Figure 1: Propane Tank Cutaway View
This is the solution for a horizontally-mounted tank. If your tank stands vertically, click here.
Several years ago I created a
Calculus tutorial, and I included a number of sample equations to show various properties of Calculus. One of the equations computed partial volumes for a cylindrical tank with spherical end caps (see figure 1 on this page). As it turns out, computing partial volumes for a horizontal tank is a common problem and, since the publication of the tutorial, I have gotten quite a few inquiries about it. So I have decided to return to this problem, add a few things, and make it a usable application of Calculus.
Many of my correspondents have relatively simple cylindrical tanks lying on their sides, and they need to know the relationship between the content height (as measured on a vertical axis in the tank) and the volume of the contents (the quantity held by the tank). Others have a tank like the one pictured on this page, with spherical end caps, typically used to store propane or natural gas, and they have the same requirement — what is the relationship between content height and the content volume?
Other correspondents have written to describe a third design, one I didn't know about, that has tank end caps that are curved but not spherical in cross section — I'll be calling this the "elliptical end cap" tank design. This sort of tank represents a compromise between fully spherical end caps (greatest resistance to bursting under pressure), and a simple cylinder with flat ends (most efficient use of space in a storage yard filled with tanks). The elliptical end cap design requires an equation I had not tried to write until now.
This article describes a general equation that can be used for all these tank types, and this page includes a calculator to produce practical results for user-entered tank measurements. This article is not a
tutorial that teaches Calculus, instead it is an
application of Calculus to a real-world problem. Those readers who would prefer an exposition on the topic of Calculus itself should read the
original tutorial before reading this article.
(Since first publication, this method has been rewritten to eliminate an unnecessary variable)
Examine Figure 2, a diagram of a typical storage tank. Notice that the diagram consists of a central cylindrical section and two elliptical end caps. This diagram contains most of the variable names and descriptions that will be used in this derivation:
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L
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The length of the cylindrical section.
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R
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The radius of the cylinder and the major radius of the elliptical end caps.
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r
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The minor radius of the elliptical end caps.
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y
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The height of the tank's contents.
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While reading what follows, remind yourself that this is a flat diagram of a three-dimensional object. In particular note that
R, the radius of the cylinder and the major radius of the ellipses, describes a circular cross-section extending into a third dimension not shown in this diagram.
This article describes a procedure to compute a tank content volume if given
y, the content height within the tank. The first step is to write equations that can produce horizontal plane areas for the ellipses and for the cylinder, for a given
y height within the tank (once we have created the area equations, we will integrate them to get volumes).
Let's start with the ellipses, and for simplicity's sake, let's start out using a spherical cross-section (we convert this to an elliptical spheroid in the next step). We need to get the area of a circle in the horizontal plane that slices through a sphere at a vertical position given by
y. For a sphere of radius
R, such a function is:
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(1)
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Now let's integrate a range of circle areas to get a partial sphere volume, while at the same time multiplying the integral by the ratio
r/R, e.g. adjusting for the fact that this is an elliptical spheroid:
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(2)
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Equation (2) provides the volume of a spheroid with major radius
R and minor radius
r, between the bottom of the tank and any chosen
y value within the range 0 <=
y <=
2R. Remember that the two end caps can be calculated as a single spheroid that happens to be split in two, with half the volume at one end of the tank and half at the other.
Now let's turn to the cylindrical part of the tank. To begin this case, we need an equation that can produce a diameter line
d across the cylinder for a given
y value (note the similarity to equation (1) above):
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(3)
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Now we must integrate between the bottom of the tank and a given
y value to get the area
a of a partial circle, then multiply by the cylinder length
L to complete the partial volume of the cylinder:
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(4)
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At this point, we have created an equation (2) that will provide the volume of the two spheroidal parts of the tank (remembering that the two end caps, if combined, would equal a spheroid), and another equation (4) that provides the volume of the cylindrical part of the tank. Both equations use the same variable (
y) to determine the content height, so we should be able to combine these two equations to represent the complete tank partial volume:
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(5)
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Equation (5) will produce a partial volume for a given argument
y, for many kinds of storage tanks — from simple cylinders, through those with spheroidal end caps, to the classic spherical end caps as shown in Figure 1, even spherical tanks without any cylindrical section.
It will be seen that, for a cylindrical tank with flat end surfaces, the value of
r is set to zero. For a tank with spherical end caps such as is seen in Figure 1, the minor radius
r equals the major radius
R. This means one controls the type of tank being modeled by simply changing the input values.
Here is an online computer based on the foregoing derivation.
NOTE: This computer is meant for a horizontal tank. For a vertical tank, click here.
