More Fluid Mech


by adzp
Tags: fluid, mech
adzp
adzp is offline
#1
Aug16-06, 04:01 AM
P: 10
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
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FredGarvin
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#2
Aug16-06, 06:34 AM
Sci Advisor
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P: 5,095
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.
adzp
adzp is offline
#3
Aug17-06, 09:52 AM
P: 10
i cant calcluate the area as i do not know all the measurements

or how would i do this?

thanks
adzp

FredGarvin
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#4
Aug17-06, 11:12 AM
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P: 5,095

More Fluid Mech


You know the radius of the gate. That's all you need.
adzp
adzp is offline
#5
Aug18-06, 06:55 AM
P: 10
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
adzp
adzp is offline
#6
Aug18-06, 07:20 AM
P: 10
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
adzp
adzp is offline
#7
Aug18-06, 08:21 AM
P: 10
would the force required to open gate be pgh

1000 x 9.81 x 10?

therefore be 98100 pa?

adzp


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