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### Liquid in an Elliptical Tank

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Date: 02/28/99 at 23:46:52
From: Tom
Subject: Elliptical Tank

I am having difficulty solving problems like:

I have an elliptical tank in a horizontal position. (Tank length = 50
ft, height of ellipse = 5 ft, and width of ellipse = 10 ft.) Given any
height of liquid, say 3.0 ft, how can I calculate the volume? I found
a formula for the total volume, .785 x h x w x l = v , but that has
not helped me find the answer.

Thank you.

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Date: 03/01/99 at 08:26:40
From: Doctor Jerry
Subject: Re: Elliptical Tank

Suppose the equation of the ellipse is x^2/a^2 + y^2/b^2 = 1. For your
application, 2a = 5  and 2b = 10. The number 2a is the horizontal
distance from one side of the ellipse to the other; 2b is the vertical
distance from the top to the bottom of the ellipse.

I am assuming that the tank has flat ends. If not, that is a different
case. Suppose the height of the liquid is measured by the number h,
which can vary from -b to +b. If the tank had just a little liquid in
it, then h would be negative. You can fix this by replacing h by b+h,
which would vary from 0 to 2b.

Given h and using horizontal slices, the cross-section A of the tank
below y = h is

A = 2*int(-b, h, (a/b)*sqrt(b^2-y^2)*dy)

This works out to be

A = (a/b)[pi*b^2/2 + h*sqrt(b^2-h^2) + b^2*arcsin(h/b)].

For the volume you would multiply this by the horizontal length of the
tank.

If h = b and a = b = r, the above should reduce to pi*r^2, the area of
a circle. Let us see:

A = [pi*r^2/2 + 0 + r^2*pi/2] = pi*r^2.

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

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Date: 06/06/2007 at 11:06:43
From: Mike
Subject: Calculating partial volume of elliptical tank

Hello Dr. Math,

I have tried the formula you have published to calculate the partial
volume of a horizontal elliptical tank but it doesn't seem to work.  I
use Excel to do the calculations and I have very thoroughly checked
them for accuracy against hand performed calculations.

We are using elliptical cylinders in a horizontal plane (a truck
tank).  I was careful not to use diameters.  The basics are:

a: 35
b: 30
Length 60
depth of fluid .25"
It should be around .39 gallons.

Thanks for the help.

```

```
Date: 06/06/2007 at 23:02:08
From: Doctor Peterson
Subject: Re: Calculating partial volume of elliptical tank

Hi, Mike.

As I understand it, a vertical cross-section of your tank is an
ellipse, with a=35 inches for the horizontal semiaxis and b=30 inches
for the vertical semiaxis, and with a length of 60 inches between the
two elliptical ends.  You have only 1/4 inch of fluid in the tank.

I looked carefully to make sure of the definitions of the variables,
and wonder if you might have missed this:

Suppose the height of the liquid is measured by the number h,
which can vary from -b to +b.  If the tank had just a little
liquid in it, then h would be negative.  You can fix this by
replacing h by b+h, which would vary from 0 to 2b.

This means that in your example, h = 30 - .25 = 29.75, not .25.

When I corrected the formula by replacing h with d - b where d is the
depth above the bottom of the tank, I got 90 in^3 = 0.391 gal.  So this
seems to work.

My actual formula, then, is

V = L(a/b)[pi*b^2/2 + (d-b)*sqrt(b^2-(d-b)^2) + b^2*arcsin(d/b-1)]

Volume of horizontal elliptical cylinder tank
A  35     In
B  30     In
D  0.25   In
L  60     In

V  90.257 In^3 <-- =(B5*B2/B3)*(PI()*B3^2/2+
(B4-B3)*SQRT(B3^2-(B4-B3)^2)+B3^2*ASIN(B4/B3-1))
0.391  Gal  <-- =B7/231

If you have any further questions, feel free to write back.

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

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Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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