More Fluid Mech

  1. Aug 16, 2006 #1
    hi guys.

    got another question for ya.


    seems to of uploaded small but i explain.

    a reservoir is designed to retain water to a max depth of 10metres. if the water level exceeds 10 metres, then a hinged semi circle gate will open to release any excess water. the gate has a radius of 1m. the gate is hinged 9 metres below the free surface.

    what would be the force required to open gate?

    what is the depth from the free surface to the the centre of pressure.


    i know the second moment of area of a semi circular surface is 0.1102R4

    but not sure how to complete the question. help with working out the answer would be much appraicated so i can work back on the answer and see how to get it.

  2. jcsd
  3. Aug 16, 2006 #2


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    In a general sense, you need to calculate the pressure distribution on the gate. From there, using the hinge location, you can use the area to calculate a force on the gate.
  4. Aug 17, 2006 #3
    i cant calcluate the area as i do not know all the measurements

    or how would i do this?

  5. Aug 17, 2006 #4


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    Science Advisor

    You know the radius of the gate. That's all you need.
  6. Aug 18, 2006 #5

    i have worked out the area and it is 1.571

    once i have got the area of the gate, then how to i calculate the force required to open it?


  7. Aug 18, 2006 #6
    the 2nd moment of area Ig about axis through centroid is 0.1102Rto the power of 4

    so that would be 0.1102 x 1.571 to the power of 4 which is 0.272?

    still confused where to go next

  8. Aug 18, 2006 #7
    would the force required to open gate be pgh

    1000 x 9.81 x 10?

    therefore be 98100 pa?

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