Fluid Mechanics Chapter 3 Solution Guide

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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.

4m t3s 6m

t4s

t2s

4m

pathline 60

t0

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.

marked particle streamline 4m

streakline 60˚ t=2s (a) marked particle

streamline

streakline

60˚ 4m t=4s (b)

267

6m


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.

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.

t=0

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.

pathline t=1s streamline x

2m

first marked particle

streakline

(a) y

streamline streakline

first marked particle t=4s

4m

t=0

x t=3s 6m pathline t=4s (b)

268


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.

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.

y pathline t=2s

streamline first marked particle streakline

6m

t=0 (a) y t=4s

pathline t = 6 s streamline first marked particle

12 m

streakline

x t=0

4m (b)

269


*3–4. A two-dimensional flow field for a fluid be described by V = [(2x + 1)i - (y + 3x)j] m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at (2 m, 3 m), and its direction measured counterclockwise from the x axis.

y

3m

x

Solution

2m

The velocity vector for a particle at x = 2 m and y = 3 m is

= [2(2) + 1]i - [3 + 3(2)]j

The magnitude of V is

= 5 5i - 9j 6 m>s

V = 2V x2 + V y2 = 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

Thus,

Vx = 5 m/s

V = 5 (2x + 1)i - (y + 3x)j 6 m>s

f = tan-1 a

Vy Vx

b = tan-1 a

9 m> s 5 m>s

Ans.

270

V (a)

b = 60.95°

u = 360° - 60.95° = 299°

Vy = 9 m/s

x


3–5. A two-dimensional flow field for a liquid can be described by V = 3 ( 5y2 - x ) i + (3x + y)j 4 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.

Solution The velocity vector of a particle at x = 5 m and y = -2 m is V = = The magnitude of V is

5( 5y2 - x ) i + 3 5( - 2)2 - 5 4 i

Vy = 13 m/s

V

(3x + y)j6 m>s

+ 33(5) + ( - 2)4j

= 5 15i + 13j 6 m>s

Vx = 15 m/s

V = 2V x2 + V y2 = 2 ( 15 m>s ) 2 + ( 13 m>s ) 2 = 19.8 m>s

Ans.

x

(a)

As indicated in Fig. a, the direction of V is defined by

u = tan - 1 a

Vy Vx

b = tan - 1 a

13 m>s 15 m>s

b = 40.9°

Ans.

Ans: V = 19.8 m>s u = 40.9° 271


3–6. The soap bubble is released in the air and rises with a velocity of V = 3 (0.8x)i + ( 0.06t 2 ) j 4 m>s where x is meters and t is in seconds. Determine the magnitude of the bubble’s velocity, and its directions measured counterclockwise from the x axis, when t = 5 s, at which time x = 2 m and y = 3 m. Draw its streamline at this instant.

v u

Solution The velocity vector of a particle at x = 2 m and the corresponding time t = 5 s is

5(0.8x)i

V =

3 0.8(2)i

=

V

+ ( 0.06t ) j6 m>s + 0.06(5)2 j4

= 5 1.6i + 1.5j 6 m>s

The magnitude of V is

Vy = 1.5 m s

2

V = 2V x2 + V y2 = 2 ( 1.6 m>s ) 2 + ( 1.5 m>s ) 2 = 2.19 m>s

Vx = 1.6 m s

Ans.

x

(a)

As indicated in Fig. a, the direction of V is defined by u = tan-1 a

Vy Vx

b = tan-1 a

1.5 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 0.06t 2 v dy = ; = dx u dx 0.8x Note that since we are finding the streamline, which represents a single instant in time, t = 5 s, t is a constant. y

L3 m

dy t

2

x

=

L2 m

0.075dx x

1 x (y - 3) = 0.075 ln 2 2 t y = a0.075t 2 ln

When t = 5 s,

y(m)

x + 3b m 2

6 5 4

x y = 0.075 ( 52 ) ln a b + 3 2

3

x y = c 1.875 ln a b + 3 d m 2

2 1

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

0 0.5 1

2

3

4

5

6

x(m)

(a)

The plot of the streamline is shown in Fig. a

Ans: V = 2.19 m>s u = 43.2° 272


3–7. A flow field for a fluid described by u = (2 + y) m>s and v = (2y) m>s, where y is in meters. Determine the equation of the streamline that passes through point (3 m, 2 m), and find the velocity of a particle located at this point. Draw this streamline.

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,

y v = 2y m/s

dy = tan u dx

dy v = dx u

u = (2 y) m/s

streamline

y

dy 2y = dx 2 + y

x x

2 + y dy = dx L 2y L ln y +

V

(a)

1 y = x + C 2

At point (3 m, 2 m), we obtain ln(2) +

1 (2) = 3 + C 2

C = - 1.31 Thus, ln y +

1 y = x - 1.31 2

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>s c 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

4 m>s v u = tan-1 a b = tan-1 a b = 45° u 4 m>s

Ans.

Ans: ln y2 + y = 2x - 2.61 V = 5.66 m>s u = 45° a 273


*3–8. A two-dimensional flow field is described by u = 1 x2 + 5 2 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.

Solution y

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 = tan u dx

streamline u = (x2 + 5) m/s

dy - 6xy v = = 2 dx u x + 5 dy

L y

= -6

x dx 2 x L + 5

v = (6xy) m/s (a)

At x = 5 m, y = 1 m. Then, ln 1 = - 3 ln 3 (5)2 + 5 4 + C C = 3 ln 30

ln y = - 3 ln ( x2 + 5 ) + 3 ln 30 ln y + ln ( x2 + 5 ) 3 = 3 ln 30 ln 3 y ( x2 + 5 ) 3 4 = 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

x x

ln y = - 3 ln ( x2 + 5 ) + C

Thus

y

30 m>s v u = tan-1 a b = tan-1 a b = 45° u 30 m>s

Ans.

274

V


3–9. Particles travel within a flow field defined by V = 3 2y2i + 4j 4 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.

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,

y

v = 4 m/s

dy = tan u dx

y2 dy =

L

x

2dx

x (a)

1 3 y = 2x + C 3

At x = 1 m, y = 2 m. Then

streamline

y

dy v 4 = = 2 dx u 2y L

V

u = 2y2 m/s

1 3 (2) = 2(1) + C 3 C =

2 3

Thus, 1 3 2 y = 2x + 3 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.


v 4 u = tan-1 a b = tan-1 a b = 26.6° u 8

Ans.

Ans: y3 = 6x + 2 V = 8.94 m>s u = 26.6° a 275


3–10. A ballon is released into the air from the origin and carried along by the wind, which blows at a constant rate of u = 0.5 m>s. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = (0.8 + 0.6y) m>s. Determine the equation of the streamline for the balloon, and draw this streamline.

y

v u

x


y v = (0.8 + 0.6y) m/s V streamline

dy = tan u dx

u = 0.5 m/s

dy 0.8 + 0.6y v = = = 1.6 + 1.2y dx u 0.5

y x

Since the balloon starts at y = 0, x = 0, using these values,

x

y

lna

x dy dx = L0 1.6 + 1.2y L0 y 1 ln(1.6 + 1.2y) ` = x 1.2 0

1.6 + 1.2y 1.6 lna1 +

(a) y(m)

b = 1.2x

y=

4 1.2x ( e – 1) m 3

3 yb = 1.2x 4

1 +

3 y = e 1.2x 4 4 y = ( e 1.2x - 1 ) m 3

streamline

Ans.

Using this result, the streamline is shown in Fig. b.

x(m) (b)

Ans: y = 276

4 1.2x (e - 1) 3


3–11. A ballon is released into the air from point (1 m, 0) and carried along by the wind, which blows at a rate of u = (0.8x) m>s. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = (1.6 + 0.4y) m>s. Determine the equation of the streamline for the balloon, and draw this streamline.

y

v u

x

Solution

y

As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

v = (1.6 + 0.4y) m/s V

dy = tan u dx

streamline

dy 1.6 + 0.4y v = = dx u 0.8x

u = (0.8x) m/s

y x

The balloon starts at point (1 m, 0).

x

y

x dy dx = 1.6 + 0.4y 0.8x L0 L1

(a) y

y x 1 1 ln(1.6 + 0.4y) ` = ln x ` 0.4 0.8 0 1

y = 4(x½ – 1) m

1.6 + 0.4y 1 1 lna b = ln x 0.4 1.6 0.8 lna1 +

a1 +

1 2 yb = ln x 4

streamline

1 2 yb = x 4

x

y = 4 ( x1>2 - 1 ) m

Ans.

Using this result, the streamline is shown in Fig. b.

1m (b)

Ans: y = 4 ( x1>2 - 1 ) 277


*3–12. A flow field is defined by u = (8y) m>s, v = (6x) m>s where x and y are in meters. Determine the equation of the streamline that passes through point (1 m, 2 m). Draw this streamline.


y

dy = tan u dx dy v 6x = = dx u 8y

L

8y dy = 2

L

v = (6x) m/s

u = (8y) m/s streamline

y

6x dx

x

2

4y = 3x + C x

At x = 1 m, y = 2 m. Then 4(2)2 = 3(1)2 + C

(a)

C = 13

Thus

V

4y2 = 3x2 + 13

Ans.

278


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.


v = (6y) ft/s

dy = tan u dx dy 6y v = = dx u 3x

V

streamline

u = (3x) ft/s

y

x

dy

dx = L 2y L x

y

x (a)

1 ln y = ln x + C 2

At x = 3 ft, y = 1 ft . Then 1 ln y = ln x - ln3 2 1 x ln y = ln 2 3 x 2 ln y = lna b 3

y =

x2 9

Ans.

279

Ans: y = x2 >9


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 streamlines and pathlines for these particles.

Solution Since the velocity V is constant, Fig. a, the streamline will be a straight line with a slope. dy = tan u dx

dy v 8 = = dx u 5

y

v = 8 m/s V

streamline pathline

u = 5 m/s

y = 1.6x + C At x = 0, y = 0. Then

x

C = 0 (a)

Thus

Ans.

y = 1.6x

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.

Ans: y = 1.6x 280


3–15. A flow field is defined by u = 3 8x> ( x2 + y2 ) 4 m>s and v = 3 8y> ( 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.


y

dy = tan u dx

v=

) x 8y+ y ) m/s V 2

2

8y> ( x2 + y2 )

y dy v = = = dx u x 8x> ( x2 + y2 )

u=

y

) x 8y+ y ) m/s 2

2

streamline x

dy dx = L y L x y ln x = C

x (a)

y = C′ x At x = 1 m, y = 1 m. Then C′ = 1 Thus,

y = 1 x y = x

Ans.

Ans: y = x 281


*3–16. A velocity components of fluid has u = 3 30>(2x + 1) 4 m>s and v = 2ty m>s where x and y are in meters and t is in seconds. Determine the pathline that passes through the point (2 m, 6 m) at time t = 2 s. Plot this pathline for 0 … x … 4 m.

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 x

t

dx 30 (2x + 1)dx = 30 dt = u = ; dt 2x + 1 L2 m L2 s

( x2 + x ) `

2m

2

= 30 t `

t 2s

x + x - 6 = 30(t - 2)

x

t =

1 2 ( x + x + 54 ) 30

(1)

y t dy dy = 2 tdt = v = 2ty; dt L6 m y L2 s

ln y ` ln

y 6m

= t2 `

y(m)

2s

y = t2 - 4 6 y 2 = et - 4 6 2

y = 6e t

t

-4

50 40

(2)

Substitute Eq. (1) into Eq. (2), y = 6e 900 (x 1

2

+ x + 54)2 - 4

Ans.

30 20 10

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

282

0

1

2

3 (a)

4

x(m)


3–17. A velocity components of fluid has u = 3 30>(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.

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 dy 2ty v 1 = ; = = ty(2x + 1) dx u dx 30>(2x + 1) 15 y

x dy 1 = t (2x + 1)dx 15 L1 m L4 m y

ln y ` ln

y 4m

=

x 1 t ( x2 + x ) ` 15 1m

y 1 = t ( x2 + x - 2 ) 4 15 y = 4e t(x

+ x - 2) >15

y = 4e (x

+ x - 2) >15

2

For t = 1 s, 2

For t = 2 s,

y = 4e 2(x

2

For t = 3 s,

Ans.

+ x - 2) >15

Ans.

y(m) 45

y = 4e

(x2 + x - 2)>5

Ans.

40

The plot of these streamlines are shown in Fig. a

35

For t = 1 s

30

x(m)

0

1

2

3

4

y(m)

3.50

4

5.22

7.79

13.3

x(m)

0

1

2

3

4

y(m)

3.06

4

6.82

15.2

44.1

t=3s

t=2s

25 20

For t = 2 s

15 10

t=1s

5

For t = 3 s x(m) y(m)

0

1

2

3

4

2.68

4

8.90

29.6

146

0

1

2

3

4

x(m)

(a)

Ans: 2 For t = 1 s, y = 4e (x + x - 2)>15 2 ( For t = 2 s, y = 4e 2 x + x - 2)>15 (x2 + x - 2)>5 For t = 3 s, y = 4e 283


fluid has 3–18. A velocity components of u = 3 30>(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.

Solution y(m)

For t = 2 s

t=5s

x(m) y(m)

0 2.70

1 3.52

2 6.00

3 13.35

4 38.80

50 40

For t = 5 s

30

x(m) y(m)

0 0.812

1 1.58

2 6.00

3 44.33

t=2s

4 638.06

20 10

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 dy 2ty v 1 = ; = = ty(2x + 1) dx u dx 30>(2x + 1) 15

0

1

2

3

4

x(m)

(a)

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. y

x dy 1 = t (2x + 1)dx 15 L2 m L6 m y

ln y `

y 6m

ln

=

x 1 t ( x2 + x ) ` 15 2m

y 1 = t ( x2 + x - 6 ) 6 15 y = 6e 15 t(x 1

2

+ x - 6)

For t = 2 s,

y = 6e 15 (x 2

2

+ x - 6)

Ans.

For t = 5 s,

y = 6e 3 (x 1

2

+ x - 6)

Ans.

The plots of these two streamlines are show in Fig. a.

Ans: 2 For t = 2 s, y = 6e 21x + x - 6)>15 1x2 + x - 6) >3 For t = 5 s, y = 6e 284


3–19. A particle travels along the streamline defined by y3 = 8x - 12. If its speed is 5 m>s when it is at x = 1 m, determined the two components of its velocity at this point. Sketch the velocity on the streamline.

Solution x(m) y(m)

0 - 2.29

1 - 1.59

1.5 0

2 1.59

3 2.29

4 2.71

5 3.04

The plot of the streamline is shown in Fig. a. Taking the derivative of the streamline equation, 3y2

dy = 8 dx

dy 8 = tan u = 2 dx 3y When x = 1 m, y3 = 8(1) - 12; y = - 1.5874 Then dy 8 ` = tan u ` = ; u 0 x = 1 m = 46.62° dx x = 1 m 3( - 1.5874)2 x=1 m

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.

y(m) 3 3 3 2 1

0 –1 –2

1

2

v

3

4

5

x(m)

5 m/s u

46.62°

–3

Ans: u = 3.43 m>s v = 3.63 m>s

(a)

285


*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.

Solution Here, u =

dx . Then, dt dx = udt

Using x = 0 when t = 0 as the integration limit, L0

x

t

3 (0.8t) m>s 4 dt L0 x = 0.4t 2

dx =

Also, v =

dy . Then dt

(1)

dy = vdt Using y = 0 when t = 0 as the integration limit, L0

y

dy =

L0

t

( 0.4 m>s ) dt (2)

y = 0.4t

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 1 2 3 4 5

x 0.4 1.6 3.6 6.4 10

y 0.4 0.8 1.2 1.6 2

y(m)

A plot of the pathline is shown in Fig. a. From Eqs. (1) and (2), when t = 4 s,

y2 = 0.4x

2

x = 0.4 ( 42 ) = 6.4 m y = 0.4(4) = 1.6 m 1

Using the definition of the slope of the streamline, dy dy v 0.4 = ; = dx u dx 0.8t y

t

L1.6 m

dy =

t(y - 1.6) = y = c

0

x

1 dx 2 L6.4 m

2

4

6

–1

1 (x - 6.4) 2

–2

1 (x - 6.4) + 1.6 d m 2t

(a)

286

8

10

x(m)


*3–20. (continued)

When t = 4 s,

y(m)

1 y = ( x - 6.4) + 1.6 2(4)

y =

y=

1 x + 0.8 8

1 x + 0.8 8

0.8

The plot of the streamline is shown is Fig. b.

x(m)

(b)

287


3–21. The velocity for an oil flow is defined by V = 1 3y2 i + 8 j 2 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?

Solution Since the velocity components are a function of position only, the flow can be classified as steady nonuniform. Here, u = 1 3y2 2 m>s and v = 8 m>s . The slope of the streamline is defined by dy dy v 8 = ; = 2 dx u dx 3y y

x

L1 m

3y2dy = 8

y3 `

y

= 8x `

1m

3

dx L2 m

x 2m

y - 1 = 8x - 16 y3 = 8x - 15 (1)

Ans.

From the definition of velocity dy = 8 dt y

L1 m

dy =

y`

y 1m

L0

= 8t `

1s

8 dt 1s 0

y - 1 = 8 Ans.

y = 9 m Substituting this result into Eq. (1) 93 = 8x - 15

Ans.

x = 93 m

Ans: y3 = 8x - 15, y = 9 m x = 93 m 288


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.

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 dy v 2 = ; = dx u dx 6 - 3x L0

y

x

dy = 2

dx 6 3x L0

x 2 y = - ln (6 - 3x) ` 3 0

6 - 3x 2 y = - ln a b 3 6

y = The plot of this streamline is show in Fig. a

2 2 ln a b 3 2 - x

Ans.

y(m) 1.5

1

0.5

0

0.25

0.5

0.75

1.0

1.25

1.5

1.75

2

x(m)

(a)

Ans: y = 289

2 2 ln a b 3 2 - x


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.

Solution Using the definition of velocity, for 0 … t 6 10 s

y(m)

dx dx = u; = -2 dt dt L0

B

x

dx = - 2

L0

30

t

dt

C

20

(1)

x = ( -2t) m

10

When t = 10 s, x = -2(10) = - 20 m

A –20

dy dy = v; = 3 dt dt y

L0

dy = 3

L0

–15

–10

–5

0

5

x(m)

(a)

t

dt (2)

y = (3t) m 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). 3 y = - x Ans. 2 For 10 6 t … 15 s. dx dx = u; = 5 dt dt x

L-20 m

dx = 5

t

dt L10 s

x - ( - 20) = 5(t - 10) (3)

x = (5t - 70) m At t = 15 s, x = 5(15) - 70 = 5 m dy dy = v; = -2 dt dt y

L30 m

dy = -2

t

dt L10 s

y - 30 = -2(t - 10) (4)

y = ( -2t + 50) m When t = 15 s, y = - 2(15) + 50 = 20 m Eliminate t from Eqs. (3) and (4), 2 y = a - x + 22b 5

Ans.

The two streamlines intersect at ( -20, 30), point B in Fig. (a). The pathline is the path ABC. 290

Ans: 3 For 0 … t 6 10 s, y = - x 2 2 For 10 s 6 t … 15 s, y = - x + 22 5


*3–24. The velocity field is defined by u = (4x) m>s and v = (2t) m>s, where t is in seconds and x is in meters. Determine the equation of the streamline that passes through point (2 m, 6 m) for t = 1 s. Plot this streamline for 0.25 m … x … 4 m.

Solution x(m)

0.25

0.5

0.75

1

2

3

4

y(m)

4.96

5.31

5.51

5.65

6

6.20

6.35

Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. Using the definition of the slope of the streamline, dy dy v 2t t = ; = = dx u dx 4x 2x y

L6 m

dy =

x

t dx 2 L2 m x

t x ln 2 2 t x y = ln + 6 2 2

y - 6 =

1 x For t = 1 s, y = a ln + 6bm 2 2 The plot of this streamline is shown in Fig. a.

Ans.

y(m) 7 6 5 4 3 2 1

0 0.25 0.5 0.75 1

2

3

4

(a)

291

x(m)


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.

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, x

t

dx dx dt = u = 4x; = dt L2 m 4x L1 S x t 1 ln x ` = t` 4 2m 1s

1 x ln = t - 1 4 2 1 x ln + 1 4 2

t =

(1)

y t dy = v = 2t; dy = 2t dt dt L6 m L1 s

y - 6 = t2 `

t 1s

2

(2)

y = t + 5

Substitute Eq. (1) into (2), 2 1 x y = a ln + 1b + 5 4 2

y = a

1 x 1 2x ln + ln + 6b 16 2 2 2

Ans.

The plot of this pathline is shown in Fig. (a) y(m) 7 6 5 4 3 2 1 0 0.25 0.5 0.75 1

2

3

4

x(m)

Ans: y =

(a)

292

1 2x 1 x ln + ln + 6 16 2 2 2


3–26. The velocity field of a fluid is defined by u = ( 12 x ) m>s, v = ( 18 y2 ) m>s for 0 … t 6 5 s and by u = ( - 14 x2 ) m>s, v = ( 14 y ) m>s for 5 6 t … 10 s, where x and y are in meters. Plot the streamline and pathline for a particle released at point (1 m, 1 m) when t = 0 s.

Solution Using the definition of velocity, for 0 … t 6 5 s, dx dx 1 = u; = x dt dt 2 x

t

dx 1 = dt L1 m x L0 2 1 t 2

ln x =

1

1

When t = 5 s, x = e 2 (5) = 12.18 m

x = ae 2 t b m

(1)

dy dy 1 = v; = y2 dt dt 8 y

dy

L1 m y

2

t

=

1 dt L0 8

1 y 1 - a b` = t y 1m 8 1 -

1 1 = t y 8

y - 1 1 = t y 8 y a1 -

When t = 5 s, y =

y = a

8 = 2.667 m 8 - 5

1 tb = 1 8 8 b m t ≠ 8 s 8 - t

(2)

The equation of the streamline and pathline can be determined by eliminating t from Eqs. (1) and (2)

x(m)

1

3

5

y(m)

1

1.38

1.67

y = a

8 bm 8 - 2 ln x

7

9

11

12.18

1.95

2.22

2.50

2.67

For 5 s < t … 10 s, dx dx 1 = u; = - x2 dt dt 4 x

t

dx 1 = dt 2 4 L5 s L12.18 m x

293


3–26. (continued)

- a

x = a

When t = 10 s, x =

1 1 1 b = - (t - 5) x 12.18 4

1 t = - 1.1679 x 4

4 b m t ≠ 4.6717 s t - 4.6717

(3)

4 = 0.751 m 10 - 4.6717

dy dy 1 = v; = y dt dt 4 y

t dy 1 = dt 4 L5 s L2.667 m y

y 1 = (t - 5) 2.667 4 y 1 = e 4 (t - 5) 2.667

ln

1

y = c 2.667e 4 (t - 5) d m

1

(4)

When t = 10 s, y = 2.667e 4 (10 - 5) = 9.31 m Eliminate t from Eqs. (3) and (4),

y = 2.667e 4 34( x + 1.1679) - 54 1

1

= c 2.667e ( x - 0.08208) d m 1

x(m)

0.751

1

3

y(m)

9.31

6.68

3.43

5

7

9

11

12.18

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. y(m) 10 C 9 8 7 6 5 4 3 2 1 A 0

B

1 2 3 4 5 6 7 8 9 10 11 12 13 0.751 12.18

Ans: x(m)

For 0 … t 6 5 s, y =

8 8 - 2 ln x

For 5 s 6 t … 10 s, y = 2.67e 11>x - 0.0821) 294


3–27. A two-dimensional flow field for a liquid can be described by V = 3( 6y2 - 1 )i + (3x + 2) j4 m>s, where x and y are in meters. Determine a streamline that passes through points 16 m, 2 m2 and determine the velocity at this point. Sketch the velocity on the streamline.

Solution We have steady flow since the velocity does not depend upon time.

y

u = 6y2 - 1

20 m s

v = 3x + 2 dy v 3x + 2 = = 2 dx u 6y - 1 L2

y

( 6y2 - 1 ) dy =

L6

2

(3x + 2)dx 6

2y3 - y ` = 1.5 x2 + 2x ` 2

2y - y -

3 2(2)

3

23 m/s

x

y

3

30.5 m s

- 2 4 = 1.5x + 2x 2

2y3 - 1.5x2 - y - 2x + 52 = 0

x

x 6

3 1.5(6)2

+ 2(6) 4

Ans.

At (6 m, 2 m) u = 6(2)2 - 1 = 23 m>s S v = 3(6) + 2 = 20 m>s c V = 2 ( 23 m>s ) 2 + ( 20 m>s ) 2 = 30.5 m>s

Ans.

Ans: 2y3 - 1.5x2 - y - 2x + 52 = 0 V = 30.5 m>s 295


*3–28. A flow field for a liquid can be described by V = 5 (2x + 1) i - y j 6 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.

Solution We have steady flow since the velocity does not depend upon time.

y

u = 2x + 1 v = -y

1

dy -y v = = dx u 2x + 1 -

3

dy dx = L y L (2x + 1)

- ln y =

1 ln (2x + 1) + C 2 1

- y = (2x + 1)2 + C ′ 1

- 1 = (2(3) + 1)2 + C ′ C ′ = - 3.65 y

1

- y ` = (2x + 1) 2 ` 1

1 2

x 3

- y + 1 = (2x + 1) -

3 2(3) 1

y = 3.65 - (2x + 1)2

+ 1 42 1

Ans.

u = 2(3) + 1 = 7 m>s v = - 1 m>s V = 2 ( 7 m>s ) 2 +

7 m/s 1m s

( - 1 m>s ) 2 = 7.07 m>s

296

Ans.

7.07 m/s x


3–29. Air flows uniformly through the center of a horizontal duct with a velocity of V = ( 6t 2 + 5 ) m>s, where t is in seconds. Determine the acceleration of the flow when t = 2 s.

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 = 0. 0x 0V 0V + u 0t 0x = 12t + 0

a =

= (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.

Ans: a = 24 m>s2 297


24 in.

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. x

Solution Since the flow is along the x axis, v = w = 0 a =

0u 0u + u 0t 0x

= 4x + (4xt)(4t) = 4x + 16xt 2 = When t = 2 s, x = 16 in. Then a =

3 4(16) 3 1

3 4x ( 1

+ 4t 2 ) 4 in.>s2

+ 4 ( 22 ) 4 4 in.>s2 = 1088 in.>s2

Note: The flow is unsteady since its velocity is a function of time.

Ans: 1088 in.>s2 298


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 acceleration of a particle passing through the point (1 ft, 2 ft), when t = 1 s.

Solution For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar component of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 1 + (6y + t)(0) + (2tx)(6)

= (1 + 12tx) ft>s2

ay =

0v 0v 0v + u + v 0t 0x 0y

= 2x + (6y + t)(2t) + (2tx)(0)

= ( 2x + 12ty + 2t 2 ) ft>s2

When t = 1 s, x = 1 ft and y = 2 ft, then

ax = 31 + 12(1)(1) 4 = 13 ft>s2

ay =

3 2(1)

+ 12(1)(2) + 2 ( 12 ) 4 = 28 ft>s2

Thus, the magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 13 ft>s2 ) 2 + ( 28 ft>s2 ) 2 = 30.9 ft>s2

Ans.

Ans: 30.9 ft>s2 299


*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.

A x 0.6 m

Solution a =

0u 0u + u 0t 0x

0u 0u = 200 = 20 x 0t 0x a =

3 200

When t = 0.01 s, x = 0.6 m. a =

5 200

+

+ ( 10x2 + 200t + 6 ) (20x) 4 m>s2

3 10 ( 0.62 )

= 339 m>s2

+ 200(0.01) + 6 4 320(0.6) 4 6 m>s2

300

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).

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 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 - 2y2 + y ) (4x) + (y + xy)( - 4y + 1) At x = 2 m, y = 4 m,

ax = - 340 m>s2

ay =

0v 0v 0v + u + v 0t 0x 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 = 2ax2 + ay2 = 2 ( - 340 m>s2 ) 2 +

( - 44 m>s2 ) 2 = 343 m>s2

Ans.

Ans: V = 23.3 m>s a = 343 m>s2 301


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).

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

16 m>s v b = 79.4° uv = tan-1a b = tan-1a u 3 m>s

Ans.

For two dimensional flow, the Eulerian description is a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axis ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 5y2 - x ) ( - 1) + 4x2(10y)

= ( x - 5y2 ) + 40x2y

ay =

0v 0v 0v + u + v 0t 0x 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) 3 5 ( 12 ) - 2 4 = 48 m>s2

The magnitude of the acceleration is

a = 2ax2 + ay2 = 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

Ans.

¢ = 17.0°

302

Ans: V = 16.3 m>s uv = 79.4° a a = 164 m>s2 ua = 17.0° a


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.

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

4

dy dy v 4x - 1 = ; = dx u dx 5y2 y

y(m)

3

x

L1 m

5y2dy =

L1 m

(4x - 1)dx

2 1

1 ( 6x2 - 3x + 2 ) where x is in m 5 For two dimensional flow, the Eulerian description is y3 =

a =

ay = 20 m/s2 a = 36.1 m/s2

ax = 30 m/s2 0

0V 0V 0V + u + v 0t 0x 0y

1

2

3 (a)

4

5

x(m)

Writing the scalar components of this equation along x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 5y2(0) + (4x - 1)(10y) = 40xy - 10y

ay =

0v 0v 0v + u + v 0t 0x 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 = 2ax2 + ay2 = 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

Ans.

¢ = 33.7°

The plot of the streamline and the acceleration on point (1 m, 1 m) is shown in Fig. a. x(m) y(m)

0 0.737

0.5 0.737

1 1

2 1.59

3 2.11

4 2.58

5 3.01

Ans: a = 36.1 m>s2 u = 33.7° a 303


*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.

B A

VA 8 m/s

VB 2 m/s

x 3m

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 + a

VB - VA 2 - 8 bx = 8 + a bx = (8 - 2x) m>s LAB 3

Ans.

For one dimensional flow, the Eulerian description gives a =

0V 0V + V dt dx

= 0 + (8 - 2x)( - 2)

= 4(x - 4) m/s2

Ans.

using the definition of velocity, x t dx dx = = V = 8 - 2x; dt dt L0 8 - 2x L0 -

x 1 ln(8 - 2x) ` = t 2 0

1 8 ln a b = t 2 8 - 2x ln a

8 b = 2t 8 - 2x

8 = e2 t 8 - 2x

x = 4 ( 1 - e -2 t ) m

Ans.

304


3–37. A fluid has velocity components of u = ( 8t 2 ) m>s and v = (7y + 3x) m>s, where x and y are in meters and t is in seconds. Determine the velocity and acceleration of a particle passing through point x = 1 m, y = 1 m when t = 2 s.

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

10 m>s v b = 17.4° uv = tan-1a b = tan-1a u 32 m>s

Ans.

uv

For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 16t + 8t 2(0) + (7y + 3x)(0)

= (16t) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 8t 2 ) (3) + (7y + 3x)(7)

=

3 24t 2

When t = 2 s, x = 1 m and y = 1 m.

ax = 16(2) = 32 m>s2

+ 7(7y + 3x) 4 m>s2

ay = 24 ( 22 ) + 737(1) + 3(1) 4 = 166 m>s2


a = 2ax2 + ay2 = 2 ( 32 m>s2 ) 2 + ( 166 m>s2 ) 2 = 169 m>s2

Ans.

Its direction is

ua = tan-1a

ay ax

b = tan-1a

166 m>s2 32 m>s2

b = 79.1°

Ans.

ua

Ans: V = 33.5 m>s uV = 17.4° a = 169 m>s2 ua = 79.1° a 305


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?

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,

y

dy dy 8y y v = ; = = dx u dx 8x x

streamline v = (8y) ft/s

y

x dy dx = L1 ft y L2 ft x

ln y `

y 1 ft

= ln x `

y =

u = (8x) ft/s x

y

x

x

2 ft

x ln y = ln 2

V

(a)

1 x 2

Ans.


0V 0V 0V + u + v 0t 0x 0y


0u 0u 0u + u + v 0t 0x 0y

= 0 + 8x(8) + 8y(0)

= (64x) ft>s2 ay =

0v 0v 0v + u + v 0t 0x 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 = 2ax2 + ay2 = 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°

Ans.

u

Ans: y = x>2, a = 143 ft>s2 u = 26.6° a 306


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?

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,

y v = (8xy) m/s

dy dy 8xy 4x v = ; = = dx u dx y xy2 y

L2 m y2 y 2

`

u = (2y2) m/s

x

y dy =

2m

L1 m

streamline

4x dx

= 2x2 `

V

y

x 1m

x

y2 - 2 = 2x2 - 2 2

x (a)

y2 = 4x2 Ans.

y = 2x (Note that x = 1, y = 2 is not a solution to y = -2x.) For two dimensional flow, the Eulerian description gives. a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 2y2(0) + 8xy(4y)

= ( 32xy2 ) m>s2

ay =

0v 0v 0v + u + v 0t 0x 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 =

3 16 ( 23 )


+ 64 ( 12 ) (2) 4 = 256 m>s2

a = 2ax2 + ay2 = 2 ( 128 m>s2 ) 2 + ( 256 m>s2 ) 2 = 286 m>s2

Ans.

Its direction is

u = tan-1a

ay ax

b = tan-1a

256 m>s2 128 m>s2

b = 63.4°

Ans.

u

307

Ans: y = 2x a = 286 m>s2 u = 63.4° a


*3–40. The velocity of a flow field is defined by V = 5 4 yi + 2 xj 6 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.

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 0V 0V 0V + u + v 0t 0x 0y

a =

Writing the scalar components of this equation along the x and y axes 0u 0u 0u + u + v 0t 0x 0y

ax =

= 0 + 4y(0) + (2x)(4) = (8x) m>s2 0v 0v 0v + u + v 0t 0x 0y

ay =

= 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 = 2ax2 + ay2

= 2 ( 16 m>s ) 2 + ( 8 m>s ) 2 = 17.9 m>s2

Using the definition of the slope of the streamline, dy dy v 2x x = ; = = dx u dx 4y 2y y

x

L1 m

2y dy =

y2 `

y 1m

=

y2 - 1 =

y2 =

x dx L2 m x2 x ` 2 2m

x2 - 2 2 1 2 x - 1 2 308

Ans.


*3–40. (continued)

The plot of this streamline is shown is Fig. a x(m) y(m)

22 0

2

3

4

5

6

±1

±1.87

±2.65

±3.39

±4.12

y(m)

y(m)

4

4 V = 5.66 m/s

3

3

v = 4 m/s

2

2 45º

1

0

ay = 8 m/s2

1

u = 4 m/s 1

2

3

4

5

a = 17.9 m/s2

ax = 16 m/s2 6

x(m)

0

–1

–1

–2

–2

–3

–3

–4

–4

(a)

309

1

2

3

4

5

6

x(m)


3–41. The velocity of a flow field is defined by V = 54 xi + 2j6 m>s, where x is in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point (1 m, 2 m). Find the equation of the streamline passing through this point, and sketch these vectors on this streamline.

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

y(m)

y(m)

v = 2 m>s The magnitude of velocity is V = 2u + v = 2 ( 4 m>s ) + ( 2 m>s ) = 4.47 m>s 2

2

2

2

Ans.

For two dimensional flow, the Eulerian description gives 0V 0V 0V + u + v 0t 0x 0y

a =

V = 4.47 m/s

v = 2 m/s

3

3

2

2

u = 4 m/s

1

Writing the scalar components of this equation along the x and y axes

1

0

1

2

3

4

x(m)

5

0u 0u 0u ax = + u + v 0t 0x 0y = 0 + 4x(4) + 2(0) = 16x y(m) 0v

ay =

0t

+ u

y(m)

0v 0v + v 0x 0y= 4.47 m/s V

v = 2 m/s

3

=30 + 4x(0) + 2(0) = 0 At point (1 m, 2 m),

2

a = 16 m/s2

2

u = 4 m/s

ax = 16(1) = 16 m>s2 ay = 0

1

1

Thus, the magnitude of the acceleration is 2 a0 = ax 1= 16 m>s 3 2

4

5

x(m)

Ans.

0

1

2

3

4

5

x(m)

Using the definition of the slope of the streamline,

dy dy 2 1 v = ; = = dx u dx 4x 2x y

x

1 dx dy = 2 L1 m x L2 m y - 2 =

1 ln x 2

1 y = a ln x + 2b 2

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 Ans: V = 4.47 m>s, a = 16 m>s2 1 y = ln x + 2 2 310

0

1


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.

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.


0V 0V 0V + u + v 0t 0x 0y


0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 - y2 ) (4x) + ( - 4xy)( - 2y) = 4x ( 2x2 - y2 ) + 8xy2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x2 - y2 ) ( - 4y) + ( - 4xy)( - 4x) = - 4y ( 2x2 - y2 ) + 16x2y At point (1 m, 1 m), ax = 4(1) 3 2 ( 12 ) - 12 4 + 8(1) ( 12 ) = 12 m>s2

ay = - 4(1) 3 2 ( 12 ) - 12 4 + 16 ( 12 ) (1) = 12 m>s2


a = 2ax2 + ay2 = 2 ( 12 m/s2 ) 2 + ( 12 m/s2 ) 2 = 17.0 m>s2

Ans.

Using the definition of the slope of the streamline, dy dy 4xy v = ; = - 2 dx u dx 2x - y2

( 2x2 - y2 ) dy = - 4xydx 2x2dy + 4xydx - y2dy = 0 However, d ( 2x2y ) = 2 ( 2xydx + x2dy ) = 2x2dy + 4xydx. Then d ( 2x2y ) - y2dy = 0

311


3–42. (continued)

Integrating this equation, 2x2y -

y3

= C

3

with the condition y = 1 m when x = 1 m, 2 ( 12 ) (1) C =

13 = C 3 y(m)

5 3

y(m)

3.0

3.0

2.5

2.5

2.0

2.0 ay =

Thus, 2x2y -

y3 3

=

5 3

6x2y - y3 = 5 x2 =

1.5

y3 + 5 6y

1.36 1.0

Taking the derivative of this equation with respect to y

1.0 v = 4 m/s

6y ( 3y2 ) - ( y3 + 5 ) (6) 2y3 - 5 dx = = 2x dy 6y2 ( 6y ) 2

0

0.5

0.5

y(m) 2y3 - 5 = 0

1.5

2.0

x(m)

y(m)

y = 1.357 3.0 m

3.0

2.5

2.5

x2 = 1.3573 + 5

a = 17.0 m/s2

2 2.0 ay = 12 m/s

x = 0.960 2.0m 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 1.5 x(m) 1.83 1.31 1.10 1.00 0.963 0.965 0.993 1.04 1.10 u1.17 1.25 1.33 = 1 m/s

1.5

1.36 1.0

1.0 v = 4 m/s

ax = 12 m/s2

V = 4.12 m/s

0.5

0

1.0 0.960

dx = 0; dy

The corresponding x is

V = 4.12 m/s

0.5

2y3 - 5 dx = dy 12xy2 Set

1.5

u = 1 m/s

Ans.

0.5

0.5

1.0

1.5

0.960

2.0

x(m)

0

0.5

1.0

1.5

2.0

x(m)

Ans: V = 4.12 m>s a = 17.0 m>s2 y3 + 5 x2 = 6y 312

0


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.

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, 2m) -y 2 = - = - 0.5 m>s 4 4

u =

x 3 = = 0.3333 m>s 9 9

v = The magnitude of the velocity is

V = 2u2 + v2 = 2 ( - 0.5 m>s ) 2 + ( 0.333m>s ) 2 = 0.601 m>s

Ans.


0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y are + 2 ax = 0u + u 0u + v 0u 1S 0t 0x 0y = 0 + a

= a-

( + c ) ay =

-y x 1 b(0) + a ba - b 4 9 4

1 xb m>s2 36

0v 0v 0v + u + v 0t 0y 0y

= 0 + a = c-

-y 1 x ba b + a b(0) 4 9 9

1 y d m>s2 36

At point (3 m, 2 m),

ax = -

1 (3) = - 0.08333 m>s2 36

ay = -

1 (2) = - 0.05556 m>s2 36


a = 2ax2 + ay2

= 2 (- 0.08333 m>s2 ) 2 + (- 0.05556 m>s2 ) = 0.100 m>s2

313

Ans.


3–43. (continued)

Using the definition of slope of the streamline, x>9 dy dy v 4x = ; = = dx u dx - y>4 9y y

9

x

L2 m

ydy = - 4

xdx L3 m

x 9y2 y ` = - ( 2x2 ) ` 2 2m 3m

9y2 - 18 = - 2x2 + 18 2 9y2 + 4x2 = 72 y2 x2 + = 1 72>4 72>9 y2 x2 + = 1 (4.24)2 (2.83)2

Ans.

This is an equation of an ellipse with center at (0, 0). The plot of this streamline is shown in Fig. a

y(m) 3 2 m

y 3 2 m

ax = 0.0833 m/s2

v = 0.333 m/s

V = 0.601 m/s

ay = 0.0556 m/s2 2 2 2 m

2 u = 0.5 m/s 3

x(m)

3 a = 0.100 m/s2

2 2 m

x

(a)

314

Ans: V = 0.601 m>s a = 0.100 m>s2 x2 > 14.242 2 + y2 > 12.832 2 = 1


*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.

Solution The flow is unsteady nonuniform. For one dimensional flow, a = Here, u = (4 tx) m>s. Then

0u 0u + u 0t 0x

0u 0u = 4x and = 4t. Thus, 0t 0x

a = 4x + (4tx)(4t) = ( 4x + 16t 2x ) m>s2 Since u = 0.8 m>s when t = 0.1 s,

0.8 = 4(0.1) x

x = 2m

The position of the particle can be determined from

dx = u = 4 tx; dt

x

ln x `

t

dx t dt = 4 L2 m x L0.15 x 2m

= 2t 2 `

t 0.15

x ln = 2t 2 - 0.02 2 x 2 e 2t - 0.02 = 2

2

- 0.02

x = 2e 2t

x = 2e 2(0.8 ) - 0.02 = 7.051 m 2

Thus, t = 0.8 s, a = 4(7.051) + 16 ( 0.82 ) (7.051)

= 100.40 m>s2

= 100 m>s2

Ans.

315


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.

Solution The flow is unsteady nonuniform. For three dimensional flow, a =

0V 0V 0V 0V + u + v + w 0t 0t 0t 0t

Thus,

0u 0u 0u 0u + u + v + w 0t 0x 0y 0t

ax =

= 0 + 2x(2) + (6tx)(0) + 3y(0)

= (4x) m>s2

0v 0v 0v 0v + u + v + 0t 0x 0y 0t

ay =

= 6x + 2x(6t) + 6tx(0) + 3y(0)

= (6x + 12tx) m>s2

0w 0w 0w 0w + u + v + w 0t 0x 0y 0z

az =

= 0 + 2x(0) + 6tx(3) + 3y(0)

= (18tx) m>s2

The position of the particle can be determined from

x

ln x = 2t x = ( e 2t ) m

t

dx = 2 dt L1 m x L0

dx = u = 2x; dt

dy = v = 6tx = 6te 2t ; dt

L0

y

dy = 6

y = y =

L0

t

te 2tdt

t 3 ( 2te 2t - e 2t ) ` 2 0

3 ( 2te 2t - e 2t + 1 ) 2

dz 9 = w = 3y = 3 2te 2t - e 2t + 1 4 ; dt 2 L0

t

z

dz =

z =

9 ( 2te 2t - e 2t + 1 ) dt 2 L0

t 9 2t 1 1 c te - e 2t - e 2t + t d ` 2 2 2 0

z =

9 2t ( te - e 2t + t + 1 ) m 2

316


3–45. (continued)

When, t = 0.5 s,

x = e 2(0.5) = 2.7183 m = 2.72 m

y =

Thus, z =

Ans.

3 3 2(0.5)e 2(0.5) - e 2(0.5) + 14 = 1.5 m 2

Ans.

9 3 0.5e 2(0.5) - e 2(0.5) + 0.5 + 14 = 0.6339 m = 0.634 m 2

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

Ans.

Then a = 5 10.9i + 32.6j + 24.5k6 m>s2

Ans.

317

Ans: x = 2.72 m y = 1.5 m z = 0.634 m a = 510.9i + 32.6j + 24.5k6 m>s2


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.

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 10y + 3 = dx - (4x + 6)

dy v = ; dx u y

x dy dx = 10y + 3 4x + 6 L1 m L1 m y x 1 1 ln(10y + 3) ` = - ln(4x + 6) ` 10 4 1m 1m

10y + 3 1 1 10 lna b = lna b 10 13 4 4x + 6 1

ln a

1

4 10y + 3 10 10 b = ln a b 13 4x + 6

a

1

1

4 10y + 3 10 10 b = a b 13 4x + 6

5

2 10y + 3 10 b = a 13 4x + 6

y = c

411

Ans.

- 0.3 d m (4x + 6)5>2 For two dimensional flow, the Eulerian description gives

dV dV dV + u + v dt dx dy Writing the scalar components of this equation along the x and y axes, a =

At point (1m, 1m),

ax =

du du du + u + v dt dx dy

= 0 + 3 - (4x + 6)( - 4) 4 + (10y + 3)(0) = 34(4x + 6) 4 m>s2

ay =

dv dv dv + u + v dt dx dy

= 0 + 3 - (4x + 6)(0) 4 + (10y + 3)(10) = 310(10y + 3) 4 m>s2 ax = 434(1) + 64 = 40 m>s2 S ay = 10310(1) + 34 = 130 m>s2 c

318


3–46. (continued)

The magnitude of acceleration is

a = 2ax2 + ay2 = 2 ( 40m>s2 ) 2 + ( 130m>s2 ) 2 = 136 m>s2

Ans.


u = tan-1 a

ay ax

b = tan-1 a

130 m>s2 40 m>s2

b = 72.9°

Ans.

u

Ans: y =

411 14x + 62 5>2

a = 136 m>s2 u = 72.9° a 319

- 0.3


3–47. A velocity field for oil is defined by u = (100y) m>s, v = ( 0.03 t 2 ) 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.

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 = v = 0.03t 2; dt

L0

y

t

t 2 dt L0 y = ( 0.01t 3 ) m

dy = 0.03

When t = 0.5 s,

y = 0.01 ( 0.53 ) = 0.00125 m = 1.25 mm

dx = u = 100y = 100 ( 0.01t 3 ) = t 3; dt

L0

y

dx =

Ans. L0

t

t 3 dt

1 x = a t4b m 4

When t = 0.5 s,

x =

1 ( 0.54 ) = 0.015625 m = 15.6 mm 4

Ans.

For a two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Write the scalar components of this equation along the x and y axes,

ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + (100y)(0) + ( 0.03t 2 ) (100)

= ( 3t 2 ) m>s2

ay =

0v 0v 0v + u + v 0t 0x 0y

= 0.06t + (100y)(0) + 0.03t 2(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>s2 c

The magnitude of acceleration is

a = 2ax2 + ay2 = 2 ( 0.75 m>s2 ) 2 + ( 0.03 m>s2 ) 2 = 0.751 m>s2

Ans.


u = tan-1 a

ay ax

b = tan-1a

0.03 m>s2 0.75 m>s2

b = 2.29°

Ans.

u

320

Ans: y = 1.25 mm x = 15.6 mm a = 0.751 m>s2 u = 2.29° a


*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.

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 -y = dx 2x2

dy v = ; dx u

y

x dy 1 dx = 2 L2 m x2 L6 m y

ln y `

y 6m

ln

ln

=

1 1 x a b` 2 x 2m

y 1 1 1 = a - b 6 2 x 2 y 2 - x = 6 4x y 2-x = e 1 4x 2 6

y = c 6e 1 4x 2 d m 2-x

Ans.

The plot of this streamline is shown in Fig. a. For two dimensional flow, the Eulerian description gives. a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes,

ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 ) (4x) + ( - y)(0)

= ( 8x3 ) m>s2

ay =

dv dv dv + u + v dt dx dy

= 0 + ( 2x2 ) (0) + ( - y)( - 1)

= (y)m>s2

At point (2 m, 6 m), ax = 8 ( 23 ) = 64 m>s2 S ay = 6m>s2 c

The magnitude of acceleration is a = 2ax2 + ay2 = 2 ( 64 m>s2 ) 2 + ( 6m>s2 ) 2 = 64.3m>s2

Ans.


u = tan-1a

ay ax

b = tan-1a

6 m>s2 64 m>s2

b = 5.36°

Ans.

u

321


3–48. (continued)

Using the definition of the velocity,

y(m)

dx dx = u; = 2x2 dt dt x

7

t

dx = dt 2 L2 m 2x L0

6

1 1 x - a b` = t 2 x 2m

5 4

1 1 1 - a - b = t 2 x 2

3

x - 2 = t 4x

x = a

2

2 bm 1 - 4t

1

dy dy = v; = -y dt dt y

-

ln

y 6m

= t

6 = t y

6 = et y y = ( 6e -t ) m

Thus, Then,

u = 2x2 = 2 a

ax =

ay =

1

2

3

4 (a)

t dy = dt y L0

L6 m

- ln y `

0

2 2 8 b = c d m>s and v = - y = 1 - 4t ( 1 - 4t ) 2

( - 6e -t ) m>s

64 du = -16(1 - 4t)-3( -4) = c d m>s2 dt (1 - 4t)3 dv = ( 6e -t ) m>s2 dt

Ans.

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

322

5

6

x(m)


3–49. Air flow through the duct is defined by the velocity field u = ( 2x2 + 8 ) m>s v = ( - 8x) m>s where x is in meters. Determine the acceleration of a fluid particle at the origin (0, 0) and at point (1 m, 0). Also, sketch the streamlines that pass through these points.

y

x

x1m

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 0V 0V 0V + u + v 0t 0x 0y

a =

Writing the scalar components of this equation along the x and y axes ax =

=

= ay =

0u 0u 0u + u + v 0t 0x 0y

30

+ ( 2x2 + 8 ) (4x) + ( -8x)(0) 4

3 4x ( 2x2

+ 8 ) 4 m>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x2 + 8 ) ( - 8) + ( - 8x)(0)

=

At point (0, 0),

3 - 8 ( 2x2

ax = 4(0) 3 2 ( 02 ) + 8 4 = 0

+ 8 ) 4 m>s2

ay = - 8 3 2 ( 02 ) + 8 4 = - 64 m>s2 = 64 m>s2 T

Thus,

a = ay = 64 m>s2 T

Ans.

At point (1 m, 0), ax = 4(1) 3 2 ( 12 ) + 8 4 = 40 m>s2 S

ay = - 8 3 2 ( 12 ) + 8 4 = - 80 m>s2 = 80 m>s2 T The magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 40 m>s2 ) 2 + ( 80 m>s2 ) 2 = 89.4 m>s2

Ans.


u = tan-1 a

ay ax

b = tan-1 a

80 m>s2 40 m>s2

b = 63.4° cu

Ans.

Using the definition of the slope of the streamline,

dy v - 8x x dx = = ; dy = - 8 2 2 dx u 2x + 8 L L 2x + 8 y = - 2 ln ( 2x2 + 8 ) + C

323


3–49. (continued)

For the streamline passing through point (0, 0), 0 = - 2 ln3 2 ( 02 ) + 8 4 + C

Then y = c 2 ln a

C = 2 ln 8

8 b d m 2x2 + 8

Ans.

For the streamline passing through point (1 m, 0), 0 = - 2 ln3 2 ( 12 ) + 8 4 + C

y = c 2 ln a

For point (0, 0)

C = 2 ln 10 10 b d m 2x2 + 8

Ans.

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

y(m)

1

–4

–3

–2

2

3

4

–1

y = 2 ln

5

x(m)

10 2x2 + 8

–2 –3 y = – 2 ln

–5

1 0

–1

8 2x2 + 8

–4

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), 8 bd m 2x2 + 8 For the streamline passing through point (1 m, 0), y = c 2 ln a y = c 2 ln a 324

10 bd m 2x2 + 8


3–50. The velocity field for a fluid is defined by u = y> ( x2 + y2 ) and v = 3 4x> ( x2 + y2 ) 4 m>s, where x and y are in meters. Determine the acceleration of particles located at point (2 m, 0) and that of a particle located at point (4 m, 0). Sketch the equations that define the streamlines that pass through these points.

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 0V 0V + u + v 0t 0x 0y

Write the scalar components of this equation along x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + a = £

ay =

y 2

2

x + y

4x3 - 6xy2

( x2 + y2 ) 3

b£

( x2 + y2 ) (0) - y(2x) ( x2 + y2 ) (1) - y(2y) 4x § + a 2 b£ § 2 2 2 2 x + y (x + y ) ( x2 + y2 ) 2

§ m>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + a

= £

y 2

b£

2

x + y

4y3 - 36x2y

( x2 + y2 ) 3

( x2 + y2 ) (4) - 4x(2x) ( x2 + y2 ) (0) - 4x(2y) 4x § + a 2 b£ § 2 2 2 2 x + y (x + y ) ( x2 + y2 ) 2

§ m>s2

At point (2 m, 0) ax =

4 ( 23 ) - 6(2) ( 02 )

ay =

( 22 + 02 ) 3

= 0.5 m>s2 S

4 ( 03 ) - 36 ( 22 ) (0)

( 22 + 02 ) 3

= 0

Thus a = ax = 0.5 m>s2 S

Ans.

325


3–50. (continued)

At point (4 m, 0) ax =

ay =

4 ( 43 ) - 6(4)(0)

( 42 + 02 ) 3

= 0.0625 m>s2 S

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, 4x> ( x2 + y2 ) dy v 4x = = = ; ydy = 4 xdx 2 2 dx u y y> ( x + y ) L L y2 = 2x2 + C′ 2 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

326


3–50. (continued)

For the streamline passing through point (2 m, 0) x(m) y(m)

2 0

3 ±4.47

4 ±6.93

5 ±9.17

6 ±11.31

7 ±13.42

8 ±15.49

9 ±17.55

For the streamline passing through point (4 m, 0) x(m) y(m)

4 0

5 ±6.00

6 ±8.94

7 ±11.49

8 ±13.86

9 ±16.12

y(m) 20 15

y = 4x2 – 16

10 5

0

1

2

3

4

5

6

7

8

9

x(m)

–5 y = 4x2 – 64 –10 –15 –20

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 - 64 327


3–51. As the value is closed, oil flows through the nozzle such that along the center streamline it has a velocity of V = 6 ( 1 + 0.4x2 ) (1 - 0.5t) m>s where x is in meters and t is on seconds. Determine the acceleration of an oil particle at x = 0.25 m when t = 1 s.

A

B

x 0.3 m

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 ax = =

=

0u 0u + u 0t 0x 0 0x

3 6( 1

3 6( 1

+ 0.4x2 ) (1 - 0.5t) 4 +

+ 0.4x2 ) (0 - 0.5) 4 +

3 6( 1

3 6( 1

Evaluating this expression at x = 0.25 m, t = 1 s, we get

+ 0.4x2 ) (1 - 0.5t) 4

0 3 6 ( 1 + 0.4x2 ) (1 - 0.5t) 4 0x

+ 0.4x2 )( 1 - 0.5t ) 4 3 6(0 + 0.4(2x))(1 - 0.5t) 4

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.

Ans: - 1.23 m>s2 328


*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

Solution

n

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

A

an streamline

as = 3 m>s2 Since the streamline does not rotate, the local acceleration along the n axis is zero, 0V so that a b = 0. Therefore, the component of the particle’s acceleration along 0t n the n axes is an = a

0V V2 b + 0t n R

= 0 +

( 5 m>s ) 2

16 m Thus, the magnitude of the particle’s acceleration is

a = 2as2 + an2 = 2 ( 3 m>s2

= 1.5625 m>s2

) 2 + ( 1.5625 m>s2 ) 2

= 3.38 m>s2

Ans.

329

as S (a)

16 m


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. s

r 0.5 m u

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 3 = ab m>s r 1 - s

This is one dimensional steady flow since the velocity is along the straight radial line. The Eulerian description gives a =

0V 0V + V 0t 0s

= 0 + a= c

3 3 bcd 1 - s (1 - s)2

9 d m>s2 (1 - s)3

When 1 - s = r = 0.5 m, this equation gives

a = a

9 b m>s2 = 72 m>s2 0.53

Ans.

The positive sign indicates that a is directed towards positive s. Note there is no normal component for motion along a straightline. s r

r0 = 1 m (a)

Ans: 72 m>s2 330


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

y

V = 2 ( 4 m>s ) 2 +

a = 2 ( 2 m>s2 ) 2 + ( 8 m>s2 ) 2 = 8.246 m>s2

a = 2i + 8j

u = tan-1

( - 3 m>s ) 2 = 5 m>s

n

3 m/s

ûs

u s = cos 36.870°i - sin 36.870°j S

= 0. 8i - 0.6j

as = a # us = (2i + 8j) # (0.8i - 0.6j)

as = - 3.20 m>s2

as = 3.20 m>s2

a = 2as2 + an2

x

V

3 = 36.870° 4

4 m/s

Ans.

(8.246)2 = (3.20)2 + an2

an = 7.60 m>s2

Ans.

Ans: as = 3.20 m>s2 an = 7.60 m>s2 331


3–55. A particle moves along the circular streamline, such that it has a velocity of 2 m>s, which is increasing at 3 m>s2. Determine the acceleration of the particle, and show the acceleration on the streamline.

3 m/s

4m

Solution The normal component of the acceleration is an =

V2 = r

( 3 m>s ) 2 4m

= 36.9º

= 2.25 m>s2 4m

Thus, the magnitude of the acceleration is

a = 2as2 + an2 = 2 ( 3 m>s2 ) 2 + ( 2.25 m>s2 ) 2 = 3.75 m>s2

u = tan-1a

2.25 m>s2 an b = tan-1a b = 36.9° as 3 m>s2

a = 3.75 m/s2

Ans. (a)


as = 3 m/s2

an = 2.25 m/s2

Ans.

The plot of the acceleration on the streamline is shown in Fig. a.

Ans: a = 3.75 m>s2 u = 36.9° c 332


*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 steady motion 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=9m

Solution Using the condition at r = 3 m, V = 18 m>s . V = Then

k k ; 18 m>s = k = 54 m2 >s r 3m V = a

54 b m>s r

54 b m>s = 6 m>s . Since the velocity is constant, the streamline 9 component of acceleration is

At r = 9 m, V = a

as = 0 The normal component of acceleration is

an = a

Thus, the acceleration is

0V V2 b + = 0 + 0t n r

( 6 m>s ) 2 9m

= 4 m>s2

a = an = 4 m>s2

Ans.

333


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°.

V 4 m/s u 0.5 m

Solution The streamline component of acceleration can be determined from as = a

0V 0V b + V 0t s 0s

However, s = ru. Thus, 0 s = r 0 u = 0.5 0 u. Also, the flow is steady, a 0V 0V 0V = = 2 = 2(16 cos u) = 32 cos u. Then 0s 0.5 0 u 0u

0V b = 0 and 0t s

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 ( 8 m>s ) 2 0V V2 an = a b + = 0 + = 128 m>s2 0t n R 0.5 m

Ans.

Ans: as = 222 m>s2 an = 128 m>s2 334


3–58. Fluid particles have velocity components of u = (8y), v = (6x) m>s, where x and y are in meters. Determine the streamline and the normal component of acceleration of a particle located at point (1 m, 2 m).

Solution

us

6 m/s

At x = 1 m, y = 2 m u = 8(2) = 16 m>s

16 m/s

v = 6(1) = 6 m>s 6 = 20.55° 16 u s = cos 20.55°i + sin 20.55°j u = tan-1

= 0.9363i + 0.3511j Acceleration. With w = 0, we have

ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 8y(0) + 6x(8) = 48x

ay =

0v 0v 0v + u + v 0t 0x 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 And its magnitude is

a = 5 48i + 96j 6 m>s2

a = 2ax2 + ay2 = 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 + 96j 4 # 30.9363i + 0.3511j 4 = 78.65 m>s2 = 78.7 m>s2

Ans.

The normal component of the particle’s acceleration is an = 2a2 - as2 = 2 ( 107.33 m>s2 ) 2 - ( 78.65 m>s2 ) 2 = 73.0 m>s2

Ans.

Ans: as = 78.7 m>s2 an = 73.0 m>s2 335


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.

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 0V 0V 0V a = + u + v 0t 0x 0y Writing the scalar components of this equation along the x and y axes, 0u 0u 0u + u + v 0t 0x 0y

ax =

= 0 + 8y(0) + 6x(8) = 48x 0v 0v 0v + u + v 0t 0x 0y

ay =

= 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 = 2ax2 + ay2 = 2 ( 48 m>s2 ) 2 + ( 48 m>s2 ) 2 = 67.9 m>s2

Ans.

Its direction is

u = tan-1a

ay ax

b = tan-1a

48 m>s2 48 m>s2

Ans.

b = 45° a

The slope of the streamline is dy dy v 6x = ; = dx u dx 8y y

x

L1 m

8ydy =

4y2 `

2

(1)

y 1m 2

6xdx L1 m

= 3x2 `

x 1m

Ans.

4y - 3x = 1

Ans: a = 67.9 m>s2 u = 45° a 4y2 - 3x2 = 1 336


*3–60. A fluid has velocity components of u = ( 2y2 ) m>s and v = (8 x y) 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).

Solution dy 8xy r dy 4x = ; = = 2 dx u dx y 2y

y

x

L2 m

y dy =

4x dx L1 m

x y2 y ` = 2x2 ` 2 2m 1m

y2 - 2 = 2x2 - 2 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,

V2 = 0 R 0u 0u 0u + u + v ax = 0t 0x 0y

Ans.

an =

= 0 + 2y2(0) + (8xy)(4y) = ( 32xy2 ) m>s2

ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 2y2(8y) + (8xy)(8x)

= ( 16y3 + 64x2y ) m>s2

At (1 m, 2 m),

ax = 32(1) ( 22 ) = 128 m>s2

ay =

3 16 ( 23 )

+ 64 ( 12 )( 2 ) 4 = 256 m>s2

a = as = 2ax2 + ay2 = 2 ( 128 m>s2 ) 2 + ( 256 m>s2 )

as = 286 m>s2

337

Ans.


3–61. A fluid has velocity components of u = ( 2y2 ) m>s and v = (8x y) 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 the velocity and acceleration at point.

Solution x(m) y(m)

23>2 0

1.0

2.0

3.0

4.0

5.0

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 8xy 4x = = dx y 2y2

dy v = ; dx u y

x

L1 m

y dy =

y2 2

`

y 1m

4x dx L1 m

= 2x2 `

x

y(m)

1m

4x2 - y2 = 3 where x and y are in m

Ans. 10

For two dimensional flow (w = 0), the Eulerian description is

a =

0V 0V 0V + u + v 0t 0x 0y

9

Writing the scalar components of this equation along the x and y axes,

ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2y2 ) (0) + (8xy) (4y)

= ( 32xy2 ) m>s2

0v 0v 0v ay = + u + v 0t 0x 0y

= 0 + ( 2y2 ) (8y) + (8xy) (8x)

= ( 16y3 + 64x2y ) m>s2

7 6 5

as = 85.4 m/s2

4 3 a = 18.2 m/s2

At point x = 1 m and y = 1 m

2

a = 68.2°

ax = 32(1) ( 12 ) = 32 m/s2

ay = 16 ( 13 ) + 64 ( 12 ) (1) = 80 m>s2

1 an = 11.6 m/s2

0

Thus,

8

a = 5 32i + 80j 6 m>s2

338

1

2

3 (a)

4

5

x(m)


3–61. (continued)

The magnitude of the acceleration is a = 2ax2 + ay2 = 2 ( 32 m>s2 ) 2 + ( 80 m>s2 )

= 86.16 m>s2 = 18.2 m>s2

and its direction is

At point (1 m, 1 m),

ay 80 m>s2 ua = tan-1a a b = tan-1a b = 68.2° x 32 m>s2 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 - as2 = 2 ( 86.16 m>s2 ) 2 - ( 85.37 m>s2 ) 2

= 11.64 m>s2 = 11.6 m>s2

Ans.

Ans: 4x2 - y2 = 3 as = 85.4 m>s2 an = 11.6 m>s2 339

Fluid Mechanics Chapter 3 Solution Guide

Recommend Documents