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Fluid pressure

  1. May 7, 2006 #1
    To determine the pressure of water, you would use Force/Area. With a column of water. For example a 4 inch diameter column of water with 24 pounds of water in it. Equation would be 24/12.56637061. Would equal 1.909 psi. But if at the bottom of the column there was a 1 inch diameter tube. Would you take 24 pounds and divide it by the area of 1 inch-(0.785398163 cubic inches) to get the pressure leaving the tube.
    If that were true would the pressure be 30.55774909psi? It just doesn't seem right. Can someone help me with this?
     
  2. jcsd
  3. May 7, 2006 #2

    Doc Al

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    Staff: Mentor

    It's not right. The static pressure exerted by a fluid depends on how deep you are below its surface, not on the total weight of all the fluid above you. (For example: A small bucket with a foot of water in it and an olympic size swimming pool with a foot of water in it both have the same water pressure acting on their bottom surfaces.) You can only use "weight of the column of fluid/Area" to find the pressure if the column of fluid is straight and unobstructed. Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html
     
  4. May 7, 2006 #3
    What I would like to know is that a column of water 7 feet tall, that is staight and unobstructed. Then there would be pressure from the weight of the water above you. But does the pressure increase when it goes from a large tube 4 inches to a small tube of 1 inch diameter. Because with the equation F/A for a large tube the weight is 77.9 pounds. But with a surface area of 12.5 cubic inches. And for the small tube of 77.9 pounds above it and .785 cubic inches. Remember that the large tube flows vertically to the small tube by gravitational force.
     
  5. May 8, 2006 #4

    Doc Al

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    Once again, the answer is no. The 1 inch diameter area supports only a 1 inch diameter column of water; the rest of the 7 foot tall column is supported by the bottom of the 4 inch diameter section of tube. (Please read the link I gave in my last post.)
     
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