SciForums.com > Science > Physics & Math > Simple Tank Underground Problem Make No Sense PDA View Full Version : Simple Tank Underground Problem Make No Sense Post ReplyCreate New Thread sk8erboyla200411-07-09, 07:33 PMI have trouble making since of this but say you to compute the work to take some liquid from a tank underground to ground level laying horizontally Well work = force * distance(displacment) I have seen this basic formula done twice for and cylinder and ellopsoid to find force you need the mass * g then you partition the tank into slices which seems to be rectangular slices lets say in this example i have a cylinder lying on its side radius 4 ft, length 12 ft, 10 ft underground the formula for the rectangular slice is 2x * length * thickness but would your length be x which is 12 and the thickness is just some change in y, your left with width which seems to be would be some y value but they took the equation of a circle x^2 + y^2 = r^2 where in this case y was equal to the distance/displacment which was (14-y) i dont see how you can just plug these random things in, this post is a mess noodler11-08-09, 01:56 AMHmm, not sure if I understand what you want to do, but if you have a tank full of water and you want all the water lifted (pumped) to ground level that's equivalent to raising the entire volume of water so the "bottom" is at ground level. Work is also equal to the work a pump has to do to empty the tank. But you can skip all that and just use the weight of a given volume (in this case the tank volume) which is the 'force component', and the distance is the depth of the tank. CheskiChips11-08-09, 04:54 AMI think he's trying to find the total mass of an object using a triple integral utilizing a density function. Then using that total mass to determine the work done, but there's no way to tell because it's too convoluted. Pete11-08-09, 07:52 AMHmm, not sure if I understand what you want to do, but if you have a tank full of water and you want all the water lifted (pumped) to ground level that's equivalent to raising the entire volume of water so the "bottom" is at ground level. The water at the bottom of the tank has to be lifted more than the water at the top of the tank. noodler11-08-09, 03:27 PMPete: if you have a tank - say a swimming pool - full of water so the surface is level with the ground, that's a volume of water at ground level = 0. Say your pool happens to have a bottom which is a large plate, supported by hydraulic rams that lift the bottom of the tank to ground level. The work the plate + hydraulic system does lifting the entire volume is equivalent to work done pumping it out say, a bucketful at a time, or by using a bucket + rope, or just letting the water evaporate. The work done on the water is the same "path", and is independent of the "means" used to lift it. Different methods will be more or less efficient as work will be done in each case, that won't move any water (for instance if you use a bucket, the work you do manipulating it when it's empty is "wasted"). Maybe the question is about calculating a pumping rate, given a pipe diameter etc, but the OP is a little vague... Pete11-08-09, 06:26 PMPete: if you have a tank - say a swimming pool - full of water so the surface is level with the ground, that's a volume of water at ground level = 0.... Of course. But your previous post is misleading. "Raising the entire volume of water so the "bottom" is at ground level" implies that the entire pool is lifted while maintaining its shape, not just poured out to ground level. And your suggested solution... But you can skip all that and just use the weight of a given volume (in this case the tank volume) which is the 'force component', and the distance is the depth of the tank. ...works for that interpretation (entire pool lifted), but not for the problem at hand (poured out on the ground), because the force (the weight being lifted) is not constant. noodler11-08-09, 06:34 PMOk... "so the bottom of the liquid is at ground level, so everything above the tank bottom, i.e. the water is moved". I mistakenly assumed that my first sentence was intransitive. P.S. The "lifting rule" of entropy here is related to something else, that some other math-head might know more about than me. But work is a symmetric function. iceaura11-08-09, 06:41 PMThe horizontal "slice" of the water's volume is a rectangle, with length equal to the length of the cylinder's axis, and width equal to the horizontal chord of the circle describing the cylinder's cross section perpendicular to its axis. It varies by depth. noodler11-08-09, 06:49 PMYou want the Laplacian for the volume form on C. Pete11-08-09, 07:27 PMI have trouble making since of this but say you to compute the work to take some liquid from a tank underground to ground level laying horizontally Well work = force * distance(displacment) I have seen this basic formula done twice for and cylinder and ellopsoid to find force you need the mass * g then you partition the tank into slices which seems to be rectangular slices lets say in this example i have a cylinder lying on its side radius 4 ft, length 12 ft, 10 ft underground OK, a cylindrical tank, on it's side. The top of the tank is 10ft under. the formula for the rectangular slice is 2x * length * thickness So the cylindrical tank is partitioned into thin horizontal slices. The volume \Delta V of a slice is length x width x thickness. The length of each slice is a constant 12 ft (the length of the cylinder). The width of each slice will be different for each slice, varying from zero at the bottom of the tank to 8ft in the middle and back to zero at the top. You can express this as 2x, where x is half the width of the tank at that particular y value. The thickness of each slice is the distance from the start of one slice to the start of the next; an arbitrarily small constant \Delta y. y is the height of the slice from some vertical reference (like the middle of the tank). \Delta V=L \times 2x \times \Delta y But to make this work, you have to make y the only variable. You can't directly integrate a function involving both x and \Delta y. That means that 2x has to be expressed as a formula involving y. but would your length be x which is 12 and the thickness is just some change in y, your left with width which seems to be would be some y value but they took the equation of a circle x^2 + y^2 = r^2 Right, that's what you need to change 2x to a function of y. x = \sqrt {r^2-y^2} Now we can get the volume in terms of y: \Delta V=2L \sqrt {(r^2-y^2)} \Delta y where in this case y was equal to the distance/displacement which was (14-y) y is the distance from the middle of the tank to the slice. The displacement involved in lifting that slice to ground level is (14-y) if the middle of the tank is 14ft down. So... if my quick and very nasty manipulations are right... For one slice, we begin with: work = force x distance distance = 14-y F = mg = \rho \Delta V g (\rho = Density of water) Put them together... W = \rho \Delta V g (14-y) And remembering that... \Delta V = 2L \sqrt {(r^2-y^2)} \Delta y We get... W = 2 \rho Lg \sqrt {(r^2-y^2)} (14-y) \Delta y And for the entire cylinder, we sum all the slices: W = \int_{-4}^4 2 \rho Lg \sqrt {(r^2-y^2)} (14-y) dy I haven't given this a reality check, so treat with caution. noodler11-08-09, 08:08 PMhttp://en.wikipedia.org/wiki/Volume_form Given a volume form ω on M, one can define the divergence of a vector field X as the unique scalar-valued function, denoted by div X, satisfying (\operatorname{div} X)\omega = L_X\omega = d(X\lrcorner\omega) where LX denotes the Lie derivative along X. If X is a compactly supported vector field and M is a manifold with boundary, then Stokes' theorem implies \int_M (\operatorname{div} X)\omega = \int_{\partial M} X\lrcorner\omega, which is a generalization of the divergence theorem. The solenoidal vector fields are those with div X = 0. It follows from the definition of the Lie derivative that the volume form is preserved under the flow of a solenoidal vector field. Thus solenoidal vector fields are precisely those that have volume-preserving flows. This fact is well-known, for instance, in fluid mechanics where the divergence of a velocity field measures the compressibility of a fluid, which in turn represents the extent to which volume is preserved along flows of the fluid. In this case you have all the boundaries and M is a cylindrical volume of fluid. (the leftright-corner thingy that Tex is having trouble with is I think called a connection form) The Lie derivative of differential forms The Lie derivative can also be defined on differential forms. In this context, it is closely related to the exterior derivative. Both the Lie derivative and the exterior derivative attempt to capture the idea of a derivative in different ways. These differences can be bridged by introducing the idea of an antiderivation or equivalently an interior product, after which the relationships fall out as a set of identities. Let M be a manifold and X a vector field on M. Let \omega \in \Lambda^{k+1}(M) be a k+1-form. The interior product of X and ω is (i_X\omega) (X_1, \ldots, X_k) = \omega (X,X_1, \ldots, X_k)\, Note that i_X:\Lambda^{k+1}(M) \rightarrow \Lambda^k(M) \, and that iX is a \wedge-antiderivation. That is, iX is R-linear, and i_X (\omega \wedge \eta) = (i_X \omega) \wedge \eta + (-1)^k \omega \wedge (i_X \eta) for \omega \in \Lambda^k(M) and η another differential form. Also, for a function f \in \Lambda^0(M), that is a real or complex-valued function on M, one has i_{fX} \omega = f\,i_X\omega The relationship between exterior derivatives and Lie derivatives can then be summarized as follows. For an ordinary function f, the Lie derivative is just the contraction of the exterior derivative with the vector field X: \mathcal{L}_Xf = i_X df For a general differential form, the Lie derivative is likewise a contraction, taking into account the variation in X: \mathcal{L}_X\omega = i_Xd\omega + d(i_X \omega). This identity is known variously as "Cartan's formula" or "Cartan's magic formula," and shows in particular that: d\mathcal{L}_X\omega = \mathcal{L}_X(d\omega). The derivative of products is distributed: \mathcal{L}_{fX}\omega = f\mathcal{L}_X\omega + df \wedge i_X \omega http://en.wikipedia.org/wiki/Lie_derivative Note: Wikipedia is a good reference when decent mathematicians contribute understandable rhetoric. James R11-09-09, 12:44 AMLet me make a few assumptions here. Assume you have a full tank whose top is level with the ground and whose bottom is buried. Assume you want to stick a hose in and pump out all the water so it runs along the ground. The net amount of energy required to do that will be equal to the energy required to lift the whole weight of water in the tank by a distance equal to half the height of the tank. (i.e. if the total mass of water is m, the work required will be mgh/2, where g is the acceleration due to gravity 9.8 m/s^2 and h is the depth of the tank.) James R11-09-09, 12:45 AMLet me make a few assumptions here. Assume you have a full tank whose top is level with the ground and whose bottom is buried. Assume you want to stick a hose in and pump out all the water so it runs along the ground. The net amount of energy required to do that will be equal to the energy required to lift the whole weight of water in the tank by a distance equal to half the height of the tank. (i.e. if the total mass of water is m, the work required will be mgh/2, where g is the acceleration due to gravity 9.8 m/s^2 and h is the depth of the tank.) Pete11-09-09, 01:39 AMThat's also assuming the tank is symmetrical on a horizontal plane. Which of course it is, in this case. Post ReplyCreate New Thread