Solucionario mecanica de fluidos merle c potter david c wiggert 3ed PDF

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Solucionario de la tercera edicion del libro de potter...

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/,%52681,9(5,67$5,26 8369 N/m

2.86

V

W water

S

water

d3

x

+ 5.1

or S d3.

x

10-5

2 x

5 x

d/ 2

0.

> 0.

< 1435 N/m3 W

V

water

water

S

water

d3

S d3.

water

4 1 1 S d / 12 (d / 2 Sd / 2) d ( ). GM 3 12S 2 2 Sd If GM = 0 the cube is neutral and 6S2 – 6S + 1 = 0. 6 36 24 S = 0.7887, 0.2113. 12 The cube is unstable if 0.2113 < S < 0.7887. Note: Try S = 0.8 and S = 0.1 to see if GM 0. This indicates stability.

2.87

16 9 16 4 = 6.5 cm above the bottom edge. 16 16 4 9. 5 16 8. 5 16S A 4 = 6.5 cm. 8 2 8 SA 16 .5

As shown, y G

27

h = Sd.

G C

h

130 + 104 SA = 174 + 64 SA . 2.88

SA = 1.1.

4 8 1 8 7 = 4. x 16 1 8 4 8 16 8 8 16 8 8 12 . 16 4 . 5 8 1 15 . 8 7 For G: y = 4.682. 1.2 16 . 5 8 1. 5 8

a) y

16

x

12 . 16 .5 8 4 1.5 8 4 1. 2 16 . 5 8 1.5 8

4

= 2.5.

= 2.364. 0.136 C G

G must be directly under C. .136 . =11.3 . tan .682

1 1 2 3.5 4 2 2 2 2 2 2 b) y = 2. x = 1.25 4 2 2 4 2 2 . 2 .5 4 1.5 4 For G:y 1.2 4 2 .5 1 1 .5 7 = 2.34. x 12 = 1.182 . . . 12 4 5 2 15 2 1. 2 4 . 5 2 1. 5 2 .068 . y = 0.34, x = 0.068. tan = 11.3 . .34 4 2 2

2.89

The centroid C is 1.5 m below the water surface. Using Eq. 2.4.47: GM The barge is stable.

2.90

2.91

3 l 8 / 12 1.5 1.777 1.5 l 8 3

0.277

0.

8 .485 3.414 1697 . 1 = 1.8 m. CG 1.8 1.5 = 0.3 m. .485 . 8 16 97 3 Using Eq. 2.4.47: GM l 8.485 / 12 . 3 1.46 . 3 116 . . Stable. 34.97l y

(A)

p plug Fplug

2.92

CG = 1.5 m.

2000 0 p plug A

h

2000 0 6660 (1.2

24070

0.02

2

5 ) 9.81

24070 Pa .

30.25 N

20 H . H = 8.155 m. pmax = 9810 (8.155 + 2) = 99 620 Pa 9.81 4 b) pmax = (g + az) h = 1000 (9.81 + 20) 2 = 59 620 Pa c) pmax = 1.94 60 (–12) – 1.94 (32.2 + 60) (–6) = 2470 psf or 17.15 psi d) pmax = 1.94 (32.2 + 60) (–6) = 1073 psf or 7.45 psi

a) tan

28

0.682

2.93

2.94

z The air volume is the same before and after. A h 10 0.5 8 = hb/2. tan . .9 81 b h 9 .81 h . h = 2.856. Use dotted line. 4 B 2 10 1 2. 5w 2.5 2.452 4. w = 0.374 m. 2 a) pA = –1000 10 (0 – 7.626) – 1000 9.81 2.5 = 51 740 Pa or 51.74 kPa b) pB = –1000 10 (0 – 7.626) = 76 260 Pa or 76.26 kPa c) pC = 0. Air fills the space to the dotted line.

Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure. a) 60 000 = –1000 ax (0–8) – 1000 4=

h 2 9 .81 2 ax

9.81 0

60 = 8 ax + 24.52 – 9.81

a 2x – 10.1 ax + 19.67 = 0

8a x . 9.81

2 a x – 5.1 ax + 1.44 = 0

2.5

2 a x – 7.6 ax + 8.266 = 0

ax – 4.435 = 1.1074

ax .

8ax . 9. 81

8 ax . ax – 1.31 = 1.574 19. 81 ax = 0.25, 4.8 m/s2

c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 + 60 = 8 ax + 37.0 – 14.81

8a x 9.81

ax = 2.64, 7.46 m/s2

b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10) 60 = 8 ax + 49.52 – 19.81

2.5

ax .

8 ax ). 14. 81

8 ax . ax – 2.875 = 1.361 14.81 ax = 1.32, 6.28 m/s2

ax .

2.95

a) ax = 20 .866 = 17.32 m/s2 , az = 10 m/s2 . Use Eq. 2.5.2 with the peep hole as position 1. The x-axis is horizontal passing thru A. We have pA = –1000 17.32 (0 – 1.232) – 1000 (9.81 + 10) (0 – 1.866) = 58 290 Pa b) pA = –1000 8.66 (0 – 1.848) – 1000 (9.81 + 5) (0 – 2.799) = 57 460 Pa c) The peep hole is located at (3.696, 5.598). Use Eq. 2.5.2: pA = –1.94 51.96 (0 – 3.696) – 1.94 (32.2 + 30) (0 – 5.598) = 1048 psf d) The peep hole is located at (4.928, 7.464). Use Eq. 2.5.2: pA = –1.94 25.98 (–4.928) – 1.94 (32.2 + 15) (–7.464) = 932 psf

2.96

a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2, p(z) = –1000 10 (–7.626) – 1000 9.81(z) = 76 260 – 9810 z

29

b h

w 1

C

x

2 .5

( 76 260 9810 z)4 dz = 640 000 N or 640 kN

FAB 0

b) The pressure on the bottom BC is p(x ) = –1000 10 (x – 7.626) = 76 260 – 10 000 x . 7 . 626

(76 260 10 000 x )4dx = 1.163

FBC

106 N or 1163 kN

0

c) On the top p(x ) = –1000

10 (x – 5.174) where position 1 is on the top surface:

5 . 174

( 51 740 10 000 x) 4 dx = 5.35

Ftop

105 N or 535 kN

0

2.97

FAB

a) The pressure at A is 58.29 kPa. At B it is pB = –1000 17.32 (1.732–1.232) – 1000 (19.81) (1–1.866) = 8495 Pa. Since the pressure varies linearly over AB, we can use an average pressure times the area: 58 290 8495 1.5 2 = 100 200 N or 100.2 kN 2

z

x

b) pD = 0. pC = –1000 17.32 (–.5–1.232) 1000 19.81(.866–1.866) = 49 810 Pa. 1 FCD 49 810 1.5 2 = 74 720 N or 74.72 kN. 2 58.29 4981 . c) pA = 58 290 Pa. pC = 49 810 Pa. 1.5 = 81.08 kN. FAC 2

2.98

Use Eq. 2.5.2 with position 1 at the open end: a) pA = 0 since z2 = z1. pB = 1000 19.81 0.6 = 11 890 Pa. pC = 11 890 Pa. b) pA = –1000 10 (.9–0) = –9000 Pa. pB = –000 10 (.9)–1000 9.81(-.6) = –3114 Pa pC = –1000 9.81 (–.6) = 5886 Pa.

z 1

C

c) pA = –1000 20 (0.9) = –18 000 Pa. pB = –1000 20 0.9–1000 19.81( 0.6) = –6110 Pa. pC = 11 890 Pa 25 d) pA = 0. pB = 1.94 (32.2-60) = 112 psf. pC = –112 psf. 12 37.5 e) pA = 1.94 60 = 364 psf. 12 37.5 25 pB = 1.94 60 – 1.94 32.2 = –234 psf. 12 12

30

A

B

x

pC = –1.94

32.2

25 = 130 psf. 12

37. 5 = 182 psf. 12 37 .5 pB = –1.94(–30) – 1.94 62.2 12 25 pC = –1.94 62.2 = 251 psf. 12

f) pA = 1.94

2.99

30

25 = 433 psf. 12

Use Eq. 2.6.4 with position 1 at the open end: 50 2 = 5.236 rad/s. 60 2 a) p A 1000 5.236 (.6 1.5) 2 = 11 100 Pa. 2 1 pB 1000 5.2362 .92 + 9810 .6 = 16 990 Pa. 2 pC = 9810 .6 = 5886 Pa. 1 1000 5.2362 0.62 = 4935 Pa. 2 p B 1 1000 5.2362 0.62 + 9810 0.4 = 8859 Pa. 2 pC = 9810 0.4 = 3924 Pa.

b) p A

c) p A pB

1 2 1 2

1.94 1.94

pB

1 2 1 2

5.236

2

37. 5 12 37. 5 12

2

= 259.7 psf. 2

62.4

25 = 389.7 psf. 12

25 = 130 psf. 12

pC = 62.4 d) p A

5.236

2

2

2

1.94

5.236

1.94

5.2362

pC = 62.4

22. 5 = 93.5 psf. 12 2 15 = 171.5 psf. 22. 5 + 62 .4 12 12

15 = 78 psf. 12

31

z 1

C

A

B

r

2.100 Use Eq. 2.6.4 with position 1 at the open end. a) p A 1 1000 102 (0 – 0.92 ) = –40 500 Pa. 2 pB = –40 500 + 9810 0.6 = –34 600 Pa. pC = 9810 0.6 = 5886 Pa. b) p A 1 1000 102 (0 – 0.62 ) = –18 000 Pa. 2 pB = –18 000 + 9810 0.4 = –14 080 Pa. pC = 9810 0.4 = 3924 Pa. c) p A

1 2

102 0

1.94

1 2

102

1.94

pB = –341 + 62.4

22.52 12 2

r

C

B

= –947 psf.

25 = –817 psf. 12

pB = -947 + 62.4 d) p A

37.52 144

z A

1

pC = 62.4

25 = 130 psf. 12

pC = 62.4

15 = 78 psf. 12

= –341 psf.

15 = –263 psf. 12

2.101.1Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0. 1 1 1000 2 (0 – 0.452 ) – 9810 (0 – 0.6). = 7.62 rad/s. a) 0 = 2 z 1 2 2 1000 (0 – 0.3 ) – 9810 (0 – 0.4). = 9.34 rad/s. b) 0 = 2 2 25 . 1 18. 75 r = 7.41 rad/s. – 62.4 1.94 2 0 c) 0 = 2 12 2 12 1 d) 0 = 2

1.94

2

11.252 – 62.4 12 2

15 . 12

= 9.57 rad/s.

2.102 The air volume before and after is equal. 1 2 r02 h = 0.144. 6.2 2. . 0r h 2 a) Using Eq. 2.6.5: r02 52 / 2 = 9.81 h h = 0.428 m pA = 1 1000 52 0.62 – 9810 (–0.372) 2 = 8149 Pa. b) r02

7 2 / 2 = 9.81 h. h = 0.6 m.

pA =

1000 2

72

0.62 + 9810

0.2 = 10 780 Pa.

32

z

h

A

r0

r

c) For

= 10, part of the bottom is bared. .26 .2 1 r02 h 1 r12 h1 . 2 2 Using Eq. 2.6.5: 2 2 2 2 r0 r1 h1 . h, 2g 2g 2g 2 2g 2 h h 0 144 . or 2 2 1

h2

z

r0 h A

r

h1

0.144 10 2 . 2 9.81

h12

Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. h = 0.859 m, r1 = 0.108 m. pA = 1 1000 102 (0.62 – 0.1082 ) = 17 400 Pa. 2 0 .144 20 2 . 1.6h – .64 = 2.936. h = 2.235 m. d) Following part (c): h 2 h12 2 9.81 pA = 1 1000 202 (0.62 – 0.2652 ) = 57 900 Pa r1 = 0.265 m 2 2.103 The answers to Problem 2.102 are increased by 25 000 Pa. a) 33 150 Pa b) 35 780 Pa c) 42 400 Pa 2.104 p (r ) p (r ) p(r )

1 2r2 2 500 2 r2 2

500 (r

d) 82 900 Pa

g[ 0 (.8 h)].

dA = 2 rdr

9810(. 8 h ) 2

if h < .8. if h > .8.

r ) 2 1

dr

.6

a) F

p 2 rdr

3 12 ( 500r

2

3650r dr ) = 6670 N.

0

(We used h = .428 m) .6

b) F

p 2 rdr

r3 (24 500

2

r dr ) = 7210 N. (We used h = 0.6 m) 1962

0 .6

c) F

p 2rdr

3 (50 000r(

2

2 .108r dr ) = 9520 N. (We used r1 = 0.108 m)

. 108 .6

d) F

p 2 rdr

(200 000r( 3

2

.2652 r dr ) = 26 400 N. (r1 = 0.265 m)

. 265

33

CHA PTER 3

Introduction to Fluids in M otion 3.1 pathline streamline

3.2

Pathlines:

streakline

Release several at an instant in time and take a time expo sure of the subsequent motions of the bulbs.

Sreakline: Co ntinue to release the devises at a given location and after the last one is released, take a snapshot of the “ line” of bulbs. Repeat this for several different release locations for additional streaklines. 3.3 streakline pathli ne t =0

hose tim e t

boy

3.4 y

streakline at t = 3 hr pat hline t = 2 hr

streamline s t = 2 hr x

34

3.5

dx t2 2 dt t2 2t c 1

a) u

x

2 xy

2

b) x

t2

x

3.6

3.7

v dr

uiˆ

c1

vjˆ

2

x

4.

4) 8

y 8x

12y

parabola.

0.

v v (V dr ) z

wkˆ

udy vdx

using iˆ,

kˆ ˆj . iˆ

jˆ

kˆ

Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest.

a) At t b) At t At t c) At t

Several college students would be positioned at each intersection and quantities w ould be recorded as a function of time.

2an d ( 0, 0, 0) V

(D )

22

2m / s .

2an d ( ,1 2, 0) V 32 22 2 a nd (0,0,0 ) V 0. 2 2and ( 1 , 2, 0) V ( 2) ( 2an d ( 0, 0, 0) V

At t

( 4)2

2 d ( , 1 , 2)0 V an ( 51.4

2

3. 606m / s. 8)2

8. 246m / s.

4m / s .

2 (

)42

)24

(

6 / s. m

10 5 jˆ)

A simultaneous solution yields nx

(They must

4/5 and n3/5. y

both have the same sign. 3.10

(35, 25) 39.8o

8 , and c 2

dxiˆ dy ˆj d z ˆk

At t

3.9

(27, 21)

parabola.

4y

c1 . y

Eulerian:

3.8

streamlines t=5s

y

2

2

4 2(

2xy

v V

y

2t

y 2

dy t2 dt y t2 c

2 y

y x

v

a) cos

v V n$

nx

v V i$ / V 0.

(3$i

2 , ny 13

(1 2) /

32

2$j ) ( n x i$ n y $j )

3 or n$ 13

22

33 .69 o

0.832 . 3n x n 2x

0.

2n y 0 n y2 1

1 ( 2i$ 3 j$ ). 13

35

ny n2x

3 nx 2 9 2 nx 1 4

v V i$ / V

b) cos

v V n$

ny

1 , nx 17

c) cos

v V i$ / V

5

v a) V

89

v dr (x

2)dy

Integrate: t

2(1 2 ln3) t x 2 nl x v b) V

drv

xydy

( 8)2

0.6202.

8 $j ) ( n x i$ n y $j )

5n x n 2x

8 89

(x 2

89

( 8i$

xyi$

2y 2 j$

64 2 ny 25

0.

dy . y

2

dy j$)

C.

.0

0 or

2 )4i$ y tj$

2 (x

y 2 tdx

Integrate:

t tan 2

0 .9636 .

nx

8n y 0 n y2 1

dy 2 dx . x y ln( y / C). 2ln( 1) l n( 2 / C). 2y n y 2 x 2. l ( / ).

2 y 2 dx

.0

51.67 o

dx ( i $ dy j $ )

(dxi$

1

0

or

x 2

C

yt tan

1

(dx i$ dy j$ )

x

4 1 . y

x 2

0.

dy . y2

tdx 2

2 tan 2

0 .9636

36

2

4n y n y2 1

$j).

5j$ ).

xdx dy . t x 2ln x x 2 2 C. C = 0.8028. 2 y 0.8028

4) dy

C

1

xdx x 2

0 or t

xtdx

0.

or n$

2i) $ xtj $

x(

.0

drv

0.

52

Integrate: 2lnx C 2. l nx 2

v c) V

n y $j)

104 o nx 16n 2y

0.2425. 2 n x 8 ny 0 n 2x n y2 1

1 ( 4 $i 17

, nx

0.

( 8) 2

4 or n$ 17

5/

( 5$i

0.

ny

3.11

( 2 $i 8 $j) ( n x$i

0.

v V n$

( 2) 2

2/

1

1 2

C

1 . 2

8 ny 5 n 2y 1

3.12

v v V V u v t x

v a

(C )

2

a

3.13

v v DV V a) u Dt x v v V V b) u v x y v v V V c) u v x

2

8

v V v y

w

16

v V w z v V t v

v V w z v

y

v v V V w y z

V

V t

z

2xy ( 2yiˆ)

v

v

V

V

v

x

w

y

v

v

V

V t

z

16iˆ 8 iˆ 16 jˆ.

4x i $ 4y j $ = $i8

4$j

17.89 m/s v V =0. t x2 ( 2i $ ) 2y (2j $ ) x 2t ( 2xt $i

2 yt$j)

$) 2 xyt(2 xtj$ 2 ztk

x i ( $ 2yzj $) 2xyz 2 ( xzj $ )tz

= xi$ ( 2yz 4x 2 yz 2 = 2i$ 114 j$ 15k$

v

1 2

3.14

v a)

v b) c)

v v

d) 3.15

3.16

u x

xx

1 xy

yz

2 xyj

2 ( xyj $ tk $) zk $

2xyzt) $j ( zt 2

z) k$

w v $ 1 u w $ 1 v u $ i j k. 2 2 y z z x x y 1 u$ k 20 yk$ = 20 k$ 2 y 1 1 1 ( 0 0)i$ (0 0 ) $j (0 0 ) k$ = 0 2 2 2 1 1 1 ( 2zt 0) i$ ( 0 0) $j (2 yt 0 ) k$ = 6 i$ 2 k$ 2 2 2 1 1 1 ( 0 2xy )i$ (0 0 )j$ ( 2 yz 0 )k$ = 2i$ 3k$ 2 2 2

The vorticity v a) = 40i$ a)

x2 $i

100$ j 54$ k

68$ i

2yzk d) u

y 2 ( 2 xiˆ 2 yjˆ)

v 2 . Using the results of Problem 3.7: v v d) v = b) 0 c) = 12 i$ 4 k$

v

0, u

v

2

y

x

1 2

v

w

z

y

v y

0,

zz

20y

20,

xz

yy

w z 1 u 2 z

0. w

0 0.

rate - of strain

37

0,

x

20 0

20

0

0

0

0

0

4i$ 6k$

b)

c)

xx

2,

yy

2,

zz

0.

xy

0,

xz

0,

yz

0.

xx

2 xt

xy

8,

2 xt

yy

1 yt (2 ) 2

2,

d)

xx

xy

1,

2 0 2 xz

yy

1 ( 2 yz ) 2

8 6

3,

3

10

40 cos r2

1 10 r 10

a

r

0,

a) ar

1 (0) 2 0

t

2.

0,

yz

10

sin 2

(10

80 sin r3

1 ( 2xy ) 2

2.

40 sin r2 r

1

40 ( sin ) r2

80 cos r3

1 r v =0 z

2.5)( 1)1.25 ( 1) = 9.375 m/ s2. 10

1600 sin cos r4

0,

10

6.

2

At (4, 180 ) 3.18

zz

80 cos r3

40 cos r2

1 100 r a = 0. b)

40 r2

1 zt (2 ) 2

yz

12 2 2 2

0 a) ar

0,

4.

6 4

xz

rate-o f strain = 3

2 yt

zz

12,

1

3.17

8,

1 (0) 2 2 0

xz

8 rate-of strain =

2 0 0 rate-of strain = 0 2 0 0 0 0

10

240 cos r4

=0

40 sin r2 r