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 Subject: Fuel Tank Level Category: Science > Math Asked by: turry-ga List Price: \$10.00 Posted: 03 May 2004 08:22 PDT Expires: 02 Jun 2004 08:22 PDT Question ID: 340288
 ```I have a fuel tank that is a cylinder laying on its side. It must be filled with 10% of its volume guaranteed. To check the level of the tank, a transparent glass visor has been installed vertically (in the same line as the diameter of the circles that are at the ends of the cylinder) What percentage of the visor must be filled to achieve the 10% minimum guaranteed volume of the tank ?```
 ```Hi, turry-ga: The situation you describe, in which the axis of a partially filled cylindrical tank is horizontally aligned, is analyzed on this Web page: [Partially Filled Horizontal Cylindrical Tank] http://www.mathematic.co.uk/examples/HorizontalCylindricalTank.htm Besides the nice illustration (Fig. 1.1) at the top of the page, which shows clearly the equivalence of the fraction of volume filled in the tank with the fraction of a circular cross-section "filled" to the liquid's surface line, there is a very useful graph (Fig. 1.8) at the bottom of the page. For practical purposes your Answer may simply be "looked up" on that graph. In order to fill 10% of the volume of the tank, the vertical depth of the liquid (at center) should be something more than 0.3 times the radius, or 0.15 times the diameter if you prefer. We make these figures more precise with the computations below. A more exact approach requires a bit of mathematical formulation. From the graceful curve shown in the graph, we recognize that the relationship between the percentage of tank volume and the ratio of liquid height (depth) to radius is nonlinear. To continue with the formulation, let's define our values. Let R be the radius of the tank. You may assume whatever units of length are convenient, because we are going to use ratios to eliminate any dependence on the specific choice of units. Let H be the vertical height of the liquid (at center), so that H/R is a dimensionless ratio expressing the height of the liquid in relationship to the tank's radius. If it were more convenient for some reason to work with the diameter of the tank, rather than its radius, note that the diameter is simply 2R. Now an empty tank will have H/R = 0, while a full tank has H/R = 2. For that matter when H/R = 1 the tank is precisely half full. The graph's curve passes smoothly through these three collinear points, but as noted before, it's not flat. For convenience let's write h = H/R, so we will have a simple variable for this dimensionless value, the ratio of liquid height to radius. Now the fraction v of the tank's volume which is filled is the same as the fraction of the area of the circle which is "filled", the circle segment bounded by a chord corresponding to the liquid's surface, as shown at the right in Figure 1.1 on the Web page above. Developing a formula for the fraction v which the liquid fills in the tank is then equivalent to a formula for the fraction of area of the circle within the circle segment (bottom right of Fig. 1.1). Some analytical formulas for v in terms of h, or more literally for v in terms of R and H, are given on that Web page. However I think some greater simplicity can be preserved if we introduce some intermediate variable to which both v and h can be functionally related. As you will see, this will serve our practical needs quite well. In particular I wish to define an angle A (in radians) which is half of the central angle subtended by the chord forming the circle segment. In other words A is the measure of the angle formed by the vertical centerline of the circle and the ray that extends to the corner of the circle segment (at the point where the liquid's surface reaches the inner tank). We then have the following relations based on the tank's geometry: v = (A - 0.5 sin(2A))/pi h = 1 - cos(A) It is because of the relatively simple relationship of h to A that one is able to express v in terms of h, i.e. A = arccos(1 - h). However to answer your question of what value of h gives v = 0.1, it is still necessary to confront the difficult nonlinear expression above for v. Essentially we still need to find what angle A will satisfy: A - 0.5 sin(2A) = 0.1 pi Once this angle A is known with the required precision, it is simple to recover a corresponding value for h = 1 - cos(A). When only a single solution, e.g. for v = 0.1, is sought, it may be found with nothing more than a decent scientific calculator and a bit of trial and error. For example, bearing in mind that we are working with angles measured in radians here, I found using the Windows calculator applet: A = 0.81 gives v = (0.3106...)/pi = 0.09887... While not an especially precise solution, it would be easy to tweak A upwards to obtain even closer approximations of v to 0.1. If we consider the corresponding value for h: h = 1 - cos(A) = 0.31050... we confirm the earlier observation based on the graph (Fig. 1.8) that the Answer to your Question is found for the height to radius ratio h slightly more than 0.3. If you wish to be assured of more than 10% filled volume, then I would suggest a target value of h = 0.32. This corresponds to A = 0.8230... and thus to: v = (A - 0.5 sin(2A))/pi = 0.103... Please let me know, by Request for Clarification, if some elaboration of the formulas or other discussion would be helpful to you. regards, mathtalk-ga``` Clarification of Answer by mathtalk-ga on 03 May 2004 19:15 PDT ```Just to clarify in terms of the percentage of "visor", a value h = 0.32 amounts to 16% of the diameter or visor as you've described. regards, mathtalk-ga```
 turry-ga rated this answer: ```Completely fulfilled my needs ! Thank you very much.```