Problem 3.63
[Difficulty: 3]
Given:
Geometry of rectangular gate
F
1
D
L
y’
F
2
Find:
Depth for gate to open
Solution:
Basic equation
dp
dh
ρ
g
⋅
=
Σ
M
z
0
=
Computing equations
F
R
p
c
A
⋅
=
y'
y
c
I
xx
Ay
c
⋅
+
=
I
xx
bD
3
⋅
12
=
Assumptions:
Static fluid;
ρ
= constant; p
atm
on other side; no friction in hinge
For incompressible fluid
p
ρ
g
⋅
h
⋅
=
where p is gage pressure and h is measured downwards
The force on the vertical gate (gate 1) is the same as that on a rectangle of size h = D and width w
Hence
F
1
p
c
A
⋅
=
ρ
g
⋅
y
c
⋅
A
⋅
=
ρ
g
⋅
D
2
⋅
D
⋅
w
⋅
=
ρ
g
⋅
w
⋅
D
2
⋅
2
=
The location of this force is
y'
y
c
I
xx
c
⋅
+
=
D
2
wD
3
⋅
12
1
⋅
×
2
D
×
+
=
2
3
D
⋅
=
The force on the horizontal gate (gate 2) is due to constant pressure, and is at the centroid
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics, Gate

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