Liquid in an Elliptical TankDate: 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. 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/ 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)] My spreadsheet is 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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