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Problem *4.118
[Difficulty: 3]
Given:
Nozzle flow striking inclined plate
Find:
Mimimum gage pressure
Solution:
Basic equations: Bernoulli and y momentum
p
ρ
V
2
2
+
gz
⋅
+
const
=
The given data is
ρ
999
kg
m
3
⋅
=
q
1200
L
sm
⋅
⋅
=
W8
0
m
m
⋅
=
h
0.25 m
⋅
=
w2
0
m
m
⋅
=
H
7.5 m
⋅
=
θ
30 deg
⋅
=
For the exit velocity and nozzle velocity
V
2
q
W
=
V
2
15.0
m
s
=
V
1
V
2
w
W
⋅
=
V
1
3.75
m
s
=
Then from Bernoulli
p
1
ρ
2
V
1
2
⋅
+
p
atm
ρ
2
V
2
2
⋅
+
=
or
p
1
ρ
2
V
2
2
V
1
2
−
⎛
⎝
⎞
⎠
⋅
ρ
g
⋅
h
⋅
−
=
p
1
103 kPa
⋅
=
(gage)
Applying Bernoulli between 2 and the plate (state 3)
p
atm
ρ
2
V
2
2
⋅
+
p
atm
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Unformatted text preview: 2 ⋅ + ρ g ⋅ H ⋅ − = V 3 V 2 2 2 g ⋅ H ⋅ + = V 3 19.3 m s = For the plate there is no force along the plate (x momentum) as there is no friction. For the force normal to the plate (y momentum) we have R y V 3 − cos θ ( ) ⋅ ρ − V 3 ⋅ A 3 ⋅ ( ) ⋅ = V 3 − cos θ ( ) ⋅ ρ − q ⋅ ( ) ⋅ = R y V 3 cos θ ( ) ⋅ ρ ⋅ q ⋅ = R y 20.0 kN m ⋅ =...
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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

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