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Fluid mechanics gage pressure - did I do it right?

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    I attached the problem.

    Please check my work.
    2. Relevant equations



    3. The attempt at a solution
    Density of oil=.92*1000=920 kg/m3
    Density of vinegar=1010 kg/m3
    Height of oil=3*.0254=.0762 m
    Heigh of water=.127 m

    a) Gage pressure at bottom=(920)(9.81)(.0762)+(1010)(9.81)(.127)=1946 Pa


    b) The words "side of the bottle" kinda threw me off. It made me think they were referring to outside the bottle. I assumed it was still inside the bottle. I guess the word "side" was intended to throw you off as the horizontal direction does not matter.

    Gage pressure between oil&vinegar=(920)(9.81)(.0762)=688 Pa

    c) I'm going to guess that shaking the bottle so that the two are thoroughly mixed will still end up having the same pressure as part a) 1946 Pa? I don't know how to explain that, but it seems logical to me. If that's the answer, can someone explain why it is?
     

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  3. Sep 2, 2012 #2

    TSny

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    Hello, charlies1902.

    I believe your answers to (a) and (b) are correct. Part (c) is interesting. I think the answer will depend on the shape of the bottle. Do you know if you are suppose to assume vertical sides (so that the horizontal cross section of the bottle is the same at all heights)?
     
  4. Sep 2, 2012 #3
    Hmm I'm not sure, the professor hasn't mentioned anything, but I did get an early start on it so maybe he will say something about it later. The problem didn't include a picture either.

    I'm assuming it's a regular salad dressing bottle. I did a quick search on google images and it seems most of the bottles have different cross sections throughout the height of the bottle.
     
  5. Sep 2, 2012 #4

    TSny

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    You might consider the two shapes shown in the attachment. Would both shapes give the same answers for (a) and (b)? For (c)?
     

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  6. Sep 2, 2012 #5
    They would give the same answers for a and b.

    Hmm for c, I'm not sure about. Are you alluding to that the amount of liquid for each could change the answer for c?
     
  7. Sep 2, 2012 #6

    TSny

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    Right, as long as the fluids are separated as shown in the attachment, the answers to (a) and (b) would be the same for both shapes of bottles. But, after mixing the fluids, would the pressures at the bottom be the same for the two different shaped bottles?
     
  8. Sep 2, 2012 #7
    Hmmm I'm really not sure, I've never done a problem like this before.

    After mixing them I'd assume both of the liquid would exist along the total height of 8" of the bottle. Would you have to combine the density somehow, like taking an average, and then solve for the pressure at the bottom?

    The way they worded question c) implies that they want a number for the answer.

    I remember from thermodynamics, that you sum up pressures of individual liquids/gases to get the total pressure after they're mixed. I'm not sure if that's the same in fluid mechanics. If so I think that requires the volume of the fluids to be known.
     
  9. Sep 2, 2012 #8

    TSny

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    Sometimes when you mix two different liquids, the total volume will change a little. But I think we can ignore that effect here. So, I think you're right in taking the total height after mixing to still be 8".

    Yes, after mixing you could get the answer by finding the density of the mixture. And that's where the amount of each liquid comes into play. Without knowing the shape of the container, I don't see how you're going to get a numerical value.
     
  10. Sep 2, 2012 #9
    Hmmm, alright thanks. I'll have to ask.
     
  11. Sep 2, 2012 #10
    The oil and vinegar mass is being supported by the force resulting from the pressure x area across the bottom of the bottle. What if anything changes when you shake the bottle?
     
  12. Sep 2, 2012 #11

    TSny

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    Hi, NoPoke. I don't want to give away the answer at this point. This is one of those problems where the pleasure of figuring it out is too good to spoil :tongue2:. charlies1902 has already given a good hint.

    Of course, maybe I'm thinking I have the answer when I don't really have it!:redface:
     
  13. Sep 2, 2012 #12
    ok, I managed to misinterpret your hint. Hope mine isn't too strong or worse sending him off in the wrong direction!
     
  14. Sep 2, 2012 #13
    The shape of the bottle has no effect in part C. When you shake up the two fluids, the density of the emulsion is a weighted average of the two fluid densities. On a scale a little bigger than the emulsion droplet size, the liquid can be treated as a continuum. This is typically how mechanics problems involving composites are solved.
     
  15. Sep 4, 2012 #14
    I emailed the professor asking if this depended ont he volume.
    He replied "Think about the weight of fluid above the bottom surface"

    I'll think about this for awhile.
     
  16. Sep 4, 2012 #15
    Shaking does not change the weight of the fluid.

    I'm guesing the force acting on the bottom surface is the weight of the mixed liquid.
    The pressure on it is the weight/area.

    Since weight and area doesn't change, the pressure should be the same as part a?
     
  17. Sep 4, 2012 #16

    TSny

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    Right
    Suppose you had pure water in the two containers shown in the attachment. Assume the containers have the same base area and that the total height of the water is the same.

    How do the pressures on the bottom surfaces compare? How do the forces on the bottom surfaces compare? How does the total weight of water in the two containers compare?
     

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  18. Sep 4, 2012 #17
    The one on the left would weigh more. Pressure would be the same. Forces should be the same.

    I think this question should end up being the same if the base area and weight aren't changed in the process of mixing?
     
    Last edited: Sep 4, 2012
  19. Sep 4, 2012 #18

    TSny

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    That's right. So, the force of the fluid acting on the bottom does not necessarily equal the weight of the fluid.

    No, the mixing adds another complication.
     
  20. Sep 4, 2012 #19

    TSny

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    Try a specific example.

    Suppose the shape of the bottle is as shown in the attachment. Before mixing the vinegar fills the lower chamber and the oil fills the thin tube. We can imagine that the shape is such that there are equal volumes of oil and vinegar, but the vinegar is 1” tall and the oil is 5” tall.

    What would be the pressure at the bottom before mixing and after mixing?
     

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