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Fluid Mechanics (2)

  1. Jan 18, 2007 #1
    I think that my textbook has too little example questions on this topic illustrating the concepts, so I have some confusions when I attempt to solve the following problems for practice. I hope someone would be nice to clarify such concepts. Any help is greatly appreciated!

    1) River A with a volume flow rate of 8x10^4 L/s joins with River B, which carries 11x10^4 L/s, to form River C. River C is 120m wide and 10m deep. What is the speed of the water in River C?

    I know from the equation of continuity that Q=vA=constant where Q is volume flow rate, v is speed, A is area. Now I need Q and A in order to find v, is Q simply 19x10^4 L/s and is A simply the area of a rectange 120mx10m=1200m^2 ? Are the calculations really that simple or am I wrong?

    2) Water is flowing out of a 10mm diameter sink faucet at 0.4m/s, at what distance below the faucet has the water stream narrowed to 5mm diameter?
    0.4*pi*(0.01)^2 = v2*pi*(0.005)^2
    This is what I am thinking, but in this case the water is falling and is affected by gravity, will the equation of continuity Q=vA=constant still hold? And is the water as simple as free falling at a constant acceleration of 9.8m/s^2?
  2. jcsd
  3. Jan 18, 2007 #2
    3) An aquarium tank is 80cm long, 30cm wide, and 35cm deep and is filled to the top. What is the force of the water on the bottom of the tank, with dimensions 80cmx30cm, AND the force on the front window with dimensions 80cmx35cm, respectively?
    For the first part, is the question asking for the normal force on the tank's bottom exerted by the water or is it asking for the total weight of the water? Are these 2 values equal? If so, why?
    Also, can someone help me on the second part? I have no idea how to solve it. I know that for a liquid, the pressure at a depth d is p=p_o+(rho)(g)(d), but in this case, what is the value for p_o?
  4. Jan 18, 2007 #3
    hi kingwinner ..

    i was just wondering if you have the answer to problem 15.22 last part

    i am also trying the same above problems as you and will help you out if i get them
  5. Jan 18, 2007 #4
    For your first question, yes, it is that simple, but pay attention to your units.

    For the second question, you are correct and should assume that continuity holds.

    For the third question think about it. The water is in static equilibrium right? What does that tell you about the force on the bottom of the tank?

    Lastly, you have to remember that pressure and force are not the same thing. p_o in this case would be the atmospheric pressure, but since it acts on the top and the side, it can be ignored. Do you know how to calculate force from a triangular pressure distribution? This is what you have to do to calc the lateral force since the pressure changes with height.
  6. Jan 19, 2007 #5
    Thanks for your help, but I am still having some trouble with Q3.

    3-part1) "What is the force of the water on the bottom of the tank", is it actually asking for the normal force of the water exerted on the bottom of the tank?

    By the way, I tried to use Force = pressure x area
    =(p_o + rho*g*d) x area
    =(101300+1000*9.8*0.35) x 0.8m x 0.3m
    =25135.2N <----but this is not what I got when I calculate the weight of the water, how come their values are different? I don't see why...

    3-part2)"p_o in this case would be the atmospheric pressure, but since it acts on the top and the side, it can be ignored." <---why can it be ignored? I am really confused...and how can the sides have atmospheric pressure? Only at the top can the water and air come into contact and have atmoshperic pressure=101300 Pa, right?
    The pressure of the WATER at depth d is given by p=p_o + rho*g*d, where p_o=p_atm=101300 Pa, right? Say if you put something at the bottom of the water to measure its pressure of the water, the value will equal to the above formula with p_o=101300 Pa. If get rid of the p_o in the equation, then p=rho*g*d is no longer the pressure in the water anymore...how can you do that? p is the pressure of the WATER, right?
    And by the way I don't know how to calculate force from a triangular pressure distribution...but I know that this question requires integration
  7. Jan 19, 2007 #6
    answer to 3 part 1 is 1372N
  8. Jan 19, 2007 #7
    Anyone nice enough to explain question 3 in greater detail? I am very very very confused...
    Last edited: Jan 19, 2007
  9. Jan 19, 2007 #8


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    Hint: draw pressure diagrams.
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