## Eliptical spheres

## Eliptical spheres

(OP)

Does anybody know of a good approximation for the partial volume of an elliptical sphere?

I'm mostly interested in one where a=b=2c e.i. a ball that has been "sqeesed". the sphere is "standing" on the large diameter the other one i know Its difficult to explain but i want to calculate the volume of a vessel with spherical end caps. The cylinder is easy and i could cheat and either say the the vessel was 0.5*diameter longer and assume flat ends or that the end caps was normal spheres - but i know that they have said geometry.

The volume should then be a function of how full the shere is. Its OK if the formular only "works" (for geometrical reasons) up to half full.

Best Regards

Morten

I'm mostly interested in one where a=b=2c e.i. a ball that has been "sqeesed". the sphere is "standing" on the large diameter the other one i know Its difficult to explain but i want to calculate the volume of a vessel with spherical end caps. The cylinder is easy and i could cheat and either say the the vessel was 0.5*diameter longer and assume flat ends or that the end caps was normal spheres - but i know that they have said geometry.

The volume should then be a function of how full the shere is. Its OK if the formular only "works" (for geometrical reasons) up to half full.

Best Regards

Morten

## RE: Eliptical spheres

^{3}for a single vessel cap or half an ellipsoid (this is not a sphere!).If you are curious 0.131 is pi/24

prex

motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

## RE: Eliptical spheres

Instead try this solution:

A semi ellipsoidal head, part of a horizontal vessel, with a diameter of 1m and filled to a level of H% of the diameter will contain the following volumes:

(%fill is volume in % of total filled)

H %fill Vol in liters

0% 0.0000 0.0

5% 0.0073 0.9491

10% 0.0280 3.6657

15% 0.0606 7.9351

20% 0.1040 13.6175

25% 0.1563 20.4568

30% 0.2160 28.2797

35% 0.2819 36.8967

40% 0.3521 46.0856

45% 0.4254 55.6825

50% 0.5000 65.4509

55% 0.5746 75.2193

60% 0.6479 84.8162

65% 0.7181 94.0051

70% 0.7840 102.6221

75% 0.8437 110.4450

80% 0.8960 117.2843

85% 0.9394 122.9667

90% 0.9720 127.2362

95% 0.9927 129.9527

100% 1.0000 130.9018 (same as prex's formula)

Put these numbers in an excel chart and you will have a curve to cover also intermediate values.

For vessels with other ID's simply multiply the %fill with the volume of the 100% filled head from prex's formula.

I made these numbers by drafting the head in a 3D CAD programme, slice it up and let the programme calculate the volumes.

If someone knows how to produce a formula from curve values, (Fourier series or something like that), it could be turned into an excel function. I will be happy to establish the values for torispherical heads too.

By the way Morten, if you might be my ex-colleague, then mail me at mgp@kabelnettet.dk. I can then mail the chart.

Regards Mogens

## RE: Eliptical spheres

Thanks a lot.

Best Regards

Morten

## RE: Eliptical spheres

The formula is:

V=pi*ID^{3}*f*(2-r-r^{3})/24where

ID=diameter of circulare basef=ratio of 2*(third axis) toID(0.5 in your case)r=ratio ofunfilledheight to total height of head or cap or emiellipsoid.With

r=1 the head is unfilled and you getV=0With

r=0 the head is full and you get the formula of my previous post.prex

motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

## RE: Eliptical spheres

**

*

*

*

*

*

*****

**** <= this volume

***

**

Maybe prex has got a formula for this one too?

That would then doo a good approximation, but in any case when it comes to torispherical heads which are normally used for ASME vessels (often treated as semiellipsoids in these calculations), it starts getting hairy to solve by formula.

regards

Mogens

## RE: Eliptical spheres

mgpwas speaking about an horizontal vessel.Now the problem is to find the volume of half the segment of an ellipsoid with two equal axes, the segment being cut perpendicular to the main circle.

The formula I find is now:

V=pi*ID^{3}f(2-3r+r^{3})/48where now

ris the ratio of unfilledradiusto full radius and the formula works only up to half filling (beyond that is simply a matter of simmetry).For more clarity,

r=0 means half full,r=1 means totally unfilled.prex

motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

## RE: Eliptical spheres

The math is OK but maybe the clarity of the question is not. Im looking for any given % of filling inbetween 0-100% (0-50% will do since i will assume symmetry).

Best Regards

Morten

## RE: Eliptical spheres

I re-examined your answer. What does "f" in the formular mean?

Best Regards

Morten

## RE: Eliptical spheres

fis defined in my second post. For a common so called 2:1 emiellipsoidal head (third axis is half the other two),f=0.5.I don't understand what you mean by % filling. If you want to express the result in a fractional way, just divide the volume found with my formula in my third post by the total volume of the head given by the formula in my first post.

prex

motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

## RE: Eliptical spheres

The first is named

VesselContent(ID, L, H)

and returns the liquid volume

where

ID = vessel ID

L = length of vessel (tan to tan)

H = liquid level

(prex's formula is slightly modified so it works all the way from H=0 to H=ID)

The second is named

VesselLevel(ID, L, Vol)

and returns the actual level in the vessel

where

ID = vessel ID

L = length of vessel (tan to tan)

H = liquid volume

(this one iterates from the first and won't work on its own)

Here goes:

*****************************************

Public Function VesselContent(ID, L, H)

Dim vaHeadContent As Variant

Dim vaShellContent As Variant

vaHeadContent = Application.Pi() * ID ^ 3 / 96 * (2 + Abs(Application.RoundUp((1 - 2 * H / ID), 0)) * ((1 - 2 * H / ID) ^ 3 - 3 * (1 - 2 * H / ID)))

vaShellContent = L * (ID ^ 2) / 8 * (2 * Application.Acos((ID - 2 * H) / ID) - Sin(2 * Application.Acos((ID - 2 * H) / ID)))

VesselContent = 2 * vaHeadContent + vaShellContent

End Function

Public Function VesselLevel(ID, L, Vol)

Dim vaHtest As Variant

Dim vaVolTest As Variant

vaHtest = ID

Do

vaVolTest = VesselContent(ID, L, vaHtest)

vaHtest = vaHtest * Vol / vaVolTest

Loop While Abs(Vol - vaVolTest) > 0.000001

VesselLevel = vaHtest

End Function

regards Mogens

mgp@kabelnettet.dk

## RE: Eliptical spheres

Dan Jones

Dan.Jones@degussa.com

## RE: Eliptical spheres

A horizontally constructed 2:1SE with no contained flat ended cylinder contains half the volume of a sphere of the same vertical diameter containing the same height of liquid. The volume contained in a sphere is derived from:-

= Pi()*h^2 * (D/2 - h/3)

- - - excuse my mix of variables and s-sheet functions!! where h is liquid height and D is diameter.

Thus the volume contained in BOTH 2:1SE's can, more simply than by mgp's function, be derived (0 - D) from:-

= Pi()*h^2 /2 * (D/2 - h/3)

Despite having recently made this "discovery" I am still using the following formulae for 2:1SE's:-

SumEnds= Pi() /8 *D^2 *(((D-2*h)^3 /D^2 - D) /6 + h)

& with T as Tan Length:-

FlatEnded= T*(ACOS((D-2*h)/D)*D^2/4-SQRT(h*(D-h))*(D-2*h)/2)

TotalVol = SumEnds + FlatEnded ...results identical to mgp.

Since 1977 these formulae have been migrated from HP97Calc to Applesoft to PC to Excel but my sources have long-since been lost and forgotten. - - - regards, bjt.

## RE: Eliptical spheres

www.gowelding.btinternet.co.uk/calcs/indexc.htm

hope this helps

## RE: Eliptical spheres

http://unicon.netian.com/tank_vol_e.html

Jin Suk Lee maintains this site and includes lots of engineering information. The above URL contains MS WORD downloadable documents containing formulas for fluid volume at arbitrary levels in various horizontal and vertical tanks and some supplemental formulas for some special geometric shapes which are used for tank heads. The supplemental formulas follow readily from the general formulas in the larger MS WORD document which was published in Nov 2002 Chemical Processing Magazine and can be seen and printed at www.chemicalprocessing.com under "Monthly Articles". The graphics printing from the latter is very poor and the formulas are hard to see, so download the MS WORD document from Lee's WEB site and the formulas will be easy to see.

Dan.Jones@degussa.com

## RE: Eliptical spheres

Pi()*h^2*(3c-h)/3

Now mentally stretch the shape in one horizontal direction by a ratio a/c, and in the other horizontal direction by b/c. You have now converted the full sphere of radius c to an ellipsoid of semi-axes (a,b,c) without changing the depth of the cap, and in the process you have increased the volume of the cap by a factor of a*b/c^2. (The resulting formula is correct for the full range of physically valid depths 0<=h<=2c.)

## RE: Eliptical spheres

V(partial)= 2*pi*a^2*(y'+b)-2/3*pi*(a/b)^2*(y'^3+b^3)

whereby a= x values when y=0

b= y values when x=0

y'= depth of liquid taken from

abscissa located at center of

ellipsoid-values can be neg. or pos.

To that equation you can add the cylindrical portion of the tank and also realize that the tank would include two halves ellipsiodal heads.

## RE: Eliptical spheres