Eliptical spheres 3

[MortenA]

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

 

[prex]

Use 0.131x ID³ 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

 

[mgp]

I believe what you're looking for is how to determine the content volume of a part-filled horizontal vessel with ellipsoidal ends. I haven't seen any straight forward formula for this, but I've got a formula with pages of coefficients, which then won't help you either. Another problem is that most vessel heads are torispherical (partly a sphere, partly a torus) rather than ellipsoidal which makes the calculation even more impossible.
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

 

[MortenA]

mgp: You got it right. Thanks for the table. I will use this if no one else coems up with a formular.

Thanks a lot.

Best Regards

Morten

 

[prex]

So you need the volume of a segment of an emiellipsoid with circular base.
The formula is:
V=pi*ID³*f*(2-r-r³)/24
where
ID=diameter of circulare base
f=ratio of 2*(third axis) to ID (0.5 in your case)
r=ratio of unfilled height to total height of head or cap or emiellipsoid.
With r=1 the head is unfilled and you get V=0
With 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

 

[mgp]

prex's formula is ok, only thats for filling the lower head of a vertical vessel. The one I think Morten needed was this one (horizontal vessel):

**
*
*
*
*
*
*****
**** <= 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

 

[prex]

Excuse me all of you, but I didn't notice mgp was 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*ID³f(2-3r+r³)/48
where now r is the ratio of unfilled radius to 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

 

[MortenA]

Hello Prex

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

 

[MortenA]

Prex

I re-examined your answer. What does &quot;f&quot; in the formular mean?

Best Regards

Morten

 

[prex]

f is 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

 

[mgp]

Based on prex's excellent formula I made two user defined functions which you can copy:

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

 

[drdjones]

The volume of fluid in a partially-filled spheroid with semiaxes a, b, & c along the x, y, & z axes, respectively, where the fluid height along the z-axis is h (0<=h<=2c), is given by V=(pi)*a*b*h^2*(1-h/(3*c))/c. This formula can be extracted from my tank volume article published in Nov 2002 in Chemical Processing Magazine. The case of a torispherical head can be exactly solved also, but not as simply.

Dan Jones
Dan.Jones@degussa.com

 

[bjt]

This is rather an old thread but drdjones' recent reply brought it to my attention. Prex's formula employing f=0.5 and mgp's function apply to (what I call) 2:1Semi-E's.
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 &quot;discovery&quot; 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.

 

[iken]

Try this link, this site has other good reference material also.

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

hope this helps

 

[drdjones]

See:

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

http://www.chemicalprocessing.com

under &quot;Monthly Articles&quot;. 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

 

[Denial]

Drdjones, in his post of 21 May, is correct. This can be readily confirmed by starting with a spherical cap of radius c and waterdepth h. The exact volume of this cap, as presented in numerous texts, is
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.)

 

[chicopee]

why dont you use calculus to determine partially filled volumes of ellipsoids. I just derived such formula which does not include the cylindrical portion of the proposed tank and which assumes both heads to be semi ellipsoidal. I did not test the formula but I think that is correct:

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.

 

[25362]

An old article appearing in the Chemical Engineering issue of June 11, 1973, under the heading of Plant Notebook, titled "Quick Calculation For Holdups In Horizontal Tanks" with flat, spherical, dished and elliptical heads may be of help.