Problem *3.116
[Difficulty: 2]
Given:
Rectangular container with constant acceleration
Find:
Slope of free surface
Solution:
Basic equation
In components
x
p
∂
∂
−
ρ
g
x
⋅
+
ρ
a
x
⋅
=
y
p
∂
∂
−
ρ
g
y
⋅
+
ρ
a
y
⋅
=
z
p
∂
∂
−
ρ
g
z
⋅
+
ρ
a
z
⋅
=
We have
a
y
a
z
=
0
=
g
x
g sin
θ
()
⋅
=
g
y
g
−
cos
θ
⋅
=
g
z
0
=
Hence
x
p
∂
∂
−
ρ
g
⋅
sin
θ
⋅
+
ρ
a
x
⋅
=
(1)
y
p
∂
∂
−
ρ
g
⋅
cos
θ
⋅
−
0
=
(2)
z
p
∂
∂
−
0
=
(3)
From Eq. 3 we can simplify from
pp
x
y
,
z
,
=
to
x
y
,
=
Hence a change in pressure is given by
dp
x
p
∂
∂
dx
⋅
y
p
∂
∂
dy
⋅
+
=
At the free surface p = const., so
dp
0
=
x
p
∂
∂
dx
⋅
y
p
∂
∂
dy
⋅
+
=
or
dy
dx
x
p
∂
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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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