Solucionario - Mecanica de fluidos - Sexta edicion - Robert L Mott(1).pdf

8/11/2019 Solucionario - Mecanica de fluidos - Sexta edicion - Robert L Mott(1).pdf

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I n s t r u c t o r ' s M a n u a l

to accompany

A P P L I E D F L U I D M E C H A N I C S

Sixth Edition

RobertL .MottU n i v e r s i t y o f D ay t on

PEARSON

Upper Saddle River,NewJersey

Columbus, Ohio


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Instructors ofclassesusingMott,Applied Fluid Mechantes, SixthEdition, may reproduce material from theinstructor's manual for classroom use.

10 9 8 7 6 5 4 3 2 1

PEARSON

ISBN 0-13-172355-3
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O n l i n e I M t o A c c o m p a n y

APPLIED FLUIDMECHANICS

S i x t h E d i t i o n

Robert L.MottU n i v e r s i t y o f D a y t o n

PEARSON

Upper Saddle River, New JerseyColumbus, Ohio


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CHAPTER FIFTEEN

FLOW MEASU REMEN T 23 5

CHAPTER SIXTEEN

FORCES DUE TO FLUIDS IN MO TI ON 2 4 0

CHAPTER SEVENTEENDRAG AN D LIFT 25 1

CHAPTER EIGHTEEN

FANS , BLOW ERS, COMP RESSO RS, AN D THE FLOW OF GASES 26 0

CHAPTER NINETEEN

FL OW OF AIR IN DU CT S 2 69

SPREADSHEETS 273

iv


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CONTENTS

CHAPTER ONE

THE NAT URE OF FLUIDS AN D THE ST UDY OF FLUID ME CHA NI CS 1

CHAPTER TWO

VIS COS ITY OF FLUIDS 12

CHAPTER THREE

PRESSURE ME AS UR EM EN T 19

CHAPTER FOUR

FORCES DUE TO ST ATI C FLUIDS 25

CHAPTER FIVE

BUO YANC Y AN D STABILITY 4 4

CHAPTER SIX

FL OW OF FLUIDS 61

CHAPTER SEVEN

GENERAL ENERGY EQ UAT ION 81

CHAPTER EIGHT

REYNOLDS NUMBER, LAMINAR FLOW, AND TURBULENT FLOW

A ND ENERGY LOSS ES DUE TO FRICT ION 9 4

CHAPTER NINE

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND

FLOW FOR NONCI RCULA R SECTIONS 113

CHAPTER TEN

MIN OR LOSSES 129

CHAPTER ELEVEN

SERIES PIPE LINE SY ST EM S 141

CHAPTER TWELVE

PAR ALL EL PIPE LINE SY ST EM S 187

CHAPTER THIRTEEN

PUMP SELECTION AN D APPLI CATIO N 21 3

CHAPTER FOURTEEN

OPEN CHA NNEL FLOW 218

i i i


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A P P L I E D F L U I D M E C H A N I C SS i x t h E d i t i o n

R o b e r t L . Mott

P r e n t i c e - Ha l l P u b l i s h i n g C o m p a n y

D e s c r i p t i o n o f S p r e a d s h e e t s I n c l u d e d o n th e C D in t h e B o o k

I n t r o d u c t i o n

This book includes a CD-ROM that contains ten computat ional aids that are keyed to thebook. The files are written as Microsoft Excel spreadsheets using Versin 2002 on WindowsXP.

The ten spreadsheets are al l included in one workbook cal led Series Pipe Systems-Master.The ames of each spreadsheet described below are on the tabs at the bottom of the

workbook when i t is opened. You must choose which is appropriate for a given problem.Most ames start with either /, //, or /// indicating whether the spreadsheet is for a Class /,Class II, or Class /// pipe line system as defined in Chapter 11 of the text.

The spreadsheets are designed to faci l tate the numerous calculat ions required to solve thevariety of problems in Chapter 11 Series Pipeline Systems. Many of the spreadsheetsappear in the text. Others were prepared to produce solutions for the Solutions Manual. Thegiven spreadsheets include data and results from certain figures in the text, from exampleproblems, or in problems from the end of Chapters 8,11, and 13 containing the analysis anddesign 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 eachspreadsheet before using them for solut ions to specif ic problems.

U s i n g t h e S p r e a d s h e e t s : It is recommended that the given spreadsheets be

maintained as they initially appear on the CD. To use them for solving other

problems, cali up the master workbook in Excel and use the "Save as" command to

ame it something different. That versin can then be used for a variety of problems

ofyour o w n c h o i c e . 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.

Th e principies involved in the sprea dshee ts com e from Chapters 6 - 1 1 and you shouldstudy the concepts and the solution techniques for each type of problem before using thegiven 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. In mostspreadsheets, the data that need to be entered are identif ied by gray-shaded reas and byitalic type. Results are typically shown in bold type.

v


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/Power SI: The objective of problems of this type is to compute the amount of powerrequired to drive a pump to deliver a given amount of f luid through a given system. Energyiosses are considered. All data must be in the listed SI units. The solution procedure is for aClass I series pipe line system. The following is a summary of the steps you must complete.

1. Ente r the pr ob lem identif icat ion informat ion first. Th e give n data in the spr ead she et ar e

for example Problem 11.1 for the system shown in Figure 11.2.

2. Describe two appropr iate referenc e points for complet i ng the analysis of the gener alenergy equat ion.

3. Specify the required volume flow rate, Q, in m 3 /s.

4. Ente r the press ure s (in kPa ), velociti es (in m/s), and elevatio ns (in m) at the refer encepoints in the System Data: at th e top of the shee t. If the veloci ty at eith er ref ere nce poin tis in a pipe, you may use a computed velocity from v = Q/A that is included in the datacells for the two pipes. In such cases, you enter the Excel command "=B20" for thevelocity in pipe 1 and "=E20" for the velocity in pipe 2.

5. Enter the fluid prop ertie s of specific weig ht (in kN /m 3 ) and kinematic viscosity (in m 2 /s).

6. Enter pipe data, includi ng flow dia met er D (in m), pipe wall roug hne ss a (in m fro m Tabl e8.2), 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 ene rg y loss coeffic ients, K, for all loss-producing elements in both pipes. SeeChapters 8, 10, and 11 for the proper form for K for each element and for necessarydata. The valu for Kfor pipe friction is com pu ted automat icall y fr om known data in the"Pipe" sect ion . Spec ify the num ber of l ike ele men ts in the "Qty. "column. Enter briefdescriptions 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 eightdifferent kinds of energy loss elements. Enter zero for the valu of the K factor for thosenot needed.

8. The Results: sect ion at the bo tto m of the spr ead she et includes the total ene rgy loss h < 1 0 l g

2 (2.25 m/ s) 2 m N s 2 -N kg

, . 3 0 . ] ? m . felSN^lkgj^

w V 12 k g N

2 C h a p t e r 1


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1.31 o= j 3 g g > 2 ( 2 1 2 m N . m ) x i r ^ x l ^ x l j ^ m

V m \ 175 g m N kg s 2 - N

, . 3 2 ^ ^ , ( l B l u g ) ( 4 f l / s ) - x l l l v s V f t = 8 0 0 ] b . f t

2 2 slug

fflu2 _ w u 2 _ (80001b)(10mi)2 1 h 2 ( 5280f t ) 2

1.3J JViS ~ "~ ~ ~X X2 2 g (2)(32.2ft /s 2 )(h) 2 (3600s ) 2 m i 2

?000)(10)2(5280;

(2)(32.2)(3600)2^ - < 8 0 0 0 > < ' > ' < 5 2 8 0 > ' 1 b . f t - M 7 M I . - *

, .34 K E - ! ! * - L - < - * ea

*1

?

2 2 g (2)(32.2ft/s2 )

2(E) 2(15 Ib - f t ) , A

l b - s1.35 m = - = v , = 6 . 2 0 = 6 .20slugs

v2 ( 2 . 2 f t / s 2 ) 2 ft

2

136 w_ 2 g ( X g ) _ 2 ( 3 2 . 2 f t ) (3 8 . 6 I b - f t ) (h2 ) x 1 m i

2

x ( 3 6 0 0 s )2

^ ( 2 ) ( 3 2 .2 ) ( 3 8 .6 ) ( 3 6 0 0 ) [ b

(19.5)2(5280) 2

s 2 (19 .5mi) 2 (5280f t ) 2 h 2

2

, . 3 7 , I g g g a )2Q2.2^ X 10 l b - a) ^ . 6 3 f t / sw V 30 Ib

1.38 o= / M ^ ) ^ 2 ( 3 2 - 2 f t / s 2 ) ( 3 0 o z . 1 n ) x i f l _

w V 6.0 oz 12 in

1 cr>A 39runs 9innings1.39 E R A = x = 2.49 runs /game

141innings game

, 3.12runs lgame1.40 x x 150 mnings = 52 runs

game 9innings

, ^ lgame 9innings1.41 40 runsxs x 2 - = 129 inn ing s

2.79 runs game

, , 49 runs 9 innings1.42 E R A = x 5_ = 3.59 runs/ game

123 innings game

T h e Nature o f F l u i d s 3


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1.53

1.54

F = pA_P[7tD

2}

4

_ 500 lb(7 T)( Din)

i n 2 4

D(in) D\in2) F(lb)

1.00 1.00 393

2.00 4.00 1571

3.00 9.00 3534

4.00 16.00 6283

5.00 25.00 9817

6.00 36.00 14137

7.00 49.00 19242

8.00 64.00 25133

F F 4F 4(5000 Ib)P ~ A = TT> 2/4 7tD2 TV(Di n ) 2

D(in) Z> 2(in2) p(psi)

1.00 1.00 6366

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

392.7Dl Ib

3025

20

Forc 15(toxIOOO) 1 0

5

0

psi

Pressure(psO

6000

4000

2000

O

4 5

0 1 2 3 4 5 8

Diametw(ri)

7 8

1.55 (Variable Answers) Example: w= 160 Ib (4.448N/lb) =712 N

F 712N (103 m m ) 2

P A 7 r (20mm) 2 /4

p = 2.27 x 10 6Pa (1 psi/6895 Pa) =329 psi

= 2.77 x 10 6Pa = 2.27MPam

1.56 (Vari able Answers) us ingp =2.27 MPa

F=pA = (2.27 x 106N / m 2 )W0.250 m) 2 /4) = 111 x 10 3N = 111k N

F= 111 k N (1 lb/4.448 N) =25050 Ib

T h e Na tu re o f F l u i d s 5


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Bulk modulus

1.57 Ap =-E(AV/V) = -1300 00psi(-O.Ol)= 1300psi

Ap = -896 MPa(-O.Ol) =8.96 MPa

1.58 Ap =-E(AV/V) = -3 .59 x 10 6psi(-O.Ol) =35900psi

Ap = -24750MPa(-O.Ol)=247.5MP a

1.59 Ap =-E(AV/V) = -18900 0psi(-O.Ol)=1890psi

Ap = -1303 MPa(-O.Ol) =13.03MP a

1.60 AVIV = - 0 . 0 1 ; AV = 0.01V= 0.01 A LAssume rea o fcylinder does not change.A F= ,4(AZ,)= 0.01 A LThen AL= 0.01 L = 0.01(12.00 in) =0.120i n

1.61AV _~p _ -3000 psi

V ~ E ~~189000 psi

= -0.0159 = -1 .59%

1.62AV _ -20.0 MPa

V ~ 1303 MP a= -0.0153 =-1 .53%

1.63 Stiffness = Force/Change in Leng th = F/AL

B u l k Modulus= E = P = ~pV

AVIV AV

Butp = FIA; V=AL; AV = -A(AL)

-F AL FL

(AL)

E =-A -A(AL) A(AL)

EA 189000 Ib -(0.5 i n ) 2

L " i n 2 (42 in)4884 lb/in

1.64F _EA _ 189000 Ib g(0 .5 i n ) 2

_ _ _ _ _ m 2 ( i o . O in)(4)3711 lb/in 4.2 times higher

1.65F EA 189000 Ib (2 .00 i n ) 2

(AL) in 2 (42.0in)(4)14137 lb/in 16 times higher

1.66 Use large diameter cyl ind ers and short strokes.

Forc and mass

w 6 1 0 N l k g - m / s 2 . . . .1.67 m = = r - x " =62.2kg

g 9.81 m/ s 2 N

6 C h a p t e r 1


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w 1 . 3 5 x l 0 3N l k g - m / s 2

1.68 w = = x - = 13 8k gg 9.81 m/ s 2 N

1.69 w = mg = 825 kg x 9.81 m/s 2 = 8093kg-m/s 2 = 8093 N

1.70 w = = 450 g x - ^ - x 9.81 m/s2

= 4.41kg-m/s2

= 4.41 N10 3 g

, - , w 7.81b - , - l b - s 2 . ,1.71 w = = - =0.242 = 0.242 slugs

g 32.2 ft/s 2 ft

w 42.01b ,1.72 w = = =1 .3 04 slugs

g 32.2 ft /s 2

1 /fl-1.73 w = mg = 1.58 slugs x 32.2 ft/s 2 x ' =5 0. 9 Ib

slug

1.74 w= mg =0.258slugs x 32.2 ft/s 2 x 1 l b " s / f t =8. 31 Ib

slug

, w 1601b . ,1.75 w = = ^ =4 .9 7 slugs

g 32.2 ft/ s 2

w = 160 lb x4.448N/lb = 712 N

m = 4.97 slugs x 14.59 kg/slug= 72.5 kg

1.76 iw = = L Q Q l b

=0. 031 1 slugsg 32.2 ft/s 2

m = 0.0311 slugs x 14.59 kg/slug = 0.453 kg

w = 1.00 lb x 4.448N/lb = 4.448 N

1.77 F = w = mg = 1000 k g x 9.81 m/s 2 = 9810 kg-m/s2 = 9810 N

1.78 F = 9 8 1 0 N x 1.0lb/4.448N = 22051b

1.79 (Variable Answers) See problem 1.75 for method.

Density, specific weight, and specific gravity

1.80 yB = (sg)Byw = (0.876)(9.81kN/m3 ) = 8.59 kN/m 3

p B = (sgVw = (0.876)(1000 kg/m3 ) = 876 k g / m 3

, y 12.02N s2 1kg-m/s 2 31.81 P- = x x = 1.225 kg /m

g m3 9.81 m N

T h e Nature of F l u i d s 7


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1.82 7 = /?g = l - 9 6 4 k g / m3 x 9 . 8 1 m / s 2 x = 19.27 N/ m 3

1kg m/s

8 .860kN/m

yW@4C 9 .81kN/m1.83 sg = ^ = = 0.903 a t5 C

y = 8.483k N / m3

= 0 . 8 6 5 a t 50C

yW@4C 9.81 kN / m3

1.84 ^ = ; F = = 2 I 2 5 _ k N _ = 0 0 1 7 3 M 3

V y 130.4 kN /m 3

1.85 V= AL = KD2LIA =w(0.150 m) 2(0.100 m)/4 = 1.767 x 10"3 m 3

m 1.56 kg , , 3p = = V ~ T = 8 8 3 k g / m J

^ F 1.767 x l O ^ m 3

y = M = 883 kg /m 3 x 9.81 m/s 2 x - = 8.66x1 5 _ =8.66 ^

1 kg m/s n i3

m

3

sg = />//> @ 4C = 883 kg/m 3/1000 kg/m 3 = 0.883

1.86 y = (sg)(yw @ 4C) = 1.258(9.81kN/m3 ) = 12.34 kN/m 3 = wlV

w = yF = (12.34kN/m 3 )(0.50 m 3 ) = 6.17 k N

w 6 .17kN 10 3N 1 kg -m/s 2

?H= = x x = 629 kgg 9.81 m/s 2 k N N

1.87 w = yV= (sg)(yw)(V) = (0.68)(9.81 kN/m3)(0.095 m 3 ) = 0.634k N = 634 N

1.88 ^ =/ ?g = (1200 kg /m3

) (9 .81 m/s2

)

f I N ^

kg m/s22J

3

11.77 kN/m3

, . 1200 kg /m _

/ ? W @ 4 C 1000 kg/m3

_ _ w 2 2 . 0 N l k N , _ 3 31.89 F = = r ~ x ; = 2 . 7 2 x l 0 3 m 3

y (0.826)(9.81kN/m 3 ) 10 3 N

, 1080kg 9 .81m I N l k N 31.90 Y~Pg~ - x x - x - =10.59 kN /m

r F * m 3 s2 1kg -m/s 2 103 N

. 1080 kg /m

3

Sg = /?//? = 2 r- =1.08S ^ ^ W 1000 kg /m 3

1.91 p = (sg)(pw) =(0.789)(1000 kg/m3 ) = 789 kg/m 3

7 = (sg)Ov) = (0.789)(9.81kN/m 3 ) = 7.74 kN/m 3

8 Chapter 1


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1.92 w0 = 35.4 N - 2.25 N = 33.15 N

Va = Ad = (nD2IA)(d) = (.1 50 m) 2(.20 m)/4 = 3.53 x 10~3 m 3

W= 33 .15N = g x 1 Q 3 N / M 3 _ 9 < 3 g k N / m ^

' V 3 . 5 3 x l O " 3 m 3

^ = 9.38 kN / m3

C i - v r > - i - v T / 3

y 9.81 kN /m

3

1.93 K =A d = (nLflAXd) = (10m) 2(6.75 m)/4 = 530.1 m 3

w = yV= (0.68)(9.81 kN/m 3X530.1 m 3 ) = 3.536 x 10 3 k N = 3.536 M N

m = pV= (0.68)(1000kg/m 3 X530.1 m 3 ) = 360.5 x 10 3 kg = 360.5 Mg

1.94 wcastor o i = yco Vco = (9.42 kN/m3)(0.02 m 3 ) = 0.1884 k N

w 0.18 84kN . . i n_3 3y = = =1 . 4 2 x1 0 m

ym (13.54X9.81 k N / m3 )

1.95 W =yF=(2.32)(9.81 kN/m 3 )(1.42x 10~ 4 m 3) = 3.23 x 10~ 3 kN = 3.23N

1.96 y = (sg)(y) = 0.876(62.4 lb / 3 ) =54.7 lb/ft 3

p = (sg)Oow)= 0.876(1.94slugs/ft3) = 1.70 slugs/ft 3

y 0.0765lb/ft 3 lslu g . i n_3 ,1.97 p = - = x f = 2.38 x 10 slugs/ ft

g 32.2 ft/s 2 l lb s 2/ft

1.98 x = p g = 0.0 0381 slu g/f t 3 (32.2ft /s 2 ) 1 l b " S = 0.1227 lb /f t 3

slug

1.99 sg = yj(yw @ 4 C) = 56.4 lb/f t3/62.4 lb/f t 3 =0.904 at 40F

sg = yJYw @ 4C) = 54.0 lb/ft3/62.4 lb/ft

3

=0.865 at 120F

1.100 V= wly = 500 lb/834 lb/ft 3 = 0.600 ft 3

1.101 r ^ l ^ x I ^ =56.1 lb / f t3

V lg al ft3

y 56.1l b / f t 3 , _ lb -s 2 . , ,

y . , 5 . 6 1 1 b / f t ^ 0 W 9

yW@4C 62.4 l b / f t3

^ ( 5 0 g a l ) i ^ lf t 3 '7. 48 gal

1.102 w = = ( 1 . 2 5 8 ) ^ = ^ ( 5 0 g a l ) ^ ^ = 5251b

1 .103 w^yV = p g V = l 3 2 X h ' s 2 x = = ^ x 2 5 . 0 g a l x 1 ' = 142 lbr f t 4 s2 7.48 gal

T h e Nature of F l u i d s 9


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1.104 S g = = ^ x - ^ x l ^ x l ^ = 1 . 2 0pw cm

3 1000 kg 10 3 g m 3

p =(sg)(pw) = 1.20(1.94slugs/ft3) =2.33slugs/ft 3

7= (sg)(7w) = (1.20)(62.4 lb/f t 3 ) = 74.9lb / f t 3

i

i_ r,

w

5.01b ft

3

0.0283m

3

(1 0

2

cm)

3

_ , _ 3

1.105 V = = x x-i ' = 2 7 4 5 c m 3y (0.826)62.4 lb f t 3 3m

1.106 y= (sg)(yw) = (1.08X62.4 lb/f t3 ) =67.4lb / f t 3

1.107 p = (0.79)(1.94 slugs/ft3) =1.53 slugs/ft 3;p = 0.79g/cm 3

U 0 8 _ ( 7 . 9 5 - 0 . 5 0 ) , y}inW_ . ^ ^

" V ( ; r ( 6 .0 in ) 2 /4)(8.0in) f t

sg =7olyw= 56.9 lb/ft3/62.4lb/f t 3 =0.912

1.109 V = A d = d = ) x 22 ft = 15550 ft 3 x 7.48 gal/ft 3 = 1.16x 10 sgal

w^yV= (0.68)(62.4 lb/f t 3)( 15550 ft 3 ) =6.60x10 slb

1.110 wco =yc0V= (59.69 lb/ft3)( 5 ga l )( l ft 3/7.48 gal) =39.90 lb

w 39.90 lb f t 3 7.48 gal _ _ ,Vm = = x r ~ =0.353gal

y_ 13.54(62.4 lb ) ft3

L U I w = y V = ( 2 . 3 2 )( 6 2 - 4 1 b ) ( 8 . 6 4 i n 3 ) 0 ft3), = 0.724 lb

ft3 1728in 3

10 C h a p te r 1


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CURVE FIT FOR THE PROPERTIES OF WATER VS. TEMPERATURE

TABLE A.1Computed Computed

Temp. Sp Wt Density Sp Wt % Diff Density %Diff

0 9.81 1000 9.811 0.002 1000.2 0.020

5 9.81 1000 9.812 0.018 1000.2 0.017

10 9.81 1000 9.809 0.012 999.8 -0.015

15 9.81 1000 9.803 -0.017 999.2 -0.075

20 9.79 998 9.794 0.045 998.4 0.03925 9.78 997 9.783 0.028 997.3 0.029

30 9.77 996 9.769 -0.015 996.0 -0.002

35 9.75 994 9.752 0.022 994.5 0.047

40 9.73 992 9.734 0.037 992.8 0.077

45 9.71 990 9.713 0.032 990.9 0.090

50 9.69 988 9.691 0.007 988.9 0.088

55 9.67 986 9.667 -0.035 986.7 0.072

60 9.65 984 9.641 -0.096 984.4 0.04265 9.62 981 9.613 -0.070 982.0 0.10370 9.59 978 9.584 -0.061 979.5 0.15475 9.56 975 9.553 -0.069 976.9 0.19580 9.53 971 9.521 -0.097 974.2 0.33185 9.50 968 9.486 -0.144 971.5 0.357

90 9.47 965 9.450 -0.211 968.6 0.376

95 9.44 962 9.411 -0.302 965.7 0.388

100 9.40 958 9.371 -0.312 962.8 0.500

Specific Wt v s .Temperatura forWater

1= -1 E -09X 4j>4 E-07X3-7E-06X2+ 0 . 0 0 0 3X = 3.669m2

Fy = yAw = (9.81)(3.669)(1.50) = 54.0 kNx, = 1.20/2 = 0.60 m

x2 - 0.22347?= 0.268m [Machinery's Handbook]

_ = AX) + A2x2 _ (3.36X0.60) + (0.309)(0.268)X A 3.669

x = 0.572 mhc = h + s/2 = 2.80 + 0.60 = 3.40 m

FH = yswhc = (9.81)( 1.20)(1 .50)(3.40) = 60.0 kN

hp = h c + = 3.40 + 1.202

3.435 mI2he 12(3.40)

FR = 4FV+FH =V5 4 .0 2+60 .0 2 =80.7kN

= t a n ^ = 4 2 . 0 60.0

1.20

1

h = 2.80 m

FOTOMsrnwntctfcigonsurfaoa.

40C h a p t e r 4


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4.54

4.55

4.57

A = TT> 2/8 = = yAw = (0.79)(62.4)(3.534)(5.0) = 871 lbx = 0.212D = 0.212(36i n) = 7.63 inhc = h + s/2 = 48 + 36/2 = 66.0 in = 5.50 ft

F =yswhc = (0.79)(62.4)(3.0)(5.0)(5.50)

F =40671b

h - 4 8 l n

h= hc =\2h

= 66 + 362

12(66)67.64 in

FR= 4F + F2

H =V8712 + 4067 2 = 41591b

, _ i - ' 8 ^ 1

= tanF

tan4067

12.1

Foroshownactjngon fl uid.

(See Prob. 4.47)

r , p 7.5 k N / m2

n , rEq . Depth = ha = = =0.765 m

y 9.81 k N / m 3

h u = hi + / * a = 1.85 +0.765 = 2.615 m;hce = hXe + s/2 = 2.615 + 0.75/2= 2.990 m

Ax = (0.75)(2.615) = 1.961 m2 ;A2 =0.442m

2 ;AT =2.403 m2

Fv = yAw = (9.81X2.403)(2.000) = 47.15 kN_ i X , + A2x2 (1.961X0.375) + (0.442)(0.318) =

2.403

FH =yswhce = (9 .81X0.75X2.00X2.99) = 44.00 kN

0.365 m

h - h =0.752

'pe = 0.016 m = 16 mm\2hce 12(2.99)

FR= 4FV+FH = V47.15 2 + 44.00 2 =64.49 kN

^ = t a n " ' 5 L tan47.15

44.0047.0

4.56 (See Prob. 4.48)

Eq . Depth = ha = =4.65 kN/ m 2

= 0.574 my (0.826)(9.81kN/m 3)

hu = hi + ha = Q.62m +0.574m = 1.194 m

A = (1.194X1.25) + 7t(1.25)2/8 = 2.106 m 2

Fv=yAw = (0.826X9.81 kN/m3)(2.106 m 2)(2.50 m) =42.66kN = FR

Ne t hori zontal forc = 0

From Section 4.11, net verticalforc equals the weight of the displaced fluid acting upwardand the weight of the cylinder acting downward.

W/= yfVd = (62A lb/ft3)(0.164 ft3) = 10.2 lb

Vd = A-L-TTD2

L =*(6.00 in) 2

10.0 in = 282.7 i n 3xft3

4 4 1728 i n 3

wc = JcV= (0.284 lb/i n3)(282.7 in 3 ) = 80.3 lb

Ne t forc on bottom =wc - Wj=80.3 - 10.2 = 70.1 lb down

0.164 ft 3

4.58 See Prob . 4.57 . wf= 10.2 lb

wc =y cV= (0.100 lb/ in3)(282.7 in 3 ) =28.27 lb

i V , = wc- wf= 28.27- 10.2 = 18.07 lb down

F o r c e s Due to S t a t i c F l u i d s 41


56/298

4.59 See Prob . 4.57. wf= 10.2 lb

wc =ycV= (30.0 lb/ft3)(0.164 ft 3 ) = 4.92 Ib

Fnet= wc- w / = 4.92 - 10.2= 5.28 Ib up

Bu t this indicates that the cylinde rwould float, as expected. Then, the forc exerted by thecylinderon the bottom of the tank iszero.

4.60 The specific wei ght o f the cyli nder must be lessthan or equal to that of the fluid i fno forc isto be exerted on the tank bottom.

4.61 (See Prob. 4.57 .) Because the depth of the fluid doesnot affect the result, Fnet = 70.1 Ibdown. Th is is true as lo ng as the fluid depth isgreater than or equal to the diameter of thecylinder.

4.62 (See Prob. 4.57. ) wc = 80.3 lbForc (downward) on upper part of cylinder =wt. of volume ofcross-hatchedvolume. Forc(upward) on lowe r part o f cylinder = wt. ofentire displaced volumeplus that ofcross-

hatched volume. Then net forc is wt. o fdisplaced vo lume (upward).

wf=yfVd = y4dL

Ad= + - (2 x) (2 .0 ) =Ai+A24 360 2 V A '

= g ( 6 . 0 i n )2

263.6' 4 360

+ (2.236)(2.0) = 25.18 in2

wf= y4dL =62.4 lb/f t3 25.18 i n 2 10.0 i n

Fnet= wc- w / = 80.3 - 9.09 = 71.21 Ib down

l f t 3

e-$m-(2.0/3.0)-41.8*x- fleo a - 3.0 co ax -238 lnB -180* +2 -283.8*

1728 in 3= 9.09 lb

42 C h a p t e r 4


57/298

4.63 (See Prob . 4.57) For any depth > 6.00 in , F n c t = 70.1 lb down

Methodof Prob. 4.62 used forh 3.00 in. For any depth < 3.00 in, use figurebe lo w.

3 4 5 6 7 8

RutiDepth ,/>0n)

4.64 Centroid:y = 0.212 D= 0.212(36 in ) = (7.63 in ) ( l ft/12 in) = 0.636 ft

x= (20 in)/sin 25 = 47.32 in

L c = 60 in+ x - y = 60 + 47.32 - 7.632 = (99.69 in ) ( l ft/12 in) = 8.308f thc =L csin 25 = (8.308 ft)(sin 25 ) = 3.511 ft

A = TTD2/8 =7i(3.0 ft ) 2/8 = 3.534 ft 2

FR yhcA= (1.06)(62.4lb/ft3)(3.511 ft)(3.534 ft 2 ) = 820 lb = FR

1= 6.86 x 10 _ 3 D 4 = 6.86 x 10~3(3.00 ft ) 4= 0.556 ft4

Lp-Lc =I/[LcA] = (0.556 ft4)/[(8.308 ft)(3.534f t 2 )] = 0.0189 ft

L P =L C+0.0189 ft =8.308ft + 0.0189 ft =8.327 ft =L p

F o r c e s Du e to S t a t i c F l u i d s43


58/298


59/298

5.8 w - Fb - FSP = O= W-y0Vd - FSP

FSP = w- y0Vd = 14.6 lb - (0.90)(62.4 lb/ft3)(40 in 3 )

FSP= 13.3 lb

1 ft3

1728 in 3

5.9 ~LFV = O= wF + ws - Fb - Fbs

Fbs =ywVs = 9.81 kN / m3 x (0.100 m) 3

0 = y F J ^ + 8 0 N - y w F / ? - 9 . 8 1 N

0 = Vp(yF - y w ) + 70.19 N

y = -70-19N _ - 7 0 . 1 9 NF rF~rw ( 4 7 0 - 9 8 1 0 ) N / m

3

= 7.515 x 1( T 3 m 3

9.81 N

5.10 wc

+ wA

=Fb

= yw

Vc

= 62A lb/ft

.3

3 tf = 588.1 lb

wA =Fb-wc =588.1 - 3 0 = 558.1 lb = y / F / load - F^ - F^ - 0

w d r u m s = 4(30 l b) = 120 lb (Prob. 5.31)

= 57.4 in

2

See Prob. 4.63 for method

o f computing As.

w w 0 0 d = 178.5 lb (Prob. 5.32)

Fbn =1 801 lb (Prob. 5.31)

FK = ywVw = 62.4 lb/ft3 x 4.464 f t 3 =278.6 lb (Prob. 5.32)

Wioad= Fh + Fb -wD-ww= 1801 + 2 7 8 . 6 - 12 0- 178.5 = 1781 1b

B u o y a n c y and S tab i l i t y49


64/298

5.35 Given: yF = 12.00 lb/f t3 , yc = 150 lb/f t

3 , wc = 600 lb

Find: Tens in in cable

Floatonly: Z f > = 0 =wF + T- Fbp

T= Fbr ~wF

Bu t wF = yFVF = 12.0 lb/f t3 x 9.0 ft 3= 108 lb

(18 .0 in)2

(48in)VF =1728 i n 3 / f t 3

= 9.00 ft3

Fbp =ywVd = (64.0 lb/ ft3)(6.375 ft 3 ) = 408 lb

( 1 8 . 0 , n ) ' ( 3 4 , n ) ^ 3 7 5 f t ,

1728 i n 3 / f t 3

T= 4 0 8 - 108 = 3001b

Check concrete block: F n c t = w c - Fb j^unstable

w= Fb = yswAX

v _ w 130lb(12 m/ft)

YSWA (64.01b/ft3 )(3ft)(4ft)

= 2.03 in

ych=X/2 = 1.016 in

ycs = 12 in + 34 in = 46.0in

/ = (48)(36) 3in 4/12 = 1.866 x 105i n 4

Vd = (48)(36)(2.03)in3= 3510 in 3

MB=I/Vd= 53.17 in

j -c= ^cb + M B = 1.016+ 53.17

= 54.18 in >ycgstable

X-7S0

46

34

i

/g

_ L

8.0

7 "2f tx4ft rectanc>

me

t54.1B

Inx48Inrectangle

B u o y a n c y an dS tab i l i t y 55


70/298

5.56 w - Fb = 0 =ywVtot -y0Vd =ywAH- yoAX

ywAH _ ywH = (32 lb /f t3 ) (6 in)

y0 0.90(62.4 l b / f t3 )

/Rwaangfc6irix12ln

= 3.419 iny c b =XI2= 1.709 iny c g = H/2 = 6.00/2 = 3.00 in/ = 12(6) 3/12 = 2 1 6 i n 4

Vd =(12)(6)(3.419)= 246.2 i n3

MB=I/Vd =0.877 in

Jmc= yCb + M B = 1.709 + 0.877 = 2.586 in< j c gunstable

IX-3-419

I *_ i 1

t3.00

I

m c - 5 rn c -51

c b / 1-708

r

6.00

ZS80

5.57

5.58

w-Fb =ywVd = y^AY

w 2100001b

ywA (62.4 lb / f t3)(60 ft)(20 ft)

y c b =J72= 1.402 ft ;y c g = 1.50 ft given

/ (60)(20) 3 /12

= 2.804 ft

' S I

M B =

(60)(20)(2.804)

= 11.888 ft

Jme=y* + M B = 1.402 + 11.888 ft = 13.290 ft> j c gs table

Wtotai= 210000 + 240000 = 450000 lb

w 4500001b= 6.010 ft (See Prob. 5.57)

= 5.547 ft

ywA (62.4)(60)(20)

ycb = X/2 =3.005 ft

M B - 7 - ( 6 ) ( 2 0 ) 3 / 1 2

Vd (60)(20)(6.010)

Jmc=ya>+ M B = 3.005 + 5.547 =8.552ftabove barge

cg is within bargestable

More complete solution:w

c

7ADepth of coal = dc

240000lb= 4.444 ft

(45lb/f t 3 )(60ft)(20ft)

y c =dJ2 = 2.222 ft from bot tom to cg of coalw

c y c+ w

B y B _ (240000)(2.222)+ (210000X1.50)

r~1&29

"cb

4500001.885 ft


71/298

5.59 w = Fb = yrVd = yrAdL

w = ycAtotL

1= - (.600)(.300)(1.20)(2.36)kN

= 0.255 k N

Ad

= GX+2 vovo

(B-2X)(X) + X20 - 0 0 0 mm(fl-OJSOm)

407

Equate

Ad =BX-2X2 +X2 =BX-X2 = 6 0 0 X - X 2

Ad =w 0.255k N

yTL (0.87)(9.81kN/m3)(1.20m)

Ad =0.02489m2 x ( 1 ^ " ) =2.489x 1 0 4 mm 2

m

6 0 0 X - X 2 =2.489 x 10 4

X2 - 600X+2.489 x 10 4 = 0;X= 44.8 mm by Quadratic Eq.

G = B- 2X= 600 -2 (4 4. 8) = 510 mm

7=G 3 Z,/12 = (510)3(1200)/12 = 1.329 x 10 1 0 mm 4

Vd= AdL = [600(44.8) - 44.82](1200) =2.985 x 1 0 7 mm 3

M B = / / K r f = 445 mm

v X(G + 2B) A A O 44.8( 510 + 1200) _ycb=X ^ - =44.8 - =2 1. 8m m

3(G +B) 3(510 + 600)

-ycb + MB = 21.8 + 445 = 467 mm

1>>cg= - (3 00 )= 100 mm ycg = SI2

ymc

X

x i x=^ c b + M B = + = +-12(5 2 )(X)

+ J L = ^ o r X2 + ^ -sx=o

2 \2X 2 6X2 - SX+ S2^ = 0

x=Sy/S2-4S2/6 S S r rTr

_ = V l - 2 / 3

= S - - ( 0 . 5 7 7 4 )2 2

2 2

= 0.7885' or 0.21 \S

SquanS x S

X> 0.7885"orX


72/298

5.61 Entirehull:

i

= (.72)(.40) +(2. 88)(1 .2)

3.60

ycg

= 1.040 mSubmerged Volume:

_ (.72)(0.4) + (2.16)(1.05)yCb

2.88

4?S3m

= 0.8875 m

F= (2.88)(5.5)= 15.84 m 3

7=5.5(2.4) 3 /12 = 6.336 m 4

y =ycb+ I/Vd = 0.8875 + 0.40 = 1.2875m >ycgstable

5.62 a) wc =Fb;ycVc = ywVd

_ 3 0 1 b ; r( .5ft) 2 (1.0ft)wc--

f t

3

V d ^

12

1.9631b

1.963 lb

yw 62.4 lb / f t3

= 0.03147 ft 3

3\/1 TOO , 3 , l

X= 3

_ (0.03147 ft')(1728 in') _ 3

ft3

_ 7v(Dxf(X) _ 7V(X/2)2X _TCX3

12 12 48

K

3X

4SVD _ 48(54.37) _9.40 in

71

0.75(9.40) = 7.05 in

_ nD\ _ -(9.40/2) 4 _= 23.95 in 4

64 64

M B =I/Vd =23.95/54.37 = 0.441 in

y =ycb+ M B = 7.05 + 0.441 = 7.491 in3H

y


73/298

nDx _ - ( ! ! .51/2)4

/ = 64 ~ 64 = 53.77i n 4

/ 53 77M B = = r l l = 0.539 i n

VA 99.69

y c b = 0.75X= 0.75(11.51)= 8.633i n

y m c =yCb + M B = 8.633+0.539= 9.172 inycs = 9.00 in 3/12 = (1.50 m) 3/12 = 0.8836 m 3

^cyi-d= n&hJA = (1. 50 m) 2(0.35 m)/4 =0.6185 m 3

Then Vd = Vhs + Vcyld = 0.8836+0.6185= 1.502 m3

Fb = yfVd = (1.16)(9.81 kN/m3)(1.502 m 3 ) = 17.09 k N = Wc + Wv (Answer)

(b) Find yv = Specific weight ofvessel material = WJVvT', Given Wc = 5.0 k N

From part (a), 17.09 k N = Wc + Wv; Then Wv = 17.09 -Wv= 17.09 - 5.0 = 12.09 k N

VVT Total volume ofvessel= Vhs + Vcyi.T

Vcyi-T= (Di-D2)(Q.6Q m)/4 = [(1 .50 m) 2 - (1.40 m) 2](0.60 m)/4 = 0.1367 m 3

VvT= Vhs + V^ =0.8836m3 + 0.1367m 3 = 1.020 m 3

yv = Specific weight ofvessel material = WJVvf, = (12.09 kN)/( 1.020 m3 )

= 11.85 k N / m 3= / v

1500

1400

350 Fluid Surface

Fb

B u o y a n c y and S tab i l i t y 59


74/298

(c) Evalate stability.Find metacenter, ymc. See figure for key dimensions.Given:ycg = 0.75 + 0.60 - 0.40 = 0.950 m from bot tom o f vessel.We must find: ymc =ycb +MB =ycb +II Vd

1= 7tD4/64 = (1.50 m) 4 /64 = 0.2485 m 4For circularcrosssection at fluid surface.Vd = 1.502 m

3 Fr om part (a).MB =(0.2548 m 4)/(1.502 m 3 ) = 0.1654 m

The center of buoyancy is at the centroid of the displaced volume. The displacedvolumeis a composite of a cylinder and a hemisphere. The position of its centroid must

be co mput ed from theprincipie o f composite volumes.Measure all y vales frombottom of vessel.

(ycbWd) =(y hs)(Vhs) + (ycyuWcyi-d)yc = (ys)(Vhs) + (ycyl.d)(Vcy,d)]/Vd

We know from part (a): Vd =1.502 m3; Vhs = 0.8836 m

3 ;VcyU = 0.6185 m3

yhs = D/2 -y = D/2 - 3D/16 = (1.50 m)/2 - 3(1.50 m)/1 6 = 0.4688 mycyt-d= D/2 + (0.35 m) /2 = (1.50 m)/ 2 + 0.175 m = 0.925 m

Then ycb = [(0.4688 m)(0 .8836 m3 ) + (0.925 m)(0.6185m 3 ) ] / ( l.502 m 3 ) = 0.657 m

Now, ymc = ycb + MB =0.657 m + 0.1654 m = 0.822 m From bottom of vessel.Because ymc


75/298

CHAPTER SIX

FLOW OF FLUIDS

Conversin factors

6.1 0 = 3.0 gal/min x 6.309 x 10~5m 3/s/1.0gal/min = 1.89 x 10"4 m 3/s

6.2 0 = 459 gal/min x 6.309 x 10"5 m 3/s/1.0gal/min = 2.90 x 10~2 m3/s

6.3 0 = 8720 gal/min x 6.309 x 10"5 m 3/s/1.0gal/min =0.550 m3/s

6.4 0 = 84.3 gal/min x 6.309 x 10~5m 3/s/1.0 gal/min = 5.32 x 10 3 m 3/s

6.5 0 = 125 L /m in x 1.0 m3/s/60000L /m in = 2.08 x 10~3 m3/s

6.6 0 = 4500 L /m in x 1.0 m3/s/60000 L /min = 7.50 x 10~2 m 3/s

6.7 0 = 15000L /m in x 1.0 m3/s/60000L /m in =0.250 m3/s

6.8 0 = 459 gal/min x3.785L/min/1.0 gal/min = 1737 L/min

6.9 0 = 8720 gal/min x 3.785L/min/1.0 gal/min = 3.30 x 104 L/min

6.10 0 = 23.5 cm3/s x m3/(100 cm) 3 = 2.35 x 10~5 m3/s

6.11 0 = 0.296 cm3/s x 1 m3/(100 cm) 3 = 2.96 x 10~7 m3/s

6.12 Q = 0.105 m3/s x 60000L/min/1.0 m3/s = 6300 L/min

6.13 0 = 3.58 x 10~3m3/s x 60000L/min/1.0 m3/s = 215 L/min

6.14 0 = 5.26 x 10~6m3/s x 60000L/min/1.0 m3/s = 0.316 L/min

6.15 0 = 459 gal/min x 1.0 ft3/s/449 gal/min = 1.02 ft 3/s

6.16 0 = 20 gal/min x 1.0 ft3/s/449 gal/min = 4.45 x 10~2 ft 3/s

6.17 Q = 2500 gal/min x 1.0 ft

3

/s/449 gal/min = 5.57 ft3

/s

6.18 0 = 2.50 gal/min x 1.0 ft2/s/449 gal/min = 5.57 x 10~3 ft 3/s

6.19 0 = 1.25 ft3/s x 449 gal/min/1.0 ft 3/s = 561 gal/min

6.20 0 = 0.06 ft3/s x 449 gal/min/1.0 ft3/s = 26.9 gal/min

6.21 0 = 7.50 ft3/s x 449 gal/min/1.0 ft 3/s =3368 gal/min

F l o w o f F l u i d s 61


76/298


77/298


78/298

xx, ^ t 60.61b rtrt,-2 4.50 ft 3600s _ _ , l S ,6.43 0^= y g =y,4u= 5x 0.2006 ft 2x x = 1.97 x 10 5lb/h

ft s hr

6.44 v = = 19.7L /min x -- ^ x =0.856 m/sA 60000L /m in 3.835 x l 0 ~ 4 m 2

6.45 g = 30gal/min x 1.0 ft 3/s/449gal/min= 0.0668 ft 3/s

g 0.0668f t3/s 1 r t - 3 2= = = 8.35 x 10 3 f t 2max.

v{ 8.0 ft/s1 1/4 x 0.065 tube - > , = 6.842 x 10" 3f t 2

Actual Dj = g / ^ = (0.0668ft3/s)/(6.842 x 10" 3f t 2 ) = 9.77 ft/s

g 0.0668 ft3/s 1 r t_3 . 2 ^ = = = 2.67x 10 ft mi n.u2 25.0 ft/s

7/8 x 0.065 tube - > 2 = 3.027 x 10~3ft 2

Actual D 2 =Q/A2 = (0.0668 ft3/s)/(3.027 x 10~3f t 2 ) = 22.07 ft/s

g 0.0668f t 3/s n - 2 ^ 26.46 AX = = = 3.34x 10 ft max .

, 2.0 ft/s2-in x0.065 is largest tube listed - >l , = 1.907 x 10~2ft 2

Actual i) != QIAX = (0.0668 ft3/s)/(1.907 x 10~2f t 2 ) = 3.50 ft/s

g 0.0668f t3/s . . . 1 A _ 3 - 2 .A2 = - S - = = 9.54 x 10 ft min .

v2 7.0 ft/s

1 1/2-in x0.083tube - > 2= 9.706 x 10~3ft 2

Actual D 2 = g/^j = (0.0668ft3/s)/(9.706 x 10 - 3 f t 2 ) = 6.88 ft/s

6.47 g L = 1800L /m in x = 0.030 m3/s

60000 L/min

A QL= 0-030 m /s = o.015 m2 minimum; 6-in Sc h 40;A = 1.864 x 10"2m 2

' o 2.0m/sQH = 9500/60000 = 0.1583 m

3/sO 0 1583

^ H = = = 7.916 x 10~2m 2 ;14-in Sch 40;A= 8.729 x 10~2m 2

o 2.0

6.48 ForAL = 0.015 m 2 min; 6-in Sch 80;A = 1.682 x 10~2m 2

ForAH = 7.916x 10~2m 2 min; 14-in Sc h 80;A = 7.916 x 10~2m 2

r *c Q 400L/min l m /s , , - . , . - ,6.49 D = = - x = 3.075 m/s [2-inSch 40pipe]

A 2.168x10 3m 2 60000L/mi n

6.50 2-i n Sch 80:v = - = = 3.500 m/sA (1.905x10 )(60000)

64 C h a p t e r 6


79/298

, r . Q 400gal/min 1f t3/s , N N O ^ , VA- Ot.

6.51 v = = & x = 10.08 ft/s [4-inSch 40]^ 0.0884 ft2 449 gal/min

6.52 i)= ^ = = 11.16 ft/s [4-inSch 80]A (0.07986)(449)

, r-, . Q 2.80L /m in l m 3 / s , r r ^ , , 2 .6.53 A=- = x = 1.556 x 10 4 m 2 min.

u 0.30 m/s 60000 L/min3/4 x0.065steel tube,A = 1.948 x 10~4m 2

Q6 95gal/min l f t3 / s ^

6.54 D 6 _ I N = = e x = 1.055 ft/s

4 0.2006 ft2 449 gal/min

1 A = a , Q j a = (Q.5X95)

4 4 (0.05132)(449)

6.55 0 = 800gal/min(l ft 3/s/449 gal/min) = 1.782 ft 3/sSuctionpipe : 5 in Sch 40;As = 0.1390 ft

2 ;vs = QIA= 12.82 ft/s6 in Sch 40;As = 0.2006 f t

2 ; v, = = 8.88 ft/sDischargepipe: 3 1/2 in Sch 40;Ad = 0.06868 ft

2 ;v=QIA= 25.94 ft/s4i n Sch 40; = 0.08840 ft 2 ;v=QIA= 20.15 ft/s

6.56 Q= 2000 gal/min(l ft 3/s/449 gal/min)= 4.454 ft 3/sSuctionpipe: 6 in Sch 40;A,= 0.2006 ft 2 ; =QIA= 22.21 ft/s

8 in Sch 40;As = 0.3472 ft2 ; v, =QIA= 12.83 ft/s

Dischargepip e: 5 in Sch 40;Ad = 0.1390 ft2; v=QIA= 32.05 ft/s

6 i n Sch 40; ^ = 0.2006 f t 2 ;vd = QIA= 22.21 ft/s

6.57 Q= 60m3

/h ( l h/3600 s) = 0.01667 m

3

/sSuctionpipe: 3 i n Sch 40;As = 4.768 x 10~3m 2 ; v, =QIA= 3.73 m/s

3 1/2 in Sch 40;As = 6.381 x 10~3m 2 ;vs = QIA= 2.61 m/s

Discharge pipe: 2 in Sch 40; ^ = 2.168 x 10~3m 2 ;vd = QIA= 7.69 m/s

2 1/2 in Sch 40;Ad = 3.090 x 10~3m 2 ;vd= 04 = 5.39 m/s

3\6.58 Q = Av = (7.538 x 10~ Jm') (3 .0 m/s) = 2.261 x 10""2m7s

A=Q = 2 - 2 6 1 x i r 2 m V s = 1.508 x lo"3 m 2= *ti>/A 15.0m/s

A = V 4 ^ / - = ^ 4 ( 1 . 5 0 8 x l 0 ^ 3 m 2 ) / - =4.38 x 10 " 2 m x 1 0 ^ =43.8 mmm

6.59 7 - 5 0 V ^ = 7.98 ft/s in pipeP Af 0.9396 ft

2

Q 7.50ft 3/sD N = = ; = 65.0 ft/s in nozzle

4 , -(4.60i n ) 2 l f t 2

4 X 1 4 4 i n 2

F l o w o f F l u i d s


80/298


81/298

2 2

6.62 Pt. A at gage;Pt. B outsid enozzl e: A - + Z A + ^A - = ^ f i - + z B + ^ - ; pB =0

7 2g y

565 kN2 2UA ~ U B = / _ Z X A = 3 6 5 M

2g B A y " m 2 (9.81 k N / m 3 )

DB

=vA

(AA

/AB

) =vA

(DA

IDB

f =D

a

( 7 0 / 3 5 )

2

= 4u A ; u

2

= 16u

2

;^ A - ^ B = - 1 5 ^

2g

=-53.94 m

4 5 u 2 = 2g(-53.94m)

u A =2g( 53.9 4m) 2(9.81 m/s 2)(53.94 m)

Q = AAvA =

15

-(.070m ) 215

8.40 m/s

x 8.40 m/s =0.0323m 7s = 3.23 x 10~ 2 m 3/s

o.6.63 Pt. A before nozzle; Pt. B outside nozzle: *- + zA+ - = - + zB + ^ - ;pB = 0, z A = zB

PA = y2 2o-o

2g

60.61bf t 3

y

(7 52 - 4 2 . 1 9 2 ) f t 2 / s 2

2(32.2 ft/s 2 )

2g y

l f t2

3 ,

2 g

u B = 75 ft/s; u A = u B 75Dp

= 75 21Vl.0y

144 in 2

42.19 ft/s

25.1 psig

6.64 Q= 10 gal /min x1 ft 3/s

449 gal/min

0.0223 f t 3 Q 0.0223ft 3/s 3.71 ft

0.0060 ft 2

u 2g _ 0.0223 _ 0.955 ft./>A + " 1 _ / > B + 7 . . 7D + ZA + + + , z A Z B

4 , 0.02333 s yK 2g yK 2 g '

PA-PB = JK2 2Un -O.

2g

501b

f t 3(0.955 2 - 3 . 7 1 2 )f t

2(32.2 ft/s2)

l f t2

144 in 2-0.0694 psi

6.65 Pt. 1 at water surface; Pt. 2 outside nozzle .

Pi \ V.

y 2 g y+ z 2 + ; P l = 0,vl =0,p2 =0

2 g

u2 =j2g(z} -z2) =^2(9.81 m/s2)(6.0 m) = 10.85 m/s

g = ^ 2 l ) 2= 7 r ( Q - Q 5 Q m ) 2 x 10.85 m/s = 2.13 x 1 0 2 m 3/s

U A = Q _ 0.0213 m3

/sAA 7 r ( .15m)

2 / 4= 1.206 m/s; u

2

_ 1.1206

2

m

2

/ s

2

2 g 2(9.81 m/ s 2 )= 0.0741 m

P x + z x + ^ - - + zA+~- ; P i = 0,v=0y

PA = yK

2 g y

( * . - * A ) - 2 g

2g

9.81 k N

m[6.0 m - 0.0741 m] =58.1 kPa



82/298

6.66 Pt. 1 at o i l surface; Pt. 2 outside nozzle.

Yo+ z . + - J - = + z 2 +-2- ; ^ , = 0,^1= 0 , ^2 = 0

2g Yo 2g

x>2= ^J2g(zl- z 2 ) =72(9.81m/ s2 ) (3 .0m) =7 .67 m/s

^ , ( .035m ) 2 7.67 m 7.38 x 10~3 m 3

Q =A2x>2 = x4 s s-3 3iQ _ 7 . 3 8 x 1 0 - i m7s_ 0.940m oA _ oB _ 0.94

2 m 2 / s 2

A ( 0 . 1 0 m )2 / 4 2g 2g 2(9.81m / s 2 )

0.0450 m

r 2g x0 2g

y*( z , - z A ) - - ^

2g

h

=(0.85)9.81m

[4 .0 -0 .045]m =33.0kPa

v s y

+ z. + - = ^ B - + z B + -J L ; pt = 0, u , = 0

2g r D

2g

2g

(0.85)(9.8 1)[3.0 - 0.045] = 24.6kP a

6.67o, + z. + = + z 2 + : Pt. 1 at water surface; Pt. 2 outside nozzle;vx = 0 ,p 2 = 0

r 2g ^ 2g

u 2 = y2g(p/r + - z 2 ) = 1232.2 ft

v s y2^ , ( 3 i n ) 2 59.06 ft ft

{?=A2v2 = x x144in 2

201b ft 3 144 i n 2

i n 2 62.4lb f t 2

: 2.90 ft7s

+ 8.0 ft 59.06 ft/s

6.68 -+ z, +-2g Y

(z2-zl) +

+ z 2 + : Pt. 1 at water surface; Pt. 2 outside nozzle;v= 0,p2 = 0

2g

2g

62.41b

f t 3-10ft +

(20) 2 ft2/s2

2(32.2 ft/s 2)

l f t 2

144 i n 2-1.64psig

6.69o:

+ z. + = + z 2 + : Pt. 1 at water surface; Pt. 2 outside nozzle;2g Y 2g

P\ =Pi=0, u , = 0

u2 =

y

j2g(z]

- z2

) = ^2(9 .81 m/s2

(4.6 m) =9.50 m/s

(2= ^ 2 u 2= ? r ( ' ' Q 2 5 m ) x 9.50 m/s =4.66 x 10~3m 3/s

68 C h a p t e r 6


83/298

P\ U\ + z,+

_ A + Z A + ^ A _ : ^ , = 0, = 0 , Z I = Z A2 g r 2 g

4

2g9.81kN

m

= v2/4 =2.375m/s

(2 .375)

2

m

2

/ s

2

2(9.81m/s 2 )2.82 kPa

r

PB= y

z , + - ^ -= - ^ +z B + - S - ;/>,=0,i) ,= 0

2g r 2 g

(*1 - Z B ) -2g

=9.81 -0.90-(2.375)2

2(9.81)-11.65kPa

6.70 (SeeProb. 6.69) v2 == 7 - l x l m / s = 14.46m/s

, - r - ^2_ ( 1 4 . 4 6 ) 2 m 2 / s 2

A, "(.025 m ) 2 / 4

10.66 m2g [2(9.81m /s 2 ) ]

6.71 (SeeProb. 6.69) vB = =5 - 6 x 1 Q ~ 3 m V s =2 .8 5 m/s

4 , ; r ( .05m) 2 /4

Mnimum pressure exists atB , highest po int in system.

A + Z I + ^ B _ + Z B + ; / , I = 0 J 1 ) ] = 0

y 2g y 2 g

Z - Z _ Y~Pn l _ - ( - 1 8 k N ) ( 2 . 85 )2 m 2 / s 2 _Z

ZX

y 2g m2

(9.81 k N / m3

) 2(9.81 m/s2

)

6.72 Ana lysi s for u 2 , Q,PA,PB sameas Prob. 6.69.

1.42 m

u 2 = ^2g(zx-z2) =72(9.81)00) = 14.01m/s

= 2 2= ; r ( ' 2 5 ^ > x 14.01= 6.88 x 10""3m 3/s

4

A = l ^ = C = U b = l V 4

A =3.502 m/s

/ A= r 02 g

= (0.86)(9.81)-(3.502)2

2(9.81)= (8.437)[-0.625]=-5.27 kP a

( z , - * , ) - - * -2g

(8.437)[-3.0- 0.625]=-30.58kPa

/>C=/>A = -5. 27 kPa



84/298

Yo

PD= YO

2g Yo 2g

2g= (8.437)[10.0 - 0.625] = 79.1 kPa

+ z A + - + z B + , z A ZBr 2g / 2g

^ - l _ /> B- / \ _ ( 4 2 - 5 0 )l b f t3 ( 1 4 4 i n 2 )

2g TV

An

i n 2 (62.41b)ft2=-18.46 ft

- | = 0.25 uB; v{ =0.0625u2

0.0625u2 -ul =2g( -18. 46 ft)

-0.9375u 2 =2g( -18 .46f t )

2(32.2 ft)(-l 8.46 ft)

' s2 (-0.9375) 35.6 ft/s

PA , v , l _ P* , _ , U B . _ _ 7r Z A 4- r ZB +- , Z A Zg

Yo 2g Yo 2g

\ _ PB -PA _ (28 .2- 25.6 ) lb ft 3 (144 in 2 )

2g i n2 (0.90)(62.4 lb)f t 2

= 6.667ft

vA = v B ^ - = oBAA v D A y

= 2.56vB; u A = 6.55Ug

6 . 5 5 u 2 - u 2 =2g(6.667ft)

5.55u 2 = 2g(6.667 ft)

2(32.2)(6.667) = , J 9 m

5.55

^ , - ( 8 i n ) 2 8.79ft ft 2 , 3 ,Q =ABx>B = - i - x x = 3 .0 7f t

3 / s4 s 144 in


85/298

2g K 2g

2g 2g

1 6 u A - u A_ 1 5 u A

2g

2g 2g

Manometer:pA +ywy +yji - ymh - ywy =pB [cancel termsw i t hy]

PA ~PB=ymh ~yji =Kym - yw) =A(13.54yH,- yw) =A(12.54yw)

PA -PB = 12.54/;=15u

2g

2 g q 2 . 5 4 ) W 2(9.81m X 1 254X0250m ) = ^ ^

15 V s 2 ( 1 5 )

Q = v A T) A = [TT(.050m) 2/4](2.025 m/s) = 3.98 x 10"3 m3/s

N 2

D,6.76 (SeeProb. 6.75)u A= u B

AB

\AAJ

= 0.25u B=0.25(10)=2.50m/s

12.54/2_ 15uA _15(2 .50)

2 m 2 / s 2 _2î2g 2(9.81m/s 2 )

/i=4.778m/12.54=0.381m

;4.778m

6.77^ + z A + ^= + z B + ^ -7D 2g yo 2g

2 2

2gPB-PA

7o(ZB~ZA)

AA (DA]

2

l 0 0 l~ O, = ÂAB AKDB) A L 50J

u2

= 1 6 u A

ol - u 2 = u 2 -16u 42 = - 1 5 u 2

Manometer: pA +y(0.35m)- y,(0.20m)- y o(0.75m)= pBPB-PA = -yw(0.20 m) - y(0.40)m

^ ^ = ^ ( Q - 2 Q m ) - 0 . 4 0 = - 9 - 8 1 ( 0 . 2 0 m ) _ a 4 Q m _ a 6 2 7 m

7o 70 8.64

z B - z A=0.60

-\5u2

A =2g[-0.627m +0.60 m]=2g(-0.027m)

u A = A /2(9.81m/s2) ( -0 .027 m) /-15 =0.188m/s

Q =AAx>A = â i ^ m ) x0.188 m/s=1.48x10~ 3m 3/s

F l o w o fF l u i d s


86/298

6.78

6.79

2g K

z B - z A=0.25m

+ (ZB~ZA)

V ^ y

( 200

v 75 y: 7.11u 4

vi =50.6u 2

u 2 - u 2 = 2 - 5 0 . 6 ^ = - 4 9 . 6 u 2

Manometer: /? A+y0>>+yo(0.60 m)- yG(0.60 m)- yay- yo(0.25 m) =pB

PB-PA =yo(0.35 m)- y G(0.60 m)

PB-PA -Q 35 ( l -40)(9 .81kN/m 3 )(0.60m) =

_0 5 8 3

y0 ' (0.90)(9.81 k N / m3 )

m

-49.6/J 2 =2g[-0.583 m + 0.25 m]=2g(-0.333m )

vA = ^2(9.81 m/s2 ) ( -0.333 m)/(-49.6) =0.363m/s

Q =AAx>A = [;r(.20 m)74](0.363 m/s)=1.14 x10~2m

3

/s

^ +zYo 2g

= ^ B + Z B +

r0 2g

Le t z A- z B=

X

Le t y =Distance fromB to^ +zYo 2g

= ^ B + Z B +

r0 2gsurfaceof Mercury.

PA -PB+ *A

2 2

_ _ ^ B - ^ A

AA^ B ^ A - T ^ A

B

^A

2

= UA 4 Y

Yo+ *A B -

2 g

AA^ B ^ A - T ^ A

B

^A 4 Y

Manometer: pB + yoy + ymh-yoh-y0y-y


87/298

6.81+ z A +

2g y 2g

PB^PA+Y2 g

u . = = ; = 20.37 ft/sA A A ; r ( 6 i n )

2 /4 ft2

/ \ 2

= 4 u A =81.49 ft/s

?B= 60.0 psig + (0.6 7)62.4 lb

ft3(20.37 2 - 8 1 .4 9 2 ) f t 2 / s 2

2(32.2 ft/s2)

l f t 2

144 in 231.94 psig

6.82

^ + zZA 2g r 2g

^B = AA A 0.08840 ft

2

n n o n

= uA - = 3.789u AA B

A 0.02333 ft2 A

PA ~PB+ ZA-

2 2

Z - U B - ^ A = ( 3 . 7 8 9 u A )2 - u 2 = 1 3 . 3 6 u A

Y+ ZA- B -

2g -ZB == -24 in

Manometer: pB +y(24in) + y 0 (6 in) +yw (8 in ) - y 0(8 in) - y c (6 in) =pAPA-PB = y 0 (16 in) + y w(8 in)

P a ~ p = 16i n + ^ ( 8 i n ) = 16in + 6 2 A l b , ( g j n ) = 25.08 inr0 r 55.0 i b / f t

3

i n 25.08 in - 24.0 in = 13 .3 6u 2/2g

^ = ^ | 2 g ( 1 . 0 8 i n ) = ^ |2(32.2f t/s2 ) ( 1.08in) = ^ ^

13.36

e =^ A w A = 0.08840 ft2

x 9

- > - ~3

13.36(12 in/ft)

= 0.0582 ft7s



88/298

6.83 Plot show n below. Data computed as follows.Pt. 1: Ta nk surface - Pt. 2: Outs ide nozzle - Ref. level at D.

2

' " EL = 0 See Probl em 6.72 fordata.2g ' r

Pt. 1: z, = 10.0 m; - - = 0;2g

vi (3.502 m/s ) zPt. A: z A = 10.0 m;

2 g 2(9.81 m/ s2

)

+0.625 m :2 2

VE _ Uc2

2g 2g 2g

7

-5.27 k N / m 2

8.437 k N / m 3= -0.625 m =

7

Pt. B: z B = 13.0 m; =0.625 m2g

7

-30.58=-3.625 m

8.437

Pt. C: Same as Pt. A

P t . D : z D = 0; ^ =0.625m; ^ = =9.375m2 g x 8.437

P t . 2 : z 2 = 0 ; i = ^ i m = 1 0 . 0 m ; ^ = 02g 2(9.81) r

Tota head T - Y

10.0 m - z,

Reference

2ff

2g

R1 C O Pt2

74 C h a p t e r 6


89/298

6.84 See Prob. 6.73 for data. L et z A = z B = 0

. pA 50.01b 144 i n2 f t 3 c t a .

P t . A : = x x = 115.4fty i n 2 ft2 62.41b

v\ _ (0 .25u B )2_ [(0.25)(35.6ft/s)]2 _

2g 2g 2(32.2 f t /s 2 ) 1.23ft

-Totalhead = 116.6 ft

P , . B : f i L = ( ^ X 1 4 4 ) = 9 ^ f t

Y 62.4

o _ (35.6) 2

2g 2(32.2)=19.7 ft

- T o t a lhead = 116.6 ft

TotalHead

" 2_ - l a -

Y Y

(

1 Referencelevei(Ptoecenterln e)



90/298


91/298

6.86 W i t h Mercury at wal lof throat, h/2 = 0.30 m;h = 0.60 m

.2 . ,2

-+ z, + = +zt2s rw 2g

u, -v.-~L= u.A

r D ^ 2 f11

25.

P - p t _ u ,2 - u 2 _(9o)

2-l2 80

K 2g 2g 2g

Manometer: pl +yw(DJ2) +yw(0.60 m) - yff l(0.60 m) - yw(D2) = pt

P\-Pt _ rm ( 0 . 6 0 m )- - 0.60 m = 13.54(0.60) - 0.60 = 7.52 m

_ feOS^n) 2(9.81 W . 5 2 m ) _

V 80 \ s 2 (80)

Q =Aix>t = 7i(0.075 m)2 /4 x 1.36 m/s = 6.00 x 10" 3 m3/s



92/298

6.87 + Z i + i ^ = l + Z 2 + - : 0 =J2g(zl -z2) =Jgh = Veloci ty of jetY 2g Y

2g

(ft) u(ft/s)

10 25.4

8 22.7

6 19.7

4 16.1

2 11.4

1.5 9.83

1.0 8.02

0.5 5.67

0 0

25

20

o


93/298

6.94 9.50 m=h +p/y = 1.50m + pdy9 81k N

Pt =y[9.50 - 1.50] = " 3 x 8.0 m =78.48kPam

Flow due to afalling head

6 . 95 t 2 - t i =2 i A ,

r l ^j ) ( h l l 2 - h l n ) = 2 ( 4 Q Q ) (2 .68 1 / 2 - 0 ) =296s[4min,56 s]

A, =7r(3.00m) 2 /4 = 7.07m 2 ;Aj=TT(.15 m) 2 /4 =0.0177m 2

AJAj =7.07/0.0177 =400

6.96 AJAj =(D/Djf =(300/20) 2= 225

h _ t = 2 ( 2 2 5 ) (,Q55 1/ 2 - 0 ) =23.8s

^ 2 ( 9 l l ) V '

6.97 4,-= (DJDjf = (12/0.50) 2= 576

t2 - h = , 2 ( 5 7 6 ) (15.0 ft,/2 - 0) =556s(9min, 16 s)

^2(32.2 ft/s 2) V 7

6.98 ^//4 7-= (PJDjf =(22.0/0.50)2= 1936

g= (32.2 ft/s2)(12 in/ft)= 386 in/s 2

_ 2 ( 4 - ^ ) , , 2 ( 1 9 3 6 ) = , 1 8 5 , / 2 . n v= ^ s ^ 5 9

V 2 g V 7 7 2 (386in/s 2 ) Vfe - fi=

6.99 4A47-= (A/ A)2 = (2-25 m/0.05 m ) 2=2025

2(2025) , m U 2t2 - tx =

^2(9.81)

(2 .68 1 / 2 - 1 . 1 8 " 2 ) =504s (8min,24 s)

6.100 AJAj = (A/ A) 2=(1-25 m/0.025m ) 2= 2500

t 2 _ f l = 2 ( 2 5 Q Q > ( l .38 1 / 2 -1 .155 1 / 2 ) =113s (1 min,53 s)

72(9.81)V ;

6.101 SeeProb.6.98: g= 386 in/s

\ 2

2

2

=,44007 7 l 0.625in i

2 - r, =2(14400)

72(386)

( 3 8 , / 2 - 2 5 . 5 1 / 2 ) = 1155s(19min,15 s)

6.102 At/Aj = (Dt/Dj)2 =

r 46.5 ft V

8.75in (l ft/12 in )J=4067

2 _ = 2 ( 4 Q 6 7 > (23 .0172-2.0172) =3427s (57min,7 s)72(32.2 f t /s 2 )



94/298

6.103 See Prob. 6.97.

= 5 . 0 1 b f t 144 m = U 5 4 f t h = 15.0 ft + 11.54 ft = 26.54 fty i n 2 62.41b ft 2

h2 = 11.54 ft= -Y

2 ( 5 7 6 ) (26.54 1 / 2 - 1 1 . 5 4 1 / 2 ) = 252 s(4 min, 12 s)72(32.2)v '

6.104 See Prob . 6.1 01.

p 2.81b f t 3 144i n 2 . . 12 in __ . ._= = 6.46 ft x = 77.5 i nY i n 262.4 lb ft 2 ft

A,= 38 in + 77.5 in = 115.5 in;h2 =25.5 in + 77.5 in = 100 in

2(14400)

72(386)

6.105 See Prob. 6.96. .2

t2-U = ( l l 5 . 51 / 2 - 1 0 0 1 / 2 ) = 774 s(12 mi n, 54 s)

p 20 k N m 2

= = 2.039 mY m 2 9.81kN

hx =0.055m + 2.039 m = 2.094 m;h2 = 2.039 m

2 - f , =^ S L ( 2 . 0 9 4l / 2 - 2 . 0 3 9 l / 2 ) = 1 .

72(9.81)V '.94 s

6.106 See Prob . 6.100.

/ ? _ 3 5 k N m 3= 3.57m

Y m 2 9.81 k N

A, = 1.38 + 3.57 = 4.95 m;h2

= 1.155 m + 3.57 m = 4.722 mh - u = j g ^ l ( 4 . 9 5 ^ - 4 . 7 2 2 '

/ 2 ) =57.8s

80 C h a p t e r 6


95/298

CHAPTER SEVEN

GENERAL ENERGY EQUATION

7.1

7.2

+ z,Yo 2g Yo 2g

; Zj= z 2 and vx = Oj

P l - p 2 (74.6 -62.2 ) lb f t3 144 i n 2

Yo m

^ + z A + i - / ^2 g 7,

(0.83)(62.4 lb) ft234.5 lb ft/lb

7

PB=PA+Y,

pB = 60 psi g

2g

2 2U A ~On

2g

= oA

oB =1 0 ft/s

62.4 lbf t 3

3 0 f t +

( l ^ C ^ _ 2 5 f t

2(32.2 ft/s 2 )1 f t

2

144 i n 2

' v 2

= 40 ft/s

52.1 psig

7.3

Yw

z ] + ^ - h L ^2g 2g

Pt. 1 at surface of water. 0\= 0Pt. 2 in stream outside nozzle. p2 = 0

7.4

o2= 2gY

(z, -z2)-hL2(9.81 m) 140 k N m 3

m 2 9.81 kN+ 2.4 m - 2.0 m

= 17.0 m/s\2//1 . . 1-7 A _ / - - - . . 1 ( 1 - 1 _ 3 /g =A2u2 =40.05 m)74 x 17.0 m/s = 3.33 x 10

m/s

Pt vf r Pl 2zx+ -hL=

i-L +z2+ 2g ' yw 2g

Yw zg 7,

P\=p2 = 0and Oi = 0

hL = (z! - z 2 ) - r2 -2g

hL= 10 m(4.56 m/s) 2

2(9.81 m / s 1 )= 8.94 m =

8.94 N m

N

Pt. 1 at water surface.Pt. 2 in stream outside pipe.

Q 0.085 m 3/s

A2 1 . 8 6 4 x l 0 ^2 m 2

02= 4.56 m/s

7.5

EA 2gJ 4 L2g

2 2

o -ov2g

o.

Q _ 0.20 flVs

A. 0.02333 ft 2 = 8.57 ft/s

e L = ^ 2 o _ = 2 2 6 / s

4 , 0.0884

G e n e r a l E n e r g y E q u a t i o n81


96/298

Manometer:pA +y w (10 in)- y m (14 in)- y w (44 in) =pB

PA-PB _ 7 m ( 1 4 i n )+ ( 3 4 i n ) _ 1 3 . 5 4 ^ ( 1 4 i n )+ ( 3 4 i n ) = 2 2 3 6 i n x J j L

= 18.6ft

L - 1 M ft+ (-4.0ft)+ ~ ni2g

^ | + 1 0 f t - 1 . 5 4 f t - 6 f t ] - l A2

ft3 144 in 2

84 Chap t e r 7


99/298

b) pc 21.8 psig (same as Prob. 7.14)

c) hA = (z D - Z A ) + hh+hLo = 30 ft + 6 ft + 12 ft = 48 ft

d) P = ^ Y w 2 = (48 ft)(62.4 lb/ft3)(20.0 ft3/s)/550 = 10.9 hp

7.16 a)

b)

+ Z.+-

Yo 2g

hA^ ^ + {z2-z,) + hL =Yo

Yo ' 2

S825 k N m

m 2 (0.85)(9.81kN)

Pt .1at lower tank surface. />, = 0

Pt. 2 atuppertank surface. o, = o2

+ 14.5 m +4 .2 m = 117.6 m

^ = A ^ o e = (117.6m)(0.85)

= 13.73 kWy

P

9.81kN

m

840L /min ( l m 3/s ) _ 13.73 k Nm

60000 L/min

l - i r Pi vi+ z, +-

7 0 " 2g

P3= 7 0

_ (0.85)(9.81 kN )

7 02

2gPt. 3 at pump inlet.

m-3.0 m -

(4.53 m/s) 2 1.4 N - m

2(9.81m/s 2 )

= 6 _ 8 4 0 L/min ( lm 3 /s) 10 3 4 60000 L /min 3.090x 10~3 m 2

N

:4.53 m/s

= -45.4 kPa

7.17 - ^ + z, +A - A , = ^ ~ 2

>7 2 g >7+ z 2 +

2 g

Pt.1at lower pump surface. />, = 0

Pt. 2 outside pipe at cutter. vx = 0

A= ( z2

- z i ) + + AL

=1 .2 5 m + (2

-9 1 m / s

) +3 .0 m = 4.68m2g 2(9.81 m/ s 2 )

= 6 _ 60L/ min ( l m7s) 1

^ ^ 60000 L /min X 3.437x l 0 ~ 4 m 22.91 m/s

PA ~ hAjjQ (4.68m)(0.95)/ 9 . 8 1 k N V 60 m 3 A

m 60000s

0.0436kN-m 103 N 43 .6 N- mx=

s k N s= 43.6W

7.18 Tub Vol ume = (xD2/4)(d) = [ TI (0.525m) 2/4](0.25 m) = 0.0541 m 3

Q=V/t = 0.0541 m790 s = 6.013 x 10~4

m3

/sn , Q 6 .013xl0"

4 m 3 / s . .Outlet 2= = ; = 2.36 m/s

4, ; r ( .018m)2 /4

U 2 , , p2 v1

-+ z, + - 1 - +A, - A, = + z, +2g 7 2g

Pt. 1 at tub surface.Pt. 2 in outlet stream.

P\ = Pz = 0; T>, = 0

A= ( z 2 - z , ) + ^ + A= (1.00- 0.375)m + (2 - 3 6 m / s ) 2

W U 2g 2(9.81 m/s 2)+ 0.22m =1.13m

G e n e r a l E n e r g y E q u a t i o n


100/298

7.19 Weight Flo wRate= W= w/t = 556 lb/10 s = 55.6 lb/s

_ W 55.61b/s n 8 Q 1 f t 3 / Q 0.891 ft3/s 1 f t 0 1 f t /

Q= = - = 0 . 8 9 1 ft/s: u A = - = ^ = ; = 10.21 ft/sX, 62.4 lb/f t 3 A ; r (4 /12 )

2 / 4 f t 2

= 10.21 ft/s/ 4 V

v3y= 18.15 ft/s at outlet ofupperpipe (Pt. B)

^ + z A A + A r A l = ^ . + z B A ; A = 0 a n d ^ = 0

^ = ( z B _ Z A ) + ^ Z _ ^ = 2 0 f t + O 8 . 1 52 - 1 0 . 2 1 2 ) f t 2 / s 2 -2.01bft 3 144m 2

2g y w 2(32.2 ft/s2) in 2(62.41b) ft

^ = 20 ft + 3.50 ft + 4.62 ft = 28.11 ft

PA = hAW= 28.11 ft(55.6 lb/s) = l* >3 fM b/ s( lh p) = 1 M ^550 ft-lb/s

7.20 g = 9 - 1 g a l / m i n ( l f t 3 / s > = 0.0203 ft3/s449 gal/min

PA = hAy0Q = (257 ft)(0.90)(62.4 lb /ft3)(0.0203 ft3/s) =292.5 ft-lb/s/550 = 0.532 hp

e M = ^ ^ * L =0.626 = 62.6%P 0.850 hp

K 1.0 L l m7.21 0 = - = ^ x - ^ - = 2 . 5 0 x 10" 5m 3/s s~\ -

r 40s 10 3 L i ' C > ' '

Pi = YmA= h x (-0.15 m) = -20.0 kPa3 AV - U . U I I I ; v . v is j r a 0 , - 0m

1 r\ > Z, 2 ~

rs 2 Y, 2g

.2

X g 6.67 kN / m3

^ = hAjgQ = (7.50 m)(6.67 kN/m 3)(2.50 x 10~5m 3/s)(10 3 N /k N) = 1.25 N-m/s

P 1 25WP = L= 1 !L = 2 .0 8^

e 0.60

x x . . stroke -(5.0i n ) 2 ( l f t 2 ) 20 in ft n ,7.22 a) Q = , C V | x = - i - -- 'x = 0.01515 ft

3/stime 4(144in 2 ) 15 s 12 in

b) Poyi= = 1 1 Q Q Q l b , = (560 lb / in 2 ) (144 in 2 / f t 2 ) = 80672 lb/f t 2

Acyi ; r (5 .0)2 / (4) in 2

86 C h a p t e r 7


101/298

3 ) A + b + + i

Yo 2g Yo 2g

PB=PC+YO ( z c - z B ) +2 2

T _ " D2g

Pt. A at tank surface.

Pt. B at pump outlet.

Pt.Cincylinder.

20 i n ftoc = x15s 12 in s

(62.41b)

n , , , , f t Q 0.01515 ft3/s=0.1111: u n =-^ - = : = 15.52 ft/s

/> B = 560 psig + 0.90-

= 576.1 psig

f t 3

A B 0.000976f t2

i o f t + ( n i l ' - 1 5 - 5 2 3 > + 3 5 . o f t

2(32.2)

l f t 2

144in 2

o* Pt. D at pump inl et.d) EA +z + ^ - h , = ^ - + z D + ^

2g y0 2g / > A = 0 , u A = 07.

/>D = YO2g

Vo= B = 15.52 ft/s

(0.90)(62.41b)PD

f t 3

P D = -4.98 psig

-5.0 ft-(15.52 f t /s ) 2

2(32.2 ft/s 2)-11.5 ft

l f t 2

144 in 2

2 2

e) A + Z a + ^ + A a - A - A = c + Z c + i ^ : ^ = 0 , ^ = 0

X 0 2g

A = + (z - z ) + + /! + AA X0

A 2g h 0 ft 2(0.9)(62.41b)

806721b-ft3 + 1 5 f t + ( . H l l )2

2g

+ 11.5 ft + 35 ft

7.23

h = 1498 ft

PA = hAyAQ == (1498 ft)(0.90)(62.4 1 ^ X 0 . 0 1 5 1 5 ftVs) = 1 2 7 5 ^ l b / s = 2 .32 hp

2g-+ z B +

2g

^ ^ - . ( z A - z l > ) + ^ - ^2g

550

4 , , c 1.772 x l 0 ~3 m 2

U a = U B - -b - = 1.5 -

A 3.835x l 0 ~ 4m 2

u A = 6.93 m/s; assumehL = 0

. ffj.-34)0tf)lft.- ( 6 . 9 3 ' - l . S t f > . W = 3 9 ( ) m

0.90(9810N / m 3 ) 2(9.81 m/ s 2 )

P= hRyQ = /IHYOABOB = (390 m)(0.90)(9.81 kN/m3)(1.772 x 10~3m 2)(1.5 m/s)

= 9.15 k W

G e n e r a l E n e r g y E q u a t i o n 87


102/298

7.24

H R = FAZPR. + (z A - z B ) - ^ *

Yw 2g

g = 3400gal/mm(lft3 /s) = ? 5 ? ft3/s

449 gal/min

(A7.57 ft 3/s

0.7771f t 2

_ 6 _ 7.57 _

= 9.74 ft/s

o, = - - = = 2.71 ft/s4 2.792

_ o n ^ A _2 .0 (9 .74 ) 2_

[ 2 1 . 4 - ( - 5 ) ] l b - f t 3 ( 1 4 4 i n 2 ) ^ o a (9.742 - 2 . 7 4 2 )

in 2 (62 .41b) l f t 2+ 3ft

2(32.2)

h, =2.0 2g 2(32.2)

ft -2 .9 5 ft = 62.3ft

= 2.95 ft

PR = hRJwQ = (62.3 ft)(62.4 lb/f t3)(7.57 ft3/s) =

29454 ft-lb/s

(550 ft-lb/s)/hp53.6 hp

7.25 ^ + z x + ^ - h R - h L = ^ + z2+^-Yo 2g yo 2g

Pt . 1at oilsurface. px = 0 and vx = 0

Pt. 2 in outlet stream. p2=0

hR = (zx -z2)- -hL= 10 m - 0.638m - 1.40 m = 7 .9 6m2g

_ e 0.25 m3/s , ol 3.54 2 m 2 / s 2

= 3.54 m/s: ~~ 0.638 m4 >r(0.30m)74 2g 2(9.8 l) m/ s 2

P = = (7.96 m)(0.86) (9.81 kN/m 3)(0.25 m 3/3) = 16.79 kN-m/s = 16.79 kW

P=PR-eM= 16.79 k W x 0.75 = 12.60 kW

7.26

Yf

1 = ^

hA =

2g

Pi -Pi

Yf

Yf

+ (z 2 - z , ) -2 2

U2 - ,

2g

2g

+ hL

Q=40 gal/min/449 =0.0891 ft 3/s

Q 0.0891 ft 3/s

4 0.05132f t 2

Q _ 0.0891

= 1.74 ft/s

4 0.02333 = 3.82 ft/s

5 0 . 0 - ( - 2 . 3 0 ) l l b - f t 3 1 4 4 i n 2 (3.822 - 1 .74 2 ) f t 2

hA= i - + 25f t + - ^ . \ + 3.4ft = 154.1 ftin 2(60.01b)ft 2 2(32.2 ft/s 2 )s 2

PA = hAyjQ = (154.1 ft)(60.0 lb/ft3)(0.0891 f t 3 / s) ( l hp/550 lb-ft/s) = 1.50 hp

7.27 P, =PA/eM= 1.50 hp/0.75= 2.00 hp

P\ i i/ . 8 2 g 7 2g

Pt . 1atwatersurface. pl~0,ul=0

Pt. 2 in outlet stream. p2 = 0

/i = (z, - z 2 ) - ^ = 4 . 0 m - (5.14 m/s)2/2(9.81 m/s2) = 2.65 m

02

2g

Q 600L/min( lm 3 /s)

4 60000L /m in 1.945 x 10~3 m 2= 5.14 m/s

88 C h a p t e r 7


103/298

7.2977 2g x K 2g

PB= 7K (ZI~^)r-~hL 2g

Pt . 1at tank surface. />, = O, u, = O

_ Q _ 0.060 m 3/s

AB 4.768x l 0 ~3 m 2

12.58 m/s

/> B = (0.823)(9.81 kN/m3

) 17.0 m -(12.58m/s)2

2(9.81 m/s 2 )- 4 . 6 0 m 35.0 kN/m

2

= 35.0 kP a

7.301

+ Z l + . n _ _ A R - A L = - + z 2 + ~2

2g 2g

Pt . 1at reservoirsurface. px = 0,ux = 0

Pt. 2 in outlet stream. p2=0

hL = (Z ] - z 2 ) - -r2- - A*

2g

O 1000 gal/min l ft 3 / st>2= = 5 x = 6.41 ft/s

A2 0.3472f t2 449 gal/ min

h, = 165 ft

h ~ P" -nR

(6.41 ft/s)2

2(32.2 ft/s 2)

37.0 hp(550ft- lb/s)

rw g 1hp(62.4lb/f t3)(2.227 ft 3/s)

146.4 ft = 17.9 ft = 17.9 ftlb/lb

= 146.4 ft

7.31 ^ + z , + ^ - A , = ^ + z 1 +2g 2gr

w

g a rw

h = (z}-zA) = P + ^ + hL

rw 2g

Q = 1500 gal/min x 1f t3

/s/449 gal/min = 3.341 ft3

/sQ _ 3.341 ft 3/s _

Pt . 1at reservoirsurface. p = 0,m = 0

oA =AK 0.5479f t

2

= 6.10 ft/s

A = 5.01b(ft3 X144m 2 ) + (6-lOft/s) + ft = ^

in 2 (62.41b)(ft 2 ) 2(32.2 f t /s 2)

7.32 ^ + z . + ^ - + A - A , = ^ B - + z + ^ D -

A,, =

A ' ' ' " A "X N B

2g rw 2g

Q _ 3.341 ft 3/s

4 0.3472f t 2

EAZJIA. + ( z z ) + + f = ( 8 5 - 5 ) l b ( f t3 ) 1 4 4 m 2 +

rw 2g in 2(62.41b)ft2

+ (9.622 - 6 . 1 0 2 ) f t 2 / s 2 + 2 8 f t = m 5 f t

= 9.62 ft/s

25 ft

2(32.2 f t /s 2 )

^ = = (238.5 ft)(62.4 lb/ft 3)(3.341 ft3/s)/550 = 90.4 hp

G e n e r a l E n e r g y E q u a t i o n


104/298


105/298

7.36A o,

+ Z , + -= -

r 2 8

P2 = y

y 2gPi =O, o, =O

2g

Q 0.390 ft 3/sUb= = r- = 7.59 it /s

4 0.05132 f t 2

J22 _ (7.59) 2= 0.896 ft

2g 2(32.2)

p 2 = 58 .0 3[ -4 .0 ft -0 .8 96 ft-2.80 ft]f t 3

l f t 2

144 in 2-3.10 psig

7.37o:.3 .

y 2g

P3 = y

y 2g

(Z

l - Z 3 ) - ^ + k

A

2g

: pl = 0,oi=0

lb,

=5 8. 03 -^ -[ -4 .0 - 2.13 - 2.80 + 552.51

'

f t

.f t3 L J

1 4 4 i n2

Pi = 219.1 psig

7.38y

z 3 + ^ - h L = A2g y 2g

p4 = j P 3 _ i =2 19 .1 psig - 58.03 ^ (28.5 ft) 207.6 psig

7.39 A + Z 5 + Ay 2g y

o:6_.

lb,/>5 = r[(z6-z5) + h J = 58. 03[ -1. 0 ft + 3.50 ft ]

l f t 2

144in 21.01 psig

7.40 For Q= 175 gal/min, Fig. 6.2 suggestsusing either a 2 Vi-in or 3-inSchedule 40 steel pipe for

the suction line.The given 3-in pipe is satisfactory. However, noting that the pressure at the

inlet to the pump is -3.10 psig, a larger pipe may be warranted todecreasethe energy Iosses

in the suction line and increase the pump inletpressure.See Chapters 9-13, especially

Section 13.12 on net positive suctionhead(NPSH).

Fig. 6.2 suggestseither a 2-in or 2 Vi-inpipe for the discharge line.The given 2 Vi-in pipe size

is satisfactory.



106/298

7.41 P+ Z i + ^ h L = l . + Z 2 + ^ : ^ = oy 2g y 2g

P = 7

2 2

2g

Y= 0.76(62.4 lb/f t 3 ) = 47.42 lb/f t 3

z2 - zi = -22 i n ( l ft/12 in ) = -1.8 33 ft

f t 3

O

8s 7.48 gal

_ Q

= 0.668 ft 3/s

0.668ft 3/s 144 i n 2

x : = 30.64 ft/s4 ^(2.0i n ) 2 / 4 ft2

0.668f t 3 / s 144 i n 2 _ 0 A /x = 0.378 ft/s

A, 7z-(18.0in) 2/4

Pi = 47.42Jb

ft:

/>, = 5.76 psig

(30.642 - 0 . 3 7 8 2 ) f t 2 / s 2

-1.833 ft+ r + 4.75 ft

2(32.2 ft/s2

)

l f t 2

144 i n2

7.42 + z, +- 1

r 2gh, +hA = ^

2 - + z, +

hA = ^ + (z2-z) + hL--r

r301b

v\ Pt. 1at creek surface. px = 0, L>,=0

Pt . 2 at tank surface. u, = 02g

f t 3 144 i n 2

i n 2 62.41b ft2+ 220 ft +15.5 ft = 304.7 ft

r 62.41b 40gal/min (l ft 3 /s)hpP


107/298

.2 . . 2

7.44 + Z i + A _ ^ = A + Z 2 + A . : Z l = Z 2Yo 2S Y 2g

R +

Yo 2S

Manometer: px + y0y +yQ(38.5 i n) - ym(38.5 in) - =p2Pi -Pi = ym(38.5 in) - y e(38.5 in)

Px-Pi = j ^ L ( 3 8 , 5 i n ) - 3 8 . 5 i n= U M y (38.5 in ) - 38.5 i nYo Yo -9Y

540.7 i n x - ^ _ = 45.06 ftYo 1 2 i n

Q 135 gal/m in l f tV s 0.3007 f t 3 / su = = e x = - = 24.50 ft/s

4 0.01227 f t 2 449gal/m in 0.01227ft 2

Q 0.3007 ft 3/s^ = J L = : r = 10.21 fiVs

4 0.02944 ft2

hR = 45.06 ft + v n , 2

= 5 2 J 6 f t(24.52

- 10 .212

) f t2

/ s2

2(32.2 ft/s 2 )

PR =Ayg = (52.76 ft)(0.90)(62.4 lb/ft 3)(0.3007 ft3/s) = 891.0 ft-lb/s

PR = 891.0 ft-lb/s x ^ = 1.62 hp550 ft-lb/s

7.45 P0 =PRxeM=\ .62 hp x 0.78 = 1.26 hp



108/298

CHAPTER EIGHT

REYNOLDS NUMBER, LAMINAR FLOW,

TURBULENT FLOW, AND ENERGY LOSSESDUE TOFRICT ION

U v = Q =J ^ ^ x l4 ! ^ ! =2 .2 9 ft/s;D =4.0 i n ( l ft/12 in) =0.333 ft

A TT(4.0i n ) 2 / 4 f t 2

oDp. (2.29)(0.333)(1.26)(1.94) _

/ / 7.5 x l O " 3

L from Ap p. D

5.2 LetNR =4000 = ZJ>/V: v = 4.38 x 10~6 ft 2/sApp. A;D = (2/12)ft

= = j V = 4000(4 .38x10^) = ( U Q s ft x 0.3048m = Q ^ m

D 2/ 12 s ft s

5.3 Le t = 2000 = vDplp

j V , = 2000(4.0x10 -)

(0.10)(0.895)(1000)

Q = i> = ^ C 0 - 1 0 " 1 ) , x 0.894 m/s = 7.02 x 10~3 m 3/s

.4 u= Q/A = 0 , 2 5 / S 9 = 10.72 ft/s;D =0.1723 ft0.02333 f t 2

a) NR = = < ! ^ ^ > = , . 5 3 x ,0* (v from App. A)v 1.21x10

u D / ^ (10.72X0.1723)(1.53) t n 5

/ / 6.60 x lO"b ) A ^ = ^ = ^ ^ ^ 4 , 2 8 x l o 5 ( p ^ f r o m A p p . B )

, A r u Z V (10.72)(0.1723)(1.86) , . . ,c) NR = ^ = i - J - '- = 253 (/>,// from App. B)

/ / 1.36 x 10 XT vDp (10.72X0.1723)(0.87)(1.94) , 4 , ,

d) Afo= ^ = ^ '.^ '- = 3.28 x 10

4

(//f rom App. D)v 9.5x10

_ oDp_QDP_ QDp 4Qp 4Qp _ 4Q1\R ; . Dmm~~ ~jU AjU 71D TTjLlD ftMNR TZNRV

Q = 4.0 L /min x = 6.667 x 10~5m 3/s: LetNR =200060000 L/min

94 C h a p t e r 8


109/298

, _ 4(6.667 x l O " 5 ) 4.244x10" 8 4.244 x lO" 8a) - D m i n =

L = T~ = 0.0647 m = 64.7 m m;r(2000)(v) v 6.56x10~ 7

3-in Type K copper tubeD - 73.8 mm

b ) > m i n = 4 2 4 4 * 1 Q " 8 ( f8 0 ) = 0.101 m;5-in tube,D = 122 mm

2.87 x 10~4

c) =4 , 2 4 4 *1 0 8 (J 9 Q ) = 0.0186 m; - - i n tube,D = 18.9 mm1.8 x 10~3 4

d) = 4 - 2 4 4 * 1 0 8 f 6 ) = 3.59 x 10~4m ; - - i n t ube,D =4.57 m m1.07x10"' 8

Smallest listed

8.6 NR = M= = ( 2 - 9 7 ) ( Q - 7 7 4

9 ) ( 8 9 0 ) - 4.12 x 10- Pa-s

fi NR 5 x l 04

Q 8.50 L /m in l m 3 / s _ .v - = r x =2.9 7 m/s

A 4.768x l 0 " 3 m 2 60000 L/minFrom App. D ,o i lmust be heated to 10 0C for SAE 10oil .

8.7 Auto.Hydraulic Oi l M d i u m HydraulicOi l

i ,r oD (10X0.4011) e 1 f t 4 i u 10(0.4011) r = = - x =3.0 6 m/s

A 1.772 x 10~3 m 2 60000 L/min

g 9 u ^ = (0.899)(0.0243)(860)3.95x10

O 25L /m in 1m 3/s . ___ ,u= = x =0. 899 m/s

A 4 . 6 3 6 x l O ^ m 2 60000 L/min

8.10 Afr = ^ = ( 1 - 7 8 ) ( - Q 1 3 74 ) = 6.62 x 10 4 Turbulent

v 3.60 x l 0 ' 7

Q 15.0L /m in 1 m3/s , . = = x = 1.78 m/s

A 1.407 x 10" 4 m 60000 L/min

R E Y N O L D S N U M B E R , L A M I NA R F L O W , T U R B U L E N T F L O W ,

A N D E N E R G Y L O S S E S D U ET O F R I C T I O N 95


110/298

^ 8 . 5 9 X ^ 5 6 3 ) = 9 5 9 > l o 5

v 1.40 x l O - 5

Q 16.5 ft 3/s 0 c n A .u = = = 8.59 ft/s

A 1.920 f t 2

^ G = 0 ^ x _ ^ x J h r _ x L _ = o.732 ft/s^ hr 7.48 gal 3600 s 2.029x l O - 5 f t 2

jO g _ (0.732)(0.00508)(0.88)(1.94) _

/ / 6 . 2 x l 0 - 3

A r (0.732)(0.00508)(0 .88)(1 .94) __ . _ .3.13 NR = - = A = 33.4 Laminar

H 1.90 x l O - 4

Not e: sg o foi l may be slightly lower at 160F.

U 4 N r = P: 1 ) = ^ = ( 4 0 0 0 ) ( 4 . 0 1 X 1 0 - )

H Dp (0.2423)(1.56)

Q = A v =4.609 x 10~2 f t 2x0.424 ft/s = 1.96 x 10~ 2 ft 3/s

, , .


111/298

Changing from laminar flow, through criticalzone, into turbulentf low couldcauseerraticperformance. Also, L>= 14.65 ft/s is quitehigh, causing largepressuredrops through the

system.

n ~, > Q 500gal/min l f t3 / s n , . . . . 2 r CT ,

8.21 A = = 5 x = 0.1114 ft 2=> 5-in Sch. 40 pipe

o 10.0 ft/s 449 gal/minA = 0.1390 ft 2, > = 0.4026 ft

g (500/499)f t3 /s O A i r .

Actual L > = = - n = 8.01 ft/sA 0.1390 ft 2

^ = ( 8 . 0 1 ) ( 0 . 4 0 2 6 ) ( 2 . 1 3 ) = 2 1 2 > < 1 0 4j. 3.38x1o- 4

8.22 ^ . j ^ . 2 0 0 0 ( 1 . 2 1 x 1 0 ^ ^ ) .

D 0.0621 ft

ForNR = 4000, = 2(0.3897 ft/s) = 0.7794 ft/s

QI=AUY = (3.027 x 10"3 ft 2)(0.3897 ft/s)

-3 A3/ 449 gal/ min= 1.180 x H T ftVsx1 ft3/s

Qi =0.530gal/minLowerLimitQi = 2Qt = 1.060 gal/minUpper Limit

8.23 (See Prob. 8.22)

_ A V ^ ( 2 0 0 0 X 3 . 8 4 x 1 0 - ) _ _

Z> 0.0621

g , = = (3.027 x 10"3 ft 2)(0.1237 ft/s) x 4 4 9 ^ m " 1 = 0.1681 gal /min

g 2 = 2gi = 0.3362 gal/min

8.24 1.30 es x l ^ x l Q - f t V s = ^ x ^ 5 ft2/g1es

Q = 45 gal /min x = 0.1002 f t3/s449 gal/min

Q 0.1002 ft 3/sv - = 14.65 ft/s

A 6.842 x l O " 3 f t 2

^ 1 4 . 6 5 X 0 . 0 9 3 3 ) ^ 3 ^

v 1.40x1o-5

8.25 v = 17.0 es x 1 0 6 m / S = 1.7 x 10" 5 m2/s1 es

Q 215 L /m in 1 m3/s , ,u= = - x =7. 142 m/s

A 5.017 x l O " 4m 2 60000 L/min g _ p . l 4 2 X W 3 ) _ l i 0 6 x l a ,

1.70x10 - 5

R E Y N O L D S N U M B E R , L A M I N A R F L O W , T U R B U L E N T F L O W ,

A ND E N E R G Y L O S S E S D U E T O FRICTION 97


112/298

8.26 v = 1.20 es x 1 0 m / S = 1.20 x 10"6 m2/sles

Q 200 L /m in l m 3 /s .v = = x = 8.69 m/s

^ 3.835x l O ^ m 2 60000 L/min

^ (8 .69) (0 .0221) ,

v 1.20 x 1 o- 6

8.27 i +z = i +z 2 + - i : 0 , = ^rD 2 g 7D 2 g

/?i - ^ 2 =To[z 2 - z , + A ]

i r vDp (0.64)(0.0243)( 0.86)(1000) _ _ . . . 64AT = - = ^ J y , A = 787 (L am in ar ); /=

ju 1.70 x l O - 2 TV,

N r = ^ = V 1 " ^ v " ; " ^ ^ = 5.35 x 10 4 (turbulent)

W i L = 0 . 0 8 1 3 x ^ x < ^ = 4 . 1 9 m > 2 g 0.0243 2(9.81)

Pi ~Pi = (0.86)(9 .82 k N / m 3 ) [-60 m + 4.19 m] = -471 k N / m 2 = -471 kPa

2 2

8.28 A + Z ] +--hL= + z 2 + -?- : ot = uzi=z2: p\-pi = yJiL

Q 12.9L /m in 1 m3/s , c ^ a ,u = -= = - x = 1.528 m/s

A 1.407 x l 0 " 4 m 2 60000 L/min

uD = (1 .528)( 0.0134) _ ^ ^ 1 f t 4v 3.83x10

D/e = 0.0134/1.50 x 10~6= 8933; Th en /= 0.0205

A i = / ^ ^ = ( 0 . 0 2 0 5 ) . - ^ . ( ^ = 8 . 1 9 mD2g 0.0134 2(9.81)

Pi -Pi = JJIL = 9.56 k N / m 3 x 8.19 m = 78.3 k N / m 2 = 78.3 kPa

8.29 LetNK = 2000; /= 64/Ng = 0.032; NR =

t j_ / V j ; ^ (2000X8.3X10-4) _ 2 5 f t J s

Dp (0.3355)(0.895)(1.94)

hL = / - = (0.032) &*?L ft = i .20 ft = 1.20 f t l b / l b^ D2g 0.3355 2(32.2)

98 C h a p t e r 8


113/298

.30 EA + Z + ^ , - H = A + Z + ^ . : U a = U B

Yo 2S Yo 2S

PB=PA + JO[ZA - Z B - h L ]

J V ^ ( 800) (4x l0~ 4 ) = Q 7 1 7 f t / s

(0.2557X0.90X1.94)

> 2g 800 0.2557 2(32 .2)

pB = 50 psig + (0.90)(62.4 lb /f t3 )[-20 f t - 12.5 ft ] l , = 37.3 ps ig

144in

P, o2 , P-> v1 i L v2.31 + z, +-~-hL = + z2 +-

2-: zi = z 2: L>I= v^. px -p2 =ybhL = ybfYb 2 g Yb

2S D2g

Q 20L /min 1 m 3/su = = ; r x =0.7 19 m/s

4.636x l 0 ~ 4 m 2 60000 L/min

y 8.62k N s2

1 03

N 1kg -m/s2

3

p = = x x x = 879 kg /m

g m 3 9.81 m k N N

(0.719X0.0243X879) = 3 8 9 x l Q 4

H 3.95 x l O 4

D/e = 0.0243/4.6 x 10~5 = 528; Then/= 0.027100 (719">2

Pi-p2 =8.62 kN/m3 x 0.027 x x i - ^ >-m = 25.2 kN/m 2 = 25.2 kPa

0.0243 2(9.81)

.32 Fr om Prob. S.3\,px -p2 = yJiL, h-pi -p2lyw

(1035-669)kN/m2 . L v2

hL

= = 37.3 m = /9.81kN/m 3 D2g

f = ^ ^ 2 g = (37.3)(0.03388)(2)(9.81) = Q 0 4 8

Lo2 (30)(4.16) 2

Q 225 L /min 1m3/s A , r ,o= = - x =4 .1 6 m/s

A 9.017 x l O ^ m 2 60000 L/min

vD = (4.16)(0.03388) . ! . 0 8 x : Then ^ = 55 for/= 0.048v 1.3 0x1 0^ ?

e = >/55 = 0.03388/55 = 6.16 x 10 - 4 m


A ND E N E R G Y L O S S E S D U E T O FRICTION


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8.33A + z, +

2g

1 Pl V2 r = + z, + -2g

r 7 ul

n = z, - z2nL +2S

Q 2.50 ft 3/s . - A e : A Iv = M.= = 12.46 ft/sA 0.2006f t 2

Pt. 1 at tank surface. px = 0, vx = 0Pt. 2 in outlet stream. p 2

= 0

D = 0.5054 ftA =0.2006 f t 2

oD = (12.46X0.5054) _ B = ,05054^ . 3 3 6 9 : ^

v 9 .1 5 x 1 0 - e 1.5xHT*

, , L o2 o2 n m 550 (12.4 6)2 (12.46)2

h= f + = 0.0165x x - + - =45 .7f tD 2g 2g 0.5054 2(32.2) 2(32.2)

8.34 Fr om Prob. 8.31 ,px -p2 ~ J J I L = J W f D 2g

Q 15.0 ftVs _ . .v

= = = 8.49 f t /s

A -(1.50ft) 2 /4

uD = (8.49X1.50) _ 9 M x t f : D _ J J ^ m

v 1 .40x10- e 4 x 1 o - 4

, L v2 62.41b A A 1 c o 5280 ft (8. 49)2 f t 2 / s 2 l f t 2 A r .

P\-Pi= Y f = rx 0.0 158 x - x =30 .5 psiyi F2 rwJ D 2 g ft3 1 5 0 f t 2(32.2 ft /s

2) 144in 2 P

1 -fi- /o8.35 Q= 1500 gal /min x = 3.34 ft 3/s

449 gal/min

Q__ 3 .34f t '/s _ , _ ( 6^ 97 / . 0 .577 f

AA 0.5479f t2

2g 2(32.2)

.2 _ 2

a) A + Z + i _ _ /, - A +Z + ^A. Pt . 1 at t ank surf ace. /7j = 0, w, = 0yw

2s s rw 2s

2

z, - z A = A = + + ,

^ g . (6:097X0.835) _ x ^ D =_ 0 8 3 5 ^ - ; , 1 M , ; 5 ,

v 1.21x10- e 1 .5x1o - 4

A = / = (0.0155) x - ^ - x 0.577 ft =0.482 ft > 2 g 0.835

^ 5 . 0 l b - f t 3 144 i n 2 + 0 5 7 7 + Q - 4 8 2 ^ 1 2 . 6 0 f ti n 262.41b f t 2

100 C h a p t e r 8


115/298

7 2g 7 2g

Q 3.34 ft 3/s n e .= = = 9.62 ft/s4 , 0.3472 ft2

* _ ^ + ( z > _ Z i ) + A=A + < k , ( 8 5 - 5 ) I b l f (144 , )

r . 2 g n'(62.41b)ft>

+ ( ^ 2 ' - 6 . 0 9 7 ' ) f t ' / s ' + _

2(32.2 ft/s 2)

^ ( 9 . 6 2 X 0 . 6 6 5 ! ) tf

* u v 1.21x 10-

m = 0-6651 = 4 4 3 4 . Q 1 6

1.5 x 10 - 4

A i = / ^ = ( 0 . 0 1 6 ) x ^ l x < ^ = 8 9 . 9 f ti j D2g 0.6651 2(32.2)

= hAywQ - 300.4 ft x *L x * L = m h pft s 550 ft-lb/s

8.36 n2

+z. + +h. -h, - +z0 + y 2 g r 2g

Pt. 1 at wel l surface (p i = 0 psig).

Pt. 2 at tank surface.

O= 02= 0

hA= +(z2-zi) + HL

rwQ = & x - l A - x

1 ft3/s = 0.0277 ft 3/sh 60 mi n 449 gal/min

Q 0.0277 ft3

/s , ft/

. .u = = , = 4.61 ft/s i n pi peA 0.0060 f t 2

^ , ( 4 . 6 0 ( 0 . 0 8 7 4 ) _ D , _ O 0 8 M = 5 8 3 : / , 0 0 2 7 5

\ 2 1 x l 0 - s 1.5X10 -4

A,. f L !L = ( . 0 2 7 5 ) - ! ^ x < 2 l ft - 14.54 ftD2g 0.0874 2(32 .2)

(401b)f t(144in ' ) + 1 2 0 + 1 4 . 5 4 = 226.8 ftin 2(62.41b)ft 2

PA = ^ 7 2 = ( 2 2 6 . 8 ft)(62.4 lb/ft3)(0.0277 ft3/s)/550 ft-lb/s/hp = 0.713 hp


A ND E N E R G Y L O S S E S D U E T O FRICTION
http://21xl0-/http://21xl0-/http://21xl0-/


116/298

8.37>1 l 7 Pl 2'+ z, + - hr = +z 2 + -

1 -

Pi = y

L>2

Pl

2g 2g

Pt. 1 at tank surface. vx = O

Pt. 2 in outlet stream. p2 = 0

(z2-zl) + -2 - + hL

11.8 ft/s:v2 (11.8)2

= 2.167 ft

2g

Q 75 gal/mi n 1f t 3/s

A, ~ 0.01414 ft2 X 449 gal/min 2g 2(32.2)

^ ^ ( 1 1 . 8 ) ( 0 . 1 3 4 2 ) = 1 3 1 > < 1 Q 5 : D=_0A3^ = 8 9 5 : / = 0.0225

v 1.21x10- s \.5x\0~4

hL = f = (0.0225) 3 0 0 (2.167 ft) = 109.0 ft

D 2g 0.1342

6 2 A l b f-3 ft + 2.167 ft+ 109.0 f t l - l f t 2

f t J ' I 4 4 i n 246.9 psi

8.38+ z, + + h, -h, = + z, + o.

y 2g y 2g

Pt. 1 at tank surface. = 0; vx = 0

Pt. 2 inhose at nozzle.Pt. 3 inhose at pump outlet.

o%= O

a) hA = + ( z 2 - z ^ + + h,y 2g

95 L/min l m 3 / s

/V=

^-(0.025 m ) 2 / 4 60000 L/min

(3.23)(0.025)(1100)

= 3.23 m/s:(3.23) 2 _

M 2.0x10"

2g 2(9.81)

= 4 . 4 4 x l 0 4 : / = 0.021 (smooth)

= 0.530 m

hi= f = (0 .0 21 )- ^- (0 .5 30 )m = 37.86 mD2g 0.025

140 k N / m 2+ 7.3 m +0.530+ 37.86 m = 58.67 m

(1.10)(9.81kN/m 3)

PA = hAyQ = (58.67 m)(1.10)(9.81 kN/m3)(95/60000)m3/s = 1.00 kN-m/s = 1.00 k W

b) + z 3 + 3

y 2ghL = + z 2 + -

2 - : ^ 3 = ^ 2 + [ ( z 2 - z 3 ) + /z]yy 2g

p3 = 140 kPa + (1.10)(9.81 kN/ m3)[ 8. 5 m + 37.86 m] = 640 kPa

102 C h a p t e r 8
http://.5x/0~4http://.5x/0~4http://.5x/0~4


117/298

8.39 Q = 1200 L / m i n x 1 m3/s/60000 L / m i n = 0.02 m 3/s ^ 2 3 _

Q 0.02 m 3 /s , 1 0 n , M g t -u = = r - = 1.189 m/s Laminar Pump

1 . 6 8 2 x l 0 " 2 m 2

a) / >2 ~P3 = YA = Y O /D2g

_ ^ = 1.189(0.1463)(930) _

/ / 0.15

/ >2 - / > 3 = YA = (0.93)(9.81 kN/m3 )3 / 64 Y 3200 V l - 1 8 9 ) 2 m = 8 5 3 k p a

1079 0.1463; 2(9.81)

,2 .,2b) + z, + - L - + /^ -/z ='- + z3 + -^-: Pi =P3, ui = Oi,Zi = z 3

hj - hL r - = 93.5 m

y 2g y 2g

853 k N / m 2

(0.93)(9.81 k N / m 3 )

^ = hAyQ = (93.5 m)(0. 93)(9. 81 kN/ m3)(0.02 m3/s) = 17.1 kN-m/s = 17.1 k W

3.40 At 100C, n = 7. 9x 10^ Pa-s

a) W i t hpumping stations 3.2 kmapart:

A r vDp (1.189)(0.1463)(930) n r i n 4 , ,NR = - ~ =

v A - =2 .05 x 104 turbulentju 7.9 x lO"3

Z>/e =0.1463 m/4.6 x 10" 3m = 3180;/= 0.026

fc-W i ^ = ( 0 . 0 2 6 ) ^ - < ! ? ) ! m . 40.98 mZ>2g 0.1463 2(9.81)

PA = ^ Y 2 = (40.98)(0.93)(9.81)(0.02) = 7.48 k W

/ i)2

b) Le thL = 93.5 m (from Prob. 9.13): hL = fD 2g

L , K D ^ _ ( 9 3 . 5 m ) ( 0 . 1 4 6 3 m ) ( 2 ) ( 9 . 8 1 m / s - )

/ u 2 (0.026)(1.189m/s)2


A N D E N E R G Y L O S S E S DUE TO FR IC T ION 103


118/298

8 - 4 1 Px o2 , pB u

28 7W 2g7W

PB = 7w

NR =

2 8

uD (1.99)(0.098)

Pt. 1 at tank surface. px = 0, vx = 0

900L /m in 1 m3/sQ

60000 L/min

Q_ 0.015 m3/s

7 ~ 7.538xlO"3 m 2

0.015 m3/s

= 1.99 m/s

1.30x10"= 1 .50xl0 5 : = 0 - 0 9 8 . =6 5 3 3 3 : /= 0.0165

e 1.5x10"

L v2 80.5 (1.99)2hL= f = (0.0165)

D2g 0.098 2(9.8

"12_


119/298

8.44 P_

Y 2g Y 2g

Y 2g

Q 0.50 ft 3/s Q 0.50

u A = = - = 7 . 2 8 ft/s: U B = = = 15.03 ft/sA 0.06868f t 2 A 0.03326

uBDp _ (15.03)(0.2058)( 1.026)(1.94) _

M

D 0.2058

e 1.5x10"

4.0x10"

= 1372: / = 0.020

1.54 x 105

A = / A ^ = ( 0 . 0 2 0 ) - ^ - x ^ l ft = 27.28 ftD2g 0.2058 2(32.2)

2 5 . 0 - ( - 3 . 5 0 ) l l b f t 3 1 4 4 i n 2 (15.03)2 - ( 7 .2 8 ) 2 f t 2 / s 2

h = i ^ . + 80 ft + i } - i } + 27.28 ft = 174.1 ftin 2(1.026)(62.41b) ft 2 2(32.2 ft /s 2 )

PA = hAyQ =(174.1 ft)(1.026)(62.4 lb/ft3

)(0.50 ft3

/s)/550 = 10.13 hp

8.45 NR vDp = QDp = QDp = AQp

i Ap. nD1 nDfiA r

4gp

nNRp

4(0.90ff/s)(1.24)(l 9 4 l b . S - / f t - ) , o m ( t

7r(300)(5.0xl0"2 l b - s / f t 2 )

2 1/2-in Type K Copper Tube:D =0.2029ft :A = 0.03234 ft 2

u = g = - 9 Q f t 3 / s , =2 7. 8 ft/si 0.03234 ft2

= v D p - (27.8)(0.2029)(1.24)(1.94)

u ~ 5.0 x l O " 2= 272

/ s L 1 ) 2 m ^ o , n 6 4 5 5 (27.8) 2 lb l f t 2P\-Pi =Y A = y / = (1.24)(62.4)- x x -v

D2g 272 0.2029 2(32.2) ft 2 144 i n 2

411 psi

8.46 + z,+-Yw 2g 2g

2g

Pt .1at pump outlet i n pipe.

Pt. 2 at reservoir surface. p2 = 0,u2 = 0

_ Q _ 4.00 ft 3/s

A 0.3472 ft

2= 11.52 ft/s

^ ( 1 1 . 5 2 X 0 . 6 6 5 ! ) ^ fl=

v 1.21x l O " 5 1.5 x1o- 4

w A i = ( 0 . 0 1 S 5 ) . ^ . < I i ^ f t = 1 2 0 . , ftD 2g 0.6651 2(32.2)

62.41b

f t 32 1 0 - m ^ : + i 2 o . i

2(32.2)

ft-lft2

144in 2142.1 psi


A N D E N E R G Y L O S S E S D U E T O FRICTION 105


120/298

8.47 E+ z + ^ + h =rw 2g

[142 .1- ( -2 .36) ]lb

^ yw (62.4 l b / f t3 ) ( in 2 ) ( 1 f t 2 / 1 4 4 i n 2 )

hA =333.5 ft-lb/lb333.5f t - l b 62.41b 4.00ft 3 lhp

Pt.Oat pump inlet.

Pt.lat pump outlet.

Assume z0 =z,v0 ux

PA = hAyQ =lb ft3 550 ft-lb/s

151 hp

8.48 PK l y PB l + Z , + - / ! , = + Z +

rg g 7 g

PA=PB + Jg[(zB - zA) + hL]vDp (7.76)(0.8350)(1.32)NR =

D

M

0.8350

e 1.5x10"

7.2x10" 6

5567: /=0.0145

1.19 x 10 6

Z) 2g 0.8350 2(32.2)

^ A = 40.0 p si g+ ^ ^ [ 8 5 + 5 1 . 9 ] ^ ^ - =80.3 psigf t 3 ' 144in 2

Q 4.25ft 3/s

i4 0.5479f t 2

Assume sg = 0.68

/ /F romApp. D .

8.49 A 7 7 Pl u l

2g 7o 2g70

= ( Z 2 - Z j ) + + /

2g

> 0.668 f t 3/s 4

:

0.08840 ft2

4.3/

= 7.56 ft/s

Pt . 1 at tank surface. /?, = O,ux = 0

Pt. 2 in outlet stream from 3-in pipe.

g = 3 0 0 g a l / m m 4 f t 3 / s = 0.668 f t 3/s

449 gal/min

o i l - A p p .C

Q 0.668 f t J/s , . ,3= = - = 13.02 ft/s = i>,

4 0.05132 f t 2 2

U vi L A ul

N _u 3 Z? 3 = (13.02X0.2557)

*3 v 2.15 x l O " 3

^ = ^ A _ (7.56) (0.3355)

64= 1548 (L am in ar ): / = = 0.0413

hL =0.0413

1.0 ft +

2.15 x l O " 3

75 (13.02)2

= 1180 (Laminar):/4

:64

= 0.0543

0.2557 2(32.2)+ 0.0543-

25 (7.56) 2

0.3355 2(32.2)35.5 ft

(13.02) 2

2(32.2)ft + 35.5 ft =39.1 ft

^ = ^ y o e = (39.1ft)(0.890)(62.41b) (0.668ft3) lhp

ft 3 550 ft-lb/s2.64 hp

106 C h a p t e r 8


121/298

8.50v\ j P2 v\+ z + -

[ ~ - h L = ^ +z2+-1-: o1 = o2

2 g Yo 2gElYo S Yo

P\-P2 = lo[{Z2-Z\)+h\

vDp (3.65)(0.0189)(930) 64

3.31x10"= 1938 (L ami na r): /= = 0.0330

^ r A i ^ o x j a S O ) ^ - ^ =20 .760,D 2g 0.0189 2(9.81)

P ] -p2 = 9.12 kN /m3 [-1.88 m + 20.76 m] = 172 kPa

8.51 pi-p2 =Jg[(z2 - z,) + hL] (From 9.24)

vDp (0.701)(0.0738)(1258)NR =

0.960

NR = 67.8 (L am in

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Solucionario - Mecanica de fluidos - Sexta edicion - Robert L Mott(1).pdf