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Pressure of fluid on bottom of container

  1. Sep 18, 2007 #1
    A cylindrical container with a cross sectional area of 65.2 cm[tex]^{2}[/tex] holds a fluid of density 806kg/m[tex]^{3}[/tex]. At the bottom of the container the pressure is 116 kPa
    a) What is the depth of the fluid?
    b) Find the pressure at the bottom of the container after an additional 2.05 [tex]\times[/tex] 10[tex]^{-3}[/tex] m[tex]^{3}[/tex] of this fluid is added to the container. Assume no fluid spills out of the container.


    a) A = 65.2 cm[tex]^{2}[/tex]
    [tex]\rho[/tex] = 806 kg/m[tex]^{3}[/tex]
    P(bottom) = 116 kPa = 1.16 [tex]\times[/tex] 10[tex]^{5}[/tex] Pa

    closed manometer [tex]\rightarrow[/tex] P = [tex]\rho[/tex]gh

    h = [tex]\frac{P}{\rho g}[/tex] = [tex]\frac{1.16 \times 10^{5}}{806 \times 9.81}[/tex]
    = 14.67 m


    i'm not really sure how to do part (b), this is all that i could come up with

    P1 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v1[tex]^{2}[/tex] + [tex]\rho[/tex]gh = constant
    P2 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v2[tex]^{2}[/tex] + [tex]\rho[/tex]gh = constant
    v2 = v1 + 2.05 [tex]\times[/tex] 10[tex]^{-3}[/tex]

    P1 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v1[tex]^{2}[/tex] + [tex]\rho[/tex]gh = P2 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v2[tex]^{2}[/tex] + [tex]\rho[/tex]gh

    P1 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v1[tex]^{2}[/tex] = P2 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v2[tex]^{2}[/tex]

    P1 + [tex]\frac{1}{2}[/tex][tex]\rho[/tex]v1[tex]^{2}[/tex] = P2 + [tex]\rho[/tex](v1 + 2.05 [tex]\times[/tex] 10[tex]^{-3}[/tex])[tex]^{2}[/tex]

    P2 = P1 + [tex]\rho[/tex]v1[tex]^{2}[/tex] - [tex]\rho[/tex](v1 + 2.05 [tex]\times[/tex] 10[tex]^{-3}[/tex])[tex]^{2}[/tex]

    = 116 + 403v1[tex]^{2}[/tex] - 403(v1 + 2.05 [tex]\times[/tex] 10[tex]^{-3}[/tex])[tex]^{2}[/tex]


    would someone be able to tell me if (a) is correct and possibly a better way to solve (b)
    thanks
     
  2. jcsd
  3. Sep 18, 2007 #2

    learningphysics

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    Homework Helper

    Find out how much more depth is added to the fluid... You know the volume added... you know the cross sectional area of the cylinder... you should be able to get the new height and hence the new pressure.
     
  4. Sep 18, 2007 #3
    new pressure

    ok so
    v = 2.05 x 10[tex]^{-3}[/tex] m[tex]^{3}[/tex]
    CSA = 65.2 cm2 = 6.52 x 10[tex]^{-3}[/tex] m[tex]^{3}[/tex]

    h = [tex]\frac{v}{CSA}[/tex] =0.3144m

    P = [tex]\rho[/tex]gh
    = 806 x 9.81 x (14.67 + 0.314)
    = 806 x 9.81 x 14.985
    = 118 486.05 Pa
    = 1.18 x 10[tex]^{5}[/tex] Pa = 118 kPa

    thankyou
     
  5. Sep 18, 2007 #4

    learningphysics

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    Homework Helper

    looks good. no prob. you're welcome.
     
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