Chapter 4
•
Differential Relations for a Fluid Particle
259
Solution:
Here we have only the single ‘one-dimensional’ convective acceleration:
2
2
1.
(
a
)
o
o
V
du
u
x
uV
A
n
s
dt
x
L
L
∂
±²
³´
³ ´
==
+
=
µ¶
µ ¶
·¸
¹º
¹ º
»¼
2
o
2V
x
1
LL
2
+
2
2(10)
2
6
10
,
1
400(1 4 ),
6/12
o
ft
du
x
For L
and V
x
with x in feet
sd
t
=
+
=
+
′′
At x
=
0, du/dt
=
400
ft/s
2
(12 g’s); at x
=
L
=
0.5 ft, du/dt
=
1200
ft/s
2
(37 g’s).
Ans.
(b)
4.3
A two-dimensional velocity field is given by
V
=
(
x
2
–
y
2
+
x
)
i
– (2
xy
+
y
)
j
in arbitrary units. At (
x
,
y
)
=
(1, 2), compute (a) the accelerations
a
x
and
a
y
, (b) the
velocity component in the direction
θ
=
40
°
, (c) the direction of maximum velocity, and
(d) the direction of maximum acceleration.
Solution:
(a) Do each component of acceleration:
22
x
y
du
u
u
u
v
(x
y
x)(2x 1)
( 2xy
y)( 2y)
a
dt
x
y
dv
v
v
u
v
(x
y
x)( 2y)
( 2xy
y)( 2x 1)
a
dt
x
y
∂∂
=+=
−
+
+
+
−
−
−
=
−
+
−
+
−
−
−
−
=
At (x, y)
=
(1, 2), we obtain
a
x
=
18
i
and
a
y
=
26
j
Ans.
(a)
(b) At (x, y)
=
(1, 2),
V
=
–2
i
– 6
j
. A unit vector along a 40
°
line would be
n
=
cos40
°
i
+
sin40
°
j
. Then the velocity component along a 40
°
line is
40
V(
2
6
)
(
c
o
s
4
0
s
i
n
4
0
)
.
(
b
)
Ans
°
−
−
⋅
°
+
°
≈
40
V n
i
j
i
j
5.39 units
°
⋅
(c) The maximum acceleration is
a
max
=
[18
2
+
26
2
]
1/2
=
31.6
units at
∠
55.3
°
Ans.
(c, d)
4.4
Suppose that the temperature field T
=
4x
2
– 3y
3
, in arbitrary units, is associated
with the velocity field of Prob. 4.3. Compute the rate of change
dT/dt
at (x, y)
=
(2, 1).
Solution:
For steady, two-dimensional flow, the rate of change of temperature is
2
(
)(8 )
( 2
)( 9
)
dT
T
T
uv
x
y
x
x
x
y
y
y
dt
x
y
−
+
+
−
−
−
At
(
x
,
y
)
=
(2, 1),
dT/dt
=
(5)(16) – 5(–9)
=
125 units
Ans.