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SolutionSince the streamlines have a constant direction for the
time interval 0 t 6 3 s, the pathline and streakline coincide with
the streamline when t = 2 s as shown in Fig. a.
The pathline and streakline will coincide with the streamline
until t = 3 s, after which the streamline makes a sudden change in
direction. Thus, the streamline of the marked particle and the
streakline when t = 4 s will be as shown in Fig. b.
31. A marked particle is released into a flow when t = 0, and
the pathline for a particle is shown. Draw the streakline, and the
streamline for the particle when t = 2 s and t = 4 s.
t = 2 s
(a)
markedparticle
streamline
streakline
60
4 m
(b)
t = 4 s
markedparticle
streamline
streakline
60
4 m
6 m
t 0
t 2 s
t 3 s t 4 s
pathline
4 m
6 m
4 m
60
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SolutionSince the streamlines have a constant direction along
the positive x axis for the time interval 0 t 6 3 s, the pathline
and streakline coincide with the streamline when t = 1 s as shown
in Fig. a.
The pathline and streakline will coincide with the streamline
until t = 3 s, after which the streamline makes a sudden change in
direction. Thus, the streamline and pathline of the first marked
particle and the streakline when t = 4 s will be as shown in Fig.
b.
32. The flow of a liquid is originally along the positive x axis
at 2 m>s for 3 s. If it then suddenly changes to 4 m>s along
the positive y axis for t 7 3 s, draw the pathline and streamline
for the first marked particle when t = 1 s and t = 4 s. Also, draw
the streaklines at these two times.
(b)
(a)
streakline
t = 1 st = 0 2 m streamline
x
first markedparticle
pathline
t = 4 s
t = 4 s
t = 3 st = 0
pathline
streamline
streakline first markedparticle
y
x
6 m
4 m
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SolutionSince the streamlines have a constant direction along
the positive y axis, 0 t 6 4 s, the pathline and streakline
coincide with the streamline when t = 2 s as shown inFig. a.
The pathline and streakline will coincide with the streamline
until t = 4 s, when the streamline makes a sudden change in
direction. The pathline, streamline, and streakline are shown in
Fig. b.
33. The flow of a liquid is originally along the positive y axis
at 3 m>s for 4 s. If it then suddenly changes to 2 m>s along
the positive x axis for t 7 4 s, draw the pathline and streamline
for the first marked particle when t = 2 s and t = 6 s. Also, draw
the streakline at these two times.
(b)
(a)
6 m
t = 0
t = 2 s
pathline streamline
streaklinefirst marked particle
y
t = 4 s t = 6 s
t = 0
pathlinestreamline
streakline
first markedparticle
4 m
12 m
y
x
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SolutionThe velocity vector for a particle at x = 2 m and y = 3
m is
V = 5(2x + 1)i - (y + 3x)j6 m>s = [2(2) + 1]i - [3 + 3(2)]j =
55i - 9j6 m>sThe magnitude of V is
V = 2V 2x + V 2y = 2(5 m>s)2 + ( -9 m>s)2 = 10.3 m>s
Ans.As indicated in Fig. a, the direction of V is defined by u =
360 - f, where
f = tan-1 aVyVx
b = tan-1 a9 m> s5 m> s b = 60.95
Thus,
u = 360 - 60.95 = 299 Ans.
*34. A two-dimensional flow field for a fluid be described by V
= [(2x + 1)i - (y + 3x)j] m>s, where x and y are in meters.
Determine the magnitude of the velocity of a particle located at (2
m, 3 m), and its direction measured counterclockwise from the x
axis.
(a)
xVx = 5 m/s
VVy = 9 m/s
3 m
2 mx
y
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SolutionThe velocity vector of a particle at x = 5 m and y = -2
m is
V = 5(5y2 - x)i + (3x + y)j6 m>s = 35(-2)2 - 54 i + 33(5) +
(-2)4j = 515i + 13j6 m>s
The magnitude of V is
V = 2V 2x + V 2y = 2(15 m>s)2 + (13 m>s)2 = 19.8 m>s
Ans.As indicated in Fig. a, the direction of V is defined by
u = tan-1 aVyVx
b = tan-1 a13 m>s15 m>s b = 40.9 Ans.
35. A two-dimensional flow field for a liquid can be described
by V = 3 (5y2 - x)i + (3x + y)j4 m>s, where x and y are in
meters. Determine the magnitude of the velocity of a particle
located at (5 m, -2 m), and its direction measured counterclockwise
from the x axis.
(a)
x
VVy = 13 m/s
Vx = 15 m/s
Ans:V = 19.8 m>su = 40.9
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SolutionThe velocity vector of a particle at x = 2 m and the
corresponding time t = 5 s is
V = 5(0.8x)i + (0.06t2)j6 m>s = 30.8(2)i + 0.06(5)2 j4 =
51.6i + 1.5j6 m>s
The magnitude of V is
V = 2V 2x + V 2y = 2(1.6 m>s)2 + (1.5 m>s)2 = 2.19 m>s
Ans.As indicated in Fig. a, the direction of V is defined by
u = tan-1 aVyVx
b = tan-1 a1.5 m>s1.6 m>s b = 43.2
Using the definition of the slope of the streamline and initial
condition at x = 2 m, y = 3 m.
dy
dx=
v
u;
dy
dx=
0.06t2
0.8x
Note that since we are finding the streamline, which represents
a single instant in time, t = 5 s, t is a constant.
Ly
3 m
dy
t2= L
x
2 m
0.075dxx
1
t2 (y - 3) = 0.075 ln
x2
y = a0.075t2 ln x2+ 3b m
When t = 5 s,
y = 0.075(52) ln ax2b + 3
y = c 1.875 ln ax2b + 3 d m
x(m) 0.5 1 2 3 4 5 6
y(m) 0.401 1.700 3 3.760 4.300 4.718 5.060
The plot of the streamline is shown in Fig. a
36. The soap bubble is released in the air and rises with a
velocity of V = 3(0.8x)i + (0.06t2)j4 m>s where x is meters and
t is in seconds. Determine the magnitude of the bubbles velocity,
and its directions measured counterclockwise from the x axis, when
t = 5 s, at which time x = 2 m and y = 3 m. Draw its streamline at
this instant.
(a)
x
VVy = 1.5 m s
Vx = 1.6 m s
0 10.5 2 3 4(a)
x(m)
y(m)
5 6
1
2
3
4
5
6
v
u
Ans:V = 2.19 m>su = 43.2
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SolutionAs indicated in Fig a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u
dy
dx=
2y
2 + y
L2 + y
2y dy = Ldx
ln y +12
y = x + C
At point (3 m, 2 m), we obtain
ln(2) +12
(2) = 3 + C
C = -1.31
Thus,
ln y +12
y = x - 1.31
ln y2 + y = 2x - 2.61 Ans.
At point (3 m, 2 m)
u = (2 + 2) m>s = 4 m>s S v = 2(2) = 4 m>sc
The magnitude of the velocity is
V = 2u2 + v2 = 2(4 m>s)2 + (4 m>s)2 = 5.66 m>s Ans.and
its direction is
u = tan-1 avub = tan-1 a4 m>s
4 m>s b = 45 Ans.
37. A flow field for a fluid described by u = (2 + y) m>s and
v = (2y) m>s, where y is in meters. Determine the equation of
the streamline that passes through point (3m,2m), and find the
velocity of a particle located at this point. Draw this
streamline.
(a)
x
y
y
u = (2 y) m/s
streamline
v = 2y m/s
V
x
Ans:ln y2 + y = 2x - 2.61V = 5.66 m>su = 45 a
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SolutionAs indicated in Fig a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
-6xyx2 + 5
Ldy
y= -6L
x
x2 + 5 dx
ln y = -3 ln(x2 + 5) + C
At x = 5 m, y = 1 m. Then,
ln 1 = -3 ln3(5)2 + 54 + C C = 3 ln 30
Thus
ln y = -3 ln(x2 + 5) + 3 ln 30 ln y + ln(x2 + 5)3 = 3 ln 30
ln3y(x2 + 5)34 = ln 303 y(x2 + 5)3 = 303
y =27(103)
(x2 + 5)3 Ans.
At point (5 m, 1m),
u = (52 + 5) m>s = 30 m>s S v = -6(5)(1) = -30 m>s = 30
m>s T
The magnitude of the velocity is
V = 2u2 + v2 = 2(30 m>s)2 + (30 m>s)2 = 42.4 m>s
Ans.And its direction is
u = tan-1 avub = tan-1 a30 m>s
30 m>s b = 45 Ans.
*38. A two-dimensional flow field is described by u = 1x2 + 52
m>s and v = (-6xy) m>s. Determine the equation of the
streamline that passes through point (5 m, 1m), and find the
velocity of a particle located at this point. Draw this
streamline.
(a)
x
x
y
y
v = (6xy) m/s
u = (x2 + 5) m/s
streamline
V
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SolutionAs indicated in Fig. a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
4
2y2
Ly2 dy = L2dx
13
y3 = 2x + CAt x = 1 m, y = 2 m. Then
13
(2)3 = 2(1) + C
C = 23
Thus,
13
y3 = 2x +23
y3 = 6x + 2 Ans.
At point (1 m, 2 m)
u = 2(22) = 8 m>s S v = 4 m>s c
The magnitude of the velocity is
V = 2u2 + v2 = 2(8 m>s)2 + (4 m>s)2 = 8.94 m>s Ans.And
its direction is
u = tan-1 avub = tan-1 a4
8b = 26.6 Ans.
39. Particles travel within a flow field defined by V = 32y2i +
4j4 m>s, where x and y are in meters. Determine the equation of
the streamline passing through point (1 m, 2 m), and find the
velocity of a particle located at this point. Draw this
streamline.
(a)
xx
y
y
streamlineVv = 4 m/s
u = 2y2 m/s
Ans:y3 = 6x + 2V = 8.94 m>su = 26.6 a
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SolutionAs indicated in Fig. a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
0.8 + 0.6y0.5
= 1.6 + 1.2y
Since the balloon starts at y = 0, x = 0, using these
values,
Ly
0
dy
1.6 + 1.2y= L
x
0dx
1
1.2 ln(1.6 + 1.2y) ` y
0= x
lna1.6 + 1.2y1.6
b = 1.2x lna1 + 3
4 yb = 1.2x
1 +34
y = e1.2x
y =43
(e1.2x - 1) m Ans.
Using this result, the streamline is shown in Fig. b.
310. A ballon is released into the air from the origin and
carried along by the wind, which blows at a constant rate of u =
0.5 m>s. Also, buoyancy and thermal winds cause the balloon to
rise at a rate of v = (0.8 + 0.6y) m>s. Determine the equation
of the streamline for the balloon, and draw this streamline.
(a)
(b)
xx
y
y
streamline
V
y(m)
x(m)
y = (e1.2x 1) m
streamline
43
v = (0.8 + 0.6y) m/s
u = 0.5 m/s
y
x
u
v
Ans:
y =43
(e1.2x - 1)
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SolutionAs indicated in Fig. a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
1.6 + 0.4y0.8x
The balloon starts at point (1 m, 0).
Ly
0
dy
1.6 + 0.4y= L
x
1
dx0.8x
10.4
ln(1.6 + 0.4y) ` y0
=1
0.8 ln x ` x
1
10.4
lna1.6 + 0.4y1.6
b = 10.8
ln x
lna1 + 14
yb2 = ln xa1 + 1
4yb2 = x
y = 4(x1>2 - 1) m Ans.Using this result, the streamline is
shown in Fig. b.
311. A ballon is released into the air from point (1 m, 0) and
carried along by the wind, which blows at a rate of u = (0.8x)
m>s. Also, buoyancy and thermal winds cause the balloon to rise
at a rate of v = (1.6 + 0.4y) m>s. Determine the equation of the
streamline for the balloon, and draw this streamline.
(a)
(b)
streamline
1 m
y
x
y = 4(x 1) m
x
x
y
y
streamline
V
v = (1.6 + 0.4y) m/s
u = (0.8x) m/s
y
x
u
v
Ans:y = 4(x1>2 - 1)
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SolutionAs indicated in Fig. a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
6x8y
L8y dy = L6x dx4y2 = 3x2 + C
At x = 1 m, y = 2 m. Then
4(2)2 = 3(1)2 + C
C = 13
Thus
4y2 = 3x2 + 13 Ans.
*312. A flow field is defined by u = (8y) m>s, v = (6x)
m>s where x and y are in meters. Determine the equation of the
streamline that passes through point (1 m, 2 m). Draw this
streamline.
(a)
x
x
y
y
streamline
Vv = (6x) m/s
u = (8y) m/s
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SolutionAs indicated in Fig. a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
6y
3x
Ldy
2y= L
dxx
12
ln y = ln x + C
At x = 3 ft, y = 1 ft . Then
12
ln y = ln x - ln3
12
ln y = ln x3
ln y = lnax3b2
y =x2
9 Ans.
313. A flow field is defined by u = (3x) ft>s and v = (6y)
ft>s, where x and y are in feet. Determine the equation of the
streamline passing through point (3 ft, 1 ft). Draw this
streamline.
(a)
x
xy
y
streamline
Vv = (6y) ft/s
u = (3x) ft/s
Ans:y = x2>9
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SolutionSince the velocity V is constant, Fig. a, the streamline
will be a straight line with aslope.
dy
dx= tan u
dy
dx=
v
u=
85
y = 1.6x + C
At x = 0, y = 0. Then
C = 0
Thus
y = 1.6x Ans.
Since the direction of velocity V remains constant so does the
streamline, and the flow is steady. Therefore, the pathline
coincides with the streamline and shares the same equation.
314. A flow of water is defined by u = 5 m>s and v = 8
m>s. If metal flakes are released into the flow at the origin
(0, 0), draw the streamlines and pathlines for these particles.
(a)
V
x
y
v = 8 m/s
u = 5 m/s
streamline pathline
Ans:y = 1.6x
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SolutionAs indicated in Fig. a, the velocity V of a particle on
the streamline is always directed along the tangent of the
streamline. Therefore,
dy
dx= tan u
dy
dx=
v
u=
8y> (x2 + y2)8x> (x2 + y2) = yx
Ldy
y= L
dxx
ln yx = C
y
x= C
At x = 1 m, y = 1 m. Then
C = 1
Thus,
y
x= 1
y = x Ans.
315. A flow field is defined by u = 38x> (x2 + y2) 4 m>s
and v = 38y> (x2 + y2) 4 m>s, where x and y are in meters.
Determine the equation of the streamline passing through point (1
m, 1 m). Draw this streamline.
(a)
xx
y
y
streamline
Vv = m/s
8yx2 + y2
u = m/s8y
x2 + y2
))
))
Ans:y = x
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SolutionSince the velocity components are a function of time and
position, the flow can be classified as unsteady nonuniform flow.
Because we are finding a pathline, t is not a constant but a
variable. We must first find equations relating x to t and y to t,
and then eliminate t. Using the definition of velocity
dxdt
= u =30
2x + 1 ; L
x
2 m(2x + 1)dx = 30L
t
2 sdt
(x2 + x) ` x2 m
= 30 t ` t2 s
x2 + x - 6 = 30(t - 2)
t =130
(x2 + x + 54) (1)
dy
dt= v = 2ty; L
y
6 m
dy
y= 2L
t
2 stdt
ln y ` y6 m
= t2 ` t2 s
ln y
6= t2 - 4
y
6= et
2-4
y = 6et2-4 (2)
Substitute Eq. (1) into Eq. (2),
y = 6e 1
900 (x2+x+54)2-4 Ans.
The plot of the pathline is shown in Fig. a.
x(m) 0 1 2 3 4
y(m) 2.81 3.58 6.00 13.90 48.24
*316. A fluid has velocity components of u = 330>(2x + 1)4
m>s and v = 2ty m>s where x and y are in meters and t is in
seconds. Determine the pathline that passes through the point (2 m,
6 m) at time t = 2 s. Plot this pathline for 0 x 4 m.
0 1 2 3 4
(a)
x(m)
y(m)
10
20
30
40
50
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Ans:For t = 1 s, y = 4e(x
2 + x - 2)>15For t = 2 s, y = 4e2(x
2 + x - 2)>15For t = 3 s, y = 4e(x
2 + x - 2)>5
SolutionSince the velocity components are a function of time and
position, the flow can be classified as unsteady nonuniform. The
slope of the streamline is
dy
dx=
v
u;
dy
dx=
2ty
30>(2x + 1) = 115 ty(2x + 1)L
y
4 m
dy
y=
115
tLx
1 m(2x + 1)dx
ln y ` y4 m
=115
t(x2 + x) ` x1 m
ln y
4=
115
t(x2 + x - 2)
y = 4et(x2+x-2)>15
For t = 1 s,
y = 4e(x2+x-2)>15 Ans.
For t = 2 s,
y = 4e2(x2+x-2)>15 Ans.
For t = 3 s,
y = 4e(x2+x-2)>5 Ans.
The plot of these streamlines are shown in Fig. a
For t = 1 s
x(m) 0 1 2 3 4
y(m) 3.50 4 5.22 7.79 13.3
For t = 2 sx(m) 0 1 2 3 4
y(m) 3.06 4 6.82 15.2 44.1
For t = 3 s
x(m) 0 1 2 3 4
y(m) 2.68 4 8.90 29.6 146
317. A fluid has velocity components of u = 330>(2x + 1)4
m>s and v = (2ty) m>s where x and y are in meters and t is in
seconds. Determine the streamlines that passes through point (1 m,
4 m) at times t = 1s, t = 2 s, and t = 3 s. Plot each of these
streamlines for 0 x 4 m.
0 1 2 3 4(a)
x(m)
y(m)
5
10
15
20
25
30
35
40
45
t = 3 st = 2 s
t = 1 s
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Ans:For t = 2 s, y = 6e21x2 + x - 6)>15For t = 5 s, y = 6e1x2
+ x - 6)>3
Solution
For t = 2 s
x(m) 0 1 2 3 4y(m) 2.70 3.52 6.00 13.35 38.80
For t = 5 s
x(m) 0 1 2 3 4
y(m) 0.812 1.58 6.00 44.33 638.06
Since the velocity components are a function of time and
position the flow can be classified as unsteady nonuniform. The
slope of the streamline is
dy
dx=
v
u ;
dy
dx=
2ty
30>(2x + 1) = 115 ty(2x + 1)Note that since we are finding
the streamline, which represents a single instant in time, either t
= 2 s or t = 5 s, t is a constant.
Ly
6 m
dy
y=
115
tLx
2 m(2x + 1)dx
ln y ` y6 m
=115
t (x2 + x) ` x2 m
ln y
6=
115
t(x2 + x - 6)
y = 6e115 t(x
2+x-6)
For t = 2 s,
y = 6e215 (x
2 +x-6) Ans.
For t = 5 s,
y = 6e13 (x
2+x-6) Ans.
The plots of these two streamlines are show in Fig. a.
318. A fluid has velocity components of u = 330>(2x + 1)4
m>s and v = (2ty) m>s where x and y are in meters and t is in
seconds. Determine the streamlines that pass through point (2 m, 6
m) at times t = 2 s and t = 5 s. Plot these streamlines for 0 x 4
m.
0 1 2
(a)
3 4x(m)
y(m)
10
20
30
40
50
t = 5 s
t = 2 s
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Ans:u = 3.43 m>sv = 3.63 m>s
Solution
x(m) 0 1 1.5 2 3 4 5y(m) -2.29 -1.59 0 1.59 2.29 2.71 3.04
The plot of the streamline is shown in Fig. a. Taking the
derivative of the streamline equation,
3y2 dy
dx= 8
dy
dx= tan u =
8
3y2
When x = 1 m,
y3 = 8(1) - 12; y = -1.5874
Then
dy
dx`x = 1 m
= tan u `x = 1 m
=8
3(-1.5874)2; u 0 x = 1 m = 46.62
Therefore, the horizontal and vertical components of the
velocity are
u = (5 m>s) cos 46.62 = 3.43 m>s Ans. v = (5 m>s) sin
46.62 = 3.63 m>s Ans.
319. A particle travels along the streamline defined by y3 = 8x
- 12. If its speed is 5 m>s when it is at x = 1 m, determined
the two components of its velocity at this point. Sketch the
velocity on the streamline.
(a)
y(m)
0 1 2 3 4x(m)
5
1
1
2
3
2
3
3
3
46.62u
v5 m/s
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Solution
Here, u =dxdt
. Then,
dx = udt
Using x = 0 when t = 0 as the integration limit,
Lx
0dx = L
t
03(0.8t) m>s4dt
x = 0.4t2 (1)
Also, v =dy
dt. Then
dy = vdt
Using y = 0 when t = 0 as the integration limit,
Ly
0dy = L
t
0(0.4 m>s)dt
y = 0.4t (2)
Eliminating t from Eqs. (1) and (2)
y2 = 0.4x
This equation represents the pathline of the particle. The x and
y values of the pathline for the first five seconds are tabulated
below.
t x y1 0.4 0.42 1.6 0.83 3.6 1.24 6.4 1.65 10 2
A plot of the pathline is shown in Fig. a.
From Eqs. (1) and (2), when t = 4 s,
x = 0.4(42) = 6.4 m y = 0.4(4) = 1.6 m
Using the definition of the slope of the streamline,
dy
dx=
v
u ;
dy
dx=
0.40.8t
tLy
1.6 mdy =
12L
x
6.4 mdx
t(y - 1.6) =12
(x - 6.4)
y = c 12t
(x - 6.4) + 1.6 d m
*320. A flow field is defined by u = (0.8t) m>s and v = 0.4
m>s, where t is in seconds. Plot the pathline for a particle
that passes through the origin when t = 0. Also, draw the
streamline for the particle when t = 4 s.
(a)
y(m)
0 2 4 6 8x(m)
10
1
1
2
2 y2 = 0.4x
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*320. (continued)
When t = 4 s,
y =1
2(4) ( x - 6.4) + 1.6
y =18
x + 0.8
The plot of the streamline is shown is Fig. b.
(b)
y(m)
0.8
x(m)
y = x + 0.818
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Ans:y3 = 8x - 15, y = 9 mx = 93 m
SolutionSince the velocity components are a function of position
only, the flow can be classified as steady nonuniform. Here, u =
13y22 m>s and v = 8 m>s . The slope of the streamline is
defined by
dy
dx=
v
u ;
dy
dx=
8
3y2
Ly
1 m3y2dy = 8L
x
2 mdx
y3 ` y1 m
= 8x ` x2 m
y3 - 1 = 8x - 16
y3 = 8x - 15 (1) Ans.
From the definition of velocity
dy
dt= 8
Ly
1 mdy = L
1 s
08 dt
y ` y1 m
= 8t ` 1 s0
y - 1 = 8
y = 9 m Ans.
Substituting this result into Eq. (1)
93 = 8x - 15
x = 93 m Ans.
321. The velocity for an oil flow is defined by V = 13y2 i + 8
j2 m>s, where y is in meters. What is the equation of the
streamline that passes through point (2 m, 1m)? If a particle is at
this point when t = 0, at what point is it locatedwhen t = 1 s?
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Ans:
y =23
ln a 22 - x
b
Solution
x(m) 0 0.25 0.5 0.75 1
y(m) 0 0.089 0.192 0.313 0.462
x(m) 1.25 1.5 1.75 2
y(m) 0.654 0.924 1.386
Since the velocity component is a function of position only, the
flow can be classified as steady nonuniform. Using the definition
of the slope of a streamline,
dy
dx=
v
u;
dy
dx=
26 - 3x
Ly
0dy = 2L
x
0
dx6 - 3x
y = -23
ln (6 - 3x) ` x0
y = -23
ln a6 - 3x6
by =
23
ln a 22 - x
b Ans.The plot of this streamline is show in Fig. a
322. The circulation of a fluid is defined by the velocity field
u = (6 - 3x) m>s and v = 2 m>s where x is in meters. Plot the
streamline that passes through the origin for 0 x 6 2 m.
0 0.25 0.5 0.75 1.0
(a)
x(m)
y(m)
1.25 1.5 1.75 2
0.5
1
1.5
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Ans:
For 0 t 6 10 s, y = -32
x
For 10 s 6 t 15 s, y = -25
x + 22
SolutionUsing the definition of velocity, for 0 t 6 10 s
dxdt
= u; dxdt
= -2
Lx
0dx = -2L
t
0dt
x = (-2t) m (1)
When t = 10 s, x = -2(10) = -20 m
dy
dt= v;
dy
dt= 3
Ly
0dy = 3L
t
0dt
y = (3t) m (2)
When t = 10 s, y = 3(10) = 30 m
The equation of the streamline can be determined by eliminating
t from Eq. (1) and (2).
y = -32
x Ans.
For 10 6 t 15 s.
dxdt
= u; dxdt
= 5
Lx
-20 mdx = 5L
t
10 sdt
x - (-20) = 5(t - 10)
x = (5t - 70) m (3)
At t = 15 s, x = 5(15) - 70 = 5 mdy
dt= v;
dy
dt= -2
Ly
30 mdy = -2L
t
10 sdt
y - 30 = -2(t - 10)y = (-2t + 50) m (4)
When t = 15 s, y = -2(15) + 50 = 20 m
Eliminate t from Eqs. (3) and (4),
y = a-25
x + 22b Ans.The two streamlines intersect at (-20, 30), point B
in Fig. (a). The pathline is the path ABC.
323. A stream of water has velocity components of u = -2 m>s,
v = 3 m>s for 0 t 6 10 s; and u = 5 m>s, v = -2 m>s for 10
s 6 t 15 s. Plot the pathline and streamline for a particle
released at point (0, 0) when t = 0 s.
(a)
0 55
A
B
C
101520x(m)
y(m)
10
20
30
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Solution
x(m) 0.25 0.5 0.75 1 2 3 4
y(m) 4.96 5.31 5.51 5.65 6 6.20 6.35
Since the velocity components are a function of time and
position, the flow can be classified as unsteady nonuniform. Using
the definition of the slope of the streamline,
dy
dx=
v
u;
dy
dx=
2t4x
=t
2x
Ly
6 mdy =
t2L
x
2 m
dxx
y - 6 =t2
ln x2
y =t2
ln x2+ 6
For t = 1 s, y = a12
ln x2+ 6bm Ans.
The plot of this streamline is shown in Fig. a.
*324. The velocity field is defined by u = (4x) m>s and v =
(2t) m>s, where t is in seconds and x is in meters. Determine
the equation of the streamline that passes through point (2 m, 6 m)
for t = 1 s. Plot this streamline for 0.25 m x 4 m.
0 10.750.50.25 2 3 4
(a)
x(m)
y(m)
1
2
3
4
5
6
7
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Ans:
y =1
16 ln2
x2+
12
ln x2+ 6
Solution
x(m) 0.25 0.50 0.75 1 2 3 4
y(m) 5.23 5.43 5.57 5.68 6 6.21 6.38
Since the velocity components are a function of time and
position, the flow can be classified as unsteady nonuniform. Using
the definition of velocity,
dxdt
= u = 4x; Lx
2 m
dx4x
= Lt
1 Sdt
14
ln x ` x2 m
= t ` t1 s
14
ln x2
= t - 1
t =14
ln x2+ 1 (1)
dy
dt= v = 2t; L
y
6 mdy = L
t
1 s2t dt
y - 6 = t2 ` t1 s
y = t2 + 5 (2)
Substitute Eq. (1) into (2),
y = a14
ln x2+ 1b2 + 5
y = a 116
ln2 x2+
12
ln x2+ 6b Ans.
The plot of this pathline is shown in Fig. (a)
325. The velocity field is defined by u = (4x) m>s and v =
(2t) m>s, where t is in seconds and x is in meters. Determine
the pathline that passes through point (2 m, 6 m) when t = 1 s.
Plot this pathline for 0.25 m x 4 m.
0 10.750.50.25 2 3 4
(a)
x(m)
y(m)
1
2
3
4
5
6
7
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SolutionUsing the definition of velocity, for 0 t 6 5 s,
dxdt
= u; dxdt
=12
x
Lx
1 m
dxx
= Lt
0
12
dt
ln x =12
t
x = ae 12 tb m (1)When t = 5 s, x = e
12 (5) = 12.18 m
dy
dt= v;
dy
dt=
18
y2
Ly
1 m
dy
y2= L
t
0
18
dt
- a1yb ` y
1 m=
18
t
1 -1y
=18
t
y - 1y
=18
t
y a1 - 18
tb = 1y = a 8
8 - tb m t 8 s (2)
When t = 5 s, y =8
8 - 5= 2.667 m
The equation of the streamline and pathline can be determined by
eliminating tfrom Eqs. (1) and (2)
y = a 88 - 2 ln x
bmx(m) 1 3 5 7 9 11 12.18
y(m) 1 1.38 1.67 1.95 2.22 2.50 2.67
For 5 s < t 10 s,
dxdt
= u; dxdt
= -14
x2
Lx
12.18 m
dx
x2= -
14L
t
5 sdt
326. The velocity field of a fluid is defined by u = (12 x)
m>s, v = (18 y2) m>s for 0 t 6 5 s and by u = ( -14 x2)
m>s, v = (14 y) m>s for 5 6 t 10 s, where x and y are in
meters. Plot the streamline and pathline for a particle released at
point (1 m, 1 m) when t = 0 s.
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Ans:
For 0 t 6 5 s, y =8
8 - 2 ln x
For 5 s 6 t 10 s, y = 2.67e11>x - 0.0821)
- a1x-
112.18
b = -14
(t - 5)
1x
=t4- 1.1679
x = a 4t - 4.6717
b m t 4.6717 s (3)When t = 10 s, x =
410 - 4.6717
= 0.751 m
dy
dt= v;
dy
dt=
14
y
Ly
2.667 m
dy
y=
14L
t
5 sdt
ln y
2.667=
14
(t - 5)
y
2.667= e
14 (t-5)
y = c 2.667e14 (t-5) d m (4)When t = 10 s, y = 2.667e
14 (10-5) = 9.31 m
Eliminate t from Eqs. (3) and (4),
y = 2.667e 1434(1x+1.1679)-54
= c 2.667e(1x -0.08208) dmx(m) 0.751 1 3 5 7 9 11 12.18
y(m) 9.31 6.68 3.43 3.00 2.83 2.75 2.69 2.67
The two streamlines intersect at (12.18, 2.67), point B in Fig.
(a). The pathline is the path ABC.
326. (continued)
0 10.751 12.18
2 3 134 5 6 7 8 9 10 11 12
A
C
B
x(m)
y(m)
123456
10
789
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Ans:2y3 - 1.5x2 - y - 2x + 52 = 0V = 30.5 m>s
SolutionWe have steady flow since the velocity does not depend
upon time.
u = 6y2 - 1v = 3x + 2
dy
dx=
v
u=
3x + 26y2 - 1
Ly
2(6y2 - 1)dy = L
x
6(3x + 2)dx
2y3 - y ` y2
= 1.5 x2 + 2x ` x6
2y3 - y - 32(2)3 - 24 = 1.5x2 + 2x - 31.5(6)2 + 2(6)42y3 - 1.5x2
- y - 2x + 52 = 0 Ans.
At (6 m, 2 m)
u = 6(2)2 - 1 = 23 m>s Sv = 3(6) + 2 = 20 m>scV = 2(23
m>s)2 + (20 m>s)2 = 30.5 m>s Ans.
327. A two-dimensional flow field for a liquid can be described
by V = 3(6y2 - 1) i + (3x + 2) j4 m>s, where x and y are in
meters. Determine a streamline that passes through points 16 m, 2
m2 and determine the velocity at this point. Sketch the velocity on
the streamline.
y
23 m/s
20 m s
6
2
30.5 m s
x
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SolutionWe have steady flow since the velocity does not depend
upon time.
u = 2x + 1v = -y
dy
dx=
v
u=
-y2x + 1
-Ldy
y= L
dx(2x + 1)
- ln y =12
ln (2x + 1) + C
- y = (2x + 1)12 + C
- 1 = (2(3) + 1)12 + C
C = - 3.65
-y ` y1
= (2x + 1)12 ` x
3
-y + 1 = (2x + 1)12 - 32(3) + 1 4 12
y = 3.65 - (2x + 1)12 Ans.
u = 2(3) + 1 = 7 m>sv = -1 m>s
V = 2(7 m>s)2 + ( -1 m>s)2 = 7.07 m>s Ans.
*328. A flow field for a liquid can be described by V = 5(2x +
1) i - y j6 m>s, where x and y are in meters. Determine the
magnitude of the velocity of a particle located at points 13 m, 1
m2. Sketch the velocity on the streamline.
y
7 m/s
7.07 m/s1 m s
3
1
x
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Ans:a = 24 m>s2
SolutionSince the flow is along the horizontal (x axis) v = w =
0. Also, the velocity is a function of time t only. Therefore, the
convective acceleration is zero, so that
u 0V0x
= 0.
a =0V0 t
+ u 0V0 x
= 12t + 0= (12t) m>s2
When t = 2 s,
a = 12(2) = 24 m>s2 Ans.Note: The flow is unsteady since its
velocity is a function of time.
329. Air flows uniformly through the center of a horizontal duct
with a velocity of V = (6t2 + 5) m>s, where t is in seconds.
Determine the acceleration of the flow when t = 2 s.
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Ans:1088 in.>s2
Solution
Since the flow is along the x axis, v = w = 0
a =0u0 t
+ u 0u0 x
= 4x + (4xt)(4t)= 4x + 16xt2
= 34x(1 + 4t2) 4 in.>s2When t = 2 s, x = 16 in. Then
a = 34(16)31 + 4(22) 4 4 in.>s2 = 1088 in.>s2Note: The
flow is unsteady since its velocity is a function of time.
330. Oil flows through the reducer such that particles along its
centerline have a velocity of V = (4xt) in.>s, where x is in
inches and t is in seconds. Determine the acceleration of the
particles at x = 16 in. when t = 2 s.
24 in.
x
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Ans:30.9 ft>s2
SolutionFor two dimensional flow, the Eulerian description
gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar component of this equation along the x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 1 + (6y + t)(0) + (2tx)(6)
= (1 + 12tx) ft>s2 ay =
0v0t
+ u 0v0x
+ v 0v0y
= 2x + (6y + t)(2t) + (2tx)(0)
= (2x + 12ty + 2t2) ft>s2When t = 1 s, x = 1 ft and y = 2 ft,
then
ax = 31 + 12(1)(1)4 = 13 ft>s2 ay = 32(1) + 12(1)(2) + 2(12)
4 = 28 ft>s2Thus, the magnitude of the acceleration is
a = 2a 2x + a 2y = 2(13 ft>s2)2 + (28 ft>s2)2 = 30.9
ft>s2 Ans.
331. A fluid has velocity components of u = (6y + t) ft>s and
v = (2tx) ft>s where x and y are in feet and t is in seconds.
Determine the magnitude of acceleration of a particle passing
through the point (1 ft, 2 ft), when t = 1 s.
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*332. The velocity for the flow of a gas along the center
streamline of the pipe is defined by u = (10x2 + 200t + 6) m>s,
where x is in meters and t is in seconds. Determine the
acceleration of a particle when t = 0.01 s and it is at A, just
before leaving the nozzle.
Solution
a =0u0t
+ u 0u0x
0u0t
= 200 0u0x
= 20 x
a = 3200 + (10x2 + 200t + 6)(20x)4 m>s2When t = 0.01 s, x =
0.6 m.
a = 5200 + 310(0.62) + 200(0.01) + 64 320(0.6)4 6 m>s2 = 339
m>s2 Ans.
xA
0.6 m
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Ans:V = 23.3 m>sa = 343 m>s2
SolutionVelocity.
At x = 2 m, y = 4 m,
u = 2(22) - 2(42) + 4 = -20 m>s v = 4 + 2(4) = 12 m>sThe
magnitude of the particles velocity is
V = 2u2 + v2 = 2( -20 m>s)2 + (12 m>s)2 = 23.3 m>s
Ans.Acceleration. The x and y components of the particles
acceleration, with w = 0 are
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + (2x2 - 2y2 + y)(4x) + (y + xy)(-4y + 1)
At x = 2 m, y = 4 m,
ax = -340 m>s2 ay =
0v0t
+ u 0v0x
+ v 0v0y
= 0 + (2x2 - 2y2 + y)(y) + (y + xy)(1 + x)
At x = 2 m, y = 4 m,
ay = -44 m>s2The magnitude of the particles acceleration
is
a = 2a 2x + a 2y = 2( -340 m>s2)2 + ( -44 m>s2)2 = 343
m>s2 Ans.
333. A fluid has velocity components of u = (2x2 - 2y2 + y)
m>s and v = (y + xy) m>s, where x and y are in meters.
Determine the magntiude of the velocity and acceleration of a
particle at point (2 m, 4 m).
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Ans:V = 16.3 m>s uv = 79.4 aa = 164 m>s2ua = 17.0 a
SolutionSince the velocity components are a function of position
only the flow can be classified as steady nonuniform. At point x =
2 m and y = 1 m,
u = 5(12) - 2 = 3 m>sv = 4(22) = 16 m>s
The magnitude of the velocity is
V = 2u2 + v2 = 2(3 m>s)2 + (16 m>s)2 = 16.3 m>s Ans.Its
direction is
uv = tan-1avu b = tan-1a16 m>s3 m>s b = 79.4 Ans.For two
dimensional flow, the Eulerian description is
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axis
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + (5y2 - x)(-1) + 4x2(10y)
= (x - 5y2) + 40x2y
ay =0v0t
+ u 0v0x
+ v 0v0y
= 0 + (5y2 - x)(8x) + 4x2(0)
= 8x(5y2 - x)
At point x = 2 m and y = 1 m,
ax = 32 - 5(12) 4 + 40(22)(1) = 157 m>s2ay = 8(2)35(12) - 24
= 48 m>s2
The magnitude of the acceleration is
a = 2a 2x + a 2y = 2(157 m>s2)2 + (48 m>s2)2 = 164 m>s2
Ans.Its direction is
ua = tan-1aayax b = tan-1 48 m>s2157 m>s2 = 17.0 Ans.
334. A fluid velocity components of u = (5y2 - x) m>s and v =
(4x2) m>s, where x and y are in meters. Determine the velocity
and acceleration of particles passing through point (2 m, 1 m).
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Ans:a = 36.1 m>s2u = 33.7 a
SolutionSince the velocity components are independent of time
but are a function of position, the flow can be classified as
steady nonuniform. The slope of the streamline is
dy
dx=
v
u;
dy
dx=
4x - 15y2
Ly
1 m5y2dy = L
x
1 m(4x - 1)dx
y3 =15
(6x2 - 3x + 2) where x is in m
For two dimensional flow, the Eulerian description is
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + 5y2(0) + (4x - 1)(10y)
= 40xy - 10y
ay =0v0t
+ u 0v0x
+ v 0v0y
= 0 + 5y2(4) + (4x - 1)(0)
= 20y2
At point x = 1 m and y = 1 m,
ax = 40(1)(1) - 10(1) = 30 m>s2ay = 20(12) = 20 m>s2
The magnitude of the acceleration is
a = 2a 2x + a 2y = 2(30 m>s2)2 + (20 m>s2)2 = 36.1 m>s2
Ans.Its direction is
u = tan-1aayax
b = tan-120 m>s230 m>s2 = 33.7 Ans.
The plot of the streamline and the acceleration on point (1 m, 1
m) is shown in Fig.a.
x(m) 0 0.5 1 2 3 4 5y(m) 0.737 0.737 1 1.59 2.11 2.58 3.01
0 1 2 3 4 5(a)
x(m)
y(m)
1
2
3
4
a = 36.1 m/s2
ax = 30 m/s2
ay = 20 m/s2
335. A fluid has velocity components of u = (5y2) m>s and v =
(4x - 1) m>s, where x and y are in meters. Determine the
equation of the streamline passing through point 11 m, 1 m2 . Find
the components of the acceleration of a particle located at this
point and sketch the acceleration on the streamline.
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SolutionSince the velocity is a function of position only, the
flow can be classified as steady nonuniform. Since the velocity
varies linearly with x,
V = VA + aVB - VALAB bx = 8 + a2 - 83 bx = (8 - 2x) m>s
Ans.For one dimensional flow, the Eulerian description gives
a =0Vdt
+ V 0Vdx
= 0 + (8 - 2x)(-2)
= 4(x - 4) m/s2 Ans.
using the definition of velocity,dxdt
= V = 8 - 2x; Lx
0
dx8 - 2x
= Lt
0dt
-12
ln(8 - 2x) ` x0
= t
12
ln a 88 - 2x
b = tln a 8
8 - 2xb = 2t
88 - 2x
= e2 t
x = 4(1 - e-2 t) m Ans.
*336. Air flowing through the center of the duct has been found
to decrease in speed from VA = 8 m>s to VB = 2 m>s in a
linear manner. Determine the velocity and acceleration of a
particle moving horizontally through the duct as a function of its
position x. Also, find the position of the particle as a function
of time if x = 0 when t = 0. VA 8 m/s VB 2 m/s
3 m
x
A
B
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Ans:V = 33.5 m>suV = 17.4a = 169 m>s2ua = 79.1 a
SolutionSince the velocity components are functions of time and
position the flow can be classified as unsteady nonuniform. When t
= 2 s, x = 1 m and y = 1 m.
u = 8(22) = 32 m>sv = 7(1) + 3(1) = 10 m>s
The magnitude of the velocity is
V = 2u2 + v2 = 2(32 m>s)2 + (10 m>s)2 = 33.5 m>s
Ans.Its direction is
uv = tan-1avu b = tan-1a10 m>s32 m>s b = 17.4 uv Ans.For
two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 16t + 8t2(0) + (7y + 3x)(0)
= (16t) m>s2ay =
0v0t
+ u 0v0x
+ v 0v0y
= 0 + (8t2)(3) + (7y + 3x)(7)
= 324t2 + 7(7y + 3x)4 m>s2When t = 2 s, x = 1 m and y = 1
m.
ax = 16(2) = 32 m>s2ay = 24(22) + 737(1) + 3(1)4 = 166
m>s2
The magnitude of the acceleration is
a = 2a 2x + a 2y = 2(32 m>s2)2 + (166 m>s2)2 = 169 m>s2
Ans.Its direction is
ua = tan-1aayax b = tan-1a166 m>s232 m>s2 b = 79.1 ua
Ans.
337. A fluid has velocity components of u = (8t2) m>s and v =
(7y + 3x) m>s, where x and y are in meters and t is in seconds.
Determine the velocity and acceleration of a particle passing
through point x = 1 m, y = 1 m when t = 2 s.
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Ans:y = x>2, a = 143 ft>s2u = 26.6 a
SolutionSince the velocity components are the function of
position but not the time, the flow is steady (Ans.) but
nonuniform. Using the definition of the slope of the
streamline,
dy
dx=
v
u ;
dy
dx=
8y
8x=
y
x
Ly
1 ft
dy
y= L
x
2 ft
dxx
ln y ` y1 ft
= ln x ` x2 ft
ln y = ln x2
y =12
x Ans.
For two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + 8x(8) + 8y(0)
= (64x) ft>s2ay =
0v0t
+ u 0v0x
+ v 0v0y
= 0 + (8x)(0) + 8y(8)
= (64y) ft>s2At x = 2 ft, y = 1 ft . Then
ax = 64(2) = 128 ft>s2 ay = 64(1) = 64 ft>s2The magnitude
of the acceleration is
a = 2a 2x + a 2y = 2(128 ft>s2)2 + (64 ft>s2)2 = 143
ft>s2 Ans.Its direction is
u = tan-1aayax
b = tan-1a 64 ft>s2128 ft>s2 b = 26.6 u Ans.
338. A fluid has velocity components of u = (8x) ft>s and v =
(8y) ft>s, where x and y are in feet. Determine the equation of
the streamline and the acceleration of particles passing through
point (2 ft, 1 ft). Also find the acceleration of a particle
located at this point. Is the flow steady or unsteady?
(a)
x
xy
y
v = (8y) ft/s
u = (8x) ft/s
streamline
V
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Ans:y = 2xa = 286 m>s2u = 63.4 a
SolutionSince the velocity components are the function of
position, not of time, the flow can be classified as steady (Ans.)
but nonuniform. Using the definition of the slope of the
streamline,
dy
dx=
v
u ;
dy
dx=
8xy
xy2=
4xy
Ly
2 my dy = L
x
1 m4x dx
y2
2` y2 m
= 2x2 ` x1m
y2
2- 2 = 2x2 - 2
y2 = 4x2
y = 2x Ans.
(Note that x = 1, y = 2 is not a solution to y = -2x.) For two
dimensional flow, the Eulerian description gives.
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + 2y2(0) + 8xy(4y)
= (32xy2) m>s2 ay =
0v0t
+ u 0v0x
+ v 0v0y
= 0 + 2y2(8y) + (8xy)(8x)
= (16y3 + 64x2y) m>s2At point x = 1 m and y = 2 m,
ax = 32(1)(22) = 128 m>s2ay = 316(23) + 64(12)(2)4 = 256
m>s2
The magnitude of the acceleration is
a = 2a 2x + a 2y = 2(128 m>s2)2 + (256 m>s2)2 = 286
m>s2 Ans.Its direction is
u = tan-1aayax
b = tan-1a256 m>s2128 m>s2 b = 63.4 u Ans.
339. A fluid velocity components of u = (2y2) m>s and v =
(8xy) m>s, where x and y are in meters. Determine the equation
of the streamline passing through point (1 m, 2 m). Also, what is
the acceleration of a particle at this point? Is the flow steady or
unsteady?
(a)
x
x
y
y v = (8xy) m/s
u = (2y2) m/s
streamline
V
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SolutionThe flow is steady but nonuniform since the velocity
components are a function of position, but not time. At point (2 m,
1 m)
u = 4y = 4(1) = 4 m>sv = 2x = 2(2) = 4 m>s
Thus, the magnitude of the velocity is
V = 2u2 + v2 = 2(4 m>s)2 + (4 m>s)2 = 5.66 m>s Ans.For
two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + 4y(0) + (2x)(4)
= (8x) m>s2ay =
0v0t
+ u 0v0x
+ v 0v0y
= 0 + 4y(2) + 2x(0)
= (8y) m>s2At point (2 m, 1 m),
ax = 8(2) = 16 m>s2ay = 8(1) = 8 m>s2
The magnitude of the acceleration is
a = 2a 2x + a 2y = 2(16 m>s)2 + (8 m>s)2 = 17.9 m>s2
Ans.Using the definition of the slope of the streamline,
dy
dx=
v
u ;
dy
dx=
2x4y
=x2y
Ly
1 m2y dy = L
x
2 mx dx
y2 ` y1 m
=x2
2` x2 m
y2 - 1 =x2
2- 2
y2 =12
x2 - 1
*340. The velocity of a flow field is defined by V = 54 yi + 2
xj6 m>s, where x and y are in meters. Determine the magnitude of
the velocity and acceleration of a particle that passes through
point (2 m, 1 m). Find the equation of the streamline passing
through this point, and sketch the velocity and acceleration at the
point on this streamline.
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*340. (continued)
The plot of this streamline is shown is Fig. a
x(m) 22 2 3 4 5 6y(m) 0 1 1.87 2.65 3.39 4.12
(a)
0 1 2 3 4x(m)
y(m)
5 6
1
1
2
3
4
2
3
4
u = 4 m/s
45
v = 4 m/s
V = 5.66 m/s
0 1 2 3 4x(m)
y(m)
5 6
1
1
2
3
4
2
3
4
ax = 16 m/s2
ay = 8 m/s2
a = 17.9 m/s2
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Ans:V = 4.47 m>s, a = 16 m>s2y =
12
ln x + 2
SolutionSince the velocity components are a function of position
but not time, the flow can be classified as steady nonuniform. At
point (1 m, 2 m),
u = 4x = 4(1) = 4 m>sv = 2 m>s
The magnitude of velocity is
V = 2u2 + v2 = 2(4 m>s)2 + (2 m>s)2 = 4.47 m>s Ans.For
two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + 4x(4) + 2(0) = 16x
ay =0v0t
+ u 0v0x
+ v 0v0y
= 0 + 4x(0) + 2(0) = 0
At point (1 m, 2 m),
ax = 16(1) = 16 m>s2 ay = 0Thus, the magnitude of the
acceleration is
a = ax = 16 m>s2 Ans.Using the definition of the slope of the
streamline,
dy
dx=
v
u ;
dy
dx=
24x
=12x
Ly
2 mdy =
12L
x
1 m
dxx
y - 2 =12
ln x
y = a12
ln x + 2b Ans.The plot of this streamline is shown in Fig. a
x(m) e-4 1 2 3 4 5
y(m) 0 2 2.35 2.55 2.69 2.80
341. The velocity of a flow field is defined by V = 54 xi + 2j6
m>s, where x is in meters. Determine the magnitude of the
velocity and acceleration of a particle that passes through point
(1 m, 2 m). Find the equation of the streamline passing through
this point, and sketch these vectors on this streamline.
0 1 2 3 4 5x(m)
y(m)
1
2
3
u = 4 m/s
V = 4.47 m/sv = 2 m/s
0 1 2 3 4 5x(m)
y(m)
1
2
3
0 1 2 3 4 5x(m)
y(m)
1
2
3
0 1 2 3 4 5x(m)
y(m)
1
2
3
a = 16 m/s2
0 1 2 3 4 5x(m)
y(m)
1
2
3
u = 4 m/s
V = 4.47 m/sv = 2 m/s
0 1 2 3 4 5x(m)
y(m)
1
2
3
0 1 2 3 4 5x(m)
y(m)
1
2
3
0 1 2 3 4 5x(m)
y(m)
1
2
3
a = 16 m/s2
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SolutionSince the velocity components are a function of position
but not time, the flow can be classified as steady but nonuniform.
At point (1 m, 1 m),
u = 2x2 - y2 = 2(12) - 12 = 1 m>sv = -4xy = -4(1)(1) = -4
m>s
The magnitude of the velocity is
V = 2u2 + v2 = 2(1 m>s)2 + ( -4 m>s)2 = 4.12 m>s
Ans.For two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + (2x2 - y2)(4x) + (-4xy)(-2y) = 4x(2x2 - y2) + 8xy2
ay =0v0t
+ u 0v0x
+ v 0v0y
= 0 + (2x2 - y2)(-4y) + (-4xy)(-4x) = -4y(2x2 - y2) + 16x2y
At point (1 m, 1 m),
ax = 4(1)32(12) - 124 + 8(1)(12) = 12 m>s2ay = -4(1)32(12) -
124 + 16(12)(1) = 12 m>s2
The magnitude of the acceleration is
a = 2a 2x + a 2y = 2(12 m/s2)2 + (12 m/s2)2 = 17.0 m>s2
Ans.Using the definition of the slope of the streamline,
dy
dx=
v
u;
dy
dx= -
4xy
2x2 - y2
(2x2 - y2)dy = -4xydx2x2dy + 4xydx - y2dy = 0
However, d(2x2y) = 2(2xydx + x2dy) = 2x2dy + 4xydx. Thend(2x2y)
- y2dy = 0
342. The velocity of a flow field is defined by u = (2x2 - y2)
m>s and v = (-4xy) m>s, where x and y are in meters.
Determine the magnitude of the velocity and acceleration of a
particle that passes through point 11 m, 1 m2 . Find the equation
of the streamline passing through this point, and sketch the
velocity and acceleration at the point on this streamline.
-
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Ans:V = 4.12 m>s a = 17.0 m>s2x2 =
y3 + 56y
Integrating this equation,
2x2y -y3
3= C
with the condition y = 1 m when x = 1 m,
2(12)(1)-13
3= C
C =53
Thus,
2x2y -y3
3=
53
6x2y - y3 = 5
x2 =y3 + 5
6y Ans.
Taking the derivative of this equation with respect to y
2 xdxdy
=6y(3y2) - (y3 + 5)(6)
(6y)2=
2y3 - 56y2
dxdy
=2y3 - 512xy2
Set dxdy
= 0;
2y3 - 5 = 0y = 1.357 m
The corresponding x is
x2 = 1.3573 + 5x = 0.960 m
y(m) 0.25 0.5 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75
3.00x(m) 1.83 1.31 1.10 1.00 0.963 0.965 0.993 1.04 1.10 1.17 1.25
1.33
342. (continued)
0 0.5 1.0x(m)
y(m)
1.5 2.0
0.5
1.0
2.5
2.0
1.5
1.36
3.0
0 0.5 1.0x(m)
y(m)
1.5 2.0
0.5
1.0
2.5
2.0
1.5
3.0
u = 1 m/s
v = 4 m/s V = 4.12 m/s
0.960
ay = 12 m/s2 a = 17.0 m/s2
ax = 12 m/s2
0 0.5 1.0x(m)
y(m)
1.5 2.0
0.5
1.0
2.5
2.0
1.5
1.36
3.0
0 0.5 1.0x(m)
y(m)
1.5 2.0
0.5
1.0
2.5
2.0
1.5
3.0
u = 1 m/s
v = 4 m/s V = 4.12 m/s
0.960
ay = 12 m/s2 a = 17.0 m/s2
ax = 12 m/s2
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SolutionSince the velocity components are a function of position
but not time, the flow can be classified as steady nonuniform. At
point (3 m, 2 m)
u =-y4
= -24
= -0.5 m>sv =
x9
=39
= 0.3333 m>sThe magnitude of the velocity is
V = 2u2 + v2 = 2( -0.5 m>s)2 + (0.333m>s)2 = 0.601 m>s
Ans.For two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
are1S+ 2 ax = 0u0t + u 0u0x + v 0u0y= 0 + a -y
4b(0) + ax
9ba-1
4b
= a- 136
xb m>s2( + c ) ay =
0v0t
+ u 0v0y
+ v 0v0y
= 0 + a -y4
ba19b + ax
9b(0)
= c - 136
y d m>s2At point (3 m, 2 m),
ax = -136
(3) = -0.08333 m>s2ay = -
136
(2) = -0.05556 m>s2The magnitude of the acceleration is
a = 2ax2 + ay2 = 2(-0.08333 m>s2)2 + (-0.05556 m>s2) =
0.100 m>s2 Ans.
343. The velocity of a flow field is defined by u = (-y>4)
m>s and v = (x>9) m>s, where x and y are in meters.
Determine the magnitude of the velocity and acceleration of a
particle that passes through point (3 m, 2 m). Find the equation of
the streamline passing through this point, and sketch the velocity
and acceleration at the point on this streamline.
-
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Ans:V = 0.601 m>s a = 0.100 m>s2x2> 14.2422 + y2>
12.8322 = 1
343. (continued)
Using the definition of slope of the streamline,
dy
dx=
v
u ;
dy
dx=
x>9-y>4 = - 4x9y
9Ly
2 mydy = -4L
x
3 mxdx
9y2
2` y2 m
= - (2x2) ` x3 m
9y2
2- 18 = -2x2 + 18
9y2 + 4x2 = 72
x2
72>4 + y272>9 = 1x2
(4.24)2+
y2
(2.83)2= 1 Ans.
This is an equation of an ellipse with center at (0, 0). The
plot of this streamline is shown in Fig. a
(a)
y(m)
3
2
v = 0.333 m/s
u = 0.5 m/s
V = 0.601 m/s
x(m)
2 m3
2 m2
2 m2
2 m3
y
3
2
ax = 0.0833 m/s2
ay = 0.0556 m/s2
a = 0.100 m/s2
x
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SolutionThe flow is unsteady nonuniform. For one dimensional
flow,
a =0u0t
+ u 0u0x
Here, u = (4 tx) m>s. Then 0u0t
= 4x and 0u0x
= 4t. Thus,
a = 4x + (4tx)(4t) = (4x + 16t2x) m>s2Since u = 0.8 m>s
when t = 0.1 s, 0.8 = 4(0.1) x x = 2 m
The position of the particle can be determined from
dxdt
= u = 4 tx; Lx
2 m
dxx
= 4Lt
0.15t dt
ln x ` x2m
= 2t2 ` t0.15
ln x2
= 2t2 - 0.02
e2t2-0.02 =
x2
x = 2e2t2-0.02
x = 2e2(0.82)-0.02 = 7.051 m
Thus, t = 0.8 s,
a = 4(7.051) + 16(0.82)(7.051)
= 100.40 m>s2 = 100 m>s2 Ans.
*344. The velocity of gasoline, along the centerline of a
tapered pipe, is given by u = (4tx) m>s, where t is in seconds
and x is in meters. Determine the acceleration of a particle when t
= 0.8 s if u = 0.8 m>s when t = 0.1 s.
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SolutionThe flow is unsteady nonuniform. For three dimensional
flow,
a =0V0t
+ u 0V0t
+ v 0V0t
+ w 0V0t
Thus,
ax =0u0t
+ u 0u0x
+ v 0u0y
+ w 0u0t
= 0 + 2x(2) + (6tx)(0) + 3y(0)
= (4x) m>s2 ay =
0v0t
+ u 0v0x
+ v 0v0y
+0v0t
= 6x + 2x(6t) + 6tx(0) + 3y(0) = (6x + 12tx) m>s2 az =
0w0t
+ u 0w0x
+ v 0w0y
+ w 0w0z
= 0 + 2x(0) + 6tx(3) + 3y(0) = (18tx) m>s2The position of the
particle can be determined from
dxdt
= u = 2x; Lx
1 m
dxx
= 2Lt
0dt
ln x = 2t
x = (e2t) m
dy
dt= v = 6tx = 6te2t ; L
y
0dy = 6L
t
0te2tdt
y =32
(2te2t - e2t) ` t0
y =32
(2te2t - e2t + 1)
dzdt
= w = 3y =9232te2t - e2t + 14 ;
Lz
0dz =
92L
t
0(2te2t - e2t + 1) dt
z =92c te2t - 1
2 e2t -
12
e2t + t d ` t0
z =92
(te2t - e2t + t + 1) m
345. The velocity field for a flow of water is defined by u =
(2x) m>s, v = (6tx) m>s, and w = (3y) m>s, where t is in
seconds and x, y, z are in meters. Determine the acceleration and
the position of a particle when t = 0.5 s if this particle is at (1
m, 0, 0) when t = 0.
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Ans:x = 2.72 my = 1.5 mz = 0.634 ma = 510.9i + 32.6j + 24.5k6
m>s2
345. (continued)
When, t = 0.5 s,
x = e2(0.5) = 2.7183 m = 2.72 m Ans.
y =3232(0.5)e2(0.5) - e2(0.5) + 14 = 1.5 m Ans.
Thus, z =9230.5e2(0.5) - e2(0.5) + 0.5 + 14 = 0.6339 m = 0.634 m
Ans.
ax = 4(2.7183) = 10.87 m>s2 ay = 6(2.7183) + 12(0.5)(2.7183)
= 32.62 m>s2 az = 18(0.5)(2.7183) = 24.46 m>s2Then
a = 510.9i + 32.6j + 24.5k6 m>s2 Ans.
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SolutionSince the velocity components are the function of
position but not of time, the flow can be classified as steady but
nonuniform. Using the definition of the slope of the
streamline,
dy
dx=
v
u;
dy
dx=
10y + 3-(4x + 6)
Ly
1 m
dy
10y + 3= -L
x
1 m
dx4x + 6
110
ln(10y + 3) ` y1m
= -14
ln(4x + 6) ` x1m
110
lna10y + 313
b = 14
lna 104x + 6
bln a10y + 3
13b 110 = ln a 10
4x + 6b 14
a10y + 313
b 110 = a 104x + 6
b 1410y + 3
13= a 10
4x + 6b 52
y = c 411(4x + 6)5>2 - 0.3 dm Ans.
For two dimensional flow, the Eulerian description gives
a =dVdt
+ u dVdx
+ v dVdy
Writing the scalar components of this equation along the x and y
axes,
ax =dudt
+ ududx
+ v dudy
= 0 + 3 -(4x + 6)(-4)4 + (10y + 3)(0) = 34(4x + 6)4 m>s2 ay
=
dv
dt+ u
dv
dx+ v
dv
dy
= 0 + 3 -(4x + 6)(0)4 + (10y + 3)(10) = 310(10y + 3)4 m>s2At
point (1m, 1 m),
ax = 434(1) + 64 = 40 m>s2 Say = 10310(1) + 34 = 130
m>s2c
346. A flow field has velocity components of u = -(4x + 6)
m>s and v = (10y + 3) m>s where x and y are in meters.
Determine the equation for the streamline that passes through point
(1 m, 1 m), and find the acceleration of a particle at this
point.
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Ans:
y =41114x + 625>2 - 0.3
a = 136 m>s2u = 72.9 a
346. (continued)
The magnitude of acceleration is
a = 2a 2x + a 2y = 2(40m>s2)2 + (130m>s2)2 = 136 m>s2
Ans.And its direction is
u = tan-1 aayax
b = tan-1 a130 m>s2 40 m>s2 b = 72.9 u Ans.
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Ans:y = 1.25 mm x = 15.6 mma = 0.751 m>s2u = 2.29 a
SolutionSince the velocity components are a function of both
position and time, the flow can be classified as unsteady
nonuniform. Using the defination of velocity,
dy
dt= v = 0.03t2; L
y
0dy = 0.03L
t
0t2 dt
y = (0.01t3) mWhen t = 0.5 s,
y = 0.01(0.53) = 0.00125 m = 1.25 mm Ans.
dxdt
= u = 100y = 100(0.01t3) = t3; Ly
0dx = L
t
0t3 dt
x = a14
t4b mWhen t = 0.5 s,
x =14
(0.54) = 0.015625 m = 15.6 mm Ans.
For a two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Write the scalar components of this equation along the x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + (100y)(0) + (0.03t2)(100) = (3t2) m>s2 ay =
0v0t
+ u 0v0x
+ v 0v0y
= 0.06t + (100y)(0) + 0.03t2(0) = (0.06t) m>s2When t = 0.5
s,
ax = 3(0.52) = 0.75 m>s2 S ay = 0.06(0.5) = 0.03 m>s2cThe
magnitude of acceleration is
a = 2a 2x + a 2y = 2(0.75 m>s2)2 + (0.03 m>s2)2 = 0.751
m>s2 Ans.And its direction is
u = tan-1 aayax
b = tan-1a0.03 m>s20.75 m>s2 b = 2.29 u Ans.
347. A velocity field for oil is defined by u = (100y) m>s, v
= (0.03 t2) m>s, where t is in seconds and y is in meters.
Determine the acceleration and the position of a particle when t =
0.5 s. The particle is at the origin when t = 0.
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SolutionSince the velocity components are a function of position
but not time, the flow can be classified as steady but nonuniform.
Using the definition of the slope of the streamline,
dy
dx=
v
u;
dy
dx=
-y2x2
Ly
6 m
dy
y= -
12L
x
2 m
dx
x2
ln y ` y6 m
=12a1
xb ` x
2 m
ln y
6=
12
a1x-
12b
ln y
6=
2 - x4x
y
6= e12- x4x 2
y = c 6e12- x4x 2 d m Ans.The plot of this streamline is shown
in Fig. a.
For two dimensional flow, the Eulerian description gives.
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
= 0 + (2x2)(4x) + (-y)(0)
= (8x3) m>s2 ay =
dv
dt+ u
dv
dx+ v
dv
dy
= 0 + (2x2)(0) + (-y)(-1)
= (y)m>s2At point (2 m, 6 m),
ax = 8(23) = 64 m>s2 S ay = 6m>s2cThe magnitude of
acceleration is
a = 2a 2x + a 2y = 2(64 m>s2)2 + (6m>s2)2 = 64.3m>s2
Ans.And its direction is
u = tan-1aayax
b = tan-1a 6 m>s264 m>s2 b = 5.36 u Ans.
*348. If u = (2x2) m>s and v = (-y) m>s where x and y are
in meters, determine the equation of the streamline that passes
through point (2 m, 6m), and find the acceleration of a particle at
this point. Sketch the streamline for x 7 0, and find the equations
that define the x and y components of acceleration of the particle
as a function of time if x = 2 m and y = 6 m when t = 0.
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348. (continued)
Using the definition of the velocity,
dxdt
= u; dxdt
= 2x2
Lx
2 m
dx
2x2= L
t
0dt
-12
a1xb ` x
2 m= t
-12
a1x-
12b = t
x - 24x
= t
x = a 21 - 4t
b mdy
dt= v;
dy
dt= -y
- Ly
6 m
dy
y= L
t
0dt
- ln y ` y6 m
= t
ln 6y
= t
6y
= et
y = (6e-t) mThus,
u = 2x2 = 2 a 21 - 4t
b2 = c 8(1 - 4t)2
d m>s and v = -y = ( -6e-t) m>sThen,
ax =dudt
= -16(1 - 4t)-3(-4) = c 64(1 - 4t)3
d m>s2 Ans.ay =
dvdt
= (6e-t) m>s2 Ans.x(m) 0.5 1 2 3 4 5 6
y(m) 12.70 7.70 6.00 5.52 5.29 5.16 5.08
0 1 2 3 4
(a)
x(m)
y(m)
5 6
1
2
3
4
5
6
7
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349. Air flow through the duct is defined by the velocity field
u = (2x2 + 8) m>s v = (-8x) m>s where x is in meters.
Determine the acceleration of a fluid particle at the origin (0, 0)
and at point (1 m, 0). Also, sketch the streamlines that pass
through these points.
SolutionSince the velocity component are a function of position
but not time, the flow can be classified as steady but nonuniform.
For two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Writing the scalar components of this equation along the x and y
axes
ax =0u0t
+ u 0u0x
+ v 0u0y
= 30 + (2x2 + 8)(4x) + (-8x)(0)4 = 34x(2x2 + 8) 4 m>s2
ay =0v0t
+ u 0v0x
+ v 0v0y
= 0 + (2x2 + 8)(-8) + (-8x)(0)
= 3 -8(2x2 + 8) 4 m>s2At point (0, 0),
ax = 4(0)32(02) + 84 = 0ay = -832(02) + 84 = -64 m>s2 = 64
m>s2T
Thus,
a = ay = 64 m>s2T Ans.At point (1 m, 0),
ax = 4(1)32(12) + 84 = 40 m>s2 S ay = -832(12) + 84 = -80
m>s2 = 80 m>s2TThe magnitude of the acceleration is
a = 2a 2x + a 2y = 2(40 m>s2)2 + (80 m>s2)2 = 89.4 m>s2
Ans.And its direction is
u = tan-1 aayax
b = tan-1 a80 m>s240 m>s2 b = 63.4 cu Ans.
Using the definition of the slope of the streamline,
dy
dx=
v
u=
-8x2x2 + 8
; Ldy = -8Lx dx
2x2 + 8
y = -2 ln (2x2 + 8) + C
x 1 m
y
x
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Ans:At point (0, 0),a = 64 m>s2wAt point (1 m, 0),a = 89.4
m>s2, u = 63.4 cFor the streamline passing through point(0,
0),
y = c 2 ln a 82x2 + 8
b d mFor the streamline passing through point(1 m, 0),
y = c 2 ln a 102x2 + 8
b d m
349. (continued)
For the streamline passing through point (0, 0),
0 = -2 ln 32(02) + 84 + C C = 2 ln 8Then y = c 2 ln a 8
2x2 + 8b d m Ans.
For the streamline passing through point (1 m, 0),
0 = -2 ln 32(12) + 84 + C C = 2 ln 10y = c 2 ln a 10
2x2 + 8b d m Ans.
For point (0, 0)
x(m) 0 1 2 3 4 5
y(m) 0 -0.446 -1.39 -2.36 -3.22 -3.96
For point (1 m, 0)
x(m) 0 1 2 3 4 5
y(m) 0.446 0 -0.940 -1.91 -2.77 -3.52
0
1
2 3 4x(m)
y(m)
55 4 3 2
1
1
1
2
3
4
y = 10
2x2 + 8
2 ln
y = 82x2 + 82 ln
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SolutionSince the velocity components are a function of position
but not time, the flow can be classified as steady but nonuniform.
For two dimensional flow, the Eulerian description gives
a =0V0t
+ u 0V0x
+ v 0V0y
Write the scalar components of this equation along x and y
axes,
ax =0u0t
+ u 0u0x
+ v 0u0y
350. The velocity field for a fluid is defined by u = y> (x2
+ y2) and v = 34x> (x2 + y2) 4 m>s, where x and y are in
meters. Determine the acceleration of particles located at point (2
m, 0) and that of a particle located at point (4 m, 0). Sketch the
equations that define the streamlines that pass through these
points.
= 0 + a yx2 + y2
b (x2 + y2)(0) - y(2x)(x2 + y2)2
+ a 4xx2 + y2
b (x2 + y2)(1) - y(2y)(x2 + y2)2
= 4x3 - 6xy2
(x2 + y2)3m>s2
ay =0v0t
+ u0v0x
+ v0v0y
= 0 + a yx2 + y2
b (x2 + y2)(4) - 4x(2x)(x2 + y2)2
+ a 4xx2 + y2
b (x2 + y2)(0) - 4x(2y)(x2 + y2)2
= 4y3 - 36x2y
(x2 + y2)3m>s2
At point (2 m, 0)
ax =4(23) - 6(2)(02)
(22 + 02)3= 0.5 m>s2 S
ay =4(03) - 36(22)(0)
(22 + 02)3= 0
Thus a = ax = 0.5 m>s2 S Ans.
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350. (continued)
At point (4 m, 0)
ax =4(43) - 6(4)(0)
(42 + 02)3= 0.0625 m>s2 S
ay =4(03) - 36(42)(0)
(42 + 02)3= 0
Thus
a = ax = 0.0625 m>s2 S Ans.Using the definition of the slope
of the streamline,
dy
dx=
v
u=
4x> (x2 + y2)y> (x2 + y2) = 4xy ; Lydy = 4Lxdx
y2
2= 2x2 + C
y2 = 4x2 + C
For the streamline passing through point (2 m, 0),
02 = 4(22) + C C = -16
Then y2 = 4x2 - 16
y = {24x2 - 16 x 2m Ans.For the streamline passes through point
(4 m, 0)
02 = 4(42) + C C = -64
Then
y2 = 4x2 - 64
y = {24x2 - 64 x 4m
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Ans:For point (2 m, 0), a = 0.5 m>s2y = {24x2 - 16For point
(4 m, 0), a = 0.0625 m>s2y = {24x2 - 64
For the streamline passing through point (2 m, 0)
x(m) 2 3 4 5 6 7 8 9y(m) 0 4.47 6.93 9.17 11.31 13.42 15.49
17.55
For the streamline passing through point (4 m, 0)
x(m) 4 5 6 7 8 9y(m) 0 6.00 8.94 11.49 13.86 16.12
350. (continued)
y(m)
0 1 2 3 4 5 6 7 8x(m)
9
5
5
10
10
15
20
15
20
y = 4x2 16
y = 4x2 64
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Ans:-1.23 m>s2
SolutionHere V only has an x component, so that V = u. Since V
is a function of time at eachx, the flow is unsteady. Since v = w =
0, we have
351. As the value is closed, oil flows through the nozzle such
that along the center streamline it has a velocity of V = 6(1 +
0.4x2)(1 - 0.5t) m>s where x is in meters and tis on seconds.
Determine the accel eration of an oil particle at x = 0.25 m when t
= 1 s.
ax =0u0t
+ u0u0x
=00x
36(1 + 0.4x2)(1 - 0.5t)4 + 36(1 + 0.4x2)(1 - 0.5t)4 00x
36(1 + 0.4x2)(1 - 0.5t)4 = 36(1 + 0.4x2)(0 - 0.5)4 + 36(1 +
0.4x2)(1 - 0.5t) 4 36(0 + 0.4(2x))(1 - 0.5t)4Evaluating this
expression at x = 0.25 m, t = 1 s, we get
as = -3.075 m>s2 + 1.845 m>s2 = -1.23 m>s2 Ans.
Note that the local acceleration component ( -3.075 m>s2)
indicates a deceleration since the valve is being closed to
decrease the flow. The convective acceleration (1.845 m>s2) is
positive since the nozzle constricts as x increases. The net result
causes the particle to decelerate at 1.23 m>s2.
x
0.3 m
AB
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SolutionThe n - s coordinate system is established with origin
at point A as shown in Fig. a. Here, the component of the particles
acceleration along the s axis is
as = 3 m>s2Since the streamline does not rotate, the local
acceleration along the n axis is zero,
so that a0V 0t
bn
= 0. Therefore, the component of the particles acceleration
along
the n axes is
an = a0V 0t bn + V2R = 0 +
(5 m>s)216 m
= 1.5625 m>s2Thus, the magnitude of the particles
acceleration is
a = 2a 2s + a 2n = 2(3 m>s2 )2 + (1.5625 m>s2)2 = 3.38
m>s2 Ans.
*352. As water flows steadily over the spillway, one of its
particles follows a streamline that has a radius of curvature of 16
m. If its speed at point A is 5 m>s which is increasing at 3
m>s2, determine the magnitude of acceleration of the
particle.
(a)
streamline
an
as
n
A
S
16 mA
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Ans:72 m>s2
353. Water flows into the drainpipe such that it only has a
radial velocity component V = (-3>r) m>s, where r is in
meters. Determine the acceleration of a particle located at point r
= 0.5 m, u = 20. At s = 0, r = 1 m.
SolutionFig. a is based on the initial condition when s = 0, r =
rD. Thus, r = 1 - s. Then the radial component of velocity is
V = -3r
= a- 31 - s
b m>sThis is one dimensional steady flow since the velocity
is along the straight radial line. The Eulerian description
gives
a =0V0t
+ V 0V0s
= 0 + a- 31 - s
b c - 3(1 - s)2
d = c 9
(1 - s)3d m>s2
When 1 - s = r = 0.5 m, this equation gives
a = a 90.53
b m>s2 = 72 m>s2 Ans.The positive sign indicates that a is
directed towards positive s.Note there is no normal component for
motion along a straightline.
(a)
r
s
r0 = 1 m
r 0.5 m
s
u
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Ans:as = 3.20 m>s2 an = 7.60 m>s2
354. A par