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Height and Pressure Relating the Two

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A cylindrical bucket, open at the top, is 24.0cm high and 15.0cm in diameter. A circular hole with a cross-sectional area 1.86cm^2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.63×10−4 m^3/s


    -Not sure how to begin please, anyone could provide me a hint, guidelines, a starting point thank you :)
     
  2. jcsd
  3. Jan 23, 2009 #2

    mgb_phys

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    What is the question ?
     
  4. Jan 23, 2009 #3
    Very soory the question was;

    How high will the water in the bucket rise?
     
  5. Jan 23, 2009 #4
    From what I know the units of the flow of water out has m^3/s i think i know this

    Av=constant in our case Av=2.63E-4 m^3/s
     
  6. Jan 23, 2009 #5

    tiny-tim

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    Welcome to PF!

    Hi kurosaki_ichi! Welcome to PF! :smile:

    Hint: the water level will reach equilibrium when the flow out of the bottom equals the flow in …

    so start with …

    what is the pressure at the bottom when the height is h? :wink:
     
  7. Jan 23, 2009 #6

    mgb_phys

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    You need to calculate the rate the water flows out of the bucket - which depends on the height of the water (ie the pressure) and the size of the hole.

    Have you done Bernouilli's equation?
     
  8. Jan 23, 2009 #7
    ? I kinda have done berunillis equation, so do i use berunillis equation to cacuuluter pressure? or p=pa + rohgh?
    help someone please, still lst lol
     
  9. Jan 23, 2009 #8

    tiny-tim

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    (have a rho: ρ … and it's Bernoulli's equation :wink:)

    Yes, they're the same thing in this case: p = pa + ρgh :smile:
     
  10. Jan 23, 2009 #9
    So if am right here am guessing i must:

    pressure=Patomostpher density*g*h

    so patm is 1.00 atm

    density of water=1000kg/m^3

    g=9.8m/s^2

    height of water colum is the heigh of my cylinder? 24.0cm high

    are those my given values?
     
  11. Jan 23, 2009 #10

    tiny-tim

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    Yup! :biggrin:

    (but you won't need the value of atmospheric pressure, since it'll be the same on both sides of the equation :wink:)
     
  12. Jan 23, 2009 #11
    THank you so much for ur help i will now try and attempt the problem furthiermore i am grateful i will ask if i need any more help :) much thankz
     
  13. Jan 23, 2009 #12

    mgb_phys

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    Another way to calculate the rate of flow out is to use conservation of energy.
    A particle of water will flow out of the bottom of the bucket with a kinetic energy equal to the potential energy if it dropped from the surface of the water
    ( - it's just another way of stating Bernouilli's law - but is simpler to understand)
     
  14. Jan 23, 2009 #13
    I got my pressure to be 103.677 kPa now am not sure what to do? do i know use the continuty equation a1v1=a2v2?
     
  15. Jan 23, 2009 #14
    how can I work with it using conservation of energy if am not given a entry speed or flowing out speed?
     
  16. Jan 23, 2009 #15

    mgb_phys

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    You need to relate the water pressure at the hole to the flow speed, you can do this with Bernouilli's equation or as I suggested.

    Imagine a drop of water (mass m) in the bucket at the surface, it has a potential energy =mgh, if all this energy goes into kinetic energy as it flows out (=mv^2) then you have the speed of the drop of water flowing out.

    You don't need the input speed if it's a bucket - you assume the water stops when it is poured in (this is the same as saying there is no water pressure pushing down on the surface form the input flow).

    Now you have the output flow speed, and the hole area you can easily work out a volume output flow rate.
     
  17. Jan 23, 2009 #16
    So your telling me i should do it as Ea=Eb with potential energy at the top and kinetic energy at the bottom, cant i use the continuity equation i used it to solve for V2 i have pressure i just dont know how to relate evreything togethor i have my pressure solved also how does h play a part? is it mgh=1/2mv^2?
     
  18. Jan 23, 2009 #17

    mgb_phys

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    Yes - all you need to find the flow rate (m^3/s) out of the bucket.
    That depends on the area of the hole and the speed of the water coming out - you can get this from Bernouilli's equation and the pressure difference across the hole.

    A simpler method (IMHO) is to image you were dropping small rocks from a distance 'h' above the bottom of the bucket through a hole in the bottom of the bucket. It's very easy to work out the speed they are going at the bottom of the bucket - right?

    Now instead of rocks just picture a drop of water it starts at the surface of the water ( a distance 'h' ) from the bottom of the bucket and ends up going out of the hole with a certain speed - it's exactly the same as the rock.
     
  19. Jan 23, 2009 #18
    serisouly confused :S...lost please help
     
  20. Jan 23, 2009 #19

    mgb_phys

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    Ok - back to the start ;-)
    You have a bucket of known volume with a known flow rate in.
    To find how the water height changes with time you need to also know the flow rate out.

    To get the flow rate out - you need to know the speed of the water flowing out.
    The easiest way to do this is just conservation of energy.
    This will give you an equation for the speed of the water out of the hole only depending on the water height

    pe = ke = mgh = 1/2m v^2
     
  21. Jan 23, 2009 #20
    mgh=1/2mv^2

    okay mass cancels out

    so i have g=9.8m/s^2

    h=0.24 m

    using the above equation i get 1.08m/s^2

    so ineed the flow rate i multiply it by the area? the cross sectional area? correct
     
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