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More Fluid Mech |
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| Aug16-06, 04:01 AM | #1 |
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More Fluid Mech
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. thanks adzp |
| Aug16-06, 06:34 AM | #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.
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| Aug17-06, 09:52 AM | #3 |
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i cant calcluate the area as i do not know all the measurements
or how would i do this? thanks adzp |
| Aug17-06, 11:12 AM | #4 |
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Recognitions:
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More Fluid Mech
You know the radius of the gate. That's all you need.
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| Aug18-06, 06:55 AM | #5 |
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hi
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? thanks adzp |
| Aug18-06, 07:20 AM | #6 |
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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 adzp |
| Aug18-06, 08:21 AM | #7 |
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would the force required to open gate be pgh
1000 x 9.81 x 10? therefore be 98100 pa? adzp |
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