Instructor’s Manual
to accompany

APPLIED F LUID MECHANICS

Sixth Edition

Robert L.  Mott
University of Dayton

PEARSON
¡...
This work is protected by United States copyright laws and is provided
solely for the use of lnstructors in ‘teaching thei...
Online IM to Accompany

APPLIED FLUID MECHANICS

Sixth Edition

 

Robert L.  Mott
University of Dayton

PEARSON
i

Prenti...
CHAPTER FIFTEEN

FLOW MEASUREMENT . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .....
CONTENTS

CHAPTER ONE

THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS . ... ... ... ... ... ... ...  ..  1
CHAPTER ...
APPLIED FLUID MECHANICS
Sixth Edition

Robert L.  Mott

Prentice-Hall Publishing Company

 

Description of Spreadsheets I...
lPower Sl:  The objective of problems of this type is to compute the amount of power
required to drive a pump to deliver a...
system shown in Figure 11.17. p;  is assumed to be 100 kPa.  pg is computed to be 78.21
kPa.  Then Ap =  -21.79 kPa. 

lPr...
Problem 11.18 is solved by the given spreadsheet data.  Only pipe friction Iosses are
included and the solution computes t...
Note that no spreadsheet system using Sl Metric units for operating point is included in this
set.  It would be good pract...
HYDROFLO 2, HCALC,  and PumpBase Software
by TAHOE DESIGN SOFTWARE
Included on the CD with this book

APPLIED FLUID MECHAN...
database for later use.  System data can be entered and displayed in standard English,  Sl or
a mix of units. 
HYDROFLO al...
In the author’s own teaching of a first course in fluid mechanics,  a design project is assigned
after class coverage of C...
CHAPTER ONE

THE NATURE OF FLUIDS AND THE
STUDY OF FLUID MECHANICS

 

Conversion factors

1.1 125o mm(l 111/103 mm)= l.25...
1.18

1.19

1.20

1.21

1.22

1.23

1.24

1.25

1.26

l .27

1.28

1.29

1.30

U: 

L): 

u: 

¿:1.50km><3600s = lo3skmlh
...
1.31

1.32

1.33

1.34

1.35

1.36

1.37

1.38

1.39

1 .40

1.41

l .42

D:  j2(KE):  2(212mN-m)X]()‘3N 103gxlkg-m zls6ml...
The definition of pressure

1.43

1.44

1.45

1.46

1 .47

1.48

1 .49

1.50

1.51

1.52

p =  F/ A =  2500 lb/ [7r(3.O0 ín...
p[7rD2] _ 500 lb(7r)(D m)? 
4 m2 4

1.53 F"pA— «392.7 D210

D(in) 1275i? ) F(lb)

 
2.00 4.00 1571
3.00 9.00 3534
4.00 16....
Bulk modulus

1.57

1.58

1.59

1.60

1.61

1.62

1.63

1.64

1.65

1.66

Ap =  —E(A V/ V) =  430000 psi(—0.01)=  1300 psi...
1.68

1.69

1.70

1.71

1.72

1.73

1.74

1.75

1.76

1.77

1.78

1.79

:1.35x103NX1kg-m/ s2

3'57? N “Mg

m: 

w =  mg = ...
1.82

1.83

1.84

1.85

1.86

1.87

1.88

1.89

1.90

1.91

lN

V= pg=1.964kg/ m‘x9.81m/ s= ><  =  19.27 N/ m’
3
sg= —————...
1.92 w, , =  35.4 N — 2.25 N =  33.15 N
V, , =  Ad =  (7rDZ/4)(d) =  puso m)3(.2o m)/4 =  3.53 x 10-3 m3

yv= í=  N/ ma z ...
Z 3
1.104 sg= —p——=  ¡‘mag >< m} x “Ïg xíflo im)
pw cm 1000 kg 10 g m
p =  (sg)(pw) =  l.20(1.94 slugs/ fts) =  2.33 slugs/...
.593 B:  mo 82a 2: c223 e32: oa
ouflflonEoh w589» baena una Emma?  owzoonm Su mcouflüm - 892G “¿n88 N acoscmmmmd.  ufinnEoo

 ...
CHAPTER TWO

VISCOSITY OF FLUIDS

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

2.10

2.11

2.12

2.13

2.14

2.15

2.16

2...
2.21

2.22

2.23

2.24

2.25

2.26

2.27

2.28

2.29

2.30

2.31

2.32

2.33

2.34

2.35

2.36

2.37

2.38

2.39

2.40

2....
2.42

2.43

2.44

2.49

2.50

2.51

2.52

2.53

2.54

2.55

14

Terminal velocity is that velocity achieved by the sphere ...
2.56

2.57

2.58

2.59

2.60

2.61

2.62

2.63

2.64

2.65

17 = = 4500 cP [(1 Pa-s)/ (1000 cP)] =  4.50 Pa-s
77 =  4.50 P...
2.66

2.67

2.68

2.69

2.70

2.71

2.72

2.73

2.74

2.75

2.76

16

From Fig 2.12, kinematic viscosity =  25 7 SUS
v =  ...
Kinematic Víscosity Conversions

Problem 2.77
SAE Víscosity Grades - Engine Oils
Kinematic Viscosi at 100 de
(mmzls) sus
M...
18

Kinematic Vlscoslty Converslons

Problem 2.79
iSO Víscosity Grades

Kinematic Viscosi at 40 de

Min
1.98 2.2 32.5 33.3...
CHAPTER TH REE

PRESSURE MEASUREMENT

 

Absolute and gage pressure

3.1 Pressure =  force/ area;  p =  F/ A

3.2 Absolute...
3.21

3.22

3.23

3.24

3.25

3.26

3.27

3.28

3.29

3.30

3.31

3.32

3.33

pabs =  4.1 + 101.3 =  105.4 kPa(abs)
par” =...
3.41

3.42

3.43

3 .44

3.45

3.46

3.47

3.48

3.49

3.50

3.51

3.52

3.53

3.54

pa“ :  pA — y, ,(64 in) :  180 psig —...
3.55

3.56

pm,  = —34.0 kPa + 70h‘,  + ywhw
=  —34.0 kPa + (0.85)(9.8l kN/ m3)(0.50 m) + (9.81 kN/ m3)(0.75 m)
pbm =  -34...
3-69 PB -M2 fi) : M3 ft) +M11 ft) ZPA
PA “PB =  m9 ft) -M3 ft) =  6“ ‘b “iz 3:99AM

x9fi ><
pA —p¡;  =3.90 psi — 1.17 psi = ...
3.82

3.83

3.84

3.85

3
848.7 lb X Zgfiínx l ft

ft3 1728 in’

pam,  = ymh =  = 14.05 psia

3
848.7lbX30_65inX lft

fi? ‘ ...
CHAPTER FOUR

FORCES DUE TO STATIC FLUIDS

 

Forces due to gas pressure

4.1 F=  Ap-A;  where Ap = pm —p, ,,, ¡,, ,; A = ...
4.10

4.11

4.12

4.13

Force on valve = p-A;  A =  1c(0.095 m)2/4 =  7.088 x 104 m2

p =  ywh =  9315“ x 1.80 m =  17.66 ...
4.16

4.17

62.4 lb
fiS

FR =  y(h/2)A =  x 6 ft x (12 ft)(20 ft)

FR =  89850 lb
11,,  = 2/3 h =  2/3(12 fi) =  8.0 ft
Mome...
4.20 ha =  3.0 m;  Lc =  hR/ cos 30° =  3.464 m
IZDZ Z 7r(2.4 m)2

A =  = 4.524 m2
4 4
4 4
I:  ¡[D :   z “329 m4
64 64
FR ...
4.23 LR =  hR/ cos 40° =  0.855/cos 40°
=  1.116 m

2
”('3°) =  0.1607 m2

A =  (.300)?  +

FR =  
=  (0.90)(9.8l kN/ m3)(...
4.26

4.27

30

-2o=2s in
+25 =  50.0 in=4.167 ft

gï- LR=  40.0 in =  3.333 fl

A =  (s)(50) =  400 ín2(l ft2/144 in’)
=  ...
a =  10 in/ cos 30° =  11.55 in

LR= a+ 8 + í =  11.55 + 8.0+ 8.48 =28.03 in
hR = = LR.  cos 30° =  24.27 in [2.023 ft]

F...
4.30

4.31

4.32

32

hR =  LR

=  0.375 + 0.150 =  0.525 m
A =  (0.60)(0.30) + 1t(0.3Ü)2/4
A =  0.2507 m2

hR-L, 

“EL?  ...
4.33 Rectangular Wall

FR =  701/314
=  62.4 Ib/ ft3 x 4.0 fi x (8.0 ft)(l5.0 ft)
=  29950 lb

L,  = %h= %(8.0 ft) =  5.333...
4.37 Rectangular Wall:  FR =  yIuA
FR =  (1.1o)(9.31 kN/ m3)(4.60/2)m(4.6)(3,0)m2
=  343 kN
FR acts l/3 from bottom or 2/3...
4.40 h,  = L‘.  = 4.00 ft;  A =  (4.oo)(1.25) =  5.00 f? 

FR =  yhcA =  (62.4 lb/ fl3)(4.OO fi)(5.00 a2)

=  1243 lb
1C =  ...
4.42 hR.  = 33 +y =  38 + 5 cos 30° =  42.33 in
L,  = hR/ cos 30° =  48.88 in
FR= yhRA
A =  1r(10)2/4 =  78.54 in’
z 62.4 ...
_ p" M 2.50 lb ft’ 1728 in’ _ .  _
4.45 hR 7 m2(1.43)(62_41b) >< 1 fi;  48.41 m — 4.034 fi I See Pmh 426
h, , =  h,  + h,  =...
4.48

4.49

38

¡——vs«»—1

Fy=  yV=  yAw

FR z (0.826)(9.81 kN/ m3)(1.389 m2)(2.50 m) =  28.1 kN
A =  AR + AR =  (0.62)(1....
4.50 FV:  yRV=  yRAw

A =  A1 + A;  = (7.50)(9.50) + 7r(7.50)2/8

A =  93.34 112

FV 3 YoAW =  (0.85)(62.4)(93.34)(3.50)
F...
4.52

4.53

40

FV =  yAw

A =  A,  + A2 =  (1 .2o)(2.so) + 7:(1.20)2/4 =  4.491 m2
FV =  (9.81 kN/ m3)(4.491 m2)(1 .50 m)...
4.54

4.55

4.56

4.57

4.58

A =  TEDZ/ g =  fl(36)2/8 =  503.9 m2 =  3.534 f8
FV =  yAw =  (0.79)(62.4)(3.534)(5.0) =  s7...
4.59

4.60

4.61

4.62

42

See Prob.  4.57. w¡=  10.2 lb

w,  : yy:  (30.0 1b/ fi3)(0.l64 ft’) :  4.92 lb

Fm,  = w,  — w¡...
4.63

4.64

(See Prob.  4.57) For any depth 2 6.00 in,  Fm,  = 70.1 lb down
Method of Prob.  4.62 used for h < 6.00 in and...
Buoyancy

5.1

5.2

5.3

5.4

5.5

5.6

5.7

44

zFV: o:w+ r-F, 
F,  : yyV=  (10.05 kN/ m3)(0.45)(0.6O)(0.30)m3 :  0.814 k...
5.8

5.9

5.10

5.11

5.12

5.13

W"Fb"FsP=0= W"YoVd-Fsp
FH,  = w — y, ,V, , :  14.6 lb — (0.90)(62.4 1b/ ft’)(40 m3)
Fsp:...
5.14

5.15

5.16

5.17

5.18

5.19

46

From Prob.  5.13: wH + ws =  0.02 + 0.02485 =  0.04485 lb =  w =  Fb

2
7'00) -1.s...
5-20 W =  F1»;  YIVI =  'YSWVd

 

.  / 3
V, ¡= V,  1L- :  -   =  V,  - 0.863; 86.3% submerged;  13.7% above
Ïsw ' m
 W = ...
5.26

5 .27

5.28

5.29

48

7r(.45)2

12;, ” :  7,, V,,  : 9.81 kN/ m3 x x .03 m’
=  0.0468 kN
zac :  7,, V,,  : wc + w, ...
5.30 From Prob.  5.29, wr=  7.02 N
2 2
VS =  ¡tf .  L =  7r(.038 m)

 :  7WVS :  9.81 kN/ m3 x 9.073 x 103 m3 x 103 N/ kN ...
5.35 Given:  7; :  12.0o 11)/ ft’,  7C =  150 lb/ ft’,  wc :  600 lb
Find:  Tension in cable

Float only:  ZF7= 0 =  wp+ T...
5.38 Fs"w,  FLC-yy,  y¡Vc'-V, (yc y¡)

Vc=  ”D2-L=  ><l0.0in>< “i.  3 —
4 4 17281n

=  0.1636 ft3

0.2841b 1728 m’ _ 62.41...
5.41

5.42

5.43

5.44

5.45

52

W= Fb: yswVd: yswAX

X:  w z 4500001b
ySwA (64.0 1b/ ft’)(20 0x50 ft)
=  7.031 ft

ya,  ...
5.46 X=  8.00 ft;  yd,  =X/2 =  4.00 ft; ycg =  12.00 ft
ymc = ycb + 
Scow is stable if ymd > ycg
MBmd,  =ymd —yd,  =ydd —...
5.51

5.52

From Prob.  5.30, X=  118 mm
Vs =  Vol.  of steel bar =  9.073 >< 104 mm’

flDé ><X

Vas =  Sub.  vol.  of cup ...
5.53 From Prob.  5.26, Fig.  5.25; t=  88.4 mm
X=700+t=788.4mm

yd,  =X/2 =  394.2 mm

4 2
MB:  ¿ 7:1) /64 D

Vd Z [zD2/4:...
5-56 W—Fb= O='YWVtot"YoVd= YW-4H"%AX g_¡d¡nd, ghx1zh
X:  yWAH =  ywH =  (32 lb/ ÍÏ3)(6 in) / 
70A 7a 0.90(62.4 lb/ ÍÏ3)

=...
5.59

5.60

W =  Fb =  YTI/ d: YIAdL
W =  'YoÁtotL

=  á (.600)(.300)(1.20)(2.36)kN
=  0.255 kN

Ad =  GX+  (XXX)
= <B—2X>...
5.61 Entire hull:  9.. .3.
_ Am +412)»;  u‘ "'
ycg * fi
:  (.72)(.4o) + (2.88)(1.2)
3.60

yc =  1.040 m T
Siïbmerged Volume...
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Solucionario mecanicadefluidosr-mott-120604135953-phpapp01
Próxima SlideShare
Cargando en…5
×

Solucionario mecanicadefluidosr-mott-120604135953-phpapp01

1.522 visualizaciones

Publicado el

Publicado en: Ingeniería
0 comentarios
1 recomendación
Estadísticas
Notas
  • Sé el primero en comentar

Sin descargas
Visualizaciones
Visualizaciones totales
1.522
En SlideShare
0
De insertados
0
Número de insertados
5
Acciones
Compartido
0
Descargas
85
Comentarios
0
Recomendaciones
1
Insertados 0
No insertados

No hay notas en la diapositiva.

Solucionario mecanicadefluidosr-mott-120604135953-phpapp01

  1. 1. Instructor’s Manual to accompany APPLIED F LUID MECHANICS Sixth Edition Robert L. Mott University of Dayton PEARSON ¡ff-D Prentice Hall Upper Saddlc River, New Jersey Columbus, Ohio
  2. 2. This work is protected by United States copyright laws and is provided solely for the use of lnstructors in ‘teaching their courses and assessing student iearning. msseminatíon or sale of any part of this work (includ- ing on the World Wide web) wili desuoy the integrity of the work and is, not pormíttedïha work and maten-int: from i! should never ha made available to students except by ¡nstructors using the accom- panying taxi in diet! dasses. All recípionts of this work aro expected to ¿hide by these restríctions and to honor the intended pedagogia! pur- poses and tha needs of other instructora: who rely en these materials. Copyright © 2006 by Pearson Education, lnc. , Upper Saddle River, New Jersey 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopyíng, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice Hall“ is a trademark of Pearson Education, Inc. Pearson® is a registered trademark of Pearson plc Prentice Hall® is a registered trademark of Pearson Education, lnc. Instructors of classes using Mott, Applied F luíd Mechanics, Sixth Edition, may reproduce material from the instructor’s manual for classroom use. 10987654321 ISBN 0-l3—l72355-3
  3. 3. Online IM to Accompany APPLIED FLUID MECHANICS Sixth Edition Robert L. Mott University of Dayton PEARSON i Prentice Hall Upper Saddle River, New Jersey Columbus, Ohio
  4. 4. CHAPTER FIFTEEN FLOW MEASUREMENT . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . 235 CHAPTER SIXTEEN FORCES DUE TO FLUIDS IN MOTION . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 240 CHAPTER SEVENTEEN DRAG AND LIFT . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 251 CHAPTER EIGHTEEN FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES . ... ... ... ... ... ... ... .. 260 CHAPTER NINETEEN FLOW OF AIR IN DUCTS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 269 SPREADSHEETS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 273 iv
  5. 5. CONTENTS CHAPTER ONE THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS . ... ... ... ... ... ... ... .. 1 CHAPTER TWO VISCOSITY OF FLUIDS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . . 12 CHAPTER THREE PRESSURE MEASUREMENT . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 19 CHAPTER FOUR FORCES DUE TO STATIC FLUIDS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . . 25 CHAPTER FIVE BUOYANCY AND STABILITY . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . 44 CHAPTER SIX FLOW OF FLUIDS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 61 CHAPTER SEVEN GENERAL ENERGY EOUATION . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . 81 CHAPTER EIGHT REYNOLDS NUMBER, LAMINAR FLOW, AND TURBULENT FLOW AND ENERGY LOSSES DUE TO FRICTION . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . . 94 CHAPTER NINE VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW FOR NONCIRCULAR SECTIONS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . .113 CHAPTER TEN MINOR LOSSES . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . .129 CHAPTER ELEVEN SERIES PIPE LINE SYSTEMS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . .141 CHAPTER TWELVE PARALLEL PIPE LINE SYSTEMS . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ..187 CHAPTER THIRTEEN PUMP SELECTION AND APPLICATION . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . .213 CHAPTER FOURTEEN OPEN CHANNEL FLOW . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ..218
  6. 6. APPLIED FLUID MECHANICS Sixth Edition Robert L. Mott Prentice-Hall Publishing Company Description of Spreadsheets Included on the CD in the Book Introduction This book includes a CD-ROM that contains ten Computational aids that are keyed to the book. The files are written as Microsoft Excel spreadsheets using Version 2002 on Windows XP. The ten spreadsheets are all included in one workbook called Series Pipe Systems-Master. The names of each spreadsheet described below are on the tabs at the bottom of the workbook when it is opened. You must choose which is appropriate for a given problem. Most names start with either I, ll, or III indicating whether the spreadsheet is for a Class l, Class ll, or Class lll pipe line system as defined in Chapter 11 of the text. The spreadsheets are designed to facilitate the numerous calculations required to solve the variety of problems in Chapter 11 Series Pipeline Systems. Many of the spreadsheets appear in the text. Others were prepared to produce solutions for the Solutions Manual. The given spreadsheets include data and results from certain figures in the text, from example problems, or in problems from the end of Chapters 8, 11, and 13 containing the analysis and design procedures featured in the programs. The following sections give brief descriptions of each spreadsheet. Many are discussed in the text in more extensive detail. It is expected that you will verify all of the elements of each spreadsheet before using them for solutions to specific problems. Using the Spreadsheets: It is recommended that the given spreadsheets be maintained as they initially appear on the CD. To use them for solving other problems, call up the master workbook in Excel and use the "Save as" command to name it something different. That version can then be used for a variety of problems of your own choice. Be careful that you do not modify the contents of critical cells containing complex equations. However, you are encouraged to add additional features to the spreadsheets to enhance their utility. The principles involved in the spreadsheets come from Chapters 6 — 11 and you should study the concepts and the solution techniques for each type of problem before using the given spreadsheets. It is highly recommended that you work sample problems by hand first. Then enter the appropriate data into the spreadsheet to verify the solution. ln most spreadsheets, the data that need to be entered are identified by gray-shaded areas and by italic type. Results are typically shown in bold type.
  7. 7. lPower Sl: The objective of problems of this type is to compute the amount of power required to drive a pump to deliver a given amount of fluid through a given system. Energy Iosses are considered. All data must be in the listed Sl units. The solution procedure is for a Class l series pipe line system. The following is a summary of the steps you must complete. 1. Enter the problem identification information first. The given data in the spreadsheet are for example Problem 11.1 for the system shown in Figure 11.2. 2. Describe two appropriate reference points for completing the analysis of the general energy equation. 3. Specify the required volume flow rate, Q, in m3/s. 4. Enter the pressures (in kPa), velocities (in m/ s), and elevations (in m) at the reference points in the System Data: at the top of the sheet. If the velocity at either reference point is in a pipe, you may use a computed velocity from v = Q/ A that is included in the data cells for the two pipes. ln such cases, you enter the Excel command "= B20" for the velocity in pipe 1 and "= E20" for the velocity in pipe 2. 5. Enter the fluid properties of specific weight (in kN/ m3) and kinematic viscosity (in mz/ s). 6. Enter pipe data, including flow diameter D (in m), pipe wall roughness a (in m from Table 8.2), and length L (in m). Other pipe-related data are computed by the spreadsheet. Equation 8-7 is used to compute the friction factor. 7. Enter energy loss coefficients, K, for all loss-producing elements in both pipes. See Chapters 8, 10, and 11 for the proper form for K for each element and for necessary data. The value for K for pipe friction is computed automatically from known data in the "Pipe" section. Specify the number of like elements in the "Qty. "column. Enter brief descriptions of each element so your printout is keyed to the given problem and so you can observe the energy loss contribution of each element. Space is given for up to eight different kinds of energy loss elements. Enter zero for the value of the K factor for those not needed. 8. The Results: section at the bottom of the spreadsheet includes the total energy loss h<, the total head on the pump h¿, and the power added to the fluid by the pump P, . If you enter an efficiency for the pump, the power input to the pump P, is computed. I Power US: Same as Power Sl: , except U. S. Customary units are used. The given solution is for Problem 11.29 for the system shown in Figure 11.26. The first reference point is taken just before the pump inlet. Therefore there are no friction Iosses or minor Iosses considered in the suction pipe. The length is given ta be zero and all Kfactors are zero for Pipe 1. In Pipe 2, pipe friction, the loss in the elbow, and the loss in the nozzle are included. [Pressure Sl: Most of the layout and data entry for this spreadsheet are the same as those in the first two spreadsheets, lPower SI, and lPower US. The difference here is that the analysis determines the pressure at one point in the system when the pressure at some other point is known. Class I systems with one or two pipe sizes including minor Iosses can be analyzed. This spreadsheet uses the known pressure at some starting point and computes the pressure at a downstream point, considering changes in elevation, velocity head, pipe friction and minor Iosses. The example is the solution of Problem 11.7 for the vi
  8. 8. system shown in Figure 11.17. p; is assumed to be 100 kPa. pg is computed to be 78.21 kPa. Then Ap = -21.79 kPa. lPressure US: This spreadsheet is virtually identical to lPressure SI: except for the different unit system used. The example is the solution for Problem 11.3 using the system shown in Figure 11.13. An important difference occurs here because the objective of the problem is to compute the upstream pressure at the outlet of a pump when the desired downstream pressure at the inlet to the hydraulic cylinder is specified. You should examine the contents of cells B7 and B8 where the pressures are listed and E43 and E44 where the actual calculations for pressure change are performed. The changes demonstrated here between l Pressure SI: and I Pressure US: show how the spreadsheets can be adapted to specific types of problems. Knowledge of the fluid mechanics of the problems and familiarity with the design of the spreadsheet are required to make such adjustments accurately. ll-A & Il-B Sl: This spreadsheet solves Class Il series pipe line problems using either Method II-A or Method Il-B as described in Chapter 11 of the text. SI Metric units are used. Example Problem 11.3 using the system shown in Figure 11.7 is solved in the given sheet. You should review the details of these sheets from the discussions ¡n the book. In summary, Method lI-A is used for finding the maximum velocity of flow and volume flow rate that a given pipe can deliver while limiting the pressure drop to a specified value, without any minor Iosses. This is accomplished by the upper part of the spreadsheet only. Enter the pressures at two system reference points in the System Data: near the top of the sheet. Enter other data called for in the gray shaded cells. Refer to Example Problem 11.2 in the text for an illustration of the use of just the upper part of this sheet to solve Class ll problems without minor Iosses. The lower part of the spreadsheet implements Method Il-B for which minor Iosses are considered in addition to the friction Iosses in the pipe. The solution procedure is iterative, requiring you to assume a volume flow rate in the upper part of the lower section of the spreadsheet that is somewhat lower than the result of the Method Il-A solution directly above. Enter the minor loss coefficients in the lower part of the spreadsheet. As each estimate for flow rate is entered, the pressure at a target point, called pg, is computed. You must compare this pressure with the desired pressure. Successíve changes in the estimate can be made very rapidly until the desired pressure is acceptable within a small tolerance that you decide. II-A & Il-B US: This spreadsheet is identical to ll-A & Il-B: except for using the U. S. Customary unit system. Problem 11.10 is solved in the example. The system in this problem has no minor Iosses so the upper part of the spreadsheet shows the most pertinent data and results. The lower part has been adjusted to use the same volume flow rate as the result from the upper part and all minor Iosses have been set to zero. The result is that the pressure at the target point, pz, is very near the desired pressure. The small difference is due to rounding and a possible difference between the result from Equation 11.3 used to solve for Q in Method Il-A and the calculation of friction factor and other terms in Method lI-B. III-A & Ill-B Sl: Class III series pipe line problems require the determination of the minimum required size of pipe to carry a given volume flow rate of fluid with a limiting pressure drop. Both Method III-A and Method lll-B as described in the text are shown on this spreadsheet. lf only pipe friction loss is considered, then only the upper part using Method lll-A is pertinent. vii
  9. 9. Problem 11.18 is solved by the given spreadsheet data. Only pipe friction Iosses are included and the solution computes that the minimum acceptable pipe flow diameter is 0.0908 m (90.8 mm). But the problem statement calls for the specification of the smallest standard Type K copper tube. So the lower part of the spreadsheet shows the application of a 4-in Type K copper tube having a flow diameter of 0.09797 m (97.97 mm). The spreadsheet then computes the predicted pressure at the end of the system for the given volume flow rate, 0.06 m'/ s. Note that the result predicts that the pressure at the end of the tube would be 48.13 kPa. In reality, the pressure there is zero as the problem states that the flow exits into the atmosphere from the end of the tube. Actually, then, the 4-in tube will permita higher flow rate for the same pressure drop. You could use the spreadsheet ll-A 4 lI-B Sl: to compute the actual expected volume flow rate when using the 4-in Type K copper tube. Class III systems with minor losses are demonstrated in the next spreadsheet. lll-A & llI-B US: This spreadsheet is identical to Ill-A & lll-B Sl: described above except for the use of U. S. Customary units instead of SI Metric units. The solution to Example Problem 11.6 from the text is shown and this spreadsheet is included in the text with extensive description of how it is used. Please refer to the text. The problem includes some minor losses so that both the upper part (Method Ill-A) and the lower part (Method lll-B) are used. The result from Method Ill-A predicts that the minimum acceptable pipe flow diameter is 0.3090 ft if no minor Iosses are considered. Using a standard 4-in schedule 40 steel pipe (D = 0.3355 ft) with the minor Iosses included in Method lll-B shows that the pressure at the target point, pz, is greater than the minimum acceptable pressure. Therefore, that pipe size is satisfactory. System Curve US: This spreadsheet is the same as that shown in Figure 13.41 in Chapter 13 of the text. It is used to determine the operating point of a pump selected for the system shown in Figure 13.40 and described in Example Problem 13.4. Refer to the extensive discussion of system curves and this spreadsheet in Sections 13.10 and 13.14. The spreadsheet includes two pages. The first page is an analysis of the total head on the pump when the desired volume flow rate, 0.5011 ft3/s (225 gal/ min), flows ¡n the system. This sheet is basically the same as that in the spreadsheet called lPower US: , discussed earlier. But the final calculation of the power delivered by the pump to the fluid has been deleted. After seeing the required total head on the pump for the desired flow rate, the user has selected a centrifugal pump, model 2x3—1O with a 9-in impeller, using the pump performance curves from Figure 13.27. This pump will actually deliver somewhat more flow at this head. The spreadsheet was used to compute data for the system curve that is a plot of the total head versus the volume flow rate (capacity) delivered. Several data points for the resulting total head for different flow rates from zero to 275 gal/ min were computed. These were entered on page 2 of the spreadsheet and the system curve was plotted on the graph shown there. Simultaneously, data from the actual performance curve for the selected pump were entered and plotted on the same graph. Where the two curves intersect is the operating point, predicting that the pump will deliver approximately 240 gal/ min. Then you must go back to Figure 13.27 to determine the other performance parameters at this operating point, efficiency, power required, and NPSHR (net positive suction head required). viii
  10. 10. Note that no spreadsheet system using Sl Metric units for operating point is included in this set. It would be good practice for you to copy this given spreadsheet and convert it to Sl Metric units. You should examine the contents of each cell to determine if the equations must be modified with different conversion factors to achieve an accurate result. Friction Factor: This is a simple spreadsheet whose sole purpose is to compute the friction factor using Equation 8-7 from Section 8.8 of the text. We refer to this equation as the Swamee-Jain Equation for its developers. See Reference 3 in Chapter 8. The spreadsheet shows the computation of the friction factor for the data from Problem 8.28. Data entry is similar to that used in the other spreadsheets described above. Either Sl Metric or U. S. Customary units can be used because only dimensionless quantities are used in the equation. But units must be consistent within a given problem. You might want to use this spreadsheet to test your ability to read accurately the Moody Diagram, Figure 8.6. ix
  11. 11. HYDROFLO 2, HCALC, and PumpBase Software by TAHOE DESIGN SOFTWARE Included on the CD with this book APPLIED FLUID MECHANICS Sixth Edition Robert L. Mott This book includes a CD that contains student versions of three powerful software programs for the solution of a variety of pipeline design and analysis problems. Created by Tahoe Design Software of Nevada City, CA, HYDROFLO 2, HCALC, and PumpBase can be used for problem solutions in Chapters 8 and 10 —13 of the book. More information about Tahoe Design Software and the professional versions of these programs can be found on their website wvvw. tahoesoft. com. HYDROFLO 2 is a unique fluid conveyance system design tool for full pipe lncompressible flow conditions. lt makes it easy to model and analyze fluid transport systems found in industrial process, water supply, petroleum transport, mining de-watering and HVAC systems. During the design process, you view a vertical elevation-scaled representation of your fluid conveyance system in HYDROFLO's workspace. Elements (such as pipes, valves, etc. ) can be added to your design with drag-and-drop and cut-and-paste ease. HYDROFLO's clipboard enables near instant creation of duplicate parallel lines. Element data and analysis results can be viewed simply by placing the cursor over an element. HYDROFLO's Group Editor eliminates repef. itive and tedious editing tasks. The academic version of HYDROFLO can model liquid conveyance systems with single sources and single discharges and up to 20 pipes, 20 fittings and valves, 3 pumps, and up to nine parallels. Gauges can be placed anywhere in a line to determine the pressure head at a point of interest, to start or end parallels, or to depict elevation changes or vertical bends in a line. Many conveyance systems include pumps in series or parallel and HYDROFLO can easily analyze these systems. See Section 13.15 in the book. Many types of fluid flow problems can be solved, such as - VaIidation/ calibration of existing pipeline systems. - Modeling a proposed system’s operation. - Determination of line head Iosses at a specific flow rate. (termed a forced-flow system). - Analysis of cavitation and net positive suction head (NPSH) problems. - Comparison of equivalent SI unit to English unit designs. - Modeling of recirculating and gravity (non-pumped) flow systems. Pipe head Iosses can be calculated using the Hazen-Williams equation for water flow (Section 8.9 in the book) or the Darcy-Weisbach equation for other types of lncompressible fluids. We use the term Darcy’s equation in the book. See Section 8.5. HYDROFLO's extensive liquid property database can be accessed to obtain hundreds of liquid properties. Accurate analyses of liquid transport systems require use of precise liquid property data. Your custom liquid property descriptions can be saved in HYDROFLO's
  12. 12. database for later use. System data can be entered and displayed in standard English, Sl or a mix of units. HYDROFLO also performs NPSHA (net positive suction head available) calculations to determine possible cavitation situations. Once a pump's operating conditions are found, PumpBase can be used to find the best pumps for the application. PumpBase provides an extensive searchable database for commercially available pumps that meet the requirements for a given system. Only a few data values must be input for basic operation of the program; the operating point for a desired volume flow rate at the corresponding total dynamic head TDH and the total static head h¿, , as found from Eq. (13- 11). The program fits a second degree equation between those two points and plots that as the system curve. See Section 13.10 ¡n the book. The program will search its database, select several candidate pumps that meet the specifications, and report a list that is ordered by pump efficiency. You may select any candidate pump and call for its performance curve to be displayed along with the system curve and listing of such operating parameters as the impeller diameter, actual flow rate, power required efficiency, and NPSHR. You are advised to verify that the selected pump meets all requirements. More input data are required if the fluid is not cool water, a limit for NPSHR is to be specified, or a certain type of pump is desired. HCALC is a handy calculator tool that resides in the system tray for easy access. It performs the calculations for any variable in the flow rate equation, Q = Av, or the Darcy- Weisbach equation, hL = f(L/ D)(v2/2g), when sufficient data are entered for fluid properties, pipe dimensions, roughness, and so forth. Reynolds number and friction factor are also calculated. You may select either SI or U. S. Customary units. Suggestions for use of these programs: As with any software, it is essential that the user have a solid understanding of the principles involved in the analyses performed by the software as well as the details of data entry and interpretation of results. lt is advised that practice with hand calculations for representative problems be completed before using the software, Then use the results of known, accurately-solved problems with the software to verify that it is being used correctly and to gain confidence in its capabilities. The following types of problems and projects can be solved with these programs: Energy Iosses due to friction in straight pipes and tubes (Chapter 8) Energy Iosses due to valves and flttings (Chapter 10) Analysis of series pipeline systems (Chapter 11) Analysis of parallel pipeline systems (Chapter 12) Analysis of pumped pipeline systems (Chapters 11-13) Selection of a suitable pump for a given system (Chapter 13) Design aid for design problems such as those outlined at the end of Chapter 13. Extensive system design as a senior design project. ooooodoo xi
  13. 13. In the author’s own teaching of a first course in fluid mechanics, a design project is assigned after class coverage of Chapter 11 on Series Pipeline Systems. Each student is given a project description and data adapted from the design problems listed at the end of Chapter 13 after the Problems. They are expected to produce the design of a pumped fluid flow system, Given the need to pump a give volume flow rate of a specified fluid from a particular source to a destination, completely define the configuration of the system, including: o Pipe types and sizes - Length of pipe for all parts of the system Layout of the piping system Location of the pump Types and sizes of all valves and flttings and their placement List of materials required for the system Analysis of the pressure at pertinent points Determination of the total dynamic head on the pump Specification of a suitable pump having good efficiency and able to deliver the required volume flow rate in the system as designed - Assurance that the specified pump has a satisfactory NPSHR to prevent cavitation over the entire range of expected system operation - Written report documenting the design and analyses performed using good report writing practice The use of the Tahoe Design Software programs after learning the fundamentals of fluid system design analysis allows more comprehensive exploration of possible designs and the completion of a more optimum design. The experience is also useful for students as they move into career positions where the use of such software is frequently expected. xii
  14. 14. CHAPTER ONE THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS Conversion factors 1.1 125o mm(l 111/103 mm)= l.25 m 1.2 1600 mm2[1 m2/(103 mm)2] = 1.6 x 10* m’ 1.3 3.65 x 103 mm3[1 m3/(l03 mm)3] = 3.65 x 10*“ m3 1.4 2.05 m2[(103 mm)2/m2] = 2.05 x 10‘ mm’ 1.5 0.391 m3[(103 mm)3/m3] = 391 x 10° mmt 1.6 55.0 ga1(0.00379 m3/gal) = 0.203 m’ 80km 103m 1h X 1.7. >< h km 3600 s = 22.2 m/ s 1.8 25.3 ft(0.3048 m/ ft) = 7.71 m 1.9 1.86 mi(l .609 km/ mi)(103 m/ km) = 2993 m 1.10 8.65 ín(25.4 mrn/ ín) = 220 mm 1.11 2580 ft(0.3048 m/ ft) = 786 m 1.12 480 ft3(0.0283 m3/rt3) = 13.6 m3 1.13 739o cm3[1 m3/(1o0 cmf] = 7.39 x 10*’ m3 1.14 6.35 L(1m3/l000 L) = 6.35 x 10* m’ 1.15 6.0 ft/ s(O.3048 m/ ft) = 1.83 m/ s 3 3 r ¡‘16 250tlft X0.028ï'>m Xlmin zLlsnf/ s min ft 60 s Consistent units in an equation . 3 1.17 u= í=050kn1xl0 m =47.2m/ s t 10.6 s km The Nature of Fluids 1
  15. 15. 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 l .27 1.28 1.29 1.30 U: L): u: ¿:1.50km><3600s = lo3skmlh t 5.25 s:1000ftx lmi X3600s _48.7mm¡ t 14s 5280ft h ¿:1.0rníX36O0s 382mm‘ t 5.75 h _ 2s _ (2)(3.2 km) x 10’ m X1mm2 a t“ (4.7 mm)’ km (60 s)’ ¿EL (21920 11,3, a 9.8lm/ s2 = 8.05 x 10*‘ nn/ s‘ 3 - 2 a_2.2r_(2)(3.2.kn12)><10 mx lfi xlmmz _0'264t'_t_ t (4.7m1n) km 0.3048m (605) s = E5: j———_. (2)(53“‘2)x. ——1fi =0.524s a 32.2fi/ s 12m 2 2 _ 2 KE: I!¿1_= =10_81<8 m 40316.“, 2 2 s1 2 2 2 2 KE= —"ï’—= í36°i)xÉ16kmJxÜ°3IÏ) ><—¿h——, =35.6x10’kg}mz 2 2 h km (36005) s KE=35.6kN-m 2 2 2 KE= ÏlL=75kgx(6'85m] =1.76x10’kg'm =1.76kN-m 2 2 S s2 =2(KE)= (2)(38.6N-m)X h 2X1kg-mX(36O0s)ZX 1km’ ¿,2 1 31.5km sl-N h’ (103 m): 2 , ,,= ¡&¿ smgskg (31.5)(10) O —3 _ 3 m=2(K2E)= (2)(94.6mN2m)x10 Nxlkïg mXIO g =3Mg u (2.25m/ s) mN s -N kg 2 D: 2(KE)= 2(15Nm)X1kgm/ s zlásm/ s m l2kg N Chapter 1
  16. 16. 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1 .40 1.41 l .42 D: j2(KE): 2(212mN-m)X]()‘3N 103gxlkg-m zls6mls m 175g mN kg sz-N 2 2 _ 2 KE= EL= xL1EïÉ = g_0o¡¡, .¡t 2 2 slug _ mu’ _ wtf = (8O00lb)(10mi)2 x 1h‘ X 62800)’ KE-T’ 2g (2)(32.2 fi/ s2)(h)2 (3600s)’ mr KE: ¡(, . fi (2)(32.2)(36o0)’ = 26700 lb - ft = m_ifi= w_tfi= (1501b)(20ft/ s)’ z 9 KE 2 2 2g (2)(32.2fr/ s) 2 m: 2am): 2(151b ft)=6_201b s u’ (2.2 ft/ s’)2 = 6.20 slugs 2g(KE)=2(32.2ft)(38.61b-fi)(h’)x 1m? X(3600s)2 W: u’ 5109.5 mi)’ (5280 mt ht w: (2)(32.2)(3;8.6)(3620O)’ ¡b = 3.04 ¡b (19.5) (5280) 2 u: jzgurs): j2(32.2ft/ s)(101b-0) 4.631% w 301b 2(32.2 ft/ sz )(30 oz - in) X 1 fi u= i230“) = =5.1s ft/ s w 6.0 oz ERA = ——————-————39_ más >< —————-——-9 mnmgs = 2.49 runs/ game 141 mnmgs game m x M x 150 inníngs = 52 runs game 9 inníngs 40 runs x Éï >< “Lin” = 129 innings 2.79 runs game ERA = LWX x É = 3.59 runs/ game 123 mnmgs game The Nature of Fluids
  17. 17. The definition of pressure 1.43 1.44 1.45 1.46 1 .47 1.48 1 .49 1.50 1.51 1.52 p = F/ A = 2500 lb/ [7r(3.O0 ín)2/4] = 354 lb/ in’ = 354 psi p = F/ A = 870O1b/ [7r(1.5O in)2/4] = 4923 psi 3 3 2 15:5- 120m“, x10 Nx“ mm) —2.72x10°—IÏ =2.72MPa A 7r(75mm) /4 kN m’ m’ 3 3 2 p Ï- 381m0 ¡N x00 mm) —30.9x10°ï =30.9MPa A 7r(40mm) /4 m2 mz p z 5 = .. _._. _—6°°_° n: = 119 psi A 7r(8.01n) /4 p e 5 e ———————18°°_° n: = 3667 psi A 7r(2.501n) /4 6 2 2 F: pA_20.5><120 NXIIÜOÁnm) Xuoïm Y =4025kN m mm F = pA = (60001b/ in2)(7r[2.00in]2/4) = 18850 lb D_ F F 4F 4F p= —= 2 = Z: ThenD= w» A rrD /4 7rD íïp ' = 236 ¡n " M5000 lb/ inz) 4F ¡tp 3 — 460X121“), -50.5x10"m=50.smm 7r(15.0><10 N/ m) Chapter 1
  18. 18. p[7rD2] _ 500 lb(7r)(D m)? 4 m2 4 1.53 F"pA— «392.7 D210 D(in) 1275i? ) F(lb) 2.00 4.00 1571 3.00 9.00 3534 4.00 16.00 6283 5.00 25.0o 9817 6.00 36.00 14137 7.00 49.00 19242 8.00 64.00 25133 1.54 p_F_ F2 _ 4172 _ 4600011:) _ 63626 psi A ¡ID /4 ¡ID 7z'(Dm) D D(in) Dïfinï) p(psi) 2.00 4.00 1592 3.00 9.00 707 4.00 16.00 398 5.00 25.00 255 6.00 36.00 177 7.00 49.00 130 8.00 64.00 99 1.55 (Variable Answers) Example: w = 160 lb (4.448 N/ lb) = 712 N _ F g 712N X (103 mm)2 A 7r(20 mm)2/4 m2 p = 2.27 >< 10° Pa (1 psi/6895 Pa) = 329 psi — 2.77 x 10° Pa = 2.27 MPa 1.56 (Variable Answers) usíngp = 2.27 MPa F= pA = (2.27 x 10° N/ m2)(7r(0.250 m)2/4) = 111 x 103 N =111 kN F = 111 kN (1 lb/4.448 N) = 25050 lb The Nature of Fluids
  19. 19. Bulk modulus 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 Ap = —E(A V/ V) = 430000 psi(—0.01)= 1300 psi Ap = —896 MPa(—0.0l) = 8.96 MPa Ap = —E(A V/ V) = —3.59 x 10° psi(—0.01)= 35900 psi Ap = "24750 MPa(—0.01)= 247.5 MPa Ap = —E(A V/ V) = 4139000 psi(—0.01) = 1890 psi Ap = 4303 MPa(—0.0t) = 13.03 MPa AV/ V= —0.0l; AV= 0.01 V= 0.01 AL Assume arca of cylinder docs not change. AV= A(AL) = 0.01 AL Then AL = 0.01 L = 0.0l(l2.00 in) = 0.120 in A_l/ _'_i_ -3000 psi V E 189000 psi = ——0.0159 = —1.59% Al: _ —20.0 MPa — = -0.0l53 = —1.53% V 1303 MPa Stiffitess = Force/ Change in Length = F/ AL Bulk Modulus = = ¿’3- = 1 AV/ V AV Butp = F/ A; V= AL; AV= —A(AL) -F AL FL E = — >< = ———-—— A —-A(AL) A(AL) F _ ¿g _ 189000 lb ; :(0.5in)’ (AL) L ín2(42 ín)4 = 884 lb/ in F _ EA _ 189000 lb ¡[(0.5 inf (AL) L m2(10.0m)(4) = 3711 lb/ in F _ E4 _ 189000 lb 74200111)“ ——— — _ 2 _ = 14137 lb/ in (AL) L m (42.01n)(4) Use large diameter cylinders and short strokes. Force and mass 1.67 2 m: w_ 610N Xlkg m/ s zózlzkg g G 9.81 m/ s“ N 4.2 times higher 16 times higher Chapter 1
  20. 20. 1.68 1.69 1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 :1.35x103NX1kg-m/ s2 3'57? N “Mg m: w = mg = 825 kg >< 9.81 m/ s? = 8093 kg-m/ sz = 8093 N w= mg=450gx 1101? x9.8lm/ s2 =4.4lkg-m/ s2 = 4.41 N g Z m = ÏÏ = ————327'28g’ 2 = 0.242 u’ S = 0.242 slugs g . s m = K = »-a-——-——42'0 lb = 1.304 slugs g 32.2 ft/ sz 2 w= mg=1.s8 slugsx 32.2 ft/ sz x 33.13 = 50.9 lb slug 2 w= mg = 0.258 slugs >< 32.2 ft/ sz x É = 8.31 lb slug m = Ï = ————16° lb 2 = 4.97 slugs g 32.2 ft/ s w = 1601b x 4.448 N/ lb = 712 N m = 4.97 slugs x 14.59 kg/ slug = 72.5 kg m = 1K = ———————I'OO lb = 0.0311 slugs g 32.2 ÍÏ/ sz m = 0.0311 slugs >< 14.59 kg/ slug = 0.453 kg w = 1.00 lb x 4.448 N/ lb = 4.448 N F= w= mg= 1000 kg >< 9.81 m/ sz = 9810 kg-m/ sz = 9810 N F= 9810 N >< 1.01b/4.448N= 2205 lb (Variable Answers) See problem 1.75 for method. Density, specific weight, and specific gravity 1.80 1.81 yB = (sg)¡3yw = (0.876)(9.81 kN/ m33) = 3.59 kN/ m: pg = (sg)gpw = (0.876)(1000 kg/ m ) = 876 kg/ m _ 2 y_12.02NX s2 xlkg m/ s p-g m3 9.8lm N - 1.225 kg/ m“ The Nature of Fluids
  21. 21. 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.89 1.90 1.91 lN V= pg=1.964kg/ m‘x9.81m/ s= >< = 19.27 N/ m’ 3 sg= ——————7°° = ————-——8'86°1‘N/ “: = o.903 at s°c yw@4 c 9.81kN/ m 3 sg= ———7° o = ————8'483kN/ “: =0.865 at50°C 7W@4 c 9.81kN/ m y= Ï,-V= Ï=———————2'25"N = o.o173m’ V y 130.4 kN/ m’ V= AL = 7rD2L/4 = ¡[(0.150 m)2(0.l00 m)/4 = 1.767 >< 1o” m3 m 1.56kg 3 = —=——— =883k p” V 1.767x10"3m’ gm IN 103N kg = =883k 3 9.81 / ’ ——————=8.66 =8.66— ya pag g/ m x ms xlkgqïl/ Sz x m3 m: sg = po/ pw @ 4°c = 883 kg/ m3/1000 kg/ m3 = 0.883 y = (sg)(yw @ 4°c) = l.258(9.81 kN/ ma) = 12.34 kN/ m3 = w/ V w = yV= (12.34 kN/ m3)(0.50 m3) = 6.17 kN 3 2 = w= 6.17kN2X10 Nxlkg-nm/ s =629kg g 9.81m/ s kN m w = yV= (sg)(yw)(V) = (0.68)(9.8l kN/ m3)(0.095 m3) = 0.634 kN = 634 N IN y: pg= (1200 kg/ m’)(9.8lm/ s2)[kg_rn/ S2] = 11.77 kN/ m3 3 5g: p = l200kglm :1” pw@4°C 1oookg/ m3 V= ï=——í—22'°N x lkN =272x10"m’ y (0.826)(9.81kN/ m’) l0’N _ _1080kg 9.81m IN 1kN _ F IOSOkg/ m’ = / = í =1.os 5g p p” IOOOkg/ m’ p = (sg)(pw) = (0.789)(1000 kg/ m3) = 789 kg/ ms y = (sg)(yw) = (0.789)(9.81 kN/ m‘) = 7.74 kN/ m’ Chapter 1
  22. 22. 1.92 w, , = 35.4 N — 2.25 N = 33.15 N V, , = Ad = (7rDZ/4)(d) = puso m)3(.2o m)/4 = 3.53 x 10-3 m3 yv= í= N/ ma z kN/ m3 . X m 3 = &=———»——: :1 8;“: y‘ m 1.93 V= Ad= (rcDZ/4)(d) = nao m)3(6.75 m)/4 = 530.1 m3 w = yV= (0.68)(9.8l kN/ m3)(53o.1 m3) = 3.536 x 103 kN = 3.536 MN m = pV= (O.68)(1000 kg/ m3)(53o.1 m3) = 360.5 x 103 kg = 360.5 Mg 1.94 www, oil = yaa - V“, = (9.42 kN/ m3)(0.o2 m3) = 0.1884 kN w 0.1884 kN 1.95 w = yV= (2.32)(9.81 kN/ m3)(1.42 >< 10”‘ m3) = 3.23 x 1o’3 kN = 3.23 N 1.96 y = (sg)(yw) = 0.876(62.4 111/113) = 54.7 ¡w113 p = (sg)(pw) = 0.876(1.94 slugs/ f?) = 1.70 slugs/ ft’ 0.0765 lb / ft3 l slug 2 x —2- = 2.38 x 10*’ slugs/ ff 32.2 ft/ s 1 lb - s m 1.97 p= Z-= g 1 lb - sz/ ft slug 1.98 y = pg = 0.00381 s1ug/ f13(32.2 ft/ s’) = 0.1227 lb/ ft’ 1.99 sg = y, /(yw @ 4°c) = 56.4 lb/ ft3/62.4 113/113 = 0.904 at 40°F sg = yo/ (yw @ 4°c) = 54.0 1b/113/62.4 113/113 = 0.865 at 120°F 1.100 V= w/ y = 50o lb/834 lb/ ft’ = 0.600 113 7.50 lb X 7.48 gal Í V lgal fi3 y 56.1 113/113 lb - s3 = 56.1 113/113 = 1.74 slugs/ fé sg= ___7o___. =_'____3 = 0399 yw @4°c 62.4 lb m (62.4lb)(50ga1) (1 ftJ) 1.102 = V: 1.258 W y ( ) ft3 7.48gal = 525 lb 2 3 1.103 w= yV= pgV= —-—————l'3zlbs x32‘2fix25.oga1x l“ fi‘ s’ 7.48 gal = 142 lb The Nature of Fluids
  23. 23. Z 3 1.104 sg= —p——= ¡‘mag >< m} x “Ïg xíflo im) pw cm 1000 kg 10 g m p = (sg)(pw) = l.20(1.94 slugs/ fts) = 2.33 slugs/ ft’ y = (sg)(yw) = (1.20)(62.4 lb/ fi3) = 74.9 11,/ rr’ = 1.20 5.0 lb fi3 0.0283 m3 (102 cm)3 . ___. :_. x T (o 826)62 4 lb X a: 3 : 2745 cms . . m 1.105 V=11= 7 1.106 y = (sg)(yw) = (1.08)(62.4 113/113) = 67.4 lb/ ft’ 1.107 p = (o.79)(1.94 slugs/ ft’) = 1.53 slugs/ ft’; p = 0.79 g/ cm’ -3 “O8 7a_w_ (7.95-0.50)1b x17281n V (n(6.oin)’/4)(s. o¡n) ft’ sg = yo/ yw = 56.9 1b/ ft3/62.41b/ fi3 = 0.912 = 56.9 lb/ ft’ 2 2 1.109 V-A-d-flï -d—”(3(; fi) w = yV= (0.68)(62.41b/ ft’)(15550 ft’) = 6.60 x 105 lb x 22 ft — 15550 ft3 x 7.48 gal/ ft’ = 1.16 x 105 gal 1.110 ww = ymV= (59.69 1b/ ft3)(5 gal)(1 ft3/7.48 gal) = 39.9o lb w 39.9o lb f8 7.48ga1 V, ” = —— = ———-——— >< = 0.353 gal y", l3.54(62.41b) n’ (62.4lb) .3 (1 ft’) 1.111 = V: 2.32 8.64 =0.724lb W 7 ( ) ft’ ( m )1728¡n3 10 Chapter 1
  24. 24. .593 B: mo 82a 2: c223 e32: oa ouflflonEoh w589» baena una Emma? owzoonm Su mcouflüm - 892G “¿n88 N acoscmmmmd. ufinnEoo 1|. u 232360.. . 09 8 8 9 8 o m8 8m m8 Em m . m5 a casmofimrlol 8mm cono 38 N 3.o- E3 8o o} 2: . m8 S2. 2am Nono. :3 m8 v} 8 8m En. .. 3mm :3- onto m8 :3 8 m8 Boo m. Em «Sa- e33 8m 8.o 3 NWH. .M. «ÍI. ..-IQ- ¡I| -¡-¡. . S3 NFS 25o- ano Ko m3 8 «.89 . saco + «esoo . x88 . ammwwfll 8 3 35 m8.. .. «m3 m5 8.a 2 .582. .8. 0.32253 . m> Émcuo 3 ño 38 Food- 30m m8 m3 o» 83 cua. .. osa- 23 .8 mom 8 1 N23 «¿m maca. Som v8 8.o 8 «B. .. 2.8 3.3- Bom 8m Ba m. .. u EEK? » 89o ma». .. Boa 23m m3 mom 8 c9 s 8 9 a o c8.. . mama una. .. 22.. 8m Ed m. .. o} s Rod 38 Bo. .. :3 N8 m3 9. MM 29o meom «Nod «m; v8 m; mm Bm m. mood- c8.. . m8.? ma; 8m Fm 8 case émll 8.a m omo. .. 38 má. .. mu? 5m E m mm Bm m mmod .32. 33 «mmm m8 m; om o; m8.. .- No8 tod. no3 c2: :3 fl m3 n23. mama «Sd mona c8. 5.o E to. .. N82 29o «So oe: 8.a m 83 No2: mood :3 es: 5m o cas 52.0o c5 e. s> am émcoa s> am dEmH umSnEoo umgsaEoo 325m0 88.0! . 8.3 . .i:5.. -í. . Sul/ NH N H1». I¡ÍD. .'. II| .I¡Í. V¡I.1¡E! 1.. + gd + NxBÉ . 2o. ? + 39m v .825 B» a. ==«. _oaEo. _. . m> :5 ukowaw e. < Emi mmnhfimasm» . m> amb‘; no mmfimumoma mI. .. mou. tu. m>mao 11 The Nature of Fluids
  25. 25. CHAPTER TWO VISCOSITY OF FLUIDS 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 12 Shearing stress is the force required to slide one unit area layer of a substance over another. Velocity gradient is a measure of the velocity change with position within a fluid. Dynamic viscosity = shearing stress/ velocity gradient. Oil. It pours very slowly compared with water. It takes a greater force to stir the oil, indicating a higher shearing stress for a given velocity gradient. N-s/ mz or Pa-s lb-s/ ftz 1 poise = l dyne-s/ cmz = 1 g/ (cm-s) It does not conform to the standard SI system. It uses obsolete basic units of dynes and cm. Kínematic Víscosity = dynamic Víscosity/ density of the fluid. mz/ s ftz/ s 1 stoke = 1 cmz/ s It does not conform to the standard SI system. It uses obsolete basic unit of cm. A newtonian fluid is one for which the dynamic Víscosity is independent of the velocity gradíent. A nonnewtonian fluid is one for which the dynamic Víscosity is dependent on the velocity gradient. Water, oil, gasoline, alcohol, kerosene, benzene, and others. Blood plasma, molten plastics, catsup, paint, and others. 6.5 x 10"‘ Pa-s 1.5 x 10* Pa-s 2.o >< 10* Pa-s Chapter 2
  26. 26. 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 2.40 2.41 1.1x1o* Pa-s 3.0 x 1o" Pa-s 1.90 Pa-s 3.2 x 10" lb-s/ ft’ 8.9 x 10*1b-s/ f€ 3.6 x 10* lb-s/ ftz 1.9 x 10" lb-s/ ftz 5.0 x 10* lb-s/ ftz 4.1 x 10*1b-s/ f€ 3.3 x 10* lbs/ ft’ 2.8 x 10* lb-s/ ftz 2.1 x 10'31b-s/ ft2 9.5 x 10* lb-s/ fi’ 1.3 x 10* ib-s/ ft’ 2.2 >< 10* lb-s/ fiz Víscosity index is a measure of how greatly the Víscosity of a fluid changes with temperature. High Víscosity index (V I). Rotating drum viscometer. The fluid occupies the small radial space between the stationary cup and the rotating drum. Therefore, the fluid in contact with the cup has a zero velocity while that in contact with the drum has a velocity equal to the surface speed of the drum. A meter measures the torque required to drive the rotating drum. The torque is a function of the drag force on the surface of the drum which is a function of the shear stress in the fluid. Knowing the shear stress and the velocity gradient, Equation 2-2 is used to compute the dynamic Víscosity. The inside diameter of the capillary tube; the velocity of fluid flow; the length between pressure taps; the pressure difference between the two points a distance L apart. See Eq. (2-4). Víscosity of Fluids 13
  27. 27. 2.42 2.43 2.44 2.49 2.50 2.51 2.52 2.53 2.54 2.55 14 Terminal velocity is that velocity achieved by the sphere when falling through the fluid when the downward force due to gravity is exactly balanced by the buoyant force and the drag force on the sphere. The drag force is a function of the dynamic viscosity. The diameter of the ball; the terminal velocity (usually by noting distance traveled in a given time); the specific weight of the fluid; the specific weight of the ball. The Saybolt viscometer employs a container in which the fluid can be brought to a lmown, controlled temperature, a small standard orifice in the bottom of the container and a calibrated Vessel for collecting a 60 mL sample of the fluid. A stopwatch or timer is required to measure the time required to collect the 60 mL sample. No. The time is reported as Saybolt Universal Seconds and is a relative measure of viscosity. Kinematic viscosity. Standard calibrated glass capillary viscometer. See Table 2.4. The kinematic viscosity of SAE 20 oil must be between 5.6 and 9.3 cSt at 100°C using ASTM D 445. Its dynamic viscosity must be over 2.6 cP at l50°C using ASTM D 4683, D 4741, or D 5481. The kinematic viscosity of SAE 20W oil must be over 5.6 cSt at 100°C using ASTM D 445. Its dynamic viscosity for cranking must be below 9500 cP at —15°C using ASTM D 5293. For pumping it must be below 60,000 cP at —20°C using ASTM D 4684. SAE 0W through SAE 250 depending on the operating environment. See Table 2.4. SAE 70W through SAE 60 depending on the operating environment and loads. See Table 2.5. 100°C using ASTM D 445 testing method and at 150° C using ASTM D 4683, D 4741, or D 5481 . At —25°C using ASTM D 5293; at —30°C using ASTM D 4684; at 100°C using ASTM D 445. See Table 2.4. The kinematic viscosity of SAE 5W-40 oil must be between 12.5 and 16.3 cSt at 100°C using ASTM D 445. Its dynamic viscosity must be over 2.9 cP at l50°C using ASTM D 4683, D 4741, or D S481. The kinematic viscosity must be over 3.8 cSt at 100°C using ASTM D 445. Its dynamic viscosity for cranking must be below 6600 cP at —30°C using ASTM D 5293. For pumping it must be below 60 000 cP at —35°C using ASTM D 4684. v = SUS/4.632 = 500/4.632 = 107.9 mmZ/ s = 107.9 x 10* mz/ s v = 107.9 x 10* mZ/ s [(10.764 ft2/s)/ (m2/s)] = 1.162 x 10* ftz/ s SAE l0W-30 engine oil: Low temperature cranking viscosity at —25°C: 7000 cP = 7000 mPa s = 7.0 Pa-s maximum Low temperature pumping viscosity at —30°C: 60 000 cP = 60 000 mPa s = 60 Pa-s maximum Low shear rate kinematic viscosity at 100°C: 9.3 cSt = 9.3 mmz/ s = 9.3 x 10" mz/ Z minimum Low shear rate kinematic viscosity at 100°C: 12.5 cSt = 12.5 mmz/ s = 12.5 x 10" mz/ Z maximum High shear rate dynamic viscosity at l50°C: 2.9 cP = 2.9 mPa s = 0.0029 Pa-s minimum Chapter 2
  28. 28. 2.56 2.57 2.58 2.59 2.60 2.61 2.62 2.63 2.64 2.65 17 = = 4500 cP [(1 Pa-s)/ (1000 cP)] = 4.50 Pa-s 77 = 4.50 Pa-s [(1 lb-s/ fi2)/ (47.88 Pa-s)] = 0.0940 lb-s/ ftz v = 5.6 cSt [(1 m2/s)/ (lO6 cSt)] = 5.60 x 10* mz/ s v = 5.60 x 10* mZ/ s [(10.764 ft2/s)/ (m2/s)] = 6.03 x 10" ftz/ s From Figure 2.12: v = 15.5 mmZ/ s = 15.5 x 104 mz/ s 77 = 6.5 x 10* Pa-s [(1 lb-s/ ft2)/ (47.88 Pa-s)] = 1.36 x 10"‘ lb-s/ ft’ 17 = 0.12 poise [(1 Pa-s)/ (10 poise)] = 0.012 Pa-s = 1.2 >< 104 Pa-s. SAE 10 oil _(71-7,)D2 Ü_M 12.2-9 18D (q ) v, = 0.94(9.s1 kN/ mi) = 9.22 kN/ m3 D=1.6mm= 1.6 ><10_3 m v = s/t = .250 m/10.4 s a 2.40 x 10* m/ s m3 2 3 _ (77.0—9.22)kN(1.6><l0 m) X10 N_O.402NZs = 0.402 P - 18 m3(2.40xlO"z m/ s) kN m a s fl- Use yMercury = 132.8 kN/ m3 (App. B) — )D2 4p’ P2 (EQ-M) y, ,=0.90(9.8l kN/ m3)=8.83 kN/ m3 Ü 32uL Manometer Eq. using principles of Chapter 3: p: + voy + 10h e rmh — rav = p2 kN kN p¡ —p2 = ymh — yah = h(y, ,, — ya) = 0.177 m(132.8 « 8.83)—3 = 2194-7 m m Z 2 3 z (21.94kN/ m )(0.0025 m) =9.04X10,6 kN s x 10 N = 9.04 x 10" Pa . s 32(1.58rn/ s)(0.300 m) m2 kN 77 See Prob. 2.61. yf= 0.94(62.4 1b/ ft3) = 58.7 1b/ ft3: D = (0.063 in)(l ft/12 in) = 0.00525 11 v= s/ t = (10.0 ¡w104 s)(l 11x12 in) = 0.0801 ft/ s: y, = (0.283 1b/ in3)(1728 in3/ft3) = 489 lb/ fi’ _ (75 a mp2 (489 *58.7)lb/ ft’(0.00525 m2 18v 32(4.s2 ft/ s) 77 e 0.00207 lb s/ ftz = 2.07 x 10" lb-s/ ftz See Problem 2.62. Use n, = 844.9 lb/ ft’ (App. B): x, = (O.90)(62.4 lb/ fl’) = 56.161b/ ft3 h = (7.00 in)(1 {1/12 in) = 0.5333 a: D = (0.100 mxi 1012111): 0.00833 a p, — p; = ha, - 70) = (0.5833 ft)(844.9 — 56.16) lb/ ft’ = 460.1 lb/ ftz (p, «p2)D2 z (460.1 lb/ ftZ)(O.00833 ft): 32vL 32(4.82 ft/ s)(1.0 ft) 77 = = 0.000207 lb s/ ftz = 2.07 x 10* lb-s/ ftz From Fig. 2.12, kinematic viscosity = 78.0 SUS Víscosity of Fluids 15
  29. 29. 2.66 2.67 2.68 2.69 2.70 2.71 2.72 2.73 2.74 2.75 2.76 16 From Fig 2.12, kinematic viscosity = 25 7 SUS v = 4.632(l88) = 871 SUS v= 4.632(244) = 1130 SUS From Fig. 2.13, A = 0.996. At 100°F, v= 4.632(153)= 708.7 SUS. At 40°F, v= 0.996(708.7) = 706 SUS From Fig. 2.13, A = 1.006. At 100°F, v= 4.632(205) = 949.6 SUS. At 190°F, v= 1.006(949.6) = 955 SUS v: 6250/4.632 : 1349 mmz/ s v= 438/4.632 : 94.6 mmZ/ s From Fig. 2.12, v: 12.5 mmZ/ s From Fig 2.12, v= 37.5 mmz/ s t = 80°C = 176°F. From Fig. 2.13, A = 1.005. At 100°F, v= 4690/4.632 = 1012.5 mm2/s. At 176°F (80°C): v= 1.005(l012.5) = 1018 mmz/ s. t= 40°C = 104°F. From Fig. 2.13, A = 1.00. At 100°F, v= 526/4.632 = 113.6 mmZ/ s. At 176°F (80°C): v= 1.000(113.6) : 113.6 mmZ/ s. Chapter 2
  30. 30. Kinematic Víscosity Conversions Problem 2.77 SAE Víscosity Grades - Engine Oils Kinematic Viscosi at 100 de (mmzls) sus Min Max Min 3.8 38.9 3.8 38.9 4.1 39.8 5.6 44.6 5.6 44.6 9.3 -—- 56.8 -—- 5.6 9.3 44.6 56.8 9.3 12.5 56.8 68.3 12.5 16.3 68.3 83.2 16.3 21.9 83.2 106.6 21.9 26.1 106.6 125.1 Conversion method for both Problem 2.77 and 2.78: Used method from Section 2.7.5 in the text. 1: 100 deg C = 212 deg F. S: From Fig. 2.13, A = 1.007 3: Read SUS for 100 deg F from Fig. 2.12. 4: Multiply A times SUS at 100 deg F to get SUS at 100 deg C (212 deg F) Example: Given minimum kinematic viscosity = 21.9 mmzls for SAE 60 Read SUS at 100 deg F = 105.9 from Fig. 2.12 SUS at 100 deg C (212 deg F) = 1.007(105.9) = 106.6 SUS NOTE: Results reported here used tabular values from ASTM 2161. Values read from Fig. 2.12 may vary because of precision of graph or reading of values from scale. Problem 2.78 See Problem 2.77 for method. SAE Víscosity Grades - Automotive Gear Lubricants Kinematic Viscosi at 100 de (mmz/ s) Min Max Víscosity of Fluids
  31. 31. 18 Kinematic Vlscoslty Converslons Problem 2.79 iSO Víscosity Grades Kinematic Viscosi at 40 de Min 1.98 2.2 32.5 33.3 2.88 3.2 35.6 36.6 4.14 4.6 39.6 41.1 6.12 6.8 46.0 48.1 9.00 10 55.4 58.8 13.5 15 71.6 77.4 19.8 22 97.0 106.3 28.8 32 136.2 150.5 41.4 46 193.1 214 61.2 68 284 315 90.0 100 417 463 135 150 625 695 198 220 917 1019 288 320 1334 1482 414 460 1918 2131 612 680 2835 3150 900 100D 4169 4632 1350 1500 6253 6948 1980 2200 9171 10190 2880 3200 13340 14822 Note: Method used is same as for Problem 2.17. Temperature: t = 40 deg C = 104 deg F From Fig. 2.13, A = 1.000 Therefore, SUS values are read directly from Fig. 2.12. 34.0 37.6 42.6 50.3 62.4 83.4 1 15.9 164.9 235 347 510 764 1 121 1630 2344 3465 5095 7643 1 1209 16305 Chapter 2
  32. 32. CHAPTER TH REE PRESSURE MEASUREMENT Absolute and gage pressure 3.1 Pressure = force/ area; p = F/ A 3.2 Absolute pressure is measured relative to a perfect vacuum. 3.3 Gage pressure is measured relative to atmospheric pressure. 3.4 Atmospheric pressure is the absolute pressure in the local area. palas 2128386 +patm 3.6 True. 3.7 False. Atmospheric pressure varies with altitude and with weather conditions. 3.8 False. Absolute pressure cannot be negative because a perfect vacuum is the reference for absolute pressure and a perfect vacuum is the lowest possible pressure. 3.9 True. 3.10 False. A gage pressure can be no lower than one atmosphere below the prevailing atmospheric pressure. On earth, the atmospheric pressure would never be as high as 150 kPa. 3.11 At 4000 ft, palm t 12.7 psia; from App. E by interpolation. 3.12 At 13,500 ft, pm = 8.84 psia; from App. E by interpolation. 3.13 Zero gage pressure. 3.14 pgage = 583 — 103 = 480 kPa(gage) 3.15 pgagc = 157 -— 101 = 56 kPa(gage) 3.16 pgage = 30 — 100 = —70 kPa(gage) 3.17 pgage = 74 — 97 = :23 kPa(gage) 3.18 pg, “ = 101 — 104 = -3 kPa(gage) 3.19 pam = 284 + 100 = 384 kPa(abs) 3.20 pm = 128 + 98 = 226 kPa(abs) Pressure Measurement 19
  33. 33. 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.30 3.31 3.32 3.33 pabs = 4.1 + 101.3 = 105.4 kPa(abs) par” = —-29.6 + 101.3 = 71.7 kPa(abs) pabs = :86 + 99 = 13 kPa(abs) pgage = 84.5 — 14.9 = 69.6 psig pgage = 22.8 — 14.7 = 8.1 psíg pgage = 4.3 — 14.6 = -10.3 psig pgage = 10.8 —— 14.0 = —3.2 psig pgage = 14.7 - 15.1 = -0.4 psíg pm = 41.2 + 14.5 = 55.7 psia pal); = 18.5 + 14.2 = 32.7 psia pm = 0.6 + 14.7 = 15.3 psia pab, = —4.3 + 14.7 = 10.4 psia pm = —l2.5 + 14.4 = 1.9 psia Pressure-Elevation Relationship 3.34 3.35 3.36 3.37 3.38 3.39 3.40 20 p = yh = 1.08(9.81 kN/ m3)(0.550 m) = 5.83 kN/ mz = 5.83 kPa(gage) p = 7h = (sgh/ Ji : sg = p/vwh) 1.820 lb ft’ 144 inz sg = _———————————— >< = 1.05 inZ(62.4 lb)(4.0 ft) ft’ 3 zp: 52¿75kNm = 6'70 m y m 7.87 kN 64.00 lb l ft’ p= yh= fis x12.50fi>< 4m, =556ps1g 2 p=7h=6ïi1bx500fix lft 2 =21.67ps1g c m p : yh = (10.79 kN/ m3)(3.0 m) = 32.37 kN/ mz = 32.37 kPa(gage) p = yh = (10.79 kN/ m3)(l2.0 m) = 129.5 kPa(gage) Chapter 3
  34. 34. 3.41 3.42 3.43 3 .44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3.52 3.53 3.54 pa“ : pA — y, ,(64 in) : 180 psig — (0.9)(62.4 lb/ fl3)(64 in)(l mi 728 in3) pa“ = 180 psig — 2.08 psi = 177.9 psig p, - = yh = (l.15)(9.8l kN/ m3)(0.375 m) = 4.23 kPa(gage) pam, = 24.77 kPa(abs) By interpolation - App. E: ym = (l3.54)9.8l kN/ m3 ym : 132.8 kN/ m3 pa = patm + yh = 24.77 kPa + (132.8 m/ m3)(0.325 m) = 67.93 kPa(abs) p : yh = (0.95)(62.4 lb/ ft3)(28.5 fi)(1 32/144 in”) = 11.73 psig p = 50.0 psig + yh = 50.0 psig + 11.73 psi = 61.73 psig (See 3.44 for yh = 11.73 psig) (62.4lb) X625“ 1 n? a3 144 in’ p = —10.8 psig + yh = —10.8 psig + (0.95) p = —10.8 psig + 2.57 psi = —8.23 psig pbot — ptop 7!) 3 - 2 h: (35.5 30.0)lbft X144 m in2(0.95)(62.4 lb) fiz ptop + Yoh = Pbm Ï h = = 13.36 ft 0 + yoho + ywhw = pbot ha = pm — ywhw = 52.3 kN/ mz —(9.81kN/ m3)(2.80 m) = 234 m y, (0.86)(9.8l kN/ m’) 0 + roho + ywhw = ptm _ pm — yoh 2125.3 kN/ mz —(0.86)(9.81kN/ m’)(6.90 m) w “ 6.84 h y, 9.81 kN/ m3 m O + yohl + ywhl zpbot; but hl = " h2 Yo(1 8 " hz) + Ywhz = Pbo1 187o T 70,72 + YwhZ = Pboi = h2('yw ’ ya) + 187o _ p, “ :18y, _ 158 kN/ mz —(18m)(.86)(9.8l kN/ m’) = M7 m y, — 70 [9.81 — (0.86)(9.81)]kN / m3 p : yh = (1.80)(9.8l kN/ m3)(4.0 m) = 70.6 kN/ mz = 70.6 kPa(gage) p = yh = (0.89)(62.4 lb/ fi3)(32.0 ft)(l M144 in’) : 12.34 psig p = yh = (10.0 kN/ m’)(11.o x103 m) = 11o x 103 kN/ mz = 110 MPa o pa“, + y, ,,(.457 m) —yw(l.38l m) — ya (0.50 m) = pa¡, p, “ = (13.54)(9.8l kN/ m3)(.457 m) — (9.81 kN/ m3)(l .381 m) — (.68)(9.8l)(.50) pa“ = (60.70 — 13.55 — 3.34) kN/ mz = 43.81 kPa(gage) Pressure Measurement 21
  35. 35. 3.55 3.56 pm, = —34.0 kPa + 70h‘, + ywhw = —34.0 kPa + (0.85)(9.8l kN/ m3)(0.50 m) + (9.81 kN/ m3)(0.75 m) pbm = -34.0 kPa + 4.17 + 7.36 = —22.47 kPa(gage) pbot zpair + Yoho + ‘Ywhw = 20o kPa + [(0.80)(9.81)(1.5)+(9.81)(2.6)]kN/ m2 P1701: 200 + 11.77 + 25.51 = 237.3 kPa(gage) Manometers (See text for answers to 3.57 to 3.61.) 3.62 3.63 3.64 3 .65 3.66 3.67 3.68 22 o pa, “ — y, ,,(.075 m) — yw(O. l0 m) = pA pA = —(13.54)(9.8l kN/ m3)(0.075 m) — (9.8l)(0.l0) = —l0.94 kPa(gage) pp + y, ,(13 in) + 7w(9 in) — y, ,(32 in) = pB 62.4 lb X 9 ¡n X 1 fi3 __ (0.85)(62.4)(l9) a3 1728 in3 1728 pg —pA = 0.325 psi —— 0.583 psi = —-0.258 psi PB -PA = M9 in) - M19 in) = pe - M33 in) + 1/«(8 in) + M13 in) = PA 3 (.85)(62.4) lb X 8 ¡n X 1 a ft3 1728 in3 1728 pA —p¡; = 0.246 psi -0.722 psi = —0.477 psi PA "PB = M8 in) -vw(20 in) = pg + y, ,(. lS m) + y, ,,(.75 m) —yW(.50 m) = pA pA —pB = (.90)(9.81 kN/ m3)(.15 m) + (l3.54)(9.8l)(.75) —(9.8l)(0.50) pA —pg = (1.32 + 99.62 —4.91)kPa = 96.03 kPa p}; + yw(. l 5 m) + 7,40575 m) -—y, ,(0.60 m) = pA PA "P13 = (9.81 kN/ m )(0.l5 m) + (13.54)(9.8l)(0.75) —(0.86)(9.8l)(0.60) pA —pg = (l .47 + 99.62 — 5.06)kPa = 96.03 kPa 0 pa, " + ym(.47S m) —y, .,(.30 m) + y, ,,(.25 m) —y¿. (.375 m) = pA pA = y, ,,(.725 m) : yw(.30 m) : y,, (.375 m) pA = (13.54)(9.8l kN/ m3)(.725 m) —(9.8l)(.30) —(.90)(9.81)(.375) pA = (96.30 —2.94 — 3.3 l )kPa = 90.05 kPa(gage) ps + M6 in) + M6 in) - M10 in) + rm(8 in) -ra(6 in) = pr PA -pn = 7m(14 in) -rv(4 in) -rn(6 in) 62.4 lb . 1 ft’ (62.4)(4) (.9)(62.4)(6) n’ 1728 in’ 1728 1728 pA -p¡; = (6.85 —0.l4 —0.195) psi = 6.51 psi pp —pB = (l3.54)>< __ (62.4)(20) Chapter 3
  36. 36. 3-69 PB -M2 fi) : M3 ft) +M11 ft) ZPA PA “PB = m9 ft) -M3 ft) = 6“ ‘b “iz 3:99AM x9fi >< pA —p¡; =3.90 psi — 1.17 psi = 2.73 psi fi’ 144 inz 144 3 62.4lbX6'8ínX lfi 3.70 pa“, + yw(6.8 in) = p,, = 0+ ña 1mm, = 0.246 psi 0 3.71 pa, “ + yGph = pA : h = L sin 15° = 0.115 m sin 15° = 0.0298 m pp, = (0.87)(9.81 kN/ m3)(0.0298 m) = 0.254 kPa(gage) 0 3.72 a. pa", + y, ,,(.815 m) —yw(.60 m) = pA PA = (l3.54)(9.81 kN/ m3)(0.815 m) —(9.8l)(.60) = 102.4 kPa(gage) b. pam, = y, ,,h = (13.54)(9.8l)(.737) = 97.89 kPa pA = 102.4 + 97.89 = 200.3 kP(abs) Barometers 3.73 A barometer measures atmospheric pressure. 3.74 See Fig. 3.14 and Section 3.7. 3.75 The height of the mercury column is convenient. p, “ z 14.711; n‘ 144 ¡m2 3.76 h= —_——2—————>< 2 yw m 62.4 lb fi = 33.92 ft very large (10.34 m) 3.77 h = 29.29 in See Example Problem 3.13 3.78 h = 760 mm See Example Problem 3.11 3.79 The vapor pressure above the mercury column and the specific weight of the mercury change. Ah z -—1.01n of Mercury 3.80 x 1250 fl = —l.25 in l000ft 3.81 101.3 kPa —> 760 mm ofMercury (See Ex. Prob. 3.11) Ah : ‘s5 mm x 520o ft >< 3048 m = —134.7 mm 1000m h = 760 — 134.7 = 625.3 mm palm = ymh = 133.3 L; >< 0.6253 m = 83.35 kPa m Pressure Measurement 23
  37. 37. 3.82 3.83 3.84 3.85 3 848.7 lb X Zgfiínx l ft ft3 1728 in’ pam, = ymh = = 14.05 psia 3 848.7lbX30_65inX lft fi? ‘ 1728 ini pam, = ymh = = 15.05 psia 14.2 lb X ft’ 1728 ini h_ palm = X 848.7 lb ft3 - 2 7m 1n ‘ 28.91 in pm, = ymh : 133.3 E >< 0.745 m : 99.3 kPa(abs) 3 m Expressing Pressures as the Height of a Column of Liquid 3_86 3.87 3.88 3.89 3.90 3.91 3.92 3.93 p = 5.37 inHzO (l .0 psi/27.68 inHzO) = 0.194 psi p = 5.37 inHgO (249.1 Pa/1.0 inHzO) = 1338 Pa = 134 kPa = —3.68 inHgO (1.0 psi/27.68 inHgO) = —0.133 psi p = —3.68 inHgO (249.1 Pa/1.0 inHgO) = —917 Pa p = 3.24 mmHg (133.3 Pa/1.0 mmHg) = 431.9 Pa p = 3.24 mmHg (1.0 psi/57.7l mmHg) = 0.0627 psi p = 21.6 mmHg (133.3 Pa/1.0 mmHg) = 2879 Pa = 2.88 kPa p = 21.6 mmHg (1.0 psi/57.71 mmHg) = 0.418 psi = —68.2 kPa (1000 Pa/ kPa)(1.0 mmHg/133.3 Pa) = —512 mmHg p = —12.6 psig (2.036 ínHg/ psi) = —25.7 inHg p = 12.4 inWC = 12.4 inHyO (1.0 psi/27.68 inHyO) = 0.448 psi p = 12.4 inHgO (249.1 Pa/1.0 inHzO) = 3089 Pa = 3.09 kPa p = 115 inWC = 115 inI-IZO (l .0 psi/27.68 inHzO) = 4.15 psi p = l 15 inHyO (249.1 Pa/ l .0 inHyO) = 28 646 Pa = 28.6 kPa Pressure Gages ami Transducers (See text for answers to 3.94 to 3.97.) 24 Chapter 3
  38. 38. CHAPTER FOUR FORCES DUE TO STATIC FLUIDS Forces due to gas pressure 4.1 F= Ap-A; where Ap = pm —p, ,,, ¡,, ,; A = ÏLÏ = 113.1 m2 pam, = ymh = gmgïlb >< 30.5 in x “¿fins = 14.91 psi F= (14.91—0.12)lb/ in2 >< 113.1in2= 1673 lb 4.2 F : p-A = (14.4 lb/ in2)(7r(30 1115/4) = 10180 lb 4.3 F: Ap-A; A : 36 x 8o inz = 288o m2 6241i’ 1 f‘; : 0.0433 11)/ m? Ap = yu]: = >< 1.20 in >< . fi’ 1728 m3 F= (0.0433 1b/ in2)(288O in? ) = 125 lb 4.4 F = p-A; A = 0.9396 f6 (App. F) F= (325 1b/ in2)(0.9396 ft2)(144 inz/ ftz) = 43973 lb ; r(0.030m)’ 4.5 F= p-A; A : : 7.07 x 1o" m2 F = (3.50 x 10° N/ m2)(7.07 x 10*‘ m2) = 2.47 x 10’ N = 2.47 kN 4.6 F= p-A; A = 140.050 m)2/4 : 1.963 >< 10* m2 F= (20.5 x 10° N/ m2)(1.963 >< 10* m2) : 40.25 >< 10’ N = 40.25 kN 4.7 F= p-A; A = (0.800 m)2 = 0.640 m2 F : (34.4 x 103 N/ m2)(0.64 m2) = 22.0 >< 103 N : 22.0 kN Forces on horizontal flat surfaces under liquids 4.8 F= p-A; A : 24 >< 18 m2 =432 inz 56.78 lb 1 ft’ : h : 12.3 ft p y” a’ x X144 inz F = (4.85 lb/ in2)(432 m2) = 2095 lb : 4.85 lb/ inz 4.9 F = p-A; A = 7r(0.75 1102/4 = 0.442 m2 3 844.9 lb x 280m X l ft ft’ 1728 in’ F = (13.76 lb/ in2)(0.442 inz) = 6.05 lb = 13.69 lb/ inz I’ = rmh = Forces Due to Static Fluids 25
  39. 39. 4.10 4.11 4.12 4.13 Force on valve = p-A; A = 1c(0.095 m)2/4 = 7.088 x 104 m2 p = ywh = 9315“ x 1.80 m = 17.66 kN/ mz m F= (17.66 x 103 N/ m2)(7.088 x 10* m2) = 125 N Acts at center 2mm = o = (125 N)(47.S mm) — F0(65 mm) Fo = 5946 N-mm/65 mm = 91.5 N = Opening force FB = pB-A; A =1.2 >< 1.8 m2 = 2.16 m2 PB = Pair + 740.50 m) + y, ,(0.75 m) p}, = 52 kPa + (o. s5)(9.s1 kN/ m3)(0.S m) + (9.s1)(o.75) = 63.5 kPa FB = (63.5 x 1o’ N/ m2)(2.l6 m2) = 137 x103 N = 137 kN FB = pB-A; A = 2.o >< 1.2 m’ = 2.4 m2 p]; = 200 kPa + y, ,(l .5 m) + yw(2.6 m) p}; = 20o kPa + (o. so)(9.s1 kN/ m3)(1.5 m) + (9.81)(2.6) = 237.3 kPa FB = (237.3 x 103 N/ m2)(2.4 m2) = 569 x 103 N = 569 kN 1 Fp = Ap-A; A = 1:(0.60 m)2/8 + (0.80 m)(0.60 m) + 5 (.60 m)(0.30 In) A = 0.711 m2 : Assume std. atmosphere above water. pw = pm + yswh = 101.3 kPa + (10.10 kN/ m3)(175 m) = 1869 kPa Ap = 1869 kPa — 100 kPa = 1769 kPa Fp = (1769 x 103 N/ m2)(0.7l1 m2) = 1.257 x 10° N = 1.26 MN Forces on rectangular walls 4.14 4.15 26 F R = yW(h/2)A = (62.4 1b/ ft3)(l.8 ft)(8.0 ft)(3.6 fi) = 3235 lb F R acts J. wall, 1,20 ft from bottom of gate Part (b) ZMmng, = F¡¿(l.2 ft) — F¿; (4.0 ft) F” = F¡¿(l.2/4.0) = 3235 | b(0.30) = 970.5 lb on two latches On each latch: FL = (970.5 lb)/2 = 485 lb Length of sloped side = L = 15.5 ft/ sin 60° = 17.90 ft A = (17.90 fl)(l 1.6 ft) = 207.6 ftz FR = YUI/ ZM) z 78.5? lb X 15.511 xzomfi a 2 FR= 126300 lb hp = 2/3 h = (2/3)(15.5 fi) = 10.33 ft L, , = 2/3 L = (2/3)(17.9o fi) = 11.93 ft Chapter 4
  40. 40. 4.16 4.17 62.4 lb fiS FR = y(h/2)A = x 6 ft x (12 ft)(20 ft) FR = 89850 lb 11,, = 2/3 h = 2/3(12 fi) = 8.0 ft Moment = F R-4 fi = 359400 lb-ft FR = y, ,(h/2)A z (0_86)9.s1kN 3 m FR = 46.8 kN h, = 2/3 h = (2/3)(1.4 m) = 0.933 m L, , = 2/3 L = (2/3)(1.98 m) = 1.32 m x 0.7 m x (1.9s)(4.0) m2 Forces on submerged plane areas 4.18 4.19 Centroid is at midpoint of AB hR= 14in+4in= 18in= l.50ft AB = 10.0 in [3-4-5 triangle] L 5 5 —"= —;Lc= hc - —=22.5 in d 4 4 C Area = Z3 - 3.5 ft = 1% - 3.5 = 2.92 r8 (420 m2) FR = yhRA = 0.93(62.4 lb/ fi3)(l .50 ¡»(2.92 ft’) FR = 254 lb _ 3113 _ <42)0o)’ in‘ _ 1¿_. ——_————————-35oo in‘ 12 - 4 L —L I” 3500"’ -0.37 in ” °= ÏR7= - LR= LR+ 0.37 ín=22.5 in+ 0.37 in = 22.87 in h, = 0.825 m = 825 mm L = he =825mm C cos30° cos30° h(450 mm)2 = 953 mm A= = 1.59x 10’ mmz [.159 m2] 4 FR = yhRA = (0.85)(9.8l kN/ m3)(.825 m)(0.l59 m2) FR = 1.09 kN 4 4 L, =ï= ï =2.013 x 109mm‘ 64 64 I 2013x109 c _ 13.3 mm " °' fi LEA _ (9s3)(1.s9x10’) L, ,=LR. +13.3 mm = 953 +133 =966 mm Forces Due to Static Fluids / L= d sin 45° =1.98 m 27
  41. 41. 4.20 ha = 3.0 m; Lc = hR/ cos 30° = 3.464 m IZDZ Z 7r(2.4 m)2 A = = 4.524 m2 4 4 4 4 I: ¡[D : z “329 m4 64 64 FR = yhBA = (1.10)(9.81 kN/ m3)(3.0 m)(4.524 m2) FR = 146.5 kN 4 L —L ——ls—e———Éï—w—— =0.104 m P ° u LGA Ñ (3.464 m)(4.524 m2) = 104 mm L, = LB + 0.104 m = 3.464 + 0.104 = 3.568 m 4.21 i g-¿5g I SnAppJ la-isn Lc= a+ 1.5 +z=8.0/cos45°+ 1.5 +z = 13.50 fi hR = LR cos 45° = 9.55 ft FR ”-‘ Vhs/ I = (62.4 lb/ ft3)(9.55 ft)(3.0 ftz) = 1787 lb A = H(G + B)/2 = l.5(4.0)/2 = 3.0 ftz 1C: =0_551ft4 36(G+B) LB — LR = IB/ LBA = 0.551/(13.50)(3.0) = 0.0136 ft = 0.163 in L, = Lc + 0.0136 ft = 13.50 + 0.0136 = 13.51 ft 4.22 LR = a + 1.0 ft = 3.0/cos 30° +1.0 = 4.464 ft ha = LR cos 30° = 3.866 ft A = nos ft)2/4 = 0.196 ftz FR = yhRA = (O.90)(62.4 1b/ ft3)(3.866 ft)(0.l96 ftz) FR = 42.6 lb 4 2 1, = ”D = “o” = 0.00307 ft‘ 64 64 ¡B 0.00307 = 0.00351 ft p 4 z ÏRÏÏ L, ,= LR. + 0.00351 = 4.468 rc 28 Chapter 4
  42. 42. 4.23 LR = hR/ cos 40° = 0.855/cos 40° = 1.116 m 2 ”('3°) = 0.1607 m2 A = (.300)? + FR = = (0.90)(9.8l kN/ m3)(0.855 m)(A) FR = 1.213 kN (0.300)“ + 74.300)‘ 12 64 IR = 0.001073 m“ Lp L RR I: a : ¿MEL LRA (1.1 16)(0.1607) LR — LR = 00598 m = 5.98 mm ¡cu (R450 R, Lp= 1.122 m y-Éoosm‘ -0.346m hc -1.20m-y-0.855m 1R= 4.24 A = nD2/4 = n(2.0)2/4 = 3.142 f8 FR : = 62.4 Ib/ ft3 x 6.536 ft x 3.142 f8 FR = 1231 lb I= ”D4 =0785ft4 ° 64 ‘ ¿“LF ¡e : LRA (9.243)(3.142) LR — LR. = 0.027 ft = [0.325 m] L, , = 9.270 ft a = 3.0/cos 45° = 4.243 ft LR=5 +a=9.243 ft hR = LR. cos 45° = 6.536 ft 4.25 LR = a + 0.50 m = 0.76 m/ cos 20° + 0.50 LR = 1.309 m hR = LR cos 20° = 1.230 m A = (1 .00)(0.60) = 0.60 m2 FR = yhRA = (0.s0)(9.s1 kN/ m3)(1.23 m)(0.60 m2) FR = 5.79 kN 3 3 4 ¡a = BH z (0.60)(I .00) m z M5 m4 12 12 L, — L, = ’—C 0'05 = 0.0637 m = 63.7 mm LGA z (I.309)(0.60) L, , = LR + 0.0637 = 1.372 m Forces Due to Static Fluids
  43. 43. 4.26 4.27 30 -2o=2s in +25 = 50.0 in=4.167 ft gï- LR= 40.0 in = 3.333 fl A = (s)(50) = 400 ín2(l ft2/144 in’) = 2.773 f? F R = YhoA = (1.43)(62.4lb/ ft3)(3.333 ft)(2.778 nz) FR = 826 lb ¡’H3 = —————(8X5°)3 = s3 333 in‘ 12 12 ’ a: Lc he H í 4 a Il IR= I 83333 in‘ Lp - Lc = ° = ———*. ————. T LGA (50.01n)(400 m ) = 4.167 in L, = LR + 4.167 in = 54.167 in (4.514 n) IR-m x1o4n‘-a. nam4 a = 0.80 m/ sin 70° = 0.851 m LR= a + 0.5 +y = 0.851 + 0.50 + 0.318 =1.669 m hR = LR sin 70° = 1.569 m FR = YhcÁ = (0.88)(9.81 kN/ m3)(l.569 m)(0.884 m2) FR = 11.97 kN 4 Lp_Lcï 1a 2 Romsm LCA (1.669 m)(0.884m ) =23.5 mm L, = LR + 0.0235 m = 1.669 m + 0.0235 m = 1.693 m Chapter 4
  44. 44. a = 10 in/ cos 30° = 11.55 in LR= a+ 8 + í = 11.55 + 8.0+ 8.48 =28.03 in hR = = LR. cos 30° = 24.27 in [2.023 ft] FR = Vhs/ Í = (1.10)(62.4 lb/ fi3)(2.023 ft FR = 605.8 lb - 4 LP_LC= IC = = 0997 LGA (28.03 m)(628.3 m ) = 7.135 in L, = LR + 7.135 in = 29.03 in 144 in2 4.29 | -———aou«———-1T A = —; —BH= (asomo) = 30o in? BH; Joao): = 6667 in‘ 36 36 a = 18 in/ cos 50° = 28.0 in LR= a+6+b=28.0+6.0+ 13.33 LR = 47.34 in hR. = LR cos 50° = 30.43 in FR = YhoA FR = 62.4 lb/ ft’ x 30.43 in x 300 in2 3 x ‘Ï 3 =329.6lb 1728 m - 4 ¡v 6667 "‘ — 0.469 in Lp LC = = _ _ 2 LEA (47.341n)(300 m ) L, = LR + 0.469 in = 47.34 + 0.469 = 47.31 in IR= Forces Due to Static Fluids )(628.3 in2)(1 02) 31
  45. 45. 4.30 4.31 4.32 32 hR = LR = 0.375 + 0.150 = 0.525 m A = (0.60)(0.30) + 1t(0.3Ü)2/4 A = 0.2507 m2 hR-L, “EL? ,. __¡R FR 3 FR = (0.67)(9.81 kN/ m3)(0.525 m)(0.2507 m2) FR = 0.865 kN = 865 N 3 4 L: (0.60)(0.30) +7ï(0.30) z 0.001748 m, 12 64 4 L, LR » I“ — 0001748 m = 0.0133 m = 13.3 mm LRA (0.525 m)(0.2507 m2) L, =LR+13.3 mm = 525 +133 = 538 mm (See Prob. 4.30) h, = LR = 0.150 m FR = yhRA = (0.67)(9.81)(0.l50)(0.2507) = 0.247 kN = 247 N 4 Lp La _ I, _ 0.001748 m LGA (0.150 m)(0.2507 m2) L, = LR + 46.5 mm =150 mm + 46.5 mm = 196.5 mm “ 0.0465 m = 46.5 mm XAy z 9.0><10’3 m3 ‘ _ La Y — | ._ .30 _. ¡ A, 0.1253 m2 "' = 0.0718 m T y ¡ñ LR= hR= í =0.07l8m=7l.8mm ”'°-‘5'" A013) mm’) 14m“) Ah? 1 Rect. 0.0900 0.075 6.750 x 10-3 ¡.688 x 10-4 . 2 Semicirc. 0.0353 0.0636 2.245 x 10"’ 5.557 x 10* 0.00816 2.35 x 10'“ 0.1253 9.000 >< 10* 2.243 x 10"‘ IR = 2.276 x r0" m‘ -4 4 L, LR — ¡c — 2276 “o m = 0.0253 m = 25.3 mm LA 8' (0.0718 ¡n)(0.l253 m2) L, = LR + 25.3 mm = 97.1 mm FR = yhRA = (0.67)(9.81 kN/ m3)(0.0718 m)(0.l253 m2) = 0.059 kN = 59 N Chapter 4
  46. 46. 4.33 Rectangular Wall FR = 701/314 = 62.4 Ib/ ft3 x 4.0 fi x (8.0 ft)(l5.0 ft) = 29950 lb L, = %h= %(8.0 ft) = 5.333 r1 G= 10.0 0- 8.0 ft/ tan 60°= 5.381 ft ¿(Gia =61.52 02 4.34 A z _ z H(G +23) z y 3(G + B) FR = FR = 62416/63 x 4.40 11 x 61.52 B2 = 16894 lb RR ‘ 36(G +3) _ 315.382 +4(5.38)(10)+102] 36(5.38 +10) R, _ RR R IR R i LRA (4.40)(61.52) L, =LR +1.176 = 5.576 rc 4.4015 = hR = LR ‘318304 = 1.176 ft 4.35 Rectangular Wall AB = 8.0 fi/ sin 60° = 9.237 ft A= AB x 15 ft=138.6ft2 FR= y(h/2)A FR = 62.4 lb/ ft3 x 4.0 ft >< 138.6 B2 = 34586 lb L, = %(E) = -: —(9.237 ft) = 6.158 ft H(G + 2 B) = 4.6[l .2 + 2(3.856)] 3(G+B) 3(1.2+3.856) hR= H— j? =4.6—2.703 =1.897m= LR FR = yhRA Acts RL page H(G + B) _ 4,6(l .2 + 3.856) 4.36 E = = 2.703 m A — 2 — 11.63 m2 FR = (1 .10)(9.81 kN/ m2)(1.897 m)(11.63 m2) = 238 kN 3 2 2 IC: H (G +4GB+B)=18'62m4 36(G + B) 4 L, LR = I" 18'62 m = 0.844 m LRA F (1.897 m)(l1.63 m2) L, = LR + 0.844 m = 1.897 + 0.844 = 2.741 m Forces Due to Static Fluids í-ZJOG L; Vluvownrd ondwnl a-unuaenvnnetr-usam |12m¡ G 33
  47. 47. 4.37 Rectangular Wall: FR = yIuA FR = (1.1o)(9.31 kN/ m3)(4.60/2)m(4.6)(3,0)m2 = 343 kN FR acts l/3 from bottom or 2/3 from surface L, = É (4.60 m) = 3.067 m 4.3 8 END WALL A y 1.046 0.75 0.784 0.151 0.15 0.023 0.151 0.50 0.075 1 348 m2 2A; = 0.882 m3 z 2A} = 0.882 m’ ZA 1.343 m2 hc= 1.5m— Y =0.846m= LC l = 0.654 m FR = yhcA = (0.90)(9.81 kN/ m3)(0.846 m)(1.348 m2) = 10.07 kN Ic= I1 + A. yf+I2+ Azyzz +Is+ Aa? ¡_ z (0.697)(1.5)3 (o. so3)(o.3o)’ +(1.046)(0.096)2+ +(o.1s1)(o.5o4)’ + + (o. 1 s 1x0. 1 54V IC = 0.1960 + 0.0096 + 0.0011 + 0.0384 + 0.0030 + 0.0036 = 0.2518 m4 L, , = LL. + I" = 0,846 m + m = 0.846 + 0.221 = 1.067 m L A (0.846)(1.348) _"| "'__: _ Lp _L U 4.39 Vertical back wall is rectangular FR = Vhs/ Í z v(h/2)(A) y = 0.90(9.81 kN/ m3) = 8.829 kN/ m3 h/2 = 1.50 m/2 = 0.75 m A = (1.50 m)(l.20 m) = 1.80 m2 FR = (8.829 kN/ m3)(0.75 m)(1.8O m2) = 11.92 kN L, ,= Zxhzz x l.50m= l.00m 3 3 34 Chapter 4
  48. 48. 4.40 h, = L‘. = 4.00 ft; A = (4.oo)(1.25) = 5.00 f? FR = yhcA = (62.4 lb/ fl3)(4.OO fi)(5.00 a2) = 1243 lb 1C = 3115/12 = (1 .25)(4.oo)3/12 = 6.667 fl‘ _ 1L, 6.667 n‘ Lp * La " = ’—: ’—"’T LGA (4.00 fi)(5.00 ft ) = 0.333 a EMS = o = FR(1.667)— F, ,(4.oo) FH z (124sx1 .667) z 520 ¡b 4.00 zM, , = o = FR(2-333) — F¿(4.00) saga :7“, 4.41 Water side: FR = vw(hw/2)Aw AW = (2.50)(0.60) = 1.50 m2 FR“ = (9.221)(1.25)(1.5o) = 18.39 kN L=2 p — -2.50=1.667m w 3 Oil side: FR“ = y, ,(h, ,/2)A, , A0 = (2.00)(0.60) = 1.20 m2 FR” = (9.81)(0.9)(1.00)(1.20) = 10.59 kN Lp“ = á »2.00= 1.333 m EMS = 0 = Few (1.967) — FR (2.133)—FH(2.80) F” = [(18.39)(1.967) — (l0.59)(2.133)]/2.80 = 4.85 kN (——- ZM” = o = FR (0.833) — FR (0.667) —F¿(2.80) FS = [(l8.39)(0.833) —(10.59)(O.667)]/2.8O = 2.95 kN (—- Forces Due to Static Fluids 35
  49. 49. 4.42 hR. = 33 +y = 38 + 5 cos 30° = 42.33 in L, = hR/ cos 30° = 48.88 in FR= yhRA A = 1r(10)2/4 = 78.54 in’ z 62.4 lb . 42.33 in .7s.54 inz F” a3 1728 in’/ ft3 FR=12o.1 lb 4 16.: No) =49o.9in“ 54 L, —LR= ‘rc La” K xoïïa - 4 ¿p_¿¿_: =o.128in __/ [ (48.88m)(78.541n‘) / 5.18 Sum moments about hinge at top of gate. zMR = o = FR(5.128) e F4500) (1204 lb)(5.l28 in) Fr: Z . 5.00 m = 123.2 lb = = cable force Píezometric head La. = hRR/ cos 30° = 2.480 m/ cos 30° = 2.864 m FR = yRhRRA = (0.85)(9.81 kN/ m3)(2.480 m)(0.159 m2) = 3.29 kN 9 4 LW L“: IC 2.013><l0 mm - — — 4.42 LÜA (2.864 mm)(1.59 x105 mm2) mm ¡’a —e 25° k“ >< m3 - 2,317 m See Prob. 4.20 711/) m2 (Lloxgfil kN) for data. ha, = he + ha = 3.000 m + 2.317 m = 5.317 m L“. = hRR/ cos 30° = 5.317/cos 30° = 6.140 m FR = yRnhazi = (l. l0)(9.8l kN/ m3)(5.3 l 7 m)(4.5Z4 m2) = 259.6 kN I, 1.629 m“ Lpc Lev - L _ 2 LRA (6.140 m)(4.524 m ) 4.44 ha — 3 4.43 ha- p" — 13'81“ x m — 1.655 m See Prob. 4.19 7., m2 (0'85)(9'8l kN) for data. ha, = ha + ha = 0.825 m + 1.655 m = 2.430 m ‘ 0.0586 m = 58.6 mm 36 Chapter 4
  50. 50. _ p" M 2.50 lb ft’ 1728 in’ _ . _ 4.45 hR 7 m2(1.43)(62_41b) >< 1 fi; 48.41 m — 4.034 fi I See Pmh 426 h, , = h, + h, = 40.0111 + 48.41 in = 88.41 ín = 7.368 ft L, , = (Í) = 88.41 m? ) = 110.5 in = 9.209 ft 4 4 FR = yRhRA = (1.43)(62.41b/ ft3)(7.368 ft)(2.778 ft’) = 1826 lb - 4 L, ,, 4 L, , = Iv = = 1.885 in LGA (l 10.5 m)(40O m ) 4.46 hn - p" - 40"’ >< fi} x 1728 m3 — 100.7111 = 8.392 fi I See Prob- 423- “, ¡R2 (1.l0)(62.4 lb) R12 h, , = h, + h, = 24.27 in + 100.7 in = 124.97 in [See Prob. 4.28] Lc, = hRR/ cos 30° = 144.3 in = 12.03 fi FR = yEGhceA = (1.10)(62.4 1b/ fi’)(124.97 ín)(l ft/12 in)(628.3 in2)(l 128/144 m2) FR = 3119 lb - 4 Lp, L, , - IC — 17 562 m _ 2 — 0.194 in LmA (144.31n)(628.31n ) Forces on curved surfaces 4.47 R= O.75 m, w=2.00 m; Fy= y-V= yAw "'T* 0.75 2 Fy= 9.81 kN/ m’ x [(1.85)(0.75)+ ”( ) ] x 2.00 m’ q = 35.89 kN X, z ARx, +A2xZ A] + A2 A. = (0.75)(1.85) = 1.388 m2 A; = 7t(0.75)2/4 = 0.442 m2 í _ (1.388)(.37S) + (0.442)(0.3l8) 1.388+ 0.442 , = h. + s/2 = 1.85 + 0.75/2 = 2.225 m FH = yswh, = 9.81 kN/ m3 x (o.75)(2.oo)(2.225)m3 " 0.361 m FH = 32.74 kN R2 0.752 h, = h, + = 5 + — = 2.225 + 0.021 = 2.246 m 12 c 12(2.22s) FR = R/ F; + F; = 35.892 + 32.742 = 48.58 kN F 35.89 = -' —V = -I = 47.6° ‘á ta“ F ta" 32.74 H Forces Due to Static Fluids 37
  51. 51. 4.48 4.49 38 ¡——vs«»—1 Fy= yV= yAw FR z (0.826)(9.81 kN/ m3)(1.389 m2)(2.50 m) = 28.1 kN A = AR + AR = (0.62)(1.25) + 7r(l.25)2/8 A = 1.389 m’ F H = 0 because horíz. forces are balanced —4- F1e= FV:28.1kN {M y, =R sin 15° = 3.882 ft s= R—y¡=15.00—3.882= 1I.118ft h, = h +y¡ + s/2 =10.0 + 3.882 + 5.559 h, = 19.441 ft FR = yh, sw = (62.4)(19.441)(11.118)(5.00) FR=67,437 lb 2 h"= h‘+1;h, h, = 19.971 ft FV = 7V: A¡ = (14.489)(l0.0) = 144.89 ft: 1 1 A2 = 5 (y¡)(R cos 15°) = 5 (3.882)(l4.489) = 28.12 f8 = 19.441 + 0.530 fl 75 75 A3=7rR2- — = n(15.0 2 —- 360 360 = 147.26 ftz AR= A, +A; +A3 = 320.27 ft? ‘ FR = yAy-w = (62.4)(320.27)(5.00) = 99,925 lb x¡ = 14.489/2 = 7.245 ft x2 = —5- (14.489) = 9.659 ft x3 = b sin 37.5° where b = 38.197 Él = 9.301 ft (from Machinery’s Handbook) x3 = 9.301 sin 37.5° = 5.662 ft 55 = ————————A‘x‘ + A2” + A2“ = 6.738 ft AT FR = JF; + F} = J574372+999252 = 120550 lb FR _t _, 99925 F, , a“ 67437 = 56.0° (15: tan" Chapter 4
  52. 52. 4.50 FV: yRV= yRAw A = A1 + A; = (7.50)(9.50) + 7r(7.50)2/8 A = 93.34 112 FV 3 YoAW = (0.85)(62.4)(93.34)(3.50) Fy= 17328 lb = FR FH = 0 because horíz. forces are balanced 4.51 s= R—y=6.00—5.196=O.804 m h, = h +y + s/2 = 5.20 + 5.196 + 0.402 h, = 10.798 m FH = yswhR = (0.72)(9.81)(0.804)(4.00)(l0.798) FH = 245.3 kN 2 2 h, =h, + S =10.798+¿824— 12h 12(l0.798) c h, = 10.798 + 0.0050 m , = 10.803 m FV: 7V: A1 = (5.20)(3.00) = 15.60 m2 1 A; = 5 (3.00)(5.196) = 7.794 m2 A} = “RR _3_0_ z 7r(6.0)2 360 12 AR= AR +A2 +A3 = 32.819 m2 FR: yAw = (0.72)(9.81)(32.8l9)(4.00) = 927.2 kN x, = 3.00/2 = 1.500 m x2 = 2(3.00)/3 = 2.00 m x3 = b sin 15° where b = 38.197 = 9.425 m2 Rsin15° = 3.954 m (from Machinery/ s Handbook) x3 = (3.954)sin 15° = 1.023 m 5a = = 1.482 m A, FR = W = 245.32 +92722 = 959.1 kN F, _, 927.2 : ‘lí: Ó tan FH tan 245.3 = 75.2° Forces Due to Static Fluids 39
  53. 53. 4.52 4.53 40 FV = yAw A = A, + A2 = (1 .2o)(2.so) + 7:(1.20)2/4 = 4.491 m2 FV = (9.81 kN/ m3)(4.491 m2)(1 .50 m) = 66.1 kN x. = o.5(1.2o)= 0.60 m; x2 = 0.4240 .20) = 0.509 m _ A, xl + A2x2 (3.36)(.60) + (1.13)(o.5o9) A, 4.491 = 0.577 m ha = h + s/2 = 2.80 + 1.20/Z = 3.40 m FH = yswhc = (9.81)(1.20)(1.50)(3.40)= 60.0 kN 2 (Lzof J? 11,4“ S —3.4o+ —3.43sm 12hC 120.40) FR= JF}? +135 = 66.12 +60.02 =89.3 kN F 66.1 = 4L: -'——— =47.s° {É tan FH tan 60.0 A = 1.20 2.80 =3.36 2 ‘ ( X 2 ) m2 A=3.669m2 A2 = R2—7tR /4=0.309 m FV= yAw = (9.8l)(3.669)(1.50) = 54.0 kN x] = 1.20/2 = 0.60 m x2 = 0.223412 = 0.268 m [Machíneryiv Handbook] í z A¡x¡ + ¿x2 z (3.36)(0.60) + (0.309)(0.268) A 3.669 J? = 0.572 m ho = = h + s/2 = 2.80 + 0.60 = 3.40 m FH = yswhc = (9.8l)(1.20)(1.50)(3.40)= 60.0 kN 2 1.202 h, ,=hc+ s =3.4o+ =3.435m IZhC 120.40) FR = JF; + F3 = J54.02 + 60.0’ = 80.7 kN F 54.0 = mL: -‘—— =42.0° 415 tan F” tan 60.0 Chapter 4
  54. 54. 4.54 4.55 4.56 4.57 4.58 A = TEDZ/ g = fl(36)2/8 = 503.9 m2 = 3.534 f8 FV = yAw = (0.79)(62.4)(3.534)(5.0) = s71 lb z = 0.21213 = 0.2l2(36 in) = 7.63 in h, = h + s/2 = 43 + 36/2 = 66.0 in = 5.50 ft FH = yswh, = (0.79)(62.4)(3.0)(5.0)(5.50) FH = 4067 lb s’ 36’ . h, = h, = = 66 + = 67.64 m 12h, 12(66) FV = JF; +F, Ï = 3712 + 40573 = 4159 lb F 871 = " —V= -* =12.1° (75 tan FH tan 4067 (See Prob. 4.47) . k 2 Eq. Depth = h, = Í3= Jill/ É; = 0.765 m y 9.81kN/ m h“, = 121+ h, =1.35 + 0.765 = 2.615 m; h, = hu, + s/2 = 2.615 + 0.75/z = 2.990 m A, = (0.75)(2.615) = 1.961 m2; A; = 0.442 m2; A’, = 2.403 m2 FV: yAw = (9.81)(2.403)(2.000) = 47.15 kN _ Alx, +A, x, _ (1.96l)(0.375)+(O.442)(0.3l8) - 0.365 m x_ A, 2.403 FH = yswhc, = (9.s1)(o.75)(2.oo)(2.99) = 44.00 kN 2 2 h, ,—h, ,= S — 0'75 =0.016m= 16 mm 12h, '12(2.99) FR = JF; + F; = «147152 + 44.002 = 64.49 kN F 47.15 z —1_V= —r ‘ú ta" F, ta“ 44.0o (See Prob. 4.48) = 47.0° ¿_ 4.65kN/ m2 7 (0.826)(9.8lkN/ m3) h. ,= h. +h, ,= 0.62 m + 0.574 m = 1.194 m A = (1.194)(1.2s) + 1c(1.25)2/8 = 2.106 m2 Eq. Depth = ha = FV= yAw = (0.826)(9.81 kN/ m3)(2.106 m2)(2.50 m) = 42.66 kN = FR Net horizontal force = O = 0.574 m From Section 4.1 l, net vertical force equals the weight of the dísplaced fluid acting upward and the weight of the cylinder acting downward. w, = y¡V, ¡ = (62.4 lb/ ft3)(0.l64 ft’) = 10.2 lb V= AL; "D2 ‘ 4 w, = y, V= (0.284 lb/ ín3)(282.7 m3) = 80.3 lb Net force on bottom = w, — w¡-= 80.3 — 10.2 = 70.1 lb down See Prob. 4.57. w¡= 10.2 lb w, = ycV= (0.100 lb/ in3)(282.7 ¡m3) = 28.27 lb Fm. = w, — w¡-= 28.27 — 10.2 = 18.07 lb down Forces Due to Static Fluids - 10.0 in = 282.7 m3 x = 0.164 t? 41
  55. 55. 4.59 4.60 4.61 4.62 42 See Prob. 4.57. w¡= 10.2 lb w, : yy: (30.0 1b/ fi3)(0.l64 ft’) : 4.92 lb Fm, = w, — w¡= 4.92 — 10.2 = 5.28 lb up But this indicates that the cylinder would float, as expected. Then, the force exerted by the cylinder on the bottom of the tank is zero. The specific weight of the cylinder must be less than or equal to that of the fluid if no force is to be exerted on the tank bottom. (See Prob. 4.57.) Because the depth of the fluid does not affect the result, Fm. = 70.1 lb down. This is true as long as the fluid depth is greater than or equal to the diameter of the cylinder. (See Prob. 4.57.) w, = 80.3 lb Force (downward) on upper part of cylinder = wt. of volume of cross-hatched volume. Force - (upward) on lower part of cylinder = wt. of entire dísplaced volume plus that of cross- hatched volume. Then net force is wt. of dísplaced volume (upward). e-skr‘(2.0I3.0)-41.a’ x-Rcoso-imcoco x-zzaah una‘ +20 453.5- Wr: Y/ Vd = YfidL = rrDz a 1 Ad 4 ' % + : 141 + A2 v 2 AH = w. 2366366 + (2.236)(2.0) = 25.18 m2 . . 1 n’ w¡= y, A,, L = 62.4 lb/ fi’ - 25.18 m2 - 10.0 1n- ma; = 9.09 lb Fm = wc — w¡= 80.3 — 9.09 = 71.21 lb down Chapter 4
  56. 56. 4.63 4.64 (See Prob. 4.57) For any depth 2 6.00 in, Fm, = 70.1 lb down Method of Prob. 4.62 used for h < 6.00 in and 2 3.00 in. For any depth < 3.00 in, use figure below. «D2 ¿-1 _ 4 X360 ZQXXR h) Fm, = w, — w¡= 80.3 lb —- yfAdL AH= A1"A2= Summary of Results: h (in) Em (lb) 6.00 70.1 5.50 70.5 32 5.00 71.2 gm Z1 4.50 72.03 ug 4-°° 73-07 S” 13111111 3.50 74.12 3 ,6 Zïflïïïïï 3.00 75.19 É Xïflïïïïïï 250 76 27 c, ‘ 1111111 ' ‘ s Zïfiïïïïï 2-00 77-32 gn _-‘k_—-- 1.5o 73,30 L 11111111 ,00 7918 n 11111111 ' ' 11111111 0.00 80.30 Mmm, “ Centroid: y : 0.212 D : 0.2l2(36 in) : (7.63 in)(1 m2 m) : 0.636 e x : (2o in)/ sin 25° : 47.32 in L, = 60 in + x — y = 60 + 47.32 — 7.632 : (99.69 ín)(1 m2 in) : 8.308 ft h, : L, sin 25° : (8.308 ft)(sin 25°): 3.511 ft A : ítDz/ S : 743.0 ft)’/8 : 3.534 r8 F, : yh, A : (1.06)(62.4 lb/ fi’)(3.5I1 fi)(3.534 ft’) = 32o lb : F, 1 = 6.86 x 104D‘ : 6.86 x 1o“3(3.0o ft)‘ = 0.556 tr‘ L, — L, : I/ [L, A] : (0.556 ft4)/ [(8.308 fi)(3.534 112)] : 0.0189 ft L, : L, + 0.0189 tt = 8.308 ft + 0.0189 a : 8.327 ft : L, Forces Due to Static Fluids 43
  57. 57. Buoyancy 5.1 5.2 5.3 5.4 5.5 5.6 5.7 44 zFV: o:w+ r-F, F, : yyV= (10.05 kN/ m3)(0.45)(0.6O)(0.30)m3 : 0.814 kN = 814 N T= FH—w=814—258=556N If both concrete block and sphere are submerged: Upward forces = F H = 17,, ’ + fic = yw V, + 7,, VC = ywU/ Ï, + VC) V. =hD’/6:h(1.om)’/6=o.5236 m’ ¿v9 _ 4.10 kN 7C 23.6 kN/ m’ = V“, = 0.6973 m‘ C = o.1737 m’ FH : ywláo, = (9.81 kN/ m3)(0.6973 m3) = 6.84 kN Downward forces = F D = wc + ws = 4.1 + 0.20 = 4.30 kN }Fu > FD It wm float‘ If pipe is submerged, F H = y¡V= (l.26)(9.81 kN/ m3)(1t(0.l68 m)2/4)(1.00 m) FH = = 0.2740 kN = 274 N; because w = 277 N > FH-It will sink. w, — FH = 0; w, = FH; y, .V, = Y/ Vd = y¡0.9 V, Then, y, = 0.9'y¡= (0.90)(1.10)(62.4 lb/ flj) = 61.78 lb/ ft’ wc — Fb = 0; YcVc — YfVd V, = V, fi 7/ 2 2 = ÏÁ;3:9._"i.1_2 m. _7;? .q= o_o5g3 mu“) . X 4 9.31 3 X= =0.9664m=966 mm Iz'(0.30m) Y= 1200 — 966 = 234 mm W—F1,=0=’Y¿V¿—'YfVd y, : y, gd- : (o.9o)(9.s1 kN/ m3)(75/l00) : 6.62 kN/ m’ c Te Chapter 5 W — F1, — T: 0 = YCVC- YfVC - T: VC('YC '-'Yf) '- T ___ T = 2.67 kN 3 = 0.217 m3 (y, — y! ) [23.6 -(l.15)(9.8l)]kN/ m Vc
  58. 58. 5.8 5.9 5.10 5.11 5.12 5.13 W"Fb"FsP=0= W"YoVd-Fsp FH, = w — y, ,V, , : 14.6 lb — (0.90)(62.4 1b/ ft’)(40 m3) Fsp: 1 ft’ 1728 in’ ZFV= O=WF+WS“ E"? "’ HX F4 : y., V,: 9.81 kN/ m3 x (0.100 m)’ = 9.81 N 0=yHVH+80N—ywVp—9.81 N 0 = V, (yH— yw) + 70.19 N —70.19N —70.19N VF. “ _ y, — y, (470 — 9810)N/ m3 : 7.515 x 10* m’ 2 w, + w, : F, = vVV, : 62.41b/ ft’- “l? . 3 a3 : 588.1 lb WA= Fb‘-Wc=588.11b= 'Y¿I/ A w, _ 558.1lbin3 1a‘ V4= ——— — ————————— - —--_-—3 =3.23ft3 7A 0.100lb 1728 m w, +w4— F, — F, :0 wA: E, ‘ : F, —w, =588.l —3o:55s.1 lb (See Prob 5.10) 'Y¿VA ’ YWVA = 74 : (0.100 lb/ in3)(1728 irP/ ft’) : 172.8 lb/ fta VA = 5581“) = = 5_055 ff’ y, —- yw (l 72.8 —— 62.4)lb/ ft w, —FH=0=y, V,——y¡VH= 'y, .S'3—yyS'2X _ 7,S’=7,S 718,2 Ï¡ Wy+Ws'-Fb=0 741.0)’ has)‘ FH= yw(V, +VH)=62.4lb/ fi3[ 4 150+ 4 "1-30 = 0.04485 lb ws = FH - wH = 0.04485 — 0.020 = 0.02485 lb Buoyancy and Stability ina fi’ 1728 in’ «¿z 45
  59. 59. 5.14 5.15 5.16 5.17 5.18 5.19 46 From Prob. 5.13: wH + ws = 0.02 + 0.02485 = 0.04485 lb = w = Fb 2 7'00) -1.so+ W= Fb = yrVa = MV; + V2) = ví = 7,(7.47 x 10"‘ ft’) 0.04485 lb 7 7,: ï/ ‘ï = —————7 ¿W104 R3 = 60.03 11)/ ft’; sg = —’ = 60.03/62.40 = 0.962 , , . 77 2 2 . 3 3 V, ,= V, + V2: m”) -1.5o+”(°‘25) -o.3o m fis =6.903 x 10411’ 4 4 1728 m 7,: ï = JÏÏ-‘Ïgï = 64.97 11)/ n’; sg= 74 = 64.97/62.4= 1.041 K, 6903x104 ft3 yw wB+wchoFLa -F; ’c :0 WC: En + E7‘; “WB 1.00 2 1.00’ W3=Y5.VB=8.0Ülb/ ÍÏ3[”(4 ) '3'0+”(6 WP‘; = 23.04 1 7r(2.0)2(3.0) fi, Fbs = y¡Vc = 64.0 lb/ ff’ —í = 201.06 lb E, = 7,11,, = 64.0 1b/ ft3 a 741.0)’ .2 00+ nao)’ fi, 4 ' 12 = 117.29 lb wc= 117.29 + 201.06 — 23.04 = 295.31 lb s= 18.00 mu m2 in) = 1.50 ft w-Fb —F¿=0 Fe = W ‘Fb = YsVc“ YwVc = VdYs * Yw) Fe = (1.so)’ ft’ (491 — 62.4) lb/ fi3 = 1447 lb T ym = 844.9 lb/ ft’ > ‘yw - cube would tend to float. W - F1, + Fe = 0 2 ¡[(0.25) _ 2'30] 4 m3 fis 1728 in’ 11 = 17,, — w = 7,, ,V¿ — ysVc = V¿(y, ,, — ys) = (1 .50)’ 190344.9 —491)¡b/ fi’ = 119411: L W-Fb=0=W"YsWVd w 292x10‘ gmx 9.81m kg 1N IOJOkN/ m“ s2 X x 10’gm kgm/ S’ 75W X lkN 103 N = 283.6 m’ Chapter 5
  60. 60. 5-20 W = F1»; YIVI = 'YSWVd . / 3 V, ¡= V, 1L- : - = V, - 0.863; 86.3% submerged; 13.7% above Ïsw ' m W = Fb; ywoodI/ T: ywVd = K1. Ywood yw VT 2 2 V7= 7'? -L: ”('45°) -6.750:1.074m3 2 Vd= V? -%+%(2X)(.115)]L _ . ¡115_ _ _ ¡t .45 2 241.5 3 2d" Fix" V : - +( 1934)( 115) 675m i 10° 2° m5‘ d 4 ' ' ‘ fiáïiïf-ÉGISOJÜ - m : 0.8703 m3 Ywood = (9.81 kN/ m3)(.8703/ 1.074) = 7.95 kN/ ma W"Fb=0='}’CVC“'YKV¿ V A-60O 76:77 —4- =8.07 kN/ m3- í“ VC A-750mm : 6.46 kN/ m3 5.23 w — F7, = 0 = ycVc —ywV, ¡=y, A-L —y, ,,AX _ 3 X 7CAAL = ———-————--—: ‘:ÍkkE; m3 -750mm=513 mm Ïw . m h= L-X=750-5l3=237mm H 5.24 wc- F, ” +w¡, — F, ” =0 ycVc "’ ywVc + ‘YBVB ’ ywVB = 0 VCA-L — "mA-L + VBA-t - vwA-t = 0 tm - vw) = ¿(vw — ve) _ _ 3 t= L yw 7C = 750 mm (9.44 6.46)kN/ m (s40 944)kN/ 3 :3“, mm ÏB-yw ' _ ' rn 5.25 yw at 15°C = 9.81 kN/ m3—it would float. wc- E, ” +wg— 17,, ” =0 7t(.45)2 wn = ycVc = 6.46 kN/ m’ >< >< .75 m3 =0.77l kN ; z(.45)2 w, : 75V, : 84.0 kN/ m’ >< >< .030 m3 = 0.4008 kN Buoyancy and Stability 47
  61. 61. 5.26 5 .27 5.28 5.29 48 7r(.45)2 12;, ” : 7,, V,, : 9.81 kN/ m3 x x .03 m’ = 0.0468 kN zac : 7,, V,, : wc + w, — E, ” = 0.771+ 0.400s — 0.0468 :1.125 kN F55 = 'YwVd = YwAX 1*; _ 1.125 kN X " * 0.721 m submerged = 721 mm 77A (9.81kN/ m3)(7t(.45)2m’/4) 29 mm above surface 7a : 15.60 kN/ m3 WC’- Iïbp +WB‘— E” :0 no45)’ wc : 76V, : 6.50 kN/ m’ x >< .75 m3 : 0.7753 kN 74.45% 117,7, = 1mm: 15.60 kN/ m’ x x .70 m3 :1.737 kN — w, = 1.737 — 0.7753 = 0.9614 kN : 0.01406 m’ = At WB “ FbB = Ya VB ’ YCT VB = VBÜYB ’ YCT) = 0.9614 kN (84.0 - l5.60)kN/ m’ z 0.9614kN__ 7B ‘Ïcï 3 V : :—————————°'°14°6 m = 0.0884 m : 88.4 mm . .1}; A 7r(0.45)2/4 m2 Va t: w = F7, = 7,V, , = (1.16)(9.81 kN/ m3)(0.8836 m3) = 10.05 kN V7 : 1103/12 = u(1.50 m)3/12 = 0.8836 m3 Entire hemisphere is submerged. W = Fb = Y/ Vd = Yw'Á'X w 0.05N 4 x kN X- = 3 2 3 70A (9.81kN/ m)(7t(0.082m)) 10 N = 0.965 x 10" m : 0.965 mm 7rD’ Wt. of steel bar = ws = ysVs = ysAL = ys L 7r(.038 m)‘ 10’ N ws : 76.8 kN/ m’ - . 0.08 m- w; + wc= 6.97 + 0.05 = 7.02 N = F¡, =ywAX ï-TIMN . ——————:4 -———lkN =0.13Sm=135mm " ywA ’ (9.81kN/ m’) 740.082 m)’ 10’ N = 6.97 N chapter 5
  62. 62. 5.30 From Prob. 5.29, wr= 7.02 N 2 2 VS = ¡tf . L = 7r(.038 m) : 7WVS : 9.81 kN/ m3 x 9.073 x 103 m3 x 103 N/ kN : 0.890 N _FbE —_: o = w7— 13,5 =7.02N—0.890N=6.l3N= y,, AX 6.l3N X 4 X lkN 9.81kN/ m3 / r(0.082m)3 103N - 0.080 m = 9.073 x 10* m3 WT- F [25 “0.118m— 118 mm (z: (211u)3) (3617,) 5.31 77,: F7 : 4y, ,,V¿, ,7m : 4(62.4 1171€) 4 m8 m3 713 : 1301 1o Drums Weigh 4(30 lb) = 120 lb Wt. ofplatfonn and load = 1801 —— 120 = 1681 lb 5.32 V01. ofWood: 2(6.0 ft)(1.50 m)(5.5o m)(1 M144 m3) : 0.6875 a3 ends 4(96 — 3)ín(l.50 m)(5.50 ¡m)(1 ft3/1728 m3) : 1.776 a3 main boards (0.50 in)(6 ma; ft)(1 0x12 m) : 2.000 ft’ plywood 4.464 f? total ww : yWV= (40.0 1b/ ft3)(4.464 113) : 173.511) 5.33 wD + w, » = F7, = ywVd Vd : W” + W” = = 4.78 f? total "f": VW 62.4 lb/ ft 111-10,; VD : 4.78/4 : 1.196 113 sub. each drum I“ VD : AS ' L x 3 - 2 AS= V—°:1'196fi :0.399f13x 144m :57.41n3 —v-——- a, 3.0 11 {r3 | L ' 1: : 4. ' h A 7: .4 ' 3 By ma X 67 m W en A 57 m See Prob. 4.63 for method of Computing As. Wdrums + wwood + Wload " E, ” " E, = o wdnmb = 4(30 lb) = 120 lb (Prob. 5.31) wwmd = 178.5 lb (Prob. 5.32) E, ” : 1801 lb (Prob. 5.31) E, “ = ywVw : 62.4 lb/ ft3 x 4.464 1:3 : 278.6 lb (Prob. 5.32) wmnd = En) + — wD — ww= 1801 + 278.6 -120 — 178.5 = 1781 lb Buoyancy and Stability 49
  63. 63. 5.35 Given: 7; : 12.0o 11)/ ft’, 7C = 150 lb/ ft’, wc : 600 lb Find: Tension in cable Float only: ZF7= 0 = wp+ T— FbF T: E”: "‘ W}: But wp : ypVp : 12.0 05/03 x 9.o 03 : 108 lb VF: (18.0 1n)3(48 m) 1728 1n3/03 P}, = 1.714, : (64.0 1b/ rt3)(6.375 a3) : 408 lb Vd = (18.0m)3(341n) 1728 ¡n3/03 T= 408 -108 = 30011) Check concrete block: Fm = wc — 15;, “ -— T wc : 6001b; VC : 1L : ÉÏÏ; : 4.00 03 7C 1501170 F76 : yWVC : (64.0 1b/113)(4.00 ft’) = 256 lb Fm = 600 - 256 — 300 = 44 lb downwOK——block sits on bottom. = 9.00 03 = 6.375 03 5.36 Rise of water level by 18.00 in would tend to submerge Y entire float. But additional buoyant force on float is sufficient to lift concrete block off sea floor. With block suspended: T = wc — EH T= 600 — 256 = 344 lb (see Problem 5.35) Float: wp+ T- FbF = 0 E)’: = wp+ T= 108 lb + 344 lb = 452 lb = ywVd E”: 45211» =7O63ft3x1728in’ yw 64.01b/ fi3 a3 : 12204 m3 Vd: V4 : (18.0 mfim X: V, =122041n3 08.0111)’ 324 m3 Y= 48 — X= 10.33 in above surface With concrete block suspended, float is unrestrained and it would drift with the currents. = 37.67 in submerged EFV= O:WA“'E, W“E")*T T= wr - n, 0.l0lb in w, , :77V, : x (6.0 m)’ : 21.6 lb (6 in )2(3.0 in) 1728 in3/ft’ 5,0 = ya Vd" = 0.85 En = 3.315 lb Then: T= 21.6 — 3.90 — 3.315 = 14.39 lb F“ :7WVL : 62.4 1b/ ft3 x : 3.9010 ¡n 50 Chapter 5
  64. 64. 5.38 Fs"w, FLC-yy, y¡Vc'-V, (yc y¡) Vc= ”D2-L= ><l0.0in>< “i. 3 — 4 4 17281n = 0.1636 ft3 0.2841b 1728 m’ _ 62.4113] F5=0.1636ft3( ¡ha X ft; fi; = 70.1 lb F g acts up on cylinder; down on tank bottom Stability 5.39 W—Fb=0 Ya Vc 2 Y/ Vd = Y/ AX X: ¿VC = 7cA(l. O m) = 8.0Ol<N/ m3/(1.0m) y/ ¡A 7¡A 9.8lkN/ m3 = 0.8155 m ycg = 1.00 m/ Z = 0.500 m yd, =X/2 = 0.8155 m/ Z = 0.4077 m MB: _1_ z 7zD“/ ó4 = 7r(1.0m)4/64 l I/ d (rrD’/4)(X) [zz(1.om)*/4](o.8155m) 4 z 0.O4909m3 = O_0766m 0.6405 m ym = yd, + MB = 0.4077 + 0.0766 = 0.4844 m < ycr-unstable 5.40 w= Fb= ywVd= ywAX w 250 lb(l44 inz/ fiz) ywA (62.4 1b/ ft3)(30 in)(40 in) X= 5.77 in; ysb = )Ü2 = 2.88 in I = (40 in)(30 in)3/ 12 = 90000 in‘ Vd = AX = (30 in)(40 in)(5.77 in) = 6923 ín3 MB = I/ Vd = 90000 in“/6923 in3 = 13.0 in ym = ycb + MB = 2.88 in + 13.0 in = 15.88 in > ycg-stable “ 0.4808 ft Buoyancy and Stability 51
  65. 65. 5.41 5.42 5.43 5.44 5.45 52 W= Fb: yswVd: yswAX X: w z 4500001b ySwA (64.0 1b/ ft’)(20 0x50 ft) = 7.031 ft ya, =X/2 = 3.516 ft 1=(50)(20)3/12 = 3.333 x 10“ 04 Vd = AX= (20)(50)(7.031)ft3 = 7031 03 MB = I/ Vd = 4.741 fi y, “ = ycb + MB = 3.516 + 4.741 = 8.256 ft > ycg—stable From Prob. 5.4; X= 0.9 H= 0.9(l2) = 10.8 in ya, = X/ Z = 5.40 in yes = H/2 = 6.00 in 1 = «M64 = n(10.0 in)4/64 = 490.9 in“ Vd = (nD2/4)(X) = [n: (10.0 in)2/4](10.80 in) = 848.2 m3 MB = I/ Vd = 490.9/848.2 = 0.579 m ym = ya, + MB = 5.40 + 0.579 = 5.98 in <ycg—-unstable From Prob. 5.5: X= 966 mm yd, =X/2 = 483 mm ycg = H/2 = 1200/z = 600 mm I= 1¡: D“/64 = 1r(300 mm)4/64 = 3.976 x 10* mm“ V, , = (nD2/4)(X') = [no00 mm)Z/4](966 mm) = 6.828 >< 10’ mm3 MB = I/ Vd = 5.82 mm ym = yd, + MB = 483 + 5.8 = 488.8 mm < ycg-unstable From Prob. 5.6: X= 75 mm ycb = )Ü2 = 37.5 mm ycg = H/2 = 100/2 = 50.0 mm 1= 114/12 = (100 mm)“/12 = 8.333 x 10° mm‘ Vd = (HZXX) = (100 mm)2(75 mm) = 7.50 x 105 mm3 MB =1/V, ,=11.11mm ym= ycb+MB = 37.5 +11.11 = 48.61 mm < y, ¿—unstable W = F» = YwVd= YuAX X_ w 70.011) ywA (62.41b/ ft’)(7r(2.0ft)2/4) = 0.357 n02 in/ ft) = 4.28 in yet = X/2 = 2.14 in ycg = H/ Z = 48/2 = 24.0 in 1 = w064 = 1c(24 1n)4/M = 16286 in‘ Vd = AX= (nD’/4)(¿Y) = [1424 in)2/4](4.28 in) = 1936 m3 MB = I/ Vd = 8.41 in ym = ycb + MB = 2.14 + 8.41 = 10.55 in <yc¿—unstable Chapter 5
  66. 66. 5.46 X= 8.00 ft; yd, =X/2 = 4.00 ft; ycg = 12.00 ft ymc = ycb + Scow is stable if ymd > ycg MBmd, =ymd —yd, =ydd —yd, == 12.0 —— 4.0 = 8.0 ft 3 2 MBmin: izLW / l2= W LWX 12X V Wddn = .}l2(X)(MBm¡__) = /12(8)(8)ftZ = 27.71 ft d 5.47 X= 16.0 ft; yd, =X/2 = 8.00 ftydg =13.50 ft MBmm = ymc ‘ya, =ycg -yc¡, z ft " fl = ft Wdd, = . /12(X)(MBddd) = i12(16.0)(5.50)ft2 = 32.50 ft 5.48 yd, = 0.70 m from bottom For hemispherefi submerged— y = 3ïlá from día. _ D 3D 5D 5(l.50 m) C -_—R _ = _____= ___= _________ y " y 2 16 16 16 = 0.469 m I= 0134/64 = «(1.50 m)4/64 = 0.2485 m‘ Vd = 7093/12 = n(1.50 m)3/12 = 0.8836 m3 MB = I/ Vd = 0.281 m ym = yd, + MB = 0.469 + 0.281 = 0.750 m > ydg-stable 5.49 From Prob. 5.28: X= 0.965 mm; D = 82 mm; H= l5O mm = _I_= nD“/64 = D2 = s2’ K, (7rD2/4)(X) 16X l6(0.965) =435 mm ymc = ycb + MB = )Ü2 + MB = .965/2 + 435 = 436 mm mc is well above cup; cg is within cup It is stable 5.50 From Prob. 5.29,X= 135 mm yd, =}02=67.5 mm MB: ¿zii/ JL = _D_2_ = .322. ¿“mm V‘, (7rD2/4)(X) 16X 16(135) ymd = yd, + MB = 67.5 + 3.11= 70.61 mm Because wt. of bar is >> wt. of cup, yd, z L Bar/2 = 80/2 = 40 mm ym > ydf-stable Buoyancy and Stability 53
  67. 67. 5.51 5.52 From Prob. 5.30, X= 118 mm Vs = Vol. of steel bar = 9.073 >< 104 mm’ flDé ><X Vas = Sub. vol. of cup = 2 “S” (118) = 6.232 x 105 mm’ Vcs = Vd= Vs+ Vd= 7.139 x105 mm’ ys= cb ofsteel bar = 225- = 19 mm ya; = cb of sub. vol. of cup ycs= Ds+ 3; =38+ L? =97mm ___ ysVs +ycsVcs ycb V (I z (19x9073 >< 10‘) + (97)(6.232 x105) 7139x105 4 MB = —I—= mm=3.11 mm V, 7.139x10 ymc = ycb + MB = 90.2 mm yd, is very low because wt. of bar is >> wt. of cup——stable From Prob. 5.22, Fig. 5.23: X = 600 mm yd, =X/2 = 300 mm yd = H/2 = 750/2 = 375 mm I 7rD‘/ 64 D’ ym = yd, + MB = 300 + 21.1 = 321.1 mm < ydg-unstable = 87.1 mm z _ __ _ (450)2 = V, (7rD’/4)(X) 16X 16(600) Chapter 5
  68. 68. 5.53 From Prob. 5.26, Fig. 5.25; t= 88.4 mm X=700+t=788.4mm yd, =X/2 = 394.2 mm 4 2 MB: ¿ 7:1) /64 D Vd Z [zD2/4:I(X) _ 16X 2 B z __íi1__ = 16 1 16(788.4) ym = yet + MB = 394.2 + 16.1 = 410.3 mm ycgwmt = ydwd + ygwg composite cyl. wd = ydVd = 0.7753 kN yd = 750/2 + t = 375 + 88.4 = 463.4 mm 840.0 kN .45 2 . ... .3.. . x M (0.0884 m) m 4 = 1.181 kN yd; = t/2 = 44.2 mm _ ycwc + yBwB _ (463.4)(0.7753) + (44.2)(l. l81) wm (0.7753 +1181) mm WB= YBVB= yd, — 210.3 mm <yme= stable w= Fb= ySWAX X_ w 3840lb - — 3 — 7.50 0 75,, A (64.0 lb/ 0 x2 0x4 0) yd, =X/2 = 3.75 0 I= 4(2)’/12 = 2.667 04 Vd = AX= (2x4x7.5) = 60 03 MB = I/ Vd = 2.667/60 = 0.0444 0 ym = yd, + MB = 3.75 + 0.0444 = 3.794 0 ym = H/2 = 8.00/2 = 4.00 ft > ym-unstable W= Fb= YSWAX X: w = 130 11x12 11110) ySWA (64.01b/ ft’)(3 0x4 0) = 2.03 in ycb = IÜZ = 1.016 in yd = 12 m + 34 in = 46.0 in I= (48)(36)3in4/12 = 1.866 x 10’ in“ Vd = (48x36x2.03)1n’ = 3510 m3 MB =1/Vd= 53.17 in ym = ycb + MB = 1.016 + 53.17 = 54.18 in >ydg—stable Buoyancy and Stability 55
  69. 69. 5-56 W—Fb= O='YWVtot"YoVd= YW-4H"%AX g_¡d¡nd, ghx1zh X: yWAH = ywH = (32 lb/ ÍÏ3)(6 in) / 70A 7a 0.90(62.4 lb/ ÍÏ3) = 3.419 in yd, = X/2 = 1.709 in ycg = H/2 = 6.00/2 = 3.00 in 1= 12(6)3/12 = 216 in“ Vd = (12)(6)(3.4l9) = 246.2 m3 MB = I/ Vd = 0.877 in ym = yd, + MB = 1.709 + 0.877 = 2.586 in < ycg-unstable 5.57 w —Fb = ywVd= ywAX w z 2l0O001b =2_804 fl me 7WA (62.41b/ ft )(60 ft)(20 fl) yd, =X/2 = 1.402 ftgycg = 1.50 ft given _. d, ¡ 3 B = L = ——-———————-—: (6O)(20) H2 =11.888 ft K, (60)(20)(2.804) yme = ycb + MB = 1.402 + 11.888 fi = 13.290 ft >ycg—stable 5.58 www = 210000 + 240000 = 450000 lb X= JL: M- = 6.010 ft (See Prob. 5.57) ywA (62.4)(60)(20) ya. = X/2 = 3.005 fi 1 (60)(20)’/12 —————-— = 5.547 ft Vd (60)(20)(6.010) ym. = yd, + MB = 3.005 + 5 .547 = 8.552 ft—-—above barge cg is within barge-—stable More complete solution: Depth of coal — dc — —W= — = 240000 n’ = 4.444 fi ¿A (451b/ fi’)(60 fi)(20 fi) yc = dc/2 = 2.222 ft from bottom to cg of coal 24 .22 21 O l. = = = 1m ¡«yfitafle w 450000 56 Chapter 5
  70. 70. 5.59 5.60 W = Fb = YTI/ d: YIAdL W = 'YoÁtotL = á (.600)(.300)(1.20)(2.36)kN = 0.255 kN Ad = GX+ (XXX) = <B—2X><X>+X2 w 0.255 kN Equate = ——- = —————————————-3ï— m, (o. s7)(9.s1 kN/ m )(1.2o m) I’ Ad= BX—-2X2+X2=BX-—X2=600X—X2 I Ad 3 2 i Ad = o.o24s9 m2 >< m =2.489><104mm2 600X— X2 = 2.489 x 104 X2 — 600X+ 2.489 x 104 = 0; X= 44.8 mm by Quadratic Eq. G= B — 2X= 600 — 2(44.s)= 51o mm 1= G3L/12 = (51o)3(12oo)/12 = 1.329 x 1o‘° mm4 V) = AyL = [600(44.8) — 4432x1200) = 2.985 x 10’ m3 MB = 1/V¿, = 445 mm X(G+2B) = 448 _ 44.s(51o+12o0) = 3(G + B) 3(51o + 600) ym = yc. , + MB = 21.8 + 445 = 467 mm 21.8 mm ycbzX- l yes = 5 (300) = 100 mm < ymr-stable a) Cube ís stable if ym > ycg = S/2 4 , m=ycb+MB= ¿+¿= ¿+_g_ 2 Vd 2 l2(S)(X) 2 2 ï+ S :5 orX2+—'g———SX=0 2 12X 2 6 X2—SX+S2/6=0 2_ 2 = Si/ S 4S/6 = gis Fm 2 2 2 = sB: %(o.5774)] = 0.788S or 0.21 lS X X > 0.788S or X < 0.21 lS will result in stable cube. b) ForS= 75 mm X> 0.788S= 59.2 mm, X< 0.21 IS = 15.8 mm Buoyancy and Stability 57
  71. 71. 5.61 Entire hull: 9.. .3. _ Am +412)»; u‘ "' ycg * fi : (.72)(.4o) + (2.88)(1.2) 3.60 yc = 1.040 m T Siïbmerged Volume: —“‘ "“*“' ““7Ó'“' = (.72)(0.4)+(2.16)(l. O5) m 2.88 = 0.8875 m V) = (2.88)(5.5) = 15.84 m3 1= 5.5(2.4)’/12 = 6.336 m‘ ym = ya, + I/ Vy = 0.8875 + 0.40 = 1.2875 m > ycg—stable ycb 5-62 a) WC = FB; YCVC z YwVd z 3o lb X ; :(.5 ft)2(l.0 fi) fi’ 12 = 1.963 lb WC Vd= (o. o3147 ft’)(l728 m’) ff _ 7r(D4,, )2(X) _ 7r(X/2)2X _7rX’ = 54.37 m3 Vd 12 12 48 X: 3i48V z 3i48(54.37) =9A0m 72' 7L’ 3X ya, = —2- = 0.75(9.40) = 7.05 in Figure for part (a) only. no; z n(9.4o/2)“ 64 64 MB = I/ Vd = 23.95/54.37 = 0.441 in ymc = ycb + MB = 7.05 + 0.441 = 7.491 in ycg = 221-1- = O.75(l2) = 9.00 in 1= = 23.95 in‘ ymc < ycg-unstable b) yc = 55.0 lb/ ft3 wc = 1.963(55/30) = 3.600 lb w 3.600 lb _ 0.0577 030778 ins) Vd — C _ 3 3 VW 62.4 lb/ ft ft X: 3f48Vd =3’48(99.69) 211.51 in 7!" 7L’ — 99.69 in’ 58 Chapter 5

×