Problem 4.72
[Difficulty: 4]
Given:
Gate held in place by water jet
Find:
Required jet speed for various water depths
Solution:
Basic equation: Momentum flux in x direction for the wall
Note: We use this equation ONLY for the jet impacting the wall.
For the hydrostatic force and location we use computing equations
F
R
p
c
A
⋅
=
y'
y
c
I
xx
A y
c
⋅
+
=
Assumptions:
1) Steady flow
2) Incompressible flow 3) Uniform flow
Hence
R
x
V
ρ
⋅
V
−
A
jet
⋅
(
)
⋅
=
ρ
−
V
2
⋅
π
D
2
⋅
4
⋅
=
This force is the force generated by the wall on the jet; the force of the jet hitting the wall is then
F
jet
R
x
−
=
ρ
V
2
⋅
π
D
2
⋅
4
⋅
=
where D is the jet diameter
For the hydrostatic force
F
R
p
c
A
⋅
=
ρ
g
⋅
h
2
⋅
h
⋅
w
⋅
=
1
2
ρ
⋅
g
⋅
w
⋅
h
2
⋅
=
y'
y
c
I
xx
A y
c
⋅
+
=
h
2
w h
3
⋅
12
w h
⋅
h
2
⋅
+
=
2
3
h
⋅
=
where h is the water depth and w is the gate width
For the gate, we can take moments about the hinge to obtain
F
jet
−
h
jet
⋅
F
R
h
y'
−
(
)
⋅
+
F
jet
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Lear
 Fluid Dynamics, Fluid Mechanics, Flux, Gate, Jet, 0.5 m, 0.25 m, Fjet

Click to edit the document details