Resolução capítulo 3 hibeller

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267
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Solution
Since 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.
3–1. 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)
marked
particle
streamline
streakline
60˚
 4 m
(b)
t = 4 s
marked
particle
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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Solution
Since 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.
3–2. 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 s
t = 0 2 m
streamline
x
first marked
particle
pathline
t = 4 s
t = 4 s
t = 3 s
t = 0
pathline
streamline
streakline first marked
particle
y
x
6 m
4 m
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Solution
Since 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 
in Fig. 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.
3–3. 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
streakline
first marked particle
y
t = 4 s t = 6 s
t = 0
pathline
streamline
streakline
first marked
particle
4 m
12 m
y
x
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ea
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Solution
The 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>s
The magnitude of V is
 V = 2V 2
x + V 2
y = 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 a
Vy
Vx
b = tan-1 a9 m> s
5 m> s
b = 60.95°
Thus,
 u = 360° - 60.95° = 299° Ans.
*3–4. A two-dimensional flow field for a fluid can be 
described by V = 5(2x + 1)i - (y + 3x)j6 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)
x
Vx = 5 m/s
�
�
VVy = 9 m/s
3 m
2 m
x
y
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Solution
The 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 2
x + V 2
y = 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 a
Vy
Vx
b = tan- 1 a13 m>s
15 m>s
b = 40.9° Ans.
3–5. A two-dimensional flow field for a liquid can be 
described by V = 5 (5y2 - x)i + (3x + y)j6 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>s
u = 40.9°
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Solution
The 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 2
x + V 2
y = 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 a
Vy
Vx
b = tan-1 a1.5 m>s
1.6 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.
 L
y
3 m
dy
t2 = L
x
2 m
0.075dx
x
 
1
t2 (y - 3) = 0.075 ln 
x
2
 y = a0.075t2 ln 
x
2
+ 3b m
When t = 5 s,
 y = 0.075(52) ln ax
2
b + 3
 y = c 1.875 ln ax
2
b + 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
3–6. The soap bubble is released in the air and rises with a 
velocity of V = 5(0.8x)i + (0.06t2)j6 m>s, where x is 
meters and t is in seconds. Determine the magnitude of the 
bubble’s velocity, and its direction 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>s
u = 43.2°
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Solution
As 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
 L
2 + y
2y
 dy = Ldx
 ln y +
1
2
 y = x + C
At point (3 m, 2 m), we obtain
 ln(2) +
1
2
 (2) = 3 + C
C = -1.31
Thus,
 ln y +
1
2
 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 av
u
b = tan-1 a4 m>s
4 m>s
b = 45° Ans.
3–7. A flow field for a fluid is 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 
(3 m, 2 m), 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 - 2.61
V = 5.66 m>s
u = 45° a
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ud
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Solution
As 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
=
-6xy
x2 + 5
 L
dy
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 av
u
b = tan-1 a30 m>s
30 m>s
b = 45° Ans.
*3–8. A 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, 1 m), 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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ud
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Solution
As 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
 
1
3
 y3 = 2x + C
At x = 1 m, y = 2 m. Then
 
1
3
 (2)3 = 2(1) + C
C = 
2
3
Thus,
 
1
3
 y3 = 2x +
2
3
 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 av
u
b = tan-1 a4
8
b = 26.6° Ans.
3–9. Particles travel within a flow field defined by 
V = 52y2i + 4j6 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)
x
x
y
y
�
streamlineVv = 4 m/s
u = 2y2 m/s
Ans:
y3 = 6x + 2
V = 8.94 m>s
u = 26.6° a
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uc
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ud
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Solution
As 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 + 0.6y
0.5
= 1.6 + 1.2y
Since the balloon starts at y = 0, x = 0, using these values,
 L
y
0
dy
1.6 + 1.2y
= L
x
0
dx
 
1
1.2
 ln(1.6 + 1.2y) `
y
0
= x
 lna1.6 + 1.2y
1.6
b = 1.2x
 lna1 +
3
4
 yb = 1.2x
 1 +
3
4
 y = e1.2xy =
4
3
 (e1.2x - 1) m Ans.
Using this result, the streamline is shown in Fig. b.
3–10. A balloon 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)
x
x
y
y
�streamline
V
y(m)
x(m)
y = (e1.2x – 1) m
streamline
4
3
v = (0.8 + 0.6y) m/s
u = 0.5 m/s
y
x
u
v
Ans:
y =
4
3
 (e1.2x - 1)
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rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
277
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
As 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 + 0.4y
0.8x
The balloon starts at point (1 m, 0).
L
y
0
dy
1.6 + 0.4y
= L
x
1
dx
0.8x
1
0.4
 ln(1.6 + 0.4y) `
y
0
=
1
0.8
 ln x `
x
1
1
0.4
 lna1.6 + 0.4y
1.6
b =
1
0.8
 ln x
lna1 +
1
4
yb
2
= ln x
a1 +
1
4
yb
2
= x
 y = 4(x1>2 - 1) m Ans.
Using this result, the streamline is shown in Fig. b.
3–11. A balloon 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, where x is in meters. Also, buoyancy and 
thermal winds cause the balloon to rise at a rate of 
v = (1.6 + 0.4y) m>s, where y is in meters. Determine the 
equation of the streamline for the balloon, and draw this 
streamline. 
(a)
(b)
streamline
1 m
y
x
y = 4(x – 1) m½
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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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
ot 
pe
rm
itte
d. 
278
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Solution
As 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
=
6x
8y
L8y dy = L6x dx
4y2 = 3x2 + C
At x = 1 m, y = 2 m. Then
4(2)2 = 3(1)2 + C
 C = 13
Thus
 4y2 = 3x2 + 13 Ans.
*3–12. A flow field is defined by u = (8y) m>s and
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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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
279
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
As 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
L
dy
2y
= L
dx
x
 
1
2
 ln y = ln x + C
At x = 3 ft, y = 1 ft . Then
1
2
 ln y = ln x - ln3
1
2
 ln y = ln 
x
3
ln y = lnax
3
b
2
 y =
x2
9
 Ans.
3–13. 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
x
y
y
�
streamline
Vv = (6y) ft/s
u = (3x) ft/s
Ans:
y = x2>9
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
280
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Since the velocity V is constant, Fig. a, the streamline will be a straight line with 
a slope.
dy
dx
= tan u
 
dy
dx
=
v
u
=
8
5
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.
3–14. 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 streamline and pathline 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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s 
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
281
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
As 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)
=
y
x
L
dy
y
= L
dx
x
ln 
y
x = C
y
x
= C′
At x = 1 m, y = 1 m. Then
C′ = 1
Thus,
y
x
= 1
 y = x Ans.
3–15. 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)
x
x
y
y
streamline
V
v = m/s
8y
x2 + y2
u = m/s
8y
x2 + y2
))
))
�
Ans:
y = x
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s 
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s p
rovide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
282
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Since 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
dx
dt
= u =
30
2x + 1
 ; L
x
2 m
(2x + 1)dx = 30L
t
2 s
dt
 (x2 + x) `
x
2 m
= 30 t `
t
2 s
 x2 + x - 6 = 30(t - 2)
 t =
1
30
 (x2 + x + 54) (1)
 
dy
dt
= v = 2ty; L
y
6 m
dy
y
= 2L
t
2 s
tdt
ln y `
y
6 m
= t2 `
t
2 s
ln 
y
6
= t2 - 4
y
6
= et2 - 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
*3–16. 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 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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law
s 
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
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e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
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itte
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
For t = 1 s, y = 4e(x2 + x - 2)>15
For t = 2 s, y = 4e2(x2 + x - 2)>15
For t = 3 s, y = 4e(x2 + x - 2)>5
Solution
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)
=
1
15
ty(2x + 1)
L
y
4 m
dy
y
=
1
15
 tL
x
1 m
(2x + 1)dx
ln y `
y
4 m
=
1
15
 t(x2 + x) `
x
1 m
ln 
y
4
=
1
15
 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 s
x(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
3–17. 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 s
t = 2 s
t = 1 s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
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e i
nte
gri
ty 
of 
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ork
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itte
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
For t = 2 s, y = 6e21x2 + x - 6)>15
For t = 5 s, y = 6e1x2 + x - 6)>3
Solution
For t = 2 s
x(m) 0 1 2 3 4
y(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)
=
1
15
 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.
L
y
6 m
dy
y
=
1
15
 tL
x
2 m
(2x + 1)dx
ln y `
y
6 m
=
1
15
 t (x2 + x) `
x
2 m
 ln 
y
6
=
1
15
 t(x2 + x - 6)
y = 6e
1
15 t(x2 + x - 6)
For t = 2 s,
 y = 6e
2
15 (x2 + x - 6) Ans.
For t = 5 s,
 y = 6e
1
3 (x2 + x - 6) Ans.
The plots of these two streamlines are show in Fig. a.
3–18. 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 4
x(m)
y(m)
10
20
30
40
50
t = 5 s
t = 2 s
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
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eb
) 
will 
de
str
oy
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e i
nte
gri
ty 
of 
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ork
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Ans:
u = 3.43 m>s
v = 3.63 m>s
Solution
x(m) 0 1 1.5 2 3 4 5
y(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.
3–19. A particle travels along a streamline defined by 
y3 = 8x - 12. If its speed is 5 m>s when it is at x = 1 m, 
determine the two components of its velocity at this point. 
Sketch the velocity on the streamline.
(a)
y(m)
0 1 2 3 4
x(m)
5
1
–1
–2
–3
2
3
3
3
46.62°
u
v
5 m/s
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tat
es
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
286
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Solution
Here, u =
dx
dt
. Then,
dx = udt
Using x = 0 when t = 0 as the integration limit,
L
x
0
dx = L
t
0
3(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,
L
y
0
dy = 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 y
1 0.4 0.4
2 1.6 0.8
3 3.6 1.2
4 6.4 1.6
5 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
=
0.4
0.8t
tL
y
1.6 m
dy =
1
2 L
x
6.4 m
dx
t(y - 1.6) =
1
2
 (x - 6.4)
y = c 1
2t
 (x - 6.4) + 1.6 d m
*3–20. 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 8
x(m)
10
1
–1
–2
2 y2 = 0.4x
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
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n t
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 W
orl
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ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
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ot 
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rm
itte
d. 
287
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*3–20. (continued)
When t = 4 s,
y =
1
2(4)
 ( x - 6.4) + 1.6
 y =
1
8
 x + 0.8
The plot of the streamline is shown is Fig. b.
(b)
y(m)
0.8
x(m)
y = x + 0.81
8
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
288
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
y3 = 8x - 15, y = 9 m
x = 93 m
Solution
Since 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
L
y
1 m
3y2dy = 8L
x
2 m
dx
y3 `
y
1 m
= 8x `
x
2 m
y3 - 1 = 8x - 16
y3 = 8x - 15 (1) Ans. 
From the definition of velocity
dy
dt
= 8
L
y
1 m
dy = L
1 s
0
8 dt
y `
y
1 m
= 8t `
1 s
0
y - 1 = 8
y = 9 m Ans.
Substituting this result into Eq. (1)
93 = 8x - 15
 x = 93 m Ans.
3–21. The velocity for an oil flow is defined by 
V = 53y2 i + 8 j6 m>s, where y is in meters. What is the 
equation of the streamline that passes through point 
(2 m, 1 m)? If a particle is at this point when t = 0, at what 
point is it located when t = 1 s?
This
 w
ork
 is
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ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
289
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Ans:
y =
2
3
 ln a 2
2 - 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
=
2
6 - 3x
L
y
0
dy = 2L
x
0
dx
6 - 3x
y = -
2
3
 ln (6 - 3x) `
x
0
y = -
2
3
 ln a6 - 3x
6
b
y =
2
3
 ln a 2
2 - x
b Ans.
The plot of this streamline is show in Fig. a
3–22. 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 This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
290
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
For 0 … t 6 10 s, y = -
3
2
 x
For 10 s 6 t … 15 s, y = -
2
5
x + 22
Solution
Using the definition of velocity, for 0 … t 6 10 s
dx
dt
= u; 
dx
dt
= -2
L
x
0
dx = -2L
t
0
dt
x = (-2t) m (1)
When t = 10 s, x = -2(10) = -20 m
dy
dt
= v; 
dy
dt
= 3
L
y
0
dy = 3L
t
0
dt
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 = -
3
2
 x Ans.
For 10 6 t … 15 s.
dx
dt
= u; 
dx
dt
= 5
L
x
-20 m
dx = 5L
t
10 s
dt
x - (-20) = 5(t - 10)
x = (5t - 70) m (3)
At t = 15 s, x = 5(15) - 70 = 5 m
dy
dt
= v; 
dy
dt
= -2
L
y
30 m
dy = -2L
t
10 s
dt
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-
2
5
x + 22b Ans.
The two streamlines intersect at (-20, 30), point B in Fig. (a). The pathline is the 
path ABC.
3–23. 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 5–5
A
B
C
–10–15–20
x(m)
y(m)
10
20
30
This
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ork
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cte
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ed
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tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
291
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
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 componentsare 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
=
2t
4x
=
t
2x
L
y
6 m
dy =
t
2 L
x
2 m
dx
x
y - 6 =
t
2
 ln 
x
2
y =
t
2
 ln 
x
2
+ 6
For t = 1 s, y = a1
2
 ln 
x
2
+ 6bm Ans.
The plot of this streamline is shown in Fig. a.
*3–24. A 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
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
292
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
y =
1
16
 ln2 
x
2
+
1
2
 ln 
x
2
+ 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,
dx
dt
= u = 4x; L
x
2 m
dx
4x
= L
t
1 S
dt
1
4
 ln x `
x
2 m
= t `
t
1 s
1
4
 ln 
x
2
= t - 1
t =
1
4
 ln 
x
2
+ 1 (1)
dy
dt
= v = 2t; L
y
6 m
dy = L
t
1 s
2t dt
y - 6 = t2 `
t
1 s
y = t2 + 5 (2)
Substitute Eq. (1) into (2),
y = a1
4
 ln 
x
2
+ 1b
2
+ 5
y = a 1
16
 ln2 
x
2
+
1
2
 ln 
x
2
+ 6b Ans.
The plot of this pathline is shown in Fig. (a)
3–25. 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
This
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ork
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ote
cte
d b
y U
nit
ed
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tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
293
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Using the definition of velocity, for 0 … t 6 5 s,
dx
dt
= u; 
dx
dt
=
1
2
 x
L
x
1 m
dx
x
= L
t
0
1
2
 dt
 ln x =
1
2
 t
x = ae
1
2 tb m (1)
When t = 5 s, x = e
1
2 
(5) = 12.18 m
dy
dt
= v; 
dy
dt
=
1
8
y2
L
y
1 m
dy
y2 = L
t
0
1
8
 dt
- a1
y
b `
y
1 m
=
1
8
t
1 -
1
y
=
1
8
t
y - 1
y
=
1
8
t
y a1 -
1
8
tb = 1
y = a 8
8 - t
b 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 
t from Eqs. (1) and (2)
y = a 8
8 - 2 ln x
bm
x(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,
dx
dt
= u; 
dx
dt
= -
1
4
x2
L
x
12.18 m
dx
x2 = -
1
4 L
t
5 s
dt
3–26. The velocity field of a fluid is defined by u = (1
2 x) m>s, 
v = (1
8 y2) m>s for 0 … t 6 5 s and by u = ( -1
4 x2) m>s, 
v = (1
4 y) m>s for 5 s 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.
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
294
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
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)
- a1
x
-
1
12.18
b = -
1
4
(t - 5)
1
x
=
t
4
- 1.1679
x = a 4
t - 4.6717
b m t ≠ 4.6717 s (3)
When t = 10 s, x =
4
10 - 4.6717
= 0.751 m
dy
dt
= v; 
dy
dt
=
1
4
y
L
y
2.667 m
dy
y
=
1
4 L
t
5 s
dt
 ln 
y
2.667
=
1
4
(t - 5)
y
2.667
= e
1
4 (t- 5)
y = c 2.667e
1
4 
(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(1
x + 1.1679) - 54
= c 2.667e(1
x - 0.08208) dm
x(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.
3–26. (continued)
0 1
0.751 12.18
2 3 134 5 6 7 8 9 10 11 12
A
C
B
x(m)
y(m)
1
2
3
4
5
6
10
7
8
9
This
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ork
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ote
cte
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ed
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tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
295
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
2y3 - 1.5x2 - y - 2x + 52 = 0
V = 30.5 m>s
Solution
We have steady flow since the velocity does not depend upon time.
u = 6y2 - 1
v = 3x + 2
dy
dx
=
v
u
=
3x + 2
6y2 - 1
L
y
2
(6y2 - 1)dy = L
x
6
(3x + 2)dx
2y3 - y `
y
2
= 1.5 x2 + 2x `
x
6
2y3 - y - 32(2)3 - 24 = 1.5x2 + 2x - 31.5(6)2 + 2(6)4
2y3 - 1.5x2 - y - 2x + 52 = 0 Ans.
At (6 m, 2 m)
u = 6(2)2 - 1 = 23 m>s S
v = 3(6) + 2 = 20 m>sc
V = 2(23 m>s)2 + (20 m>s)2 = 30.5 m>s Ans.
3–27. A two-dimensional flow field for a liquid can be 
described by V = 5(6y2 - 1) i + (3x + 2) j6 m>s, where x 
and y are in meters. Find the streamline that passes through 
point 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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str
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Solution
We have steady flow since the velocity does not depend upon time.
u = 2x + 1
v = -y
dy
dx
=
v
u
=
-y
2x + 1
- L
dy
y
= L
dx
(2x + 1)
- ln y =
1
2
 ln (2x + 1)
-y `
y
1
= (2x + 1)
1
2 `
x
3
-y + 1 = (2x + 1)
1
2 - 32(3) + 1 4 1
2
y = 3.65 - (2x + 1)
1
2 Ans.
u = 2(3) + 1 = 7 m>s
v = -1 m>s
V = 2(7 m>s)2 + ( -1 m>s)2 = 7.07 m>s Ans.
*3–28. 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/s
1 m s
3
1
x
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ly 
for
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e u
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 in
str
uc
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s i
n t
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ing
 
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s a
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ss
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) 
will 
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nte
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of 
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Ans:
a = 24 m>s2
Solution
Since 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 
0V
0x
= 0.
a =
0V
0 t
+ u 
0V
0 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.
3–29. 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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 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
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s a
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 as
se
ss
ing
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ud
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ing
. D
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n o
r 
sa
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y p
art
 of
 th
is 
work
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 W
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) 
will 
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Ans:
1088 in.>s2
Solution
Since the flow is along the x axis, v = w = 0
a =
0u
0 t
+ u 
0u
0 x
= 4x + (4xt)(4t)
= 4x + 16xt2
= 34x(1 + 4t2) 4 in.>s2
When t = 2 s, x = 16 in. Then 
a = 34(16)31 + 4(22) 4 4 in.>s2 = 1088 in.>s2
Note: The flow is unsteady since its velocity is a function of time.
3–30. 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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e u
se
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ea
ch
ing
 
the
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s a
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Ans:
30.9 ft>s2
Solution
For two dimensional flow, the Eulerian description gives
a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar component of this equation along the x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 1 + (6y + t)(0) + (2tx)(6)
 = (1 + 12tx) ft>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 2x + (6y + t)(2t) + (2tx)(0)
 = (2x + 12ty + 2t2) ft>s2
When 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>s2
Thus, the magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(13 ft>s2)2 + (28 ft>s2)2 = 30.9 ft>s2 Ans.
3–31. 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 the acceleration of a 
particle passing through the point (1 ft, 2 ft) when t = 1 s.
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e u
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str
uc
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s i
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ing
 
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s a
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n o
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art
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work
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oy
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nte
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*3–32. 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 =
0u
0t
+ u 
0u
0x
0u
0t
= 200 
0u
0x
= 20 x
a = 3200 + (10x2 + 200t + 6)(20x)4 m>s2
When 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.
x
A
0.6 m
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e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
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rse
s a
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ss
ing
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ing
. D
iss
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n o
r 
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of 
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art
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is 
work
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oy
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Ans:
V = 23.3 m>s
a = 343 m>s2
Solution
Velocity.
At x = 2 m, y = 4 m,
 u = 2(22) - 2(42) + 4 = -20 m>s
 v = 4 + 2(4) = 12 m>s
The magnitude of the particle’s 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 particle’s acceleration, with w = 0 are
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
= 0 + (2x2 - 2y2 + y)(4x) + (y + xy)(-4y + 1)
At x = 2 m, y = 4 m,
 ax = -340 m>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
= 0 + (2x2 - 2y2 + y)(y) + (y + xy)(1 + x)
At x = 2 m, y = 4 m,
ay = -44 m>s2
The magnitude of the particle’s acceleration is
 a = 2a 2
x + a 2
y = 2( -340 m>s2)2 + ( -44 m>s2)2 = 343 m>s2 Ans.
3–33. 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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d s
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for
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e u
se
 of
 instr
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
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rse
s a
nd
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se
ss
ing
 st
ud
en
t le
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ing
. D
iss
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ina
tio
n o
r 
sa
le 
of 
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art
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 th
is 
work
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din
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 W
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) 
will 
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Ans:
V = 16.3 m>s 
uv = 79.4° a
a = 164 m>s2
ua = 17.0° a
Solution
Since 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>s
v = 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-1av
u
b = tan-1a16 m>s
3 m>s
b = 79.4° Ans.
For two dimensional flow, the Eulerian description is
a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axis
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + (5y2 - x)(-1) + 4x2(10y)
 = (x - 5y2) + 40x2y
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 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>s2
ay = 8(2)35(12) - 24 = 48 m>s2
The magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(157 m>s2)2 + (48 m>s2)2 = 164 m>s2 Ans.
Its direction is
 ua = tan-1a
ay
ax
b = tan-1° 48 m>s2
157 m>s2¢ = 17.0° Ans.
3–34. 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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uc
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ch
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ing
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n o
r 
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is 
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Ans:
a = 36.1 m>s2
u = 33.7° a
Solution
Since 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 - 1
5y2
L
y
1 m
5y2dy = L
x
1 m
(4x - 1)dx
y3 =
1
5
(6x2 - 3x + 2) where x is in m
For two dimensional flow, the Eulerian description is
a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + 5y2(0) + (4x - 1)(10y)
 
= 40xy - 10y
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + 5y2(4) + (4x - 1)(0)
 = 20y2
At point x = 1 m and y = 1 m,
ax = 40(1)(1) - 10(1) = 30 m>s2
ay = 20(12) = 20 m>s2
The magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(30 m>s2)2 + (20 m>s2)2 = 36.1 m>s2 Ans.
Its direction is
 u = tan-1a
ay
ax
b = tan-1°20 m>s2
30 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 5
y(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
3–35. 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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se
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 in
str
uc
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n t
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ch
ing
 
the
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ss
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 st
ud
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ing
. D
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Solution
Since 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 - VA
LAB
bx = 8 + a2 - 8
3
bx = (8 - 2x) m>s Ans.
For one dimensional flow, the Eulerian description gives
a =
0V
dt
+ V 
0V
dx
 = 0 + (8 - 2x)(-2)
 = 4(x - 4) m/s2 Ans.
using the definition of velocity,
dx
dt
= V = 8 - 2x; L
x
0
dx
8 - 2x
= L
t
0
dt
-
1
2
 ln(8 - 2x) `
x
0
= t
1
2
 ln a 8
8 - 2x
b = t
ln a 8
8 - 2x
b = 2t
8
8 - 2x
= e2 t
x = 4(1 - e-2 t) m Ans.
*3–36. 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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for
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e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
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s a
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se
ss
ing
 st
ud
en
t le
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ing
. D
iss
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tio
n o
r 
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le 
of 
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art
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is 
work
 (in
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 W
orl
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) 
will 
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oy
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nte
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ty 
of 
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Ans:
V = 33.5 m>s
uV = 17.4°
a = 169 m>s2
ua = 79.1° a
Solution
Since 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>s
v = 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-1av
u
b = tan-1a10 m>s
32 m>s
b = 17.4° uv Ans.
For two dimensional flow, the Eulerian description gives
a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 16t + 8t2(0) + (7y + 3x)(0)
 = (16t) m>s2
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + (8t2)(3) + (7y + 3x)(7)
 = 324t2 + 7(7y + 3x)4 m>s2
When t = 2 s, x = 1 m and y = 1 m.
ax = 16(2) = 32 m>s2
ay = 24(22) + 737(1) + 3(1)4 = 166 m>s2
The magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(32 m>s2)2 + (166 m>s2)2 = 169 m>s2 Ans.
Its direction is
 ua = tan-1a
ay
ax
b = tan-1a166 m>s2
32 m>s2 b = 79.1° ua Ans.
3–37. 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 (1 m, 1 m) when t = 2 s.
This
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n t
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the
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s a
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Ans:
y = x>2, a = 143 ft>s2
u = 26.6° a
Solution
Since 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
L
y
1 ft
dy
y
= L
x
2 ft
dx
x
ln y `
y
1 ft
= ln x `
x
2 ft
ln y = ln 
x
2
 y =
1
2
x Ans.
For two dimensional flow, the Eulerian description gives
a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + 8x(8) + 8y(0)
 = (64x) ft>s2
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + (8x)(0) + 8y(8)
 = (64y) ft>s2
At x = 2 ft, y = 1 ft . Then
ax = 64(2) = 128 ft>s2 ay = 64(1) = 64 ft>s2
The magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(128 ft>s2)2 + (64 ft>s2)2 = 143 ft>s2 Ans.
Its direction is
 u = tan-1a
ay
ax
b = tan-1a 64 ft>s2
128 ft>s2 b = 26.6° u Ans.
3–38. 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
x
y
y
v = (8y) ft/s
u = (8x) ft/s
streamline
V
�
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d i
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d s
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for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
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 W
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 W
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) 
will 
de
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nte
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ty 
of 
the
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 an
d i
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ot 
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rm
itte
d. 
307
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
y = 2x
a = 286 m>s2
u = 63.4° a
Solution
Since 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 =
4x
y
L
y
2 m
y dy = L
x
1 m
4x dx
y2
2
`
y
2 m
= 2x2 `
x
1m
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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + 2y2(0) + 8xy(4y)
 = (32xy2) m>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + 2y2(8y) + (8xy)(8x)
 = (16y3 + 64x2y) m>s2
At point x = 1 m and y = 2 m, 
ax = 32(1)(22) = 128 m>s2
ay = 316(23) + 64(12)(2)4 = 256 m>s2
The magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(128 m>s2)2 + (256 m>s2)2 = 286 m>s2 Ans.
Its direction is
 u = tan-1a
ay
ax
b = tan-1a256 m>s2
128 m>s2 b = 63.4° u Ans.
3–39. 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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d i
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ide
d s
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ly 
for
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e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
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 W
eb
) 
will 
de
str
oy
 th
e i
nte
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ty 
of 
the
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ork
 an
d i
s n
ot 
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rm
itte
d. 
308
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Solution
The 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>s
v = 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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + 4y(0) + (2x)(4)
 = (8x) m>s2
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + 4y(2) + 2x(0)
 = (8y) m>s2
At point (2 m, 1 m),
ax = 8(2) = 16 m>s2
ay = 8(1) = 8 m>s2
The magnitude of the acceleration is
a = 2a 2
x + a 2
y
 = 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
=
2x
4y
=
x
2y
L
y
1 m
2y dy = L
x
2 m
x dx
 y2 `
y
1 m
=
x2
2
`
x
2 m
 y2 - 1 =
x2
2
- 2
 y2 =
1
2
x2 - 1
*3–40. 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.
This
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ork
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es
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d s
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e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
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ide
 W
eb
) 
will 
de
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oy
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ty 
of 
the
 w
ork
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ot 
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309
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*3–40. (continued)
The plot of this streamline is shown is Fig. a
x(m) 22 2 3 4 5 6
y(m) 0 ±1 ±1.87 ±2.65 ±3.39 ±4.12
(a)
0 1 2 3 4
x(m)
y(m)
5 6
1
–1
–2
–3
–4
2
3
4
u = 4 m/s
45º
v = 4 m/s
V = 5.66 m/s
0 1 2 3 4
x(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
This
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tat
es
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rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
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 W
orl
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ide
 W
eb
) 
will 
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str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
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rm
itte
d. 
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Ans:
V = 4.47 m>s, a = 16 m>s2
y =
1
2
 ln x + 2
Solution
Since 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>s
v = 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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
= 0 + 4x(4) + 2(0) = 16x
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
= 0 + 4x(0) + 2(0) = 0
At point (1 m, 2 m),
ax = 16(1) = 16 m>s2 ay = 0
Thus, 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
=
2
4x
=
1
2x
 L
y
2 m
dy =
1
2 L
x
1 m
dx
x
y - 2 =
1
2
 ln x
y = a1
2
 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
3–41. The velocity of a flow field is defined by 
V = 54xi + 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 5
x(m)
y(m)
1
2
3
u = 4 m/s
V = 4.47 m/s
v = 2 m/s
0 1 2 3 4 5
x(m)
y(m)
1
2
3
0 1 2 3 4 5
x(m)
y(m)
1
2
3
0 1 2 3 4 5
x(m)
y(m)
1
2
3
a = 16 m/s2
0 1 2 3 4 5
x(m)
y(m)
1
2
3
u = 4 m/s
V = 4.47 m/s
v = 2 m/s
0 1 2 3 4 5
x(m)
y(m)
1
2
3
0 1 2 3 4 5
x(m)
y(m)
1
2
3
0 1 2 3 4 5
x(m)
y(m)
1
2
3
a = 16 m/s2
This
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s 
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ide
d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
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Solution
Since 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>s
v = -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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + (2x2 - y2)(4x) + (-4xy)(-2y)
 = 4x(2x2 - y2) + 8xy2
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 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>s2
ay = -4(1)32(12) - 124 + 16(12)(1) = 12 m>s2
The magnitude of the acceleration is
a = 2a 2
x + a 2
y = 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 = -4xydx
2x2dy + 4xydx - y2dy = 0
However, d(2x2y) = 2(2xydx + x2dy) = 2x2dy + 4xydx. Then
d(2x2y) - y2dy = 0
3–42. 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.
This
 w
ork
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ote
cte
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y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
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Ans:
V = 4.12 m>s 
a = 17.0 m>s2
x2 =
y3 + 5
6y
Integrating this equation,
2x2y -
y3
3
= C
with the condition y = 1 m when x = 1 m,
2(12)(1)-
13
3
= C
C =
5
3
Thus,
2x2y -
y3
3
=
5
3
6x2y - y3 = 5
 x2 =
y3 + 5
6y
 Ans.
Taking the derivative of this equation with respect to y
2 x
dx
dy
=
6y(3y2) - (y3 + 5)(6)
(6y)2
=
2y3 - 5
6y2
dx
dy
=
2y3 - 5
12xy2
Set 
dx
dy
= 0;
2y3 - 5 = 0
y = 1.357 m
The corresponding x is
x2 = 1.3573 + 5
x = 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 3.00
x(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
3–42. (continued)
0 0.5 1.0
x(m)
y(m)
1.5 2.0
0.5
1.0
2.5
2.0
1.5
1.36
3.0
0 0.5 1.0
x(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 1.0
x(m)
y(m)
1.5 2.0
0.5
1.0
2.5
2.0
1.5
1.36
3.0
0 0.5 1.0
x(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
This
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tat
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law
s 
an
d i
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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313
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Solution
Since 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 =
-y
4
= -
2
4
= -0.5 m>s
v =
x
9
=
3
9
= 0.3333 m>s
The 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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y are
1S+ 2 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
= 0 + a -y
4
b(0) + ax
9
ba-
1
4
b
= a-
1
36
xb m>s2
( + c ) ay =
0v
0t
+ u 
0v
0y
+ v 
0v
0y
= 0 + a -y
4
ba1
9
b + ax
9
b(0)
= c -
1
36
y d m>s2
At point (3 m, 2 m),
ax = -
1
36
 (3) = -0.08333 m>s2
ay = -
1
36
 (2) = -0.05556 m>s2
The magnitude of the acceleration is
a = 2ax2 + ay2
 = 2(-0.08333 m>s2)2 + (-0.05556 m>s2)
 = 0.100 m>s2 Ans.
3–43. 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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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir cou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
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ot 
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rm
itte
d. 
314
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Ans:
V = 0.601 m>s 
a = 0.100 m>s2
x2> 14.2422 + y2> 12.8322 = 1
3–43. (continued)
Using the definition of slope of the streamline,
dy
dx
=
v
u
 ; 
dy
dx
=
x>9
-y>4
= -
4x
9y
9L
y
2 m
ydy = -4L
x
3 m
xdx
9y2
2
`
y
2 m
= - (2x2) `
x
3 m
9y2
2
- 18 = -2x2 + 18
9y2 + 4x2 = 72
x2
72>4
+
y2
72>9
= 1
x2
(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
This
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 S
tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
315
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Solution
The flow is unsteady nonuniform. For one dimensional flow,
a =
0u
0t
+ u 
0u
0x
Here, u = (4 tx) m>s. Then 
0u
0t
= 4x and 
0u
0x
= 4t. Thus,
a = 4x + (4tx)(4t) = (4x + 16t2x) m>s2
Since 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
 
dx
dt
= u = 4 tx; L
x
2 m
dx
x
= 4L
t
0.15
t dt
 ln x `
x
2m
= 2t2 `
t
0.15
 ln 
x
2
= 2t2 - 0.02
 e2t2 - 0.02 =
x
2
 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.
*3–44. 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.
This
 w
ork
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cte
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nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
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Solution
The flow is unsteady nonuniform. For three dimensional flow,
a =
0V
0t
+ u 
0V
0t
+ v 
0V
0t
+ w 
0V
0t
Thus,
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
+ w 
0u
0t
 = 0 + 2x(2) + (6tx)(0) + 3y(0)
 = (4x) m>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
+
0v
0t
 = 6x + 2x(6t) + 6tx(0) + 3y(0)
 = (6x + 12tx) m>s2
 az =
0w
0t
+ u 
0w
0x
+ v 
0w
0y
+ w 
0w
0z
 = 0 + 2x(0) + 6tx(3) + 3y(0)
 = (18tx) m>s2
The position of the particle can be determined from
 
dx
dt
= u = 2x; L
x
1 m
dx
x
= 2L
t
0
dt
 ln x = 2t
 x = (e2t) m
 
dy
dt
= v = 6tx = 6te2t ; L
y
0
dy = 6L
t
0
te2tdt
y =
3
2
(2te2t - e2t) `
t
0
y =
3
2
(2te2t - e2t + 1) 
dz
dt
= w = 3y =
9
2
32te2t - e2t + 14 ;
L
z
0
dz =
9
2 L
t
0
(2te2t - e2t + 1) dt
z =
9
2
c te2t -
1
2
 e2t -
1
2
 e2t + t d `
t
0
z =
9
2
 (te2t - e2t + t + 1) m
3–45. 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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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
317
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Ans:
x = 2.72 m
y = 1.5 m
z = 0.634 m
a = 510.9i + 32.6j + 24.5k6 m>s2
3–45. (continued)
When, t = 0.5 s,
 x = e2(0.5) = 2.7183 m = 2.72 m Ans.
 y =
3
2
32(0.5)e2(0.5) - e2(0.5) + 14 = 1.5 m Ans.
Thus, z =
9
2
30.5e2(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>s2
Then
a = 510.9i + 32.6j + 24.5k6 m>s2 Ans.
This
 w
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law
s 
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d i
s p
rov
ide
d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
318
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Since 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)
L
y
1 m
dy
10y + 3
= - L
x
1 m
dx
4x + 6
1
10
 ln(10y + 3) `
y
1m
= -
1
4
 ln(4x + 6) `
x
1m
1
10
 lna10y + 3
13
b =
1
4
 lna 10
4x + 6
b
ln a10y + 3
13
b
1
10
= ln a 10
4x + 6
b
1
4
a10y + 3
13
b
1
10
= a 10
4x + 6
b
1
4
10y + 3
13
= a 10
4x + 6
b
5
2
 y = c 411
(4x + 6)5>2 - 0.3 dm Ans.
For two dimensional flow, the Eulerian description gives
a =
dV
dt
+ u 
dV
dx
+ v 
dV
dy
Writing the scalar components of this equation along the x and y axes,
 ax =
du
dt
+ u
du
dx
+ v 
du
dy
 = 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>s2
At point (1m, 1 m),
ax = 434(1) + 64 = 40 m>s2 S
ay = 10310(1) + 34 = 130 m>s2c
3–46. 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.
This
 w
ork
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e use
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
319
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Ans:
y =
411
14x + 625>2 - 0.3 
a = 136 m>s2
u = 72.9° a
3–46. (continued)
The magnitude of acceleration is
 a = 2a 2
x + a 2
y = 2(40m>s2)2 + (130m>s2)2 = 136 m>s2 Ans.
And its direction is
 u = tan-1 a
ay
ax
b = tan-1 a130 m>s2 
40 m>s2 b = 72.9° u Ans.
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 w
ork
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Ans:
y = 1.25 mm 
x = 15.6 mm
a = 0.751 m>s2
u = 2.29° a
Solution
Since 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
0
dy = 0.03L
t
0
t2 dt
 y = (0.01t3) m
When t = 0.5 s,
 y = 0.01(0.53) = 0.00125 m = 1.25 mm Ans.
 
dx
dt
= u = 100y = 100(0.01t3) = t3; L
y
0
dx = L
t
0
t3 dt
 x = a1
4
 t4b m
When t = 0.5 s,
 x =
1
4
(0.54) = 0.015625 m = 15.6 mm Ans.
For a two dimensional flow, the Eulerian description gives
 a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Write the scalar components of this equation along the x and y axes,
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + (100y)(0) + (0.03t2)(100)
 = (3t2) m>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0.06t + (100y)(0) + 0.03t2(0)
 = (0.06t) m>s2
When t = 0.5 s,
 ax = 3(0.52) = 0.75 m>s2 S
 ay = 0.06(0.5) = 0.03 m>s2c
The magnitude of acceleration is
 a = 2a 2
x + a 2
y = 2(0.75 m>s2)2 + (0.03 m>s2)2 = 0.751 m>s2 Ans.
And its direction is
 u = tan-1 a
ay
ax
b = tan-1a0.03 m>s2
0.75 m>s2 b = 2.29° u Ans.
3–47. 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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ork
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
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321
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Since 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
=
-y
2x2
 L
y
6 m
dy
y
= -
1
2 L
x
2 m
dx
x2
 ln y `
y
6 m
=
1
2
a1
x
b `
x
2 m
 ln 
y
6
=
1
2
 a1
x
-
1
2
b
 ln 
y
6
=
2 - x
4x
 
y
6
= e12 - x
4x 2
 y = c 6e12 - x
4x 2 d m Ans.
The plot of this streamline is shown in Fig. a.
For two dimensional flow, the Eulerian description gives.
 a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes,
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + (2x2)(4x) + (-y)(0)
 = (8x3) m>s2
 ay =
dv
dt
+ u 
dv
dx
+ v 
dv
dy
 = 0 + (2x2)(0) + (-y)(-1)
 = (y)m>s2
At point (2 m, 6 m), 
ax = 8(23) = 64 m>s2 S
 ay = 6m>s2c
The magnitude of acceleration is
 a = 2a 2
x + a 2
y = 2(64 m>s2)2 + (6m>s2)2 = 64.3m>s2 Ans.
And its direction is
 u = tan-1a
ay
ax
b = tan-1a 6 m>s2
64 m>s2 b = 5.36° u Ans.
*3–48. 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, 6 m), 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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ou
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ud
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em
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3–48. (continued)
Using the definition of the velocity,
dx
dt
= u; 
dx
dt
= 2x2
L
x
2 m 
dx
2x2 = L
t
0
dt
-
1
2
 a1
x
b `
x
2 m
= t
-
1
2
 a1
x
-
1
2
b = t
x - 2
4x
= t
x = a 2
1 - 4t
b m
dy
dt
= v; 
dy
dt
= -y
- L
y
6 m
dy
y
= L
t
0
dt
- ln y `
y
6 m
= t
 ln 
6
y
= t
6
y
= et
y = (6e-t) m
Thus,
 u = 2x2 = 2 a 2
1 - 4t
b
2
= c 8
(1 - 4t)2
d m>s and v = -y = ( -6e-t) m>s
Then,
 ax =
du
dt
= -16(1 - 4t)-3(-4) = c 64
(1 - 4t)3 d m>s2 Ans.
ay =
dv
dt
= (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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3–49. Airflow 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.
Solution
Since 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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y= 30 + (2x2 + 8)(4x) + (-8x)(0)4
 = 34x(2x2 + 8) 4 m>s2
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + (2x2 + 8)(-8) + (-8x)(0)
 = 3 -8(2x2 + 8) 4 m>s2
At point (0, 0),
ax = 4(0)32(02) + 84 = 0
ay = -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>s2T
The magnitude of the acceleration is
 a = 2a 2
x + a 2
y = 2(40 m>s2)2 + (80 m>s2)2 = 89.4 m>s2 Ans.
And its direction is
 u = tan-1 a
ay
ax
b = tan-1 a80 m>s2
40 m>s2 b = 63.4° cu Ans.
Using the definition of the slope of the streamline,
 
dy
dx
=
v
u
=
-8x
2x2 + 8
; Ldy = -8L
x dx
2x2 + 8
 y = -2 ln (2x2 + 8) + C
x � 1 m
y
x
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uc
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ea
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ing
 
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ir c
ou
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s a
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ud
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Ans:
At point (0, 0),
a = 64 m>s2w
At point (1 m, 0),
a = 89.4 m>s2, u = 63.4° c
For the streamline passing through point
(0, 0),
y = c 2 ln a 8
2x2 + 8
b d m
For the streamline passing through point
(1 m, 0),
y = c 2 ln a 10
2x2 + 8
b d m
3–49. (continued)
For the streamline passing through point (0, 0),
 0 = -2 ln 32(02) + 84 + C C = 2 ln 8
Then y = c 2 ln a 8
2x2 + 8
b d m Ans.
For the streamline passing through point (1 m, 0),
 0 = -2 ln 32(12) + 84 + C C = 2 ln 10
y = c 2 ln a 10
2x2 + 8
b 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 4
x(m)
y(m)
5–5 –4 –3 –2
–1
1
–1
–2
–3
–4
y = 
10
2x2 + 8 ��
2 ln
��
y = – 82x2 + 8
2 ln
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uc
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ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
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ud
en
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ing
. D
iss
em
ina
tio
n o
r 
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is 
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Solution
Since 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 =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Write the scalar components of this equation along x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
3–50. The velocity field for a fluid is defined by 
u = 3y> (x2 + y2) 4 m>s and v = 34x> (x2 + y2) 4 m>s, 
where x and y are in meters. Determine the acceleration of 
a particle located at point (2 m, 0) and a particle located at 
point (4 m, 0). Sketch the equations that define the 
streamlines that pass through these points.
= 0 + a y
x2 + y2 b £
(x2 + y2)(0) - y(2x)
(x2 + y2)2
§ + a 4x
x2 + y2 b £
(x2 + y2)(1) - y(2y)
(x2 + y2)2
§
= £ 4x3 - 6xy2
(x2 + y2)3
§m>s2
 ay =
0v
0t
+ u
0v
0x
+ v
0v
0y
 = 0 + a y
x2 + y2 b £
(x2 + y2)(4) - 4x(2x)
(x2 + y2)2
§ + a 4x
x2 + y2 b £
(x2 + y2)(0) - 4x(2y)
(x2 + y2)2
§
 = £ 4y3 - 36x2y
(x2 + y2)3
§m>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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se
ss
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ud
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iss
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tio
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3–50. (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)
=
4x
y
 ; 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 Ú 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 Ú 4m
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uc
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ch
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ir c
ou
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ud
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iss
em
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Ans:
For point (2 m, 0), 
a = 0.5 m>s2
y = {24x2 - 16
For point (4 m, 0), 
a = 0.0625 m>s2
y = {24x2 - 64
For the streamline passing through point (2 m, 0)
x(m) 2 3 4 5 6 7 8 9
y(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 9
y(m) 0 ±6.00 ±8.94 ±11.49 ±13.86 ±16.12
3–50. (continued)
y(m)
0 1 2 3 4 5 6 7 8
x(m)
9
5
–5
–10
10
15
20
–15
–20
y = �4x2 – 16
y = �4x2 – 64
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uc
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ea
ch
ing
 
the
ir c
ou
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s a
nd
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se
ss
ing
 st
ud
en
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iss
em
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Ans:
-1.23 m>s2
Solution
Here V only has an x component, so that V = u. Since V is a function of time at 
each x, the flow is unsteady. Since v = w = 0, we have
3–51. As the valve is closed, oil flows through the nozzle 
such that along the center streamline it has a velocity of 
V = 36(1 + 0.4x2)(1 - 0.5t)4 m>s, where x is in meters 
and t  is in seconds. Determine the accel eration of an oil 
particle at x = 0.25 m when t = 1 s.
ax =
0u
0t
+ u
0u
0x
=
0
0x
 36(1 + 0.4x2)(1- 0.5t)4 + 36(1 + 0.4x2)(1 - 0.5t)4 0
0x
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)4
Evaluating 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
A
B
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Solution
The n - s coordinate system is established with origin at point A as shown in Fig. a. 
Here, the component of the particle’s acceleration along the s axis is
as = 3 m>s2
Since the streamline does not rotate, the local acceleration along the n axis is zero, 
so that a0V 
0t
b
n
= 0. Therefore, the component of the particle’s acceleration along 
the n axes is
an = a0V 
0t
b
n
+
V2
R
 = 0 +
(5 m>s)2
16 m
= 1.5625 m>s2
Thus, the magnitude of the particle’s acceleration is
a = 2a 2
s + a 2
n = 2(3 m>s2 )2 + (1.5625 m>s2)2
 = 3.38 m>s2 Ans.
*3–52. 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 m
A
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Ans:
72 m>s2
3–53. 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.
Solution
Fig. a is based on the initial condition when s = 0, r = rD. Thus, r = 1 - s. Then the 
radial component of velocity is
V = -
3
r
= a-
3
1 - s
b m>s
This is one dimensional steady flow since the velocity is along the straight radial line. 
The Eulerian description gives
 a =
0V
0t
+ V 
0V
0s
 = 0 + a-
3
1 - s
b c -
3
(1 - s)2 d
 = c 9
(1 - s)3 d m>s2
When 1 - s = r = 0.5 m, this equation gives
 a = a 9
0.53 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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 S
tat
es
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rig
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law
s 
an
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s p
rov
ide
d s
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ly 
for
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e u
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str
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s i
n t
ea
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the
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s a
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se
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en
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Ans:
as = 3.20 m>s2 
an = 7.60 m>s2
3–54. A particle located at a point within a fluid flow has 
velocity components of u = 4 m>s and v = -3 m>s, and 
acceleration components of ax = 2 m>s2 and ay = 8 m>s2. 
Determine the magnitude of the streamline and normal 
components of acceleration of the particle.
Solution
 V = 2(4 m>s)2 + ( -3 m>s)2 = 5 m>s
 a = 2(2 m>s2)2 + (8 m>s2)2 = 8.246 m>s2
 a = 2i + 8j
 u = tan-1 
3
4
= 36.870°
 us = cos 36.870°i - sin 36.870°j
 = 0. 8i - 0.6j
 as = a # us = (2i + 8j) # (0.8i - 0.6j)
 as = -3.20 m>s2
 as = 3.20 m>s2 Ans.
 a = 2a 2
s + a 2
n
 (8.246)2 = (3.20)2 + a 2
n
 an = 7.60 m>s2 Ans.
x
y
4 m/s
ûs
S
n
3 m/s
V
�
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s a
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Ans:
a = 3.75 m>s2 
u = 36.9° c
Solution
3–55. A particle moves along the circular streamline, such 
that it has a velocity of 3 m>s, which is increasing at 3 m>s2. 
Determine the acceleration of the particle, and show the 
acceleration on the streamline.
The normal component of the acceleration is
an =
V2
r
=
(3 m>s)2
4 m
= 2.25 m>s2
Thus, the magnitude of the acceleration is
 a = 2a 2
s + a 2
n = 2(3 m>s2)2 + (2.25 m>s2)2 = 3.75 m>s2 Ans.
And its direction is
 u = tan-1aan
as
b = tan-1a2.25 m>s2
3 m>s2 b = 36.9° Ans.
The plot of the acceleration on the streamline is shown in Fig. a.
(a)
a = 3.75 m/s2
as = 3 m/s2
 = 36.9º
an = 2.25 m/s2
4 m
�
4 m
3 m/s
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uc
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n t
ea
ch
ing
 
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s a
nd
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se
ss
ing
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ing
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work
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will 
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Solution
Using the condition at r = 3 m, V = 18 m>s .
V =
k
r
; 18 m>s =
k
3 m
 k = 54 m2>s
Then
V = a54
r
b m>s
At r = 9 m, V = a54
9
b m>s = 6 m>s . Since the velocity is constant, the streamline 
component of acceleration is
as = 0
The normal component of acceleration is
 an = a0V
0t
b
n
+
V2
r
= 0 +
(6 m>s)2
9 m
= 4 m>s2
Thus, the acceleration is
a = an = 4 m>s2 Ans.
*3–56. The motion of a tornado can, in part, be described 
by a free vortex, V = k>r where k is a constant. Consider 
the steadymotion at the radial distance r = 3 m, where 
V = 18 m>s. Determine the magnitude of the acceleration 
of a particle traveling on the streamline having a radius of 
r = 9 m. r = 9 m
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e u
se
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uc
tor
s i
n t
ea
ch
ing
 
the
ir c
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rse
s a
nd
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se
ss
ing
 st
ud
en
t le
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ing
. D
iss
em
ina
tio
n o
r 
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le 
of 
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work
 (in
clu
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 W
eb
) 
will 
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oy
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nte
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ty 
of 
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ork
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Ans:
as = 222 m>s2
an = 128 m>s2
Solution
The streamline component of acceleration can be determined from
as = a0V
0t
b
s
+ V 
0V
0s
However, s = ru. Thus, 0 s = r 0 u = 0.5 0 u. Also, the flow is steady, a0V
0t
b
s
= 0 and 
0V
0 s
=
0V
0.5 0 u
= 2 
0V
 0 u
= 2(16 cos u) = 32 cos u. Then
as = 0 + 16 sin u(32 cos u) = 512 sin u cos u = 256 sin 2u
When u = 30°,
as = 256 sin 2(30°) = 221.70 m>s2 = 222 m>s2 Ans.
V = (16 sin 30°) m>s = 8 m>s
The normal component of acceleration can be determined from
an = a0V
0 t
b
n
+
V2
R
= 0 +
(8 m>s)2
0.5 m
= 128 m>s2 Ans.
3–57. Air flows around the front circular surface. If 
the steady-steam velocity is 4 m>s upstream from the 
surface, and the velocity along the surface is defined 
by V = (16 sin u) m>s, determine the magnitude of the 
streamline and normal components of acceleration of a 
particle located at u = 30°.
0.5 m
V
4 m/s
u
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e u
se
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str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
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rse
s a
nd
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se
ss
ing
 st
ud
en
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arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
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is 
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 (in
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din
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n t
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 W
orl
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ide
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eb
) 
will 
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oy
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e i
nte
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ty 
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ork
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Ans:
as = 78.7 m>s2
an = 73.0 m>s2
Solution
At x = 1 m, y = 2 m
u = 8(2) = 16 m>s
v = 6(1) = 6 m>s
u = tan-1 6
16
= 20.56°
us = cos 20.56° i + sin 20.56° j
= 0.9363i + 0.3511j
Acceleration. With w = 0, we have
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + 8y(0) + 6x(8) = 48x
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + 8y(6) + 6x(0) = 48y
At x = 1 m, and y = 2 m, 
ax = 48(1) = 48 m>s2
ay = 48(2) = 96 m>s2
Therefore, the acceleration is
a = 548i + 96j6 m>s2
And its magnitude is
a = 2a 2
x + a 2
y = 2(48 m>s2)2 + (96 m>s2)2 = 107.33 m>s2
Since the direction of the s axis is defined by us, the component of the particle’s 
acceleration along the s axis can be determined from
as = a # us = 348i + 96j4 # 30.9363i + 0.3511j4
= 78.65 m>s2 = 78.7 m>s2 Ans.
The normal component of the particle’s acceleration is
an = 2a2 - a 2
s = 2(107.33 m>s2)2 - (78.65 m>s2)2 = 73.0 m>s2 Ans.
3–58. Fluid particles have velocity components of 
u = (8y) m>s v = (6x) m>s, where x and y are in meters. 
Determine the magnitude of the streamline and the normal 
components of acceleration of a particle located at point 
(1 m, 2 m).
�
6 m/s
16 m/s
us
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se
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uc
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ea
ch
ing
 
the
ir c
ou
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s a
nd
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se
ss
ing
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ud
en
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ing
. D
iss
em
ina
tio
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r 
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Ans:
a = 67.9 m>s2 
u = 45° a
4y2 - 3x2 = 1
Solution
Since the velocity components are independent of time, but a function of position, 
the flow can be classified as steady nonuniform. For two dimensional flow, (w = 0), 
the Eulerian description is
a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes,
ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
= 0 + 8y(0) + 6x(8)
= 48x
ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
= 0 + 8y(6) + 6x(0)
= 48y
At point x = 1 m and y = 1 m
ax = 48(1) = 48 m>s2
ay = 48(1) = 48 m>s2
The magnitude of the acceleration is
a = 2a 2
x + a 2
y = 2(48 m>s2)2 + (48 m>s2)2 = 67.9 m>s2 Ans.
Its direction is
u = tan-1a
ay
ax
b = tan-1a48 m>s2
48 m>s2 b = 45° a Ans.
The slope of the streamline is 
dy
dx
=
v
u
 ; 
dy
dx
=
6x
8y
 (1)
L
y
1 m
8ydy = L
x
1 m
6xdx
4y2 `
y
1 m
= 3x2 `
x
1 m
4y2 - 3x2 = 1 Ans.
3–59. Fluid particles have velocity components of 
u = (8y) m>s and v = (6x) m>s, where x and y are in 
meters. Determine the acceleration of a particle located at 
point (1 m, 1 m). Determine the equation of the streamline, 
passing through this point.
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ea
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ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
 st
ud
en
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ing
. D
iss
em
ina
tio
n o
r 
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le 
of 
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art
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is 
work
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Solution
 
dy
dx
=
r
u
; 
dy
dx
=
8xy
2y2 =
4x
y
L
y
2 m
y dy = L
x
1 m
4x dx
 
y2
2
`
y
2 m
= 2x2 `
x
1 m
 
y2
2
- 2 = 2x2 - 2
 y2 = 4x2
 y = 2x
(Note that x = 1, y = 2 is not a solution y = -2x.) Two equation from streamline 
is y = 2x (A straight line). Thus, R S ∞ and since the flow is steady,
 an =
V2
R
= 0 Ans.
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
= 0 + 2y2(0) + (8xy)(4y) = (32xy2) m>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + 2y2(8y) + (8xy)(8x)
 = (16y3 + 64x2y) m>s2
At (1 m, 2 m),
 ax = 32(1)(22) = 128 m>s2
 ay = 316(23) + 64(12)(2) 4 = 256 m>s2
 a = as = 2a 2
x + a 2
y = 2(128 m>s2)2 + (256 m>s2)
 as = 286 m>s2 Ans.
*3–60. A fluid has velocity components of u = (2y2) m>s 
and v = (8xy) m>s, where x and y are in meters. Determine 
the magnitude of the streamline and normal components of 
acceleration of a particle located at point (1 m, 2 m).
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e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
 st
ud
en
t le
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ing
. D
iss
em
ina
tio
n o
r 
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le 
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art
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is 
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nte
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Solution
x(m) 23>2 1.0 2.0 3.0 4.0 5.0
y(m) 0 1.0 3.61 5.74 7.81 9.85
Since the velocity component is independent of time and is a function of position, 
the flow can be classified as steady nonuniform. The slope of the streamline is
 
dy
dx
=
v
u
 ; 
dy
dx
=
8xy
2y2 =
4x
y
 L
y
1 m
y dy = L
x
1 m
4x dx
 
y2
2
`
y
1 m
= 2x2 `
x
1 m
 4x2 - y2 = 3 where x and y are in m Ans.
For two dimensional flow (w = 0), the Eulerian description is
 a =
0V
0t
+ u 
0V
0x
+ v 
0V
0y
Writing the scalar components of this equation along the x and y axes,
 ax =
0u
0t
+ u 
0u
0x
+ v 
0u
0y
 = 0 + (2y2)(0) + (8xy) (4y)
 = (32xy2) m>s2
 ay =
0v
0t
+ u 
0v
0x
+ v 
0v
0y
 = 0 + (2y2)(8y) + (8xy) (8x)
 = (16y3 + 64x2y) m>s2
At point x = 1 m and y = 1 m
 ax = 32(1)(12) = 32 m/s2
 ay = 16(13) + 64(12)(1) = 80 m>s2
Thus,
 a = 532i + 80j6 m>s2
3–61. A fluid has velocity components of u = (2y2) m>s 
and v = (8xy) m>s, where x and y are in meters. Determine 
the magnitude of the streamline and normal components of 
the acceleration of a particle located at point (1 m, 1 m). 
Find the equation of the streamline passing through this 
point, and sketch the streamline and normal components of 
acceleration at this point.
0 1 2 3 4
(a)
x(m)
y(m)
5
1
2
3
4
5
6
7
8
9
10
 a = 68.2°
as = 85.4 m/s2
a = 18.2 m/s2
an = 11.6 m/s2
�
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uc
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ch
ing
 
the
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rse
s a
nd
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se
ss
ing
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ud
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Ans:
4x2 - y2 = 3 
as = 85.4 m>s2
an = 11.6 m>s2
3–61. (continued)
The magnitude of the acceleration is
a = 2a 2
x + a 2
y = 2(32 m>s2)2 + (80 m>s2) = 86.16 m>s2 = 18.2 m>s2
and its direction is
ua = tan-1a
ay
ax
b = tan-1a80 m>s2
32 m>s2 b = 68.2°
At point (1 m, 1 m),
tan u =
dy
dx
`
x = 1 m
y = 1 m
=
4(1)
1
= 4; u = 75.96°
Thus, the unit vector along the streamline is
u = cos 75.96°i + sin 75.96°j = 0.2425i + 0.9701j
Thus, the streamline component of the acceleration is
as = a # us = (32i + 80j) # (0.2425i + 0.9701j)
 = 85.37 m>s2 = 85.4 m>s2 Ans.
Then
 an = 2a2 - a 2
s = 2(86.16 m>s2)2 - (85.37 m>s2)2
 = 11.64 m>s2 = 11.6 m>s2 Ans.
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uc
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the
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ou
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s a
nd
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se
ss
ing
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ud
en
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. D
iss
em
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