)) ? Note: to answer Katmar's earlier question, the length of the horizontal pipe is L = 8m.
Let me illustrate what I've done with the Bernoulli equation so far. For simplicity, I've taken the 9700mm as the figure to be used (we could argue that 10000mm or even the 10,300mm could be the better number).
Calculation done as follows:
http://i7
My question is:
a) Have I used the numbers correctly, example: Z1=9.7m?
b) Where does the frictional losses come into the above calculation/equation?
My own answer (Correct me if I'm wrong):
- Use Darcy-Weisbach equation and somehow, by making some arbitrary assumptions, obtain the Pressure Loss
- Put this estimated Pressure Loss value into the Bernoulli equation in the P2 term; this means that P1-P2 becomes negative, since P1=atmospheric.
- Recalculate iterative until all the values can match each other
Looks really far-fetched!
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
Further, the velocity you got seems to be ridiculously high and it would take a 250 mm pipe to cancel the frictional loss by taking credit of the 10m pipe elevation.
First, you have to calculate the flowrate with whatever small head that is available above the overflow pipe bottom (the maximum is 300 mm). On a safe side, I would take 300mm as the head. This is about 1.94 m/s. If you go with a 3" pipe, the pressure drop is approximately 0.8 meters and you have no problem.
Then using the calculated flowrate, size the piping based on 10m (or 9.7m elevation)elevation. You can reduce the pipe size as long as the frictional losses equal the elevation of 9.7 meters.
But, apart from the available static head of liquid in tank, you have another potential(that causes flow) acting here i.e pumping. So, if you are flowing in the tank at a cosntant rate of 60cu.m/hr, the 3" pipe can take care of it by just consuming 2.2 meters of elevation, leaving out 7.8 meters of available head untouched. Go with a 2.5" piping and you will have 4 meters of available head untouched. 2" is too tight for this application.
If subscript 2 is the end state then you have to add the frictional loss to the RHS of the equation.
RE: Pressure Drop - how do you "guess" the value?
For a standard overflow- where the tank has an internal weir or a standard nozzle followed by a vertical downleg the calculation falls into two parts.
The first one determines the head required to push the flow over a weir (if it is present)- in your case this would be compared to the 300mm clearance you have).
The second calculation (and the only one if you don't have a weir) is to determine the head loss in the exit nozzle and first bend from the tank- again this is compared to the 300mm of clearance you have available to drive it.
Thats it- assuming that the outlet piping (which is vertical) is the same size as your outlet nozzle- if you can get it out of the tank it'll fall vertically no problems.
The full 10000mm is NOT available to drive the flow unless you've formed a siphon as noted above- i.e. fully flooded pipe. Designing overflow piping using the full 10000mm will give you grief- in particular you'll find that air locks, air entrainment and surging will reduce the capacity of the line.
In your case I'd check the capacity of the outlet nozzle and bend with respect to your 300mm. Then do another calc based on the same sized piping with the full 9700mm as the driving force to the overflow collection pot. In the abscence of a software package that can perform calculations for launder flow assume the pipe is half full (i.e. double the velocity for your calculation). Unless your overflow collection pot is a looong way away you should have plenty of head available (the outlet nozzle on the tank should be the point that dictates the line size). If the pipe/nozzle size you calculate is a bit close- go up one size.
Lastly- I'm assuming the purpose of the excercise is to ensure the tank can overflow safely in an emergency and be contained. If it is designed to be a continuous process be aware that you are going to get significant air entrainment in the overflow line.
Cheers.
RE: Pressure Drop - how do you "guess" the value?
1. Pressure change - zero in your case as you have atmospheric both sides
2. Velocity head - to be calculated
3. Static head - 9.7 m in your case
4. Friction head - to be calculated
5. Internal energy change - negligible
6. Shaft work - not applicable to your system
7. PxV work - not applicable to liquids
Ignoring the zeroes and the N/A's you have
Static head + velocity head + friction head = 0
The math to solve this is ugly (for an old man like me) so that is why I said that once you have understood it, you just wack it with the software.
As the tank fills up, the level should never get above the top of the overflow pipe. That is why I said earlier that the actual nozzle on the side of the tank should be one size larger than the rest of the overflow piping. Itdepends has highlighted a very important aspect here and the outlet nozzle probably needs to be designed with more attention than just to say "one size bigger".
I agree completely with Itdepends that there are two separate calculations - one for the outlet nozzle and one to drive the liquid to the collection pot. This is probably what has caused the confusion over whether we should use the 300 mm or the 9700 mm as the driving force. When you separate it into the two components then it makes sense. 300 mm is the maximum driving for to get the water through the outlet nozzle, and 9700 mm is the maximum driving force to get the water to the collection pot. The Bernoulli analysis above applies to the second part of the problem.
However, I disagree that you can use all of the 300 mm as driving force to get the liquid through the exit nozzle. If the level in the tank gets above the top of the outlet nozzle then you will start to get air locks and surging.
In my previous posts I have deliberately ignored the aspect of making the vertical section of pipe self-venting because we have enough confusion as it is. But once you have sorted out what the situation is in terms of pressures, flows and velocities you will realize that the vertical leg is not running full if you select a standard size pipe. The section of pipe above the liquid level in the vertical leg must be wide enough to allow any entrained gas to escape back up the pipe, or it will reduce the density in the vertical leg and lower the driving force and therefore also lower the flow. If the exit nozzle runs flooded it will not be possible for this air to escape easily - hence my argument that the liquid level should be below the top of the nozzle. If the air cannot get out it would lead to oscillating flowrates as the vertical leg gradually filled up and then syphoned out.
By my calculations an 8" outlet and vertical leg down to the liquid level would be about the smallest you could use, but a 10" would be safer. The lower part of the vertical leg and the horizontal section could be 3". It is the requirement for self-venting that makes the pipe so big. If you don't require steady flow and you could put up with the surging you could make the whole line 3", but it would be hard to predict how much of the 300 mm ullage you would use during the air lock phase of the surge cycle.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Pressure Drop - how do you "guess" the value?
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
However- you may find it very difficult to get the required flow through an outlet nozzle of a reasonable size without using the available ullage in the tank (in your case 300mm) as the driving force- in fact you may need to increase the ullage. This is one of the common reasons for installing a weir- it allows you to install the outlet pipe at a lower level in the tank- thus giving you more driving force through the nozzle. The top of the weir is relatively high- giving you more live volume in the tank. The crest height over a weir will be small relative to the height required to drive the flow through the overflow nozzle.
RE: Pressure Drop - how do you "guess" the value?
simplifies to
gz1 = (fL/D)u22/2 + (∑K)u22/2 + u22/2gc
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
1) Latex, in your equation above, you have a "sum of K". What is that referring to?
2) In my sample calculation http://i7
---engineering your life---
RE: Pressure Drop - how do you "guess" the value?
2) Yes, u1 is zero for this calculation.
3) If you are asking questions this basic then you need to go back to your books and do a bit of revision of what you studied at college.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Pressure Drop - how do you "guess" the value?
I agree with katmar on all points. I also recommend you buy Crane's Technical Paper 410 and study that. It can be purchased here:
http://www.cranevalve.com/tech.htm
Good luck,
Latexman
RE: Pressure Drop - how do you "guess" the value?
In your post of 15 Apr 06 6:51, you said:
"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."
Would you be so kind as to write completely your modified expression for A in Churchill's equation? This may hold the key to removing some discrepancies between the Colebrook and Churchill formulas. (You might recall we had exchanged some remarks in another forum about the desirability of using Churchill's formula to avoid discontinuities in iterative calculations).
Thanks very much.
RE: Pressure Drop - how do you "guess" the value?
For the turbulent regime Churchill (Chem Eng. Nov 7, 1977) estimated the friction factor with his equation (14)
f -1/2 = 2.457 * Ln(1/((7/Re)0.9 + 0.27e/D))
Note that in his original paper Churchill was using the Stanton form of the fiction factor, which is 1/8 of the more frequently used Darcy-Weisbach form.
This expression, raised to the value of 16, is what Churchill terms "A" in his overall equation.
Colebrooks estimate for the friction factor, also in the turbulent regime, is
f -1/2 = -5.65685 * log10( (e/D)/3.7 + (0.88742 * f -1/2 / Re ));
Note that I have converted this to the Stanton form as well, so the constants will be a bit different from the commonly seen ones.
The basic forms of these two equations are very similar, except that Churchill's form is explicit while the Colebrook formulation has f -1/2 on both sides of the equation.
In pseudo-code my calculation for "A" is simply
A := "initial guess" {I use 14.14 as initial guess}
For I := 1 to 5 do
A := -5.65685 * log10( (e/D)/3.7 + (0.88742 * f -1/2 / Re ))
A := A^16
And then I use this "A" in Churchill's equation (18). By experimenting, I found that 3 or 4 iterations was sufficient to get this trial and error equation to converge to an acceptable degree. I simplified my code by leaving out the convergence test and always doing 5 iterations. I'm sure the purists will find it a bit crude, but it works very well.
I have made plots of the friction factor against Re and e/D and it is remarkable just how smoothly Churchill's method works. A true work of genius in my opinion. All I have done is take advantage of the speed of modern PC's to bring Churchill's method a bit closer to the Colebrook numbers. As various people have stated, Colebrook may not be perfectly accurate but it is the standard that others are judged against. If you would like a copy of this plot, contact me via my signature below and I will email it to you with pleasure.
regards
Harvey
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Pressure Drop - how do you "guess" the value?