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Fluid statics - two water tanks

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data

    A water tank, 3.7 m deep and 2 m in diameter, is connected at the bottom to a second water tank by using a 2 cm diameter pipe (see the figure). The tank A is open to the air and tank B is filled to the top and sealed.

    attachment.php?attachmentid=72826&d=1410109551.png

    a) what is the mass of the water in tank A?
    b)what is the gauge pressure at the bottom of tank A?
    c) what is the absolute pressure at the bottom of tank B?
    d) what is the net force on the top of tank B?

    2. Relevant equations

    P=ρgh



    3. The attempt at a solution

    a) to find the mass of the water in tank A
    m= ρv = ρAh = ρ(PiD^2/4)h = ρ(Pi(2)^2/4)h = ρPih = (1000)(3.14)(3.7) = 12000
    m = 12000

    b) to find the gauge pressure at the bottom of tank A
    P = ρgh = (1000)(9.8)(3.7) = 3600
    P = 3600

    c) absolute pressure at bottom of tank B
    gauge pressure = P = ρgh
    h = 3-1 = 2 , 3.7-2 = 1.7, 1.7+3 = 4.7
    h= 4.7
    P = ρgh = (1000)(9.8)(4.7) = 46000
    absolute pressure = gauge pressure + atmospheric pressue
    AP = 46000 + (1.013x10^5) = 147300
    absolute pressure = 147300

    d) net force at top of tank B. the water here isnt moving so the net force must be = 0.


    I think my solution might be wrong because it makes no use of the info about the connecting tube. Not sure where I went wrong
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2014 #2

    SteamKing

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    You need to show units for all the quantities you calculated. If you don't get into this habit now, you will surely lose marks on future HW and exams. It's also less confusing to a second party who picks up and checks your work.
     
  4. Sep 8, 2014 #3
    ahh right sorry about that.

    a) to find the mass of the water in tank A
    m= ρv = ρAh = ρ(PiD^2/4)h = ρ(Pi(2m^2)^2/4)h = ρ(1m^2)Pih = (1000 kg/m^3)(1m^2)(3.14)(3.7m) = 12000 kg
    m = 12000 kg

    b) to find the gauge pressure at the bottom of tank A
    P = ρgh = (1000 kg/m^3)(9.8m/s^2)(3.7m) = 3600 kgm/m^2s^2 = 3600 N/m^2
    P = 3600 Pa

    c) absolute pressure at bottom of tank B
    gauge pressure = P = ρgh
    h = 3m -1m = 2m , 3.7m - 2m = 1.7m, 1.7m + 3m = 4.7m
    h= 4.7m
    P = ρgh = (1000 kg/m^3)(9.8 m/s^2)(4.7m) = 46000 Pa
    absolute pressure = gauge pressure + atmospheric pressue
    AP = 46000 Pa + (1.013x10^5)Pa = 147300 Pa
    absolute pressure = 147300 Pa
     
  5. Sep 8, 2014 #4

    SteamKing

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    This calculation looks OK. I wouldn't apply sig. figs. until stating the final answer, though.

    You might want to check this calculation again. Pay attention to how the units cancel.


    For parts c) and d), remember the pressure at both ends in the connecting line must be the same, otherwise, there will be flow between the two tanks.
     
  6. Sep 8, 2014 #5
    For the second part i need units of pressure.
    Pa = N/m^2 = kgm/s^2m^2 = kg/s^2m
    My calculation used density(kg/m^3) height(m) and acceleration(m/s^2)
    Multiplying the units together
    Kgm^2/m^3s^2 = kg/ms^2 = Pa

    I cant figure out where i went wrong:(

    As for part c ... I need to find the pressure at the height of the connecting tube and multiply it by the area of tank a to get the force and then divide this by the area of the connecting tube to get the pressure there. Is this correct so far? Im not sure what to do with the info after that. They give me no info about tank bs area so im not sure how to find the pressure at the bottom other than dgh. Im am lost
     
  7. Sep 8, 2014 #6

    SteamKing

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    ρ has units of kg/m^3

    g has units of m/s^2

    h has units of m

    N has units of kg-m/s^2

    so ρgh has units of

    [itex]\frac{kg-m^{2}}{m^{3}s^{2}} = \frac{kg}{m s^{2}}[/itex]

    Pa has units of N/m^2, so

    Pa = [itex]\frac{kg-m/s^{2}}{m^{2}} = \frac{kg}{m s^{2}}[/itex]

    so the units check.

    However, make sure you multiplied the numbers together correctly and got the correct answer.

    That's not necessary. The hydrostatic pressure, which is determined by ρgh, is the same everywhere at that particular level of the water in Tank A (Pascal's Law). All you need to know is the h at the mouth of the tube in order to calculate the pressure there. By extension, the pressure at the mouth of the tube in Tank B will be the same.
     
  8. Sep 13, 2014 #7
    ahh i see in part b it should be 36000. I cant figure out what is wrong with part c. if all that matters is the height , the bottom of tank b is 4.7m under the surface of the water so that calculation should be correct right? as for part d the top of the tank is 1.7m below the surface of the water so it would be dgh = (100)(9.8)(1.7) for the pressure right? but if i have no info about the area how can i find the net force?
     
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