Gas in a Cylindrical Tank with Hemispheric Caps

Date: 02/09/2010 at 10:55:36
From: Matt
Subject: Volume of tank with hemisphere caps

I work with my dad, and we have a tank on our farm.  It is a cylinder 
with hemispheric caps that looks like a large pill:


  _ _
 /   \
 |   |
 |   |
 \ _ /


It stands upright, on end, and holds gas for the tractor.  If I look 
into the tank I can estimate the gas level height.  Beyond that, 
though, it is hard to tell how much gas is left in the tank.  It's 
too heavy to lift and make calculations based on differences in 
weight, for example.

I took calculus in high school, and I know this problem can be solved 
with integrals.  I have the following variables:

   height of tank = htank
   radius of tank = rtank
    height of gas = hgas

With these three variables, how can I set up an integral to figure 
out the volume of the gas in the tank?

I just don't know how I can incorporate the hemispheric bottom and 
top into the integral.  I have tried to use the disc method, and I 
understand that I can revolve a line around an axis; but I don't know 
how to order the variables, and am extremely confused about finding 
the volume of only the gas rather than the entire tank.

Date: 02/15/2010 at 22:33:12
From: Doctor Jeremiah
Subject: Re: Volume of tank with hemisphere caps

Hi Matt,

First let me be sure I understand.  The assumptions I am making:

  1) htank is the height of the entire tank, including the hemispheres
  2) the hemispheres each have height rtank
     (they are actual hemispheres - half a sphere each)
  3) htank and rtank are the interior distances
     (they don't include the thickness of the wall of the tank)

We need to break this into three equations because there are three 
pieces to the tank: the bottom hemisphere, the middle cylinder and 
the top hemisphere.  So there will be three equations:

  case a) for when hgas is less than rtank
  case b) for when hgas is between rtank and
                   the difference (htank - rtank)
  case c) for when hgas is more than the difference (htank - rtank)

To make the equations easier to read and manipulate, I am going to 
shorten your variabes and then switch back at the end:

  height of tank (htank) = H
  radius of tank (rtank) = R
  height of gas  (hgas)  = z

If you don't want to see all the calculations, just skip to the end.  
I have used calculus to determine the spherical cap caused by 
partially filling a hemisphere.  If that is too complicated for you, 
please ask me to explain in a bit more detail.

In case a), when 0 < z < R, the volume can be derived with:

       /
  V =  |  A dz  where A = cross sectional area of sphere (a circle)
       /

  A = Pi x^2    where x is:

    z
    |
 2R +++
    |   ++
    |      +
    |       +
    |        +
 R --        +       ---
    |        +        |
    |<--x-->+     range of
    |      +     0 < z < R
    |   ++            |
  --+++--------- x   ---
    |

The equation of this circle is

  R^2 = x^2 + (z - R)^2

which means

  x^2 = R^2 - (z - R)^2

Back to the integral:

      /        /
  V = | A dz = | Pi x^2 dz
      /        /

      /
  V = | Pi ( R^2 - (z - R)^2 ) dz
      /

      /
  V = | Pi ( R^2 - z^2 + 2 R z - R^2 ) dz
      /

      /
  V = | Pi ( 2 R z - z^2 ) dz
      /

             /           /
  V = 2 Pi R | z dz - Pi | z^2 dz
             /           /

  V = 2 Pi R z^2/2 - Pi z^3/3

  V = Pi R z^2 - Pi z^3/3

Let's check our calculation.

  When z = 0 (empty)

  V = (Pi R 0^2 - Pi 0^3/3)
  V = 0

  When z = R (full hemisphere)

  V = (Pi R^3 - Pi R^3/3)
  V = 2/3 Pi R^3
  V = 1/2 * 4/3 Pi R^3   (half a sphere)

So when z is between 0 and R, the volume in the tank is:

  V = Pi R z^2 - Pi z^3/3  when  0 < z < R

In case b), when R < z < (H - R), the volume is a full hemisphere 
plus the cross sectional area of the cylindrical part times the 
height of the cylindrical part (liquid height - tank radius). So when 
the level is between R and H - R, the volume in the tank is:

  V = 1/2 * 4/3 Pi R^3  +  Pi R^2 (z - R)  when  R < z < H - R

In case c), when H - R < z < H, the volume is a full hemisphere plus 
the full cylindrical part of the tank plus the partially filled top 
half of a sphere of radius R:

  V = 1/2 * 4/3 Pi R^3

    + Pi R^2 (H - 2R)

      /
    + | A dz  where A = cross section of sphere
      /

  A = Pi x^2    where x is:

        z
        |
   H   +++               ---
        |   ++            |
        |      +      range for
        |<--x-->+   H - R < z < H
        |        +        |
 H - R --        +       ---
        |        +
        |       +
        |      +
        |   ++
 H - 2R +++
        |
        |
        |
      --+----------- x
        |

The equation of the sphere is:

  R^2 = x^2 + (z - (H - R))^2

But that would be complicated to integrate, so let's substitute

  Z = z - (H - R) to get:

  R^2 = x^2 + Z^2   where Z is the liquid height (z) minus
                                          the difference (H - R)

Back to the integral:

                                           /
  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + | A dZ
                                           /

                                             /
  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) +   | Pi x^2 dZ
                                             /

                                           /
  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + | Pi ( R^2 - Z^2 ) dZ
                                           /

                                                  /         /
  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 | dZ - Pi | Z^2 dZ
                                                  /         /

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 Z - Pi Z^3/3

But remember I said "where Z is the liquid height (z) minus the 
difference (H - R)"? Now is when we put it back in:

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R)
    + Pi R^2 (z - (H - R)) - Pi (z - (H - R))^3/3

Let's check our calculation.

  When z = H - R (full hemisphere + cylindrical part)

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R)

  When z = H (both hemispheres + cylindrical part)

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R)
    + Pi R^2 (H - (H - R)) - Pi (H - (H - R))^3/3

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R)
    + Pi R^3 - Pi R^3/3

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R)
    + 1/2 * 4/3 Pi R^3

So when z is (H - R) or more, the volume in the tank is

  V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R)
    + Pi R^2 (z - (H - R)) - Pi (z - (H - R))^3/3  when  H - R < z < H

So, to calculate the total volume of liquid in the tank:

 - if the height of the liquid is less than rtank deep

      V = Pi rtank hgas^2 - Pi hgas^3/3
      when  0 < hgas < rtank

 - if the height of the liquid is between rtank
                             and (htank - rtank) deep

      V = 2/3 Pi rtank^3 + Pi rtank^2 (hgas - rtank)
      when  rtank < hgas < htank - rtank

 - if the height of the liquid is more than (htank - rtank) deep

      V = 2/3 Pi rtank^3 + Pi rtank^2 (htank - 2rtank)
        + Pi rtank^2 (hgas - htank + rtank)
        - Pi (hgas - htank + rtank)^3/3

      when  htank - rtank < hgas < htank

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

Date: 02/16/2010 at 09:58:18
From: Matt
Subject: Thank you (Volume of tank with hemisphere caps)

Thank you so much for your time!!!  I really appreciate what you guys 
do.  It's nice to know that people are out there who use their 
talents to help others.  

Thanks again!