Fluid Mechanics Chapter 3 Solution Guide

267

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.

SolutionSince the streamlines have a constant direction for the time interval 0 t 6 3 s, the pathline and streakline coincide with the streamline when t = 2 s as shown in Fig. a.

The pathline and streakline will coincide with the streamline until t = 3 s, after which the streamline makes a sudden change in direction. Thus, the streamline of the marked particle and the streakline when t = 4 s will be as shown in Fig. b.

31. A marked particle is released into a flow when t = 0, and the pathline for a particle is shown. Draw the streakline, and the streamline for the particle when t = 2 s and t = 4 s.

t = 2 s

(a)

markedparticle

streamline

streakline

60

4 m

(b)

t = 4 s

markedparticle

streamline

streakline

60

4 m

6 m

t 0

t 2 s

t 3 s t 4 s

pathline

4 m

6 m

4 m

60

268


SolutionSince the streamlines have a constant direction along the positive x axis for the time interval 0 t 6 3 s, the pathline and streakline coincide with the streamline when t = 1 s as shown in Fig. a.

The pathline and streakline will coincide with the streamline until t = 3 s, after which the streamline makes a sudden change in direction. Thus, the streamline and pathline of the first marked particle and the streakline when t = 4 s will be as shown in Fig. b.

32. The flow of a liquid is originally along the positive x axis at 2 m>s for 3 s. If it then suddenly changes to 4 m>s along the positive y axis for t 7 3 s, draw the pathline and streamline for the first marked particle when t = 1 s and t = 4 s. Also, draw the streaklines at these two times.

(b)

(a)

streakline

t = 1 st = 0 2 m streamline

x

first markedparticle

pathline

t = 4 s

t = 4 s

t = 3 st = 0

pathline

streamline

streakline first markedparticle

y

x

6 m

4 m

269


SolutionSince the streamlines have a constant direction along the positive y axis, 0 t 6 4 s, the pathline and streakline coincide with the streamline when t = 2 s as shown inFig. a.

The pathline and streakline will coincide with the streamline until t = 4 s, when the streamline makes a sudden change in direction. The pathline, streamline, and streakline are shown in Fig. b.

33. The flow of a liquid is originally along the positive y axis at 3 m>s for 4 s. If it then suddenly changes to 2 m>s along the positive x axis for t 7 4 s, draw the pathline and streamline for the first marked particle when t = 2 s and t = 6 s. Also, draw the streakline at these two times.

(b)

(a)

6 m

t = 0

t = 2 s

pathline streamline

streaklinefirst marked particle

y

t = 4 s t = 6 s

t = 0

pathlinestreamline

streakline

first markedparticle

4 m

12 m

y

x

270


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

V = 5(2x + 1)i - (y + 3x)j6 m>s = [2(2) + 1]i - [3 + 3(2)]j = 55i - 9j6 m>sThe magnitude of V is

V = 2V 2x + V 2y = 2(5 m>s)2 + ( -9 m>s)2 = 10.3 m>s Ans.As indicated in Fig. a, the direction of V is defined by u = 360 - f, where

f = tan-1 aVyVx

b = tan-1 a9 m> s5 m> s b = 60.95

Thus,

u = 360 - 60.95 = 299 Ans.

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

(a)

xVx = 5 m/s

VVy = 9 m/s

3 m

2 mx

y

271


SolutionThe velocity vector of a particle at x = 5 m and y = -2 m is

V = 5(5y2 - x)i + (3x + y)j6 m>s = 35(-2)2 - 54 i + 33(5) + (-2)4j = 515i + 13j6 m>s

The magnitude of V is

V = 2V 2x + V 2y = 2(15 m>s)2 + (13 m>s)2 = 19.8 m>s Ans.As indicated in Fig. a, the direction of V is defined by

u = tan-1 aVyVx

b = tan-1 a13 m>s15 m>s b = 40.9 Ans.

35. A two-dimensional flow field for a liquid can be described by V = 3 (5y2 - x)i + (3x + y)j4 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at (5 m, -2 m), and its direction measured counterclockwise from the x axis.

(a)

x

VVy = 13 m/s

Vx = 15 m/s

Ans:V = 19.8 m>su = 40.9

272


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

V = 5(0.8x)i + (0.06t2)j6 m>s = 30.8(2)i + 0.06(5)2 j4 = 51.6i + 1.5j6 m>s

The magnitude of V is

V = 2V 2x + V 2y = 2(1.6 m>s)2 + (1.5 m>s)2 = 2.19 m>s Ans.As indicated in Fig. a, the direction of V is defined by

u = tan-1 aVyVx

b = tan-1 a1.5 m>s1.6 m>s b = 43.2

Using the definition of the slope of the streamline and initial condition at x = 2 m, y = 3 m.

dy

dx=

v

u;

dy

dx=

0.06t2

0.8x

Note that since we are finding the streamline, which represents a single instant in time, t = 5 s, t is a constant.

Ly

3 m

dy

t2= L

x

2 m

0.075dxx

1

t2 (y - 3) = 0.075 ln

x2

y = a0.075t2 ln x2+ 3b m

When t = 5 s,

y = 0.075(52) ln ax2b + 3

y = c 1.875 ln ax2b + 3 d m

x(m) 0.5 1 2 3 4 5 6

y(m) 0.401 1.700 3 3.760 4.300 4.718 5.060

The plot of the streamline is shown in Fig. a

36. The soap bubble is released in the air and rises with a velocity of V = 3(0.8x)i + (0.06t2)j4 m>s where x is meters and t is in seconds. Determine the magnitude of the bubbles velocity, and its directions measured counterclockwise from the x axis, when t = 5 s, at which time x = 2 m and y = 3 m. Draw its streamline at this instant.

(a)

x

VVy = 1.5 m s

Vx = 1.6 m s

0 10.5 2 3 4(a)

x(m)

y(m)

5 6

1

2

3

4

5

6

v

u

Ans:V = 2.19 m>su = 43.2

273


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

dy

dx= tan u

dy

dx=

v

u

dy

dx=

2y

2 + y

L2 + y

2y dy = Ldx

ln y +12

y = x + C

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

ln(2) +12

(2) = 3 + C

C = -1.31

Thus,

ln y +12

y = x - 1.31

ln y2 + y = 2x - 2.61 Ans.

At point (3 m, 2 m)

u = (2 + 2) m>s = 4 m>s S v = 2(2) = 4 m>sc

The magnitude of the velocity is

V = 2u2 + v2 = 2(4 m>s)2 + (4 m>s)2 = 5.66 m>s Ans.and its direction is

u = tan-1 avub = tan-1 a4 m>s

4 m>s b = 45 Ans.

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

(a)

x

y

y

u = (2 y) m/s

streamline

v = 2y m/s

V

x

Ans:ln y2 + y = 2x - 2.61V = 5.66 m>su = 45 a

274


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

dy

dx= tan u

dy

dx=

v

u=

-6xyx2 + 5

Ldy

y= -6L

x

x2 + 5 dx

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

At x = 5 m, y = 1 m. Then,

ln 1 = -3 ln3(5)2 + 54 + C C = 3 ln 30

Thus

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

ln3y(x2 + 5)34 = ln 303 y(x2 + 5)3 = 303

y =27(103)

(x2 + 5)3 Ans.

At point (5 m, 1m),

u = (52 + 5) m>s = 30 m>s S v = -6(5)(1) = -30 m>s = 30 m>s T


V = 2u2 + v2 = 2(30 m>s)2 + (30 m>s)2 = 42.4 m>s Ans.And its direction is

u = tan-1 avub = tan-1 a30 m>s

30 m>s b = 45 Ans.

*38. A two-dimensional flow field is described by u = 1x2 + 52 m>s and v = (-6xy) m>s. Determine the equation of the streamline that passes through point (5 m, 1m), and find the velocity of a particle located at this point. Draw this streamline.

(a)

x

x

y

y

v = (6xy) m/s

u = (x2 + 5) m/s

streamline

V

275


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

dy

dx= tan u

dy

dx=

v

u=

4

2y2

Ly2 dy = L2dx

13

y3 = 2x + CAt x = 1 m, y = 2 m. Then

13

(2)3 = 2(1) + C

C = 23

Thus,

13

y3 = 2x +23

y3 = 6x + 2 Ans.

At point (1 m, 2 m)

u = 2(22) = 8 m>s S v = 4 m>s c


V = 2u2 + v2 = 2(8 m>s)2 + (4 m>s)2 = 8.94 m>s Ans.And its direction is

u = tan-1 avub = tan-1 a4

8b = 26.6 Ans.

39. Particles travel within a flow field defined by V = 32y2i + 4j4 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m), and find the velocity of a particle located at this point. Draw this streamline.

(a)

xx

y

y

streamlineVv = 4 m/s

u = 2y2 m/s

Ans:y3 = 6x + 2V = 8.94 m>su = 26.6 a

276



dy

dx= tan u

dy

dx=

v

u=

0.8 + 0.6y0.5

= 1.6 + 1.2y

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

Ly

0

dy

1.6 + 1.2y= L

x

0dx

1

1.2 ln(1.6 + 1.2y) ` y

0= x

lna1.6 + 1.2y1.6

b = 1.2x lna1 + 3

4 yb = 1.2x

1 +34

y = e1.2x

y =43

(e1.2x - 1) m Ans.

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

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

(a)

(b)

xx

y

y

streamline

V

y(m)

x(m)

y = (e1.2x 1) m

streamline

43

v = (0.8 + 0.6y) m/s

u = 0.5 m/s

y

x

u

v

Ans:

y =43

(e1.2x - 1)

277



dy

dx= tan u

dy

dx=

v

u=

1.6 + 0.4y0.8x

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

Ly

0

dy

1.6 + 0.4y= L

x

1

dx0.8x

10.4

ln(1.6 + 0.4y) ` y0

=1

0.8 ln x ` x

1

10.4

lna1.6 + 0.4y1.6

b = 10.8

ln x

lna1 + 14

yb2 = ln xa1 + 1

4yb2 = x

y = 4(x1>2 - 1) m Ans.Using this result, the streamline is shown in Fig. b.

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

(a)

(b)

streamline

1 m

y

x

y = 4(x 1) m

x

x

y

y

streamline

V

v = (1.6 + 0.4y) m/s

u = (0.8x) m/s

y

x

u

v

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

278



dy

dx= tan u

dy

dx=

v

u=

6x8y

L8y dy = L6x dx4y2 = 3x2 + C

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

4(2)2 = 3(1)2 + C

C = 13

Thus

4y2 = 3x2 + 13 Ans.

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

(a)

x

x

y

y

streamline

Vv = (6x) m/s

u = (8y) m/s

279



dy

dx= tan u

dy

dx=

v

u=

6y

3x

Ldy

2y= L

dxx

12

ln y = ln x + C

At x = 3 ft, y = 1 ft . Then

12

ln y = ln x - ln3

12

ln y = ln x3

ln y = lnax3b2

y =x2

9 Ans.

313. A flow field is defined by u = (3x) ft>s and v = (6y) ft>s, where x and y are in feet. Determine the equation of the streamline passing through point (3 ft, 1 ft). Draw this streamline.

(a)

x

xy

y

streamline

Vv = (6y) ft/s

u = (3x) ft/s

Ans:y = x2>9

280


SolutionSince the velocity V is constant, Fig. a, the streamline will be a straight line with aslope.

dy

dx= tan u

dy

dx=

v

u=

85

y = 1.6x + C

At x = 0, y = 0. Then

C = 0

Thus

y = 1.6x Ans.

Since the direction of velocity V remains constant so does the streamline, and the flow is steady. Therefore, the pathline coincides with the streamline and shares the same equation.

314. A flow of water is defined by u = 5 m>s and v = 8 m>s. If metal flakes are released into the flow at the origin (0, 0), draw the streamlines and pathlines for these particles.

(a)

V

x

y

v = 8 m/s

u = 5 m/s

streamline pathline

Ans:y = 1.6x

281



dy

dx= tan u

dy

dx=

v

u=

8y> (x2 + y2)8x> (x2 + y2) = yx

Ldy

y= L

dxx

ln yx = C

y

x= C

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

C = 1

Thus,

y

x= 1

y = x Ans.

315. A flow field is defined by u = 38x> (x2 + y2) 4 m>s and v = 38y> (x2 + y2) 4 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 1 m). Draw this streamline.

(a)

xx

y

y

streamline

Vv = m/s

8yx2 + y2

u = m/s8y

x2 + y2

))

))

Ans:y = x

282


SolutionSince the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform flow. Because we are finding a pathline, t is not a constant but a variable. We must first find equations relating x to t and y to t, and then eliminate t. Using the definition of velocity

dxdt

= u =30

2x + 1 ; L

x

2 m(2x + 1)dx = 30L

t

2 sdt

(x2 + x) ` x2 m

= 30 t ` t2 s

x2 + x - 6 = 30(t - 2)

t =130

(x2 + x + 54) (1)

dy

dt= v = 2ty; L

y

6 m

dy

y= 2L

t

2 stdt

ln y ` y6 m

= t2 ` t2 s

ln y

6= t2 - 4

y

6= et

2-4

y = 6et2-4 (2)

Substitute Eq. (1) into Eq. (2),

y = 6e 1

900 (x2+x+54)2-4 Ans.

The plot of the pathline is shown in Fig. a.

x(m) 0 1 2 3 4

y(m) 2.81 3.58 6.00 13.90 48.24

*316. A fluid has velocity components of u = 330>(2x + 1)4 m>s and v = 2ty m>s where x and y are in meters and t is in seconds. Determine the pathline that passes through the point (2 m, 6 m) at time t = 2 s. Plot this pathline for 0 x 4 m.

0 1 2 3 4

(a)

x(m)

y(m)

10

20

30

40

50

283


Ans:For t = 1 s, y = 4e(x

2 + x - 2)>15For t = 2 s, y = 4e2(x

2 + x - 2)>15For t = 3 s, y = 4e(x

2 + x - 2)>5

SolutionSince the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. The slope of the streamline is

dy

dx=

v

u;

dy

dx=

2ty

30>(2x + 1) = 115 ty(2x + 1)L

y

4 m

dy

y=

115

tLx

1 m(2x + 1)dx

ln y ` y4 m

=115

t(x2 + x) ` x1 m

ln y

4=

115

t(x2 + x - 2)

y = 4et(x2+x-2)>15

For t = 1 s,

y = 4e(x2+x-2)>15 Ans.

For t = 2 s,

y = 4e2(x2+x-2)>15 Ans.

For t = 3 s,

y = 4e(x2+x-2)>5 Ans.

The plot of these streamlines are shown in Fig. a

For t = 1 s

x(m) 0 1 2 3 4

y(m) 3.50 4 5.22 7.79 13.3

For t = 2 sx(m) 0 1 2 3 4

y(m) 3.06 4 6.82 15.2 44.1

For t = 3 s

x(m) 0 1 2 3 4

y(m) 2.68 4 8.90 29.6 146

317. A fluid has velocity components of u = 330>(2x + 1)4 m>s and v = (2ty) m>s where x and y are in meters and t is in seconds. Determine the streamlines that passes through point (1 m, 4 m) at times t = 1s, t = 2 s, and t = 3 s. Plot each of these streamlines for 0 x 4 m.

0 1 2 3 4(a)

x(m)

y(m)

5

10

15

20

25

30

35

40

45

t = 3 st = 2 s

t = 1 s

284


Ans:For t = 2 s, y = 6e21x2 + x - 6)>15For t = 5 s, y = 6e1x2 + x - 6)>3

Solution

For t = 2 s

x(m) 0 1 2 3 4y(m) 2.70 3.52 6.00 13.35 38.80

For t = 5 s

x(m) 0 1 2 3 4

y(m) 0.812 1.58 6.00 44.33 638.06

Since the velocity components are a function of time and position the flow can be classified as unsteady nonuniform. The slope of the streamline is

dy

dx=

v

u ;

dy

dx=

2ty

30>(2x + 1) = 115 ty(2x + 1)Note that since we are finding the streamline, which represents a single instant in time, either t = 2 s or t = 5 s, t is a constant.

Ly

6 m

dy

y=

115

tLx

2 m(2x + 1)dx

ln y ` y6 m

=115

t (x2 + x) ` x2 m

ln y

6=

115

t(x2 + x - 6)

y = 6e115 t(x

2+x-6)

For t = 2 s,

y = 6e215 (x

2 +x-6) Ans.

For t = 5 s,

y = 6e13 (x

2+x-6) Ans.

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

318. A fluid has velocity components of u = 330>(2x + 1)4 m>s and v = (2ty) m>s where x and y are in meters and t is in seconds. Determine the streamlines that pass through point (2 m, 6 m) at times t = 2 s and t = 5 s. Plot these streamlines for 0 x 4 m.

0 1 2

(a)

3 4x(m)

y(m)

10

20

30

40

50

t = 5 s

t = 2 s

285


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

Solution

x(m) 0 1 1.5 2 3 4 5y(m) -2.29 -1.59 0 1.59 2.29 2.71 3.04

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

3y2 dy

dx= 8

dy

dx= tan u =

8

3y2

When x = 1 m,

y3 = 8(1) - 12; y = -1.5874

Then

dy

dx`x = 1 m

= tan u `x = 1 m

=8

3(-1.5874)2; u 0 x = 1 m = 46.62

Therefore, the horizontal and vertical components of the velocity are

u = (5 m>s) cos 46.62 = 3.43 m>s Ans. v = (5 m>s) sin 46.62 = 3.63 m>s Ans.

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

(a)

y(m)

0 1 2 3 4x(m)

5

1

1

2

3

2

3

3

3

46.62u

v5 m/s

286


Solution

Here, u =dxdt

. Then,

dx = udt

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

Lx

0dx = L

t

03(0.8t) m>s4dt

x = 0.4t2 (1)

Also, v =dy

dt. Then

dy = vdt

Using y = 0 when t = 0 as the integration limit,

Ly

0dy = L

t

0(0.4 m>s)dt

y = 0.4t (2)

Eliminating t from Eqs. (1) and (2)

y2 = 0.4x

This equation represents the pathline of the particle. The x and y values of the pathline for the first five seconds are tabulated below.

t x y1 0.4 0.42 1.6 0.83 3.6 1.24 6.4 1.65 10 2

A plot of the pathline is shown in Fig. a.

From Eqs. (1) and (2), when t = 4 s,

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

Using the definition of the slope of the streamline,

dy

dx=

v

u ;

dy

dx=

0.40.8t

tLy

1.6 mdy =

12L

x

6.4 mdx

t(y - 1.6) =12

(x - 6.4)

y = c 12t

(x - 6.4) + 1.6 d m

*320. A flow field is defined by u = (0.8t) m>s and v = 0.4 m>s, where t is in seconds. Plot the pathline for a particle that passes through the origin when t = 0. Also, draw the streamline for the particle when t = 4 s.

(a)

y(m)

0 2 4 6 8x(m)

10

1

1

2

2 y2 = 0.4x

287


*320. (continued)

When t = 4 s,

y =1

2(4) ( x - 6.4) + 1.6

y =18

x + 0.8

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

(b)

y(m)

0.8

x(m)

y = x + 0.818

288


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

SolutionSince the velocity components are a function of position only, the flow can be classified as steady nonuniform. Here, u = 13y22 m>s and v = 8 m>s . The slope of the streamline is defined by

dy

dx=

v

u ;

dy

dx=

8

3y2

Ly

1 m3y2dy = 8L

x

2 mdx

y3 ` y1 m

= 8x ` x2 m

y3 - 1 = 8x - 16

y3 = 8x - 15 (1) Ans.

From the definition of velocity

dy

dt= 8

Ly

1 mdy = L

1 s

08 dt

y ` y1 m

= 8t ` 1 s0

y - 1 = 8

y = 9 m Ans.

Substituting this result into Eq. (1)

93 = 8x - 15

x = 93 m Ans.

321. The velocity for an oil flow is defined by V = 13y2 i + 8 j2 m>s, where y is in meters. What is the equation of the streamline that passes through point (2 m, 1m)? If a particle is at this point when t = 0, at what point is it locatedwhen t = 1 s?

289


Ans:

y =23

ln a 22 - x

b

Solution

x(m) 0 0.25 0.5 0.75 1

y(m) 0 0.089 0.192 0.313 0.462

x(m) 1.25 1.5 1.75 2

y(m) 0.654 0.924 1.386

Since the velocity component is a function of position only, the flow can be classified as steady nonuniform. Using the definition of the slope of a streamline,

dy

dx=

v

u;

dy

dx=

26 - 3x

Ly

0dy = 2L

x

0

dx6 - 3x

y = -23

ln (6 - 3x) ` x0

y = -23

ln a6 - 3x6

by =

23

ln a 22 - x

b Ans.The plot of this streamline is show in Fig. a

322. The circulation of a fluid is defined by the velocity field u = (6 - 3x) m>s and v = 2 m>s where x is in meters. Plot the streamline that passes through the origin for 0 x 6 2 m.

0 0.25 0.5 0.75 1.0

(a)

x(m)

y(m)

1.25 1.5 1.75 2

0.5

1

1.5

290


Ans:

For 0 t 6 10 s, y = -32

x

For 10 s 6 t 15 s, y = -25

x + 22

SolutionUsing the definition of velocity, for 0 t 6 10 s

dxdt

= u; dxdt

= -2

Lx

0dx = -2L

t

0dt

x = (-2t) m (1)

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

dy

dt= v;

dy

dt= 3

Ly

0dy = 3L

t

0dt

y = (3t) m (2)

When t = 10 s, y = 3(10) = 30 m

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

y = -32

x Ans.

For 10 6 t 15 s.

dxdt

= u; dxdt

= 5

Lx

-20 mdx = 5L

t

10 sdt

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

x = (5t - 70) m (3)

At t = 15 s, x = 5(15) - 70 = 5 mdy

dt= v;

dy

dt= -2

Ly

30 mdy = -2L

t

10 sdt

y - 30 = -2(t - 10)y = (-2t + 50) m (4)

When t = 15 s, y = -2(15) + 50 = 20 m

Eliminate t from Eqs. (3) and (4),

y = a-25

x + 22b Ans.The two streamlines intersect at (-20, 30), point B in Fig. (a). The pathline is the path ABC.

323. A stream of water has velocity components of u = -2 m>s, v = 3 m>s for 0 t 6 10 s; and u = 5 m>s, v = -2 m>s for 10 s 6 t 15 s. Plot the pathline and streamline for a particle released at point (0, 0) when t = 0 s.

(a)

0 55

A

B

C

101520x(m)

y(m)

10

20

30

291


Solution

x(m) 0.25 0.5 0.75 1 2 3 4

y(m) 4.96 5.31 5.51 5.65 6 6.20 6.35

Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. Using the definition of the slope of the streamline,

dy

dx=

v

u;

dy

dx=

2t4x

=t

2x

Ly

6 mdy =

t2L

x

2 m

dxx

y - 6 =t2

ln x2

y =t2

ln x2+ 6

For t = 1 s, y = a12

ln x2+ 6bm Ans.

The plot of this streamline is shown in Fig. a.

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

0 10.750.50.25 2 3 4

(a)

x(m)

y(m)

1

2

3

4

5

6

7

292


Ans:

y =1

16 ln2

x2+

12

ln x2+ 6

Solution

x(m) 0.25 0.50 0.75 1 2 3 4

y(m) 5.23 5.43 5.57 5.68 6 6.21 6.38

Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. Using the definition of velocity,

dxdt

= u = 4x; Lx

2 m

dx4x

= Lt

1 Sdt

14

ln x ` x2 m

= t ` t1 s

14

ln x2

= t - 1

t =14

ln x2+ 1 (1)

dy

dt= v = 2t; L

y

6 mdy = L

t

1 s2t dt

y - 6 = t2 ` t1 s

y = t2 + 5 (2)

Substitute Eq. (1) into (2),

y = a14

ln x2+ 1b2 + 5

y = a 116

ln2 x2+

12

ln x2+ 6b Ans.

The plot of this pathline is shown in Fig. (a)

325. The velocity field is defined by u = (4x) m>s and v = (2t) m>s, where t is in seconds and x is in meters. Determine the pathline that passes through point (2 m, 6 m) when t = 1 s. Plot this pathline for 0.25 m x 4 m.

0 10.750.50.25 2 3 4

(a)

x(m)

y(m)

1

2

3

4

5

6

7

293


SolutionUsing the definition of velocity, for 0 t 6 5 s,

dxdt

= u; dxdt

=12

x

Lx

1 m

dxx

= Lt

0

12

dt

ln x =12

t

x = ae 12 tb m (1)When t = 5 s, x = e

12 (5) = 12.18 m

dy

dt= v;

dy

dt=

18

y2

Ly

1 m

dy

y2= L

t

0

18

dt

- a1yb ` y

1 m=

18

t

1 -1y

=18

t

y - 1y

=18

t

y a1 - 18

tb = 1y = a 8

8 - tb m t 8 s (2)

When t = 5 s, y =8

8 - 5= 2.667 m

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

y = a 88 - 2 ln x

bmx(m) 1 3 5 7 9 11 12.18

y(m) 1 1.38 1.67 1.95 2.22 2.50 2.67

For 5 s < t 10 s,

dxdt

= u; dxdt

= -14

x2

Lx

12.18 m

dx

x2= -

14L

t

5 sdt

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

294


Ans:

For 0 t 6 5 s, y =8

8 - 2 ln x

For 5 s 6 t 10 s, y = 2.67e11>x - 0.0821)

- a1x-

112.18

b = -14

(t - 5)

1x

=t4- 1.1679

x = a 4t - 4.6717

b m t 4.6717 s (3)When t = 10 s, x =

410 - 4.6717

= 0.751 m

dy

dt= v;

dy

dt=

14

y

Ly

2.667 m

dy

y=

14L

t

5 sdt

ln y

2.667=

14

(t - 5)

y

2.667= e

14 (t-5)

y = c 2.667e14 (t-5) d m (4)When t = 10 s, y = 2.667e

14 (10-5) = 9.31 m

Eliminate t from Eqs. (3) and (4),

y = 2.667e 1434(1x+1.1679)-54

= c 2.667e(1x -0.08208) dmx(m) 0.751 1 3 5 7 9 11 12.18

y(m) 9.31 6.68 3.43 3.00 2.83 2.75 2.69 2.67

The two streamlines intersect at (12.18, 2.67), point B in Fig. (a). The pathline is the path ABC.

326. (continued)

0 10.751 12.18

2 3 134 5 6 7 8 9 10 11 12

A

C

B

x(m)

y(m)

123456

10

789

295


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

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

u = 6y2 - 1v = 3x + 2

dy

dx=

v

u=

3x + 26y2 - 1

Ly

2(6y2 - 1)dy = L

x

6(3x + 2)dx

2y3 - y ` y2

= 1.5 x2 + 2x ` x6

2y3 - y - 32(2)3 - 24 = 1.5x2 + 2x - 31.5(6)2 + 2(6)42y3 - 1.5x2 - y - 2x + 52 = 0 Ans.

At (6 m, 2 m)

u = 6(2)2 - 1 = 23 m>s Sv = 3(6) + 2 = 20 m>scV = 2(23 m>s)2 + (20 m>s)2 = 30.5 m>s Ans.

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

y

23 m/s

20 m s

6

2

30.5 m s

x

296


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

u = 2x + 1v = -y

dy

dx=

v

u=

-y2x + 1

-Ldy

y= L

dx(2x + 1)

- ln y =12

ln (2x + 1) + C

- y = (2x + 1)12 + C

- 1 = (2(3) + 1)12 + C

C = - 3.65

-y ` y1

= (2x + 1)12 ` x

3

-y + 1 = (2x + 1)12 - 32(3) + 1 4 12

y = 3.65 - (2x + 1)12 Ans.

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

V = 2(7 m>s)2 + ( -1 m>s)2 = 7.07 m>s Ans.

*328. A flow field for a liquid can be described by V = 5(2x + 1) i - y j6 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at points 13 m, 1 m2. Sketch the velocity on the streamline.

y

7 m/s

7.07 m/s1 m s

3

1

x

297


Ans:a = 24 m>s2

SolutionSince the flow is along the horizontal (x axis) v = w = 0. Also, the velocity is a function of time t only. Therefore, the convective acceleration is zero, so that

u 0V0x

= 0.

a =0V0 t

+ u 0V0 x

= 12t + 0= (12t) m>s2

When t = 2 s,

a = 12(2) = 24 m>s2 Ans.Note: The flow is unsteady since its velocity is a function of time.

329. Air flows uniformly through the center of a horizontal duct with a velocity of V = (6t2 + 5) m>s, where t is in seconds. Determine the acceleration of the flow when t = 2 s.

298


Ans:1088 in.>s2

Solution

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

a =0u0 t

+ u 0u0 x

= 4x + (4xt)(4t)= 4x + 16xt2

= 34x(1 + 4t2) 4 in.>s2When t = 2 s, x = 16 in. Then

a = 34(16)31 + 4(22) 4 4 in.>s2 = 1088 in.>s2Note: The flow is unsteady since its velocity is a function of time.

330. Oil flows through the reducer such that particles along its centerline have a velocity of V = (4xt) in.>s, where x is in inches and t is in seconds. Determine the acceleration of the particles at x = 16 in. when t = 2 s.

24 in.

x

299


Ans:30.9 ft>s2

SolutionFor two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

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

= (1 + 12tx) ft>s2 ay =

0v0t

+ u 0v0x

+ v 0v0y

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

= (2x + 12ty + 2t2) ft>s2When t = 1 s, x = 1 ft and y = 2 ft, then

ax = 31 + 12(1)(1)4 = 13 ft>s2 ay = 32(1) + 12(1)(2) + 2(12) 4 = 28 ft>s2Thus, the magnitude of the acceleration is

a = 2a 2x + a 2y = 2(13 ft>s2)2 + (28 ft>s2)2 = 30.9 ft>s2 Ans.

331. A fluid has velocity components of u = (6y + t) ft>s and v = (2tx) ft>s where x and y are in feet and t is in seconds. Determine the magnitude of acceleration of a particle passing through the point (1 ft, 2 ft), when t = 1 s.

300


*332. The velocity for the flow of a gas along the center streamline of the pipe is defined by u = (10x2 + 200t + 6) m>s, where x is in meters and t is in seconds. Determine the acceleration of a particle when t = 0.01 s and it is at A, just before leaving the nozzle.

Solution

a =0u0t

+ u 0u0x

0u0t

= 200 0u0x

= 20 x

a = 3200 + (10x2 + 200t + 6)(20x)4 m>s2When t = 0.01 s, x = 0.6 m.

a = 5200 + 310(0.62) + 200(0.01) + 64 320(0.6)4 6 m>s2 = 339 m>s2 Ans.

xA

0.6 m

301


Ans:V = 23.3 m>sa = 343 m>s2

SolutionVelocity.

At x = 2 m, y = 4 m,

u = 2(22) - 2(42) + 4 = -20 m>s v = 4 + 2(4) = 12 m>sThe magnitude of the particles velocity is

V = 2u2 + v2 = 2( -20 m>s)2 + (12 m>s)2 = 23.3 m>s Ans.Acceleration. The x and y components of the particles acceleration, with w = 0 are

ax =0u0t

+ u 0u0x

+ v 0u0y

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

At x = 2 m, y = 4 m,

ax = -340 m>s2 ay =

0v0t

+ u 0v0x

+ v 0v0y

= 0 + (2x2 - 2y2 + y)(y) + (y + xy)(1 + x)

At x = 2 m, y = 4 m,

ay = -44 m>s2The magnitude of the particles acceleration is

a = 2a 2x + a 2y = 2( -340 m>s2)2 + ( -44 m>s2)2 = 343 m>s2 Ans.

333. A fluid has velocity components of u = (2x2 - 2y2 + y) m>s and v = (y + xy) m>s, where x and y are in meters. Determine the magntiude of the velocity and acceleration of a particle at point (2 m, 4 m).

302


Ans:V = 16.3 m>s uv = 79.4 aa = 164 m>s2ua = 17.0 a

SolutionSince the velocity components are a function of position only the flow can be classified as steady nonuniform. At point x = 2 m and y = 1 m,

u = 5(12) - 2 = 3 m>sv = 4(22) = 16 m>s


V = 2u2 + v2 = 2(3 m>s)2 + (16 m>s)2 = 16.3 m>s Ans.Its direction is

uv = tan-1avu b = tan-1a16 m>s3 m>s b = 79.4 Ans.For two dimensional flow, the Eulerian description is

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

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

= (x - 5y2) + 40x2y

ay =0v0t

+ u 0v0x

+ v 0v0y

= 0 + (5y2 - x)(8x) + 4x2(0)

= 8x(5y2 - x)

At point x = 2 m and y = 1 m,

ax = 32 - 5(12) 4 + 40(22)(1) = 157 m>s2ay = 8(2)35(12) - 24 = 48 m>s2

The magnitude of the acceleration is

a = 2a 2x + a 2y = 2(157 m>s2)2 + (48 m>s2)2 = 164 m>s2 Ans.Its direction is

ua = tan-1aayax b = tan-1 48 m>s2157 m>s2 = 17.0 Ans.

334. A fluid velocity components of u = (5y2 - x) m>s and v = (4x2) m>s, where x and y are in meters. Determine the velocity and acceleration of particles passing through point (2 m, 1 m).

303


Ans:a = 36.1 m>s2u = 33.7 a

SolutionSince the velocity components are independent of time but are a function of position, the flow can be classified as steady nonuniform. The slope of the streamline is

dy

dx=

v

u;

dy

dx=

4x - 15y2

Ly

1 m5y2dy = L

x

1 m(4x - 1)dx

y3 =15

(6x2 - 3x + 2) where x is in m

For two dimensional flow, the Eulerian description is

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

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

= 40xy - 10y

ay =0v0t

+ u 0v0x

+ v 0v0y

= 0 + 5y2(4) + (4x - 1)(0)

= 20y2

At point x = 1 m and y = 1 m,

ax = 40(1)(1) - 10(1) = 30 m>s2ay = 20(12) = 20 m>s2


a = 2a 2x + a 2y = 2(30 m>s2)2 + (20 m>s2)2 = 36.1 m>s2 Ans.Its direction is

u = tan-1aayax

b = tan-120 m>s230 m>s2 = 33.7 Ans.

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

x(m) 0 0.5 1 2 3 4 5y(m) 0.737 0.737 1 1.59 2.11 2.58 3.01

0 1 2 3 4 5(a)

x(m)

y(m)

1

2

3

4

a = 36.1 m/s2

ax = 30 m/s2

ay = 20 m/s2

335. A fluid has velocity components of u = (5y2) m>s and v = (4x - 1) m>s, where x and y are in meters. Determine the equation of the streamline passing through point 11 m, 1 m2 . Find the components of the acceleration of a particle located at this point and sketch the acceleration on the streamline.

304


SolutionSince the velocity is a function of position only, the flow can be classified as steady nonuniform. Since the velocity varies linearly with x,

V = VA + aVB - VALAB bx = 8 + a2 - 83 bx = (8 - 2x) m>s Ans.For one dimensional flow, the Eulerian description gives

a =0Vdt

+ V 0Vdx

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

= 4(x - 4) m/s2 Ans.

using the definition of velocity,dxdt

= V = 8 - 2x; Lx

0

dx8 - 2x

= Lt

0dt

-12

ln(8 - 2x) ` x0

= t

12

ln a 88 - 2x

b = tln a 8

8 - 2xb = 2t

88 - 2x

= e2 t

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

*336. Air flowing through the center of the duct has been found to decrease in speed from VA = 8 m>s to VB = 2 m>s in a linear manner. Determine the velocity and acceleration of a particle moving horizontally through the duct as a function of its position x. Also, find the position of the particle as a function of time if x = 0 when t = 0. VA 8 m/s VB 2 m/s

3 m

x

A

B

305


Ans:V = 33.5 m>suV = 17.4a = 169 m>s2ua = 79.1 a

SolutionSince the velocity components are functions of time and position the flow can be classified as unsteady nonuniform. When t = 2 s, x = 1 m and y = 1 m.

u = 8(22) = 32 m>sv = 7(1) + 3(1) = 10 m>s


V = 2u2 + v2 = 2(32 m>s)2 + (10 m>s)2 = 33.5 m>s Ans.Its direction is

uv = tan-1avu b = tan-1a10 m>s32 m>s b = 17.4 uv Ans.For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

= 16t + 8t2(0) + (7y + 3x)(0)

= (16t) m>s2ay =

0v0t

+ u 0v0x

+ v 0v0y

= 0 + (8t2)(3) + (7y + 3x)(7)

= 324t2 + 7(7y + 3x)4 m>s2When t = 2 s, x = 1 m and y = 1 m.

ax = 16(2) = 32 m>s2ay = 24(22) + 737(1) + 3(1)4 = 166 m>s2



ua = tan-1aayax b = tan-1a166 m>s232 m>s2 b = 79.1 ua Ans.

337. A fluid has velocity components of u = (8t2) m>s and v = (7y + 3x) m>s, where x and y are in meters and t is in seconds. Determine the velocity and acceleration of a particle passing through point x = 1 m, y = 1 m when t = 2 s.

306


Ans:y = x>2, a = 143 ft>s2u = 26.6 a

SolutionSince the velocity components are the function of position but not the time, the flow is steady (Ans.) but nonuniform. Using the definition of the slope of the streamline,

dy

dx=

v

u ;

dy

dx=

8y

8x=

y

x

Ly

1 ft

dy

y= L

x

2 ft

dxx

ln y ` y1 ft

= ln x ` x2 ft

ln y = ln x2

y =12

x Ans.

For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y


ax =0u0t

+ u 0u0x

+ v 0u0y

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

= (64x) ft>s2ay =

0v0t

+ u 0v0x

+ v 0v0y

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

= (64y) ft>s2At x = 2 ft, y = 1 ft . Then

ax = 64(2) = 128 ft>s2 ay = 64(1) = 64 ft>s2The magnitude of the acceleration is

a = 2a 2x + a 2y = 2(128 ft>s2)2 + (64 ft>s2)2 = 143 ft>s2 Ans.Its direction is

u = tan-1aayax

b = tan-1a 64 ft>s2128 ft>s2 b = 26.6 u Ans.

338. A fluid has velocity components of u = (8x) ft>s and v = (8y) ft>s, where x and y are in feet. Determine the equation of the streamline and the acceleration of particles passing through point (2 ft, 1 ft). Also find the acceleration of a particle located at this point. Is the flow steady or unsteady?

(a)

x

xy

y

v = (8y) ft/s

u = (8x) ft/s

streamline

V

307


Ans:y = 2xa = 286 m>s2u = 63.4 a

SolutionSince the velocity components are the function of position, not of time, the flow can be classified as steady (Ans.) but nonuniform. Using the definition of the slope of the streamline,

dy

dx=

v

u ;

dy

dx=

8xy

xy2=

4xy

Ly

2 my dy = L

x

1 m4x dx

y2

2` y2 m

= 2x2 ` x1m

y2

2- 2 = 2x2 - 2

y2 = 4x2

y = 2x Ans.

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

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

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

= (32xy2) m>s2 ay =

0v0t

+ u 0v0x

+ v 0v0y

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

= (16y3 + 64x2y) m>s2At point x = 1 m and y = 2 m,

ax = 32(1)(22) = 128 m>s2ay = 316(23) + 64(12)(2)4 = 256 m>s2



u = tan-1aayax

b = tan-1a256 m>s2128 m>s2 b = 63.4 u Ans.

339. A fluid velocity components of u = (2y2) m>s and v = (8xy) m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m). Also, what is the acceleration of a particle at this point? Is the flow steady or unsteady?

(a)

x

x

y

y v = (8xy) m/s

u = (2y2) m/s

streamline

V

308


SolutionThe flow is steady but nonuniform since the velocity components are a function of position, but not time. At point (2 m, 1 m)

u = 4y = 4(1) = 4 m>sv = 2x = 2(2) = 4 m>s

Thus, the magnitude of the velocity is

V = 2u2 + v2 = 2(4 m>s)2 + (4 m>s)2 = 5.66 m>s Ans.For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y


ax =0u0t

+ u 0u0x

+ v 0u0y

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

= (8x) m>s2ay =

0v0t

+ u 0v0x

+ v 0v0y

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

= (8y) m>s2At point (2 m, 1 m),

ax = 8(2) = 16 m>s2ay = 8(1) = 8 m>s2


a = 2a 2x + a 2y = 2(16 m>s)2 + (8 m>s)2 = 17.9 m>s2 Ans.Using the definition of the slope of the streamline,

dy

dx=

v

u ;

dy

dx=

2x4y

=x2y

Ly

1 m2y dy = L

x

2 mx dx

y2 ` y1 m

=x2

2` x2 m

y2 - 1 =x2

2- 2

y2 =12

x2 - 1

*340. The velocity of a flow field is defined by V = 54 yi + 2 xj6 m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point (2 m, 1 m). Find the equation of the streamline passing through this point, and sketch the velocity and acceleration at the point on this streamline.

309


*340. (continued)

The plot of this streamline is shown is Fig. a

x(m) 22 2 3 4 5 6y(m) 0 1 1.87 2.65 3.39 4.12

(a)

0 1 2 3 4x(m)

y(m)

5 6

1

1

2

3

4

2

3

4

u = 4 m/s

45

v = 4 m/s

V = 5.66 m/s

0 1 2 3 4x(m)

y(m)

5 6

1

1

2

3

4

2

3

4

ax = 16 m/s2

ay = 8 m/s2

a = 17.9 m/s2

310


Ans:V = 4.47 m>s, a = 16 m>s2y =

12

ln x + 2

SolutionSince the velocity components are a function of position but not time, the flow can be classified as steady nonuniform. At point (1 m, 2 m),

u = 4x = 4(1) = 4 m>sv = 2 m>s

The magnitude of velocity is

V = 2u2 + v2 = 2(4 m>s)2 + (2 m>s)2 = 4.47 m>s Ans.For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y


ax =0u0t

+ u 0u0x

+ v 0u0y

= 0 + 4x(4) + 2(0) = 16x

ay =0v0t

+ u 0v0x

+ v 0v0y

= 0 + 4x(0) + 2(0) = 0

At point (1 m, 2 m),

ax = 16(1) = 16 m>s2 ay = 0Thus, the magnitude of the acceleration is

a = ax = 16 m>s2 Ans.Using the definition of the slope of the streamline,

dy

dx=

v

u ;

dy

dx=

24x

=12x

Ly

2 mdy =

12L

x

1 m

dxx

y - 2 =12

ln x

y = a12

ln x + 2b Ans.The plot of this streamline is shown in Fig. a

x(m) e-4 1 2 3 4 5

y(m) 0 2 2.35 2.55 2.69 2.80

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

0 1 2 3 4 5x(m)

y(m)

1

2

3

u = 4 m/s

V = 4.47 m/sv = 2 m/s

0 1 2 3 4 5x(m)

y(m)

1

2

3

0 1 2 3 4 5x(m)

y(m)

1

2

3

0 1 2 3 4 5x(m)

y(m)

1

2

3

a = 16 m/s2

0 1 2 3 4 5x(m)

y(m)

1

2

3

u = 4 m/s

V = 4.47 m/sv = 2 m/s

0 1 2 3 4 5x(m)

y(m)

1

2

3

0 1 2 3 4 5x(m)

y(m)

1

2

3

0 1 2 3 4 5x(m)

y(m)

1

2

3

a = 16 m/s2

311


SolutionSince the velocity components are a function of position but not time, the flow can be classified as steady but nonuniform. At point (1 m, 1 m),

u = 2x2 - y2 = 2(12) - 12 = 1 m>sv = -4xy = -4(1)(1) = -4 m>s


V = 2u2 + v2 = 2(1 m>s)2 + ( -4 m>s)2 = 4.12 m>s Ans.For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y


ax =0u0t

+ u 0u0x

+ v 0u0y

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

ay =0v0t

+ u 0v0x

+ v 0v0y

= 0 + (2x2 - y2)(-4y) + (-4xy)(-4x) = -4y(2x2 - y2) + 16x2y

At point (1 m, 1 m),

ax = 4(1)32(12) - 124 + 8(1)(12) = 12 m>s2ay = -4(1)32(12) - 124 + 16(12)(1) = 12 m>s2


a = 2a 2x + a 2y = 2(12 m/s2)2 + (12 m/s2)2 = 17.0 m>s2 Ans.Using the definition of the slope of the streamline,

dy

dx=

v

u;

dy

dx= -

4xy

2x2 - y2

(2x2 - y2)dy = -4xydx2x2dy + 4xydx - y2dy = 0

However, d(2x2y) = 2(2xydx + x2dy) = 2x2dy + 4xydx. Thend(2x2y) - y2dy = 0

342. The velocity of a flow field is defined by u = (2x2 - y2) m>s and v = (-4xy) m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 11 m, 1 m2 . Find the equation of the streamline passing through this point, and sketch the velocity and acceleration at the point on this streamline.

312


Ans:V = 4.12 m>s a = 17.0 m>s2x2 =

y3 + 56y

Integrating this equation,

2x2y -y3

3= C

with the condition y = 1 m when x = 1 m,

2(12)(1)-13

3= C

C =53

Thus,

2x2y -y3

3=

53

6x2y - y3 = 5

x2 =y3 + 5

6y Ans.

Taking the derivative of this equation with respect to y

2 xdxdy

=6y(3y2) - (y3 + 5)(6)

(6y)2=

2y3 - 56y2

dxdy

=2y3 - 512xy2

Set dxdy

= 0;

2y3 - 5 = 0y = 1.357 m

The corresponding x is

x2 = 1.3573 + 5x = 0.960 m

y(m) 0.25 0.5 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00x(m) 1.83 1.31 1.10 1.00 0.963 0.965 0.993 1.04 1.10 1.17 1.25 1.33

342. (continued)

0 0.5 1.0x(m)

y(m)

1.5 2.0

0.5

1.0

2.5

2.0

1.5

1.36

3.0

0 0.5 1.0x(m)

y(m)

1.5 2.0

0.5

1.0

2.5

2.0

1.5

3.0

u = 1 m/s

v = 4 m/s V = 4.12 m/s

0.960

ay = 12 m/s2 a = 17.0 m/s2

ax = 12 m/s2

0 0.5 1.0x(m)

y(m)

1.5 2.0

0.5

1.0

2.5

2.0

1.5

1.36

3.0

0 0.5 1.0x(m)

y(m)

1.5 2.0

0.5

1.0

2.5

2.0

1.5

3.0

u = 1 m/s

v = 4 m/s V = 4.12 m/s

0.960

ay = 12 m/s2 a = 17.0 m/s2

ax = 12 m/s2

313


SolutionSince the velocity components are a function of position but not time, the flow can be classified as steady nonuniform. At point (3 m, 2 m)

u =-y4

= -24

= -0.5 m>sv =

x9

=39

= 0.3333 m>sThe magnitude of the velocity is

V = 2u2 + v2 = 2( -0.5 m>s)2 + (0.333m>s)2 = 0.601 m>s Ans.For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y

Writing the scalar components of this equation along the x and y are1S+ 2 ax = 0u0t + u 0u0x + v 0u0y= 0 + a -y

4b(0) + ax

9ba-1

4b

= a- 136

xb m>s2( + c ) ay =

0v0t

+ u 0v0y

+ v 0v0y

= 0 + a -y4

ba19b + ax

9b(0)

= c - 136

y d m>s2At point (3 m, 2 m),

ax = -136

(3) = -0.08333 m>s2ay = -

136

(2) = -0.05556 m>s2The magnitude of the acceleration is

a = 2ax2 + ay2 = 2(-0.08333 m>s2)2 + (-0.05556 m>s2) = 0.100 m>s2 Ans.

343. The velocity of a flow field is defined by u = (-y>4) m>s and v = (x>9) m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point (3 m, 2 m). Find the equation of the streamline passing through this point, and sketch the velocity and acceleration at the point on this streamline.

314


Ans:V = 0.601 m>s a = 0.100 m>s2x2> 14.2422 + y2> 12.8322 = 1

343. (continued)

Using the definition of slope of the streamline,

dy

dx=

v

u ;

dy

dx=

x>9-y>4 = - 4x9y

9Ly

2 mydy = -4L

x

3 mxdx

9y2

2` y2 m

= - (2x2) ` x3 m

9y2

2- 18 = -2x2 + 18

9y2 + 4x2 = 72

x2

72>4 + y272>9 = 1x2

(4.24)2+

y2

(2.83)2= 1 Ans.

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

(a)

y(m)

3

2

v = 0.333 m/s

u = 0.5 m/s

V = 0.601 m/s

x(m)

2 m3

2 m2

2 m2

2 m3

y

3

2

ax = 0.0833 m/s2

ay = 0.0556 m/s2

a = 0.100 m/s2

x

315


SolutionThe flow is unsteady nonuniform. For one dimensional flow,

a =0u0t

+ u 0u0x

Here, u = (4 tx) m>s. Then 0u0t

= 4x and 0u0x

= 4t. Thus,

a = 4x + (4tx)(4t) = (4x + 16t2x) m>s2Since u = 0.8 m>s when t = 0.1 s, 0.8 = 4(0.1) x x = 2 m

The position of the particle can be determined from

dxdt

= u = 4 tx; Lx

2 m

dxx

= 4Lt

0.15t dt

ln x ` x2m

= 2t2 ` t0.15

ln x2

= 2t2 - 0.02

e2t2-0.02 =

x2

x = 2e2t2-0.02

x = 2e2(0.82)-0.02 = 7.051 m

Thus, t = 0.8 s,

a = 4(7.051) + 16(0.82)(7.051)

= 100.40 m>s2 = 100 m>s2 Ans.

*344. The velocity of gasoline, along the centerline of a tapered pipe, is given by u = (4tx) m>s, where t is in seconds and x is in meters. Determine the acceleration of a particle when t = 0.8 s if u = 0.8 m>s when t = 0.1 s.

316


SolutionThe flow is unsteady nonuniform. For three dimensional flow,

a =0V0t

+ u 0V0t

+ v 0V0t

+ w 0V0t

Thus,

ax =0u0t

+ u 0u0x

+ v 0u0y

+ w 0u0t

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

= (4x) m>s2 ay =

0v0t

+ u 0v0x

+ v 0v0y

+0v0t

= 6x + 2x(6t) + 6tx(0) + 3y(0) = (6x + 12tx) m>s2 az =

0w0t

+ u 0w0x

+ v 0w0y

+ w 0w0z

= 0 + 2x(0) + 6tx(3) + 3y(0) = (18tx) m>s2The position of the particle can be determined from

dxdt

= u = 2x; Lx

1 m

dxx

= 2Lt

0dt

ln x = 2t

x = (e2t) m

dy

dt= v = 6tx = 6te2t ; L

y

0dy = 6L

t

0te2tdt

y =32

(2te2t - e2t) ` t0

y =32

(2te2t - e2t + 1)

dzdt

= w = 3y =9232te2t - e2t + 14 ;

Lz

0dz =

92L

t

0(2te2t - e2t + 1) dt

z =92c te2t - 1

2 e2t -

12

e2t + t d ` t0

z =92

(te2t - e2t + t + 1) m

345. The velocity field for a flow of water is defined by u = (2x) m>s, v = (6tx) m>s, and w = (3y) m>s, where t is in seconds and x, y, z are in meters. Determine the acceleration and the position of a particle when t = 0.5 s if this particle is at (1 m, 0, 0) when t = 0.

317


Ans:x = 2.72 my = 1.5 mz = 0.634 ma = 510.9i + 32.6j + 24.5k6 m>s2

345. (continued)

When, t = 0.5 s,

x = e2(0.5) = 2.7183 m = 2.72 m Ans.

y =3232(0.5)e2(0.5) - e2(0.5) + 14 = 1.5 m Ans.

Thus, z =9230.5e2(0.5) - e2(0.5) + 0.5 + 14 = 0.6339 m = 0.634 m Ans.

ax = 4(2.7183) = 10.87 m>s2 ay = 6(2.7183) + 12(0.5)(2.7183) = 32.62 m>s2 az = 18(0.5)(2.7183) = 24.46 m>s2Then

a = 510.9i + 32.6j + 24.5k6 m>s2 Ans.

318


SolutionSince the velocity components are the function of position but not of time, the flow can be classified as steady but nonuniform. Using the definition of the slope of the streamline,

dy

dx=

v

u;

dy

dx=

10y + 3-(4x + 6)

Ly

1 m

dy

10y + 3= -L

x

1 m

dx4x + 6

110

ln(10y + 3) ` y1m

= -14

ln(4x + 6) ` x1m

110

lna10y + 313

b = 14

lna 104x + 6

bln a10y + 3

13b 110 = ln a 10

4x + 6b 14

a10y + 313

b 110 = a 104x + 6

b 1410y + 3

13= a 10

4x + 6b 52

y = c 411(4x + 6)5>2 - 0.3 dm Ans.

For two dimensional flow, the Eulerian description gives

a =dVdt

+ u dVdx

+ v dVdy


ax =dudt

+ ududx

+ v dudy

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

dv

dt+ u

dv

dx+ v

dv

dy

= 0 + 3 -(4x + 6)(0)4 + (10y + 3)(10) = 310(10y + 3)4 m>s2At point (1m, 1 m),

ax = 434(1) + 64 = 40 m>s2 Say = 10310(1) + 34 = 130 m>s2c

346. A flow field has velocity components of u = -(4x + 6) m>s and v = (10y + 3) m>s where x and y are in meters. Determine the equation for the streamline that passes through point (1 m, 1 m), and find the acceleration of a particle at this point.

319


Ans:

y =41114x + 625>2 - 0.3

a = 136 m>s2u = 72.9 a

346. (continued)

The magnitude of acceleration is

a = 2a 2x + a 2y = 2(40m>s2)2 + (130m>s2)2 = 136 m>s2 Ans.And its direction is

u = tan-1 aayax

b = tan-1 a130 m>s2 40 m>s2 b = 72.9 u Ans.

320


Ans:y = 1.25 mm x = 15.6 mma = 0.751 m>s2u = 2.29 a

SolutionSince the velocity components are a function of both position and time, the flow can be classified as unsteady nonuniform. Using the defination of velocity,

dy

dt= v = 0.03t2; L

y

0dy = 0.03L

t

0t2 dt

y = (0.01t3) mWhen t = 0.5 s,

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

dxdt

= u = 100y = 100(0.01t3) = t3; Ly

0dx = L

t

0t3 dt

x = a14

t4b mWhen t = 0.5 s,

x =14

(0.54) = 0.015625 m = 15.6 mm Ans.

For a two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

= 0 + (100y)(0) + (0.03t2)(100) = (3t2) m>s2 ay =

0v0t

+ u 0v0x

+ v 0v0y

= 0.06t + (100y)(0) + 0.03t2(0) = (0.06t) m>s2When t = 0.5 s,

ax = 3(0.52) = 0.75 m>s2 S ay = 0.06(0.5) = 0.03 m>s2cThe magnitude of acceleration is

a = 2a 2x + a 2y = 2(0.75 m>s2)2 + (0.03 m>s2)2 = 0.751 m>s2 Ans.And its direction is

u = tan-1 aayax

b = tan-1a0.03 m>s20.75 m>s2 b = 2.29 u Ans.

347. A velocity field for oil is defined by u = (100y) m>s, v = (0.03 t2) m>s, where t is in seconds and y is in meters. Determine the acceleration and the position of a particle when t = 0.5 s. The particle is at the origin when t = 0.

321


SolutionSince the velocity components are a function of position but not time, the flow can be classified as steady but nonuniform. Using the definition of the slope of the streamline,

dy

dx=

v

u;

dy

dx=

-y2x2

Ly

6 m

dy

y= -

12L

x

2 m

dx

x2

ln y ` y6 m

=12a1

xb ` x

2 m

ln y

6=

12

a1x-

12b

ln y

6=

2 - x4x

y

6= e12- x4x 2

y = c 6e12- x4x 2 d m Ans.The plot of this streamline is shown in Fig. a.

For two dimensional flow, the Eulerian description gives.

a =0V0t

+ u 0V0x

+ v 0V0y


ax =0u0t

+ u 0u0x

+ v 0u0y

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

= (8x3) m>s2 ay =

dv

dt+ u

dv

dx+ v

dv

dy

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

= (y)m>s2At point (2 m, 6 m),

ax = 8(23) = 64 m>s2 S ay = 6m>s2cThe magnitude of acceleration is

a = 2a 2x + a 2y = 2(64 m>s2)2 + (6m>s2)2 = 64.3m>s2 Ans.And its direction is

u = tan-1aayax

b = tan-1a 6 m>s264 m>s2 b = 5.36 u Ans.

*348. If u = (2x2) m>s and v = (-y) m>s where x and y are in meters, determine the equation of the streamline that passes through point (2 m, 6m), and find the acceleration of a particle at this point. Sketch the streamline for x 7 0, and find the equations that define the x and y components of acceleration of the particle as a function of time if x = 2 m and y = 6 m when t = 0.

322


348. (continued)

Using the definition of the velocity,

dxdt

= u; dxdt

= 2x2

Lx

2 m

dx

2x2= L

t

0dt

-12

a1xb ` x

2 m= t

-12

a1x-

12b = t

x - 24x

= t

x = a 21 - 4t

b mdy

dt= v;

dy

dt= -y

- Ly

6 m

dy

y= L

t

0dt

- ln y ` y6 m

= t

ln 6y

= t

6y

= et

y = (6e-t) mThus,

u = 2x2 = 2 a 21 - 4t

b2 = c 8(1 - 4t)2

d m>s and v = -y = ( -6e-t) m>sThen,

ax =dudt

= -16(1 - 4t)-3(-4) = c 64(1 - 4t)3

d m>s2 Ans.ay =

dvdt

= (6e-t) m>s2 Ans.x(m) 0.5 1 2 3 4 5 6

y(m) 12.70 7.70 6.00 5.52 5.29 5.16 5.08

0 1 2 3 4

(a)

x(m)

y(m)

5 6

1

2

3

4

5

6

7

323


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

SolutionSince the velocity component are a function of position but not time, the flow can be classified as steady but nonuniform. For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y


ax =0u0t

+ u 0u0x

+ v 0u0y

= 30 + (2x2 + 8)(4x) + (-8x)(0)4 = 34x(2x2 + 8) 4 m>s2

ay =0v0t

+ u 0v0x

+ v 0v0y

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

= 3 -8(2x2 + 8) 4 m>s2At point (0, 0),

ax = 4(0)32(02) + 84 = 0ay = -832(02) + 84 = -64 m>s2 = 64 m>s2T

Thus,

a = ay = 64 m>s2T Ans.At point (1 m, 0),

ax = 4(1)32(12) + 84 = 40 m>s2 S ay = -832(12) + 84 = -80 m>s2 = 80 m>s2TThe magnitude of the acceleration is

a = 2a 2x + a 2y = 2(40 m>s2)2 + (80 m>s2)2 = 89.4 m>s2 Ans.And its direction is

u = tan-1 aayax

b = tan-1 a80 m>s240 m>s2 b = 63.4 cu Ans.

Using the definition of the slope of the streamline,

dy

dx=

v

u=

-8x2x2 + 8

; Ldy = -8Lx dx

2x2 + 8

y = -2 ln (2x2 + 8) + C

x 1 m

y

x

324


Ans:At point (0, 0),a = 64 m>s2wAt point (1 m, 0),a = 89.4 m>s2, u = 63.4 cFor the streamline passing through point(0, 0),

y = c 2 ln a 82x2 + 8

b d mFor the streamline passing through point(1 m, 0),

y = c 2 ln a 102x2 + 8

b d m

349. (continued)

For the streamline passing through point (0, 0),

0 = -2 ln 32(02) + 84 + C C = 2 ln 8Then y = c 2 ln a 8

2x2 + 8b d m Ans.

For the streamline passing through point (1 m, 0),

0 = -2 ln 32(12) + 84 + C C = 2 ln 10y = c 2 ln a 10

2x2 + 8b d m Ans.

For point (0, 0)

x(m) 0 1 2 3 4 5

y(m) 0 -0.446 -1.39 -2.36 -3.22 -3.96

For point (1 m, 0)

x(m) 0 1 2 3 4 5

y(m) 0.446 0 -0.940 -1.91 -2.77 -3.52

0

1

2 3 4x(m)

y(m)

55 4 3 2

1

1

1

2

3

4

y = 10

2x2 + 8

2 ln

y = 82x2 + 82 ln

325


SolutionSince the velocity components are a function of position but not time, the flow can be classified as steady but nonuniform. For two dimensional flow, the Eulerian description gives

a =0V0t

+ u 0V0x

+ v 0V0y

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

ax =0u0t

+ u 0u0x

+ v 0u0y

350. The velocity field for a fluid is defined by u = y> (x2 + y2) and v = 34x> (x2 + y2) 4 m>s, where x and y are in meters. Determine the acceleration of particles located at point (2 m, 0) and that of a particle located at point (4 m, 0). Sketch the equations that define the streamlines that pass through these points.

= 0 + a yx2 + y2

b (x2 + y2)(0) - y(2x)(x2 + y2)2

+ a 4xx2 + y2

b (x2 + y2)(1) - y(2y)(x2 + y2)2

= 4x3 - 6xy2

(x2 + y2)3m>s2

ay =0v0t

+ u0v0x

+ v0v0y

= 0 + a yx2 + y2

b (x2 + y2)(4) - 4x(2x)(x2 + y2)2

+ a 4xx2 + y2

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

= 4y3 - 36x2y

(x2 + y2)3m>s2

At point (2 m, 0)

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

(22 + 02)3= 0.5 m>s2 S

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

(22 + 02)3= 0

Thus a = ax = 0.5 m>s2 S Ans.

326


350. (continued)

At point (4 m, 0)

ax =4(43) - 6(4)(0)

(42 + 02)3= 0.0625 m>s2 S

ay =4(03) - 36(42)(0)

(42 + 02)3= 0

Thus

a = ax = 0.0625 m>s2 S Ans.Using the definition of the slope of the streamline,

dy

dx=

v

u=

4x> (x2 + y2)y> (x2 + y2) = 4xy ; Lydy = 4Lxdx

y2

2= 2x2 + C

y2 = 4x2 + C

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

02 = 4(22) + C C = -16

Then y2 = 4x2 - 16

y = {24x2 - 16 x 2m Ans.For the streamline passes through point (4 m, 0)

02 = 4(42) + C C = -64

Then

y2 = 4x2 - 64

y = {24x2 - 64 x 4m

327


Ans:For point (2 m, 0), a = 0.5 m>s2y = {24x2 - 16For point (4 m, 0), a = 0.0625 m>s2y = {24x2 - 64

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

x(m) 2 3 4 5 6 7 8 9y(m) 0 4.47 6.93 9.17 11.31 13.42 15.49 17.55

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

x(m) 4 5 6 7 8 9y(m) 0 6.00 8.94 11.49 13.86 16.12

350. (continued)

y(m)

0 1 2 3 4 5 6 7 8x(m)

9

5

5

10

10

15

20

15

20

y = 4x2 16

y = 4x2 64

328


Ans:-1.23 m>s2

SolutionHere V only has an x component, so that V = u. Since V is a function of time at eachx, the flow is unsteady. Since v = w = 0, we have

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

ax =0u0t

+ u0u0x

=00x

36(1 + 0.4x2)(1 - 0.5t)4 + 36(1 + 0.4x2)(1 - 0.5t)4 00x

36(1 + 0.4x2)(1 - 0.5t)4 = 36(1 + 0.4x2)(0 - 0.5)4 + 36(1 + 0.4x2)(1 - 0.5t) 4 36(0 + 0.4(2x))(1 - 0.5t)4Evaluating this expression at x = 0.25 m, t = 1 s, we get

as = -3.075 m>s2 + 1.845 m>s2 = -1.23 m>s2 Ans.

Note that the local acceleration component ( -3.075 m>s2) indicates a deceleration since the valve is being closed to decrease the flow. The convective acceleration (1.845 m>s2) is positive since the nozzle constricts as x increases. The net result causes the particle to decelerate at 1.23 m>s2.

x

0.3 m

AB

329


SolutionThe n - s coordinate system is established with origin at point A as shown in Fig. a. Here, the component of the particles acceleration along the s axis is

as = 3 m>s2Since the streamline does not rotate, the local acceleration along the n axis is zero,

so that a0V 0t

bn

= 0. Therefore, the component of the particles acceleration along

the n axes is

an = a0V 0t bn + V2R = 0 +

(5 m>s)216 m

= 1.5625 m>s2Thus, the magnitude of the particles acceleration is

a = 2a 2s + a 2n = 2(3 m>s2 )2 + (1.5625 m>s2)2 = 3.38 m>s2 Ans.

*352. As water flows steadily over the spillway, one of its particles follows a streamline that has a radius of curvature of 16 m. If its speed at point A is 5 m>s which is increasing at 3 m>s2, determine the magnitude of acceleration of the particle.

(a)

streamline

an

as

n

A

S

16 mA

330


Ans:72 m>s2

353. Water flows into the drainpipe such that it only has a radial velocity component V = (-3>r) m>s, where r is in meters. Determine the acceleration of a particle located at point r = 0.5 m, u = 20. At s = 0, r = 1 m.

SolutionFig. a is based on the initial condition when s = 0, r = rD. Thus, r = 1 - s. Then the radial component of velocity is

V = -3r

= a- 31 - s

b m>sThis is one dimensional steady flow since the velocity is along the straight radial line. The Eulerian description gives

a =0V0t

+ V 0V0s

= 0 + a- 31 - s

b c - 3(1 - s)2

d = c 9

(1 - s)3d m>s2

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

a = a 90.53

b m>s2 = 72 m>s2 Ans.The positive sign indicates that a is directed towards positive s.Note there is no normal component for motion along a straightline.

(a)

r

s

r0 = 1 m

r 0.5 m

s

u

331


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

354. A par

Fluid Mechanics Chapter 3 Solution Guide

Documents

Transcript of Fluid Mechanics Chapter 3 Solution Guide