Talk:Venturi tube: Difference between revisions
imported>Milton Beychok m (→Bernoulli equation: More dialogue) |
imported>Milton Beychok m (→Bernoulli equation: Still more dialogue) |
||
| Line 14: | Line 14: | ||
:::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? [[User:Milton Beychok|Milton Beychok]] 15:25, 29 March 2010 (UTC) | :::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? [[User:Milton Beychok|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. [[User:Milton Beychok|Milton Beychok]] 16:00, 29 March 2010 (UTC) | |||
== Beta == | == Beta == | ||
Revision as of 11:01, 29 March 2010
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 h1 is canceled against h2 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 ρgh1 against ρgh2 you make an extra approximation (on top of assuming that ρ1=ρ2). --Paul Wormer 15:16, 29 March 2010 (UTC)
- 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)
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 A1 and A2. 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)
- 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)