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Topic: Volume Calculation for Liquid Inside Cylindrical Container (2 cases)
Replies: 8   Last Post: Jan 13, 1999 10:34 PM

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Lee Crandell

Posts: 35
Registered: 12/6/04
Re: Volume Calculation for Liquid Inside Cylindrical Container (2 cases)
Posted: Jan 10, 1999 2:30 AM
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Hi,
I come up with volume of liquid in the tank as:
V = L*(r^2*ACOS((r-d)/r)-(r-d)*SQRT(2*d*r-d^2))
where L = length of tank; r = radius of cylindrical tank; d = depth
of liquid in the tank
(The format should be easily cut and pasted into most BASIC and
spreadsheet programs with perhaps minor alteration).

Derived thus:

Imagine the circle area at the end of a horizontal cylinder filled to a
given level. The level of the liquid at the edges of the circular form an
isosceles triangle with the center of the circle. From the center drop a
vertical plumb line (perpendicular to the liquid level) to the bottom of the
circle. The altitude a of this triangle is also a common leg of the two
mirror-image congruent right triangles. With a little bit of trigonometry,
geometry, and algebra we should be able to demonstrate the derivation of the
formula.

r = radius
d = depth of liquid
a = r - d (the altitude of the isosceles triangle so formed and also the
common leg of the 2 right triangles).

AREA OF SECTOR
A(sector) = (pi * r^2 * theta)/(2 * pi) = (theta * r^2)/2
Theta is the angle of the sector at the center of the circle.
We have:
(1/2) * theta = ACOS(a/r) = ACOS((r-d)/r) or
theta = 2 * ACOS((r-d)/r)
A(sector) = 2 * ACOS((r-d)/r) * (1/2) * r^2 or
A(sector) = r^2 * ACOS((r-d)/r)

AREA OF TRIANGLE
A(triangle) = (1/2) * a * 2 * sqrt(r^2 - a^2)
= a * sqrt(r^2 - a^2)
= (r - d) * sqrt((r^2 - (r - d)^2))
= (r - d) * sqrt((r^2 - (r^2 - 2dr + d^2)))
= (r - d) * sqrt(r^2 - r^2 + 2dr - d^2)
A(triangle) = (r - d) * sqrt(2dr - d^2)

AREA OF SEGMENT
A(segment) = A(sector) - A(triangle)
= r^2 * ACOS((r-d)/r) - (r - d) * sqrt(2dr - d^2)

Now we can multiply this by L, the length of the cylindrical tank so:

V(liquid in tank) = L*(r^2 * ACOS((r - d)/r) - (r - d) * sqrt(2dr - d^2))


--
Bye
Best wishes
Lee
Panteli Poulikakos wrote in message <36945f6b.4514136@news.wenet.net>...
>I am in the process of writing a simple program for the company I work
>for that calculates the volume of a liquid inside a cylindrical
>container. Now if the container is placed vertically, the volume is
>simply pi * r^2 * h where r is the radius of the circle and h is the
>height of the fill level inside the cylindrical container. However, I
>am not sure how to handle the case where the container is placed
>horizontally. One might say simply take the container and place it
>vertically. Well, that is not really possible since these are
>cylindrical tanks with weights in excess of 5,000 lbs. When the
>container is placed horizontally, the radius of the circle doesn't
>really play a role in the volume. At first I thought that may be some
>calculus might help, however there is really no dimension here to
>integrate over. For that reason and through talking to some colleagues,
>
>I am under the impression that this problem involves possibly some
>advanced geometry and trig. If anyone has any suggestions on how to go
>about solving this problem, I would appreciate it if you could email me
>at ppouli1@uic.edu.
>
>Thanks in advance,
>
>Panteli Poulikakos
>



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