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Whirlpool action in a tank created by inlet water

  1. Jan 21, 2007 #1
    I would like to determine what should be the size and number of pipes that are attached tangent to the cone of a tank. I will very much appreciate it that some one comment on how to approach and calculate it.

    I am trying to create a whirlpool action with an angular velocity of 1 rpm in a tank with cone bottom. The tank is 6 meter in diameter and 2 meters in height. The cone height is 1 meter. Initially, the tank is filled with water. Water is, then, pumped into the tank through the pipes ( the size and number are to be determined) that are attached tangent to the cone with a constant flow of 0.6 cubic meter / minute. The over-flow water exits at the top of the tank.
     
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  3. Jan 22, 2007 #2

    russ_watters

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    Velocity divided by circumference...?
     
  4. Jan 22, 2007 #3
    Thank you. but ..

    Could you please elaborate it for me to understand?
     
  5. Jan 22, 2007 #4

    russ_watters

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    The velocity of the water coming out of the pipe divided by the circumference is equal to the angular speed in rpm. Perhaps if you are trying to spin a large volume of water you'll want the jets to have somewhat higher speed, but it probably won't need to be much higher. There isn't much friction to slow the water down as it goes around the tank.
     
  6. Jan 24, 2007 #5
    Thank you for explantion...

    However, if I place the inlet pipes at two levels of the cone, the circumference at the lower level is smaller than that on the top level. Does this mean that the inlet pipes at the lower should have a lowered flow rate to match the same angular velocity of the whirlpool?

    What will be the resultant angular velocity of the whirlpool if the incoming flow rate from the two levels is the same?
     
  7. Jan 24, 2007 #6

    russ_watters

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    If the jets are the same speed, you'll have a whirlpool with a speed that varies with height. If, for example, the lower jet is halfway down the cone, the angular velocity is doubled at that level (because the diameter is cut in half).... except that water is viscous enough it'll want to spin at a constant rate. You'll end up closer to the higher rate than the lower one.

    Is water also exiting the bottom of the cone? May I ask what the purpose of this is? It makes a difference, because as water travels up the cone, its angular velocity will decrease and if it travels down, its angular velocity will increase (due to conservation of angular momentum). Obviously, this gets to be a very complex problem very quickly, so it would be helpful to know what the constraints really are.
     
    Last edited: Jan 24, 2007
  8. Jan 25, 2007 #7
    Thank you. If I understand correctly, the lower jet halfway down the cone; generates an angular velocity that is doubled. However, as water travels up the top of cone, its angular velocity decreases to the same velocity as that generated by the jet on the top of the cone.

    In this system, water comes in from bottom and exits from top. The purpose of this design is to prevent clogging of the suspended solid in the tank. Since a uniform distribution of solid in the tank is not an issue and the density of the solid is only slightly higher than water, I try to save energy by using whirlpool action instead of an agitator. Do you think this is feasible system?

    Does the direction of the whirlpool (clockwise or counter clockwise) make any differences?
     
  9. Feb 19, 2007 #8
    There's a great Mythbusters episode on whirlpools -- they actually build a whirlpool tank and create rotation at various speeds. Maybe you could find it on YouTube?
     
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