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pressure drop due to obstruction in pipe 1

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DYV1973

Mechanical
Aug 9, 2004
75
LS,

If a small cylinder is placed in a pipe, what will be the pressure drop.

The force acting on the cylinder will be:

F=Cw*A*0,5*rho*v^2

Flows are turbulent (Re>10000).

Should I use an energy approach?

Thanks DYV
 
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Treat it as a reduction in area, followed by a length of the same reduced area pipe, followed by an expansion.

Good luck,
Latexman
 
except that it is an annular ring, the wetted perimeter is much higher so friction will be greater.

this has been discussed before:

thread378-236832
 
Annular, but for near 0 distance, ie. not long.

Let your acquaintances be many, but your advisors one in a thousand’ ... Book of Ecclesiasticus
 
I might have been not clear enough in explaining the situation.

The flow is normal to the surface of the cylinder. The cylinder is fixed to the piping wall.

Piping is DN50, and the cylinder has a diameter of 20 mm and a length of 35 mm.

On the internet I found a picture of how the flow is around the cylinder in this existing situation.

Flow will be approx. 18 m3/h (is approx. 2,5 m/s) and the fluid is light viscous (estimated at 10 mPas), which makes flow turbulent.

Pressure drop is very low, but it needs to be quantified.

DYV
 
 http://files.engineering.com/getfile.aspx?folder=59d20879-f474-423b-827b-266d403a7afb&file=Fig_1.jpg
It looks like a CFD model - maybe thats the way? Is it an annubar or similar - if so - then the vendor may have info.

Best regards

Morten
 
Won't the spaghetti get damaged going thru the pump?

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
 
You can check this thread378-19596. I agree with majority of members in the thread and ignore such pressure drops.





 
If it's a real CFD work that you have, then you have the answer. define 2 planes right before and after the cylinder and calculate (well, ask the program to calculate!) the average pressure on each plane. The difference is your pressure drop.
 
Unfortunately I do not have CFD. The picture I just found on the internet, to illustrate what I meant.
 
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