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Math Help - Work, Setting up integral

  1. #1
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    Question Work, Setting up integral

    I have this really tough project to do, and I could use some help getting started... What I need to do is find the surface area of this "filter" in the picture. Then I also need to find the energy requirements to force the water through the system. I think with this I'll just need to figure out the work integral for the pump that would be driving the system. Any ideas/suggestions on how to do this would be GREATLY appreciated as I'm trying to finish this by tomorrow (Thursday) morning. Thanks!!!

    Stephanie
    Attached Thumbnails Attached Thumbnails Work, Setting up integral-chemglobal2.gif  
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  2. #2
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    If you stretch the surface out so that it lays "flat" you can see that it is a rectangle. So the surface area is just the width (.5 m) times the length. The length of the rectangle is the arc length of the function which describes the shape of the surface in the z/x plane (I am taking the positive z-axis to point "up", positive y-axis points "out" of the image, positive x-axis in the direction of the water flow)

    If you are not given the function I guess we are to assume it behaves like a cosine function (seems so). It osccilates 3.5 periods in 2 meters and has an amplitude of .25m, with this information we can construct the cosine function that describes the shape.

    h(x) = .25\cos(3.5\pi x)

    Then the length of this curve is given by:

    \int_0^2 \sqrt{1+h'(x)^2} = \int_0^2 \sqrt{1+(-.875\pi \sin(3.5 \pi x))^2} = \int_0^2 \sqrt{1+.765625 \sin^2(3.5 \pi x)}

    \approx 2.34038637151

    So the area of the upper surface is approximately 1.170193185755 \text{ m} ^2. It appears that the same object is the lower surface as well, so the total area of the upper and lower surfaces is just 2.34038637151 \text{ m} ^2.

    Are we to assume there are edges as well? If so it appears that we are assuming a constant "height" (distance between the two surfaces) of .02 meters? Then the area of each edge is just (.02 \text{m})(2.34038637151 \text{m}) = 0.0468077274302 \text{ m}^2

    Which leads to a total surface area of 2.4340018263704 \text{ m}^2. What do you think?
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  3. #3
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    Maybe?

    I believe you're onto something, I thought about using the arclength integral, but had no idea about a formula... the only part of your proposed solution that might be tricky is assuming that it's a cos function, although it does look like one, and appear to behave like one, so I'd be willing to bet you're right about that as well...

    For the force/work to push the water through the filter/conduit, could I just put together a work integral? I know that work is equal to the force times the distance, and since we now know the "length" of the conduit, that would be the distance... I'm unsure about the rest of the force equation though, just that I'll need to use an integral to find the work...
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  4. #4
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    Well without a function being given explicitly I don't see how you can determine the surface area. The picture gives general dimensions but what can one say from a drawing? Is it drawn to scale? You weren't given specific information about the nature of the shape of this object?

    Well work is force times distance. If force is being applied along some path you can integrate the force over the path to determine work done on the object, but what about a fluid in a pipe? Seems more complicated. Does it depend on the density of the liquid? In other less difficult work integral problems you are often asked to determine the work required to pump a liquid from say a conical tank to a height x-distance above the tank. This requires you know the density of the liquid to detemine the weight of the liquid and from this you can determine the work. But in your case how is this pipe oriented? If it is oriented as in the picture (gravity in the negative z-direction) then it seems to me the pump would only be doing work to move the liquid horizontally since the net work in the vertical direction is 0 (work is path dependent and no work is done on an object if the path ends where it begins). But I am hardly an expert in physical laws. What resources is this project meant to exploit, what results do you have at your disposal and is your post a complete explanation of the project as it was given to you?
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  5. #5
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    Quote Originally Posted by UDaytonFlyer View Post
    I have this really tough project to do, and I could use some help getting started... What I need to do is find the surface area of this "filter" in the picture. Then I also need to find the energy requirements to force the water through the system. I think with this I'll just need to figure out the work integral for the pump that would be driving the system. Any ideas/suggestions on how to do this would be GREATLY appreciated as I'm trying to finish this by tomorrow (Thursday) morning. Thanks!!!

    Stephanie
    As to the amount of work to push a fluid through this, if you have an "irrotational non-viscous incompressible perfect fluid" (this is one of my favorite obfuscational phrases) then you can use the baby version of the Bernoulli equation. This leads to the questions
    1) What is the height difference between input and output?
    2) What is the initial fluid pressure and at what pressure is the output at?
    (There is the third question about flow speed, but I'm presuming that the input cross-sectional area is the same as the output.)

    If you do have this kind of fluid, the the answer is very nice: The work done on the fluid will simply be
    W = \rho g \Delta h + \Delta P

    If you don't have this kind of fluid you are likely stuck with the full out Navier-Stokes equation and I am not the one to ask about help with this one. (Though you could probably still approximate a solution using Bernoulli.)

    -Dan
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  6. #6
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    More details...

    I just re-read the instructions and found the following more information. At first I didn't think it'd be that important to the overall problem, but now that I look at it, it could come in handy:

    "The purification of the water is accomplished by forcing it through the system shown in figure 1. As it passes through the system, it is exposed to the chemically active surface of the conduit, which reacts with the dissolved toxins in the water to produce a solid precipitate. This precipitate is filtered out by the filters shown in the figure. Because the efficacy of this process is dependent on the (chemically active) surface area of the conduit, we need an accurate assessment of this area. This is the first information we are contracting with you to determine. Additionally, we are concerned with the energy requirements to force the water through the system. The second item in your contract is therefore to determine the work that will be required of the pump we will use to drive the system."

    It's kind of like a really really long word problem, where you're solving a problem for a fictional company. The whole thing is like, 2 pgs long and the solution has to be typed and 2.5-5 pages, which I'm fine with the rest of it, it's just the calculations where I'm coming up short. I basically need the surface area and the work done in pushing the liquid through the conduit.

    I still need to figure out the work that is done. Thanks in advance for your input!!!
    Last edited by UDaytonFlyer; March 20th 2008 at 04:20 AM.
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  7. #7
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    Question work

    Any ideas for the work?
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  8. #8
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    As topsquark illuminated the work calculation of such a problem is considerably more complicated than force times distance. I worked with some other students a couple summers ago who did research on the motion of incompressible fluids in a disk with a rotating boundry and they were deep into Navier-Stokes but I didn't get to fully read their results or understand this mathematics. (I was too busy trying to prove the Collatz conjecture! ) Anyways this post is merely a courtesy. Sorry and good luck.
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