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

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

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

(OP)
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?

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

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.

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

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.

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

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)=πHD2(h/D)2(1-2(h/D)/3)

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

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

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.     

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

(OP)
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.

http://vps.arachnoid.com/TankCalc/   

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