# Aerospace Flow rate through a squre tube

1. Jan 21, 2007

### blaster

Find: Flow rate of water through square tube

Given:

Cross-sectional wall and area of square tube, A=w^2
Length of square tube, L
Pressure at start, P1
Pressure at finish, P2
Water

***
If you need specifics I am pushing water (room temp) at 1 bar through
1200mm of square tube that is 1mm x 1mm. The end of the tube is open to
atmosphere.

What is the flow rate in liters per hour that will flow through this
tube?

2. Jan 22, 2007

### Clausius2

My question is: how is it possible to push water at 1 bar being the end of the pipe open to the atmosphere (1 bar)?. In order to generate flow you need a pressure gradient. There is only one case in which you wouldn't need a pressure gradient. And that case is one that belongs only to the literature. That will be an ideal pipe with no friction. If you are assuming an ideal straight pipe with no friction then any mass flow fits with the solution of your problem. I hope that you are not assuming an ideal pipe, since your dimensions indicate that the flow surely would be a low Reynolds flow. The velocity profile inside your tube is going to be some kind of Poiseuille flow corrected for the square section. If you know the pressure at the pipe inlet then you know the mass flow by only applying Poiseuille flow relations. If the pipe inlet is connected to a pump, then the pressure at the inlet is given by the jump of pressures of the pump, whereas if the pipe inlet is connected to a reservoir, the pressure is given by the pressure of the reservoir adecuately corrected.

Here are some dimensional analysis that an engineer should know how to do:

Reynolds Number:
$$Re=\frac{\rho Ua}{\mu}\sim \frac{\rho \Delta P a^3}{L\mu^2}$$

where $$a$$ is the pipe hydraulic diameter and $$L$$ is the pipe length.

Last edited: Jan 22, 2007
3. Jan 22, 2007

### blaster

Let me clarify P2=0bar.

How do you do the "corrected for a square section" part

4. Jan 22, 2007

### Clausius2

You didn't clarify anything. You don't know what intake pressure you have. Think about it a little bit more and come out with the Reynolds Number by yourself. It will give you what kind of flow best addresses your problem. I won't be helping you farther.

5. Jan 22, 2007

### Q_Goest

Hi Blaster,
Is this homework? I assume it is. Have you reveiwed "hydraulic diameter" yet? You can equate a square tube to a round one using hydraulic diameter. Once you do that, you can apply the Darcey-Weisbach equation directly.

If this isn't homework, I'd be glad to run through this for you.

6. Jan 22, 2007

### FredGarvin

I am still confused about the pressure. If the end of the pipe is open to atmosphere, there is no way that P2=0 bar.

7. Jan 22, 2007

### Q_Goest

Hi Fred. The way I read this is there is a long square tube, open to atmosphere. The pressure at the outlet is therefore at atmospheric pressure as it opens to atmosphere (ie: 0 barg). At a point 1200 mm upstream of the open end the pressure is 1 barg.

Note: barg = bar guage pressure

8. Jan 22, 2007

### blaster

Actually this is not homework. I am a mechanical engineer designing irrigation products. And yes my pressures are gauge pressures.

Thanks for your help. I am excited to see you run through the solution.

9. Jan 22, 2007

### FredGarvin

Hey Q. Good to see you again. I completely missed the 1200 mm upstream part. That makes some sense now.

In regards to the OP, you do indeed need to use the hydraulic diameter in your calcs. The hydraulic diameter is defined as:

$$D_h = \frac{4*\mbox{cross sect area}}{\mbox{wetted perimeter}}$$

Last edited: Jan 22, 2007
10. Jan 22, 2007

### Q_Goest

Hi Fred. Nice award, I think you earned it!

Hi Blaster,
Equations for fluid flow, such as Darcy-Weisbach, Hagen-Poiseuille, Colebrook equations, the Moody chart and similar flow equations all assume flow in circular pipes. These equations can be used for other cross sectional shapes such as square cross sections, but these cross sections have to be equated to a circular cross section that produces the same restriction. That cross section is called the "hydraulic diameter". Note that the hydraulic diameter is not the same as an "equivalent diameter" that results in the areas being equal.

The hydraulic diameter is can be calculated from:

Dh = 4A/U
Where A = cross sectional area
U = wetted perimeter
Ref: "http://en.wikipedia.org/wiki/Hydraulic_diameter" [Broken]

This is the equation Fred provided above. For example, a 1 mm square tube (inner dimensions) has a hydraulic radius of 1 mm.

Once you've found the hydraulic diameter, you can attack this just like any other circular pipe flow problem. For incompressible flow, I'd suggest applying Darcy-Weisbach directly.

Pressure drop is called head loss or frictional head loss and a few other names too. That equation is:

h = f L V^2 / ( 2 D g)

where f = friction factor
L = pipe length
V = fluid velocity
D = hydraulic diameter
g = constant (acceleration due to gravity = 32.174 ft/s2 = 9.806 m/s2)
Ref: "http://www.lmnoeng.com/darcy.htm" [Broken]- I'm using this reference because they have a calculator at this site. You may not want to use it, but it would serve as a check of your own numbers. I'd suggest creating your own program using a spreadsheet.

The only variable above you won't have is friction factor, which is a function of Reynolds number. Wikipedia has a good article on Reynolds number here. It's the same equation provided by Clausius above.
"http://en.wikipedia.org/wiki/Reynolds_Number" [Broken]

Of course, you still need kinematic or absolute viscosity. The values for viscosity can be found in any text book or on the web, for example here:
"http://www.engineeringtoolbox.com/water-dynamic-kinematic-viscosity-d_596.html" [Broken]

Once you do that, you still need to find friction factor. There are many ways to calculate that, including taking it directly off of the Moody diagram. I'd recommend you create a spreadsheet that does all this for you, so I'd suggest using an equation as described at Engineering Tips Forum here:
"http://www.eng-tips.com/faqs.cfm?fid=1236" [Broken]

Note that for the above friction factor you'll need pipe roughness, which is dependant on your actual hardware. If you don't know what it is for your square tube, I'd suggest using 0.00015 which is commonly used for clean pipe. Note also, D in these equations is the hydraulic diameter.

The last thing to do is determine flow as a function of pressure drop per the Darcy-Weisbach equation. Note that head (h) is the pressure created by a given column of the fluid in question. Velocity is a function of flow rate. You'll need to separate out those portions and treat them separately, or possibly use the calculator given on the internet.

~

If I do these calculations, I come up with a Reynolds number of 2577, which is somewhere in the transition zone. It may be turbulent, and it may be laminar.
- If I assume it is laminar, the flow rate for your case will be .0324 gallons per minute. Convert that how you need to.
- If I assume the flow is turbulent, I get .0178 GPM.

Couple things to note here. I've neglected exit losses, which in this case are fairly small, but to explain how to calculate exit losses would take another post this size. I've also assumed there are no elbows or other restrictions in this line. Finally, I've assumed the line is horizontal. If there is any elevation change in the line, you can correct for that using Bernoulli's equation.

This is a lot to cover in a single post, so I may have been overly brief. Feel free to ask questions, I'm sure myself or others here can help out. Hope that helps.

Last edited by a moderator: Apr 22, 2017 at 3:18 PM
11. Jan 22, 2007

### Clausius2

It seems that we have forgotten the lesson about viscous flows. I hope you are not planning to calculate the mass flow using the -Bernouilli Corrected equation with losses- or whatever is the name. Looking at the left hand part of the Moody chart the only thing that you find for Low Re is the friction coefficient that by the way is automatically and analitically included in the Poseuille relations, so there is no need of Moody Chart. The first thing that I said is that he should get a Reynolds Number, and I still don't see it. If the Reynolds Number is large (and I don't think it's gonna be large given the 1mm section), then the Moody Chart is gonna be helpful and the Bernouilli-corrected equation is helpful too. If the Re is small then it is misleading to employ the Bernouilli-corrected equation that assummes fully turbulent flow with a laminar friction coefficient.

One more time for the OP, please provide Reynolds Number. That is the FIRST thing you should know when dealing with pipes and pressure losses. If you don't know that, then you know nothing, and we cannot help you.

The thing of the Hydraulic diameter is a good advice though.

12. Jan 22, 2007

### Q_Goest

I assume you're refering to the use of Bernoulli's equation to resolve pressure changes along a pipe where pressure drop due to friction is incorporated. I'll have to reference "http://www.flowoffluids.com/tp410.htm" [Broken]on this:
~
Per the Crane paper:
So you are correct that the Poiseuille law is applicable where laminar flow is concerned and one wishes to determine frictional pressure losses. But as you can see, this law is simply a very specific case of the Darcy-Weisbach equation. There is no difference. That is why I would not suggest using the P' equation. It is too specific. The D-W equation is much more general.

~
The "Bernoulli corrected equation" doesn't assume fully turbulent flow. Nor does the Darcy-Weisbach equation. Friction factor is a function of whether or not the flow is laminar or turbulent, and friction factor is the only correction that needs to be made to account for the two types of flow. The type of flow (ie: laminar or turbulent) has nothing to do with the Bernoulli corrected equation you're refering to.

~
How is the OP going to provide Reynolds number without velocity? And of course, you can't determine velocity without flow rate.

One thing I skipped over in explaining all this is that one needs to iterate to determine frictional pipe loss, which is why a spread sheet or other computer program is so valuable.

One input to the calculation is a guess at the flow rate. With this 'guess' one can determine velocity. With velocity you can then determine Reynolds number so that a friction factor can be determined. Only then can you calculate dP. So it's not a simple task. It requires iteration, but of course a computer can do all these iterations in a fraction of a second.

Last edited by a moderator: Apr 22, 2017 at 3:18 PM
13. Jan 22, 2007

### FredGarvin

I have not seen the energy equation being restricted to fully turbulent flow. The only restrictions I have seen are fully developed, steady state, incompressible.

Last edited: Jan 22, 2007
14. Jan 22, 2007

### Clausius2

The holly molly, it is impossible to deal with Q_Goest. It seems we are coming from Universes totally different, where the physics laws are completely different. I wish I could know the other Universe, if it exists. I will try to answer you rather than to answer the OP later, now I have work to do. I honestly get dissapointed and discouraged to talk with someone that uses only the reference of the "Cranepaper". I think that in these world there are better references to cite. But anyways, I don't feel like discussing with a wall again. So I may be not replying to your considerations. I will try to make an effort though, but not for wanting you to change your mind, rather for not letting the rest of the people (included students) to believe that you're right.

Last edited: Jan 22, 2007
15. Jan 22, 2007

### Q_Goest

Hi Clausius. People often resort to insults when disagreements arise. I'm not a wall. I don't come from another universe. In fact, aren't you still a student? I graduated almost 20 years ago and since then have done flow analysis, and more complex thermodynamic, heat transfer and two phase flow analysis on piping systems ever since. When working for General Dynamics Space Systems Division, I wrote a paper that was used as a standard that covered all of the piping flow analysis we've discussed in this thread. If I didn't understand this stuff, almost every launch pad in the US would have serious problems. I also hold a number of patents which required this knowledge. And I continue to analyze complex piping systems today, working in the air separation and chemical processing industry.

I'd apreciate it if you'd use thoughtful arguments instead of resorting to insults.

~

Regarding the Crane paper, if you do any piping analysis you'll find the Crane paper is the formost authority on the subject. If you don't believe that I can also reference others.

Regarding the Poiseuille law for example:
Ref: Wikipedia[/PLAIN] [Broken]

If you don't understand the use of Bernoulli's equation and how frictional pressure drop is also accounted for, but you don't like the Crane paper reference you can try this reference also:
"http://www.lmnoeng.com/DarcyWeisbach.htm" [Broken]

The analysis of pressure drop through a straight, horizontal pipe with no fittings, valves, or other restrictions is the most simple of all analysis. Until you understand this basic analysis, you won't be able to move on to how other restrictions affect flow. Then there are thermodynamic and heat transfer considerations which also must rely on understanding these basics. Two phase fluid flow then relies on understanding all of this.

Last edited by a moderator: Apr 22, 2017 at 3:18 PM
16. Jan 22, 2007

### Clausius2

Here we go, on the contrary I am not going to throw here my CV, that is probably much better than yours one 20 years ago. I will use scientific arguments instead of medals:

Even though the Crane paper says that, that makes no sense sir!!!. What a naive generalization!!!. What a reduction of the spectra of the fluid dynamics science!!!.

What is a Darcy Weisbach equation???. You mean the experimental correlations written on the Moody Chart??. Or you mean the Darcy Weisbach
coefficient??. Do you mean as DW equations the correlations of Colebrook, Prandtl or Von Karman?. There is no exact solution for the DW coefficient for the whole range of Re, so your comments about the DW equation are naive and shows that your rockets may not be that good designed.

Those correlations of Prandtl, Von Karman and Colebrook DON'T match with the laminar solution of 64/Re at low Re, so I don't see the generality of something that cannot be handled generally.

That is not true sir. And in particular reveals that you don't know where those correlations of the Moody Chart are coming from. In particular those correlations such that the Colebrook one come from the Turbulent Flow in a Channel analysis. Do you know what is the defect law of Von Karman?. Here there is an excellent ppt from a guy of the U of Taiwan (not that prestigious as the Crane paper) but it may be helpful for us, especially from slide 48 on:

www.esoe.ntu.edu.tw/courses/50522100/Chapter_08(compact%20version).ppt [Broken]

There you are gonna find familiar comments.

This is the culmination. I gave you a way of calculating the Reynolds number in my first thread. The velocity is an OUTCOME of an external force, so that the Reynolds number is intrinsically defined even if you don't know the solution. That's the task of an engineer and you should know it. To guess the unknown variable estimating it by using the flow equations (NAVIER STOKES EQUATIONS).

I'm going to finish with you saying something. I appreciate your experience and your background, and I am pretty sure that you have developed an excellent job. But don't cover your ears never when hearing something that does not fit with your stuff, despites how old are you.

Now I will answer Fred:
My questions are:
Given this equation (the Bernouilli corrected or whatever Crane calls it):

$$P_1+ \rho g z_1+ \rho \frac{U_1^2}{2}=\lambda L \rho \frac{U_1^2}{2D}+P_2+ \rho g z_2+ \rho \frac{U_2^2}{2}$$

or the similar version you use, that is coming from a differential version:

First:
How are you integrating for obtaining U_1 and U_2 constants when the profiles are not constants? (recall Poseuille profile in viscous flow)

Second and much more important:
How do you know that the energy dissipated on the RHS is proportional to the Kinetic Energy?.

The answer to both of them is that it does not make sense to use it in a Low Re flow. On the other hand, we do know that energy dissipation is proportional to the kinetic energy of the large scales in the bulk of a turbulent flow. As we approach to the wall, the small scales are dominant and the roughness of the wall starts to be of the order of the viscous length. There the energy dissipation is proportional to the viscous stresses, but such contribution is of second order in a fully turbulent flow.

Last edited by a moderator: Apr 22, 2017 at 3:18 PM
17. Jan 22, 2007

### Q_Goest

Hi Clausius.
I mean the Darcy-Weisbach equation. Not the Darcy friction factor f, the equation. Please see my post above and the reference to the LMNO web site. You can also find it on Wikipedia, here:
http://en.wikipedia.org/wiki/Darcy-Weisbach_equation
and here:
http://biosystems.okstate.edu/darcy/DarcyWeisbach/Darcy-WeisbachHistory.htm

Note also this equation is listed on your own reference from the gentleman in Tiawan on sheet 23 and again on 24, except he fails to mention the equation by name. He does however, mention the "Darcy-Weisbach Friction Factor" though this is generally simply called a "friction factor" or at best, Darcy friction factor. See also this Wikipedia article which correctly defines the "Darcy friction factor".
http://en.wikipedia.org/wiki/Darcy_friction_factor

Please refrain from insults, Clausius. If you don't, I'll have to mention this to one of the moderators.

Regarding the DW equation, it is valid for laminar and turbulent flow. It is not a friction factor equation. It is an equation for calculating pressure (ie: head) loss. Friction factor is a variable in it.

I suspect however, you've confused the Darcy friction factor with the DW equation. The friction factor may be obtained from the Moody diagram. Unfortunately, charts such as this must be interpreted mathematically for use in computer programs. I'd strongly recommend the use of explicit equations as given by the Engineering Tips forum FAQ I posted above. From that post:
Reference: http://www.eng-tips.com/faqs.cfm?fid=1236

From this remark, it appears obvious you've mistaken the DW equation for the "Darcy" friction factor. Von Karman for example has an equation that is only applicable to highly turbulent flow where Reynolds number aproaches infinity. This isn't applicable to any of my comments thus far.

Here's another good reference regarding the Darcy-Weisbach equation and friction factors.
http://biosystems.okstate.edu/darcy/DarcyWeisbach/Darcy-WeisbachHistory.htm
From that reference:
So yes, it seems some work was done using open channels early on, but that's not where the relationships for friction factor come from, which is from pipe as this reference continues on:
So it seems most of the data is from "sand lined pipes".

Are you refering to the von Karman equation? It's given on slide 60 of the reference you provided!

Regarding the aproximation you used for Re, I've honestly not seen that one before. The Crane paper has a list of 14 different variations on Re, but none of them match the aproximation you gave, nor do any of them provide a method of determining Re without knowing either velocity or flow rate in some way. Please provide a reference to the aproximation you gave.

Using some aproximation when it is simple enough to iterate however, seems like a poor method of doing any analysis.

I'm not covering my ears, Clausius. I'll give you a chance to show me what you've learned, but you need to prove to me you really understand all this.

Regarding the equation you left for Fred, have you checked also the reference you provided? Check out slides 27 and 77. They show Bernoulli's equation modified to handle head losses as well as pump head increases. There is no indication anywhere in his presentation that indicates this equation is invalid for laminar flow. Fred is correct. The energy equation is not restricted to fully turbulent flow, and this also agrees with how the Crane paper handles things.

Note also slides 67 through 77. Much of this data and the concepts of resistance coefficient come from the Crane paper which is updated yearly for practicing engineers. Yes, the presentation you provided is a very good one, and I only spotted the one minor oversight I mentioned above.

18. Jan 23, 2007

### Clausius2

Allright, thanks for not feeding the flames, sometimes I take it more seriously than what I should. Anyways I hope the reader is learning something out of all of this. At least we agree that the stuff of my link has been useful, and still it's gonna be the workhorse of my answer, since you have pointed out partial truths here.

My mistake. I have seen many times that equation but I didn't recall that such a trivial equation may have a name.

Sure it is. The friction coefficient there reaches a certain particular states of similarity when the flow is fully turbulent, not depending on the Re, and when the flow is viscous, not depending on the rugosity.

I meant Turbulent flow in Channels, no in Open Channels. A turbulent flow in a Channel is an essential problem in Turbulence (see Pope's book on Turbulence Flows). From the analysis of the Channel flow it emerges the Von Karman defect law (in slide 50) which is the theoretical basis for the experimentally corrected equation of the slide you posted.

References: G.K. Batchelor 'Introduction to Fluid Dynamics'. IF the flow is viscous (the starting pre-assumption of your virtual iterative process), then the velocities expected are gonna be of order $$U\sim \Delta P a^2/\mu L$$ which comes naturally from the Poiseuille profile:

$$u(x,z)=-\frac{1}{2\mu}\frac{\partial P}{\partial x}z(a-z)$$

Substitute that characteristic velocity on the Reynolds number and you will obtain what I showed. If the flow is viscous for sure that my velocities are gonna be of that order (sorry man that's physics of fluids) and everything would be coherent. If turns out that the Re so calculated is very large then my assumption is wrong and we should consider other regime (still waiting for the input here of the OP).

Man, don't be like that. I don't know where you been working, but I am working in maybe the best department of Mechanical and Aerospace Engineering as far as Fluid Dynamics is concerned on the whole earth, and you will find that many grants here are provided by powerful institutions of U.S. So don't diminish what I have could learned here about simple pipe flows.

Again you are citing partial truths. Take a look at that slide and tell me what are the $$\alpha$$. They come from the Integral Momentum Equation and the Integration of a Velocity profile averaged. Even for viscous flows, it represents something less accurated than the exact Poiseuille theory and the equation of energy for viscous flows. Take a look at slide 67. My question is, again, (you didn't answer), how is it possible that in general (for the whole range of Re) the ENERGY loss is gonna be directly proportional to the Kinetic ENERGY!!! (where is the viscosity playing at low Re???). The key, Q-Goest, is that the equation I posted is the result of an integration over a fluid volume, each one of the variables are averaged. And now, how am I going to integrate an hypothetical ENERGY loss which coefficient depends itself on the fluid velocity for not Fully turbulent flow??? (If the flow is fully turbulent the DW coefficient reaches a similarity regime in which it does not depend on Re but on rugosity). To my point of view, you are right about the DW equation, but your jump from there to your energy equation is mistaken.

I hope you read carefully this last paragraph, it is my main conclusion.

Last edited: Jan 23, 2007
19. Jan 23, 2007

### Clausius2

And I am going to add an addendum:

Even though you may have much more experience than me on all these stuffs, I think I am satisfied enough if a simple grad student 24 years old with only the experience of the books has been able to challenge your knowledge of several years in well known companies here and to make you develop the best of you in this and past threads.

20. Jan 23, 2007

### Q_Goest

Hi Clausius,
sure it is? The Darcy-Weisbach equation is for calculating pressure drop. The friction factor in it changes depending on flow regime and Re. Flow regime is dependant on surface roughness (what you're calling rugosity*). Are we in agreement on this? And once the flow is compleltely turbulent, friction factor no longer is dependant on Re, right? But friction factor is still dependant on surface roughness, right? Let's verify we at least agree on these basics.

$$U\sim \Delta P a^2/\mu L$$
$$Re=\frac{\rho Ua}{\mu}\sim \frac{\rho \Delta P a^3}{L\mu^2}$$