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Volume of a (horizontal) 2:1 ellipitical head at various depths 2

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ironstar

Civil/Environmental
Oct 2, 2006
2
Can anyone give me a formula to determine the volume of a horizontal 2:1 elliptical tank head at various levels of fluid fill?

The total volume is easily calculated, but I cannot find any source to determine the volume of a horizontal segment. E.G., If I have an 2200mm 2:1 tank head, the depth of the head is 550mm. What volume would be present in the head at say, 385mm?
 
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0.785 m^3.

I used a 3D cad program to build it and calculated the volume. The more scientific way is to write the ellipse equaition and use tripple integration.
 
Remember that the CL is assumed horizontal for this problem. 8<)

Did the same thing one (used a 3D CAD program to calc the volume of a comparable tank head) but it tried to calculate the "volume filled" by the tank walls, not the "volume between the tank wall and a vertical surface."

Took a few minutes to figure out why the volume was so far off.
 
Look in the first site below, under Vessels -> Volumes -> Horiz. -> Ell. , then choose Options -> Show formulae in the menu.
BTW here is the (quite simple) formula (for the two heads):
V(h)=[&pi;]HD2(h/D)2(1-2(h/D)/3)

prex
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I think there are a couple of unknowns that will affect that equation:

He's looking for the volume of a partially-filled tank to calibrate his level guage: So the formula needs a "HT" or "Depth" value.

He will need to add the length between welds for each elliptical head: L = ?

and

for what length "tangent length" (TL = ?) to account for the "straight leg length" (usually between 2 - 3 inches) on each elliptical head between the small radius bend ("d" I assume) and the head weld.

So the "horizontal straight length" of his tank will be: TL + L + TL, filled to some height HT.
 
Thanks to all for your input. The link below advised by hydtool is excellent giving me exactly what I was after. If you work with process tanks in any way, save this.

 
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