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Need desperate help on fluid dynamics

  1. Nov 18, 2006 #1
    Hi, I have been doing a simple experiment with a 2L bottle and a cut out slot near the bottom. I have gathered data regarding the time it takes the water level in the bottle to go from 13cm above the slot to the slot itself, versus the area of the slot. My data shows a perfect inverse relationship between the time for the water level to reach the slot with a constant of 0.0169. I am working in standard units and my functions looks as follows T=0.0169/A. Now I need to determine what the 0.0017 is so i can derive a formula that can determine the time it takes to empty the bottle depending on initial volume, bottle radius, and slot area. I have been searching everytwhere for relevent information but I honestly cannot find anything I can use to help me. I am desperate to know because it would solve all my problems. The next part of my experiment is to derive a formula for the volume in the bottle after a certian time has elasped. The data for this experiment fits a parabola which I also cannot explain. If someone can please clear this up for me I would be so grateful, but please don't think I am asking someone to do my homework, I have doen the experiments and sat down since 5:00 (and now its 12:30) with a pencila nd aper searching online for info and taking derivatives and integrals and using pressure and potential energy and reading up on CV factors, etc. Please, if anyone can help I would be so relieved.

    Thanks,
    Mike.
     
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 18, 2006 #2
    This is going to involve differential equations. Awk.

    You are going to be sitting around for another couple of hours I'm afraid.

    Cant help you right now, but look into differential equations. This is a pretty classic example.

    Sorry, im no help to you right now.
     
  4. Nov 18, 2006 #3
    I can do differentials just fine, but I need to know what initial formula I need to use and where I'm headed. Personally I am shocked that such a simple experiemnt isn't well documented on. if anyone else has any ideas I am all ears.

    Thanks in advance,
    Mike.
     
  5. Nov 19, 2006 #4

    vanesch

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    If some approximations are allowed, then this is an application of Bernouilli's law. Mind you, it will not be 100% exact, but the idea is the following:

    An opening of area A at a depth d will put the fluid just "inside" at the outside pressure (1 atmosphere), simply because the wall is removed and cannot provide the necessary counter force to have a higher pressure.
    But a bit "further inside" the bottle, where the water is at rest, the pressure is given by P = atmospheric pressure + g d rho
    (g = grav. acc 9.81 m/s^2 and rho = 1000 kg/m^3, the density of water).

    This means that the water "just in front of the window" must be moving, if we apply Bernouilli's law, which will give you a certain velocity at atmospheric pressure. If we now assume that the velocity is normal to the opening, then we know the fluid flux through the hole (velocity x area = fluid flux). However this is the instantaneous flux, because of course the contents of the bottle diminishes which diminishes d.

    But all this is explained here:

    http://plus.maths.org/issue2/bottle/index.html
     
  6. Nov 19, 2006 #5
    If your data fits a parabola you'd expect an x2 term. But you also said your data shows time to be inversely proportional to area, have you got other data or something?

    If you want to know how much water will be in your bottle at some chosen time you should do the experiment again and moniter the height of the water at intervals. I would expect that the water will drain more slowly with time.

    Darcy's law: Q=KAi
    Q= volume flux
    K= hydraulic conductivity (this is your constant of proportionality)
    A= area of slot
    i= pressure gradient

    warning: I don't think this law holds if you're dealing with turbulence!!

    you need to establish the pressure gradient at different water clomun heights and try to tie that together with time. That's how I would approach it anyway!
     
  7. Nov 19, 2006 #6
    Thank you vanesch, I will read up on that, although i have already seen similar info I have disregarded it becuase I didn't think I could apply the same law or principals to tank with no input velocity, but I guess I could look at it and see how I can use it.

    Sorry billiards if my initial post was too messy. What I meant to say is I have data for total time taken for the bottle to the level of the slot versus area of the slot and I also have data for height iof water in the bottle versus time elasped. So I do have the rquired information which fits the formula H = (8E-05)*t^2 - 0.0062*t + 0.13, which can easily be manipulated to be either volume or mass versus time, however I have been unsucessful in finding ways to derive a formula for instantaneous flow rate in either m^3/s or kg/s. Once I have the flow rate I would need to create a function of height versus velocity probably with an integral (please correct me if I'm wrong) from initial flow rate to final. I am still unsure of that though.

    Thanks.
     
  8. Nov 19, 2006 #7
    Alright, I have used Bernouilli's law to find the general function of water level height versus time and have compared it to the function I have found in my experiment for water level height versus time and by comparing them they match very well and I solve for the constant lambda in my equation (one in front of my t^2 term and one in front of my t term and I found it to be 4.039240E-3 and 3.882726E-3 respectvely, which are pretty close. Now I have to somehow manipulate one or several equations to determine flow rate as a function of time.

    EDIT: Also, how to I calculated what lambda should be depending on the variables of the experiment such as slot cross section area, luquid density, etc?
     
    Last edited: Nov 19, 2006
  9. Nov 19, 2006 #8
    You might find it helpful to say that:

    flux rate is proportional to the area of the slot.
    fluid velocity is proportional to the specific weight of the fluid (density*gravity)
    fluid velocity is inversely proportial to the dynamic viscosity of the fluid.

    Then bung the constants of proportionality into one big constant of proportionality and combine those statements.

    v=C*A*density*g*pressure gradient/(dynamic)viscosity.

    where C is your dimensionless constant of proportionality. This constant will depend largely on the shape of your slot.

    Edit: I changed the last statement to viscosity. sorry about that!
     
    Last edited: Nov 19, 2006
  10. Nov 19, 2006 #9
    That does help, but isn't statement 2 and 3 contradicting in that if velocity is already proportional to density * gravity, then how can velocity be also inversely proportional to density. Maybe I am just not getting it but I think it's best if i explain myself.

    This experiment is part of an Electricity and Magnetism course in which the professor decided the best thing for us to do to study for his exam on cicuits, capacitors, magnetism, induction, electric fields, etc is to give us a radndom do at home lab on a topic that was never and will never be covered in this course. The test is Monday and this lab is due on the same day so I am working back and forth trying to cover two totally different topics and techniques today. Now that I have explained this I hope everyone can see that I do not have any prior knowledge of viscosity (dynamic or I'm guessing static), specific weight of fluids, pressure gradients, etc. I hate to be a burden, but I don't know how to link these properties since I don't know what these properties are to begin with.

    I thank you billards and everyone else who contributed and I will try to make sense of the information in the posts and online and I will get back to you on my progress.

    Thank you.
     
  11. Nov 19, 2006 #10

    vanesch

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    ??? :confused:

    If you use Bernoulli's law in this case, we simply have:

    v^2/2 + h g = constant
    (the density of the fluid drops out).
    If we are at the surface, v = 0, and h = d, so constant = g d
    If we are at the hole, h = 0 and hence v = sqrt(2 d g)

    So the velocity of the fluid that squirts out of a hole at depth d from the surface of the fluid, equals v = sqrt(2 d g)

    This is independent of the area of the hole or even the density of the fluid.

    However, the fluid flux will be equal to this velocity times the area of the hole A:
    phi = v.A = A sqrt(2 d g).

    Now, in a time dt, a fluid flux phi will correspond to a volume dV = phi.dt of water that has escaped. This will then lower the surface of the water with an amount - d d (silly notation, meaning that depth d has diminished by d d). If the section of your bottle equals S, we obtain that when a volume dV has been taken away, we have that d d . S = - dV

    So: d d. S = - dV = - phi.dt = - A.sqrt(2 d g).dt

    which gives you a differential equation for the function d(t):

    (hum, I did a bit too much here, you should work this out yourself...)
     
    Last edited: Nov 19, 2006
  12. Nov 19, 2006 #11
    Hmm, well earlier today I got up to "fluid flux phi" or what i like to call volume per second :D is equal to (Area of slot)(sqrt[2*height*g]) but for the life of me I cannot figure out how to get the sum of all volumes lost at their respective fluid fluxes (of fluxi) to figure out the total vloume lost. What I have now after vanesch's guidance is

    V = A*sqrt(2*height*g)*t = -(surface area of bottle)*h

    where A = area of slot, V = escaped volume (or volume flowed through slot), and h = height lost (expressed as a positive number) correct? Let me see if it fits my data.

    Edit: Wait so this function is linear, how can that be, i simply intergrated each of the terms vanesch laid out, unless my "height" (or rather vanesch's "d" in "2 d g") must also be a dynamic variable. And now I have brought myslef right back to being confused... partly.
     
    Last edited: Nov 19, 2006
  13. Nov 19, 2006 #12
    Wow, alright I have found a small calculation error where my lot area should have been shifted one decimal place over to the left. Now I have used the equation and plugged these values in

    A= 0.00005 m^3)
    height= 0.130m which is my initial height
    g= 9.80665 m/s^2
    t= 5 s

    I get 0.0003992 m^3 lost after 5 seconds and since I took my first reading of new vloume after 5 seconds I compared and found that by doing initial minus lost volume found in the formula I get 0.000853633 m^3 left compared to my data of 0.000932879 m^3 left. However, for 10 seconds the formula indicates I have 0.0004544395 m^3 left when I really had 0.000723752 m^3. Clearly, the funstion should not have been linear.

    Anyone have an idea how to incorporate a changinf "height" with time?
     
  14. Nov 19, 2006 #13
    Please anyone, I give up, I have been trying to solve the relation between total time requiore to empty a bottle versus the slot size and the relation between volume in the bottle versus time elasped after draining started. I cannot do it. I simply can't. If anyone can please provide the answer I would be so garteful. I have tried for the past two days and I give up. I'm done.

    Thanks anyway,
    Mike.
     
  15. Nov 19, 2006 #14
    I'll concede that my first statement that vanesch picked up on was a bit iffy. I should really have said flux rate was proportional to area, and I also put density when I meant to put viscosity. Doh!! I've changed it now anyway.

    But I really don't think the Bernoulli equation is the right approach. This is because the Bernouuli equation assumes no friction and no viscosity. If this were true your velocity would be constant and you would indeed have a linear pattern for height drop against time. The fact is that you need to account for viscous forces, both intrinsic to the fluid and by friction with the side of the container, mechanical energy is lost and you'd expect the flux rate to drop as a result. Actually this friction is analagous to resistance in your electrical circuit.

    What you are looking for is the friction factor. good luck.
     
  16. Nov 19, 2006 #15
    Perhaps you should show some work. I don't have the time to do it for you, but someone can check your work.

    If you quit, you suck.
     
  17. Nov 20, 2006 #16
    Show some work... I have in front of me 10 sheets of paper with different concepts and different integrals or differntial equations, all leading nowhere or in a loop. I have 4 graphs drawn out and their function and a whole bunch of info to try to link to them. I have been on this for 2 days and If you wnat me to scan my papers and post them I will. You think I suck because I give up trying to link water level and volume escape rates to time and area of a hole when I am in an Electricity and Magnetism class? I've tried and I cannot figure it out and I did not come here because I could do it myself but figured you people would do it for me, I came here because I can't do it and I was wondering if this phenomena or formula or differential equation or anything is available online or in someone's head that they've picked up along the way. I am not asking someone to derive and equation for me, i am asking if anyone knows the exact answer. When someone posts what the equation or way to determine the minimum stall speed of an aircraft they don't ask you to derive it, they are asking where to find it and what it is because hopefully, like I have, they have searched and come back empty handed. Anyway, I can't solve this as I have not yet taken differential equations and know absolutely nothing of fluid dynamics or viscocity, etc. If someone does and can help before the end of tonight since I have to hand my report tomorow then thank you, if not, I will hand in what I have and deal.
     
    Last edited: Nov 20, 2006
  18. Nov 20, 2006 #17
    Arg, stop whining.

    Show us some work so we can help you! :smile:

    You're not allowed to quit.

    Edit:
    Oh, crap, well then...your screwed. :frown:

    Now I must ask, WHY are you given this assignment if you are not doing fluid mechanics? :confused: That's a bit unfair.
     
    Last edited: Nov 20, 2006
  19. Nov 20, 2006 #18
    EXACTLY MY POINT. My professor, for lack of a better phrase, is an unfair jokester who thinks its funny and intelligent to give us a lab to do at home about a topic not covered in his or another class while a test for the same class about actual covered information will be held on the same due date. I have to do it so I am asking here, and trust me, when I actually take fluid dynamics in University I will do it on my own as i have never asked for online help when doing something I have already learned about or should know. That's why my title is "NEED DESPERATE HELP".

    So please, if you can find somewhere what the actual equation is for water level height vs time and/or require complete drainage time vs area of slot please link to it because I cannot find this same experiment of phenomena anywhere. Why isn't this experiment or equation documented on if its seems like suck a fundamental application of fluid dynamics and/or differential equations?
     
    Last edited: Nov 20, 2006
  20. Nov 20, 2006 #19
    You "NEED NEW PROFESSOR" :rofl:
     
  21. Nov 20, 2006 #20
    And the funniest thing is, our proffessor actually emailed the teachers of the college informing them NOT TO HELP US!
     
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