What is the Flow Rate of Water Through a Square Tube?

[quark]

Q_Goest,

There are two excellent papers that give you the theory behind development of explicit equations of friction factor. One is by Churchill (Friction-factor equation spans all fluid-flow regimes) appeared in November, 77 issue of Chemical Engineering Journal. The other one is by Chandra Verma (Solve pipe flow problems directly) appeared in August, 79 issue of Hydrocarbon Processing.

When I was comparing various friction factors, Churchill's was deviating significantly from Colebrook's at low Reynolds numbers. One member(katmar) at eng-tips modified Churchill's equation by using the method given by Verma and this single equation is in excellent agreement with both Poiseuille's as well as Colebrook's friction factor. Further, plotting of this equation gives you a continuous curve over a wide range of Reynolds numbers. That is why, I presumed this one single equation can give us correct friction factor in Laminar, Transient and Turbulent regimes as well.

I checked my pressure drop calculator with other commercially available software and freeware. The values are matching at higher Reynolds numbers. Please note that, so far, I didn't see any commercial software available based on 2-K and 3-K methods. So, it is difficult to check the results at lower Reynolds numbers. However, the papers of Hooper and Darby show excellent curve fit of the experimental data.

 

[Q_Goest]

Hi Quark. OMG, is this the same quark I've been quoting from EngTips? LMAO If so, thanks for the great FAQ post. Regardless, yes I'd like to see a good correlation for the transient zone. Was this one of the equations posted on the FAQ?

I've been using an explicit equation handed down to me from a guy I worked with about 10 years ago, and I've cheked the accuracy of it but of course it's only valid in the fully turbulent zone. I'll send you a PM with my email - I'd love to see the paper your referring to. Thanks

 

[blaster]

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.

I have been trying to use this but the dimensions do not cancel out at the end.

using SI units:

for $$\rho$$ I have $$\frac{kg}{m^3}$$

for $$\Delta P$$ I have $$\frac{N}{m^2}=\frac{kg m}{m^2}=\frac{kg}{m}$$

for $$a^3$$ I have $$m^3$$

for $$L$$ I have $$m$$

for $$\mu^2$$ I have $$Poise^2 = Pa^2 s^2 = \frac{N^2 s^2}{m^4}= \frac{kg^2 s^2}{m^2}$$

I always end up with s^2 as the units of this equation. Please help me see where I have gone wrong.

everything cancels out but the s^2

*** Nevermind. Clausius2 pointed out that Newton is not a kgm it is a kgm/s^2

 

[KR79]

I actually needed to do the same calculation for my research and came across this thread. Very helpful information from the person who posted the "exact solution" from the text. Now

P2 - P1 = Q * R

I did the calculation from the textbook equation 3-48 for the square tube case and solved for the R. Here's the resutling equation

P2 - P1 = Q * 1.7784 (l*u/a^4)

where l is the length of your tube, a is the width of your channel divided by 2, and u is the dynamic viscosity of your fluid. Plug in your flow rate (volume/time) and you have your pressure drop.

Enjoy. This equation only applies for laminar flow, so do your Reynold's # calculation first.