Problem 6.1
[2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,1)
Solution:
NOTE: Units of B are s
-1
not ft
-1
s
-1
Basic equations
For this flow
u x y
,
(
)
A
y
2
x
2
−
(
)
⋅
B x
⋅
−
=
v x y
,
(
)
2 A
⋅
x
⋅
y
⋅
B y
⋅
+
=
a
x
u
x
u
∂
∂
⋅
v
y
u
∂
∂
⋅
+
=
A
y
2
x
2
−
(
)
⋅
B x
⋅
−
⎡
⎣
⎤
⎦
x
A
y
2
x
2
−
(
)
⋅
B x
⋅
−
⎡
⎣
⎤
⎦
∂
∂
⋅
2 A
⋅
x
⋅
y
⋅
B y
⋅
+
(
)
y
A
y
2
x
2
−
(
)
⋅
B x
⋅
−
⎡
⎣
⎤
⎦
∂
∂
⋅
+
=
a
x
B
2 A
⋅
x
⋅
+
(
)
A x
2
⋅
B x
⋅
+
A y
2
⋅
+
(
)
⋅
=
a
y
u
x
v
∂
∂
⋅
v
y
v
∂
∂
⋅
+
=
A
y
2
x
2
−
(
)
⋅
B x
⋅
−
⎡
⎣
⎤
⎦
x
2 A
⋅
x
⋅
y
⋅
B y
⋅
+
(
)
∂
∂
⋅
2 A
⋅
x
⋅
y
⋅
B y
⋅
+
(
)
y
2 A
⋅
x
⋅
y
⋅
B y
⋅
+
(
)
∂
∂
⋅
+
=
a
y
B
2 A
⋅
x
⋅
+
(
)
B y
⋅
2 A
⋅
x
⋅
y
⋅
+
(
)
⋅
2 A
⋅
y
⋅
B x
⋅
A
x
2
y
2
−
(
)
⋅
+
⎡
⎣
⎤
⎦
⋅
−
=
Hence at (1,1)
a
x
1
2 1
⋅
1
⋅
+
(
)
1
s
⋅
1 1
2
⋅
1 1
⋅
+
1 1
2
⋅
+
(
)
×
ft
s
⋅
=
a
x
9
ft
s
2
⋅
=
a
y
1
2 1
⋅
1
⋅
+
(
)
1
s
⋅
1 1
⋅
2 1
⋅
1
⋅
1
⋅
+
(
)
×
ft
s
⋅
2 1
⋅
1
⋅
1
s
⋅
1 1
⋅
1
1
2
1
2
−
(
)
⋅
+
⎡
⎣
⎤
⎦
×
ft
s
⋅
−
=
a
y
7
ft
s
2
⋅
=
a
a
x
2
a
y
2
+
=
θ
atan
a
y
a
x
⎛
⎜
⎝
⎞
⎟
⎠
=
a
11.4
ft
s
2
⋅
=
θ
37.9 deg
⋅
=
For the pressure gradient
x
p
∂
∂
ρ
g
x
⋅
ρ
a
x
⋅
−
=
2
−
slug
ft
3
⋅
9
×
ft
s
2
⋅
lbf s
2
⋅
slug ft
⋅
×
=
x
p
∂
∂
18
−
lbf
ft
2
ft
⋅
=
0.125
−
psi
ft
⋅
=
y
p
∂
∂
ρ
g
y
⋅
ρ
a
y
⋅
−
=
2
slug
ft
3
⋅
32.2
−
7
−
(
)
×
ft
s
2
⋅
lbf s
2
⋅
slug ft
⋅
×
=
y
p
∂
∂
78.4
−
lbf
ft
2
ft
⋅
=
0.544
−
psi
ft
⋅
=

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