How Do You Calculate the Speed of Air Over a U-Tube?

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Homework Help Overview

The discussion revolves around calculating the speed of air over a U-tube that is partially filled with water and oil. The problem involves understanding the relationship between air pressure and fluid heights in the tube, with specific measurements provided for the oil height and air density.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to establish an equation to relate the pressures and heights in the U-tube, questioning how to incorporate the speed of air into their calculations. Other participants express uncertainty and seek clarification on the relationship between air pressure and fluid height.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of pressure equations, but there is no explicit consensus on how to proceed with the calculations.

Contextual Notes

Participants are working with specific measurements and assumptions, including the density of air and the height of the oil. There is an indication of confusion regarding the effects of air pressure on the calculations, which may impact the approach taken.

thenewbosco
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A problem i have encountered:

diagram at http://img.photobucket.com/albums/v11/biggm/dia.jpg

Here is a u-tube open at both ends partially filled with water and oil with given density is poured in the right column so the height of the oil is 5.0 cm.

i have calculated the difference in height to be 1.25 cm.
in the second picture, air motion on the right is shielded while air is blown over the left opening until the surfaces are at the same height.
i am to calculate the speed of the air taking the density of air to be
1.29kg/[tex]m^3[/tex].

I do not know how to create the equation needed to solve this.
i am thinking i need to find the external pressure minus the effect of the blown air on the left since it is less than the right which is just the external atmospheric pressure.

thanks for help on this
 
Last edited by a moderator:
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anyone... .
 
someone ...anyone
 
right..the equation should be the same as last time for the two depths, but how does the speed of the air factor into the Po. On the side with the shield it will be just Po + density*g*h and on the other side it will be P + density*g*h. How can i factor the air into the equation for P on the left?

it must be Po - effect of air...but how
 
Last edited:

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