Gas in a Cylindrical Tank with Hemispheric CapsDate: 02/09/2010 at 10:55:36 From: Matt Subject: Volume of tank with hemisphere caps I work with my dad, and we have a tank on our farm. It is a cylinder with hemispheric caps that looks like a large pill: _ _ / \ | | | | \ _ / It stands upright, on end, and holds gas for the tractor. If I look into the tank I can estimate the gas level height. Beyond that, though, it is hard to tell how much gas is left in the tank. It's too heavy to lift and make calculations based on differences in weight, for example. I took calculus in high school, and I know this problem can be solved with integrals. I have the following variables: height of tank = htank radius of tank = rtank height of gas = hgas With these three variables, how can I set up an integral to figure out the volume of the gas in the tank? I just don't know how I can incorporate the hemispheric bottom and top into the integral. I have tried to use the disc method, and I understand that I can revolve a line around an axis; but I don't know how to order the variables, and am extremely confused about finding the volume of only the gas rather than the entire tank. Date: 02/15/2010 at 22:33:12 From: Doctor Jeremiah Subject: Re: Volume of tank with hemisphere caps Hi Matt, First let me be sure I understand. The assumptions I am making: 1) htank is the height of the entire tank, including the hemispheres 2) the hemispheres each have height rtank (they are actual hemispheres - half a sphere each) 3) htank and rtank are the interior distances (they don't include the thickness of the wall of the tank) We need to break this into three equations because there are three pieces to the tank: the bottom hemisphere, the middle cylinder and the top hemisphere. So there will be three equations: case a) for when hgas is less than rtank case b) for when hgas is between rtank and the difference (htank - rtank) case c) for when hgas is more than the difference (htank - rtank) To make the equations easier to read and manipulate, I am going to shorten your variabes and then switch back at the end: height of tank (htank) = H radius of tank (rtank) = R height of gas (hgas) = z If you don't want to see all the calculations, just skip to the end. I have used calculus to determine the spherical cap caused by partially filling a hemisphere. If that is too complicated for you, please ask me to explain in a bit more detail. In case a), when 0 < z < R, the volume can be derived with: / V = | A dz where A = cross sectional area of sphere (a circle) / A = Pi x^2 where x is: z | 2R +++ | ++ | + | + | + R -- + --- | + | |<--x-->+ range of | + 0 < z < R | ++ | --+++--------- x --- | The equation of this circle is R^2 = x^2 + (z - R)^2 which means x^2 = R^2 - (z - R)^2 Back to the integral: / / V = | A dz = | Pi x^2 dz / / / V = | Pi ( R^2 - (z - R)^2 ) dz / / V = | Pi ( R^2 - z^2 + 2 R z - R^2 ) dz / / V = | Pi ( 2 R z - z^2 ) dz / / / V = 2 Pi R | z dz - Pi | z^2 dz / / V = 2 Pi R z^2/2 - Pi z^3/3 V = Pi R z^2 - Pi z^3/3 Let's check our calculation. When z = 0 (empty) V = (Pi R 0^2 - Pi 0^3/3) V = 0 When z = R (full hemisphere) V = (Pi R^3 - Pi R^3/3) V = 2/3 Pi R^3 V = 1/2 * 4/3 Pi R^3 (half a sphere) So when z is between 0 and R, the volume in the tank is: V = Pi R z^2 - Pi z^3/3 when 0 < z < R In case b), when R < z < (H - R), the volume is a full hemisphere plus the cross sectional area of the cylindrical part times the height of the cylindrical part (liquid height - tank radius). So when the level is between R and H - R, the volume in the tank is: V = 1/2 * 4/3 Pi R^3 + Pi R^2 (z - R) when R < z < H - R In case c), when H - R < z < H, the volume is a full hemisphere plus the full cylindrical part of the tank plus the partially filled top half of a sphere of radius R: V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) / + | A dz where A = cross section of sphere / A = Pi x^2 where x is: z | H +++ --- | ++ | | + range for |<--x-->+ H - R < z < H | + | H - R -- + --- | + | + | + | ++ H - 2R +++ | | | --+----------- x | The equation of the sphere is: R^2 = x^2 + (z - (H - R))^2 But that would be complicated to integrate, so let's substitute Z = z - (H - R) to get: R^2 = x^2 + Z^2 where Z is the liquid height (z) minus the difference (H - R) Back to the integral: / V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + | A dZ / / V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + | Pi x^2 dZ / / V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + | Pi ( R^2 - Z^2 ) dZ / / / V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 | dZ - Pi | Z^2 dZ / / V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 Z - Pi Z^3/3 But remember I said "where Z is the liquid height (z) minus the difference (H - R)"? Now is when we put it back in: V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 (z - (H - R)) - Pi (z - (H - R))^3/3 Let's check our calculation. When z = H - R (full hemisphere + cylindrical part) V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) When z = H (both hemispheres + cylindrical part) V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 (H - (H - R)) - Pi (H - (H - R))^3/3 V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^3 - Pi R^3/3 V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + 1/2 * 4/3 Pi R^3 So when z is (H - R) or more, the volume in the tank is V = 1/2 * 4/3 Pi R^3 + Pi R^2 (H - 2R) + Pi R^2 (z - (H - R)) - Pi (z - (H - R))^3/3 when H - R < z < H So, to calculate the total volume of liquid in the tank: - if the height of the liquid is less than rtank deep V = Pi rtank hgas^2 - Pi hgas^3/3 when 0 < hgas < rtank - if the height of the liquid is between rtank and (htank - rtank) deep V = 2/3 Pi rtank^3 + Pi rtank^2 (hgas - rtank) when rtank < hgas < htank - rtank - if the height of the liquid is more than (htank - rtank) deep V = 2/3 Pi rtank^3 + Pi rtank^2 (htank - 2rtank) + Pi rtank^2 (hgas - htank + rtank) - Pi (hgas - htank + rtank)^3/3 when htank - rtank < hgas < htank - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 02/16/2010 at 09:58:18 From: Matt Subject: Thank you (Volume of tank with hemisphere caps) Thank you so much for your time!!! I really appreciate what you guys do. It's nice to know that people are out there who use their talents to help others. Thanks again! |
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