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Pressure Drop

Pressure Drop

Pressure Drop

(OP)
This one goes towards the same redesign from my previous question on axial force from torquing a screwed in piece (though unrelated).
A picture of what I'm working on:
http://img13.imageshack.us/img13/1653/32593323.jpg

I have water flowing at a specified GPM between the green "rod" and the I.D. of the glass. It eventually reaches this unique designed end piece. Because the end piece is going to be a bottle neck (smaller area), how would I go about calculating pressure drop? The geometry is a bit difficult to measure exactly (and is changing on this redesign), so I am currently just estimating the area of the end piece's cutouts.

Since there is no change in height, should I just be using something like Bernoulli's equation? I know the flow rate, so I could divide by the area between the rod and I.D. to get one velocity, and divide by the estimated area of the orifices to get the other velocity, then use something like:

Pressure Drop=((V_1^2-V_orifice^2)/2)*rho
where V_1 = area between rod and I.D., and V_orifice is the total cutout area (estimate).

I also found this equation in my fluid dynamics book...


Pressure Drop=(FlowRate^2)*rho*(1-(Area_orifice/Area_flow))/(2*Area_orifice^2)

Any input would be appreciated!
 

RE: Pressure Drop

Hi KevinH

There are better qualified people than me to answer this on here however I would use the Bernoulli equation along with the continuity equation.
How accurate does your answer need to be because using the above won't give any losses for friction unless you include them in the above when setting the equation out.
Also you would need to know what the pressure was on one side of the resriction.

regards

desertfox

RE: Pressure Drop

you need the cross-section of the flow path i.e. change in diameter with axial distance,to better estimate the discharge coefficient
 

RE: Pressure Drop

(OP)
hacksaw, I have both the cross section between the rod and glass tube's inside diameter, as well as a fairly good approximation of the cutouts on the metal piece.

We have two different scenario's we are designing for, a 12 mm I.D., and a 13 mm I.D., with a 7.5 mm rod. The old design was a 10 mm rod, and a 15 mm I.D. on the glass. Using approximations for what I can use for the cutouts on the end piece, which is 32.16 mm^2 for a 13 mm ID, 18.79 mm^2 for a 12 mm ID, and the old design, which was about 30.63 mm^2.

Using the first pressure drop equation:
13 mm ID: 9.148 psi
12 mm ID: 12.74 psi
Old Design: 17.37 psi

Using the second equation:
13 mm ID: 7.0 psi
12 mm ID: 23.42 psi
Old Design: 8.3 psi

The reason for the huge pressure drop with a 12mm ID tube is that the area for the cutouts on the end piece is reduced.

If the flowrate is around 7 GPM, do these numbers seem reasonable?

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