Register to reply 
Flow rate through a squre tube 
Share this thread: 
#1
Jan2107, 10:09 PM

P: 11

Find: Flow rate of water through square tube
Given: Crosssectional 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
Jan2207, 12:08 AM

Sci Advisor
PF Gold
P: 1,479

Here are some dimensional analysis that an engineer should know how to do: Reynolds Number: [tex] Re=\frac{\rho Ua}{\mu}\sim \frac{\rho \Delta P a^3}{L\mu^2}[/tex] where [tex]a[/tex] is the pipe hydraulic diameter and [tex]L[/tex] is the pipe length. 


#3
Jan2207, 12:19 AM

P: 11

Let me clarify P2=0bar.
How do you do the "corrected for a square section" part 


#4
Jan2207, 12:26 AM

Sci Advisor
PF Gold
P: 1,479

Flow rate through a squre tube
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
Jan2207, 06:33 AM

Sci Advisor
HW Helper
PF Gold
P: 2,906

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


#6
Jan2207, 06:41 AM

Sci Advisor
P: 5,095

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
Jan2207, 07:23 AM

Sci Advisor
HW Helper
PF Gold
P: 2,906

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
Jan2207, 09:43 AM

P: 11

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


#9
Jan2207, 11:08 AM

Sci Advisor
P: 5,095

In regards to the OP, you do indeed need to use the hydraulic diameter in your calcs. The hydraulic diameter is defined as: [tex]D_h = \frac{4*\mbox{cross sect area}}{\mbox{wetted perimeter}}[/tex] 


#10
Jan2207, 11:59 AM

Sci Advisor
HW Helper
PF Gold
P: 2,906

Hi Fred. Nice award, I think you earned it!
Hi Blaster, Equations for fluid flow, such as DarcyWeisbach, HagenPoiseuille, 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: Wikipedia 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 DarcyWeisbach 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: LMNO Engineering  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. Wikipedia Reynolds Number 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: Viscosity for Water 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: Friction Factor at Eng Tips 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 DarcyWeisbach 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. 


#11
Jan2207, 01:35 PM

Sci Advisor
PF Gold
P: 1,479

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
Jan2207, 02:27 PM

Sci Advisor
HW Helper
PF Gold
P: 2,906

~ ~ 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. 


#13
Jan2207, 03:12 PM

Sci Advisor
P: 5,095

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.



#14
Jan2207, 03:19 PM

Sci Advisor
PF Gold
P: 1,479

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.



#15
Jan2207, 04:57 PM

Sci Advisor
HW Helper
PF Gold
P: 2,906

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: 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: LMNO Engineering 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. 


#16
Jan2207, 06:14 PM

Sci Advisor
PF Gold
P: 1,479

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. http://www.esoe.ntu.edu.tw/courses/5...20version).ppt There you are gonna find familiar comments. 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: Given this equation (the Bernouilli corrected or whatever Crane calls it): [tex]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}[/tex] 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. 


#17
Jan2207, 08:12 PM

Sci Advisor
HW Helper
PF Gold
P: 2,906

Hi Clausius.
http://en.wikipedia.org/wiki/DarcyWeisbach_equation and here: http://biosystems.okstate.edu/darcy/...achHistory.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 "DarcyWeisbach 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 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: http://biosystems.okstate.edu/darcy/...achHistory.htm From that reference: Using some aproximation when it is simple enough to iterate however, seems like a poor method of doing any analysis. 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
Jan2307, 12:25 AM

Sci Advisor
PF Gold
P: 1,479

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.
[tex] u(x,z)=\frac{1}{2\mu}\frac{\partial P}{\partial x}z(az)[/tex] 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). I hope you read carefully this last paragraph, it is my main conclusion. 


Register to reply 
Related Discussions  
Ppm and flow rate  General Physics  4  
Flow Rate  Mechanical Engineering  6  
Gas Flow through a tube  Classical Physics  15  
Liquid flow down a small tube  General Physics  1  
Time of water flow in a tube  Advanced Physics Homework  0 