# Talk:Venturi tube

## This is a new article

This articles was written from scratch. Milton Beychok 15:30, 20 March 2010 (UTC)

## Bernoulli equation

I have a (small) problem with the height *h* in the Bernoulli equation. Now height is defined of a **point**, which implies that there may a difference in gravitational attraction over the cross section (as opposed to the length) of a tube. Later *h*_{1} is canceled against *h*_{2} for the gravitational terms in a horizontal tube, which implies that *h* is assumed constant over the cross-sectional dimension. It seems to me that *h* is (approximated as) a function of *one* dimension. In cylinder coordinates with the axis of the tube as *z*-axis, my guess is that *h* is a function of *z* only and not of *r* and θ. This is physically reasonable and allows cancellation of the gravitational terms in a horizontal tube. --Paul Wormer 07:49, 29 March 2010 (UTC)

- Paul, if you believe that the Bernoulli equation should use
*z*rather than*h*, feel free to change it. Milton Beychok 15:04, 29 March 2010 (UTC)

- My problem is not with notation
*h*versus*z*, but with the dependence of gravitational force on position. The Bernoulli equation as you give it is correct but it seems to me that when you cancel*ρgh*_{1}against*ρgh*_{2}you make an extra approximation (on top of assuming that*ρ*_{1}=*ρ*_{2}). --Paul Wormer 15:16, 29 March 2010 (UTC)

- My problem is not with notation

- I confess that you have lost me. If we use z and define z as the centerline (or axis), would that clear up the problem? Milton Beychok 15:25, 29 March 2010 (UTC)

- I'm okay with the way you changed the definition of h. The greater potential energy of any point in the fluid above the centerline would be balanced by the lesser potential energy of the corresponding point below the centerline. Milton Beychok 16:00, 29 March 2010 (UTC)

- How can gravitational forces cancel? They all point downward. Moreover, in the derivation you don't cancel points below or above the centerline, but of point 1 in the inlet against point 2 in the throat. What one would have to do is the following: Take an infinitesimally thin slice in the throat (at point 2) with diameter of the throat and calculate its volume. Cut out a similar slice in the inlet (at point 1) and calculate its volume. Since the diameters of these two slices are different their volumes are too. If the density is the same in throat as in the inlet, the masses contained in the respective slices are different and you cannot cancel them, but strictly speaking have to introduce some sort of correction. --Paul Wormer 16:29, 29 March 2010 (UTC)

- I did not say the gravitational forces "cancel". What I was trying to say was that points above the centerline would have a higher potential energy than the corresponding points below the centerline ... and thus, the potential energy of the corresponding point on the centerline represents the average potential energy in that vertical slice (or cross-sectional area). That is what I meant by "balanced". I hope that I am making sense now. Am I?? Milton Beychok 17:25, 29 March 2010 (UTC)

- Yes, the volumes and masses at the inlet and the throat are different ... but so are the linear velocities. Thus, the volumetric flow rates are the same and it is the volumetric flow rate (Q) that is being derived for incompressible fluids. And if the density of an incompressible fluid is constant, than the mass flow rate is also constant. Or am I missing your point? Milton Beychok 17:25, 29 March 2010 (UTC)

[unindent]

Milt, I finally see your point. I mistakenly thought that you canceled the gravitational term on the lhs of the Bernoulli equation against the term on the rhs, but now I see that for a horizontal tube the terms themselves are zero. But since this is not completely self-evident, I suggest that we add the following paragraph (or something like it):

- The Bernoulli equation contains gravitational terms of the form
*ρgh*. For a horizontal tube these terms vanish. To see this we recall that the zero level of gravitational energy can be chosen freely at any height. Take a horizontal plane passing through the centerline of the Venturi tube and use this plane as zero level, then volumes of mass above the plane have positive gravitational energy and volumes below the plane have negative energy. Because the amount of mass above the plane is equal to the amount below it, the total gravitational energy adds up to zero.

This paragraph summarizes how I understand your reasoning. --Paul Wormer 07:17, 30 March 2010 (UTC)

- Paul, I've said it before and I say again that you continue to amaze me! I have added essentially your explanation as a referenced footnote. I did not want it to interrupt the flow of the text of the "Incompressible fluid" section. Thanks for your help.Milton Beychok 20:45, 30 March 2010 (UTC)

## Beta

I added another definition for β, but on second thought it seems that the first definition (**d/D**) is superfluous. By referring to *diameters* it is assumed that tubes are cylindrical and then **d** and **D** do not need to be introduced in addition to the respective cross sections **A _{1}** and

**A**. For non-cylindrical ducts one needs more dimensions, e.g., width and height for rectangular shaped ones.--Paul Wormer 08:01, 29 March 2010 (UTC)

_{2}- I will revise the the definition of beta to use the areas as you suggest, and thanks for your comments. Milton Beychok 15:04, 29 March 2010 (UTC)