Volume of a Horizontal Cylindrical Tank with Elliptic Heads

Date: 11/15/2003 at 15:15:48
From: Francois
Subject: Volume of a horizontal tank with elliptic heads

I need an equation that will relate the height of liquid and the 
volume of liquid partially filling a horizontal tank with elliptic
heads.  

The main part is cylindrical, and with elliptic heads on both sides,
that could be considered as half an oblate spheroid with semi-axes a,
b, c with b = c = 2a and 2b = diameter of the tank.

The volume of a half-ellipsoid is simple : V = 2/3*pi*h*(e/2)^2, with 
h the height of liquid in the bottom head of the tank and e the 
diameter of the liquid surface at h, and can be used for a vertical 
tank.  But what about a partially filled horizontal tank?

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Thanks for your help.

Date: 11/15/2003 at 19:03:39
From: Doctor Jeremiah
Subject: Re: Volume of a horizontal tank with elliptic heads

Hi Francois,

Half the ellipsoid is on one end and half is on the other.  Together
they make a whole ellipsoid.  So the problem boils down to a partially
filled cylinder and a partially filled ellipsoid.

The ellipsoid has radii in three different directions: in the vertical
direction the radius is z = R and in the horizontal direction along
the tank the radius is x = a and in the other horizontal direction the
radius is y = R.  So we can use the formula for a partially filled
sphere (a spherical cap) and modify the result by a factor of a/R to
compensate for the one shorter/longer radii.

A spherical cap has a volume of:

   SphericalCapVolume = (Pi/3)(3hR - h^2)h
   where h is the depth of fluid

But when we modify it to he an ellipsoid we get:

   EllipsoidalCapVolume = (a/R)SphericalCapVolume
   where h is the depth of fluid
   and a is the thickness of the ellipsoidal ends

Which comes out to:

   Volume = (a/R)(Pi/3)(3hR - h^2)h
   where h is the depth of fluid
   and a is the thickness of the ellipsoidal ends

Here is an answer from the archives that talks about this:

  Variable Volumes in an Oblate Spheroid
    http://mathforum.org/library/drmath/view/61901.html 

Now, if you search the Dr. Math archives for the words horizontal tank
you will find that the volume of a partially filled horizontal
cylinder is:

   Volume = L(R^2*acos((R-h)/R) - (R-h)*sqrt(2Rh-h^2))
   where h is the depth of fluid

So all we need to do to get the volume of the entire tank at any level
is add the volume of the partially filled cylindrical part and the
partially filled ellipsoidal part that adds on the ends.

   Volume = L(R^2*acos((R-h)/R) - (R-h)*sqrt(2Rh-h^2))
          + (a/R)(Pi/3)(3hR - h^2)h
   where h is the depth of fluid
   and a is the thickness of the ellipsoidal ends

When the tank is full h = 2R and:

   Volume @ h=2R = L(R^2*acos((R-2R)/R)-(R-2R)*sqrt(2R(2R)-(2R)^2))
         + (a/R)(Pi/3)(3(2R)R - (2R)^2)(2R)

   Volume @ h=2R = L(R^2*acos(-1)+R*sqrt(4R^2-4R^2))
         + (a/R)(Pi/3)(6R^2 - 4R^2)(2R)

   Volume @ h=2R = L(R^2*Pi) + (a/R)(Pi/3)(4R^3)

   Volume @ h=2R = L Pi R^2 + (4/3) Pi a R^2

At this point its clear that the first term is the volume of a 
cylinder and the second is the volume of a sphere with a modifier to
make it into an ellipsoid.  So the formula is correct.

Let me know if you have other questions or need more details about
this answer.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/