Applied Fluid Mechanics-Mott-Sol 6

Upper Saddle River, New Jersey Columbus, Ohio

PEARSON

Prentice Hall

Robert L. Mott University of Dayton

Sixth Edition

APPLIED FLUID MECHANICS

Instructor's Manual to accompany

ISBN 0-13-172355-3

PEARSON

Prentice Hall

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posos and the needs of other instructor. who rely on

Upper Saddle River, New Jersey Columbus, Ohio

PEARSON

Prentice Hall

Robert L. Mott U niversity of Dayton

Sixth Edition

Online IM to Accompany

APPLIED FLUID MECHANICS

ááá

CHAPTER FOURTEEN OPEN CHANNEL FLOW 218

CHAPTER THIRTEEN PUMP SELECTION AND APPUCATION 213

CHAPTER TWEL VE PARALLEL PIPE UNE SYSTEMS 187

CHAPTER ELEVEN SERIES PIPE UNE SYSTEMS 141

CHAPTER TEN MINOR LOSSES 129

CHAPTER NINE VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW FOR NONCIRCULAR SECTIONS 113

CHAPTER EIGHT REYNOLDS NUMBER, LAMINAR FLOW, AND TURBULENT FLOW AND ENERGY LOSSES DUE TO FRICTION 94

CHAPTER SEVEN GENERAL ENERGY EOUATION 81

CHAPTER SIX FLOW OF FLUIDS 61

CHAPTER FIVE BUOY ANCY ANO ST ABIUTY 44

CHAPTER FOUR FORCES DUE TO STA TIC FLUIDS 25

CHAPTER THREE PRESSURE MEASUREMENT 19

CHAPTER TWO VISCOSITY OF FLUIDS 12

CHAPTER ONE THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS 1

CONTENTS

iv

SPREADSHEETS 273

CHAPTER NINETEEN FLOW OF AIR IN DUCTS 269

CHAPTER EIGHTEEN FANS, BLOWERS, COMPRESSORS, ANO THE FLOW OF GASES 260

CHAPTER SEVENTEEN DRAG ANO LIFT 251

CHAPTER SIXTEEN FORCES DUE TO FLUIDS IN MOTION 240

CHAPTER FIFTEEN FLOW MEASUREMENT 235

V

The principies 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. lt is highly recommended that you work sample problems by hand first. Then enter the appropriate data into the spreadsheet to verify the solution. In most spreadsheets, the data that need to be entered are identified by gray-shaded areas and by itallc type. Results are typically shown in bold type.

Using the Spreadsheets: /t is recommended that the given spreadsheets be maintained as they initially appear on the CD. To use them for solving other problems, cal/ up the master workbook in Exce/ 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 critica/ ce/Is containing complex equations. However, you are encouraged to add additiona/ features to the spreadsheets to enhance their utility.

The following sections give brief descriptions of each spreadsheet. Many are discussed in the text in more extensive detail. lt is expected that you will verify all of the elements of each spreadsheet before using them for solutions to specific problems.

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 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 /, 11, or 111 indicating whether the spreadsheet is for a Class t, Cfass JI, or Class /// pipe line system as defined in Chapter 11 of the text.

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.

lntroduction

Description of Spreadsheets lncluded on the CD in the Book

Prentice-Hall Publishing Company

APPLIED FLUID MECHANICS Sixth Edition

Robert L. Mott

vi

I Pressure SI: Most of the layout and data entry for this spreadsheet are the same as those in the first two spreadsheets, I Power SI, and I Power US. The difference here is that the analysis determines the pressure at one point in the system when the pressure at sorne other point is known. Class 1 systems with one or two pipe sizes including minor losses can be analyzed. This spreadsheet uses the known pressure at sorne starting point and computes the pressure at a downstream point, considering changes in elevation, velocity head, pipe friction and minor losses. The example is the solution of Problem 11. 7 for the

I Power US: Same as Power SI:, 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 losses or minor losses considered in the suction pipe. The length is given ta be zero and all K factors are zero for Pipe 1. In Pipe 2, pipe friction, the loss in the elbow, and the loss in the nozzle are included.

8. The Results: section at the bottom of the spreadsheet includes the total energy loss h<, the total head on the pump hà, and the power added to the fluid by the pump P,. lf you enter an efficiency for the pump, the power input to the pump P, is computed.

7. Enter energy loss coefficients, K, for all loss-producing elements in both pipes. See Chapters 8, 1 O, 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.

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.

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. lf the velocity at either reference point is in a pipe, you may use a computed velocity from v = QIA that is included in the data cells for the two pipes. In such cases, you enter the Excel command "=820" 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 m2/s).

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.

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.

/ Power SI: 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 losses are considered. All data must be in the listed SI units. The solution procedure is for a Class 1 series pipe line system. The following is a summary of the steps you must complete.

vii

111-A & 111-8 SI: Class 111 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. 8oth Method 111-A and Method 111-8 as described in the text are shown on this spreadsheet. lf only pipe friction loss is considered, then only the upper part using Method 111-A is pertinent.

11-A & 11-8 US: This spreadsheet is identical to 11-A & 11-8: except for using the U.S. Customary unit system. Problem 11.1 O is solved in the example. The system in this problem has no minor losses 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 ali minor losses have been set to zero. The result is that the pressure at the target point, P2, 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 11-A and the calculation of friction factor and other terms in Method 11-8.

The lower part of the spreadsheet implements Method 11-8 for which minor losses are considered in addition to the friction losses 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 11-A solution directly above. Enter the minor loss coefficients in the lower part of the spreadsheet. As each est²mate for flow rate is entered, the pressure ata target point, called p2, is computad. You must compare this pressure with the desired pressure. Successive changes in the estimate can be made very rapidly until the desired pressure is acceptable within a small tolerance that you decide.

You should review the details of these sheets from the discussions in the book. In summary, Method 11-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 losses. 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 11 problems without minor losses.

11-A & 11-8 SI: This spreadsheet solves Class 11 series pipe line problems using either Method 11-A or Method 11-8 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.

I Pressure US: This spreadsheet is virtually identical to I Pressure 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 87 and 88 where the pressures are listed and E43 and E44 where the actual calculations for pressure change are performed. The changes demonstrated here between I 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.

system shown in Figure 11.17. p2 is assumed to be 100 kPa. P2 is computed to be 78.21 kPa. Then 11p = -21.79 kPa.

viii

After seeing the required total head on the pump for the desired flow rate, the user has selected a centrifuga! pump, model 2x3-1 O 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).

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 in the system. This sheet is basically the same as that in the spreadsheet called I Power US:, discussed earlier. But the final calculation of the power delivered by the pump to the fluid has been deleted.

System Curve US: This spreadsheet is the same as that shown in Figure 13.41 in Chapter 13 of the text. lt 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 13.1 O and 13.14.

The problem includes sorne minor losses so that both the upper part (Method 111-A) and the lower part (Method 111-B) are used. The result from Method 111-A predicts that the minimum acceptable pipe flow diameter is 0.3090 ft if no minor losses are considered. Using a standard 4-in schedule 40 steel pipe (O = 0.3355 ft) with the minar losses included in Method 111-B shows that the pressure at the target point, p2, is greater than the minimum acceptable pressure. Therefore, that pipe size is satisfactory.

111-A & 111-B US: This spreadsheet is identical to 111-A & 111-B SI: 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 lt is used. Please refer to the text.

Class 111 systems with minor losses are demonstrated in the next spreadsheet.

Problem 11.18 is solved by the given spreadsheet data. Only pipe friction losses 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 11-A 4 11-B SI: to compute the actual expected volume flow rate when using the 4-in Type K copper tube.

ix

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 SI 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.

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.

Note that no spreadsheet system using SI Metric units for operating point is included in this set. lt would be good practice for you to copy this given spreadsheet and convert it to SI 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.

X

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

Pipe head losses can be calculated using the Hazen-Williams equation for water flow (Section 8.9 in the book) ar the Darcy-Weisbach equation for other types of incompressible fluids. We use the term Darcy's equation in the book. See Section 8.5.

Validation/calibration of existing pipeline systems. Modeling a proposed system's operation. Determination of line head losses at a specific flow rate. (termed a forced-flow system).

Å Analysis of cavitation and net positive suction head (NPSH) problems. Å Comparison of equivalent SI unit to English unit designs. Å Modeling of recirculating and gravity (non-pumped) flow systems.

Many types of fluid flow problems can be solved, such as

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, orto 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.

HYDROFLO 2 is a unique fluid conveyance system design tool for full pipe incompressible 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.

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 1 O -13 of the book. More information about Tahoe Design Software and the professional versions of these programs can be found on their website www.tahoesoft.com.

HYDROFLO 2, HCALC, and PumpBase Software by TAHOE DESIGN SOFTWARE lncluded on the CD with this book

APPLIED FLUID MECHANICS Sixth Edition Robert L. Mott

xi

Å Energy losses due to friction in straight pipes and tubes (Chapter 8) Energy losses dueto valves and fittings (Chapter 1 O)

Å Analysis of series pipeline systems (Chapter 11) Å 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 far design problems such as those outlined at the end of Chapter 13.

Å Extensive system design as a senior design project.

The fallowing types of problems and projects can be solved with these programs:

As with any software, it is essential that the user have a so/id understanding of the principies involved in the analyses performed by the software as we// as the details of data entry and interpretation of resu/ts. lt is advised that practice with hand calculations for representative problems be completed befare using the software, Then use the resu/ts of known, accurately-so/ved problems with the software to verify that it is being used correctly and to gain confidence in its capabilities.

Suggestions for use of these programs:

HCALC is a handy calculator tool that resides in the system tray for easy access. lt performs the calculations for any variable in the flow rate equation, Q =Av, or the Darcy- Weisbach equation, bá = f(L/D)(v2/2g), when sufficient data are entered for fluid properties, pipe dimensions, roughness, and so farth. Reynolds number and friction factor are also calculated. You may select either SI or U.S. Customary units.

More input data are required if the fluid is not cool water, a limit for NPSHR is to be specified, ora certain type of pump is desired.

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 h0, 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 13.1 O in the book. The program will search its database, select severa! 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.

database far later use. System data can be entered and displayed in standard English, SI 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 far the application.

xii

The use of the Tahoe Design Software programs after learning the fundamentals of fluid system design analysis allows more comprehensiva 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.

Å Pipe types and sizes Length of pipe for all parts of the system Layout of the piping system

Å Location of the pump Å Types and sizes of all valves and fittings and their placement Å Ust of materials required for the system Analysis of the pressure at pertinent points

Å 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

Å 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 practica

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:

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,

1 The Nature of Fluids

V=:!_= 0.50kmx103 m =47.2m/s t 10.6 s km

1.17

Consistent units in an equation

2500 ft' 0.0283 rrr' 1 min .118 3á ---X X--=, m S min ft" 60 s

6.0 ft/s(0.3048 m/ft) = 1.83 mis

7390 cm'[l m3/(100 cm)"] = 7.39 x 10-3 m3

6.35 L(l m3/1000 L) = 6.35 x 10-3 m3

2580 ft(0.3048 m/ft) = 786 m

480 ft\0.0283 m3/ft3) = 13.6 m3

8.65 in(25.4 mm/in)= 220 mm

25.3 ft(0.3048 m/ft) = 7.71 m

1.86 mi(l.609 km/mi)(l03 m/km) = 2993 m

80kmx103 m x~ = 22.2 mis h km 3600 s

55.0 gal(0.00379 m3/gal) = 0.208 m3

1250 mm(l m/103 mm)= 1.25 m

1.16

1.15

L14

1.13

1.12

1.11

1.10

1.9

1.8

1.7.

1.6

1.5

1.4

1.3

1.2

1.1

Conversion factors

Chapter 1

u= J2(KmE) = 2(15 N Ŀ m) x 1 kg ~m/s2 = 1.58 mis 12kg

m=2(KE)=(2)(94.6mNĿm)xl0-3 NxlkgĿmx103 g =37.4 u2 (2.25 mis )2 mN s2 Ŀ N kg g

m = 2(KE) = (2)(38.6N Ŀm) x( h )2 x lkgĿm x (3600s)2 x 1 km2 u2 1 31.5km s2ĿN h2 (103 m)"

m = (2)(38.6)(3600)2 k = 1.008 k (31.5)2(103)2 g g

KE =muz = 75 kg x (6.85m)z=1.76x 103 kg Ŀm2 = 1.76 kN. m 2 2 s s2

kg m" = 35.6 X 103 "2 s

KE= mv2 = (3600kg) x(16km)2 x (103m)2 x 1 h2 2 2 h km2 (3600s)2

KE = 35.6 kN Ŀ m

KE =muz= (15 kg)(l.2m/s)2=10.8 kg . mz = 10.8 N. m 2 2 82

t = Õ:;- = (2)(53 in) X___!!!__ :::: 0.524 S v-;; 32.2 ft/ s2 12 in

a= 2s = (2)(3.2 km) x 103 m x 1 ft x 1 min2 = 0.264 f! 2 (4.7min)2 km 0.3048m (60s)2 s

t = {2;" = (2)(13 m) = 1.63 s ~r:; 9.81m1s2

a= 2s = (2)(3.2 km) x 103 m x 1 min2 = 8.05 x 10-2 m/s2 t2 (4.7min)2 km (60s)2

u= s = 1.0 mi x 3600 s = 632 mi/h t 5.7 s h

s 1000 ft 1 mi 3600 s U=-= X X-- =48.7mi/h t 14 s 5280 ft h

U=~= 1.50 km X 3600 S = 1038 km/h t 5.2 s h

2

1.30

1.29

1.28

1.27

1.26

1.25

1.24

1.23

1.22

1.21

1.20

1.19

1.18


ERA= 49 runs x 9 innings = 3.59 runs/game 123 innings game

40 1 game 9 innings 129 . . runs x x = mnmgs 2. 79 runs game

3Ŀ 12 runs x 1 game x 150 innings = 52 runs game 9 innings

ERA= 39 runs x 9 innings = 2.49 runs/game 141 innings game

u= ~2g(wKE) = 2(32.2 ft/s2)(30 ozĿin) x __!_!!__ = 5.18 ft/s 6.0oz 12in

u=~ 2g(wKE) = 2(32.2 ft/ s2 )(10 lbĿ ft) = 4Å63 ft/s 30lb

2g(KE) 2(32.2ft)(38.6lbĿft)(h2) 1 mi" (3600s)2 w= = X X--- U2 s2(19.5 m²)2 (5280 ft)2 h2

w= (2)(32.2)(38.6)(3600)2 lb = 3.041b (19.5)2(5280)2

m = 2(KE) = 2(15 lb. ft) = 6.20 lb. s2 = 6.20 slu s u2 (2.2ft/s2)2 ft g

KE=mu2 = wu2 =(150lb)(20ft/s)2 =932lbĿft 2 2g (2)(32.2 ft/ s2)

KE= mu2 = wv2 = (8000lb)(10mi)2 x 1 h2 x (5280ft)2 2 2g (2)(32.2 ft/s2)(h )2 (3600 s )2 mi"

KE = (8000)(l 0)\5280)2 lbĿ ft = 26700 lb Ŀ ft (2)(32.2)(3600)2

KE = mu2 = (1 slug)(4 ft/s )2 x 1 lbĿ s2 /ft = 8.00 lb. ft 2 2 slug

u=~2(KE)= 2(212mNĿm)x10-3Nxl03gxlkgĿm =1.56m/s m 175 g mN kg s2 Å N

1.42

1.41

1.40

1.39

1.38

1.37

1.36

1.35

1.34

1.33

1.32

1.31

Chapter 1

D = [!!p = 4(30x103 N) = 50.5 x 10-3 m = 50.5 mm 4iJ;p 1l'(15.0x106 N/m2)

.,: =__!_= 4F: ThenD= {4F A 1l'D214 1l'D2 4iJ;p

D = 4(20000 lb) = 2Å26 in Jr(5000 lb/in")

F = pA = (6000 lb/in2)( Jr[2.00in J2 / 4) = 18850 lb

F = A= 20.5x106 N x .7l'(50 mm)2 x 1 m2 = 40.25 kN p m2 4 (103 mm)"

F 18000lb p - - - = 3667 psi - A - .7r(2.50in)2 / 4

F 6000lb p - - - = 119 psi - A - .7r(8.0in)2/4

p = FIA = 2500 lb/[n{3.00 in)2/4] = 354 lb/in2 = 354 psi

p = FIA = 8700 lb/[n{l.50 in)2/4] = 4923 psi

4

1.52

1.51

1.50

1.49

1.48

1.47

1.46

1.45

1.44

1.43

The definition of pressure


1.56 (Variable Answers) usingp = 2.27 MPa F= pA = (2.27 x 106 N/m2)(n(0.250 m)2/4) = 111x103 N = 111 kN F = 111 kN (1 lb/4.448 N) = 25050 lb

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

= F = 712 N x (l03 mm/ = 2.77 x 106 Pa = 2.27 MPa p A Jr(20 mm )214 m2 p = 2.27 x 106 Pa (1 psi/6895 Pa) = 329 psi

0o 1 2 3 4 s s 1 a

Oiamet8r (rt)

eooo ' \ \ \ ...

1

7 8

30 25 20 15 10 s 0o 1 2 3 4 s a

aam.r-(ln)

1

1 / /

/ _.,.,.., L...-f--

D(in) D!(in!) 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

1.53 F = pA = p[JrD2] = 500 lb(Jr)(D in)" = 392.7 D2 lb 4 in ' 4

D~(in!) D(in) F(lb)

1.00 1.00 393

2.00 4.00 1571 Force 3.00 9.00 3534

(" )( 1000)

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

1.54 F F 4F 4(5000 lb) 6366 . p=-;á= JrD214 = JrD2 = Jr(Din)2 =7psi

Chapter 1

16 times higher

4.2 times higher

6

m= w= 610N xlkgĿm/s2 =62.2kg g 9.81 m/s" N

1.67

Force and mass

1.66 Use large diameter cylinders and short strokes.

_!_=EA= 189000lb²'l'(2.00in)2 = 14137 lb/in (L\L) L in2(42.0 in)(4)

1.65

_!_=EA 189000 lb ²'l'(0.5 in)2 = 3711 lb/in (L\L) L in2(10.0 in)(4)

1.64

1.63 Stiffness = Force/Change in Length = FI L\L

Bulk Modulus =E= _::_p_ = - p V !iV/V !iV

Butp=F/A; V=AL;!iV=-A(L\L) E=-Fx AL =~

A -A(L\L) A(L\L)

_!_=EA= 189000 lb ²'l'(0.5 in)2 = 884lb/in (L\L) L in2(42 in)4

1.62 !iV = -20.0MPa =-0.0153 =-1.53% V 1303MPa

1.61 !iV = -p = -3000 psi = -0.0159 = -1.59% V E 189000 psi

1.60 !iVIV= -0.01; !iV = 0.01V=0.01 AL Assume area of cylinder does not change. !iV=A(L\L) = 0.01 AL Then L\L = 0.01 L = 0.01(12.00 in)= 0.120 in

1.59 Sp = -E(!iVIV) = -189000psi(-0.01)=1890 psi Sp = -1303MPa(-0.01)=13.03 MPa

1.57 Sp = -E(!iV/V) = -130000psi(-0.01)=1300 psi Sp = -896 MPa(-0.01) = 8.96 MPa

1.58 Sp = -E(!iV/V) = -3.59 x 106 psi(-0.01) = 35900 psi Sp = -24750 MPa(-0.01) = 247.5 MPa

Bulk modulus


y 12.02 N 52 1 kg Ŀ m/s" p=-= x--x = 1.225 kg/m3 g m ' 9.81m N

YB = (sg)BYw = (0.876)(9.81kN/m3)=8.59 kN/m3 PB = (sg)BPw = (0.876)(1000 kg/nr') = 876 kg/nr'

1.81

1.80

Density, specific weight, and specific gravity

(Variable Answers) See problem 1.75 far method.

F= 9810 N X 1.0 lb/4.448 N = 2205 lb

F = w = mg = 1000 kg x 9.81m/s2=9810 kg-m/s' = 9810 N

w 1.00 lb m = - = = 0.0311 slugs g 32.2 ft/s2

m = 0.0311 slugs x 14.59 kg/slug = 0.453 kg w = 1.00 lb X 4.448 N/lb = 4.448 N

w 160lb m = - = = 4.97 slugs g 32.2 ft/s"

w = 160 lb X 4.448 N/lb = 712 N m = 4.97 slugs x 14.59 kg/slug = 72.5 kg

1 lb. 82 /ft w= mg = 0.258 slugs x 32.2 ft/s" x = 8.31 lb slug

1 lbĿs2/ft w = mg = 1.58 slugs x 32.2 ft/s" x = 50.9 lb slug

w 42.0lb m = - = = 1.304 slugs g 32.2 ft/s2

m = w = 7 Ŀ81b = 0.242 lbĿ 82 = 0.242 slugs

g 32.2 ft/s" ft

1 kg 2 2 w= mg=450 gx--x9.81m/s =4.41kgĿmls = 4.41 N l03g

w = mg = 825 kg x 9.81m/s2=8093 kg-m/s' = 8093 N

_ w _ 1.35 x 103 N 1 kg Ŀ m/s2 _ 138 k m--- X - g

g 9.81 m/s2 N

1.79

1.78

1.77

1.76

1.75

1.74

1.73

1.72

1.71

1.70

1.69

1.68

8 Chapter 1

1.91 p = (sg)(pw) = (0.789)(1000 kg/nr') = 789 kg/nr' y= (sg)(Yw) = (0.789)(9.81kN/m3)=7.74 kN/m3

1080 kg 9.81 m 1 N 1 kN = 10.59 kN/m3 y= pg= m3 x-s-2-X1kgĿm/s2X103 N

8 = 1 = 1080 kg/m3 = 1.08 g p p w 1000 kg/nr'

V_ w _ 22.0 N 1 kN _

2 72 10_3 3 --- X-- - , X ID y (0.826)(9.81kN/m3) 103 N

1.90

1.89

1.88 r=pg=(1200kg/m3)(9.81m/s2)( IN 2) =ll.77kN/m3 kgĿm/s

8 = p = 1200 kg/m3 = 1.20 g Pw@4ÜC 1000kg/m3

w = yV= (sg)(yw)(V) = (0.68)(9.81 kN/m3)(0.095 nr') = 0.634 kN = 634 N

y= (sg)(Yw@4ÜC) = 1.258(9.81kN/m3)=12.34 kN/m3 = w!V w = yV= (12.34 kN/m3)(0.50 rrr') = 6.17 kN m = w = 6.17 kN x 103 N x 1 kg Ŀ m/s2 = 629 kg g 9.8lm/s2 kN N

V= AL= nD2L/4 = n(0.150 m)2(0.100 m)/4 = 1.767 x 10-3 m'

= m = 1.56 kg = 883 kg/m" Po V 1.767xl0-3m3

3 1 N 103 N kg y0 = P$=883kg/m x9.81m/s2 x 2 =8.66x--=8.66- 1 kg-rn/s m ' m ' sg = pofpw@4ÜC = 883 kg/m3/1000 kg/rn' = 0.883

- w ĿV- w - 2.25kN =0.0173 m3 r= v' ---;;- I30.4kN/m3

1.87

1.86

1.85

1.84

1.83 ro 8.860kN/m3 = 0.903 at 5ÜC sg= rw@4ÜC 9.81 kN/m3

Yo 8.483kN/m3 = 0.865 at 50ÜC sg= rw@4ÜC 9.81kN/m3

lN r=pg=l.964kg/m3x9.81m/s2x 2 =19.27N/m3

Lkg -m/s 1.82


V V 1.32 lbĿ s2 32.2 ft 25 0 1 1 ft3 __ 142 lb w= Y = pg = X --2 - X Å ga X ft4 8 7.48 gal

1.103

1.102 w= yV = (1.258) (62.4 lb) (50 gal) (1 ft3) = 525 lb ft3 7.48 gal

Y= w = 7.50 lb x 7.48 gal = 56.l Ib/ft3 V 1 gal ft3

- r - 56. llb /ft3 -1 74 lb. 82 - 1 74 I /fi 3 p- - - - . -- - . s ugs t g 32.2 ft/s" ft 4

sg= Ya 5.611b/ft3 =0.899 r; @4ÜC 62.4 lb/ft3

1.101

1.99 sg = yj(yw@4ÜC) = 56.4 lb/ft3/62.4 lb/ft3 = 0.904 at 40ÜF sg = yj(yw@4ÜC) = 54.0 lb/ft3/62.4 lb/ft3 = 0.865 at 120ÜF

1.100 V= wly = 500 lb/834 lb/ft3 = 0.600 ft3

y= pg = 0.003818lug/ft3 (32.2 ft/s") l lb. 82 /ft = 0.1227 lb/ft3

slug

- r - 0.0765 lb/ft3 l slug - 2 38 10-3 l /f 3 p - - - x - Å x s ugs t g 32.2 ft/s" 1 lbĿ 82 /ft

w = yV= (2.32)(9.81 kN/m3)(1.42 x 10-4 m") = 3.23 x 10-3 kN = 3.23 N

y= (sg)(yw) = 0.876(62.4 lb/ft3) = 54.7 lb/ft3

p = (sg)(pw) = 0.876(1.94 slugs/ft') = 1.70 slugs/ft3

Wcastoroil = Yco. Veo= (9.42 kN/m3)(0.02 m') = 0.1884 kN = ~ = 0.1884 kN = 1.42 X 10_3 m3

V m y m (13.54)(9.81 kN /m")

V=Ad= (rcD2/4)(d) = n(lO m)2(6.75 m)/4 = 530.l rrr' w = yV= (0.68)(9.81 kN/m3)(530.l rrr') = 3.536 x 103 kN = 3.536 MN m = pV = (0.68)(1000 kg/m3)(530.1m3)=360.5 x 103 kg = 360.5 Mg

W0 = 35.4 N - 2.25 N = 33.15 N V0 =Ad= (rcD2/4)(d) = rc(.150 m)2(.20 m)/4 = 3.53 x 10-3 m'

=w= 33.l5N =9.38xl03N/m3=9.38kN/m3 Yo V 3.53x10-3 rrr'

sg =y o= 9.38 kN/m3 = 0.956 y w 9.81 kN/m3

1.98

1.97

1.96

1.95

1.94

1.93

1.92

Chapter 1

w= yV = (2.32) (62.4 lb) (8.64 in") (1 ft3) = 0.724 lb ft3 1728 in '

ll'D2 ll'(30 ft)2 s V=AĀd=--Ād= x22ft=15550 ft3 x7.48gal/ft3=1.16x10 gal 4 4 w = yV== (0.68)(62.4 lb/ft3)(15550 ft3) = 6.60 X 105 lb

Wco = YcoV= (59.69 lb/ft3)(5 gal)(l ft3/7.48 gal)= 39.90 lb

=~= 39.901bft3 x 7.48gal =0.353 al r, r; 13.54(62.4lb) ft3 g

= 56.9 lb/ft3 w (7.95-0.50)lb 1728 in3

ro= V= ( ll'(6.0 in )2 / 4 )(8.0 in) X ft3

sg = yofyw = 56.9 lb/ft3/62.4 lb/ft3 = 0.912

10

1.111

1.110

1.109

1.108

1.106 y= (sg)(Yw) = (1.08)(62.4 lb/ft3) = 67.4 lb/ft3

1.107 p =(O. 79)(1.94 slugs/fr') = 1.53 slugs/fr'; p =O. 79 g/cm3

V= w = 5.0 lb ft3 x 0.0283 m3 x (102 cm)3 = 2745 cm' r (0.826)62.4 lb ft3 m '

sg =_E_= 1.20 g x m3 x 1 kg x (102 cm)J = 1.20 Pw cm" 1000 kg 103 g m '

p = (sg)(pw) = 1.20(1.94 slugs/fr') = 2.33 slugs/²r' y= (sg)(Yw) = (1.20)(62.4 lb/ft3) = 74.9 lb/ft3

1.105

1.104


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12 Chapter 2

2.20 2.0 x 10-5 PaĿs

2.19 1.5 x 10-3 Pa-s

2.17 Blood plasma, molten plastics, catsup, paint, and others.

2.16 Water, oil, gasoline, alcohol, kerosene, benzene, and others.

2.15 A nonnewtonian fluid is one for which the dynamic viscosity is dependent on the velocity gradient.

2.14 A newtonian fluid is one for which the dynamic viscosity is independent ofthe velocity gradient.

2.13 It does not conform to the standard SI system. It uses obsolete basic unit of cm.

2.12 1stoke=1 cm2/s

2.9 Kinematic viscosity = dynamic viscosity/density ofthe fluid.

2.8 It does not conform to the standard SI system. It uses obsolete basic units of dynes and cm.

2.7 1 poise = 1 dyne-s/cm' = 1 g/(cms)

2.6 lb-s/tr'

2.5 NĿs/m2 or PaĿs

2.4 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.

2.3 Dynamic viscosity = shearing stress/velocity gradient.

2.2 Velocity gradient is a measure of the velocity change with position within a fluid.

2.1 Shearing stress is the force required to slide one unit area layer of a substance over another.

13 Viscosity of Fluids

2.41 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).

2.40 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 viscosity.

2.39 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.

2.38 Rotating drum viscometer.

2.3 7 High viscosity index (VI).

2.36 Viscosity index is a measure of how greatly the viscosity of a fluid changes with temperature.

2.35 2.2 X 10-4 Ib-s/ft'

2.34 1.3 X 10-2 lb-s/ft'

2.33 9.5 x 10-5 lbĿs/ft2

2.32 2.1 X 10-3 Ib-s/ft'

2.31 2.8 X 10-S Ib-s/ft'

2.30 3.3 x 10-5 lb-s/fr'

2.29 4.1 x 10-3 lb-s/fr'

2.28 5.0 X 10-2 lb-s/ft'

2.27 1.9 X 10-7 lb-s/ff

2.26 3.6 X 10-7 lb-s/ft'

2.25 8.9 X 10-6 lb-s/ft'

2.24 3.2 x 10-5 lbĿs/ft2

2.23 1.90 Pa-s

2.22 3.0 x 10-1 PaĿs

2.21 1.1 x 10-5 Pa-s

Chapter 2 14

2.55 SAE lOW-30 engine oil:

Low temperature cranking viscosity at-25ÜC: 7000 cP = 7000 rnPa s = 7.0 Pa-s maximum Low temperature pumping viscosity at-30ÜC: 60 000 cP = 60 000 rnPa s = 60 Pa-s maximum Low shear rate kinematic viscosity at lOOÜC: 9.3 cSt = 9.3 mm'/s = 9.3 x 10-6 m2/2 minimum Low shear rate kinematic viscosity at lOOÜC: 12.5 cSt = 12.5 mm2/s = 12.5 x 10-6 m2/2 maximum High shear rate dynamic viscosity at 150ÜC: 2.9 cP = 2.9 rnPa s = 0.0029 Pa-s minimum

2.54 v = SUS/4.632 = 500/4.632 = 107.9 mmvs = 107.9 x 10-6 m2/s v = 107.9 x 10-6 m2/s [(10.764 ft2!s)/(m2/s)] = 1.162 x 10-3 ft2!s

2.53 See Table 2.4. The kinematic viscosity of SAE 5W-40 o²l must be between 12.5 and 16.3 cSt at lOOÜC using ASTM D 445. Its dynamic viscosity must be over 2.9 cP at 150ÜC using ASTM D 4683, D 4741, or D 5481. The kinematic viscosity must be over 3.8 cSt at lOOÜC using ASTM D 445. Its dynamic viscosity for cranking must be below 6600 cP at-30ÜC using ASTM D 5293. For pumping it must be below 60 000 cP at-35ÜC using ASTM D 4684.

2.52 At -25ÜC using ASTM D 5293; at -30ÜC using ASTM D 4684; at lOOÜC using ASTM D 445.

2.51 lOOÜC using ASTM D 445 testing method and at 150Ü C using ASTM D 4683, D 4741, or D 5481.

2.50 SAE 70W through SAE 60 depending on the operating environment and loads. See Table 2.5.

2.49 SAE OW through SAE 250 depending on the operating environment. See Table 2.4.

2.48 See Table 2.4. The kinematic viscosity of SAE 20 o²l must be between 5.6 and 9.3 cSt at lOOÜC using ASTM D 445. Its dynamic viscosity must be over 2.6 cP at 150ÜC using ASTM D 4683, D 4741, or D 5481. The kinematic viscosity of SAE 20W o²l must be over 5.6 cSt at lOOÜC using ASTM D 445. Its dynamic viscosity for cranking must be below 9500 cP at -15ÜC using ASTM D 5293. For pumping it must be below 60,000 cP at -20ÜC using ASTM D 4684.

2.47 Standard calibrated glass capillary viscometer.

2.46 Kinematic viscosity.

2.45 No. The time is reported as Saybolt Universal Seconds and is a relative measure of viscosity.

2.44 The Saybolt viscometer employs a container in which the fluid can be brought to a known, controlled temperature, a small standard or²fice 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.

2.43 The diameter of the hall; the terminal velocity (usually by noting distance traveled in a given time); the specific weight of the fluid; the specific weight of the hall.

2.42 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.


2.65 From Fig. 2.12, kinematic viscosity = 78.0 SUS

= (pá - P2 )D2 = ( 460. l lb/ft2 )(0.0Á833 ft)2 = 0.000207 lb s/ft2 = 2.07 X 10-4 lbĿs/ft2 rá 32vL 32(4.82 ft/s)(l.O ft)

2.64 See Problem 2.62. Use y.11 = 844.9 lb/ft3 (App. B): y0 = (0.90)(62.4 lb/ft3) = 56.16 lb/ft3 h = (7.00 in)(l ft/12 in)= 0.5833 ft: D = (0.100 in)(l ft/12 in)= 0.00833 ft Pi - P2 = h(rm - ro) = (0.5833 ft)(844.9 - 56.16) lb/ft3 = 460.1 lb/ft2

= (r, - Yr )D2 = ( 489 - 58Ŀ 7)lb/ft3 (O.oo525 ft)2 = 0.00207 lb s/ft' = 2.07 x 10-4 lbĿs/ft2 rá 18v 32(4.82 ft/s)

2.63 See Prob. 2.61. n= 0.94(62.4 lb/ft3) = 58.7 lb/ft3: D = (0.063 in)(l ft/12 in)= 0.00525 ft v= s/t = (10.0 in/10.4 s)(l ft/12 in)= 0.0801 ft/s: Ys = (0.283 lb/in3)(1728 in3/ft3) = 489 lb/ft3

Manometer Eq. using principles of Chapter 3: Pi + YoY + Yoh - Ymh - YQ)l = P2

kN kN Pi =P: = Ymh - Yoh = h(ym - y0) = 0.177 m(132.8 - 8.83)-- = 21.94- 3 2 m m

= (21.94 kN/mz)(0.0025 m)2=9.04x10-6 kN. s x 103 N = 9.04 x 10-3 Pa- s rá 32(1.58 m/s)(0.300 m) m2 kN

Use YMercury = 132.8 kN/m3 (App. B) Yo= 0.90(9.81kN/m3)=8.83 kN/m3 rá = (P1 - P2)D2 (Eq. 2-4)

32vL

2.62

v = slt = .250 m/10.4 s = 2.40 x 10-2 mis = (77.0-9.22)kN(l.6x 10Ŀ3m)2 x 103 N =0.402 N Ŀs = 0.402 Pa- s Õ 18m3(2.40x10-2m/s) kN m '

n= 0.94(9.81kN/m3)=9.22 kN/m3 D = 1.6 mm= 1.6 x 10-3 m rá = (rs - y á)D2 (Eq. 2-9)

18v

2.61

2.60 rá = 0.12 poise [(1 PaĿs)/(10 poise)] = 0.012 Pa-s = 1.2 x 10-2 Pa-s, SAE 10 oil

2.59 rá = 6.5 x 10-3 PaĿs [(1 lbĿs/ft2)/(47.88 PaĿs)] = 1.36 x 10-4 lbĿs/ft2

2.58 From Figure 2.12: v = 15.5 mm2/s = 15.5 x 10-6 m2/s

2.57 v = 5.6 cSt [(1 m2/s)/(106 cSt)] = 5.60 x 10-6 m2/s v = 5.60 x 10-6 m2/s [(10.764 ft2/s)/(m2/s)] = 6.03 x 10-5 ft2/s

2.56 n = 4500 cP [(1 PaĿs)/(1000 cP)] = 4.50 PaĿs rá = 4.50 Pa-s [(1 lbĿs/ft2)/(47.88 Pa-sj] = 0.0940 lbĿs/ft2

16 Chapter 2

2.76 t = 40ÜC = 104ÜF. From Fig. 2.13, A= 1.00. At lOOÜF, v= 52614.632 = 113.6 mmvs. At 176ÜF (80ÜC): v= 1.000(113.6) = 113.6 mmvs.

2.75 t = 80ÜC = l 76ÜF. From Fig. 2.13, A= 1.005. At lOOÜF, v= 469014.632 = 1012.5 mmvs, At 176ÜF (80ÜC): v= 1.005(1012.5) = 1018 mm'/s.

2.74 From Fig 2.12, v= 37.5 mm2/s

2. 73 From Fig. 2.12, v= 12.5 mm2/s

2.72 v= 438/4.632 = 94.6 mmvs

2.71 v= 625014.632 = 1349 mm2/s

2.70 From Fig, 2.13, A= 1.006. At lOOÜF, v= 4.632(205) = 949.6 SUS. At 190ÜF, v= 1.006(949.6) = 955 SUS

2.69 From Fig. 2.13, A= 0.996. At lOOÜF, v= 4.632(153) = 708.7 SUS. At 40ÜF, v= 0.996(708.7) = 706 SUS

2.68 v= 4.632(244) = 1130 SUS

2.67 V= 4.632(188) = 871 SUS

2.66 From Fig 2.12, kinematic viscosity = 257 SUS


Problem 2.78 (See Problem 2.77 for method.) SAE Viscosity Grades - Automotive Gear Lubricants

Kinematic Viscositv at 100 deo C (mm2/s) sus

SAE No. Min Max Min Max 70W 4.1 --- 39.8 --- 75W 4.1 --- 39.8 --- 80W 7.0 --- 49.1 --- 85W 11.0 --- 62.8 --- 80 7.0 11.0 49.1 62.8 85 11.0 13.5 62.8 72.1 90 13.5 24.0 72.1 115.8 140 24.0 41.0 115.8 192.6 250 41.0 --- 192.6 ---

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.

Example: Given minimum kinematic viscosity = 21.9 mm2/s 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

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)

SAE Viscosity Grades - Engine Oils Kinematic Viscosity at 100 deg C (mm2/s) sus

SAE No. Min Max Min Max ow 3.8 --- 38.9 --- 5W 3.8 --- 38.9 --- 10W 4.1 _,.._ 39.8 --- 15W 5.6 --- 44.6 --- 20W 5.6 --- 44.6 --- 25W 9.3 --- 56.8 --- 20 5.6 9.3 44.6 56.8 30 9.3 12.5 56.8 68.3 40 12.5 16.3 68.3 83.2 50 16.3 21.9 83.2 106.6 60 21.9 26.1 106.6 125.1

Problem 2.77

Kinematic Viscosity Conversions

Chapter 2 18

Note: Method used is same as for Problem 2.77. 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.

ISO Viscosity Grades Kinematic Viscositv at 40 dec C

(mm2/s) sus ISOVG Min Nom Max Min Nom Max 2 1.98 2.2 2.40 32.5 33.3 34.0 3 2.88 3.2 3.52 35.6 36.6 37.6 5 4.14 4.6 5.06 39.6 41.1 42.6 7 6.12 6.8 7.48 46.0 48.1 50.3 10 9.00 10 11.0 55.4 58.8 62.4 15 13.5 15 16.5 71.6 77.4 83.4 22 19.8 22 24.2 97.0 106.3 115.9 32 28.8 32 35.2 136.2 150.5 164.9 46 41.4 46 50.6 193.1 214 235 68 61.2 68 74.8 284 315 347 100 90.0 100 110 417 463 510 150 135 150 165 625 695 764 220 198 220 242 917 1019 1121 320 288 320 352 1334 1482 1630 460 414 460 506 1918 2131 2344 680 612 680 748 2835 3150 3465 1000 900 1000 1100 4169 4632 5095 1500 1350 1500 1650 6253 6948 7643 2200 1980 2200 2420 9171 10190 11209 3200 2880 3200 3520 13340 14822 16305

Problem 2.79

Kinematic Viscosity Conversions

19 Pressure Measurement

3.20 Pabs = 128 + 98 = 226 kPa(abs)

3.19 Pabs = 284 + 100 = 384 kPa(abs)

3.18 pgage= 101-104=-3kPa(gage)

3.17 Pgage = 74 - 97 = -23 kPa(gage)

3.16 Pgage = 30 - 100 = -70 kPa(gage)

3.15 Pgage = 157 - 101=56 kPa(gage)

3.14 Pgage = 583 - 103 = 480 kPa(gage)

3 .13 Zero gage pressure.

3.12 At 13,500 ft, Patm = 8.84 psia; from App. E by interpolation.

3.11 At 4000 ft, Patm = 12.7 psia; from App. E by interpolation.

3 .1 O 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.

3.9 True.

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. 7 False. Atmospheric pressure varies with altitude and with weather conditions.

3.6 True.

3.5 Pabs = Pgage + Patm

3 .4 Atmospheric pressure is the absolute pressure in the local area.

3 .3 Gage pressure is measured relative to atmospheric pressure.

3 .2 Absolute pressure is measured relative to a perfect vacuum.

3.1 Pressure = force/area; p = FIA

Absolute and gage pressure

CHAPTER THREE

PRESSURE MEASUREMENT

Chapter 3

p = yh = (10.79 kN/m3)(3.0 m) = 32.37 kN/m2 = 32.37 k.Pa(gage)

p = yh = (10.79 kN/m3)(12.0m)=129.5 k.Pa(gage)

62.4 lb 1 ft2 . p = rh = x 50.0 ft x . 2 = 21.67 ps1g ft3 144 m

h 64Ŀ00 lb 12 50 f 1 ft2 5 56 . p = r = x . t x = . ps1g ft3 144 in2

h= p = 52.75kNm3 = 6.70 m r m2 7.87kN

p = yh = (sg)y,.h: sg = plywh) 1.820 lb ft3 144 in2

sg = in2(62.4 lb)(4.0 ft) x ft2 = 1.05

p = yh = 1.08(9.81 kN/m3)(0.550 m) = 5.83 kN/m2 = 5.83 k.Pa(gage)

20

3.40

3.39

3.38

3.37

3.36

3.35

3.34

Pressure-Elevation Relationship

3.33 Pabs = -12.5 + 14.4 = 1.9 psia

3.32 Pabs = -4.3+14.7=10.4 psia

3.31 Pabs = 0.6 + 14.7 = 15.3 psia

3.30 Pabs = 18.5 + 14.2 = 32.7 psia

3.29 Pabs = 41.2 + 14.5 = 55.7 psia

3.28 Pgage = 14.7 - 15.1 = -0.4 psig

3.27 Pgage = 10.8 - 14.0 = -3.2 psig

3.26 pgage = 4.3 - 14.6 = -10.3 psig

3.25 Pgage = 22.8 - 14.7 = 8.1 psig

3.24 Pgage = 84.5 - 14.9 = 69.6 psig

3.23 Pabs = -86 + 99 = 13 k.Pa(abs)

3.22 Pabs = -29.6 + 101.3 = 71.7 kPa(abs)

3.21 Pabs = 4.1+101.3=105.4 k.Pa(abs)


o 3.54 pi\m + Ym(.457 m) -yw(l.381 m) -ye; (0.50 m) = Pair

Pair = (13.54)(9.81 kN/m3)(.457 m)- (9.81 kN/m3)(1.381 m) - (.68)(9.81)(.50) Pair = (60.70 - 13.55 - 3.34) kN/m2 = 43.81 kPa(gage)

3.53 p = yh = (10.0 kN/m3)(1l.0x103 m) = 110 x 103kN/m2=110 MPa

3.51 p = yh = (l.80)(9.81 kN/m3)(4.0 m) = 70.6 kN/m2 = 70.6 kPa(gage)

3.52 p = yh = (0.89)(62.4 lb/ft3)(32.0 ft)(l ft2/144in2)=12.34 psig

=4.47m

O+ Yohl + Ywh2 = Pbot; but h, = 18.0 h2 Yo(18 - h2) + Ywh2 = Pbot 18yo - Yoh2 + Ywh2 = Psç = h2(Yw -yo)+ l 8ya h _ Pbot l8y0=158kN/m2 -(18m)(.86)(9.81kN/m3) 2- Yw-Ya [9.81-(0.86)(9.81)]kN/m3

3.50

3.49 O+ Yoho + Ywhw = Pbot = Pbot - y,A = 125.3 kN /m2 -(0.86)(9.81 kN /m3)(6.90 m) = 6 84 m h; r ; 9.8lkN/m3 Å

3.48 O+ Yoho + Ywhw = Pbot = pb01-ywhw _ 52.3kN/m2-(9.81kN/m3)(2.80m) =2.94m

ha r, (0.86)(9.81 kN /m")

3 47 h _ . h _ Pbot - Ptop Ā Ptop+yo -Pbot Ŀ - --~ Yo

h=(35.5-30.0)lb ft3 xl44in2 =l3.36ft in2(0.95)(62.4 lb) ft2

p = -10.8 psig + 2.57 psi= --8.23 psig

p = -10.8 psig + yh = -10.8 psig + (0.95) (62.4 lb) x 6.25 ft x 1 ~2 2 ft3 144 m

p = 50.0 psig + yh = 50.0 psig + 11.73 psi= 61.73 psig (See 3.44 for yh = 11.73 psig)

Patm = 24.77 kPa(abs) By interpolation - App. E: Ym = (13.54)9.81 kN/m3 Ym = 132.8 kN/m3 PB = Patm + yh = 24.77 kPa + (132.8 kN/m3)(0.325 m) = 67.93 kPa(abs)

p = yh = (0.95)(62.4 lb/ft3)(28.5 ft)(l ft2/144 in2) = 11.73 psig

pá = yh = (1.15)(9.81 kN/m3)(0.375 m) = 4.23 kPa(gage)

Pair = PA -yo(64 in)= 180 psig- (0.9)(62.4 lb/ft3)(64 in)(l ft3/l 728 irr') Pair = 180 psig - 2.08 psi= 177.9 psig

3.46

3.45

3.44

3.43

3.42

3.41

22 Chapter 3

3.68 PB + Yw(6 in)+ Ym(6 in) -yw(lO in)+ Ym(8 in) -yo(6 in)= PA PA =Pe = Ym04 in) -yw(4 in) -yo(6 in) - = (13.54)x 62.4 lb x(l4in x 1 ft3 - (62.4)(4) - (.9)(62.4)(6)

PA PB ft3 ) 1728 in ' 1728 1728 PA -pu= (6.85 -0.14 -0.195) psi= 6.51 psi

o 3.67 rltm + Ym(.475 m) -yw(.30 m) + Ym(.25 m) -yo(.375 m) = PA

PA = Ym(.725 m) -yw(.30 m) -yo(.375 m) PA = (13.54)(9.81 kN/m3)(.725 m) -(9.81)(.30) -(.90)(9.81)(.375) PA = (96.30 -2.94 -3.3 l)kPa = 90.05 kPa(gage)

3.66 PB + Yw(.15 m) + Ym(0.75 m) -yo(0.60 m) = PA PA =Pè = (9.81 kN/m3)(0.15 m) + (13.54)(9.81)(0.75) -(0.86)(9.81)(0.60) PA =Pe " (1.47 + 99.62 -5.06)kPa = 96.03 kPa

3.65 PB + Yo(.15 m) + Ym(.75 m) -yw(.50 m) = PA PA =Pè = (.90)(9.81 kN/m3)(.15 m) + (13.54)(9.81)(.75) -(9.81)(0.50) PA =Pè = (1.32 + 99.62 -4.91)kPa = 96.03 kPa

3.64 PB -yw(33 in)+ Ya(8 in)+ yw(13 in)= PA

(8 . ) (20 . ) (.85)(62.4) lb 8. 1 ft3 (62.4)(20) PA-pB=Yo m -yw m = X mx . ft3 1728 m3 1728

PA -pu= 0.246 psi -0.722 psi= -0.477 psi

3.63 PA + Yo(13 in)+ Yw(9 in) - Yo(32 in)= PB

(9 . ) (19. ) 62.4 lb 9. 1 ft3 (0.85)(62.4)(19) PB - PA = Yw m -yÜ m = ft3 x m x 1728 in ' 1728

Pu - PA = 0.325 psi -0.583 psi= -0.258 psi

o 3.62 P~m -ym(.075 m)-yw(0.10 m) = PA

PA = -(13.54)(9.81 kN/m3)(0.075 m)-(9.81)(0.10) =-10.94 kPa(gage)

Manometers (See text for answers to 3.57 to 3.61.)

3.56 Pbot = Pair + Yoho + Ywhw = 200 kPa + [(0.80)(9.81)(1.5) + (9.81)(2.6)]kN/m2

Pbot = 200 + 11.77 + 25.51=237.3 kPa(gage)

3.55 Pbot = -34.0 kPa + v.h; + Ywhw = -34.0 kPa + (0.85)(9.81 kN/m3)(0.50 m) + (9.81 kN/m3)(0.75 m)

Pbot = -34.0 kPa + 4.17 + 7.36 = -22.47 kPa(gage)


3.81 101.3 kPa ~ 760 mm ofMercury (See Ex. Prob. 3.11)

Ah= -85 mm x 5200 ft x 3048 m = -134.7 mm 1000 m 1 ft

h = 760 -134.7 = 625.3 mm kN

Patm = Ymh = 133.3 -3 X 0.6253 m = 83.35 kPa m

3.80 Ah= - l .O in of Mercury x 1250 ft = -1.25 in 1000 ft

3.79 The vapor pressure above the mercury column and the specific weight ofthe mercury change.

3.78 h = 760 mm See Example Problem 3.11

3. 77 h = 29.29 in See Example Problem 3 .13

very large (10.34 m) h= Patm = 14.7lb ft3X144 inz =33.92 ft r w in262.4 lb ft2

3.76

3.75 The height of the mercury column is convenient.

3.74 See Fig. 3.14 and Section 3.7.

3. 73 A barometer measures atmospheric pressure.

Barometers

o 3.72 a. p;(m + Ym(.815 m) -yw(.60 m) = PA

PA = (13.54)(9.81 kN/m3)(0.815 m) -(9.81)(.60) = 102.4 kPa(gage)

b. Patm = Ymh = (13.54)(9.81)(.737) = 97.89 kPa PA = 102.4 + 97.89 = 200.3 kP(abs)

o Pltm + YGFh = PA: h = L sin 15Ü = 0.115 m sin 15Ü = 0.0298 m PA = (0.87)(9.81 kN/m3)(0.0298 m) = 0.254 kPa(gage)

3.71

. ) o 62.4 lb . 1 ft3 o Patm + Yw(6.8 m = PA = + X 6.8 in X Å 3 = .246 psi ft3 1728 m

3.70

1 ft2 (.90)(62.4)(3) X ------ 144 in2 144

PB -yw(2 ft) -yo(3 ft) + Yw(l l ft) = PA 62.4 lb

PA =Pè = yw(9 ft) -y0(3 ft) = 3 X 9 ft ft

PA =Pè." 3.90 psi -1.17 psi= 2.73 psi

3.69

Chapter 3 24

Pressure Gages and Transducers (See text for answers to 3.94 to 3.97.)

3.93 p = 115 inWC = 115 inH20 (1.0 psi/27.68 inH20) = 4.15 psi p = 115 inH20(249.l Pa/1.0 inH20) = 28 646 Pa = 28.6 kPa

3.92 p = 12.4 inWC = 12.4 inH20 (1.0 psi/27.68 inH20) ""0.448 psi p = 12.4 inH20 (249.1 Pa/1.0 inH20) = 3089 Pa = 3.09 kPa

3.91 p = -12.6 psig (2.036 inHg/psi) = -25.7 inHg

3.90 p = -68.2 kPa (1000 Pa/kPa)(l.O mmHg/133.3 Pa) = -512 mmHg

3.89 p = 21.6 mmHg (133.3 Pa/1.0 mmlIg) = 2879 Pa = 2.88 kPa p = 21.6 mmHg (1.0 psi/57.71mmHg)=0.418 psi

3.88 p = 3.24 mmHg (133.3 Pa/1.0 mmHg) = 431.9 Pa p = 3.24 mmHg (1.0 psi/57.71mmHg)=0.0627 psi

3.87 p = -3.68 inH20 (1.0 psi/27.68 inH20) = -0.133 psi p = -3.68 inH20 (249.1 Pa/1.0 inH20) = -917 Pa

3.86 p = 5.37 inH20 (1.0 psi/27.68 inH20) = 0.194 psi p = 5.37 inH20 (249.1 Pa/1.0inH20)=1338Pa=134 kPa

Expressing Pressures as the Height of a Column of Liquid

kN Patm = Ymh = 133.3 -3 x 0.745 m = 99.3 kPa(abs)

m 3.85

h = Patm = 14.2 lb X ft3 x 1728 in3 = 28.91 in r m in" 848.7 lb ft3 3.84

848.7 lb . 1 fr' Patm = Ymh = X 30.65 m X . 3 = 15.05 psia ft3 1728 m

3.83

848.7 lb . 1 ft' Patm = Ymh = x 28.6 m X . 3 = 14.05 psia ft3 17281n

3.82

25 Forces Dueto Static Fluids

4.9 F= p-A; A= n(0.75 in)2/4 = 0.442 in2

= h = 844Ŀ9 lb x 28.0 in x 1 ft3 = 13.69 lb/in2 P Ym ft3 1728 irr' F= (13.76 lb/in2)(0.442 irr') = 6.05 lb

4.8 F=pĀA;A=24x 18in2=432in2

= h = 56Ŀ78}bX12.3 ft X l ft2 = 4.85 lb/in2 p YA ft3 144 in2 F = (4.85 lb/in2)(432 in2) = 2095 lb

Forces on horizontal flat surfaces under liquids

4.6 F = p-A; A= n(0.050 m)2/4 = 1.963 x 10-3 m2 F = (20.5 x 106 N/m2)(1.963 x 10-3 nr') = 40.25 x 103 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 x 103 N = 22.0 kN

4.5 F = p-A; A= 7r(0.030 m )2 = 7.07 x 10-4 m2 4

F = (3.50 x 106 N/m2)(7.07 x 10-4 nr') = 2.47 x 103 N = 2.47 kN

4.3 F = 11p-A; A = 36 x 80 in2 = 2880 in2 11 = h = 62.4 lb x 1.20 in x 1 ft3 = 0.0433 lb/in2 .p y11 ft3 1728 in" F = (0.0433 lb/in2)(2880 irr') = 125 lb

4.4 F = p-A; A = 0.9396 ft2 (App. F) F= (325 lb/in2)(0.9396 ft2)(144 in2/ft2) = 43973 lb

4.1 _ . _ . _ 7r(l2 in )2 _ . 2 F-11pĀA, where 11p - Patm - Pinside, A - - 113.1 m

4

h 844.9lb 3 . 1 fr' . Patm=Ym =

3 X 0.5mx . 3 = 14.91 psi ft 1728 m

F = (14.91 - 0.12)lb/in2 x 113.1 in2 = 1673 lb

4.2 F= p-A = (14.4 lb/in2)(n(30 in2)/4) = 10180 lb

Forces dueto gas pressure

CHAPTER FOUR

FORCES DUETO STATIC FLUIDS

Chapter 4

T 15.5 ft

l

Length of sloped side = L = 15 .5 ft/sin 60Á = 17.90 ft

A= (17.90ft)(l1.6 ft) = 207.6 ft2 FR = y(h/2)(A) = 78.50 lb X 15.5 ft X 207.6 ft

ft3 2 FR = 126300 lb hp = 2/3 h = (2/3 )(15 .5 ft) = 10.33 ft Lp = 213 L = (2/3)(17.90 ft) = 11.93 ft

Part (b) l:Mhinge = FR(l.2 ft) - Fn(4.0 ft) Fu= FR(I.214.0) = 3235 lb(0.30) = 970.5 lb on two latches On each latch: Fà = (970.5 lb )/2 = 485 lb

26

4.15

4.14 Fn = Yw(h/2)A = (62.4 lb/ft3)(1.8 ft)(8.0 ft)(3.6 ft) = 3235 lb

Fn acts J_ wall, 1.20 ft from bottom of gate

Forces on rectangular walls

1 Fp = '1.pĀA; A = n(0.60 m)218 + (0.80 m)(0.60 m) + - (.60 m)(0.30 m)

2 A= 0.711 m": Assume std. atmosphere above water. Pw = Patm + Yswh = 101.3 kPa + (10.10 kN/m3)(175 m) = 1869 kPa Sp = 1869 kPa- 100 kPa = 1769 kPa Fp = (1769 x 103 N/m2)(0.711 m2) = 1.257 x 106 N = 1.26 MN

Fa= paĀA; A= 2.0 x 1.2 m2 = 2.4 m2 Pa = 200 kPa + Yo(l .5 m) + Yw(2.6 m) Pè = 200 kPa + (0.80)(9.81 kN/m3)(1.5 m) + (9.81)(2.6) = 237.3 kPa Fa= (237.3 x 103 N/m2)(2.4 m2) = 569 x 103 N = 569 kN

Fa= paĀA; A= 1.2 x 1.8 m2 = 2.16 m2 Pa = Pair + Yo(0.50 m) + Yw(0.75 m) Pa = 52 kPa + (0.85)(9.81 kN/m3)(0.5 m) + (9.81)(0.75) = 63.5 kPa Fa= (63.5 x 103 N/m2)(2.16 m2) = 137 x 103N=137 kN

Force on valve = pĀA; A= n(0.095 m)2/4 = 7.088 x 10-3 m2 9.81 kN 2 p=v.h=

3 x l.80m= 17.66kN/m

m F = ( 17 .66 x 103 N/m2)(7 .088 x 10-3 m2) = 125 N Acts at center l:Mhinge =O= (125 N)(47.5 mm) -Fo(65 mm) F0 = 5946 NĿmm/65 mm= 91.5 N = Opening force

4.13

4.12

4.11

4.10

27

~4 hà Å.825 .A5

4\\Ŀ 1 .375 .75

1.5

)< R

+ .30

t= -

L= dsin 45Ü =1.98 m

1.4mÅh

-....,,.,,...- Å t--.r--------""-- M ~

12.0ft hp = 8.0 ft

__..__Jjá. 4.0ft

Forces Due to Static Fluids

4.19 he= 0.825 m = 825 mm Le = he 825 mm = 953 mm

cos 30Á cos 30Á

A= n(450 mm)2 = 1.59 x 105 mm2 [.159 m2] 4

FR = yhcA = (0.85)(9.81 kN/m3)(.825 m)(0.159 m2)

FR= 1.09 kN

l.= nD4 = n(450)4 = 2.013 x 109 mm4 e 64 64

L -L =ĉ= 2.013xl09 = 13.3 mm p e Le A (953)(1.59X105)

Lp =Le+ 13.3 mm= 953 + 13.3 = 966 mm

- 10 2 Å 2 Area= AB Ā 3.5 ft= - Ŀ 3.5 =2.92 ft (420 m) 12

FR = yhcA = 0.93(62.4 lb/ft3)(1.50 ft)(2.92 ft2) FR=254 lb 1 = BHJ = (42)(10)3 in4 =3500 in" e 12 12

L -L =ĉ= 3500 in4 = 0.37 in p e LCA (22.5 in)(420 in") Le= Le+ 0.37 in= 22.5 in+ 0.37 in= 22.87 in

4.18 Centroid is at midpoint of AB he= 14 in + 4 in = 18 in= 1.50 ft AB = 10.0 in [3-4-5 triangle]

Le=~Ā L =h. Å ~=22.5 in d 4' e e 4 e

Forces on submerged plane areas

4.17 FR = Yo(h/2)A 9.8lkN = (0.86) 3 x 0.7 m x (1.98)(4.0) m2 m

FR=46.8 kN hp = 213 h = (2/3)(1.4 m) = 0.933 m LP = 213 L = (2/3)(1.98 m) = 1.32 m

FR = 89850 lb hp = 213 h = 2/3(12 ft) = 8.0 ft Moment = FRĀ4 ft = 359400 lb-ft

FR = y(h/2)A = 62.4 lb X 6 ft X (12 ft)(20 ft) ft3 4.16

Chapter4

8.0 tt

28

4.22 Le= a+ 1.0 ft = 3.0/cos 30Á + 1.0 = 4.464 ft he= Le COS 30Ü = 3.866 ft A= n(.5 ft)2/4 = 0.196 ft2

FR = yhlA = (0.90)(62.4 lb/ft3)(3.866 ft)(0.196 ft2)

FR = 42.6 lb 1 = 1²D4 = 1²(0.5)2 = 0.00307 ft4 e 64 64

L - L = _!_:__ = 0.00307 = 0.00351 ft p e LcA (4.46)(0.196) LP =Le+ 0.00351 = 4.468 ft

1 GÅ1.5ft

Le= a+ 1.5 + z = 8.0/cos 45Á + 1.5 + z = 13.50 ft

he= Le COS 45Á = 9.55 ft FR = yhlA = (62.4 lb/ft3)(9.55 ft)(3.0 ft2)

= 1787 lb A= H(G + B)/2 = 1.5(4.0)/2 = 3.0 ft2 1 = H3(Gz +4GB+Bz) = 0.551 ft4 e 36(G+B)

Lp - Le= Ic!LcA = 0.551/(13.50)(3.0) = 0.0136 ft = 0.163 in LP =Le+ 0.0136 ft = 13.50 + 0.0136 = 13.51 ft

zÅHĿyÅ0.888ft

1 S..App.J -..-.;,..------i-- 4.21

4.20 he= 3.0 m; Le= hJcos 30Ü = 3.464 m A= 1²D2 = 1²(2.4 m )2 = 4.524 m2

4 4 != 1rD4 = 1²(2.4m)4 = 1.629 m"

64 64 FR = yheA = (1.10)(9.81 kN/m3)(3.0 m)(4.524 m2) FR = 146.5 kN L -L =ĉ= 1.629 m4 = 0.104 m p e LCA (3.464 m)(4.524 m")

=104mm Lp =Li.+ 0.104 m = 3.464 + 0.104 = 3.568 m

29 Forces Due to Static Fluids

4.25 Le= a+ 0.50 m = 0.76 micos 20Ü + 0.50 Le= 1 .309 m he= Le cos 20Ü = 1.230 m A = (1.00)(0.60) = 0.60 m2 FR = yheA = (0.80)(9.81 kN/m3)(1.23 m)(0.60 m2) FR= 5.79 kN

1 = BH3 = (0.60)(1.00)3 m4 = 0_05 m" e 12 12

- I e - 0.05 - O 0637 = 63 7 L - Le - - - - Å m . mm p LCA (1.309)(0.60) Lp =Le+ 0.0637 = 1.372 m

a= 3.0/cos 45Á = 4.243 ft Le= 5 +a= 9.243 ft

he= Le COS 45Á = 6.536 ft

ACÅ0.450m yÅ AC cos 40Å Å 0.346 m he Å 1.20 m-yè 0.855 m

1.2m he

4.24 A= rrD2/4 = n(2.0)2/4 = 3.142 ft2 FR =yh(A = 62.4 Jb/ft3 X 6.536 ft X 3.142 ft2

FR = 1281 lb 1 = JTD4 = 0.785 ft4 e 64

L -L =!.s_= 0.785 p e LcA (9.243)(3.142) Lp - Le= 0.027 ft = [0.325 in] Lp = 9.270 ft

4.23 Le= helcos 40Á = 0.855/cos 40Ü = l.116m

A= (.300)2 + JT(.30)2 = 0.1607 m2 4

FR = yh(A = (0.90)(9.81 kN/m3)(0.855 m)(A)

FR= 1.213 kN (0.300)4 JT(.300)4

1= +--- e 12 64 le= 0.001073 m"

I 0.001073 L -L =_e =----- P e LcA (1.116)(0.1607) Lp - Le= .00598 m = 5.98 mm Lp = 1.122 m

Chapter4 30

a= 0.80 m/sin 70Á = 0.851 m Le= a+ 0.5 +y= 0.851+0.50 + 0.318 = 1.669 m he= Le sin 70Á = 1.569 m rè =vh.A = (0.88)(9.81 kN/m3)(1.569 m)(0.884 m2)

FR= 11.97 kN

L -L = ĉ 0.0347m4 =0.0235m P e LcA (1.669 m)(0.884 m")

=23.5 mm Lp =Le+ 0.0235 m = 1.669 m + 0.0235 m = 1.693 m

e 1Å0.212DÅ0.318m

ái....Åi----- 1.50 m 4.27

4.26 a = ~ Ŀ 20 = 25 in 4

Le= a+ 25 = 50.0 in= 4.167 ft

he= Ñ -Làè 40.0 in= 3.333 ft 5

A= (8)(50) = 400 in\l ft2/144 in2) = 2.778 ft2

Fn=YhcA = (1.43)(62.4 lb/ft3)(3.333 ft)(2.778 ft2)

FR= 826 lb

1 = BH3 = (8)(50)3 = 83 333 in" e 12 12 '

L -L = ĉ= 83333 in4 p e LCA (50.0 in)(400 in")

= 4.167 in Lp =Li+ 4.167 in= 54.167 in (4.514 ft)


A= lBH= (1)(30)(20) = 300 in2

1 = BH3 = 30(20)3 = 6667 in" e 36 36 a= 18 in/cos 50Á = 28.0 in Le= a+ 6 + b = 28.0 + 6.0 + 13.33 Le= 47.34 in he= Le COS 50Á = 30.43 in FR=yhcA FR = 62.4 lb/ft3 x 30.43 in x 300 in2

ft3 X 3 = 329.6 lb 1728 in

L - L.= .!s._= 6667 in4 = 0.469 in P e LcA (47.34in)(300 in") Lp =Le+ 0.469 in= 47.34 + 0.469 = 47.81 in

.Y- 20l3 Å 6.87 In

:=.,. ==-30-ln====ĿI T e ----.~~ ~--'""...,.__...__

f

a= 10 in/cos 30Á = 11.55 in Le= a+ 8 + y = 11.55 + 8.0 + 8.48 = 28.03 in he= Le COS 30Á = 24.27 in [2.023 ft] FR =vh.A

= (l.10)(62.4 lb/ft3)(2.023 ft) (628Ŀ3 in~)(l ft2) 144 in '

FR= 605.8 Ib

L -L =.!s._= 17 562in4 =0.997 p e LCA (28.03 in)(628.3 in")

= 7.135 in Lp =Le+ 7.135 in= 29.03 in

le = (6.86x10"3)D4 = (6.86x10"3)(40)4 = 17 562 in4

~1n----1

4.29

4.28

Chapter4 32

T L ~ 2.27: IO' m4

L _L.= is= 2Ŀ276x10-4 m4 = 0.0253 m = 25.3 mm P e LA (0.0718m)(0.1253 m") Lp =Lc+ 25.3 mm= 97.1 mm FR = yhcA = (0.67)(9.81 kN/m3)(0.0718 m)(0.1253 m2) = 0.059 kN = 59 N

A(m2) j²(m) Aj²(m3) l;(m4) h(m) Ah2

1 Rect. 0.0900 0.075 6.750 X 10-3 } .688 X 10-4 0.00324 9.4 X 10-7

2 Semicirc. 0.0353 0.0636 2.245 X 10-3 5.557 X 10-5 0.00816 2.35 X 10-6

0.1253 9.000 X 10-3 2.243 X 10-4 3.30 X 10-6

Li=hà> Y =0.0718m=71.8mm

y= L:Ay = 9.0xl0-3 m3 Ar 0.1253 m"

= 0.0718 m

4.32

4.31 (See Prob. 4.30) he= Le= 0.150 m FR = yhcA = (0.67)(9.81)(0.150)(0.2507) = 0.247 kN = 247 N

Lp L.=~= 0.001748 m4 = 0.0465 m = 46.5 mm e LcA (0.150 m)(0.2507 m")

Lp =Le+ 46.5 mm = 150 mm+ 46.5 mm = 196.5 mm

F11 = yhcA F11 = (0.67)(9.81 kN/m3)(0.525 m)(0.2507 rrr') F11 = 0.865 kN = 865 N le= (0.60)(0.30)3 + JT(0.30}4 = 0.001748 m"

12 64

L-L=~ 0.001748m4 =0.0133m=13.3mm P e LcA (0.525 m)(0.2507 m") Lp =Ls.+ 13.3 mm= 525 + 13.3 = 538 mm

f o.30m

1 f.

.50m

1 o.37Sm

1

a

4.30 he= Le = 0.375 + 0.150 = 0.525 m

A= (0.60)(0.30) + n(0.30)214 A= 0.2507 m2

33

B Å 1.2 m + 4.6 ml!an ea-- 3.858 m

Vlew1DWWd endwal

sg Å 1.10

EndVIN

--- t t hĿ8.0ft .L ~

Forces Due to Static Fluids

= H(G+2B)= 4.6(1.2+2(3.856)) =2_703m y 3(G + B) 3(1.2 + 3.856) hc=H- y =4.6-2.703 = l.897m=Lc FR = yh"A Acts J_ page A= H(G + B) == 4.6(1.2 + 3.856) = l l.63 m2

2 2 FR = (1.10)(9.81 kN/m2)(1.897 m)(l l.63 m2) =238 kN

L = H3(G2 + 4GB + B2) = 18.62 m" e 36(G+B)

L -L = _!_s_= 18Ŀ62 m4 = 0.844 m P e LcA (1.897m)(l1.63 m") Lp =Le+ 0.844 m = 1.897 + 0.844 = 2.741 m

4.36

4.35 Rectangular Wall AB = 8.0 ft/sin 60Ü = 9.237 ft A= AB X 15 ft = 138.6 ft2 FR = y(h/2)A F11 = 62.4 lb/ft3 X 4.0 ft X 138.6 ft2 = 34586 lb

2 - 2 i, = 3(AB) = 3(9.237 ft) = 6.158 ft

L = H3(G2 + 4GB + B2) e 36(G+B)

= 83(5.382 +4(5.38)(10)+102] = 318.3 ft4 36(5.38+10)

L-L=_!_s_= 318Ŀ3 =1.176ft p l LCA (4.40)(61.52) Lp =Le+ 1.176 = 5.576 ft

4.34 G = 1 O.O ft- 8.0 ft/tan 60Ü == 5.381 ft A= H(G + B) = 61.52 ft2

2 - = H ( G + 2B) = 4.40 ft = h = L. Y 3(G+B) e e

FR == yhlA F11 = 62.4 lb/ft3 X 4.40 ft X 61.52 ft2 = 16894 lb

4.33 Rectangular Wall F11 = y(h/2)A = 62.4 lb/ft3 X 4.0 ft X (8.0 ft)(l5.0 ft) = 29950 lb 2 2 Lp= -h=-(8.0ft) =5.333ft 3 3

Chapter4

Bad<Wall --

T 1..50m

34

Vertical back wall is rectangular FR = yhcA = y(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

Fè = (8.829 kN/m3)(0.75 m)(l.80 m2) = 11.92 kN 2 2

Lp= 3xh=J x L50m=l.00m

4.39

FR = yhcA = (0.90)(9.81 kN/m3)(0.846 m)(l .348 m2) = 10.07 kN le= 11 + A1y~+ ]z + ĆiY~ + l3 + A3y;

le= (0.697)(1.5)3 + (1.046)(0.096)2 + (0.503)(0.30)3 + (0.151)(0.504)2 12 12

+ (0.503)(0.60)3 + (0.151)(0.154)2 36

le= 0.1960 + 0.0096 + 0.0011+0.0384 + 0.0030 + 0.0036 = 0.2518 m' 0.2518 Lp =Li+ = 0.846 m + m = 0.846 + 0.221=1.067 m

LCA (0.846)(1.348) e:._ L - L p e

T 4.8m

BlldcWd

f = Z::Ay = 0.882 m3 = 0.654 m IA 1.348 m2

he= 1.5 m - Y = 0.846 m =L;

A Ay - - y yá= y-Y

(.!) 1.046 0.75 0.784 0.096

0 0.151 0.15 0.023 -0.504 G) 0.151 0.50 0.075 -0.154

LA= 1.348 m2 LAY= 0.882 m3

ENDWALL 4.38

4.37 Rectangular Wall: FR = yhlA FR = (1.10)(9.81 kN/m3)(4.60/2)m(4.6)(3.0)m2

=343 kN FR acts 113 from bottom or 213 from surface

2 Lp = J (4.60 m) = 3.067 m


Oil side: F'áá0 = Yo( hof 2 )Ao Ćo = (2.00)(0.60) = 1.20 m2 F'áá0=(9.81)(0.9)(1.00)(1.20)=10,59 kN

2 Lr = - Ŀ 2.00 = 1.333 m o 3 IMs =O

= F'ááw (J.967) - F'áá0 (2.J 33) - FH(2.80) Fu= [(18.39)(1.967) - (10.59)(2.133)]/2.80 = 4.85 kN ~ LM H =O= F'ááw (0.833) - F'áá0 (0.667) - Fs{2.80) Fs = [(18.39)(0.833) - (10.59)(0.667)]/2.80 = 2.95 kN ~

G.80 á~ UIJ7 dw Å 1.25 ---~ 1

i 0.887 ------- __ ...__/), \H

Fs --4'1'-----~

1 2.00

FH_......,._J_hcĀ~ T H 1 4-00

2.00 1 2.33S

e:! i '+ 1,..___,.._ FR

0.331 1.llIT

s 1

---

4.41 Water side: FRw = Yw(hwf2)Aw .r, = (2.50)(0.60) = 1.50 m2 FRW = (9.81)(1.25)(1.50) = 18.39 kN

2 L1, = - Ŀ 2.50 = 1.667 m

w 3

4.40 he= Le= 4.00 ft; A= (4.00)(1.25) = 5.00 ft2 FR = yhcA = (62.4 lb/ft3)(4.00 ft)(5.00 ft2) = 1248 lb

le= BH/12 = (1.25)(4.00)3112 = 6.667 ft4 L _ L = .!s._= 6.667 ft4 p e LcA (4.00ft)(5.00ft2)

= 0.333 ft 2.Ms =O= Fáá(l .667) - F H(4.00) F = (1248)(1.667) = 520 lb H 4.00 2.MH =D= Fu(2.333) - Fc\{4.00) Fs = (1248)(2.333) = 728 lb

e 4

Chapter4

See Prob. 4.20 for data.

See Prob. 4.19 for data.

h = Pa =25.0kN x m3 =2.317m a YoD m2 (1.10)(9.81 kN) h.; =he+ ha= 3.000 m + 2.317 m = 5.317 m Lee= hcefcos 30Á = 5.317/cos 30Á = 6.140 m F11 = YoDhceA = (1.10)(9.81kN/m3)(5.3l7 m)(4.524 m2) = 259.6 kN

- le - 1.629 m" - o 0586 - 8 Lpe - Lee - -- - 2 - Å m - 5 .6 mm Lc,Ć (6.140 m)(4.524 m )

h = Pa 13.8kN x m3 = 1.655 m Û r ; m ' (0.85)(9.81 kN) h.; =h-+ ha= 0.825 m + 1.655 m = 2.480 m Lee= heelcos 30Á = 2.480 micos 30Á = 2.864 m F11 = y,AeA = (0.85)(9.81 kN/m3)(2.480 m)(0.159 m2) = 3.29 kN

Lpe - Lee= ~ = 2.013x109 mm4 = 4.42 mm LcA (2.864 mm)(l.59 x 105 mm")

36

4.44

4.43

Piezometric head

1:,Má{ =O= FR(5.128) Fc(5.00) Fe= (l20.llb)(5.t28in) = 123.2 lb=cableforce

5.00in

Sum moments about hinge at top of gate.

-.è. 5.128_)

--:z-- 4.42 he= 38 +y= 38 + 5 cos 30Á = 42.33 in

Le= h.lcoç 30Á = 48.88 in FR=yhcA A= n(I0)2/4 = 78.54 in2 F _ 62.4 lb 42.33 inĿ 78.54 in2 R - ft3 Å 1728 in3 /ft3

FR = 120.1 lb le= Jr(l0)4 = 490.9 in4

64

L -L =~ P e LA

e

L -L = 490.9in4 =O 12sĿ p e , Å lll (48.88 in)(78.54 in-)


,á, = _1 Fv _ _1 35.89 = 47 60 'f' tan FH =tan 32.74 .

_ Ć1X1 + ĆzX2 X=----

Ćá +Ćz A1 = (0.75)(1.85) = 1.388 m2 A2 = n"(0.75)214 = 0.442 m2 x = (1.388)(.375) + (0.442)(0.318) = 0.361 m

1.388 + 0.442 he= h, + s/2 = 1.85 + 0.75/2 = 2.225 m FH = yswhc = 9.81 kN/m3 x (0.75)(2.00)(2.225)m3

Fn= 32.74 kN 2 o 752 hp =he+ _s_ = 2.225 + Ŀ = 2.225 + 0.021=2.246 m

12hc 12(2.225)

FR= ~Fá; +F~ =.J35.892 +32.742 =48.58kN

.375 á-:e, R = 0.75 m, w = 2.00 m; Fv= yĀV= yAw

F,~ 9.81 kN/m' x [(1.85)(0.75)+ 7r(0~75)'] x 2.00 m'

= 35.89 kN

4.47

Forces on curved surfaces

ha = p Û = 4.0 lb x ft3 x 1728 in3 = 1OO.7 in= 8.392 ft 1 See Prob. 4.28. Y EG in 2 (1.10)(62.4 }b) ft3

h.; =hi+ ha= 24.27 in+ 100.7 in= 124.97 in [See Prob. 4.28] Lee= hcefcos 30Á = 144.3 in= 12.03 ft FR = YEahceA = (1.10)(62.4 lb/ft3)(124.97 in)(l ft/12 in)(628.3 in2)(1 ft2/144 in2) FR = 3119 lb L -L = ĉ= 17 562 in4 = 0.194 in pe ce LeeA (144.3 in)(628.3 in")

FR = YesheeĆ = (1.43)(62.4 lb/ft3)(7.368 ft)(2.778 fr') = 1826 lb 83333 in" = 1.885 ²n Lpe -Lee= 2

LcA (110.5 in)(400 in )

1 See Prob. 4.26. ha= Pa = 2.50 lb ft3 X 1728 in3 = 48.41 in= 4.034 ft

Yes in2(1.43)(62.4 lb) 1 ft3 h.; =he+ ha= 40.0 in+ 48.41 in= 88.41 in= 7.368 ft

Lee= hce ( %) = 88.41 in(%) = 110.5 in= 9.209 ft

4.46

4.45

Chapter4 38

FR = ~F~ + Fi =../674372+999252 = 120550 lb ,á,= _1 Fv _ _1 99925 = 56 OÜ 'f' tan F - tan 67437 .

H

. R sin(37.5Ü) x3 = b sm 37.5Á where b = 38.197 = 9.301 ft (from Machinery's Handbook)

37.5Á X3 = 9.301 sin 37.5Á = 5.662 ft X = AáXá + A2X2 + ~X3 = 6.738 ft

Ar

4.49 y1 = R sin 15Á = 3.882 ft s = R - y1 = 15.00 - 3.882 = 11.118 ft he = h +yá + s/2 = 10.0 + 3.882 + 5.559

he= 19.441 ft FH = yhcSW = (62.4)(19.441)(11.118)(5.00) Fn = 67,437 lb

2 s hp =he+ - = 19.441+0.530 ft 12hc

hp = 19.971 ft Fv=yV=yA1w Ćá = (14.489)(10.0) = 144.89 ft2

1 1 A2 = - (y1)(R cos 15Á) = - (3.882)(14.489)

2 2 = 28.12 ft2

A3=nR2Ā }2_ =n(15.0)2 75 360 360

= 147.26 ft2

Ar= A1 + A2 + A3 = 320.27 ft2 Fv= yA1w = (62.4)(320.27)(5.00) = 99,925 lb

Xá= 14.489/2 = 7.245 ft 2

X2 = (14.489) = 9.659 ft 3

4.48 Fv= yV= yAw Fv= (0.826)(9.81 kN/m3)(1.389 m2)(2.50 m) = 28.1 kN A= A1 + A2 = (0.62)(1.25) + n(l.25)2/8 A= 1.389 m2 F H = O because horiz. forces are balanced FR = Fv= 28.1 kN

l-1.25 '"'m----' 1


,/.- -i Fá, -1 927.2 - 7- 2Ü 'f'- tan -=tan -- - !:'l.

FH 245.3

Forces shown acUng on ftuid.

5.20 m = h

hp = 10.798 + 0.0050 m hp = 10.803 m Fv=yV=yAw A1 = (5.20)(3.00) = 15.60 m2

A2 = _!_ (3.00)(5.196) = 7.794 m2 2

Ć3 = nR2 Ā 30 = 1f(6Ŀ0)2 = 9.425 m2 360 12

Ar=A1 +A2+A3=32.819m2

Fv= yAw = (0.72)(9.81)(32.819)(4.00) = 927.2 kN

Xá= 3.00/2 = 1.500 ID

X2 = 2(3.00)/3 = 2.00 ID

. Rsin15Ü x3 = b sm 15Ü where b = 38.197 = 3.954 m (from Machinery's Handbook) 15

x3 = (3.954)sin 15Á = 1.023 m .X = A,x, + A2x2 + A3x3 = 1.482 m

Ar

Fu= ~ F~ + F;} = .J245.32 + 927 .22 = 959.1 kN

14-1Å----.Å I R sin 309 Å 3.00 m s = R- y= 6.00 - 5.196 = 0.804 m he= h +y+ s/2 = 5.20 + 5.196 + 0.402 he= 10.798 m FH = vswh; = (0.72)(9.81)(0.804)(4.00)(10.798) F9=245.3 kN

h = h + _i_ = 10.798 + 0Ŀ8042 p e 12hc 12(10.798)

4.51

4.50 Fv= YoV= y,Aw IĿ 7.SOft --l A = A1 + A2 = (7.50)(9.50) + n(7.50)2/8

f IT A= 93.34 ft2

Fv= y,Aw = (0.85)(62.4)(93.34)(3.50) Fv= 17328 lb =FR 9.50 F H = O because horiz. forces are balanced

+

Chapter 4

l uo 1

"" áx, h = 2.80 m

+

j1.2omj

Å. [ ~1~ -h-= ..... I.-80 m

_-1-_á ....___ RÅÅÅ1.20m F,, " ~~ jxl

40

Fv= yAw = (9.81)(3.669)(1.50) = 54.0 kN Xá = 1.20/2 = 0.60 m x2 = 0.2234R = 0.268 m [Machinery's Handbook] - Ć1X1 + ĆiX2 (3.36)(0.60) + (0.309)(0.268) X = =---------

Ć 3.669 x =0.572 m he= h + s/2 = 2.80 + 0.60 = 3.40 m FH = yswhc = (9.81)(1.20)(1.50)(3.40) = 60.0 kN

s2 1.202 h = h + - = 3 .40 + = 3.435 m p e 12hc 12(3 .40)

Fáá= ~Fá:+F~ =.J54.02+60.02 =80.7kN

d.- -1 Fv -1 54.0 - 42 OÜ r- tan -=tan - - Ŀ FH 60.0

A1 =(1.20)(2.80)=3.36 m2} 2 A=3.669 m Ći =R2-JrR214=0.309 m2

4.53

4.52 Fv=yAw A= A1 + A2 = (1.20)(2.80) + n(l .20)2/4 = 4.491 m2 Fv= (9.81 kN/m3)(4.491 m2)(1.50 m) = 66.1 kN x1 = 0.5(1.20) = 0.60 m; x2 = 0.424(1.20) = 0.509 m - Ć1X1 + ĆiX2 (3.36)(.60) + (1.13)(0.509) X= = -

AT 4.491 = 0.577 m

he= h + s/2 = 2.80 + 1.20/2 = 3.40 m F H = yswhc = (9 .81)(1.20)( 1.50)(3 .40) = 60.0 kN

h = h + L = 3.4o + ci.2Ü)2 = 3.435 m p e 12hc 12(3 .40)

Fáá = ~Fá: + F~ = ..)66.12 + 60.02 = 89.3 kN

,á, -1 Fv -1 66.1 47 8Ü r= tan -=tan - = Ŀ FH 60.0


4.58 See Prob. 4.57. w1= 10.2 lb Wc = YcV= (0.100 lb/in3)(282.7 in') = 28.27 lb Fnet = Wc - w1= 28.27 - 10.2=18.07 lb down

4.57 Net horizontal force = O From Section 4.11, net vertical force equals the weight of the displaced fluid acting upward and the weight of the cylinder acting downward. w1= YfVd = (62.4 lb/ft3)(0. l 64 ft3) = 10.2 lb

Vd=AĀL= 7rD2 ĀL= 7r(6.00in)2Ŀ10.0in=282.7in3x ft3 =0.164ft3 4 4 1728in3

Wc = YcV= (0.284 lb/in3)(282.7 irr') = 80.3 lb Net force on bottom = Wc -w1= 80.3 - 10.2 = 70.1 lb down

,.á,= _1 Fv _ _1 47.15 = 47 OÜ r tan FH - tan 44.00 .

4.56 (See Prob. 4.48)

E. De th=h = p = 4Ŀ65kN/m2 =0.574m q p Û y (0.826)(9.81 kN/m3) h-à = h, +b;è 0.62 m + 0.574 m = 1.194 m A = ( 1.194)(1.25) + n( 1.25)2 /8 = 2.106 m2 Fv= yAw = (0.826)(9.81 kN/m3)(2.106 m2)(2.50 m) = 42.66 kN = F11

(See Prob. 4.47)

E .De th=h = p= 7Ŀ5kN/m2 =0.765m q p Û y 9.81 kN/m3 Iu, = h, +Iu= 1.85 + 0.765 = 2.615 m; hce = h.: + s/2 = 2.615 + 0.75/2 = 2.990 m A 1 = (O. 75)(2.615) = 1.961 m2; A2 = 0.442 m2; Ar= 2.403 m2 Fv= yAw = (9.81)(2.403)(2.000) = 47.15 kN x = A1x1 + A2x2 = (1.961)(0.375) + (0.442)(0.318) = 0.365 m

AT 2.403 FH = yswhce = (9.81)(0.75)(2.00)(2.99) = 44.00 kN

s2 0.752 hpe-hce= --= =0.016m=16mm 12hce 12(2.99)

~---- Fáá = ~Fá] + F; = .J47.152 + 44.002 = 64.49 kN

hp = 67.6 in

A= 1²D2/8 = n(36)2/8 = 508.9 in2 = 3.534 ft2 Fv= yAw = (0.79)(62.4)(3.534)(5.0) = 871 lb x = 0.212D = 0.212(36 in)= 7.63 in he= h + s/2 = 48 + 3612 = 66.0 in= 5.50 ft FH = yswhc = (0.79)(62.4)(3.0)(5.0)(5.50) Fn= 4067 lb

s2 362 Å h =h = -=66+ -- =67.64m p e 12hc 12(66)

F11= ~Fv2+Fi =J8712+40672 =4159lb

,.á, -1 Fv -1 871 2 1Ü r= tan -=tan -- = 1 . Fái 4067

4.55

4.54

Chapter 4 42

Fnet = Wc - wá= 80.3 - 9.09 = 71.21 lb down

wá= yáVd = yáAdL TrD2 a 1 Ad= -- Ā - + -(2x)(2.0) = A1 + A2 4 360 2

Ad= Tr(6.0 in )2. 263.6 + (2.236)(2.0) = 25.18 in2 4 360

w1= yáAdL = 62.4 lb/ft3 Å 25.18 in2 Ŀ 10.0 inĿ 1 ft3. 3 = 9.09 lb 1728 m

eÅ slrr' (2.0f,1.0) Å '41.S- XÅ ReosÅÅ 3.0 coaÅ XÅ2.2381n 11 - 1 eo- + 2e - 263.S"

5.0 h

1

4.62 (See Prob. 4.57.) Wc = 80.3 lb Force ( downward) on upper part of cylinder = wt. ofvolume of cross-hatched volume. Force (upward) on lower part of cylinder = wt. of entire displaced volume plus that of cross- hatched volume. Then net force is wt. of displaced volume (upward).

4.61 (See Prob. 4.57.) Because the depth of the fluid <loes not affect the result, Fnet = 70.1 lb down. This is true as long as the fluid depth is greater than or equal to the diameter of the cylinder.

4.60 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.

4.59 See Prob. 4.57. w1= 10.2 lb Wc = YcV= (30.0 lb/ft3)(0.164 ft3) = 4.92 lb Fnet = Wc - wá= 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.


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 25Á = 47.32 in Le= 60 in+ x - y= 60 + 47.32 - 7.632 = (99.69 in)(l ft/12 in)= 8.308 ft he= Le sin 25Á = (8.308 ft)(sin 25Ü) = 3.511 ft A= JCD2/8 = .n(3.0 ft)2/8 = 3.534 ft2 FR = yh,A = (l .06)(62.4lb/ft3)(3.5l1 ft)(3.534 ft2) = 820 lb= FR I = 6.86 X 10-3 D4 = 6.86 X 10-3(3.00 ft)" = 0.556 ft4 LP - Le= Il[L,A] = (0.556 ft4)/[(8.308 ft)(3.534 ft2)] = 0.0189 ft LÕ =Lc+ 0.0189 ft = 8.308 ft + 0.0189 ft = 8.327 ft = LÕ

1 1 2 3 4 5 8 Fluid Depth. h (In)

1

~ ~- LY' / h Å

""'-.. i ' - '\ r-, \

" <, ~ <,

f"".,....

1

Fnet = We - wá= 80.3 lb - yáAdL

Summary of Results:

h (in) r: (lb) 6.00 70.1 5.50 70.5 82

5.00 71.2 leo 4.50 72.08 14.! 4.00 73.07 - 78 i 3.50 74.12 c8 78 3.00 75.19 ..-: i 2.50 76.27 1- 74 2.00 77.32 8

~ 72 1.50 78.30 11.. 1.00 79.18 70 0.50 79.89 0.00 80.30

4.63 (See Prob. 4.57) For any depth 2 6.00 in, Fnet = 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.

JCD2 /3 l Ad=A1 -A2 = --x---(2x)(R-h) 4 360 2

Chapter 5 44

w=Fà - T= O =ycVe-yáVe-T= Vc(ye-yá)-T Ve= T 2.67 kN = 0.217 m3 (re - r1) [23.6-(l.15)(9.81)]kN/m3

5.7

5.6 w-Fb=O=ycVc-yfVd

Ye= yá Vd = (0.90)(9.81kN/m3)(75/100)=6.62 kN/m3 V:

Y= 1200-966 = 234 mm

Wc - F, =O; YcVc -yáVd

V -V Ye d- e - Y1

= JT(.30m)zĿ1.2 m- 7.90 = 0.0683 m3 = JTD2 ÅX 4 9.81 4

X= 4(0.0683) m3 = 0.9664 m = 966 mm JT(0.30 m)2

5.5

5.4 Wc - Fb =O; Wc = Fb; Ye Ve= yáVd = yá0.9 Ve Then, Ye= 0.9yá= (0.90)(1.10)(62.4 lb/ft3) = 61.78 lb/ft3

5.2 Ifboth concrete block and sphere are submerged: Upward forces= Fu= F,,, + Fhc = YwV, + YwVc = yw(V, +Ve)

V, =nD3 /6=Jr(l.Om)3/6=0.5236 m3}- _ 3 We 4.1 O kN 3 - v;01 - 0.6973 m Ve=-= =0.1737 m

, Ye 23.6 kN/m3

Fu= YwVioi = (9.81 kN/m3)(0.6973 nr') = 6.84 kN }Fu> Fn It will float. Downward forces= Fo= we + ws = 4.1 + 0.20 = 4.30 kN

5.3 Ifpipe is submerged, Fb = y1V= (1.26)(9.81 kN/m3)(n(0.168 m)2/4)(1.00 m) Fb = 0.2740 kN = 274 N; because w = 277 N > Fb-lt will sink.

5 .1 ²:F V= o = w + T - Fb Fb = yáV= (10.05 kN/m3)(0.45)(0.60)(0.30)m3 = 0.814 kN = 814 N T= Fs - w = 814-258 = 556 N

Buoyancy

CHAPTER FIVE

BUOYANCY ANO STABILITY

45

T i 3.0ft -wÛ j_

Buoyancy and Stability

wH+ws-Fb =O

[J²(l.0)2 Ŀ 1 50 l²(.25)2 1 30] in3 ft3

Fb = Yw(Vi + Vi)= 62.4 lb/ft3 4 Ŀ + 4 x Ŀ 1728 in3

= 0.04485 lb Ws = Fb - WH = 0.04485 - 0.020 = 0.02485 lb

Wc =F, =O= YcVc-YáVd = Ycs3 -yáSX X= YcS3 =Yes

y1S2 r,

YAVA -ywVA = 558.1 lb YA= (0.100 lb/in3)(1728 in3/ft3) = 172.8 lb/ft' VA = 558.1 lb = 558.1 lb = 5.055 ft3 r A - Yw (172.8-62.4)lb/ft3

WA - F;, = F;, - Wc = 588.1 - 30 = 558.1 lb (See Prob 5.10) A e

O= YFVF+ 80 N -ywVF- 9.81 N O= VF(yF-Yw)+70.19N

V_ -70.19N -70.19N F-

YF =r; (470-9810)N/m3 = 7.515 X 10-3 m3

Foam

l:Fv=O=wF+ws-F. -F. bp bs

F;, = YwVs = 9.81 kN/m3 x (0.100 m)" = 9.81 N s

5.13

5.12

5.11

5.10

5.9

5.8 w-Fb -Fsp= O= W-y0Vd-Fsp 3 . 3 1 ft3 Fsp = w -y0Vd = 14.6 lb - (0.90)(62.4 lb/ft )(40 m) . 3 1728 m

Fsp= 13.3 lb

Chapter 5

w+ Fs =O= w-rswVd Vd=~= 292xl06gmx9.8lmx~x lN x lkN =283.6m3

Ysw 10.10 kN/m3 s2 103 gm kg m/s2 103 N

Ym = 844.9 lb/ft3 > Yw - cube would tend to float. w-Fb+Fe=O Fe =F, - w =ymVc-ysVc= Vc(ym -ys) =(1.50)3ft3(844.9-491)1b/ft3=1194 Ib.J,

T .._w~l ___, I"

Å 1 áá

46

5.19

5.18

5.17 S = 18.00 in(l ft/12 in)= 1.50 ft w-Fb-Fe=O Fe= w-Fb = YsVc-ywVc = Vc(ys-Yw) Fe= (1.50)3 ft3 (491 - 62.4) lb/ft3 = 1447 lb t

= 23.04 lr 7r(2.0)2(3.0) 3]

F,,c = yáVc = 64.0 lb/ft3 12 ft = 201.06 lb

[ 7r(l.0)2 7r(l.0)3] 3

F',,n = yáVd = 64.0 lb/ft3 4 . 2.00 + 12 ft

= 117.29 lb Wc = 117.29 + 201.06 - 23.04 = 295.31 lb

2.00 ft _J_

Sea WaJet

r Å64.o.Br lt

Wc = F. + F. - Ws bn be

[ 7r(l.OO) 2

3 O 7r(l.00)3 l 3 Ws =ys Ā Vs = 8.00 lb/ft3 4 Ŀ Ŀ + 6 ft

1.50 tt Water SW1ace 1

Vd= Vá + V2 = [ JT(l .0)2 Ŀ 1.50 + JT(0.25)2 Å 0.30] in 3 ft3 = 6.903 x 10-4 ft3

4 4 1728in3

w 0.04485 lb r y - - = 64.97 lb/ft3; sg = _L = 64.97/62.4 = 1.041 r: Vd - 6.903X10-4 ft3 Yw

From Prob. 5.13: WH + Ws = 0.02 + 0.02485 = 0.04485 lb= w = Fb

w=Fb=yFVd=y_t(Vá + Vi)=yá[JT(l.0)2. 1.50+ 7r(0.25)2. 2.30] in3 ft3 4 4 1728 in'

= y_t(7.47 X 10-4 ft3)

w O.o4435 lb = 60.03 lb/ft'; sg = Yá = 60.03/62.40 = 0.962 ri= vd = 7.47x10-4 ft3 r

\V

5.16

5.15

5.14

47 Buoyancy and Stability

.75m

.ĉ. 1----1

T .030m

3 1C(.45)2 3 Ws = ysVs = 84.0 kN/m x x .030 m 4

= 0.4008 kN

lĿ~I Surfaca

Yw at 15ÜC = 9.81 kN/m3-it would float. w-F. +ws-F. =O e b, b8

Wc =Ye Ve= 6.46 kN/m3 X 1C(.45)2 X .75 m' . 4 = 0.771 kN

5.25

=30.0mm

Wc- Fi, +ws- Fbn =O Ye Ve -ywVc + YnVB -ywVB =O ycAĀL-y.,AĀL + ysAĀt-y.,AĀt =O t(ys -yw) = L(Yw -ye) t = L y w - Ye = 750 mm (9.44-6.46)kN/m3 r, =r; (84.0-9.44)kN/m3

5.24

--=SwtlC9=..=.....-,,,,___~b- Tw Å r L 95Ŀc- ...... _ .... Ll_x 9.44kN/m1 -

5.23 w - Fb =O> Ye Ve -ywVd = ycAĀL -y.,A X

X= YcA ĀL = 6.46 kN/m3 . 750 mm= 513 mm YwA 9.44kN/m3

h =L-X= 750-513 = 237 mm

5.22 w=Fá =O =ycVc-YKVd

Ye= YK Vd = 8.07 kN/m3 Å AĀ 600 mm = 6.46 kN/m3 Ve AĀ750mm

115 8 Å lir'l"1$Å 30.74Å , .180 + 28 Å 241.S- XÅ Rcoa 8 Å .225 coa 30.74Å XÅ0.1934m

1 110

1 115

o- '4SO ---

' RÅ 225

Vr= 1CD2 ĀL=1C(.450)2 Ŀ6.750=1.074m3 4 4

Vd= [1CD2 .L+.!_(2X)(.115)]L 4 360 2

Vd= [JC(.45)2Å 241.5 + (.1934)(.115)]6.75 m3 4 360

= 0.8703 rrr' Ywood = (9.81kN/m3)(.8703/1.074)=7.95 kN/m3

5.20 w = Fi; Y1Vi = YswVd

Vd= Vá -1L = VáĀ 8Ŀ 72 kN/ m3 = ViĀ 0.863; 86.3% submerged; 13. 7% above Ysw 10.10 kN/m3

Chapter 5 48

s D" Wt. of steel bar= Ws = YsVs = ysAL = Ys--L 4

Ws = 76.8 k:N/m2 Å ll'(.038 m)2 Å 0.08 m Ŀ 103 N = 6.97 N 4 kN

Ws + Wc = 6.97 + 0.05 = 7.02 N = Fb = y,AX V- WT - 7.02N 4 lkN -0135 -13- A--- Ā---. m- ~mm

rwA (9.81 k:N/m3) ll'(0.082 m)" 103 N

5.29

w = Fb = yáVd = YwĀAĀX X=~= 0.05N 4 x~ =0.965x10-3m=0.965mm

YwA (9.81 k:N/m3)( ll'(0.082 m)2) 103 N

5.28

w = Fb = yáVd = (1.16)(9.81 k:N/m3)(0.8836 nr') = 10.05 kN Vd= rrD3/12 = n(l.50 m)3/12 = 0.8836 m3 Entire hemisphere is submerged.

5.27

5.26 YcT = 15.60 k:N/m3

w-F. +ws-F. =O e b, b8

- - 3 ll'(.45)2 3 - Wc -ycVc - 6.50 k:N/m X X .75 m - 0.7753 k:N 4

3 ll'( .45) 2 3 F,, = YcTVd = 15.60 k:N/m x x .70 m = 1.737 kN

' 4 Ws - F,, = YsVs -YcTVB = Vs(Ys -YcT) = F,, - Wc = 1.737 - 0.7753

B e

=0.9614 kN Vs = 0.9614 kN 0.9614 kN = 0_01406 m' = At

y8 -rcT (84.0-15.60)k:N/m3

t = Vs = 0Ŀ01406 m3 = 0.0884 m = 88.4 mm A ll'(0.45)2 / 4 m2

29 mm above surface

- - 3 ll'(.45)2 3 F.h -ywVB - 9.81 k:N/m X X .03 m B 4

=0.0468 kN F,,c = YwVd = Wc + Ws - F,,. =O. 771 + 0.4008 - 0.0468 = 1.125 kN

F,,c = YwVd = y,AX F,, 1.125 kN

X= _e = ( ) = 0.721msubmerged=721 mm YwA (9.81 kN/m3) ll'(.45)2m2/4


Wctrums = 4(30 lb)= 120 lb (Prob. 5.31) Wwood = 178.5 lb (Prob. 5.32) F;, = 1801 lb (Prob. 5.31)

D

F'á,w = YwVw = 62.4 lb/ft3 X 4.464 ft3 = 278.6 lb (Prob. 5.32) W1oad = F;, + F;, - Wn - Ww = 1801 + 278.6 - 120 - 178.5 = 1781 lb

D w

5.34 Wctrums + Wwood + W1oad - Fb - F;, =O /) w

See Prob, 4.63 for method of computing As.

á 1

RÅ10.S

5.33 Wn + Wp = Fb = YwVd V = Wv + wP = 120lb+178.5 lb = 4.78 ft3 total d r; 62.4 lb/ft3 Vn = 4.78/4 = 1.196 ft3 sub. each drum Vn=AsĀ L As= Vv=l.196ft3=0.399ft2x144in2 =57.4in2

L 3.0 ft ft2 By trial: X= 4.67 in when As= 57.4 in2

= 0.6875 ft3 ends = l. 77 6 ft3 main boards = 2.000 ft3 plywood 4.464 ft3 total

Vol. ofWood: 2(6.0 ft)(l.50 in)(5.50 in)(l ft2/144 in2) 4(96- 3)in(l.50 in)(5.50 in)(l ft3/1728 in3) (0.50 in)(6 ft)(8 ft)(l ft/12 in)

Ww = YwV= (40.0 lb/ft3)(4.464ft3)=178.5 Ib

5.32

3 (7r(21in)2) (36in) wr= Fb = 4ywVdrum = 4(62.4 lb/ft) . 3 ft3 = 1801 lb 4 1728 m Drums Weigh 4(30 lb)= 120 lb Wt. of platform and load= 1801 - 120=1681 lb

5.31

5.30 From Prob. 5.29, Wr = 7.02 N

Vs = 7rD2 Å L = 7r(.038 m)2 Å 0.080 m = 9.073 x 10-5 m' 4 4

F;,, = YwVs = 9.81 kN/m3 x 9.073 x 10-5 m3 x 103 N/kN = 0.890 N

Wr - F;,s - F;,c = O F;,, = Wr- F;,s = 7.02 N - 0.890 N = 6.13 N = y,AX

X- 6.13 N 4 1 kN _O 118 _ 118 - x x-- - m- mm 9.81 kN/m3 Jr(0.082 m)2 103 N Ŀ

Chapter 5

ater

3.00

OI .... T

50

T=wA- F. - F. bw bo

wA=yAVr= O.lOlb x(6.0in)3=21.6lb . 3 m

F. = =62.4lb/ft3x (6in)2(3.0in) =3.90lb b" Yw Vdw 1728 irr' /ft3 F,,0 =Yo~!,, = 0.85 F,,w = 3.315 lb Then: T= 21.6- 3.90- 3.315=14.39 lb

5.37

Vd= (18.0 in)2(X) V 12204 irr' . X= d 2 2

= 37.67 m submerged (18.0 in) 324 in

Y= 48-X= 10.33 in above surface With concrete block suspended, float is unrestrained and it would drift with the currents.

/y ) ~

Å 1 iic

= 12204 ²n'

5.36 Rise of water level by 18.00 in would tend to submerge entire float. But additional buoyant force on float is sufficient to lift concrete block off sea floor. With block suspended: T = Wc - F,, e

T= 600- 256 = 344 lb (see Problem 5.35) Float: wF + T - F. = O bF

F,,,. = wF+ T= 108 lb+ 344 lb= 452 lb= YwVd

V: _ F,,,. _ 452 lb = 7_063 ft3 X 1728 in3 a : Yw - 64.0 lb/ft" ft3

5.35 Given: yF= 12.00 Ib/fr', ye= 150 Ib/fr', wc= 600 lb Find: Tension in cable Float only: 2:.F v = O = wF + T - F,,F T= Fbá. -WF But wF= YFVF= 12.0 lb/ft3 x 9.0 ft3 = 108 lb V: = (18.0in)2(48in) = 9_00 fr' F 1728 in ' /ft3

F,, = YwVd = (64.0 lb/ft3)(6.375 ft3) = 408 lb F

V: = (18.0in)\34in) = 6_375 ft3 d 1728 in3/ft3 T= 408 - 108 = 300 lb Check concrete block: Fnet = Wc - F,,c - T

W = 600 lbĿ V. = Wc = 600 lb = 4.00 ft3 e 'e Ye 150lb/ft3 F,,c = YwVc = (64.0 lb/ft3)(4.00 ft3) = 256 lb Fnet = 600 - 256 - 300 = 44 lb down-OK-block sits on bottom.


--- ç>in---

me+ TT 221n -r cri+1ao 1s 10.50

c:D+ X

w =F; = YwVd = y,4% X=~= 250lb(144in2/ft2) =0.4808ft

rwA (62.4lb/ft3)(30in)(40in) X= 5.77 in; Ycb = X/2 = 2.88 in I= (40 in)(30 in)3/12 = 90000 in" Vd= AX= (30 in)(40 in)(5.77 in)= 6923 in3 MB =l/Vs= 90000 in4/6923 in3 = 13.0 in Ymc = Ycb + MB = 2.88 in+ 13.0 in

= 15.88 in> Ycg-stable

5.40

Ycg = 1.00 m/2 = 0.500 m Ycb =X/2 = 0.8155 m/2 = 0.4077 m MB = _!_ = 1²D4164 7²(l.Om)4/64

Vd (1²D214)(X) [7²(l.Om)2!4J(0.8155m)

0.04909 m4 = 0.0766 m 0.6405 m'

Ymc = Ycb + MB = 0.4077 + 0.0766 = 0.4844 m < Ycg-unstable

= 0.8155 m B

--::=--me

-----+-"----' rr 1.0mdla

5.39 w-Fb=O Ye Ve= yáVd = yáAX X= rJ';, = rcA(l.O m) = 8.00 kN/ m3 /(1.0 m) y1A y1A 9.8lkN/m3

Stability

= 70.1 lb Fs acts up on cylinder; down on tank bottom

5.38 Fs = Wc - Fb, = YcVc -yáVc =V/ye-Yá)

V= 1²D2 ĀL= 7²(6.0in)z xlO.Oinx ft3 e 4 4 1728 in ' = 0.1636 ft3

(0.2841b 1728in3 62.41b)

fi3 X - Fs=0.1636 t in ' ft3 ft3

Chapter 5

24.0

me

300mma 1

12

.__~_~....._mc T á 10.0ln dla

52

5.45 w = Fb = YwVd = yu,AX X=~= 70.0lb

YwA (62.4 lb/ft3)( ;ir(2.0 ft)" / 4) = 0.357 ft(12 in/ft) = 4.28 in

Ycb =X/2 = 2.14 in Ycg = H/2 = 48/2 = 24.0 in 1 = nD4/64 = n(24 in)4/64 = 16286 in4 Vd= AX= (nD214)(X) = [n(24 in)2/4](4.28 in)

= 1936 in3 MB =l/Vàè 8.41 in Ymc = Ycb + MB = 2.14+8.41=10.55 in <ycg-unstable

_J_

From Prob. 5.6: X= 75 mm Ycb =X/2 = 37.5 mm Ycg = H/2 = 100/2 = 50.0 mm 1 = F/12 = (100 mm)4/12 = 8.333 x 106 mm" Vd= (H2)(X) = (100 mm)2(75 mm)= 7.50 x 105 mm' MB =l/Và> 11.11 mm Ymc=Ycb+MB=37.5+ 11.11

= 48.61 mm< Ycc-unstable

'00 mm t1qUm9

_.....__-----'-~

;;Y-r r'² 5.44

5.43 FromProb. 5.5:X=966mm Ycb = X/2 = 483 mm Ycg = H/2 = 1200/2 = 600 mm != nD4/64 = n(300 mm)4/64 = 3.976 x 108 mm" Vd= (nD214)(X) = [n(300 mm)2/4](966 mm)= 6.828 x 107 mm' MB =l/Và= 5.82 mm Ymc = Ycb + MB = 483 + 5.8 = 488.8 mm <ycg-unstable

FromProb. 5.4;X=0.9 H=0.9(12)= 10.8in Ycb = X/2 = 5.40 in Ycg = H/2 = 6.00 in 1 = nD4/64 = n(lO.O in)4/64 = 490.9 in4 Vd= (nD2/4)(X) = [n(lO.O in)2/4](10.80 in)= 848.2 in' MB =!/Vd= 490.9/848.2 = 0.579 in Ymc = Ycb + MB = 5.40 + 0.579 = 5.98 in <ycg-unstable

~--~~~----+---<i--Û~.258--~rr Sea W cb 3.StS X 8.00

Pldomt t 1

:\_ ....á,_ _ _.../20 IC 50 ft MB

w = Fb = YswVd = YsuĆX X=~= 450000lb

YswĆ (64.0 lb/ ft3)(20 ft)(50 ft) = 7.031 ft

Ycb =X/2 = 3.516 ft I = (50)(20)3 /12 = 3 .333 X 104 ft4 Vd=AX= (20)(50)(7.03l)ft3 = 7031 ft3 MB=I/Vd=4.741 ft Ymc = Ycb + MB = 3.516 + 4.741

= 8.256 ft > Ycc-stable

5.42

5.41

53

--t 82 nvn dla. 1--

T H

me

T L

i.. ~;~

y


5.50 From Prob. 5.29, X= 135 mm Ycb =X/2 = 67.5 mm MB = !_= trD4 /64 = D2 = 822 = 3.11 mm

v;1 (trD2 l 4)(X) 16X 16(135) Ymc = Ycb + MB = 67.5 + 3.11=70.61 mm Beca use wt. of bar is>> wt. of cup,

Ycg ~ L Bar/2 = 80/2 = 40 mm Ymc > Ycg-Stable

5.49 FromProb. 5.28: X= 0.965 mm; D = 82 mm; H= 150mm MB = !_= trD4164 = D2 = 822

Vd (trD2 I 4)(X) 16X 16(0.965) =435 mm

Ymc = Ycb + MB = X/2 + MB = .965/2 + 435 = 436 mm me is well above cup; cg is within cup .'. It is stable

Ycg = 0.70 m from bottom

For hemisphere- submerged- y = 3D from dia. 16

= R __ = D _ 3D _ 5D _ 5(1.50 m) Ycb y 2 16 - 16 - 16 = 0.469 m

I = rcD4/64 = rc(l .50 m)4/64 = 0.2485 m" Vd= rcD3/12 = rc(l.50 m)3/12 = 0.8836 m3 MB =l/Vàè 0.281 m Ymc = Ycb + MB = 0.469 + 0.281=0.750 m > Ycg-stable

5.48

5 .4 7 X= 16.0 ft; Ycb = X/2 = 8.00 ft; Ycg = 13 .50 ft MBmin = Ymc - Ycb = Ycg - Ycb = 13.50 ft - 8.00 ft = 5.50 ft

Wmin = J12(X)(MBmin) =~12(16.0)(5.50)ft2 = 32.50 ft

W min = Jl 2( X)(MB min) = ~l 2{8)(8)ft2 = 27. 71 ft

5.46 X= 8.00 ft; Ycb = X/2 = 4.00 ft; Ycg = 12.00 ft Ymc =ycb+ MB Scow is stable if Ymc > Ycg MBmin = Ymc - Ycb = Ycg - Ycb = 12.0 - 4.0 = 8.0 ft MB . = Ir _ LW3 /12 = W2

mm V - LWX 12X d

Chapter 5 54

5.52 FromProb. 5.22, Fig. 5.23: X= 600 mm Ycb = X/2 = 300 mm Ycg = H/2 = 75012 = 375 mm MB = !_= 1iD4/64 = D2 = (450)2 = 21.1 mm

Vd (7iD214)(X) 16X 16(600) Ymc = Ycb + MB = 300 + 21.l

= 321.1 mm< Ycg-unstable

= (19)(9.073xl04)+(97)(6.232xl05) = 87.l mm 7.139x 105

MB = !_ = 7i(82)4/64 mm= 3.11 mm V:i 7.139xl05

Ymc = Ycb + MB = 90.2 mm Ycg is very low beca use wt. of bar is >> wt. of cup---stable

1 111-x mi-

SI ,_.97

l 31_L l uaĿ y~ r 38 mm dla. bar

5.51 FromProb. 5.30,X= 118mm Vs = Vol. of steel bar = 9 .073 x 104 mm'

7iD2 Ves= Sub. vol. of cup = __ e x X

4

Ves= 1i(82)2 (118) = 6.232 x 105 mm' 4

Vd= Vs+ Ves= 7.139 x 105 mm' D

Ys = cb of steel bar= _s = 19 mm 2

Yes = cb of sub. vol. of cup X 118

Yes= Ds + - = 38 + - = 97 mm 2 2

_ YsVs +Yes Ves Ycb- V

d

55

XÅ1.50 lIT~ 8.0


5.55 w = Fb = YswAX X=~ == 130lb(12in/ft)

YswA (64.0 lb/ft3)(3 ft)(4 ft) == 2.03 in

Ycb =X/2 == 1.016 in Ycg = 12 in+ 34 in= 46.0 in != (48)(36)3in4/12 = 1.866 x 105 in4 Vd= (48)(36)(2.03)in3 == 3510 in3 MB =l/Và> 53.17 in Ymc =ycb + MB == 1.016 + 53.17

= 54.18 in> Ycg-stable

w=Fb=YswAX X=~ = 3840lb = 7.50 ft

YswA (64.0 lb/ft3)(2 ft)(4 ft) Ycb = X/2 = 3.75 ft 1=4(2)3/12 = 2.667 ft" Vd= AX= (2)(4)(7.5) = 60 ft3 MB =l/Và= 2.667160 == 0.0444 ft Ymc == Ycb + MB == 3.75 + 0.0444 = 3.794 ft Ycg == H/2 == 8.0012 == 4.00 ft > Ymc-unstable

5.54

Yn = t/2 = 44.2 mm - Ycwc+YsWs = (463.4)(0.7753)+(44.2)(1.181) =210.3mm< -stable

Ycg- w001 (0.7753+1.181) Ymc

,-=- XÅ788.4

= 1.181 kN

5.53 From Prob. 5.26, Fig. 5.25; t = 88.4 mm X= 700 + t = 788.4 mm Ycb = X/2 = 394.2 mm

1 1rD4/64 D2 MB=-- =-- ~1 -[7rD2!4J{X) 16X

MB = 4502 = 16.1 mm

16(788.4) Ymc == Ycb + MB == 394.2 + 16.1 = 410.3 mm YcgWtot == YcWc + YnWn composite cyl. Wc = YcVc = 0.7753 kN Ye= 750/2 + t = 375 + 88.4 = 463.4 mm - V - 840.0 kN 7r(.45 m)2 (O 0884 ) Wn - Yn B - 3 x . m m 4

56 Chapter 5

5.58 Wtotal = 210000 + 240000 = 450000 lb X= ~ = 450000 lb = 6.010 ft (See Prob. 5.57)

rwA (62.4)(60)(20) Ycb = X/2 = 3.005 ft MB = _!_ = (60)(20//l2 = 5.547 ft

Vd (60)(20)(6.010) Ymc = Ycb + MB = 3.005 + 5.547 = 8.552 ft-above barge cg is within barge--stable More complete solution:

w 240000lb Depth of coal = d = _e = = 4.444 ft e rcA (45 lb/ft3)(60 ft)(20 ft)

Ye= dj2 = 2.222 ft from bottom to cg of coal Y = WcYc + WsYs = (240000)(2.222) + (210000)(1.50) = 1.885 ft < Ymc-stable cgtot wtot 450000

5.57 w-Fb=ywVd=y,,AX

X= r:A = (62.4 lb~;t~~(~~l~)(20 ft) = 2.804 ft ~me: 1----1'..---- ..... I ~3r Ycb = X/2 = 1.402 ft; Ycg = 1.50 ft given -Ê .... k!X.,....,._"- __ .....,._._..,._ --:t--

MB = _!_ = (60)(20)3 /12 = 11.888 ft Vd (60)(20)(2.804)

Ymc = Ycb + MB = 1.402 + 11.888 ft = 13.290 ft > Ycg-stable

Ycb =X/2 = 1.709 in Ycg = H/2 = 6.00/2 = 3.00 in I= 12(6)3/12 = 216 in" Vd= (12)(6)(3.419) = 246.2 in' MB =!/Vd= 0.877 in Ymc = Ycb + MB = 1.709 + 0.877 = 2.586 in <ycg-unstable

w - Fb =O= YwViot -yo Vd= YwAH - yoA}{ X= rwAH = rwH = (32 lb/ft3)(6 in) r0A r; 0.90(62.4 lb/fr')

= 3.419 in

5.56


b) For S= 75 mm X> 0.788S = 59.2 mm, X< 0.21IS=15.8 mm

X> O. 788S or X< 0.211S will result in stable cube.

5.60 a) Cube is stable if Ymc > Ycg = S/2 X 1 X S4

Ymc = Ycb + MB = 2 +Vd = 2 + 12(S2)(X) X S2 S S2 -+-=- orX2+ - -SX=O 2 12X 2 6 x2 - sx + s216 =o X= S Ñ .J S2 - 4S216 = S Ñ S .JI_ 213

2 2 2

= s[lÑl(0.5774)] =0.788Sor0.211S

600X -X2 = 2.489 X 104 X- - 600X + 2.489 x 104 =O; X= 44.8 mm by Quadratic Eq. G = B- 2X= 600-2(44.8) = 510 mm != c3L!12 = (510)3(1200)/12 = 1.329 x 1010 mm" Vd= AcJ, = [600(44.8)- 44.82](1200) = 2.985 x 107 mm" MB =IiVà= 445 mm =X- X(G+2B) = 44_8 _ 44.8(510+1200) = 21.8 mm

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

Ymc = Ycb + MB = 21.8 + 445 = 467 mm 1 Ycg = - (300) = 100 mm < Ymc-stable 3

r Ad =BX-2X2 +X2 =BX-X2 =600X-X2

w 0.255 kN IL A,~ y,L ~ (0.87X9.8 l kN/m')(l.20 m)

Ad= 0.02489 m" x (103n:m)2=2.489x104mm2 m

Equate

mo.----- .... A- 11

á fr'.a-!~~ + i r/J ;;m j\jj 1 l:L s-ann .l;

(BÅo.eGm)

5.59 w = Fb = YrVd = YrAdL W =ycAtotL

= _!_ (.600)(.300)(1.20)(2.36)kN 2

= 0.255 kN

Ad= GX + 2(l) (X)(X)

= (B - 2X)(X) + X2

58 Chapter 5

b) Ye= 55.0 lb/ft' Wc = 1.963(55/30) = 3.600 lb

Vd= Wc = 3.600 lb = 0.0577 ft3(1778 in3) = 99_69 in3 r w 62.4 lb/ rt3 rt3 X= v48v;1 = 3 48(99.69) = 11.51 in

1T 1T

Figure for part (a) only.

5.62 a) Wc = Fi: Ye Ve= YwVd Wc = 30 lb X 1T(.5 ft)z(l .0 ft) = 1.963 lb

ft3 12

V. = Wc = 1.963 lb = 0.03147 ft3 d r w 62.41b/ft3

Vd= (0.03147 ft3)(1728 in3) = 54_37 in3 ft3

V. = JT(Dx )2(X) = JT(Xl2)2X _ JTX3 d 12 12 - 48

X= v48Vd = 3 48(54.37) = 9.40 in 1T 1T

Ycb = 3X = 0.75(9.40) = 7.05 in 4

I= JTD~ = 1T(9.40/2)4 = 23.95 in4 64 64

MB =l/V,= 23.95/54.37 = 0.441 in Ymc = Ycb + MB = 7.05 + 0.441=7.491 in

Ycg = 3H = 0.75(12) = 9.00 in 4

Ymc < Ycg-unstable

Ycg = 1.040 m Submerged Volume:

(.72)(0.4) + (2.16)(1.05) .ao Ycb = 2.88 = 0.8875 m

Vd= (2.88)(5.5) = 15.84 m' I= 5.5(2.4)3/12 = 6.336 m" Ymc = Ycb +Irt/à> 0.8875 + 0.40 = 1.2875 m > Ycg-stable

5.61 Entire hull: _ Ć1Y1 + ĆzY2

Ycg- Ćr

(.72)(.40) + (2.88)(1.2) 3.60


Fb

750 \Je IJv

@~:Ŀ ____,-fÜ 950=ycg -r--- 1 822=yMd

657=ycb

Fluid Surfo.ce 250

F :::: ===²l ,

(b) Find Yv = Specific weight of vessel material= WvfVvr; Given Wc = 5.0 kN From part (a), 17.09 kN = Wc + Wv; Then Wv = 17.09 - Wv = 17.09 - 5.0 = 12.09 kN Vvr= Total volume ofvessel =Vis+ Vcyl-T

Vcy1-r = ( D; - Dn (0.60 m)/4 = n[(l.50 m)2 - (1.40 m)2](0.60 m)/4 = 0.1367 m3 Vvr = Vis + Vcyl- = 0.8836 m3 + 0.1367 m' = 1.020 m3 Yv = Specific weight of vessel material= WvfVvr; = (12.09 kN)/(1.020 nr') = 11.85 kN/m3 = Yv

Wc = Weight of contents; Wv = Weight of vessel; Find W, + Wi:. Wc + Wv = Fb; But F, = rtVi Vd = Vis + Vcyl-d; Where Vis = Vol. of hemisphere; VcyI-d = Vol. of cyl. below surface Vis= nD3/12 = ll(l.50 m)3/12 = 0.8836 m3 Vcyl-d = nD2h,á/4 = ll(l.50 m)2(0.35 m)/4 = 0.6185 m3 Then Vd= Vis+ Vcyl-d = 0.8836 + 0.6185 = 1.502 m' Fb = rtVi = (1.16)(9.81 kN/m3)(1.502 m') = 17.09 kN = Wc + Wv (Answer)

""Vp =O=Fb-W-W, à V C V (a) 5.63

nD~ n(l 1.51/2)4

I= 64= 64 = 53.77 in"

MB = _!_ = 53Ŀ77 = 0.539 in V" 99.69

Ycb = 0.75X= 0.75(11.51) = 8.633 in Ymc = Ycb + MB = 8.633 + 0.539 = 9.172 in Ycg = 9.00 in < Ymc-stable

Chapter 5 60

Fb = YwVa1; W = Ya1Va1; Where Va1 = Volume of aluminum club head Va1 = Wlra1 = (0.500 lb)/(0.100 lb/in3) = 5.00 in3 F, = W - Fà = 5.00 lb - (62.4 lb/ft3)(5.00 in3)(1 ft3)/(l 728 in') r, = 0.319 lb

rJ l ____ .) c __ !_w _J i 'Å

Let F, be the supporting force acting vertically upward when the club head is suspended in the water. "f\' F = O = F + Fb - WĀ Then F = W - Fb L....i V S ' S

5.64

We know from part (a): Vd= 1.502 rrr'; Vis= 0.8836 m'; Vcyl-d = 0.6185 m' Yhs = D/2 - y= D/2 - 3D/16 = (1.50 m)/2 - 3(1.50 m)/16 = 0.4688 m Ycyt-d = D/2 + (0.35 m)/2 = (1.50 m)/2 + 0.175 m = 0.925 m

Then Ycb = [(0.4688 m)(0.8836 rn') + (0.925 m)(0.6185 m3)]/(l.502 m') = 0.657 m Now, Ymc = Ycb + MB = 0.657 m + 0.1654 m = 0.822 m From bottom of vessel. Because Ymc < Ycg' vessel is unstable.

(ycb)(Vd) = (yhs)(Vis) + (ycyt-d)(VcyJ-d) Ycb = [(yhs)(Vis) + (ycyt-d)(Vcyt-d)]/Vd

(e) Evaluate stability. Find metacenter.jc., See figure for key dimensions. Given: Ycg = 0.75 + 0.60 - 0.40 = 0.950 m from bottom of vessel. We must find: Ymc =ycb + MB = Ycb +!/Vd

I = 1CD4/64 = n(l .50 m)4/64 = 0.2485 m" For circular cross section at fluid surface. Vd= 1.502 m' From part (a). MB = (0.2548 m4)/(1.502 m') = 0.1654 m

The center ofbuoyancy is at the centroid ofthe displaced volume. The displaced volume is a composite of a cylinder and a hemisphere. The position of its centroid must be computed from the principle of composite volumes. Measure ali y values from bottom of ves sel.

61 Flow of Fluids

Conversion factors

6.1 Q = 3.0 gal/min x 6.309 x 10-5 m3/s/1.0 gal/min = 1.89 x 10-4 m3/s

6.2 Q = 459 gal/min x 6.309 x 10-5 m3!s/1.0 gal/min = 2.90 x 10-2 m3/s

6.3 Q = 8720 gal/min x 6.309 x 10-5 m3/s/1.0 gal/min = 0.550 m3/s

6.4 Q = 84.3 gal/min x 6.309 x 10-5 m3/s/l.O gal/min = 5.32 x 10-3 m3/s

6.5 Q = 125 L/min x 1.0 m' /s/60000 L/min = 2.08 x 10-3 nr' /s

6.6 Q = 4500 L/min x 1.0 m' /s/60000 L/min = 7.50 x 10-2 m3 /s

6. 7 Q = 15000 L/min x 1.0 m' /s/60000 L/min = 0.250 nr' /s

6.8 Q = 459 gal/min x 3.785 L/min/1.0gal/min=1737 L/min

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

6.10 Q = 23.5 cm'zs x m3/(100 cmj' = 2.35 x 10-5 m3/s

6.11 Q = 0.296 cm3/s x 1 m3/(100 cm)' = 2.96 x 10-7 m3/s

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

6.13 Q = 3.58 x 10-3 m3/s x 60000 L/min/1.0 m3/s = 215 L/min

6.14 Q = 5.26 x 10-6 m3/s x 60000 L/min/1.0 m3/s = 0.316 L/min

6.15 Q = 459 gal/min x 1.0 ft3 /s/449 gal/min = 1.02 ft3 Is

6.16 Q = 20 gal/min x 1.0 ft3 /s/449 gal/min = 4.45 x 10-2 ft3 /s

6.17 Q = 2500 gal/min x 1.0 ft3/s/449 gal/min = 5.57 ft3/s

6.18 Q = 2.50 gal/min x 1.0 ft2 /s/449 gal/min = 5.57 x 10-3 ft3 /s

6.19 Q = 1.25 ft3/s x 449 gal/min/1.0 ft3/s = 561 gal/min

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

6.21 Q = 7.50 ft3/s x 449 gal/min/1.0 ft3/s = 3368 gal/min

Chapter 6

M Q 1.20 kg 640 ft3 1.0 slug 1 min 0.0283 m' =p = X X X--X--- m' min 14.59 kg 60 s ft3

= 2.48 x 10-2 slug/s M=pQ=yQlg= Wlg

W = gM = 32.2 ft X 2.48X10-2 slug X 1 lbĿ S2 /ft X 3600 S = 2878 lb/hr s2 s slug hr

Q _ W _ 28.5 N m' 1.0 kN 1.0 h _

7 47 10-1 3á - -- X X--X-- - Å X ID S y h 1.08(9.SlkN) 103N 3600s

M = pQ = (1.08)(1000 kglm3)(7.47 x 10-7 m3/s) = 8.07 x 10--4 kg/s

W= yQ = (9.81 kNlm3)(0.075 m3/s) = 0.736 kNls(l03 N/kN) = 736 Nis M = pQ = (1000 kglm3)(0.075 m31s) = 75.0 kgls

W= yQ = (0.90)(9.81 kNlm3)(2.35 x 10-3 m3/s) = 2.07 x 10-2 kN/s = 20.7 Nis M = pQ = (0.90)(1000 kglm3)(2.35 x 10-3 m31s) = 2.115 kg/s

62

6.33

6.32

6.31

6.30

6.29

Fluid jlow rates

Q = 19.5 mL x 1.0 L x 1.0 m3 Is = 3Å25 x 10-1 m3/s min 103 mL 60000 L/min

Q = 11.4 gal x 1 h x 1.0 ft3 Is = 1.76 x 10-s ft3/s 24 h 60 min 449 gal/min

Q = 0.85 gal x 1 h x 1.0 ft3 Is = 3.16 x 10-s ft3ls h 60 min 449 gallmin

Q = 745 gal x 1 h x 1.0 ft3 Is = 2.77 x 10-2 ft3/s h 60 min 449 gal/min

Q = 3.0 gallmin x 1.0 ft31sl449 gallmin = 6.68 x 10-3 ft31s Q = 30.01449 = 6.68 X 10-2 ft3/s Q = 3.0 gallmin x 6.309 x 10-5 m3/sll.O gal/min = 1.89 x 10--4 m3/s Q = 30(6.309 x 10-5) = 1.89 x 10-3 m31s

Q = 500 gallmin x 1.0 ft31sl449 gallmin = 1.11 ft3/s Q = 25001449 = 5.57 ft3/s Q = 500 gallmin x 6.309 x 10-5 m3lsll.O gal/min = 3.15 x 10-2 m3/s Q = 2500(6.309 x 10-5) = 0.158 m3/s

Q = 0.008 ft31s x 449 gallmin/1.0 ft31s = 3.59 gal/min

6.28

6.27

6.26

6.25

6.24

6.23

6.22

63 Flow of Fluids

Ćmin = Q = 1 O gallmin x 1 ft3 Is x -1- = 0.02227 ft2: 2-in Sch. 40 pipe u 449 gal/min 1.0 ftls

Q1 = Ć1V1 = Ć2V2 + Ć3V3 = Qz + Q3 JT(0.050 m)2 3 Qz = A2u2 = x 12.0 mis= 0.0236 m Is

4 Q3 = Qi - Qz = 0.072 - 0.0236 = 0.0484 m31s v = Q3 = 0.0484 m3 Is = 6.17 m/s 3

~ JT(.100 m):' /4

A (D )2 (150)2 A,v, = Aiv2; V2 =u, -1 =u, -' = 1.20 mis - = 0.300 mis Ći D2 300

Q = 2000 L/min x l .O m3 Is = 0.0333 m' Is 60000L/min

v, = il_ = 0.0333 m3 Is = 0.472 mis A, TC(.30 m)" /4 _ _f?_ _ 0.0333 rrr' /s = 1 89 1 u- - . ms 2 Ći TC(.15m)214

( )2 ( )2 A, D, 1.65 ft 12

Ć1V1 =A2V2" V2 =Vá-= V - =-- - = 26.4 ft/s ' Ći ' D2 s 3

Q =Av: A= Q = 75.0 ft3 /s = 7.50 ft2 = TCD2 V 10.0 ftls 4

D= .)4AIJT=~4(7.50)!JT =3.09ft

6.42

6.41

6.40

6.39

6.38

6.37

Continuity equation

W- Q- 62.4lb 1.65gal 1.0 ft3 _ 13 76lb/ . _ w -y - x x - . mm- - ft3 min 7.48 gal t

w 7425 lb min 1 hr t = = X = 8.99 br W 13.76 lb 60 min

Q= W =1200lbx ft3 x~ =5.38ft3/s r hr 0.062 lb 3600 s

W = yQ = (0.075 lb/ft3)(45700 ft3/min) = 3428 lb/min

M - Q - rQ - w - 3428 lblmin - 106 lb. s2 /ft - 106 l / . - p - -- - - - - - s ugs mm g g 32.2 ft/s" min

6.36

6.35

6.34

Chapter 6

2-in Sch 80: u = Q = 40~ = 3.500 m/s

A (1.905X10- )(60000)

Q 400 L/min 1 m ' /s u = - = x = 3.075 mis [2-in Sch 40 pipe]

A 2.168 x 10-3 rrr' 60000 L/min

For A1, = 0.015 m2 ruin; 6-in Sch 80; A= 1.682 x 10-2 m2 For AH= 7.916 x 10-2 m2 min; 14-in Sch 80; A= 7.916 x 10-2 m2

. l rn 'Zs 3 QL = 1800 L/mm x = 0.030 m Is 60000L/min

A QI 0.030m31s 0015 z . . 6. S 40 z 2 1,= -~ = . m mnumum; -m ch ;A= 1.864 x 10- m u 2.0mls

QH = 9500160000 = 0.1583 m3ls

AH= QH = 0Ŀ 1583 = 7.916 x 10-2 m'; 14-in Sch 40; A= 8.729 x 10-2 m2 u 2.0

A _ Q _ 0.0668ft31s _ 9

54 10_3 ~2 .

2 - - - - Å x tt mm. u2 7.0 ftls

11/2-in x 0.083 tube ~ A2 = 9.706 x 10-3 ft2 Actual u2 = QIA2 = (0.0668 ft31s)l(9.706 x 10-3 ft2) = 6.88 ftls

A _ Q _ 0.0668ft31s _

3 34 10_2 ~2

1 - - - - Å x tt max. u, 2.0 ftls

2-in x 0.065 is largest tube listed ~ A1 = 1.907 x 10-2 ft2 Actual v1 = QIA1 = (0.0668 ft31s)/(1.907 x 10-2 ft2) = 3.50 ftls

A _ Q _ 0.0668 ft3 Is _ 2 67 10_3

~ 2 . 2 - - - - . x 1t mm.

U2 25.0 ft/s 7/8 x 0.065 tube ~ A2 = 3.027 x 10-3 ft2 Actual v2 = QIA2 = (0.0668 ft31s)l(3.027 x 10-3 fr') = 22.07 ftls

Q = 30 gallmin x 1.0 ft31sl449 gal/min = 0.0668 ft3ls

A _ Q _ 0.0668 fr' Is _ 8 35 10_3

~ 2 1 - - - - Å x tt max.

U1 8.0 ftls 1 1/4 X 0.065 tube ~ Al = 6.842 X 10-3 ft2 Actual u1 = QIA1 = (0.0668 ft3/s)l(6.842 x 10-3 ft2) = 9.77 ft/s

Q . 1 m" /s 1 v = - = 19.7 L/mm x x = 0.856 mis A 60000 L/min 3.835x10-4m2

W = yQ =y Av= 60Ŀ6 lb X 0.2006 ft2 X 4Ŀ5Ü ft X 3600 S = 1.97 X 105 lb/h ft3 s hr

64

6.50

6.49

6.48

6.47

6.46

6.45

6.44

6.43

65 Flow of Fluids

= 65.0 ft/s in nozzle u = 2_= 7.50ft3/s n ~ tr(4.60in)2 1 ft2 --~-x--

4 144in2

Q 7.50 ft3 /s 7 98 f't/ . . v = - = = . s m pipe p ~ 0.9396 ft2

Q =Av= (7.538 x 10-3 m3)(3.0 mis)= 2.261 x 10-2 m3/s

A = Q = 2.261x10-2 m3 /s = 1.508 x 10-3 mz = trD,2 / 4 V 15.0mls

Di= .J4Altr=~4(1.508xl0-3m2)1tr =4.38x10-2mx l03mm =43.8mm m

Q = 800 gal/min(l ft3/s/449 gal/min) = 1.782 ft3/s Suction pipe: 5 in Sch 40; As= 0.1390 ff; v, = Q/A = 12.82 ft/s

6 in Sch 40; As = 0.2006 ft2; v, = QIA = 8.88 ft/s Discharge pipe: 3 1/2 in Sch 40; Ad= 0.06868 ft2; vd = Q/A = 25.94 ft/s

4 in Sch 40; Ad= 0.08840 ft2; vd = Q/A = 20.15 ft/s

Q = 2000 gal/min(l ft3/s/449 gal/min) = 4.454 ft3/s Suction pipe: 6 in Sch 40; A,= 0.2006 ft2; v, = Q/A = 22.21 ft/s

8 in Sch 40; As= 0.3472 ft2; Vs = Q/A = 12.83 ft/s Discharge pipe: 5 in Sch 40; Ad= 0.1390 ft2; vd = Q/A = 32.05 ft/s

6 in Sch 40; Ad= 0.2006 ft2; vd = Q/A = 22.21 ft/s

Q = 60 m3/h(l h/3600 s) = 0.01667 m3/s Suction pipe: 3 in Sch 40; As= 4.768 x 10-3 m2; v, = Q/A = 3.73 mis

3 112 in Sch 40; As= 6.381 x 10-3 m2; Vs = Q/A = 2.61 mis Discharge pipe: 2 in Sch 40; Ad= 2.168 X 1 o' m2; Vd= Q/A = 7.69 mis

2 1/2 in Sch 40; Ad= 3.090 x 10-3 m2; vd = Q/A = 5.39 mis

u6-in = Q6 = 95 gal/min x 1 ft3 /s = 1.055 ft/s A6 0.2006 ft2 449 gal/min

u _. = Q3 = 0.5Q6 = (0.5)(95) = 2.061 ft/s 3 m ~ ~ (0.05132)(449)

A = Q = 2.80 L/min x 1 m' /s 4 2 ---- = 1.556 x 10- m min. u 0.30 mis 60000 L/min

3/4 X 0.065 steel tube, A = 1.948 X 10-4 m2

u= Q = 4oo = 11.16 ft/s [4-in Sch 80] A (0.07986)(449)

u= Q = 4oo gal/min x 1 ft3 /s = 10.08 ft/s [4-in Sch 40] A 0.0884 ft2 449 gal/min

6.59

6.58

6.57

6.56

6.55

6.54

6.53

6.52

6.51

Chapter 6

Ps=PA +yw [czA -za)+ v12~u~] = 66.2 kPa+ 9.81 kN[-4.5+ (5.232 -l.312)m2/s2]

rrr' 2(9.81 m/s") Ps = 66.2 kPa - 31.3 kPa = 34.9 k:Pa

s 5.23m PA +z + v1 = Ps +z +u~ Āu =iL= 0.37m3/s

Yw A 2g Yw 8 2g' A AA JT(0.3m)214

u =u (DA)2=1.31m B A D

B S

p u2 p u2 -1 +z +-1 =-2 +z +-2 Å z1 =z2

1 2 2 2 ' Ya g Ya g

u =Q_= 0.11 m3/s = 6.22m. u =u (D, )2 = 24.90m 1 A1 JT(0.15m)2/4 s ' 2 1 D2 s

= +[u?-uJ] =4l5kPa+[6.222-24.902]m2/s2 x0.67(9.81kN) Pz r. 2g YG 2(9.81m/s2) nr' = 220 kPa

66

6.61

6.60

Bernoulli's equation

67 Flow of Fluids

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

P v2 P v2 __L + z + -1 = -2 + z + -2 Å pá == o Uá == o P2 == o r l 2g r z 2g , , ,

V2 =..}2g(z1 -z2) =~2(9.81m/s2)(6.0m) == 10.85 mis

Q == A2u2 == n(0.050 m)2 x 10.85 mis== 2.13 x 10-2 m3/s 4

uA == JL= 0.0213 m3/s =1.206mls; v1==1.12062 m2/s2 = 0.0741 m AA ²'Z"(.15m)2/4 2g 2(9.81mls2)

P vz P v2 __L + Z + -1 = tè. + Z + zs. . pá == 0 Uá = 0 r 1 2g r A 2g , ,

[ ) v;] 9.81kN[ ] PA= Yw (z1-zA -- = 3 6.0m-0.0741m =58.lkPa 2g m

Q == 10 gal/min x lft3 Is = 0.0223 ft3. v = JL = 0.0223 ft3/s == 3.71 ft 449 gal/min s ' A AA 0.0060 ft2 s

_ Q _ 0.0223 _ 0.955 ft PA v1 _ p8 vi VB - - - - , - +ZA+ - - - +z, +-; ZA= ZB AB 0.02333 s YK 2g h 2g

_ [v~-v;]_50lb[(0.9552-3.7l2)ft] lft2 =-0.0694psi PA-PB-YK 2g -7 2(32.2ft/s2) 144in2

P v2 p v2 Pt. A before nozzle; Pt. B outside nozzle: ____1!_ + z A + zs. == -1L + zB + -1L; PB ==O, ZA= zB r 2g r 2g

= [ vi - v;] == 60.6 lb [ (752 - 42.192 )ft2 /s2] 1 ft2 == 25Å1 psig PA r 2g ft3 2(32.2ft/s2) 144in2

uB=75 ft/s;uA=uB Aa =75(DÛ)2 =75(Ŀ75)2 =42.19ft/s AA DA 1.0

P u2 p u2 Pt. A at gage; Pt. B outside nozzle: ; + z A +

2~ ": + zB +

2; ; PB = O

v1-ui =(z -z )-PA =3.65m- 565kN =-53.94m 2g B A Y ID2(9.81k:N/m3)

l)B = VA(AA/AB) = VA(DAIDB)2 = uA(70/35)2 = 4uA; vi= 16u1; u2 -v2 =-15v2 A B A

-l5v1 = 2g(-53.94 m) .----------

VA= 2g(53.94 m) = 2(9.81 mls2)(53.94 m) = 8.40 mis 15 15

Q ==AA u A= 1t(.070 m)2 x 8.40 mis== 0.0323 m3/s = 3.23 x 10-2 m3/s 4

6.65

6.64

6.63

6.62

Chapter 6

P u2 P uz __!_ + z, + -1 = -2 + z2 + -2 : Pt. 1 at water surface; Pt. 2 outside nozzle; Y 2g r 2g

pá =e= 0, Uá= 0 u2 = ~2g( z1 - z2) = ~~2(-9-.8-1 _ml_s_2 (-4-.6-m-) = 9 .50 mis

Q = A2u2 = JT(.025 m )2 x 9.50 mis= 4.66 x 10-3 m3/s 4

p u2 p u2 __!_ + z1 + --1 = -2 + z2 + -2 : Pt. 1 at water surface; Pt. 2 outside nozzle; u1 =O, p2 = O y 2g y 2g

= [cz - z ) + u; ] = 62.4 lb [-10 ft + (20)2 ft2 /s2 ] 1 ftz = -1.64 si Pi y 2 1 2g ft3 2(32.2 ft/s") 144 in2 p g

u2 p uz Pi + z1 + --1 = -2 + z2 + -2 : Pt. 1 at water surface; Pt. 2 outside nozzle; u1 =O, pi = O r 2g Y 2g

(32.2 ft)[20 lb ft3 144 irr' ] Uz = ,,j2g(p1 f y+ Z1 - Z2) = 2 --2 - Å 2 2 + 8.0 ft = 59.06 ft/s s in 62.4 lb ft

Q =A u = JT(3 in )2 x 59.06 ft x ft2 = 2.90 ft3/s 22

4 s 144in2

68

6.69

6.68

6.67

6.66 Pt. 1 at o²l surface; Pt. 2 outside nozzle.

P u2 p u2 __!_ + z + -1 = -2 + z + -2 Å pá =O Uá =D P2 = o

l 2 2 2 ' ' ' Ya g Yo g u2 = ~2g(z1 -z2) = ~~2(-9-.8-l_ml_s-2)-(3 0_m_) = 7.67 mis

Q = ĆzUz = .n"(.035 m)2 x 7.67 m = 7.38 x 10-3 m3 4 s s

uA =un= i?.._= 7.38xl0-3 m3/s = 0.940m: v1 =u~= 0.942 m2/s2 = 0_0450 m AA JT(0.10m)2/4 s 2g 2g 2(9.81mls2)

P u2 p u2 __!_ + Z + --1 =_A_+ Z + zs. . p, = 0 Uá = 0

1 2 A 2 ' ' Ya g Yo g

PA = Yo[(zá -ZA)-~~]= (0.85{ 9-~~

m )4.0-0.045]m = 33.0 kPa

P1 + z + y~ = Pa + z + u~ . Pi = o Uá = o Yo 1 2g Yo B 2g ' '

Pe= y 0[ (zá - Z8) - ~!] = (0.85)(9.81)[3.0 - 0.045] = 24.6 kPa

69 Flow of Fluids

Pe= PA = -5.27 kPa

Ps = Y0 [ (z, -z8)- ~;] = (8.437)[-3.0 - 0.625] = -30.58 kPa

PA = Yo [-u~]= (0.86)(9.81)[-<3Ŀ5Ü2)2] = (8.437)(-0.625] = -5.27 kPa 2g 2(9.81)

uA =u8 =v- =u0 =u/4 UA = 3.502 mis

U2 = ~2g(z1 -z2) = ~2(9.81)(10) = 14.01 mis

1Z'(.025)2 - Q = A2u2 = x 14.01=6.88 x 10 3 m3!s 4

6.72 Analysis for u2, Q,pA,PB same as Prob. 6.69.

Q 5.6x10-3 m3 Is (See Prob. 6.69) uB = - = 2 = 2.85 mis

A8 1Z'(.05 m) 14 M²nimum pressure exists at B, highest point in system. P uz P uz _..!_ + Z + -1 = ___!!_ + Z + _!L . p, = 0 Uá = 0 y 1 2g y B 2g ' '

zs-z1=Y= -ps_u; = -(-18kN) (2.85)2m2/s2 =1.42m y 2g m2(9.81 kN/m3) 2(9.81 m/s")

6.71

Q 7.lxl0-3 m3/s (See Prob. 6.69) u2 = - = 2 = 14.46 mis

Ćz 1Z'(.025 m) 14

z -z =X= u; =(14.46)2m2/s2 =10.66m 1 2 2g [2(9.81 mls2)]

6.70

uz u2 J!i. + z + _1 = Ps + z + _!L . P1 =O Uá= O y l 2g y B 2g ' '

[ u2 l [ (2.375)2 l PB= y (z -z )-_!L =9.81 -0.90- =-11.65kPa

I B 2g 2(9.81)

u2 u2 Pi + z + -1 = PA + z + _,i_: pá = o Uá = o Zá =ZA y l 2g y A 2g ' '

u =u =u ~=u ( D2 )2 = u2/4 = 2.375 mis A B 2A 2 D

A A

= [-u:]=9.81kNá-(2.375)2m2/s2] =-l.SlkPa PA y 2g m" 2(9.81 m/s")

Chapter 6 70

6.74 PA+zA+v1=PÛ+z8+u~ ;zA=zs ~ 2g ~ 2g v1 -u~ = p8 - pA = (28.2-25.6) lb ft3 (144 in2) = 6_667 ft 2g Yo in" (0.90)(62.4 lb)ft2

uA = u8 Aa =u (Da )2 =u (~)2 = 2.56uBĿ v2 = 6.55v2 A ÛD Û5 'A Û A A

6.55v~ - u~ = 2g(6.667 ft)

5.55v~ = 2g(6.667 ft)

v = 2(32Ŀ2)(6Ŀ667) ft/s = 8.79 ft/s B 5.55

Q = ĆsUs = .1l"(8 in )2 x 8.79 ft x ft2 = 3.07 ft3/s 4 s 144in2

0.0625u~ -u~ = 2g(-18.46 ft) -0.9375u~ = 2g(-18.46 ft) t)B = 2(32.2 ft)(-18.46 ft) = 35.6 ft/s

s2 (-0.9375)

u2 u2 P A + z + _A_ = Pa + z + -Û- Ŀ z A = ZB Y A 2g Y Û 2g , v1 -u~= p8 - PA = (42-50)lbft3 (144 in2) =-l8.46 ft 2g Yw in" (62.4 lb)ft2

A (D )2 ( 1 )2 UA = Us __!!_=V __ B =V - = 0.25usĿ u2 = 0.0625 u2 A ÛD 82 ÅA B A A

6.73

u2 u2 P1 + z + _1 = Po + z +-Ü-- . pá = O Uá = O Yo 1 2g Yo D 2g ' '

Po= Yo [ (z, - z0)- ~;] = (8.437)[10.0 - 0.625] = 79.1 k.Pa

71 Flow of Fluids

-15ui = 2g[-0.627 m + 0.60 m] = 2g(-0.027 m)

uA = ~2(9.81 mls2)(-0.027 m)/-15 = 0.188 mis

Q=AAvA= 7r(O.lOm)2 x0.188mls=l.48xl0-3m3/s 4

ZB-ZA = 0.60

Manometer: PA + Yo(0.35 m) -yw(0.20 m) - ro(0.75 m) = PB PB - PA = -yw(0.20 m)-yo(0.40)m PB - PA = -yw(0.20m) 0.40 = -9.81(0.20m) -0.40m=-0.627m Yo Yo 8.64

v2=16v2 B A v2 v2 A - B = Pe - P A + (z _ z ) 2 B A g Yo

u =u AA =u (DA)2=v (100)2 =4v B A A A D A 50 A

B B

u2 u2 PA + z + zs. = PB + z + -1L Yo A 2g Yo B 2g

u2 - u2 = u2 - l 6u2 = -15u2 A B A A A

6.77

6.76 (See Prob. 6. 75) uA ~ u8 ( ~ ) ~v.(~:)' ~ 0.25u8 ~ 0.25( 10) ~ 2.50 mis

12.54h = 15u1 = 15(2.50)2 m2/s2 = 4.778 m 2g 2(9.81 m/s")

h = 4.778 ml12.54 = 0.381 m

Q =AA u A= [7t(.050 m)2/4](2.025 mis)= 3.98 x 10-3 m3/s

P u2 P u2 .......A-+zA +~=_.!!.+zB +-1L; zA=zB r; 2g r. 2g pA -pB =u~ -v1 =[uA(AAIAB)J2-v1 =[uA(DAIDB)2f-v1 Yw 2g 2g 2g

l6u2 -u2 15u2 = A A =--A 2g 2g

Manometer: PA + YwY + ywh -ymh-YwY = PB [cancel terms withy] PA - PB = Ymh -ywh = h(ym -yw) = h(13.54yw-Yw) = h(I2.54yw) PA - PB = 12.54h = 15v1 Yw 2g

.--~~~~~~~- u = 2g(l2.54)(h) = 2(9.81 m)(12.54)(0.250 m) = 2.025 mis A 15 S2(15)

6.75

72 Chapter6

6.80 (See also Prob. 6. 79) uA = uB/4 = 1 O.O ft/s/4 = 2.50 ft/s 14.04h = 15v; /2g

h= 15(2.50ft/s)2 =0.1037ftx 12in =1.24in 2(32.2ft/s2)(14.04) ft

u = 2g(14.04)(h) ~2(32.2 ft/s2)(14.04)(28 in) = 11.86 ft/s A 15 15(12 in/ft)

Ü=A u = ;ir(4 in)2 x 11.86 ft x ftz = 1.035 ft3/s A A 4 S 144 ĉn2

In(D: 14.04h-X+X= 16v;-v; =15v; 2g 2g

Manometer: PB + YoY + Ymh-yoh-YoY-YoX= PA [cancel terms withy]

PA -Pa =ymh _h-X=l3.54yw Āh-h-X=l4.04h-X Yo Yo 0.90yw

LetzA-ZB =X Let y = Distance from B to surface of Mercury.

u =u AA =u (DA)2 =u (i)2 =4v B AA AD A 2 A B B P P vz -vz A-B+z-z=Û A

A B 2g Yo

CD

6.79 P v2 p vz _A_+z +~=-Û-+z +---Û- Yo A 2g Yo B 2g

-49.6v; = 2g[-0.583 m + 0.25 m] = 2g(-0.333 m)

uA = ~2(9.81 m/s2)(-0.333 m)/(-49.6) = 0.363 mis Q =AA u A= [n(.20 m)2/4](0.363 mis)= 1.14 x 10-2 m3/s

Manometer: PA + YoY + Yo(0.60 m) -yG(0.60 m) -yÜy- yo(0.25 m) = PB PB - PA = Yo(0.35 m) -yG(0.60 m) p8 - PA = 035 _ (1.40)(9.81 kN/m3)(0.60 m) = _0.583 m Yo (0.90)(9.81 kN/m3)

u; - u~ =u; - 50.6v; = - 49 .6v;

V~ =50.6u;

(DA )2 (200)2 u =u - =u - =7.llv B A D A 75 A

B

6.78

73 Flow of Fluids

in§) 25.08 in-24.0 in= 13.36v~ /2g

uA = 2g(l.08 in)= 2(32.2ft/s2)(1.08in) = 0_659 ft/s 13.36 13.36(12 in/ft)

Q = AAUA = 0.08840 ft2 X 0Ŀ659 ft = 0.0582 ft3/s s

Manometer: PB + Yo(24 in)+ Yo(6 in)+ Yw (8 in) -yo(8 in) -yo(6 in)= PA PA - PB = Yo(16 in)+ Yw(8 in)

PA - Ps = 16 in+ Yw (8in)=16 in+ 62.4 lb/ft3 (8 in) = 25.08 in Yo Yo 55.0 lb/fr'

P P v2 vz A-B+z-z=B-A A B ~ 2g

u =u . AA =v . 0.08840 ft2 =3.789v B A AB A 0.02333 ft2 A

u~ -u~ =(3.789vA)2-v~ =13.36v~ ZA - ZB = -24 in

=60.0 si +(0.67)(62.4lb)[(20.372-81.492)ft2/s2] lft2 =31.94 si PB p g ft3 2(32.2 ft/s") 144 in ' p g

vA =g= 4.0ft3 Is _ 144 in2 =20_3?ft/s AA 7r(6 in)2 / 4 ft2

u, ~u,{~:)' ~4u, ~8I.49ftfs P v2 P v2 -A+z +_A._=_.!i.+z +_!L Y A 2g Y Û 2g

6.82

6.81

Chapter 6 74

10.0mÅz1 ZA z, Zc !D. u2 ..:.L. y 2g 2

Reference 2!J2.. lfMlt 2tl

Pt 1 A B e D Pt2 1.ro-01

!L y

,_,.......,'""' u2 :.rà_ 211 &;._ / t p- y

2 .!L 2g

6.83 Plot shown below. Data computed as follows. Pt. 1: Tank surface - Pt. 2: Outside nozzle - Ref. level at D.

u2 P Pt. 1: z1=10.0 m; -1 =0; -1 =O See Problem 6.72 for data.

2g r Pt. A: zA = 10.0 m; u~ = (3-5Ü2 m/s)2 = +0.625 m = uii = u~= u~

2g 2(9.81 m/s") 2g 2g 2g

PA = -5.27kN/mz =-0.625m= Pe r 8.437 kN/m3 r

u2 Pt. B: zB = 13.0 m; --1L = 0.625 m

2g

PB = -30.58 - -3.625 m r 8.437 Pt. C: Same as Pt. A

u2 p 79.1 Pt. D: Zn =O; _Q_ = 0.625 m; _Q_ = -- = 9.375 m 2g r 8.437

Pt. 2: z =OĿ u~ = (l4.0l)2 m = 10.0 m: P2 =O 2 ' 2g 2(9.81) ' r

75 Flow of Fluids

A centerllne) erenoe level

.!.B.. 2g

&.. y

!!A. y

TccmHead

Pt. B: Pa == (42.0)(144) == 96.9 ft r 62.4

-Total head == 116.6 ft

u~== (35.6)2 -I9.7ft 2g 2(32.2)

-Total head = 116.6 ft v1 = (0.25v8)2 = [(0.25)(35.6ft/s)]2 =1.23ft 2g 2g 2(32.2 ft/ s2)

6.84 See Prob. 6.73 for data. Let ZA= zB =O

Pt.A: PA = 50.0lb x 144 in2 x ft3 =115.4ft r in2 ft2 62.4 lb

Chapter 6 76

2

Point F: Pr = (zA - zF) - ~ = 30.0 ft - 4.0 ft = 26.0 ft r 2g PF = y(26.0 ft) = (62.4)(26.0)/144 = 11.27 psig

Point E: PE = (zA - zE) - uil2g = 9 ft - 4.0 ft = 5.0 ft r PE= y(5.0 ft) = (62.4)(5.0)/144 = 2.17 psig

uz Point C: Pe = (zA - zc) - _f.. = 15.0 ft- 0.054 ft = 14.95 ft r 2g Pe= y(14.95 ft) = (62.4 lb/ft3)(14.95 ft)(l ft2/144 in2) = 6.48 psig

Same values at point D.

= 4.77 psig

p uz p uz PointB: _!1..+z +~=-1L+z +_!!... pA=O uA=O r A 2g r Û 2g, ,

[ u~ ] 62.4 lb 1 ft2 pn=r (zA-z8)-- = [15.0ft-4.0ft] . 2g ft3 144m2

P u2 _.!L =(ZA - Zn) - _!L = 15 ft - 4.0 ft = 11.0 ft r 2g

u~ /2g = (43.95 ft/s)2/2(32.2 ft/s2) = 30.0 ft u; l 2g = (16.06)2/64.4 = 4.00 ft = u~/2g = u~/2g = uil2g = uil2g uà/ 2g = (1.867)2/64.4 = 0.054 ft = uà 12g

V elocity in 6-in Sch 40 pipe: u6 = uG x 4, = ( 43.95) 8Ŀ522x10-3 = 1.867 ft/s A6 0.2006

P uz P uz _!1_ + z + ~ =-Ü- + z +-Ü- (Neglecting energy losses); PA = PG =O, uA =O r A 2g r G 2g UG = ~2g(zA -zG) = ~2(32.2 ft/s2)(30 ft) = 43.95 ft/s

A = ;r(l.25 in )2 x ft2 = 8.522 x 10-3 ft2 G 4 144 in2

1 . . 2 . S h 40 . 4, (43 95) 8Ŀ522 X l0-3 16 06 ft/ Ve ocity m -m e pipe: u2 =uG x -= . ---- = . us Ćz 0.02333

6.85

77 Flow of Fluids

u = 2g(7.52 m) = 2(9.81)m(7.52 m) = 1.36 mis 1

80 s2(80) Q = A1u1 = n(0.075 m)2/4 x 1.36 mis= 6.00 x 10-3 m3/s

P1 - P1 v12 - v12 (9v1) 2 - v12 = 80v12

Yw 2g 2g 2g Manometer: pá + y,v(D/2) + Yw(0.60 m) - Ym(0.60 m) - Yw(D/2) = pá Pi - Pi = Ym (0.60 rn) -0.60rn=13.54(0.60)-0.60 = 7.52 m r; r:

vz vz li.+z +-1 = P1 +z +-1

1 t 2 r; 2g r; g

6.86 With Mercury at wall ofthroat, h/2 = 0.30 m; h = 0.60 m

ea.o y

-~:L. z !A. !A...:o

y 2

Chapter 6

28.0 ft = h + ply = 4.50 ft + pá/y

= (28 - 4.50] = 62.4 lb X 23.5 ft X l ft2 = 10.18 psig Pi y ft3 144 in2

h = 3 .50 ft - 1.0 ft = 2.50 ft = Depth of fluid above nozzle.

h = h + P = 2 50 ft + 12Ŀ0 lb ft3 144 in2 = 2.50 + 27.69 = 30.19 ft total r . in ' 62.4 lb ft2

To the level of the fluid surface in the tank. 1.675 m above the outlet nozzle.

h = uJ = (9.07)2 ft2/s)2 = 1.28 ft 2g 2(32.2 ft/s")

P u2 p u2 -:+z1 +

2~ =-;+z2 +

2~: v2 ~2g(z1 -z2)=~2gh

Ref. Fig. 6.37; Pt. 1 at tank surface; Pt. 2 in jet outside or²fice

78

6.93

6.92

6.91

6.90

6.89

Torricelli's theorem

. lft3/s Q = 200 gal/mm x = 0.445 ft3/s 449 gal/min

U. __ Q __ 0.445 fr' Is 144 in2 _ 9.07 ft ~2gh (See Prob 6.87)

'} A JT(3.00in)2/4(ft2) s

h= uJ = (9.07ft/s)2 =1.28ft 2g 2(32.2 ft/s")

20

u(ftls) 15

10

5

o o 2 4 8 8 10 h(ft)

-: V"

// / / /

h(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

o o

6.88

79 Flow of Fluids

6.102 AIAĀ=(DIDĀ)2= ( 46Ŀ5ft )2 =4067 1 1 1 1 8.75 in(l ft/12 in)

t2 - t1 = 2(4067) (23.0112 -2.0112) = 3427 s (57 min, 7 s) ~2(32.2 ft/s")

See Prob. 6.98: g = 386 in/s2

A /A.= (D/DĀ)2 = ( 6.25 ft(12 in/ft))2 = 14400 1 1 1 0.625 in

t2-t1= 2(l4400)(38112-25.5112)=1155s(19min 15s) ,J2Q8~ '

A/Aj= (D/Dj)2 = (1.25 m/0.025 m)2 = 2500 t2 - t1 = 2C2500) (1.38112-1.155112)=113 s (1min,53 s)

,J2(9.81)

A/Aj= (D/Dá)2 = (2.25 m/0.05 m)2 = 2025 tz - t, = 2(2025) ( 2.68112 -1.18112) = 504 s (8 min, 24 s)

,J2(9.81)

A/Aj= (D/Dá)2 = (22.0/0.50)2 = 1936 g = (32.2 ft/s2)(12 in/ft) = 386 in/s'

t2 - ti= 2(4 -Aá\1iá112 -h~12)= 2(1936) (18.5112 in-o)= 599 s (9 min, 59 s) ~ ~2(386in/s2)

A/Aj= (D/D;)2 = (12/0.50)2 = 576 2(576) ( i12 ) . t2-ti = ~ 15.0ft -O =556s(9mm, 16s) 2(32.2 ft/s")

12-ti= 2(Aá/Aj)(há112-h112)= 2(400) (2.68112-0) =296s[4min,56s] ~

2 ,J2(9.81) A1 = n(3.00 m)2/4 = 7.07 m2; Aj= n:(.15 m)2/4 = 0.0177 m2

A/Aá= 7.07/0.0177 = 400

A/Aj= (D/Dj)2 = (300/20)2 = 225

t2 - ti = 2(225) ( .055l/2 - o) = 23.8 s ,J2(9.81)

6.101

6.100

6.99

6.98

6.97

6.96

6.95

Flow due to a falling head

6.94 9.50 m = h + ply = 1.50 m +pi/y 9.81kN Pi = y[9.50 - 1.50] = 3 x 8.0 m = 78.48 kPa m

Chapter 6 80

6.106 See Prob. 6.100. p=35kN m3 =3.57m r m2 9.81kN h1 = 1.38 + 3.57 = 4.95 m; h2 = 1.155 m + 3.57 m = 4.722 m

f2 - t, = 2(25ÜÜ) ( 4.95112 -4.722112) = 57.8 s ~2(9.81)

6.105 See Prob. 6.96. p - 20kN m2 = 2.039 m r m" 9.81kN h, = 0.055 m + 2.039 m = 2.094 m; h2 = 2.039 m

t2 - t, = 2(225) ( 2.094112 - 2.039112) = 1.94 s ~2(9.81)

6.104 See Prob. 6.101. p = 2.8 lb ft3 144in2 = 6.46 ft x 12 in = 77.5 in r in2 62.4 lb ft2 ft h, = 38 in+ 77.5 in= 115.5 in; h2 = 25.5 in+ 77.5 in= 100 in

tz - t, = 204400) (115.5112 -100112) = 774 s (12 min, 54 s) ~2(386)

hi = 11.54 ft = p r

t2 - t1 = 2(576) ( 26.54112 - l l.54u2) = 252 s (4 min, 12 s) ~2(32.2)

6.103 See Prob. 6.97.

p = 5Ŀ0 lb ft3 144 in2 = 11.54 ft; h, = 15.0 ft + 11.54 ft = 26.54 ft r in ' 62.4 lb ft2

81 General Energy Equation

1 _ Q _ 0.20 ft3 Is _ 8 57 f1/ VA---

2-. t/s

AA 0.02333 ft Q 0.20

Va=-= =2.26ft/s Aa 0.0884

P uz h = __!!_ + z + ---Û- L r; a 2g

u2 u2 h = P A Pa + (z _ z ) + A - a L A a 2 Yw g

Q 0.085m3/s Uz = Ćz = 1.864X10-2 rrr'

Uz = 4.56 mis

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

P u2 p u2 _1 +z +-1 -h =-2 +z +-2

1 2g L 2 2g rw rw p1=p2=0andu1=0

v2 hL = (Zá - Z2) - -2

2g

h = 10 m- (4.56mls)2 = 8.94 m = 8.94NĿm L 2(9.81 m/s") N

2(9.81m)[140 kN m3 + 2.4 m _ 2.0 m] s2 m2 9.8lkN

=17.0mls Q = A2¿2 = n{0.05 m)2/4 x 17.0 mis= 3.33 x 10-2 m3/s

Pt. 1 at surface of water. u1 = O Pt. 2 in stream outside nozzle. p2 = O

u2 p u2 f!J_+z +-1 =h =-2 +z +-2 r w 1 2g L r w 2 2g

u =u AA =u (DA )2 a A A A D

a a

rç = PA + rw[(zA -Za)+ v;2~u~ -hL] Ua =lOft/s(~)2 =40ft/s

= 60 si + 62.4 lb [30 ft + (l 02 - 402 )ft2 /s2 - 25 ft] 1 ft2 = 52.1 si

Pa p g ft3 2(32.2 ft/s2) 144 in ' p g

P u2 P u2 __!_ + zl +-1 -hL =-2 + Z2 +-2 ; Zá = Z2 and Uá= Uz r. 2g ~ 2g

h =Pi P2 = (74.6-62.2) lb x ft3 x 144 in2 = 34.5 lb. ft/lb L Yo in" (0.83)(62.4 lb) ft2

7.5

7.4

7.3

7.2

7.1

CHAPTER SEVEN

GENERAL ENERGY EQUATION

Chapter 7

P - p (500 - 68)kN h = 2 1 = = 44.04 m (See Prob. 7.8 p = 68 kPa) A r; m29.81 kN/m3 '

1

PA = hA'YwQ = (44.04)(9.81)(0.0375) = 16.20 kW

PA = hA'YwQ = 50.97 m(9.81 kN/m3)(0.0375 m3/s) = 18.75kNĿm/s=18.75 kW

Q = 2250 L/min(l m3 /s) = 0.0375 m' /s 60000L/min

= P2 = 500 kN mJ = 50.97 m h , r ; m2 9.81kN

Ref. pts. at tank surfaces. Assume hi. =O. P v2 p v2 -' +z +-1 +h -h =-2 +z +-2

l 2g A l 2 2 r: r; g

v2 v2 p p p, +z, +-1 -hL = P2 +z2 +-2 : Vá= Vi: hL = i - 2 +(z, -z2) Yo 2g Yo 2g Yo Manometer: pá + y0(X) + y0(0.38 m) -ym(0.38 m) -y0(X) -y0(l.O m) = P2 P1 - P2 =y m(0.38 m) +y Ü(0.62 m) = 13.54yw(0.38 m) + 0.62 m = 6.337 m YÜ YÜ YÜ o.so- w

hL = 6.337 m + (-1.0 m) = 5.337 m = K( ~~) K= 2ghL = 2(9.81 m/s2)(6.337 m) = 72_7 u; (1.20 m/s)2

hL = K u; : K = 2ghL = 2(32.2 ft/s2)(0.32 ft) = 5_43 = K 2g u; (l.95 ft/s)"

Vi= 2_ = 0.10 ft3 /s = 1.95 ft/s Ći 0.05132 ft2

=0.32 ¶=hà

v2 v2 P -p Ji+ z + -1 - h = P2 + z + -2 Ŀ z = z and u =u Ā h = ' 2

1 2 l 2 2. 1 2 1 2'l r; g r, g r, Manometer: Pi + Yw(X) + Yw(6.4 in) -YcT(6.4 in) - Yw(X) = P2 p, - p2 _y ci6.4 in)- Yw(6.4in) = l.60yw(6.4 in)- Yw(6.4in) = 3.84 in Yw y w r; 12 in/ft

h = 18.6 ft + (-4.0 ft) + (8Ŀ572 - 2Ŀ262)ft2182 = 15.7 ft L 2(32.2 ft/s2 )

Manometer: PA + YwOO in) -ym(14 in) -yw(44 in)= Ps PA -p8 =Ym(14in)+yw(34in)=l3.54yw(14in)+yw(34in) =223.6inx JI!_ Yw Yw ~ 12~

= 18.6 ft

82

7.9

7.8

7.7

7.6


P v2 p v2 -1 +z +-1 +h -h =-2 +z +-2 Å Vá= Uz Yw 1 2g A L Yw 2 2g .

h = P2-P1+(z -z)=[520-(-30)]kN +0.75m A y w 2 l m29.81 kN/m3 = 56.82 m

Q = 75 L/min x 1 m3 /s = 1.25 x 10-3 m3/s 60000L/min

PA = hAYwQ = (56.82 m)(9.81 kN/m3)(1.25 x 10-3 m3/s) = 0.697 kNĀmls = 0.697 kW

2

eM= PA = 0.700hp =0.700=70.0% Pá 1.0hp

b)

p - h Q - 222.8 lb. ft 62.4 lb 0.0277 ft3 - 348 lb. ft 1 hp A- AYw - lb x~x s - lb 550lbĿft/lb

= 0.700 hp

a)

p u,2 Pi u~ Pt.1 at well surface. Pi =O and u1 =O -1 +z +-+h -h =-+z +- Yw 1 2g A L Yw 2 2g Pt.2attanksurface. V2 =O

p 40 lb ft3 144 in 2

hA = -2 + (z, -z1) + hL = . 2 2 + 120ft+10.5 ft = 222.8 lb-ft/lb r; m 62.4lbft

Q = 745 ga1/h x 1 ft3/s = 0.0277 ft3/s 60 min/h 449 gal/min

eM = PA = 0.254 hp = 0.508 = 50.8% Pá 0.50hp

b)

PA = hAYwQ = (21.55 ft)(62.4 lb/ft3)(0.1039 ft3/s) = 139.8 lb-ft/s x 1 hp

550 lbĿ ft/s = 0.254 hp

a)

20.0ft

_L 11/4 Sc:h40

pipe

u2 p v2 fi_+z +-1 +h -h =-2 +z +-2 Å

12 AL 22' r; g r, g Pi = Pz = O; Vá = O

u~ (10 ft/s)2 hA=(z2-z1)+ -=20ft+ =21.55 ft

2g 2(32.2 ft/s")

Q = 2800 gal/hr x 1 ft3 /s = 0.1039 ft3 Is 60 min/hr 449 gal/min

= 2_= 0.1039ft3/s = lO.O ft/s Vz Ći 0.01039 ft2

7.12

7.11

7.10

Chapter 7

[ ) V~ h ] 62.4 lb f. f 1 ft2 PB = Yw (z , -Z8 --- L = 3 [+10 t-1.54 t-6 ft) . 2 2g ft' 144 m

a)

84

7.15 (See Prob. 7.14)

PA = hAywQ = (68.0 ft)(62.4 lb/ft3)(2.0 ft3/s) = 8486 lb. ft hp = 15.4 hp lb 550 lb. ft/lb

d)

e)

1Jo=O 62.4 lb 1 ft2 .

Pe= 3 (40ft-l.54ft+12ft] . 2 =21.9ps1g ft 144m

Po=O P = y [cz - z ) - v~ + h J e w o e 2g 4,

Pt. D at upper tank surface. v0 = O Ve= VB = 9.97 ft/s

Pt. C at pump outlet. p o: p v2 ___f_+z +_s;_-h =__Q_+z +-Ü-

e 2 ~ D 2 r. g r; g b)

= -7.60 psig

Pt. A at lower tank surface. p A = O Pt. B at pump inlet. vA =O

v = g = 2Ŀ0 ft3 /s = 9.97 ft/s B AB 0.2006 ft2

P v2 p v2 ~+z +_A_-h =_.!!_+z +...JL r; A 2g Ls Yo B 2g

P = y [cz - z ) - v~ - h J B w A B 2g l

= 62.4 lb [- l 0 ft _ (9 .97 ft/s )2 _ 6 ft] lft2

Pç ft3 2(32.2 ft/s") 144in2

a)

PA = hAy0Q + (37.46 m)(0.85)(9.81 kN/m3)(1.25 x 10-3 m3/s) = 0.39 kNĿmls =0.390kW

hA = 35.38 m + 1.20 m + 0.24 m + 0.64 m = 37.46 m

V = g_ = 1.25 X l0-3 = 0.577 mis A AA 2.168X10-3

V = -9._ = 1.25 X l0-3 = 2.243 mis B AB 5.574X10-4

P v2 p v2 ~+z +_A_+h -h =_.!!_+z +...JL

A 2g A L B 2 ~ ~ g P P v2 v2

h = B - A + (z - z ) + B - A + h A B A 2 L

Ya g

hL =2.s( ~;) h = [275-(-20)]kN + l.20m + (2.2432-0.5772}m2/s2 +2_5(2.243mls)2

A m2(.85)(9.81 kN/m3) 2(9.81 mis ) 2(9.81 m/s")

Q = 75 L/m²n = 1.25 x 10-3 m3/s (Prob. 7.12)

7.14

7.13


Pt. 2 in outlet stream. P1 = P: = O; u1 = O

Pt. 1 at tub surface.

Tub Volume = (rrD2/4)(d) = [rr(0.525 m)2/4](0.25 m) = 0.0541 m' Q = V/t = 0.0541m3/90s=6.013 x 10-4 m3/s

O 1 _ Q _ 6.013x10-4 m' is _ 2 36 mi ut et Vi - - - 2 - Å s Ćz 7i(.018m) /4

P v2 p v2 _1 +z +-1 +h -h =-2 +z +-2

1 2 A L 2 2 r; g Y g v; (2.36 mis )2 hA = (z2 -z1) + - + hL = (1.00 - 0.375)m + 2 +0.22 m = 1.13 m 2g 2(9.81 mis )

vl =O P2 =O

v; (2.91 m/s)" hA = (z2 - z1) + - + hL = 1.25 m + Ŀ 2 + 3.0 m = 4.68 m 2g 2(9.81 mis )

Q 60 L/min(l m3 /s) 1 = 2.91 mis Vi= Ćz = 60000 L/min x 3.437x10-4 m2

PA = hAY/2 = (4.68m)(0.95)(9Ŀ~:N)( 6~~;s)

0.0436 kN Ŀm x 103 N = 43.6 N Ŀm = 43.6 W s kN s

Pt.1 at lower pump surface. P. =O P v2 p v2 -1 +z +-1 +h -h =-2 +z +-2 Pt.2outsidepipeatcutter. Yá 1 2g A L Yá 2 2g

b)

P v2 p v2 Pt.1 at lower tank surface. P, =O .....1..+z +-1 +h -h =-2 +z +-2 y0 1 2g A L y0 2 2g Pt. 2 at upper tank surface. V1 = v2 =O P2 825 kNm3 h s= -+(z2-z1)+hL= 2 +14.5m+4.2m=ll7.6m Yo ID (0.85)(9.81 kN)

p = h Q = (l17.6 m)(0.85)(9.81kN)840 L/min(l m3/s) = 13.73 kN Ŀ m A AYo m" 60000 L/min s

= 13.73 kW Pi v12 P3 vi 1 -+z1 +--hls =-+z3 +- Pt. 3 atpump inlet. Yo 2g Yo 2g

p3 = Y0[(zá -Z3)- ~~ -hls] = (0.85)(9.81 kN)[-3.0m- (4.53mls)2 _ 1.4 NĿm] =-45.4 kPa

rrr' 2(9.81 m/s") N

VJ = ..fl._ = 840 L/min(l m3 /s). 1 = 4_53 mis ~ 60000 L/min 3.090x10-3 rrr'

a)

b) Pe= 21.8 psig (same as Prob. 7.14) e) hA = (zo - ZA)+ hls + hL0 = 30 ft + 6 ft + 12 ft = 48 ft

d) PA = hAYwQ = (48 ft)(62.4 lb/ft3)(20.0 ft3/s)/550 = 10.9 hp

7.18

7.17

7.16

Chapter 7

__ ll_O_OO_l_b_ = (560 lb/in2)(144 ²n2/ft2) = 80672 lb/ft2 ²T(5.0)2 /(4) in2

F Pcy² =

Ćcyl

b)

a)

Q _ V _ 1.0 L 1 nr' _ 2 50 10_5 3/ - ----X-- - . X ID S t 40 s 103 L

133.4kN pá=ymh= 3 x(-0.15m)=-20.0kPa m

P vi vi _1 +z +-1 +h =¶ = P2 +z +-2 rg 1 2g A L rg 2 2g p2 - P. (30-(-20))kN/m2 tu= = 3 =7.50m r, 6.67kN/m

PA = hAygQ = (7.50 m)(6.67 kN/m3)(2.50 x 10-5 m3/s)(l03 N/kN) = 1.25 N'rn/s P1 = PA = 1.25W = 2.0SW

eM 0.60

Q= 9.lgal/m²n(lft3/s) =0.0203 ft3/s 449gal/m²n

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

eM = PA = 0.532 hp = 0.626 = 62.6% P, 0.850hp

-2.0lbft3 144in2

in2(62.4 lb) ft2

86

7.22

7.21

7.20

7.19 Weight Flow Rate = W= wlt = 556 lb/10 s = 55.6 lb/s Q= W = 55Ŀ6lb/s =0.891 ft3/s: vA= g= 0Ŀ891ft3/s = 10.21 ft/s

Yw 62.4lb/ft3 AA ²T(4/12)2/4ft2

LB~ vA( ~: )' ~ 10.21 fl/s( ~ )' ~ 18.15 ft/s at outlet ofupper pipe (Pt. B)

P vi P vi _!:... + z + _!:... + h -h = -1t + z + ....JL Ŀ h Ā =O andp =O A2 AL B2,L B r; g r; g

h -( ) u~-v! PA _20f1 (18.152-10.212)ft2/s2 A- Zs-ZA + -- - t+ -------

2g r; 2(32.2 ft/s") hA = 20 ft + 3.50 ft + 4.62 ft = 28.11 ft P =h W=28.ll ft(55.6lb/s)= 1563ftĿlb/s(lhp) =2.84h A A 550ftĿ}b/s p


=9.15kW

h = (6.8-3.4)(106)N/m2 + 3.0 m + (6.932 -l.502)m2/s2 = 390 m R 0.90(9810 N/m3) 2(9.81 m/s")

u = . A8 = 1.5 l.772xl0-3m2 A Va AA 3.835x10-4m2 uA = 6.93 mis; assume hL =O

hA = 1498 ft

PA = hAYoQ = (1498 ft)(0.90)(62.4 lb/ft2)(0.01515 ft3/s) = 1275 ft Ŀ lb/s = 2.32 hp 550

u2 P u2 PA +z +_!!..+h -h -h =....s_+z +i: PA:::;:O uA=O

A2 A Ls Ln c2 ' ~ g ~ g

h =Pe + (z -z ) + V~ + h + h = 80672 lbĿ ft3 + 15 ft + (.1111)2 A y0 e A 2g Ls Ln ft2 (0.9)( 62.4 lb) 2g

+ 11.5 ft + 35 ft

e)

P A + z + u1 _ h = Po + z + u~ . Pt. D at pump inlet. A 2Ŀ Lg D 2g ' 0 0 Yo g Yo PA= ,VA=

po= Yo[zA-zD- ~;-hls]

Vo = Va= 15.52 ft/s = (0.90)(62.4lb)[_5.0ft- (15.52ft/s)2 -ll.5ft] 1 ft2

Po ft3 2(32.2 ft/s") 144 in ' Po = -4.98 psig

d)

P ui p ui ----Û- + z + ----Û- - h = ....s._ + z + i Pt. A at tank surface. Yo B 2g Lo Yo e 2g Pt. B at pump outlet.

+ [( ) +u~ -u~+ h ) Pt.Cincylinder. Pa =r: Yo Zc - Za 2g Lo

Ve= 20 in x _!!.._ = 0.1111 ft: Va= g = 0.01515 ft3 Is = 15.52 ft/s 15 s 12 in s A8 0.000976 ft2

= 560 si +0.90(62.4lb)[10ft+(O.lllI2-l5.522)+35.0ft] lft2 PB p g fr' 2(32.2) 144in2 = 576.1 psig

e)

7.23

Chapter 7

2

hà = (zá - z2) - ~~ = 4.0 m- (5.14 mls)2/2(9.81mls2)=2.65 m

_ Q _ 600 L/min(l m3 /s) = 5.14 mis Vi - Az - 60000L/min1.945x10-3 rrr'

Pt.1 at water surface. P. =O, v1 =O Pt. 2 in outlet stream. P: =O

P v1 p v2 _!_+z +-1 -h =-2+z +-2 Y 1 2g l Y z 2g

88

7.28

[50.0-(-2.30)]lbĿft3144 in ' (3.822 -1.742)ft2 h = +25ft+ +3.4ft=154.lft A in2 ( 60 .O lb) ft2 2(32.2 ft/s" )s2 PA = hAyJ2 = (154.l ft)(60.0 lb/ft3)(0.0891 ft3/s)(l hp/550 Ib-ft/s) = 1.50 hp

Q = 40 gal/mini 449 = 0.0891 fr' /s

u = _g_ = 0.0891 ft3/s=1.74 ft/s 1 Aá 0.05132 ft2

V = _9_ = 0Ŀ0891 = 3.82 ft/s 2 Ćz 0.02333

2

hè = (z, -z2) - 22_ - he = 10 m- 0.638 m - 1.40 m = 7.96 m 2g

= 2._ = 0.25 m3 /s = 3.54 mis: u: = 3.542 m2 /s2 = 0.638 m Vi Az Jr(0.30 m)2/4 2g 2(9.81)mls2

PR = hRy0Q = (7.96 m)(0.86)(9.81 kN/m3)(0.25 m3/3) = 16.79 kNĿmls = 16.79 kW Po= PR. eM = 16.79 kW X 0.75 = 12.60 kW

Pt. l at oil surface. P, =O and v1 =O Pt. 2 in outlet stream. p2 = O

2 2 P1 + z + !!J_ - h - h = P2 + z + !:!L

1 2g R L 2 2 Ya Yo g

P v2 p u: -A+z +---A-h -h =......l!.+z +-Û- Yw A 2g R L r; B 2g

P P v2 v2 h = A - B + (z - z ) + A - B - h R A B 2 L r: g Q = 3400 gal/min(l ft3 /s) = 7.57 ft3/s

449 gal/min [21.4-(-5)]lbĿft3(144 in2) (9.742 -2.742)

hR = 2 2 + 3 ft + ft - 2.95 ft = 62.3 ft in (62.4 lb) lft 2(32.2)

PR = hRYwQ = (62.3 ft)(62.4 lb/ft3)(7.57 ft3/s) = 29454 ft Ŀ lb/s = 53.6 hp (550 ft Ŀ lb/s)/hp

u = il = 7Ŀ57 ft3/s -9.74 ft/s A AA 0.7771ft2 Q 7.57

V8 =-=--=2.71ft/S AB 2.792

hL = 2.0 V~ = 2.0(9.74)2 = 2.95 ft 2g 2(32.2)

7.26

7.25

7.24


PA +z +v~ +h -h =Pa +z +v~ v =2_=3.34lft3ls=9.62ftls A 2g A lo B 2 B A 0.3472 ft2 r: r, g B

h _ p8-pA ( ) u~-v~ h _(85-5)lb(ft3)144in2 25ft A - + Z -z + + - + Yw 8

A 2g L" in2(62.4 lb)ft2

(9 622 6 102 \f1t2 / 2 + Å - Å V ,~ 's + 28 ft = 238.5 ft 2(32.2 ft/s")

PA = hA"fwQ = (238.5 ft)(62.4 lblft3)(3.341ft31s)l550=90.4 hp

p u2 p v2 -1 + z + -1 - h - _A_ + z + _A_ Pt.1 at reservoir surface. P, = O, v1 = O

1 2g ls - A 2 r; r; g p v2

h = (z, - ZA) = _A_+~+ hL r ; 2g

Q = 1500 gallmin x 1 ft31sl449 gal/min = 3.341 ft31s

u = 2_ = 3.341 ft3 Is = 6.10 ftls A AA 0.5479 ft2

h = 5.0 lb(ft3)(144 in2) + (6.10 ftls )2 + 0.65 ft = 12.8 ft in2(62.4 lb)(ft2) 2(32.2 ft/s")

2

hi. = (z, - z2) .!!1__ - hR 2g

Uz = S2__ = 1000 gallmin x 1 ft3 Is = 6.4 l ftls Ćz 0.3472 ft2 449 gal/min

(6.41 ft/s)" hL=165ft- 2 -146.4ft=17.9ft= 17.9fHbllb

2(32.2 ft/s )

h = PR = 37.0 hp(550 ft ĿIbis) = 146.4 ft R r WQ 1 hp(62.4 Iblft3)(2.227 ft3 Is)

Pt.1 at reservoir surface. P, =O, u1 =O

Pt. 2 in outlet stream. p2 =O

2 u2 J!.i + z + l _ h _ á,, = P2 + z + _2 12 R"L 22 r; g r; g

(l2.58m/s)2 -4.60m] = 35.0 kNlm2 = 35.0 kPa 2(9.81 m/s")

Pt.1 at tank surface. P, =O, u1 =O

u = g = 0.060 m3 Is = 12.58 mis 8 A8 4.768x10-3 m2

7.32

7.31

7.30

7.29

Chapter 7 90

Zá - Z6 = -1.0 ft Q 0.390 ft3 Is

VG = - = = 11.72 ftls = lh = V4 = Vs A6 0.03326 ft2

vá (ll.72)2 =2.13 ft 2g 2(32.2) hL,_6 = 2.80 + 28.50 + 3.50 = 34.8 ft

p = h Q: h = PA = 12496 ft ĿIbis = 552_5 ft A AY A rQ (58.03 lbl ft3)(0.390 ft3 Is) hR = -1.0-2.13 - 34.8 + 552.5 = 514.5 ft p = h Q = (514.5 ft)(58.03 lblft3)(0.390 ft31s) = 21.16 h R RY 550 ft Ŀ lbls/hp p

Pi + z + v12 - h + h -h = P6 + z + vá : Pi =p6 =O V1 =O r 1 2g uà A R r 6 2g ,

v2 hR = (z1 - z6) - -6- - h + h 2g 4-6 A

7.35

Yo= sgoYw = 0.93(62.4 lblft3) = 58.03 lblft3

Q = 175 gallmin x 1 ft31sl449 gallmin = 0.390 ft3ls

PA =Pà x eM = 28.4 hp(0.80) = 22.7 hp x 550 ft-lb/s/hp = 12496 ft-lb/s

General datafor Problems 7.35 through 7.40:

7.34 (See Prob. 7.33) Vi= 2(4.28 mis)= 8.56 mis

v; = (8Ŀ56mls)2 =3.74m:ThenhL=6.88mx 3Ŀ74m =27.52m 2g 2(9.81 m/s") 0.936m

[ u; h] 8.07k:N[ ] P1 = h (z, -z1)+-+ L = 3 5.0+3.74+27.52 m=292.6kPa 2g m

p, = 292.6 kPa x 1.0 psil6.895 kPa = 42.4 psig

Vi = 2._ = 500 L/min x 1 m3 Is = 4_28 mis Ćz 1.945x10-3m2 60000 L/min

p1 = 15.0 psi x 6895 Palpsi = 1.034 x 105 Nlm2 x 1k:Nll03N=103.4 kNlm2

L=o.936m

Pt .1 at tank surface. v1 = O Pt. 2 in outlet stream. p 2 = O

- 5.0 m- (4Ŀ28 mls)2 = 6.88 m 2(9.81 m/s")

P v2 p v2 _1 +z +-1--h =-2 +z +-2 rK 1 2g L rK 2 2g be= li+(z -z )-v; =103.4k:Nlm2 h ' 2 2g 8.07 kNlm3

7.33


Fig. 6.2 suggests either a 2-in or 2 Yz-in pipe for the discharge line. The given 2 Yz-in pipe size is satisfactory.

7.40 For Q = 175 gal/min, Fig, 6.2 suggests using either a 2 Yz-in or 3-in Schedule 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 .1 O psig, a larger pipe may be warranted to decrease the energy losses in the suction line and in crease the pump inlet pressure. See Chapters 9-13, especially Section 13.12 on net positive suction head (NPSH).

P u2 p u2 -5 + z + -5 - h = -6 + z + -6 Å us = u6, P6 = O r 5 2g lw r 6 2gÜ

[ J lb 1 ft2 Ps = y (z, -z5) + hl = 58.03-(-1.0 ft + 3.50 ft] . 2 = 1.01 psig

s-s ft3 144 m

. fu 1~ p4=p3- yhL,_, =219.lpsig-58.03 3 (28.5ft) . 2 =207.6psig ft 144m

P u2 p u2 U3 = u4 _3 +z +-3 -h =-4 +z +-4 .

3 l 4 " r 2g ,_, r 2g z3 = z,

Ps = 219.1 psig

u2 u2 12 + z + -1 - h + h = P3 + z + -3 : PI = o Uá = o

l 2 L1_, A 3 2 ' r g - r g

[ u2 ] lb ft Ŀ ft2 Ps =y (zá -z3)--3 -hl, +h, =58.03-(-4.0-2.13-2.80+552.5] . 2 2g -z ft3 144 m

P u2 P u2 _..!_+z +-1 -h =-2 +z2 +-2: PI =O, Vi =O r 1 2g i,_, r 2g

Pz = r[(z1 -z2)- ~~ -hi,_,]

= j{ = 0.390 ft3 /s = 7.59 ft/s Uz Ćz 0.05132 ft2

u~ = (7.59)2 = 0.896 ft 2g 2(32.2) lb 1 ft2 Å

p2= 58.03-3[-4.0ft-0.896ft-2.80ft] . 2 =-3.lOpsig ft 144 m

7.39

7.38

7.37

7.36

Chapter 7 92

P1 = PA = 3.08 hp = 4.28 hp eM 0.72

7.43

p1 + z + u12 -h + h = p2 + z + u~ Pt.1 at creek surface. P, =O, u1 =O r 1 2g L A r 2 2g Pt.2attanksurface. U2 =O

b,= p2 +(z -z)+h =30lb ft3 144!n2 +220ft+15.5ft=304.7ft y 2 1

L in" 62.4 f¿ ft

PA = hAyQ = 304.7 ft x 62.4 lb x 40 gallmin (1 ft31s)hp = 3.08 h ft' 449 gal/min 550 ft -Ib Is p

7.42

y= 0.76(62.4 lblft3) = 47.42 lblft3 z2 - zá = -22 in(l ft/12 in)= -1.833 ft

Q = 40 gal x ft3 0.668 ft' Is 8 s 7.48 gal

Vz = _fl = 0.668 ft3;s X 144 in2 _ 30.64 ft/s Ćz 7²(2.0 in) I 4 ft2

_ Q _ 0.668 ft3 Is 144 in" _ 0 378 fil Uá - -~ 2 X - Å t S A1 JT(18.0in) 14 ft2

i=47.42 ~[-1.833ft+(30.642-0.3782)ft2ls2 +4.75ft] lft2 p ft3 2(32.2 ft/s") 144 in2 Pt = 5.76 psig

u2 u2 P1 + Zá + _l - hL = P2 + Z2 + _2 : P2 = o r 2g r 2g

[ u2 -u2 ]

Pi = r (z, Zá)- 2 2g 1 + hL

7.41


7.45 P0 = PR X eu= 1.62 hp x 0.78 = 1.26 hp

h =45.06ft+ (24Ŀ52-10Ŀ2i2)ft2/s2 =52.76ft R 2(32.2 ft/s")

PR = hRyQ = (52.76 ft)(0.90)(62.4 lb/ft3)(0.3007 ft3/s) = 891.0 ft-lb/s lhp PR = 891.0 ft-lb/s x = 1.62 hp

550 ft -Ib/s

2 2 hR = Pi - Pz + u, - Vz

Yo 2g Manometer: pá + y0y + Yo(38.5 in) -ym(38.5 in) -yo))= P2

Pi - P2 = y,,,(38.5 in) - y0(38.5 in)

p, - Pz = Ym (38.5 in)-38.5 in = 13Ā54Yw (38.5 in) - 38.5 in Yo Yo 0.9yw

Pi - Pz = 540.7 in x __!_!!_ = 45.06 ft Yo 12in

Vi = S1_ = 135 gal/min x lft3 /s = 0.3007 ft3 Is = 24.50 ft/s Aá 0.01227 ft2 449 gal/min 0.01227 ft2

= 2_= 0.3007ft3/s =l0.2lft/s Vi Ćz 0.02944 ft2

7.44

Chapter 8 94

NR= vDp =QDp = QDp = 4Qp: Dmin= 4Qp =~ Õ AÕ ²'TD2 ²'TÕD ²'TÕNR ²'TNRv -Õ

4

Q = 4.0 L/min x 1 m3 /s = 6.667 x 10-5 m' /s: Let NR = 2000 60000 L/min

8.5

d)

e)

b)

N - vD - (10.72)(0.1723) - 1 53 lOs ( fr A A) R - - - . 5 - Å X V om pp. V 1.21X10-

N = vD p = (10. 72)(0.1723)(1.53) = 4 28 105 (p ti A ) R 6 Å x , Õ rom pp. B Õ 6.60x10-

N _ vDp _ (10.72)(0.1723)(1.86) _253 .(p f A B) R- --- Ā Ā - ,Õ rom pp.

Õ 1.36 X 10-2

N. _ vDp _(10.72)(0.1723)(0.87)(1.94) _328 104 (Õfi A D) R - -- - 5 - Å x rom pp. V 9.5X10-

a)

0.25 ft3 /s v=QIA= 2 =10.72ft/s;D=0.1723ft

0.02333 ft 8.4

8.3 Let NR = 2000 = vDp/Õ

v.= NRÕ= 2000(4.0xl¿-2) =0.894m/s max Dp (0.10)(0.895)(1000)

Q=Av= ²'T(O.l¿m)z x0.894m/s=7.02xl0-3m3/s 4

8.2 Let NR = 4000 = vD!v: V= 4.38 X 1 o-6 ft2/s-App. A; D = (2/12)ft o= NRv = 4000(4.38x10-6) =0.105 ft x 0.3048 m =0.03204 m

D 2112 s ft s

N = vD p = (2.29)(0.333)(1.26)(1.94) = 249 Laminar R Õ 7.5xl0-3

L fromApp.D

Q 0.20ft3/s 144in2 . . u= - = 2 x 2 = 2.29 ft/s; D = 4.0 m(l ft/12 in) = 0.333 ft A JZ"( 4.0 in) / 4 ft

8.1

REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, ANO ENERGY LOSSES DUE TO FRICTION

CHAPTER EIGHT

95 REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, ANO ENERGY LOSSES DUETO FRICTION

N = uD = (l.78)(0.0l34) = 6.62 x 104 Turbulent R V 3.60X10-?

u= Q = 15.0L/min x 1 m31s = 1.78 mis A 1.407 X 10-4 m 2 60000 Llmin

8.10

N _ uDp (0.899)(0.0243)(860) _ 4 76 104 T b I

R - -- - Å x ur u ent Õ 3.95X10-4

Q 25 L/min 1 m' Is u= - x = 0.899 mis A 4.636xl0-4m2 60000 L/m²n

8.9

8 8 .,., _ uD _ (3.06)(0.0475) _ 112 105 T b l t Å 1v R - - - -6 - Å x ur u en

V l.30x10

u= Q = 325 L/min x 1 m31s = 3.06 mis A 1.772x 10-3 m2 60000 L/min

At 104ÜF

8.7 Auto. Hydraulic Oil Medium Hydraulic Oil

N _ uD _ (10)(0.4011) = 5 ll 104 t b .,., = 10(0.4011) = 511 104 R - - 5 Å X ur . lVR 5 Å X turb. V 7.85 X 10- 7.85 X 10-

N = (10)(0.40ll) = 9328 turb. N = 10(0.40l l) = 5563 turb. R 4.30 X 10-4 R 7.21 X 10-4

At 212ÜF

NR = uD p: Õ = uD p = (2.97)(0.0779)(890) = 4.12 X 1 o-3 Pa-s Õ NR 5X104

u= Q = 8.50 L/min x 1 m3 Is = 2.97 mis A 4.768x10-3 m2 60000 L/min

From App. D, oil must be heated to lOOÜC for SAE 10 oil.

8.6

Smallest listed

D . = 4.244 X 10-8(906) = 3 59 10-4 . .!__. b D = 4 57 mm 1 . X m, lll tu e, . mm 1.07X10- 8

d)

4.244 X 10-8(790) 3 . Dmin = 3 = 0.0186 m; --m tube, D = 18.9 mm

1.8 X 10- 4 e)

D . = 4Å244 X 10-8(680) =O 101 . 5-' t b D = 122 mm 4 Å m, lll u e, mm 2.87 X 10-

b)

Dmin = 4(6.667 x 10-5) = 4.244x10-s = 4.244 x 10-s = 0.0647 m = 64.7 mm n(2000)(v) v 6.56 x 10-1

3-in Type K copper tube--D = 73.8 mm

a)

Chapter 8

NR = uD = (14.65)(0.01788) = 6237 Turbulent V 4.20X10-5

u= Q 1.65 gallmin x 1 ft3 Is = 14_65 ft/s A 2.509x10-4 ft2 449 gallmin

N _ uD (14.65)(0.01788) _ 1105 L . R - - = - ammar

V 2.37 xl0-4

(0.423)(0.0475)(890) 5 97 l L . = . very ow- ammar 3.0

u= Q = 45 L/min x 1 m3 Is = 0.423 mis A 1. 772X10-3 m2 60000 L/min

NR = uDp (0.423)(0.0475)(890) 2237 e . . 1 z - -'----'-'---3-'--'-__:_ = r²tlca one Õ 8xl0-

u _ uDp (2.67)(0.01892)(890) _ 15 O l L . 1v R - -- = - . very ow- ammar

Õ 3.0

Q 45 L/min 1 m' Is U= - = X = 2.67 mis

A 2.812x10-4 m" 60000 L/min NR = uDp = (2.67)(0.01892)(0.89)(1000) = 5Å61x103 Turbulent

Õ 8xl0-3

Note: Õ from App. D.

Q =Av= 4.609 x 10-2 ft2 x 0.424ftls=1.96 x 10-2 ft3/s

96

8.20

8.19

8.18

8.17

8.16

8.15

8.14 NR = uDp: u= NRÕ = (4000)(4.01x10-s) = 0.424 ftls Õ Dp (0.2423)(1.56)

NR = uDp = (0.732)(0.00508)(~.88)(1.94) = 33.4 Laminar Õ l.90xl0-

Note: sg of oil may be slightly lower at 160ÜF.

_ Q _ 0.40 gal ft3 1 hr 1 _ 0 732 "~' U- -- X X--X - . u1S A hr 7.48 gal 3600 s 2.029X1 o' ft2

NR = uDp = (0.732)(0.00508)(?.88)(1.94) = 1.02 Laminar Õ 6.2x 10-

NR = uD = (8.59)(1.563) = 9.59 x 10s V l.40x 10-5

u= Q = 16.5 ft3 Is = 8.59 ft/s A l.920 ft2

8.13

8.12

8.11


10-6 m2/s v = 17.0 es x = 1.7 x 10-5 m2/s

1 es

u= Q = 215 L/min x 1 m3/s = 7.142 mis A 5.017x10-4m2 60000 L/min

NR = uD = (7.142)(0.0253) = l.06 x 104 V l.70xl0-5

u= 1.30 es x 1.076x10-s ft2 Is = 1.40 x 10-5 ft2/s 1 es

. 1 ft3 Is Q = 45 gal/mm x = 0.1002 ft3/s 449 gal/min

u=Ü= 0.1002ft3/s =14.65ft/s A 6.842X10-3 ft2

NR = uD = (14.65)(0.0933) = 9Å78 x 104 v l.40x 10-s

8.25

8.24

8.23 (See Prob. 8.22)

u1 = NRv = (2000)(3Ŀ84x10-6) = 0.1237 ft/s: Vz = 2u1 = 0.2473 ft/s D 0.0621 '

-3 2 449 gal/min . Q1=Au1=(3.027x10 ft)(0.1237ft/s)x lft3/s =0.1681gal/mm

Qz = 2Q1 = 0.3362 gal/mio

Q1 = 0.530 gal/min-Lower Limit Q2 = 2Q1 = 1.060 gal/min-Upper Limit

Uá= N Rv = 2000(1.21x10-s ft2/s) = 0.3897 ft/s D 0.0621ft

For NR = 4000, Vz = 2(0.3897 ft/s) = 0.7794 ft/s Q1 =A u1 = (3.027 x 10-3 ft2)(0.3897 ft/s)

= 1.180 x 10-3 ft3/s x 449 ga;/min 1 ft Is

A _ Q _ 500 gal/min 1 ft3 Is _ 0 1114 f12 5 . S h 40 . - - - x - . t => -m e . pipe u 10.0 ft/s 449 gal/min

A = 0.1390 fr', D = 0.4026 ft

A 1 _ Q_(500/499)ft3/s _801f~1 ctua u - - - - . u s A 0.1390 ft2

NR = uDp = (8.01)(0.4026)(2.13) = 2.12 x 104 Õ 3.38xl0-4

Changing from laminar flow, through critica} zone, into turbulent flow could cause erratic performance. Also, u= 14.65 ft/s is quite high, causing large pressure drops through the system.

8.22

8.21

Chapter 8 98

Let NR = 2000;/ = 64/NR = 0.032; NR = uD p Õ

u= NRÕ = (2000)(8.3x10-4) = 2.85 ft/s D p (0.3355)(0.895)(1.94) L u2 100 (2.85)2 hL = j - - = (0.032) Ŀ Ŀ ft = 1.20 ft = 1.20 ft-Jb/Jb D 2g 0.3355 2(32.2)

8.29

P u2 p u2 -1 +z +-1 -hl =-2 +z2 +-2 : Vá= Vi; zá =z2: Pi =P: =ywhL Yw 1 2g r; 2g u= Q = 12.9L/min x 1 m3/s = 1.528 mis A 1.407 x 10-4 m2 60000 L/min

N _ uD _ (1.528)(0.0134) _ 5 35 104 ( b l ) R - - - - Å x tur u ent

V 3.83xl0-7 Die= 0.0134/1.50 x 10-6 = 8933; Then/ = 0.0205

h = f ~~=(0.0205)Ā 45 Ŀ (1.528)2 =8.19m L D 2g 0.0134 2(9.81) Pi - P2 = YwhL = 9.56 kN/m3 x 8.19 m = 78.3 kN/m2 = 78.3 kPa

8.28

P u2 P u2 -1+z +-1 -hL=-2+z2+---1....: u1=LJi Yo 1 2g Yo 2g P1 -p2 =ro[z2 -z, +hL] NR = uDp = (0.64)(0.0243)(0.86)(1000) = 787 (Laminar);/= 64 = 0.0813

Õ 1.70X10-2 NR

h = f ~~=0.0813x 60 x (0.64)2 =4.19m L D 2g 0.0243 2(9.81) P1 -p2 = (0.86)(9.82 kN/m3)[-60m +4.19 m] = -471 kN/m2 = -471 kPa

8.27

v = 1.20 es x 10-6 m2/s = 1.20 x 10-6 m2/s 1 es

u= Q = 200L/min x lm3/s = 8.69 mis A 3.835x10-4m2 60000 L/min

NR = uD = (8.69)(0.0221) = 1.60 x 10s V 1.20X10-·

8.26

99 REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, ANO ENERGY LOSSES DUE TO FRICTION

8.32 From Prob. 8.31, pá - P2 = yjiL; hL = pi - p2IYw h = (1035 - 669)kN/m2 = 3 7 _3 m = f !::__ !!._ L 9.81 kN/m3 D 2g f = hLD2g = (37.3)(0.03388)(2)(9.81) = 0.048

Lv2 (30)(4.16)2

u= Q = 225 L/min x 1 m3 Is = 4.16 mis A 9.017x10-4m2 60000 L/min

NR= vD =(4.l6)(0.03388)=1.08xl05:Then D =55for/=0.048 V 1.30X10--6 E

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

v2 p u2 L z Pi + z1 + -1 -hL = -2 + z2 +-2 : Zá = z2: Vá = l}z: P1 - P2 = YbhL = Yb f - ~ yb 2g Yb 2g D 2g

u= Q= 20L/min x lm3/s =0.719m/s A 4.636x10-4 nr' 60000 L/min

p = I_= 8.62 kN x-s_2 -x 103 N x 1 kg Ŀmls2 = 879 kg/m' g m' 9.81m kN N

NR = vDp = (0.719)(0.0243)(879) = 3_89 x 104 Õ 3.95X10-4

Die= 0.0243/4.6 x 10-5 = 528; Then/ = 0.027 3 100 (0.719)2 2

p, -p2 = 8.62 kN/m x 0.027 x x m = 25.2 kN/m = 25.2 kPa 0.0243 2(9.81)

8.31

v2 p vz p A +ZA + ~ - hl = _..!! + Z8 + _.!L : VA = LJs Yo 2g Yo 2g PB = PA + Yo[zA -z8 -hL] v=NRÕ= (800)(4x10-4) =0.717ft/s

D p (0.2557)(0.90)(1.94)

h = f !::__ s: = 64 . 5000 . (0.717)2 = 12.5 ft L D 2g 800 0.2557 2(32.2)

PB = 50 psig + (0.90)(62.4 lb/ft3)[-20 ft - 12.5 ft] 1 f~2 2 = 37.3 psig 144m

8.30

Chapter 8

p u2 p u2 -1 + z + -1 - h = s:ç + z + _.!:,_ Pt. 1 at tank surface. p1 = O, t>1 = O

1 2g Ls A 2 Yw Yw g

zá - zA = h = PA + v! + hL Yw 2g

NR = uAD = (6.097)(0.835) = 4.21 x l05: D = 0.835 = 5567:f= 0.0155 V 1.21X10-5 e 1.5 X }0-4 L u2 45

he = j - - == (0.0155) X -- X 0.577 ft = 0.482 ft D 2g 0.835

h = S.O lbĿ ft3 144 in2 + 0.577 + 0.482 = 12.60 ft in262.4lb ft2

a)

. 1 ft3 /s 3 Q = 1500 gal/mm x = 3.34 ft /s 449 gal/min

= JL = 3.34 ftJ /s = 6.097 ft/s: u! = (6.097)2 = 0.577 ft VA AA 0.5479 ft2 ' 2g 2(32.2)

L u2 FromProb. 8.31,pá -p2=YwhL=Ywf -- D 2g

_ Q_ 15.0ft3/s _849f1/ v- -- - . t s A ²'Z"(I .50 ft )2/4

NR = uD = (8.49)(I.50) = 9.09 x 105: D I.50 = 3750Ŀ f= 0.0158 V 1.40 X 10-5 e 4 X 10-4 '

- = f !:_ ':!___ = 62.4 lb X 0.0158 X 5280 ft X (8.49)2 ft2 /s2 X 1 ft2 = 30.5 psi p, P2 r; D2g ft' l.50ft 2(32.2ft/s2) 144in2

NR = uD = (12.46)(0.5054) = 6.88 x 105: D = 0.5054 = 3369: f= 0.0165 v 9.15x10--6 e 1.5x10-4

h = f !:_ tj__ + tj__ = 0.0165 X 550 X (12.46)2 + (l2.46)2 = 45.7 ft D 2g 2g 0.5054 2(32.2) 2(32.2)

Pt. 1 at tank surface. p1 =O, u1 =O Pt. 2 in outlet stream. p2 = O D = 0.5054 ft = 0.2006 ft2

P u2 P u2 -1 +z +-1 -h =-2 +z +-2

1 2g l 2 2 Yw Yw g u2

h = Zá - Zz = he + -2 2g

v= Q = 2Ŀ50ft3/s = 12.46 ft/s A 0.2006 ft2

100

8.35

8.34

8.33


hA = P2 +(z2 -z1)+hL Yw

Q= 745 gal x 1 h x 1 ft3/s = 0.0277 ft3/s h 60 min 449 gal/min

Q 0.0277ft3/s 461 ft/s ² Å

u= A = 0.0060 ft2 = Ŀ t s m pipe

NR = uD = (4.61)(0.0874) = 3.33 x 104: D = 0.0874 = 583: f= 0.0275 V 1.21X10-s 8 1.5X10-4

h =f !:~=(0.0275) 140 x(4.61)2 ft=14.54ft L D 2g 0.0874 2(32.2)

h = (40lb)ft3(l44in2) +120+14.54=226.8ft A in2(62.4 Jb)ft2 PA = hAyQ = (226.8 ft)(62.4 lb/ft3)(0.0277 ft3/s)/550 ft-lb/s/hp = 0.713 hp

Pt. 1 at well surface (p1 = O psig). Pt. 2 at tank surface. Uá=t>i=O

8.36 p u2 p u2 -1.+z +-1 +h -h =-2 +z +-2 r 1 2g A L r 2 2g

p A + z + v1 -h + h =Pe + z + u~ r A 2g Ld A r B 2g = Jl = 3Ŀ34 ft3 /s = 9.62 ft/s

Va AB 0.3472 ft2

hA= Pa-PA +(zs-zA) + u~-v1 +h =(85-5) lbft3(144in2) +25 2 LJ in2(62.4 lb)ft2 r; g

+ (9.622 -6.0972)ft2/s2 + 89.9 = 300.4 ft 2(32.2 ft/s")

N = u8D8 = (9.62)(0.6651) = 5.29 x 10s Rn V l.21xl0-5

DIE= 0Ŀ6651 = 4434: f= 0.016 1.5 X 10-4

h =f!:~=(0.016)x 2600 X (9Ŀ62)2 =89.9ft Ld D 2g 0.6651 2(32.2)

p = h Q = 300.4 ft X 62,4 }b X 3Ŀ34 ft3 hp = 113.8 h

A AYw ft' S 550 ft Ŀ lb/s p

b)

Chapter 8

2 2 p 3 V3 h p 2 V2 (( ) h ] - + Z3 + - - L = - + Z2 + - : P3 = P2 + Zz - Z3 + L Y r 2g r 2g Ps = 140 kPa + (1.10)(9.81kN/m3)[8.5m+37.86 m] = 640 kPa

b)

2

hA = Pz + (z - z ) + !:i + h 2 1 2 L r g

v = Q = 95 L/min x 1 m3 /s = 3.23 mis: !!._ = (3.23)2 - 0.530 m A JT(0.025m)2/4 60000L/min 2g 2(9.81)

NR = vDp = (3.23)(0.025)~1100) = 4.44x 104: f= 0.021 (smooth) Õ 2.ox10- L v2 85

hr = f - - = (0.021)--(0.530)m = 37.86 m D 2g 0.025 140kN/m2 hA = 3 + 7.3 m + 0.530 + 37.86 m = 58.67 m (1.10)(9.81 kN/m )

PA = hAyQ = (58.67 m)(l.10)(9.81 kN/m3)(95/60000)m3/s = 1.00kNĿmls=1.00 kW

a)

Pt. 1 at tank surface. Pi = O; v1 = O Pt. 2 in hose at nozzle. Pt. 3 in hose at pump outlet. IJ:>=Uz

102

8.38 p v2 p v2 --1+z +-' +h -h =-2 +z +-2 r 1 2g A L r 2 2g

è. = Yw[(Zz -zá)+ ~~ +hr]

Vz = 2.._ = 75 gaVmin x 1 ft3 Is = 11.8 ft/s: s: = (1 l .8)2 = 2.167 ft Ćz 0.01414 ft2 449 gal/min 2g 2(32.2)

NR = vD = (11.8)(0.1342) = 1.31x105: D = 0.1342 = 895:f = 0.0225 V 1.21 X 10-5 e 1.5 X 10--4

hr = f ~!!._=(0.0225) 300 (2.167 ft) = 109.0 ft D 2g 0.1342

p1 = 62Ŀ~ lb [-3 ft + 2.167 ft + 109.0 ft] 1 f~2 2 = 46.9 psi

ft 1441n

Pt. 1 at tank surface. v1 = O Pt. 2 in outlet stream. P2 = O

8.37 p v2 p v2 _1+z +-1 -h =-z+z +-2 á 2g L 2 2 Yw Yw g


L v2 Let hL = 93.5 m (fromProb. 9.13): hi. = f -- D 2g

L = hLD(2g) = (93.5 m)(0.1463 m)(2)(9.81 m/s2) = 8682 m = 8.68 km fu2 (0.026)(1.189 m/s)"

b)

With pumping stations 3.2 km apart:

N _ uDp _ (1.189)(0.1463)(930) _ 2 05 104 b l R - -- - - Å x tur u ent Õ 7.9xl0-3

Dle = 0.1463 m/4.6 X 1 o' m = 3180;! = 0.026

h = h = f ~ _:C = (0.026) 3200 (1.189)2 m = 40.98 m A L D2g 0.14632(9.81) PA = hAyQ = (40.98)(0.93)(9.81)(0.02) = 7.48 kW

a)

8.40 At 1 OOÜC, Õ = 7 .9 X 10-3 Pa-s

u2 p u2 ~á + Z1 +

2~ + hA -hL = / + Z3 +

2~ : pá = p3, Uá= U:J, Zá = Z3

h = h = 853 k:N/m2 = 93.5 m A L (0.93)(9.81 kN/m3) PA = hAyQ= (93.5 m)(0.93)(9.81 k:N/m3)(0.02m3/s)=17.1 k:NĿmls=17.1 kW

b)

L u2 a) P2-p3=yohL=Yof -- D 2g

NR = uDp = 1.189(0.1463)(930) = 1079 Laminar Õ 0.15

- = h = (0.93)(9.81 k:N/m3)(~)( 3200 ) (1.189)2 m = 853 kPa Pz p3 Yo L 1079 0.1463 2(9.81)

8.39 Q = 1200 L/min x 1 m3/s/60000 L/min = 0.02 m3/s u= Q = 0.02 m3/s = 1.189 mis

A 1.682x10-2 m2

Chapter 8 104

2 2 Pi Vá h P2 Vz h - + Zá + 2- L = - + Zz + 2: PI - P2 =Yo L Yo g Yo g Q 60 gallmin 1 ft3 Is u= - = x = 9.45 ft/s A 0.01414 ft2 449 gal/min

N _ vDp _ (9.45)(0.1342)(0.94)(1.94) _

272 L .

R- --- - ammar Õ 8.5X10-3

h = j s ç. 64 X 40 X (9.45)2 ft = 97.23 ft L D 2g 272 0.1342 2(32.2)

(62.4 lb) 1 ft2 . pá - P2 = YohL = (0.94) 3 (97.23 ft) . 2 = 39.6 psi ft 1441n

8.43

(b) Increase the pipe size to 1 112-in Schedule 40. Results: u= 7.88 ftls; NR = 8.74 x 104;

DI e = 895;f = 0.0232; Then, hL = 37.5 ft; hA = 349.8 ft; Power = PA = 4.42 hp,

Q = 50 gallmin x 1 ft3 Is = 0.1114 ft31s: u= Q =O. l l14 ft3 Is = 18.56 ft/s 449gaVmin A 0.0060ft2

h+z + v12 +h -è =p2+z +u;Ā pá =O Vá= Vz=O 1 2g A L 2 2 " ' r; r; g

b,= Pz +(z2 -z1)+hi Yw

NR= vD =(18.56)(0.0874)=1.34xl05: D = 0.0874 =583:f=0.0243 V 1.21X10-5 S 1.5X10-4

h = f ~ ~ = (0.0243) 225 X (l 8Ŀ56)2 ft = 335 ft L D 2g 0.0874 2(32.2)

h = (40lb)ft3 x144in2 +220ft+335ft=647ft A in" 62.4 lb ft2 PA = hAYwQ = (647 ft)(62.4lblft3)(0.l114 ft31s)l550 = 8.18 hp

8.42

Pt. 1 at tank surface. p1 = O, u1 = O

Q = 900L/min1 m3 Is = 0_015 m3ls 60000L/min

u= Q = 0.015 m3 Is = 1.99 mis A 7.538xl0-3 m2

NR = vD = (1.99)(0.098) = 1.50x105: D = 0.098 = 65333:f = 0.0165 V 1.30 X 10-6 S 1.5 X 10-6

h = f ~~=(0.0165) 80Ŀ5 X (1.99)2 =2.735 fil L D 2g 0.098 2(9.81)

[ (1.99)2 ]

ps=9.81kNlm3 12-2(9.81) -2.735 m=89.9kPa

8.41


J!i + z + v12 -h = p2 + z + u~ Pt. l at pump outlet in pipe. Yw 1 2g L Yw 2 2g Pt.Zat reservoir surface.jr, =O,v2 =O

[ v2 ] Q 4.00 ft3/s p, = Yw (z - z ) - -1 + h V= - = 2 = 11.52 ft/s 2 1 2g L .__ __ A __ o_.3_47_2_ft _

NR = vD = (11.52)(0.6651) = 6.33 x 105: D = 0.6651 = 4434:/= 0.0155 V 1.21X10-5 e 1.5X10-4

h = f ~ ~ = (0.0155). 2500 . (l l.52)2 ft = 120.l ft L D 2g 0.6651 2(32.2)

= 62.4lb[210- (lt.52)2 +120.1] ft-lft2 = 142.1 si Pi ft3 2(32.2) 144 irr' p

2 1/2-in Type K Copper Tube: D = 0.2029 ft: A = 0.03234 ft2

u= Q = 0.90 ft3 /s = 27.8 ft/s A 0.03234 ft2

NR = vDp = (27.8)(0.2029)(1.24)(1.94) = 272 Õ 5.0X10-2

PI = h = t~ ~ = (1 24)(62 4) 64 X 55 X (27.8)2 ~ 1 ft2 Pz yg L Ygá D 2g Ā Ā 272 0.2029 2(32.2) ft2 144in2

= 411 psi

NR= vDp =QDp = QDp = 4Qp: Dmin= 4Qp Õ AÕ JCD2 JCDÕ 7CNRÕ -Õ

4 D.= 4(0.90ft3/s)(l.24)(1.941bĿs2/ft4)=0.l84ft mm 7C(300)(5.0x10-2 lb-s/ft")

+ 27.28 ft = 174.l ft

P v2 p vi ~+z +_!!_+h -h =-1!+z +-1L r A 2g A L r B 2g

P P v2 v2 h = B - A +z -z + B - A +h A B A 2 L r g

Q 0.50 ft3/s Q O 50 u = -= = 7.28 ft/s: Vs = -= Ŀ = 15.03 ft/s A AA 0.06868 ft2 AB 0.03326 N = VB D p = (15.03)(0.2058)(1.026)(1.94) = 1.54 X 105 Re Õ 4.0X10-5

D = 0.2058 = 1372: f= 0.020 e 1.5X10-4

h = j ~ ~ = (0.020) 80 X (15.03)2 ft = 27.28 ft L D 2g 0.2058 2(32.2)

h = [25.0-(-3.50)]lbft3144in2 + 80 ft + (15.03)2 -(7.28)2ft2/s2

A in2(1.026)(62.4lb) ft2 2(32.2 ft/s") PA = hAyQ = (174.1 ft)(l.026)(62.4 lb/ft3)(0.50 ft3/s)/550 = 10.13 hp

8.46

8.45

8.44

Chapter 8

hA= 1.0ft+ (13.02)2 ft+35.5ft =39.1 ft 2(32.2)

p = h Q = (39.1 ft)(0.890) (62.4 Ib) (0.668 ft3) 1 hp = 2.64 h A AYo fr' s 550 ft -Ib/s P

oil-App.C

p v2 p v2 -1.+z +-1 +h -h =-2+z +-2 1 2g A L 2 2 ro ro g

v2 hA = (z2 -Zá) +

2~ +hL

u = J;?_ = 0.668 ft3 Is = 7.56 ftls 4

Ć4 0.08840 ft2

= J;?_ = 0.668 ft3 Is = 13.02 ft/s =u V:i Ći 0.05132 ft2 2

he = h + h = + ~ V~ + + L4 u: L, L, J 3 D3 2g J 4 D4 2g

N = V3D3 = (l3.02)(o.2557) = 1548 (Laminar):h = ~ = 0.0413 R, v 2.15x10-3 Nn

N = V4D4 = (7.56)(o.3355) = 1180 (Laminar):.f4 = ~ = 0.0543 R, V 2.15X10-3 NR

h = 0.0413 . 75 . (l3.02)2 + 0.0543. 25 . (7Ŀ56)2 = 35.5 ft L 0.2557 2(32.2) 0.3355 2(32.2)

Pt. 1 at tank surface. P, = O, v1 = O Pt. 2 in outlet stream from 3-in pipe. p2 = O

Q = 300gallminĿ1 ft3 Is= 0.668 ft3 Is 449 gal/min

u= Q = 4.25 ft3 Is = 7.76 ft/s A 0.5479ft2

Assume sg = 0.68 ÕFromApp.D.

Pt. O at pump inlet.

Pt.1 at pump outlet.

Assume z0 = z1 , V0 = V1

P A + z + v1 -h = PB + z + u~ A 2 L B 2g r. g r

PA = PB + Yg[(zs - ZA)+ hL] NR = vDp = (7.76)(0.8350)(1.32) = 1.l9 x 106

Õ 7.2x 10-6

D = 0Ŀ8350 = 5567: f= 0.0145 e 1.5X10--4

h = f ~~=0.0145Ā 3200 X (7Ŀ76)2 =51.9ft

L D 2g 0.8350 2(32.2)

= 40.0 psig + 42.4 lb [85 + 51.9] ft Ŀ l ft2 = 80.3 psig PA ft3 144in2

p v2 p v2 -0 +z +-0 +h --1 +z +-1

O 2g A - 1 2 r; r, g h _ p1 - Po_ [142.1-(-2.36))1b A - rw - (62.4 lblft3)(in2)(1 ft2/144 in") hA = 333.5 ft-lb/lb p = h Q = 333.5 ft Ŀlb . 62.4 lb. 4.00 ft3 . 1 hp = 151 h A AY lb ft3 s 550 ft -Ib/s p

106

8.49

8.48

8.47


8.53 Benzene at 60ÜC: p = 0.88(1000) = 880 kg/nr'; Õ = 3.95 x 10-4 Pa-s Q "Ü LI . 1 3 I u=-= ~ ,mm . m ,s = 0.719 mis A 60000 L/min 4.636x10-4 m2

NR = uDp = (0.719)(0.0243)(880) = 3.89 x 104 Õ 3.95X10-4

Die= 0.0243/4.6 x 10-5 = 528;/ = 0.0273

8.52 Water at 75ÜC; v = 3.83 x 10-7 m2/s u= Q = 12.9 L/min . 1 m3/s = l.528 mis

A 60000 L/min 1.407x10-4 m2 NR = uD = (1.528)(0.0134) = 5.34 x 104

Õ 3.83X10-7

Die= 0.0134/1.5 x 10-6 = 8933;/= 0.0209

For each problem, the calculation ofthe Reynolds number and the relative roughness are shown followed by the result of the calculation for f

f = 0.25 2

[l ( 1 5.74J] og 3.7(DI s) + NR ÜĿ9

(8-7)

NOTE: For problems 8.52 through 8.62 the objective is to compute the value of the friction factor,f, from the Swamee-Jain equation (8-7) from Section 8.8, shown below:

Q = 180L/minĿ1 m3 /s = 0.003 m3/s 60000L/min

u= Q = 0.003 =0.701mls A 4.282x1 o-J

8.51 P1 - Pi = yg[(z2 - z1) + hL] (From 9.24) NR = uDp = (0.701)(0.0738)(1258)

Õ 0.960

NR = 67.8 (Laminar):/=~ = 0.944 NR

h = f ~ ~ = (0.944) Ŀ 25Ŀ8 Ŀ (0.70l)2 = 8.27 m L D 2g 0.0738 2(9.81) p, - P: = 12.34 kN/m3[0.68 m + 8.27 m] = 110 kPa

P u2 p u2 --1.+z1 +-1 -hL =~+Z2 +-2 : Uá= lJ:2 ro 2g ro 2g Pi - P2 = Yo[(zz - Zá) + hL] NR = uDp = (3.65)(0.0189)(930) = 1938 (Laminar):/= 64 = 0.0330

Õ 3.31xl0-2 NR

h = f ~~=(0.0330) 17Ŀ5 Ŀ (3Ŀ65)2 = 20.76 m L D 2g 0.0189 2(9.81) Pv - P: = 9.12 kN/m3[-l.88 m + 20.76m]=172 kPa

8.50

Chapter 8 108

8.58 Crude oil (sg = 0.93) at lOOÜC P = (0.93)(1000 kg/rrr') = 930 kg/m'; Õ = 7.8 x 10-3 Pa-s u= Q= 1200L/min x lm31s =1.l9m/s

A 1.682x10-2 m2 60000 L/min NR= vDp = (1.19)(0.1463)(9.30) =2.07x104

Õ 7.8x10-3

Dle= ÜĿ1463 = 3180Ŀ/= 0.0264 4.6 X 10-5 '

8.57 A= rcD214 = n(0.025)2/4 = 4.909 x 10-4 m2 u= Q = 95 L/min x 1 m31s = 3.23 mis A 4.909ft2x10-4 m' 60000 L/min

N _ vDp _ (3.23)(0.025)(1.10)(1000) _ 4 44 104. DI _ S h [L DI ] R - -- - - Å x . c. - moot arge c.

Õ 2.0X10-3

f= 0.0213

8.56 Water at 60ÜF: v = 1.21 x 10-5 ft21s u= Q = 1500 gal/min x 1 ft3 Is = 6_097 ft/s

A 0.5479 ft2 449 gal/min

NR = vD = (6.097)(0.835) = 4.21 x 105: Die= 0.835 = 5567 V 1.21 X 10-5 1.5 X 10-4

f= 0.0156

u= Q = 15.0 ft3 Is = 8.49 ft/s A 1.767 ft2

NR = vD = (8.49)(1.50) = 9.09 x 105: Dle= 1.50 = 3750 V 1.40X10-5 4X10-4

f= 0.0155

8.55 Water at 50ÜF: v = 1.40 x 10-5 ft2/s

D = 18 in(l ft/12 in)= 1.50 ft; A= 1iD2 = 1.767 ft2 4

D = 0.512 ft; A = 0.2056 ft2

u= Q = 2Ŀ50ft3/s = 12.16 ft/s A 0.2056ft2

NR = vD = (12.16)(0.512) = 6_80 x 105: Die= 0.512 = 1280 v 9.15x10-6 4x10-4

f= 0.0191

8.54 Water at 80ÜF: v = 9.15 x 10-6 ft2/s


8.63 Q = 1.5 ft3 Is, L = 550 ft, D = 0.512 ft, A = 0.2056 ft2 R = Dl4 = 0.128 ft, Ch= 140

h - LI Q ]1.852

L - l 1.32AChRÜĀ63 h = 550[ 1.5o ]1.352 = 15.22 ft L (1.32)(0.2056)(140)(0.128)¿.63

Hazen-Williams Formula

8.62 Heavy fuel o²l at 77ÜF; p = 1.76 slugs/ft' Õ = 2.24 X 10-3 lb-s/ft' u= 12 ftls; D = 0.5054 ft N = uDp = (12.0)(0.5054)(1.76) = 4.77 x 103 R Õ 2.24X10-3

Die= 0.5054 = 3369Ŀ f= 0.0388 1.5X10-4 '

8.61 Water at 70ÜF; v = 1.05 x 10-5 ft2ls

u= Q = 3Ŀ0 ft3 Is = 3.82 ftls A 0.7854 ft2

N = uD = (3.82)(1.00) = 3.64 x 10s R V 1.05 X 10-5

1.00 Die = = 2500Ŀ f = 0.0175 4.0x 10-4 '

8.60 Propyl alcohol at 25ÜC; p = 802 kg/nr' Õ = 1.92 x 10-3 Pa-s

u= Q - 0.026 m3 Is = 6.07 mis A - 4.282X10-3

N = uDp = (6.07)(0.0738)(802) = 1.87 x 10s R Õ 1.92X10-3

Die= (0.0733) = 49200Ŀ f= 0.0159 1.5X10-6 '

8.59 Water at 65ÜC; v = 4.39 x 10-7 m21s D= 0.0409 m N = uD = (10)(0.0409) = 9.32 x 10s R V 4.39X10-7

Die= 0.0409 = 889Ŀ f = 0.0206 4.6x 10-s '

Chapter 8 110

Q = 0.20 ft3/s; D = 2.469 in= 0.2058 ft; A= 0.03326 ft2 Ch= 100; R = D/4 = 0.0515 ft u= Q/A = 6.01 ft/s (OK)

r ]1.s52 . - 8~' o.20 - s 35 fi. ne - U - t l (1.32)(0.03326)(100)(0.0515)ÜÜ63 Å

8.68

. 1 m3/s 3 Q = 900 L/mm x = 0.015 m Is; L = 80 m 60000L/min

D = 97.97 mm= 0.09797 m; A= 7.538 x 10-3 m2 R = D/4 = 0.0245 m; Ch = 130

[ ]

1.852

h = 80 O.Ol5 = 3.56 m L (0.85)(7.538X10-3)(130)(0.0245)ÜÜ63

8.67

he = 31.38 ft

. 1 ft3 Is 3 Q = 1500 gal/mm x = 3.341ft/s;L=1500 ft

449 gal/min D = 10.02 in= 0.835 ft; A= 0.5479 ft2 Ch= 100; R = D/4 = 0.2088

[ ll.852

hr = 1500 3Ŀ34l (1.32)(0.5479)(100)(0.2088)ÜÜ63

8.66

Q = 7.50 ft3/s; L = 5280 ft, D = 18 in= 1.50 ft, A= 1.767 ft2 R = D/4 = 0.375 ft; Ch= 100

[ ]

1.852

he = 5280 7 Ŀ50 = 28.51 ft (1.32)(1. 767)(100)(0.375)ÜÜ63

8.65

he = 2.436 m

. 1 rrr' /s 3 Q = 1000 L/mm x = 0.0167 m Is; L = 45 m

60000L/min 4-in type K copper tube; D = 97.97 mm= 0.09797 m, A= 7.538 x 10-3 m2 R = D/4 = 0.0245 m, Ch = 130

[ ]

IB52

h - L Q L - 0.85AChR0Ü63

[ ]

1.852

- 45 0.0167 - 0.85(7.538X10-3 )(130)(0.0245)ÜÜ63

8.64

111 REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, ANO ENERGY LOSSES DUE TO FRICTION

hL = 14.72 ft

8.71 From 8.70 Q = 0.668 ft3/s D = 0.5054 ft = 6.065 in; A = 0.2006 ft2 R = D/4 = 0.1264 ft; Ch= 100

hl = 1200[ 0.668 ]1.852 1.32(0.2006)(100)(0.1264)Ü"63

[ ]

1.852 h - L Q L - 1.32AChR0.63

[ ]

1.852 h = 1200 ÁĿ668 = 9.05 ft L (1.32)(0.2006)(130)(0.1264)ÜĿ63

D = 0.5054 ft; R = D/4 = 0.1264 ft A= 0.2006 ft2

Actual he for 6-in pipe

6-in Schedule 40 steel pipe; D = 0.5054 ft

. 1 ft:3/s Q = 300 gal/mm x = 0.668 ft3/s 449 gal/min

s = 10 ft/1200 ft = 0.00833 ft/ft

D = [ 2.31(0.668) ]Ü.3so = 0.495 ft (130)(0.00833)054

8.70 Specify a new Schedule 40 steel pipe size. Use Ch= 130

b) Cement lined 8-in ductile iron pipe D = 8.23 in= 0.686 ft; A= 0.3694 ft2; Ch= 140 R=D/4=0.1715ft

h = 2500[ 2ĿÜ ]1.852 = 28.3 ft L (1.32)(0.3694)(140)(0.1715)ÜĿ63

a) 8-in Schedule 40 steel pipe; D = 0.6651ft;A=0.3472 ft2 R = D/4 = 0.1663 ft; Ch= 100

h = 2500[ 2.0 ]1.852 = 61.4 ft L (1.32)(0.3472)(100)(0.1663)ÜĿ63

8.69 Q = 2.0 ft3/s; L = 2500 ft

Chapter 8 112

he = 27.27 ft

b) 3-in pipe; D = 3.068 in= 0.2557 ft; A= 0.05132 ft2 R = D/4 = 0.0639 ft

h = 1000[ 0.2227 ]1.852 L (1.32)(0.05132)(130)(0.0639)ÜĿ63

8.72 Q = 100 gal/min x 1 ft3/s/449 gal/min = 0.2227 ft3/s L = 1000 ft; Ch = 130 (New steel)

a) 2-in pipe: D = 2.067 in= 0.1723 ft; A = 0.02333 ft2 R == D/4 == 0.0431 ft

h = 1000 ft[ 0.2227 ]1.852 L (1.32)(0.02333)(130)(0.0431)ÜĿ63

hi. = 186 ft

113 VELOCITY PROFILES FOR CIRCULAR SECTIONS ANO FLOW IN NONCIRCULAR SECTIONS

9.4 Local velocity: U= 2v[l - (r/r0)2]; Q = 25.0 L/min = 4.167 x 10-4 m3/s 2-in drawn steel tube; D = 47.50 mm= 0.00475 m; rà = D/2 = 23.75 mm A= 1.772 x 10-3 m2; v = Q/A = 4.167 x 10-4 m3/s/1.772 x 10-3 m2 = 0.2352 mis NR = vDp/1] = vD/v = (0.2352)(0.0475)/7.9 x 10-3 = 1258 Laminar See spreadsheet listing for results for U as a function of radius.

9.3 Local velocity: U= 2v[l - (r/r0)2]; Q = 3.0 L/min = 5.0 x 10-5 m3/s 4-in Type K copper tube; D = 97.97 mm= 0.09797 m; rà = D/2 = 48.99 mm A= 7.538 x 10-3 m2; v = QIA = 5.0 x 10-5 m3/s/7.538 x 10-3 m2 = 6.633 x 10-3 mis NR = vDp/l] = vD/v = (6.663 x 10-3)(0.09797)/4.22 x 10-7 = 1540 Laminar See spreadsheet listing for results for U as a function of radius.

Problem 9.1 Problem 9.2 Average velocity 10.72 ft/s Average velocity 0.3679 ft/s

Full radius 1.0338 in Full radius 0.3725 in Radius (in) Velocity, U Radius (in) Velocity, U

0.00 21.44 ft/s 0.00 0.7358 ft/s 0.20 20.64 ft/s 0.05 0.7225 ft/s 0.40 18.23 ft/s 0.10 0.6828 ft/s 0.60 14.22 ft/s 0.15 0.6165 ft/s 0.80 8.60 ft/s 0.20 0.5237 ft/s 1.00 1.38 ft/s 0.25 0.4044 ft/s

1.0338 0.00 ft/s 0.30 0.2585 ft/s 0.35 0.0862 ft/s 0.3725 0.0000 ft/s

9.2 Local velocity: U= 2v[l - (r/r0)2]; Q = 0.50 gal/min = 0.001113 ft3/s 3/4-in Type K copper tube; D = 0.745 in= 0.0621 ft; r0 = D/2 = 0.373 in; A= 0.02333 ft2; v = Q/A = 0.001113 ft3/s/0.003027 ft2 = 0.3679 ft/s NR = vDp/1] = vD/v = (0.3679)(0.0621)/1.21 x 10-5 = 1888 Laminar See spreadsheet listing for results for U as a function of radius.

9.1 Local velocity: U= 2v[l - (r/r0)2] for laminar flow 2-in Sch 40 steel pipe; D = 2.067 in= 0.1723 ft; rà = D/2 = 1.034 in; A = 0.02333 ft2 v = QIA = 0.25 ft3/s/0.02333 ft2 = 10.72 ft/s NR = vDp/1] = vD/v = (10.72)(0.1723)/7.31 x 10-3 = 253 Laminar See spreadsheet listing for results for U as a function of radius.

VELOCITY PROFILES FOR CIRCULAR SECTIONS ANO FLOW IN NONCIRCULAR SECTIONS

CHAPTER NINE

114 Chapter 9

9.7 Center of pipe is Dol2 from outside surface

D 168Ŀ3mm 84 15 hi . . O U 2 -Ü == = . mm; at t LS position, r == , = v 2 2 With r = 5.0 mm, r/r; = 5.0/73.15 = 0.06835 U= 2u[ 1-(rlrJ2 J = 2u[l - (0.06835)2] = 1.9907 u

If vis expected to be U/2,

U == 1 Ŀ9907 v = 0.9953 u 0.47% low 2 2

9.6 If probe is inserted 5.0 mm too far: r == 51.72 mm- 5.0 mm= 46.72 mm r/r; == 46.72/73.15 = 0.6387 U= 2v[l - (0.6387)2] = 1.184 u 18.4% high

If probe is inserted 5.0 mm too little: r = 51.72 mm+ 5.0 mm== 56.72 mm r/r; = 56.72/73.15 = 0.7754 U= 2u[l -(0.7754)2] = 0.798 u 20.2% low

r0 Å Q2 Å 73.15 mm ~n Sdledule 80 páptt

d Å lnaerllon dlstance

0.2352 m/s 23.75mm Velocity, U 0.470 m/s 0.457 m/s 0.417 m/s 0.350 m/s 0.257 m/s 0.137 m/s 0.000m/s

tÅ10.97mm 1 2 i=l-(rlrJ

(rlrJ2 =0.5 r = J0:5 ro = 0.707 ro= 0.707(73.15 mm)

=51.72mm d = ir, - r) + t = (73.15 - 51.72) + 10.97 = 32.40 mm

(See Section 9.4) U= Local Velocity U= 2u[l-(r/rJ2]

If U= v =average velocity v = 2u[ 1-(rlrJ2 J

9.5

6.63E-03 m/s 48.985 mm Velocity, U 0.01327 m/s 0.01291 m/s 0.01185 m/s 0.01008 m/s 0.00760 m/s 0.00442 m/s 0.00053 m/s 0.00000 m/s

Problem 9.4 Average velocity

Full radius Radius Imm)

0.000 4.00 8.00 12.00 16.00 20.00 23.75

Problem 9.3 Average velocity

Full radius Radius (in)

0.00 8.00 16.00 24.00 32.00 40.00 48.00 48.99


SAE 1 O oil: sg = 0.87; p = 0.87(1000) = 870 kg/nr' Õ = 2.20 x 10-2 Pa-s (App. D) at 40ÜC u= N11Õ = (3.60xl04)(2.20x10-2) =0.759mls

D p (1.20)(870)

Jr(l .20 m)" 3 Q =Av= x 0.759 mis= 0.858 m /s

4

D = 1.20 = 26000;/= 0.0222 e 4.6x 10-5 ¿.; =u( l + 1.43.fl) = 0.759( 1+1.43../0.0222) = 0.920 m/s

At other points: U= u [ 1 + 1.43.Jf + 2.15.fj log., (y!r0) J U= 0.759mis[1+1.43../0.0222 + 2.15../0.0222 log., (y!r0) J U= O. 759 [ 1.213 + 0.320 log., (y!r0) J mis r0 = D/2 = 1200 mml2 = 600 mm

u= Q= 12.9L/min x lm3/s =l.53mls A l .407x10-4 rrr' 60000 L/min

N = uD = (1.53)(0.0134) = 5.35 x 104 11 V 3.83X10-7

D = O.Ol34 = 8933;/= 0.0205 e 1.5X10-6 U max = u( 1 + 1.43.fJ) = 1.53( 1 + 1.43../.0205) = 1.84 mis

9.10

9.9

Velocity profile for turbulent flow

9.8 U= 2.48 mis; d = 60.0 mm (See Prob. 8.27) d=r; r+t;r=r0+t-d=73.15+10.97 60.0=24.12mm rlr; = 24.12/73.15 = 0.3297 U= 2u[1-(r!rJ2 J = 2u[l - (0.3297)2] = 1.783 u

U 2.48mls 139 mi 1 Ŀ u= -- = = . s average ve oc²ty 1.783 1.783

v = 850 es x 1.0 x 10-6 m2/s/cs = 8.50 x 10-4 m2/s

NR = uD = (L39)(0. l463) = 239 Laminar OK V 8.50x10-4

Chapter 9 116

9.12 U= u[ 1+I.43.Jf+2.15Jf Iog.à (y!r0) J Let U= u, divide both sides by u 1 = 1+1.43.Jf + 2. l 5.Jj Iog10 (ylr0)

Subtract 1.0 from both sides O = 1.43.JY + 2. l 5.Jf Iog10 (y!r0)

Divide by .J7 O= 1.43 + 2. l 5 log., (ylr0)

-1.43 log., (y!r0) = -- = -0.665

2.15 y/r0 = 10-0Ŀ665 = 0.216 or y= 0.216 r0

The lower friction factor far the hot oil results in a 3% lower maximum velocity. But the velocity in the boundary layer (y < 100 mm) is slightly higher.

Results included with Prob. 9.10.

9.11 Oil at lOOÜC: Õ = 4.2 x 10-3 Pa-s N = uDp = (0.759)(1.20)(870) = 1.89 x 105Āf= 0.0158 R Õ 4.20xl0-3 '

u-: = u[ 1+1.43.Jf] = 0.759[ 1+1.43.J0.0158 J = 0.895 mis

U= 0.759 [ 1+1.43.J0.0158 + 2.I5.J0.0158 log.à (y!r0) J U= 0.759[1.180 + 0.270 log., (y!r0)]

o o .20 .40 .60 .60 1.0

Velodty, U (m/s)

100

500

Prab a .. 11 / r\ V

/ 17

- V

800 y(mm) ylr; U(m/s) o o o o 10 0.0167 0.488 0.530 20 0.0333 0.561 0.592 30 0.050 0.604 0.628 40 0.0667 0.634 0.654 50 0.0833 0.658 0.674 100 0.1667 0.731 0.735 300 0.50 0.847 0.833 500 0.833 0.901 0.879 600 1.00 0.920 0.895

Prob. Prob. 9.10 9.11

Prob9./0


u= Q = 400 gal/min x 1ft31 s = 10.08 ft/s A 0.0884 ft2 449 gal/min

N11 = vD = (10.08)(0.3355) = 2.42 x 105: D = 0.3355 = 2237; f= 0.0182 V 1.40 X 10-5 & 1.5 X }0-4

U= u( 1+l.43.J0.0182+2.I5.J0.0182 log., (ylr0} J U= 10.08 [ 1.193 + 0.290 log., (ylr0} J rà = D/2 = 4.026 in/2 = 2.013 in

N11 f çc.: 4 X 103 0.041 0.775 1 X 104 0.032 0.796 1 X 105 0.021 0.828 1 X 106 0.0185 0.837

9.16

9.15 o.; = v(l + 1.43!.f): v!Umax = 11(1+1.43!.f)

Concrete pipe: D = (8/12) ft = 0.667 ft; DIE= 0.667/4 x 10-4 = 1670

N11 f v!Umax 4 X 103 0.040 0.778 1 X 104 0.031 0.799 1 X }05 0.0175 0.841 1 X 106 0.0118 0.866

9.14 Umax= v(l+l.43.Jf} v!Umax= 11(1+1.43!.f) Smooth pipes

9.13 y= 0.216 r: = 0.216(D/2) = 0.216(22.626 in/2) = 2.44 in _ Q 16.75ft3/s _ _ .

Vavg - A = 2. 792

ft2 - 6.00 ft/s at y - 2.44 m

At y1 = 2.44 in+ 0.50 in= 2.94 in; ylr; = 2.94/11.313 = 0.260 At j-, = 2.44 in - 0.50 in= 1.94 in;ylr0 = 1.94/11.313 = 0.172 N11 = vD = (6.00)(1.886) = 8.08x105: D = 1.886 = 12570;

V 1.40X10-5 8 1.5X10-4

f= 0.0137 U1 = 6.00ft/s[1+l.43.J0.0137+2.15.J0.0137 log10(0.260) J U1 = 6.12 ft/s (2.0% higher than Vavg) U2=6.oo[1+l.43.J0.0137+2.15~~o.-o-13-7 Iog10(0.172) J U2 = 5.85 ft/s (2.50% lower than Vavg)

Chapter 9

Q 450 L/min 1 m3 Is lJpipe = A = 1.864X10-2 m2 X 60000 L/min = 0.402 mis

2 3 60000 L/min Qduct = A0Uo = (0.0355 m )(0.402 mis)= 0.0143 m Is x 3 1 m Is A0 = (0.40 m)(0.20 m) - 2(n)(0.1683 m)2/4 = 0.0355 m2 Qduct = 857 L/min

Qs = Asvs; Qt = Atv1; But Vs = V1 specified Q, Ap, A, -=-=- Qt Apt Ćá A1 = 8.189 x 10-s m2 As= 3.059 x 10-4 m2 - 7t(0.0127m)2/4 = 1.792 x 10-4 m2

Q, = A, = l. 792X10-4 = 2.19 Q1 Aá 8.189xl0-5

h = f ~~= (0.0182)(250)(10.08)2 = 21.39 ft L D 2g (0.3355)(2)(32.2) 11p = (62.4 lb/ft3)(21.39 ft)(l ft21144 irr') = 9.27 psi h = (0.0436)(250)(10.08)2 = 51.24 ft L (0.3355)(2)(32.2) ĉ1p = yhL = (62.4)(51.24)/144 = 22.2 psi

Prob. 9.17:

118

9.20

9.19

9.18 Sp=vhà

Prob. 9.16:

D 0.3355 s = 3 = 67.1; NR = 2.42 X 10 ;f= 0.0436

8 5.0xlO-

U = 10.08 [ 1+l.43.J0.0436+2.15.J0.0436 log.à (y!r0) J U= 10.08 [ 1.299 + 0.449 log., (ylr0) J (See results for 9.16)

o o 4 8 12 U, Velodty (ftls)

1.5

y(in) ylrà U(ftls) U (Prob. 9.17)

o o o o 0.05 0.025 7.33 5.83 0.10 0.050 8.21 7.19 0.15 0.075 8.72 7.98 0.20 0.099 9.09 8.55 0.50 0.248 10.25 10.35 1.00 0.497 11.13 11.71 1.50 0.745 11.65 12.51 2.013 1.000 12.02 13.09

9~i- I ' I 9.1'1-....i

/ ' 1

i'

' 9.17 1 ' - /~ ~-" - 1

2.0

9.17


9.25 Tube: D = 10.21mm=0.01021m;A=8.189 x 10-5 m2

Q = 4. 75 gallmin x 6Ŀ309x10-s m3 Is = 3 .00 x 104 rn' Is 1 gal/min

u= Q = 3.00x10-4 m3 Is = 3.66 mis A 8.189x10-s m2

N = vD = (3.66)(0.01021) = 1.23 x 10s R V 3.04X10-7

9.24 A= (10 in)2 - n(6.625 in)2/4 = 65.53 in2(1 ft2/144 in2) = 0.455 ft2 WP = 4(10 in)+ n(6.625 in)= 60.81 in(l ft/12 in)= 5.067 ft R = AIWP = 0.455 ft2/5.067 ft = 0.0898 ft

u= Q = 4.00 ft3/s = 8.79 ft/s A 0.455 ft2

NR= u4R=(8.79)(4)(0.0898) =3.808x 10s V 8.29X10-{)

9.23 A= (12 in)(6 in)- 2n(4.0 in)2/4 = 46.87 in2(1 ft2/144 irr') = 0.3255 ft2 WP = 2(6 in)+ 2(12 in)+ 2n( 4.0 in)= 61.13 in(l ft/12 in)= 5.094 ft

R = __:!_ = 0Ŀ3255 ft2 = 0.0639 ft WP 5.094ft

u= Q = 200ft3x1 min x 1 = 10.24 ft/s A min 60 s 0.3255 ft2

=y= 0.ll4}bĿS2 X ls}ug =0.00354slu s/ft3 p g fr' (32.2 ft) 1 lbĿ s2 /ft g

NR = v(4R)p = (10.24)(4)(0.0639)(0.00354) = 2.77 x 104 Õ 3.34x 10-1

9.22 A= (0.05 m)2 + (0.5)(0.05 m)2 - n(0.025 m)2/4 = 3.259 x 10Ŀ3 m2

WP = 0.05 + 0.05 + 0.1 O + ../.052 + .052 + n(0.025) = 0.3493 m

R = __:!_ = 3.259x10-Jmz == 9.33 x 10-3 m WP 0.3493m

u= Q = 15Ü m3 x~x 1 = 12.78 mis A hr 3600 s 3.259x10-3m2

_ y_ 12.5 NĿs2 1 kgĿm/s2 _ l 274 k I 3 p- -- 3 X - Å g ID g m 9.8lm in

NR = v(4R)p = (12.78)(4)(9.33xl0-3)(1.274) == J.04x104 Jl 2.0X10-5

Noncircular cross sections

9.21 QT = ĆTV =(O.O 1414 ft2)(25 ft/s) = 0.3535 ft3/s Qs = Asv == (0.0799 ft2)(25 ftls) = 1.998 ft3/s

A = 0.1390 ft2 - 3JT(l.900 in)2 x ft2 == 0.0799 ft2 s 4 144in2

Chapter9 120

Shell: A= 0.1390 ft2 - 3(n)(0.1583 ft)214 = 0.0799 ft2

D0 = 1.900 in(l ft/12 in)= 0.1583 ft = Outside dia. of 1 _!_ l in pipe. 2

WP = n(0.4206 ft) + 3n(0.1583 ft) = 2.813 ft R = ~ 0.0799 ft2 = 0.0284 ft

WP 2.813ft N = v(4R) = (25)(4)(0.0284) = 2.35 x 10s R V 1.2lxl0-S

9.27 lnside Pipes: NR = vD = (25)(0. l342) = 1.00 x 106 V 3.35X10-6

Duct: Benzene at 70ÜC: p = sg(pw) = 0.862(1000) = 862 kg/rn" Õ = 3.5 x 10-4 Pa-s (App. D) A= (0.40 m)(0.20 m)- 2(n)(0.1683 m)214 = 0.0355 m2 WP = 2(0.40) + 2(0.20) + 2(n)(0.1683) = 2.257 m R = ~ = 0.0355 mz = 0.0157 m

WP 2.257m N = v(4R)(p). u= NRÕ = (6.08x104)(3.5x10-4) = 0_392 mis R Õ ' 4Rp 4(0.0157)(862) Q =Av= (0.0355 m2)(0.392mis)=1.39 x 10Ŀ2 m3ls

9.26 Pipe: Q = 450L/minx1 m3 Is = 0.00750 m31s 60000L/min

u= Q = 0.00750 m3 Is = 0.402 mis A 1.864x10-2 m"

N = vD = (0.402)(0.1541) =6_08xl04 R V 1.02 X 10-6

Twater at 20ÜC

Shell: A= 3.059 x 10-4 m2 - n(0.0127 m)214 = 1.792 x 10-4 m2 WP = n(0.01974 m) + n(0.0127 m) = 0.1019 m R = AIWP = 0.00176 m = 1.76 x 10-3 m

Q = 30.0 gallmin X 6Ŀ309X1 o-s m3 Is = 1.89 X 10-3 m' Is l gal/min

u= Q=l.89xlo-3m31s =10.56mls A 1.792x10-4 m2

NR= v(4R)(p)=(10.56)(4)(1.76x10-3)(1100) =5.049x 103=5049Turbulent Õ 1.62x 10-2


9.31 A= (0.45)(0.30) + rr(0.30)2/4 - 2(0.15)2 = 0.1607 m2 WP = 2(0.45) + rr(0.30) + 8(0.15) = 3.042 m R = AIWP = 0.1607 m2/3.042 m = 0.0528 m u= Q/A = 0.10 m3/s/0.1607 m2 = 0.622 mis NR= u(4R)p = (0.622)(4)(0.0528)(1260) =552

Õ 0.30 L(App. D)

u= 8.55 mis NR = u(4R)(p) = (8.55)(4)(0.00244)(1.15) = 5Å90 x 103

Õ 1.63X10-5

50m3 1 l06mm2 lh ---Ŀ--- H-5.85 H-s.30 Clrnensklns In mm m2 3600s h 1625mm2

9.30 Air flows in cross-hatched area. A= [4(5.30) + 2(5.65)](50) = 1625 mm2 WP = 12(50) + 4(5.65) + 8(5.30) = 665 mm

R = ~= 1625 mm2 = 2.444 mm WP 665mm

u= Q A

9.29 A= (0.25 in)(l.75 in)+ 4(0.25 in)(0.50 in)= 0.9375 in2(1 ft2/144 in2) = 0.00651 ft2 WP = (1.75 in)+ 7(0.25 in)+ 2(0.75 in)+ 6(0.50 in) WP = 8.00 in(l ft/12 in)= 0.667 ft R = AIWP = 0.00651 ft2/0.667 ft = 9.77 X 10-3 ft N = u4Rp: u= NRÕ = (1500)(3.38x10-4) = 6.09 ft/s R Õ 4Rp 4(9.77X10-3)(2.13) Q =A u= (0.00651 ft2)(6.09 ft/s) = 3.97 x 10-2 ft3/s

. 1 rrr' /s 3 Q = 850 L/mm x = 0.01417 m Is 60000L/min

A= Ćshell - 34ube-Do = 1.945 X 10-3 m2 - 3rr(0.0127 m)2/4 = 1.565 X 10-3 m2 WP = rr(0.04976 m) + 3rr(0.0127 m) = 0.2760 m R = ~ = 1.565x10-3 mz = 5.670 x 10-3 m

WP 0.2760m u= Q= 0.01417m3/s =9.05mls

A I .565x10-3 m2 NR = u4R = (9.05)(4)(5.670x10-3) = 1.58 x 105

V 1.30x 10-6

9.28

Chapter 9

f-150 --1

122

A= 1.089 in2(1 ft2/144 irr') = 0.00756 ft2 WP = n(0.50) + 4(0.50) + (JT)(.25)(2) + (JT)(0.75)(2)

4 4 WP = 5.142 in(l ft/12 in)= 0.428 ft R = A/WP = 0.00756 ft2/0.428 ft = 0.0177 ft u= NRv = (1.5xlOs)(l.40x10-s) = 29.66 ft/s (Hi h)

4R 4(0.0177) g Q =A u= (0.00756 ft2)(29.66 ft/s) = 0.224 ft3/s

9.34 A= (2(7²)(0.50)z + 2(0.50)(0.50) + [JT(0.75)2 - JT(0.25)2] 8 4 4

9.33 A= (0.75)(0.75) + JT(l.50)2 = 1.446 in2(1 ft2/144 in2) = 0.01004 ft2 8

WP = 0.75 + 2.25 + n(l.50)/2 = 5.356 in(l ft/12 in)= 0.4463 ft R = A/WP = 0.01004 ft2/0.4463 ft = 0.02250 ft u= Q = 78.0 gal/min x 1 ft3 /s = 17.30 ft/s

A 0.01004ft2 449 gal/min N = u(4R) = (17.30)(4)(0.02250) = 1.l12 x 10s R V 1.40 X 10-5

"' Å t 1 144.48 l

2.Tlmm 1

9.32 A= (0.14446)2 = 0.02087 m2 WP = 4(0.14446) = 0.5778 m R = ~ = 0.02087 mz = 0.0361 m

WP 0.5778m = u(4R) = (35.9)(4)(0.0361) = 1.6l x 107

NR v 3.22xl0-7

u=Ü= 0.75m3/s =35.9m/s A 0.02087m2


Sp = yJiL = (62.l lb/ft3)(1.65 ft)(l ft2/144 in2) = 0.713 psi

9.37 From Prob. 9.24, R = 0.0898 ft; NR = 3.808 x 105; v = 8.79 ft/s; Water at 90ÜF 4 R = 4(0.0898) = 4226; thenf = 0.0165 e 8.5X10-S h = f .i .ç, (0.0165)(30)(8.79)2 = 1.65 ft L 4 R 2g ( 4)(0.0898)(2)(32.2)

9.36 A = (2)(8) = 16 mnr'(I m2/106 mnr') = 1.60 x 10-5 m2 WP = 2(8) + 2(2) = 20 1ll1ll = 0.020 m R = __!!_ = 1.60x10-5 mz = 8.00 x 10-4 m

WP 0.020m Q = Av = (1.60 x 10-5 m2)(25.0 m/s) = 4.00 x 10-4 m3/s each Qtot = (4.00 x 104 m3/s)6 = 2.40 x 10Ŀ3 m3/s N = v(4R)(p) = (25)(4)(8.00x10-4)(1.20) = 6_40 x l03 R Õ l.50X10-5

R = AIWP = 0.5651 in(l ft/12 in)= 0.04710 ft u= NRÕ = (2.6xl04)(6.60x10-6) =0.595 ft/s 4Rp (4)(0.0471)(1.53)

Q =Av= (4.369in2)(0.595 ft/s) = O.OlSl ft3/s (144in2/ft2)

9.35 a= 1.50 sin 45" ~ 1.098 in } sin 1 OJÜ By Law of Sines

b = 1.50 sin 30Ü = 0.7765 in sin 105Á

PART A WP !.-- 1.00 1 0.2209 in2 1.178 in 2 0.0552 0.2945 3 0.0736 0.3927 T 4 0.0920 0.4909

2.25 5 0.5625 1.500 1.50 1.50

6 2.250 2.000 1 7 0.4118 o @ 8 0.4118 1.0981 =a 9 0.2912 0.7765 = b 1@

A= 4.3690 in2 7.7307 = WP

124 Chapter 9

9.40 Sorne data from Prob. 9.27 and Fig. 9.12. Each Pipe: Water at 200ÜF; Yw = 60.l lb/ft3; NR = 1.00 x 106

D = 0Ŀ1342 = 895: then/= 0.021 8 1.5X10-4

h = f ~~= (0.021)(50)(25)2 = 75.9 ft L D 2g (0.1342)(2)(32.2) Sp = y,)áL = (60.1 lb/ft3)(75.9 ft)(l ft2/144 in2) = 31.7 psi

Shell: Water at 60ÜF; y= 62.4 lb/ft3; NR = 2.35 x 105; R = 0.0284 ft 4 R = (4)(0.0284) = 758; then/= 0.0225 8 1.5X10-4 h = f ~. ~ = (0.0225)(50)(25)2 = 96.1 ft L 4 R 2g ( 4)(0.0284)(2)(32.2) Sp = y,)áL = (62.4 lb/ft3)(96.1 ft)(l ft2/144 in2) = 41.6 psi

9.39 Sorne data from Prob. 9.26 and Fig. 9.11. Each Pipe: Water at 20ÜC; y= 9.79 kN/m3; NR = 6.08 x 104

D = (O. l54l) = 3350Ŀ then/= 0.021 s 4.6xl0-5 '

h = f ~ ~ = (0.021)(3.80)(0.402)2 = 0.00427 m L D 2g (0.1541)(2)(9.81) Sp = y,)áL = (9.79 kN/rn3)(0.00427 rn) = 0.00418 kPa = 41.8 Pa

Duct: Benzene at 70ÜC; Ys = (0.862)(9.81 kN/rn3) = 8.46 kN/rn3

N = 6.08 x 104Å R = 0.0157 m: 4 R = (4)(0.0l57) = 1365 R ' ' 8 4.6X10-5

f= 0.023Ŀ h = f .i.,ç, (0.023)(3.80)(0.392)2 = 0.0109 m ' L 4 R 2g (4)(0.0157)(2)(9.81)

Sp = YshL = (8.46 kN/rn3)(0.0109 rn) = 0.0920 kPa = 92.0 Pa

9.38 Sorne data from Prob. 9.25 and Fig. 9.10. Tube: Water at 95ÜC; y = 9.44 kN/m3: NR = 1.23 x 105

D = (O.Ol02l) = 222; thenf = 0.030 s 4.6x 10-5 h = f ~ ~ = (0.030)(5.25)(3.66)2 = l0.53 m L D 2g (0.01021)(2)(9.81) Sp = yjlL = (9.44 kN/m3)(10.53 m) = 99.4 kPa

Shell: Ethylene glycol at 25ÜC; y = 10.79 kN/m3: v = 10.56 mis: R = 1.76 x 10-3; NR = 5049 = 5.049 x 103 4Rls= (4)(1.76 x 10-3 rn)/(4.6 x 10-5 m) = 153; Then/= 0.044

h = !~!_ = 0.044 525 10Ŀ562 = 186.5 m L 4R 2g 4(0.00176) 2(9.81) Sp = rhL = (10.79 kN/rn3)(186.5 m) = 2018 kN/rn2 = 2018 kPa [Very high]


9.45 Data from Prob. 9.33 and F²g. 9.19. NR = 1.112 x 105

R = 0.0225 ft: 4 R = (4)(0.0225) = 3600: i= 0.019 s 2.5X10-5

L = (105 ²n)(l ft/12 in)= 8.75 ft h = f .i..ç: (0.019)(8.75)(17.30)2 = 8.58 ft L 4 R 2g ( 4)(0.0225)(2)(32.2) f:...p = yhL = (62.4 lb/ft3)(8.58 ft)(l ft2/144 in2) = 3.72 psi

9.44 Data from Prob. 9.32: R = 0.0361 m: NR = 1.61 x 107 4 R = (4)(0.03:l) = 96267: f= 0.008 approx. s l.5x10

h = f __!::_. !!__ = (0.008)(22.6)(35.9)2 = 82_2 m L 4 R 2g (4)(0.0361)(2)(9.81) l).p = yhL = (9.47 kN/m3)(82.2 m) = 779 kPa

L Water at 90ÜC

9.43 DatafromProb.9.3landF²g.9.18. NR=552Laminar 64 64 f= -=- =0.116: R=0.0528m NR 552

h = f __!::_. !!__ = (0.116)(22.6)(0.622)2 = 0_244 m L 4 R 2g (4)(0.0528)(2)(9.81) Sp = YahL = (1.26)(9.81 kN/m3)(0.244 m) = 3.02 kPa

9 .42 Data from Prob. 9 .29 and F²g. 9 .16. NR = 1500 Laminar

f= ~ =~ = 0.0427: R = 9.77 x 10Ŀ3 ft: L = 57 in = 4.75 ft NR 1500 12 in/ft

h = á__!::_.!!__= (0.0427)(4.75)(6.09)2 =2.99ft L 4 R 2g (4)(9.77X10-3)(2)(32.2) Ethylene glycol at 77ÜF - y = 68.4 7 lb/ft3

Sp = yhL = (68.47 lb/ft3)(2.99 ft)(l ft2/144 irr') = 1.42 psi

9.41 Data from Prob. 9.28 and F²g. 9.15. Copper tubes. Water at lOÜC; Yw = 9.81 kN/m3; NR = 1.58 x 105: R = 5.670 x 10Ŀ3 m 4R=(4)(5.670x10-3)=15120: /=0.017 e 1.5X10-6

h = f __!::_. !!__ = (0.017)(3.60)(9.05)2 = l l.26 m L 4 R 2g (4)(5.670x10Ŀ3)(2)(9.81) Sp = YwhL = (9.81 kN/m3)(1 l.26m)=111 kPa

Chapter 9 126

9.50 Each Tube: Ethyl Alcohol at OÜF; assume p = 1.53 slugs/ft' -5 2 vDp Õ = 5 x 10 lb-s/ft (App. D); NR = --

Õ

u= NRÕ = (3.5x104)(5.0x10-s) = 26.02 ft/s Dp (0.044)(1.53)

D = 13.4 mm(l.O in/25.4 mm)= 0.5276 in x 1 ftl12 in= 0.044 ft

9.49 A= (28)(14) - 3[(8)(2) + n(2)214] = 334.6 irr'(I ft21144 irr') = 2.32 ft2 WP = 2(28) + 2(14) + 3[2(8) + n(2)] = 150.8 in(l ft/12 in)= 12.57 ft R =AIWP = 0.1848 ft NR = v(4 R)p (20)(4)(0.1848)(2.06x 10-3) = 7_36 x 104

Õ 4.14xl0-7

A= (0.100 m)" - 4(0.02)(0.03)m2 = 0.0076 m2 } R =__:!_=O.O 119 m WP=4(0.10)+8(0.03)=0.64m WP

u= Q = 3000 L/min x 1 m3 Is = 6_58 mis A 0.0076 m2 60000 L/min

N = u(4 R)p = (6.58)(4)(0.0119)(789) = 4.40 x l05Ŀ/= O.Ol35 R Õ 5.60xl0-4 '

h = f ~-~= (0.0135)(2.25)(6.58)2 = 1.41 m L 4 R 2g 4(0.0119)(2)(9.81)

9.48

2 3 449 gal/min . Q =Av= (0.01807 ft )(23.05 ftls) = 0.417 ft/s x 3 = 187 gal/mm

1 ft Is

l á~0.0325 A= (2.25)(1.50)- 7(n(0.375)214) = 2.602 in2(1 ft21144 in2) = 0.01807 ft2 WP = 2(2.25) + 2(1.50) + 7(n)(0.375) = 15.75 in(l ftl12 in)= 1.312 ft

R = __:!_ = 0.01807 ft2 = O.Ol38 ft WP 1.312 ft

N R =u 4 R p: u= NRÕ = (8000)(3.38x10-4) = 23_05 ft/s Õ 4 R p 4(0.0138)(2.13)

L = 128 in(l ftl12in)=10.67 ft: 4 R = 4(0.0l38) 11000 8 5X10-6

h = f ~. ~ = (0.0325) 10Å67 (23Ŀ05)2 = 51.9 ft L 4 R 2g 4(0.0138) 2(32.2)

9.47

Data from Prob. 9.34 and Fig. 9.20. NR = 1.5 x 105: R = 0.0177 ft; u= 29.66 ftls 4 R = (4)(0.0l 77) = 472: f = 0.025 [Steel] 8 1.5X10-4 L = (45 in)(l ft/12 in)= 3.75 ft

h = f ~-~= (0.025)(3.75)(29.66)2 = 18.1 ft L 4 R 2g (4)(0.0177)(2)(32.2) Sp = yhL = (62.4 lblft3)(18.1 ft)(l ft2/144 irr') = 7.84 psi

9.46


9.52 Given: Figure 9.28. Three semicircular sections. Velocity = v = 15 ft/s in each. Ethylene glycol at 77ÜF; p = 2.13 slugs/ft'; Õ = 3.38 x 10-4 lb s/fr'(App, B) Find: Reynolds number in each passage. NR = v(4R)p!Õ Top channel: 2-in Type K.copper tube (Half), ID= 0.1632 ft; A101 = 2.093 x 10-2 ft2 A = A1oJ2 = (2.093 x 10-2 ft2)/2 = 1.0465 x 10-2 ft2 WP =ID+ 1Z(ID)/2 = ID(I + m2) = 0.4196 ft R = A/WP = (1.0465 X 10-2 ft2)/(0.4196 ft) = 0.0249 ft NR = v(4R)plÕ= (15)(4)(0.0249)(2.13)/(3.38 x 10-4) = 9.43 x 103 = 9430 =NR Turbulent

Energy Loss for 92 in (7.667 ft) of drawn steel: hL = f{L/4R)(v/2g) 4Rle= 4(0.0206 ft)/(5.0 x 10-6 ft) = 16 480 Thenf = 0.023 hi. = (0.023) [7.667/(4)(0.0206))((7.09)2/(2)(32.2)] = 1.67 ft

9 51 Given: Water at 40ÜF; NR = 3.5 x 104; Figure 9.27; Section is semicircular. Find: Volume fiow rate of water. NR = v( 4R)lv; Then, v = NR vi( 4 R) = A vera ge velocity of flow v = 1.67 t: 10-5 ft2/s = Kinematic viscosity A = (1.431 X 10-2 ft2)/2 = 7.155 X 10-3 ft2 ID= 0.1350 ft; WP =ID+ 1Z(ID)/2 = ID(l + m2) = 0.347 ft R = AIWP = (7.155 X 10-3 ft2)/(0.347 ft) = 0.0206 ft Then, v = NRv/(4R) = (3.5 x 104)(1.67 x 10-5)/(4)(0.0206) = 7.09 ft/s Q =Av= (7.155 x 10-3 ft2)(7.09 ft/s) = 0.0507 ft3/s = Q

Q =A u= ;r(0.044 ft)2 x 26.02 ft = 0.0395 ft3 Is 4

Total Flow for 3 Tubes: Qr = 3(0.0395) = 0.118 ft3/s h = f.!:_.!!___= (0.0232)(10.5)(26.02)2 = 58.2 ft L D 2g (0.044)(2)(32.2)

D = 0.044 = 8793;/= 0.0232 e 5X10--6 Sp = yhL = (49.01 lb/ft3)(58.2 ft)(l ft2/144in2)=19.8 psi

Shell: Methyl Alcohol at 77ÜF (App. B) A = (2.00)(1.00) + n(l.00)2/4 - 3n(0.625)2/4 = 1.865 in\1 ft21144 in') A = 0.01295 ft2 WP = 2(2.0) + n(l.O) + 3n(0.625) = 13.03 in(l ft/12 in)= 1.086 ft R = AIWP = 0.0119 ft u= NRÕ = (3.5xl04)(1.17x10-5) = 5.61 ft/s 4 Rp 4(0.0119)(1.53)

Q =Au=(0.01295 ft2)(5.61 ft/s)=0.0727ft3/s 4 R = 4(0.0l19) = 9540;/= 0.0232 e 5X10--6 h = f _.!:__.!!___= (0.0232)(10.5)(5.61)2 = 2.50 ft L 4 R 2g 4(0.0119)(2)(32.2) Sp = yhL = (49.1 lb/ft3) X 2.50 ft X (1 ft2/144 irr') = 0.851 psi

128 Chapter 9

9.53 Given: Figure 9.29. Rectangular channel with three fins. Q = 225 L/min Brine (20% NaCI), sg =1.10 at OÜC; p = 1.10(1000 kg/nr') = 1100 kg/m' Õ = 2.5 x ro-s N s/m2 (App.D) Find: Reynolds number for the flow. NR = v(4R) piÕ A = (20)(50) - 3(5)(10) = (850 mm' [(1 m2)/(103 mmr']> 8.5 x 10-4 m2 WP = 2(20) + 2(50) + 6(10) = 200 mm R = AIWP = (850 mm2)/(200 mm)= (4.25 mm)(l m2/1000 mm)= 0.00425 m v = QIA = (0.00375 m3/s)/(8.5 x 10-4 m2) = 4.41 mis NR = v( 4R) piÕ= ( 4.41)(4)(0.00425)(1100)/(2.5 X 1o-3)=3.30 X 104 = NR Turbulent Energy Loss for 1.80 m of commercial steel: hi. = f(L/4R)(v2/2g) 4R!&= 4(0.00425 m)/(4.6 x 10-s) = 370 Then/= 0.030 hà = (0.030)(1.80/(4)(0.00425)][(4.41)2/(2)(9.81)) = 3.149 ID

Energy Loss for 54 in (4.50 ft) of drawn copper: hL = j(L/4R)(v2!2g) Top: 4RI&= 4(0.0249 ft)/(5.0 x 10-6 ft) = 19 920 Then/= 0.0315 he = (0.0315)(4.50/(4)(0.0249))((15.0)2/(2)(32.2)) = 4.97 ft Each side: R!&= 4(0.01885 ft)/(5.0 x 10-6 ft) = 15 080 Then/= 0.0340 hi. = (0.0340)[4.50/(4)(0.01S85)][(15.0)2/(2)(32.2)] = 7.09 ft (each of two sides) Total loss for all three channels: hLT = 4.97+2(7.09)=19.15 ft

Both Side Channels: 1 Yz-in Type K copper tube (Half). ID = 0.1234 ft; Ćtot = 1.196 x 10-2 ft2 A= Ć10J2 = (1.196 X 10-2 ft2)/2 = 5.98 X 10-3 ft2 WP =ID+ n(ID)/2 = ID(l + m2) = 0.3172 ft R = AIWP = (5.98 x 10-3 ft2)1(0.3 l 72 ft) = 0.01885 ft NR = v(4R)pl Õ = (15)(4)(0.01885)(2.13)/(3.38 x 10-4) = 7.127 x 103 = 7127 = NR Turbulent

129 MINOR LOSSES

10.7 K=0.7l;hL= 0.71(3.00)2/2(9.81) =0.326m

D 75 -2 = - = 3.0; K = 0.31 (Table 10.2) for B = 20Á D1 25

hL = Ku12 /2g =0.31(3.00 m/s)2/2(9.81 m/s") =0.142 m

D -2 = 3.0; K= 0.16 (Table 10.2) for B= 15Á D1

hà = K Uá2 /2g = 0.16( 4.0)2 /2(32.2) = 0.0398 ft

P =P = 62.4[0.442-4Ŀ002 +0.0398]ft X lft2 =-0.0891psi 1 2 2(32.2) 144in2

(D )2 (20)2 Vi = V1 -1 = 4.0 ft/s -Ŀ- = 0.444 ft/s D2 6.0

- = 62.4lb[0 . .4442 -4ĿÜ2 +0.194]ftx lft2 =-0.0224 si Pi p2 ft3 2(32.2) 144in2 p

[u2 -u2 ] pá - P2 = r 2 2g 1 + hL ; assume Zá = Z2

hi. = Ku12/2g = (0.78)(4.0 ft/s)2/2(32.2 ft/s") = 0.194 ft

u2 u2 D 6 O ll+z1 +-1 -hL = Pz +z2 +-2 :-2 =-Ŀ-=3.0~K=0.78(Table10.1) r 2g r 2g D1 2.0

u = Q= 0.10ft3/s -20.0ft/s:D2 = 0Ŀ2803ft 1 =3.51Ŀ K=0.73(Table10.1) 1 A1 0.00499 ft2 D1 0.07975 ft '

hL = K u12 = (0.73)(20.0 ft/s)2 = 4.55 ft 2g 2(32.2 ft/s")

Vá= Q = 3.0xl0-3 ~3/s2 =6.47 mis: D2 = 85.4 mm = 3.51 A1 4.636x1 o- m D1 24.3 mm

K= 0.73 (Table 10.1): hL =K u12=0.73(6.47)2m=1.56 m 2g 2(9.81)

D2/D1=100/50=2.00;K=0.52(Table 10.1) hi. = Ku12 /2g =0.52(3.0 m/s)2/(2)(9.81 m/s") = 0.239 m

10.6

10.5

10.4

10.3

10.2

10.1

130 Chapter 1 O

u= Q = 85 gal/min x 1 ft3 Is = 2.099 ft/s A 0.0902 ft2 449 gal/min

NR = vD = (2.099)(0.3389) = 5.88x104: D = 0.3389 = 2259: f = 0.0215 V 1.21X10-5 8 1.5 X 10--4

Prob.10.10 Prob. 10.8 Prob.10.11 e 812 sin(B/2) L(ft) bà (ft) hà (ft) hL (ft) r exp cxp

2Ü 1Ü 0.01745 9.54 0.0414 0.031 0.0724

10Ü 5Ü 0.08716 1.910 0.0083 0.082 0.0903

15Ü 7.5Á 0.1305 1.276 0.0055 0.164 0.1695

20Ü 10Ü 0.1736 0.959 0.0042 0.317 0.3212

30Ü 15Ü 0.2588 0.643 0.0028 0.491 0.4938

40Á 20Ü 0.342 0.487 0.0021 0.603 0.6051

60Ü 30Ü 0.500 0.333 0.0014 0.726 0.7274

h = f !:___!!.__ = (0.0215)(L)(2.099)2 = 0.00434(L) L.r D 2g (0.3389)(2)(32.2) D = (D2 +DI )12 (0.5054 + 0.1723)/ 2 = 0.3389 ft A= 7²D214 = 7²(0.3389)2 /4 = 0.0902 ft2

L = (D2 -D,)12 sin(B/2)

L = (0.5054-0.1723)/2 = 0.1666ft sin(B/2) sin(B/2)

10.10 and 10.11

sin(~)= (Dz

10.9 Graph shown after Problem 10.11.

hL 0.031 0.082 0.164 0.317 0.491 0.603 0.726ft

0.71 0.03 0.08 0.16 0.31 0.48 0.59 K

60Á 2Ü 10Ü 15Á 20Ü 30Á 40Ü e

D2 = 0.5054 = 2.93 D, 0.1723

Vi = 2:_ = 85 gal/min x 1 ft3 Is = 8. l l ft/s: A1 0.02333 ft2 449 gal/min

he = Kv12 /2g = K(8.l 1)2/2(32.2) = K(l.022 ft)

10.8

131 MINOR LOSSES

10.15 Exitloss: be= 1.0 v2/2g=l.0(2.146)2/2(9.81) =0.235m

v = Q = 0.04 m3 Is = 2.146 mis A 1.864x10-2m2

10.14 Same analysis as 10.13 a. e= 60Á; K= 0.71; b: = 0.899 m;pz = 503.6 kPa; 29% of ideal b. e= 30Ü; K = 0.48; he = 0.607 m; Pi= 506.4 kPa; 53% of ideal c. 8= 10Á; K= 0.08; h: = 0.101m;p2=511.2 kPa; 92% ofideal

] [

2 2 ] 503.7 kPa h 00 9 4.98 - 0.584

- L = 5 + 9.7 -0.873 = 3.7kParecov. 2(9.81)

30%of ideal

hi. = K v12 = 0.69 ( 4Ŀ98)2 = 0.873 m 2g 2(9.81)

D/D1 = 73.8 mml25.3 mm = 2.92; K = 0.69

10.13 P2 = pá + r[V12 -vi 2g

Vá = id._= 150 L/min x 1 m3 Is = 4_98 mis A1 5.017x10-4 60000 L/min

Vi= fl = 150 = 0.584 mis Ćz (4.282X10-3)(60000)

". = 512.2 kPa

12.2 kPa Recovery

Ideal diffuser - No energy loss v2 v2

Pi + z1 + -1 - hL = P2 + z2 + -2 : : hL =O, Zá = Z2 r 2g r 2g

= + [v( -vi] = 500 kPa + 9.79 kN[4.982 -0.5842 P2 Pi y 2g m3 2(9.81)

10.12

60Ü 50Ü 20Ŀ 30Á 4()- ConeAngle

10Ŀ o

Energy l.oss4 (ft) Å

.6

_,.. ... V

..........

J ... ....-Ŀ-- 1 1 V ", rob. 10 ÅÅ

Pnlb. 1p.11 / / I/

~I 1.;f Mln 1mur , lo8 IOC arn at 1 '8Å .,. _, 1

.2

.8

132 Chapter 10

10.22 Sudden contraction: D1 / D2 = 0.3188 ft/0.125 ft = 2.55

Vz = 2- = 25Ü gal/min x 1 ft3 /s = 45.38 ft/s; K = 0.31 (Table 10.3) Az 0.01227 ft2 449 gal/min

hL =Kv; /2g = (0.31)(45.38)2 /2(32.2) = 9.91 ft

D Gradual contraction: From Prob. 10.20: Vz = 4.37 mis; -1 = 2.48 D2

K=0.23;hL= (0.23)(4.37)2/2(9.81) =0.224m (Fig.10.10)0= 105Á

10.21

Sudden contraction: he = K( v; /2g) D1 = 122.3 mm = 2.48 D2 49.3mm

Vz = 2- = 5oo L/min x 1 m3 Is = 4.37 mis; K = 0.38 (Table 10.3) Az 1.905x10-3 m" 60000 L/min b: = (0.38)(4.37)2 /2(9.81) = 0.371 fil

10.20

10.19 False: Kdecreases,buthL=K(v;12g) The velocity head increases faster than K decreases. Examples: Vz = 1.2 mis; K = 0.44; hi. = 0.0323 m

Vz = 6.0 mis; K = 0.39; he = 0.716 m Vz = 12.0 mis; K = 0.33; hL = 2.42 m

P2 = 80 psig- 1.91psi=78.09 psig See Probs. 10.12, 10.13 for analysis.

[v2 -v2 ] P2 =p, + r 1 2g

2 - hL

= 80 si + (0.87)[62.4 lb][4.002 -16.002 -l.35]ftx 1 ft2 p g ft3 2(32.2) 144 in '

10.18

(A) (D )2 (4.0)2 Vz = v1 ~ = v1 D: = 4.00 ft/s 2.0

= 16.00 ft/s

For D2/D12 = 2.0, K= 0.34 (Table 10.3) hL = (0.34)(16.0 ft/s)2 /2(32.2 ft/s") = 1.35 ft

10.17 Sudden contraction: hL = Kv; /2g

10.16 Exitloss: hàè 1.0 v2/2g=l.0(7.48)2/2(32.2) =0.868ft _ Q _ 1.50 ft' /s _ 7 48 f11 v- -- - . t s A 0.2006 ft2

133

a Å 180Ü - sudden

MINOR LOSSES

0Å90Ü

a-10Á

o 3 5 10 15-40 5CMIO 78 90 105 120 150 Cone Angle (DegrMI)

2

o ,/ V

1 / Ir

s I/ V '

-, / - _/ .. o

1

10.25 Sketches of selected contractions:

150Ü 0.36 11.51 120Ü 0.28 8.95 105Á 0.23 7.35 90Á 0.19 6.08 76Á 0.135 4.32 50-60Á 0.075 2.40 15-40Ü 0.045 1.44 10Ü 0.048 1.53 5Ü 0.084 2.69 3Ü 0.109 3.49

10.24 Gradual contraction: Data from Prob. 10.22: D1/D2 = 2.55 hL =Kv~ 12g=K(45.38)2/2(32.2) =K(31.98 ft)

e K hà (ft)

10.23 Gradual contraction: Data from Prob. 10.22: K= 0.135 (Fig. 10.10) B= 76Ü he = (0.135)( 45.38)2 /2(32.2) = 4,32 ft

Chapter 10 134

10.36 Long radius elbow: LjD = 20 (See Prob. 10.34, 10.35) Sp = yhL = 9.81 [(0.018)(20)(0.834)] = 2.95 kPa Lowest !!..pis proportional to LjD.

10.35 Street elbow: LjD = 50 (See Prob. 10.34) Sp = yhL = (9.81)((0.018)(50)(0.834)] = 7.36 kPa

u= Q = 750 L/min x 1 m312 = 4.045 mis: ~ = ( 4.045)2 = 0.834 m A 3.09xl0-3m2 60000L/min 2g 2(9.81)

Elbow fr= 0.018 (Table 10.5)

[ L u2] 9.81kN !!..p = yhl = r fr ___f.__ = 3 [0.018(30)(0.834 m)] = 4.42 kPa D2g m

10.34

10.33 !!.. = = [á; Le ~]=(0.90)(62.4 lb)[(0.016)(150) (I0.4)2 ](ft)(l ft2) = 1.58 si '1J n; r r D 2g ft3 2(32.2) 144 in ' P

Angle u= Q = 650 gal/min x 1 ft3 Is = 10.4 ft/s valve A 0.1390 ft2 449 gal/min

fr= 0.016 (Table 10.5)

10.32 Ball-type check valve: Let K = fr(LelD) = fr(150) = 0.019(150) = 2.85 For 2-in Schedule 40 pipe, fr= 0.019 (Table 10.5)

10.31 Gate valve, fully open: LjD = 8 Le= (LjD) x D = 8(0.2545 m) = 2.04 m

10.30 10 in Sch. 40 pipe: D = 0.2545 m Globe valve: Lj D = 340 Le= (LjD) x D = 340(0.2545 m) = 86.53 m

V (3.0 m/s)" Entrance loss: h = K-2 =K = K(0.459 m) L 2g 2(9.81 m/s")

a. K = 1.0; hi. = 0.459 m b. K = 0.50; he = 0.229 m c. K = 0.25; hL = 0.115 m d. K = 0.04; hL = 0.018 m

_ Q _ 50 gallmin 1 fr' Is _ 5 84 fil Uz- -- X - Å t S Ć.i 1.907 X 10-2 ft2 449 gallmin

h = K( u;)= o.so[ (5.34 ft/s)2 ] = 0.265 ft L 2g 2(32.2 ft/s")

Gradual contraction, B = 120Ü D1 = 6.14 in, D2 = 3.32 in Ductile iron pipe D11D2 = 1.85; K= 0.255

10.29

10.28

10.27

135 MINOR LOSSES

See Problem 10.37: P1 Pz = r[ hl'"b' + h4c.,d] = (68.47) [18.46 + 2.07)/144 p1 - P: = 9.76 psi (Virtually equal to Prob. 10.37)

90Ü Bend: r = R0 - Dof2 = 3.50 in - 2.0012 = 2.50 in r/D = 2.50 in/0.620 in= 4.03 Use Lef D = 14 frorn Fig. 10.23

h, = 2fr Le~ = (2)(0.027)(14) (13Ŀ28)2 = 2.07 ft (2 Bends) bend D 2g 2(32.2)

h =/~~=(0.041)[ 8Ŀ50 ](13Ŀ28)2 =18.46ft l1"be D 2g 0.05167 2(32.2)

10.38 3/4-in Steel tube, 0.065 in wall thickness D = 0.620 in= 0.05167 ft; A = 2.097 X 10-3 ft2

u= Q = (12.51449) ft3/s = 13.28 ft/s A 2.097X10-3 ft2

NR= vDp =(13.28)(0.05167)(2.13) =4.32x 103 Õ 3.38X10-4

Die= 0.05167 =344Ŀ/=0.041 (1.5X10-4) '

Use/r= 0.027 because Die is the same as for the 112-in steel pipe in Problem 10.37.

Close return bend: LjD = 50, fr= 0.027 (Table 10.5) _ Q _ 12.5 gal/min 1 ft3/s _ 13 19 .1::á u~ -- X - Å tt S A 0.00211 ft2 449 gal/min

NR = vDp = (13.19)(0.0518)(2.13) = 4_31x103 Õ 3.38xl0-4

Die= 0.0518/1.5 x 10-4 = 345: f= 0.041

h =J. Le~=(0.027)(50)(13.19)2 ft=3.65ft Lb,"J T D 2g 2(32.2)

h =I. ~~=(0.041)[ 8Ŀ00 ]Ü3Ŀ19)2 = 17.12ft L,,b, T D 2g 0.0518 2(32.2) Pi - P: = (68.47)[17.12 + 3.65)/144 = 9.87 psi

10.37 Pt. 1 at inlet; Pt. 2 at outlet 2 v2

!i + z +!:l. - h - h = Pz + z + -2 Å v1 = 1J:2; assume z1 = z2 r 1 2g Lp,,~ 4.,"d r z 2g. P1 P2 = r( hLP,P' + h4,.,d )

136 Chapter 10

10.42 Pipe bend: See Prob. 10.41 for sorne data. For minimum energy loss, r/D = 3, LiD = 11.8 (Fig. 10.23)

r= (;)n=(3)(25.3mm)=76mm

h = f, Le~= (0.0115)(11.8) (3-3 l)2 = 0.477 m L T D 2g 2(9.81)

fr= 0.018 (Table 10.5)

h = I, Le ~=(0.018)(60) (4Ŀ76)2 = 1.25 m L T D 2g 2(9.81)

10.41 Pipe bend: R0 = 300 mm; r = R0 -Dj2 = 300- 28Ŀ6 = 285.7 mm 2

r/D = 0.2857 m/0.0253 m = 11.3 ~ LefD = 32 from Fig. 10.23. u= Q = 250 L/min x 1 m3 Is = 8_31 mis

A 5.017x10-4m2 60000 L/min Nç = uD = (8.31)(0.0253) = 5_84x105: D = 0.0253 = 16867:

v 3.60x10-7 e 1.5x10--6 fr= 0.0115 in zone ofcomplete turbulence

h = f, Le ~=(0.0115)(32) (3.3l)2 = 1.29 m L T D 2g 2(9.81)

10.40 Tee-flow through branch: LefD = 60 u=Q= 0.08m3/s =4.76m/s

A 0.01365 m"

Q 0.40 ft3 /s L u= A=

0_05132 ft2 = 7.79 ft/s: Tee-flow through run: Jy =20

fr= 0.018 (Table 10.5)

he = fr Le ~=(O.O 18)(20) (7. 79)2 = 0.340 ft D 2g 2(32.2)

10.39

137 MINOR LOSSES

10.45 Bend intube: !_ = 150 mm = 9.52: Le = 29 (Fig. 10.23) D 15.75mm D

u= Q = 40L/min x 1 m3/s = 3.42 mis A 1.948x10-4 m2 60000 L/min

D O.Ol575 42 áĿ O 025 . f 1 b 1 - = 5 = 3 : r = . m zone o comp ete tur u ence e 4.6x10-

h = J, Le!!___= (0.025)(29) (3.42)2 = 0.432 m L T D 2g 2(9.81)

2Ŀ59 m 3 05 , 1 1 -- = . times proposa . 0.85 m 2.59 m 2 3 . 12 -- = .3 times proposa . 1.11 m

n; =2.59m h =f !:_!!___=(0.0215)( L4o )c2.194)=1.340m ""'1 Lp;pe D 2g 0.0493

u= Q = 750 L/min x 1 m3 /s = 6.56 mis: !!___ = (6.56)2 = 2.194 m A 1.905xl0-3 m2 60000 L/min 2g 2(9.81)

NR = uDp = (6.56)(0.0493)(802) = 1.35 x 105 Õ 1.92x 10-3

DIE= 0.0493/4.6 x 10-5 = 1072: f= 0.0215: fr= 0.019

h = fr Le !!___=(0.019)(30)(2.194m)=l.25 m ldhow D 2g

10.44

0.862 m hL =1.11 m to\

h =f !:_!!___=(0.017) 1.20 (6.43)2 Ltuoo D 2g 0.0498 2(9.81)

r 150mm L Proposal 2 - Bend + tube: - = = 3.01: .zs. = 11.8 D 49.8mm D

h = J,, i.ç: (0.010)(11.8) (6.43)2 = 0.248 m Lb<md

1 D 2g 2(9.81)

10.43 Proposall-Pipebend: !_= 750mm=15.1; Le =40.5(Fig.10.23) D 49.8mm D

Q 750 L/min 1 m3/s u=-= x = 6.43 mis A 1.945x10-3 m2 60000 L/min

NR = uDp = (6.43)(0.0498)(802) = 1.34 x 105: D = 0.0498 = 33200: Õ 1.92X10-3 e 1.5X10--6

f= 0.017;fr= 0.010 in zone of complete turbulence

h = f, Le!!___= (0.010)(40.5) (6.43)2 = 0.85 m L T D 2g 2(9.81)

138 Chapter 1 O

10.49 Data from Problem 10.47. Coil with 6.0 revolutions. K = 0.5015 for one 90Á bend. n = 6.0 rev(4 90Á bends)/rev = 24 90Á bends. v2/2g = 1.692 ft. r/D = 5.41.fr = 0.0295. K8 = (n - 1) [0.25efá{r/D) + 0.5K] + K [Equation 10-10] K8 = (24 - 1)[0.25m:0.0295)(5.41) + 0.5(0.5015)] + 0.5015 = 9.15 = Kn hi. = K8(v2/2g) = (9.15)(1.692ft)=15.5 ft = /1L

10.48 90Ü Bend. Steel tube; 1 114 in OD x 0.083 in wall thickness. r == 3.25 in. D = 1.084 in= 0.09033 ft. Die= 0.09033/1.5 x 10-4 = 602. Thenfr = 0.023 in fully turbulent zone. A= 6.409 x 10-3 ft2. Q = 27.5 gal/min = 0.0612 ft3/s. v = QIA = (0.0612 ft3/s)/(0.006409 ft2) = 9.556 ft/s (v2/2g) = (9.556 ft/s)2/[2(32.2 ft/s2)] = 1.418 ft. r/D = 3.25 in/1.084 in= 3.00. LefD = 12.5 (Fig.i0.27), K = f,!(L.)D) = 0.023(12.5) = 0.288. hà = K(v2/2g) = 0.288(1.418 ft) = 0.408 ft = hL

Then hi. = 5.65 - 1.20 + 0.476 - 0.0297 = 4.90 m vz

LethL= K-i 2g

ThenK= ~= 4.90m = 10.3 u12 /2g 0.476 m

10.47 90Ü Bend. Steel tube; '/i in OD x 0.065 in wall thickness. r = 2.00 in. D = 0.370 in= 0.03083 ft. DI&= 0.030S3/1.5 x 10-4 = 206. Thenfr = 0.0295 in fully turbulent zone. A = 7.467 x 10-4 ft2. Q = 3.5 gal/min = 0.00780 ft3/s. v = Q/A = (0.00780 ft3/s)/(0.00780 ft2) = 10.44 ft2Å (v2/2g) = (10.44 ft/s)2/[2(32.2 ft/s2)] = 1.692 ft. r/D == 2.00 in/0.370 in= 5.41. LefD = 17 (Fig. 10.27). K = fr!(LefD) == 0.0295(17) = 0.5015. b: = K(v2/2g) = 0.5015(1.692 ft) = 0.849 ft = b:

Manometer: pá + Yw(0.25 m) - Ym(0.35 m) - Yw(l .1 O m) = P2 Pi - P: = Ym(0.35 m) + Yw(l .1 O - 0.25)m Pi - p2 = Ym (0.35 m) + Yw(0.85 m) = 132.8(0.35 m) + 0.85 m r: r; r; 9.69

Pi - Pz = 4.80 m + 0.85 m = 5.65 m Yw

v = Q = 6.0x10-3 m3 /s = 3.056 mis: Vi2 = (3.056)2 == 0.476 m 1 Ai 0.001963 m" 2g 2(9.81)

= Sl = 6Ŀ0x10-3 = 0.764 mis: v~ == (0.764)2 == 0.0297 m Uz Ćz 0.007854 2g 2(9.81)

z1 -z2=-1.20m

A = rcDiz = rc(0.050 m)2 = 0.001963 m2 1 4 4

A2= rcD~ = rc(O.lOm)2 =0.007854m2 4 4

2 2 Pi Vá h P2 Vz -+z +-- L =-+z2 +- r; i 2g Yw 2g

P P vz v2 h = i- 2 +z -z + i - 2 L i 2 2g Yw

10.46

139 MINOR LOSSES

10.59 Sp = sg(Q!Cv)2 = 0.981(150/170)2 = 0.764 psi sg = 61.2 lb/ft3 /62.4 lb/ft3 = 0.981

For Problems 10.59 to 10.70, values of C, are found from Table 10.6.

10.58 See Problem 10.57 for procedure and data from Problem 10.54. For Q = 7.5 gal/min: Cv = Q/ ,j!'!:..plsg = 7.5/ '110410.90 = 0.698 = Cv

For Q = 1 O.O gal/min: c. = Q / ,J !'!:..p/sg = 1 O.O/ '1175 / 0.90 =O. 717 = Cv

10.57 Use Eq. 10-10. Solving for c. = Q/ ,j!'!:..p/sg. From Problem 10.53, Q = 5.0 gal/min, !'!:..p = 60.0 psi, and sg = 0.90. Then, e, = Q/ ,j!'!:..p/sg = 5.0/ '160/0.90 = 0.612 = c.

10.56 Procedure like Prob. 10.55: For Q = 7.5 gal/min, !'!:..p = 104 psi. v = 12.5 ft/s. K = 110 For Q = 10 gal/min, Sp = 175 psi. v = 16.67 fUs. K= 104.

10.55 Find K if !'!:..p = 'J'hL = yK(v2/2g). Then, K = (!'!:..p)/[J(v2!2g)] Dimensionless. From Prob. 10.53: For Q = 5 gal/min, Sp = 60 psi= 8640 lb/ft2Å

5/8 OD tube with t = 0.065 in. A= 1.336 X 10-3 ft2Å r= 0.90(62.4 lb/ft') = 56.16 lb/ft3Å

v = QIA = [(5 gpm)/(1 ft3/s/449 gpm)]/(1.336 x 10-3 ff) = 8.34 ft/s (v2/2g) = (8.34 ft/s)2/[2(32.2ft/s2)] = 1.079 ft K= (!'!:..p)/[J(v2!2g)] = (8640 lb/ft2)/[(56.16 lb/ft3)(1.079ft)]=143 =K

10.54 Directional control valve. Figure 10.29. For Q = 7.5 gal/min, !'!:..p = 104 psi. For Q = 10 gal/min, !'!:..p = 175 psi.

10.53 Directional control valve. Figure 10.29. For Q = 5 gal/min, !'!:..p = 60 psi.

10.52 Data from Problem 10.48. 60 degree bend. K = 0.288 for one 90Ü bend. n = 60Á190Á = 0.667 90Á bends. v2/2g = 1.418 ft. r!D = 3.00.fr = 0.023. Ks = (n - l)[(0.25efi(r/D)+ 0.5K] + K [Equation 10-10] Ks = (0.667 - 1) [0.25m(0.023)(3.00) + 0.5(0.288)] + 0.288 = 0.222 = K8 hL = Ks(v2!2g) = (0.222)(1.418) = 0.315 ft = hL

10.51 Data from Problem 10.47. 145 degree bend. K = 0.5015 for one 90Á bend. n = 145Á/90Á = 1.61 90Á bends. v2/2g = 1.692 ft. r/D = 5.41.fr = 0.0295. Ks = (n - l)[0.25efi(r/D)) + 0.5K] + K [Equation 10-10] Ks = (1.61 - 1) [0.25ll(0.0295)(5.41) + 0.5(0.5015)] + 0.5015 = 0.731 = Ks hi. = Ks(v2/2g) = (0.731)(1.692ft)=1.25 ft = hL

10.50 Data from Problem 10.48. Coil with 8.5 revolutions. K = 0.288 for one 90Á bend. n = 8.5 rev(4 90Á bends)/rev = 34 90Á bends. v2/2g = 1.418 ft. r!D = 3.00.fr = 0.023. Ks = (n - 1)(0.25efi(r/D) + 0.5K] + K [Equation 10-10] Ks = (34 - 1)[0.25m(0.023)(3.00) + 0.5(0.288)] + 0.288 = 6.83 = Ks he = Ks(v2/2g) = (6.83)(1.418) = 9.68 ft = hL

Chapter 10 140

10.62 L1p = sg(QICv)2 = 1.590(60/90)2 = 0.707 psi

10.63 Sp = sg(QICv)2 = 0.68(300/330)2 = 0.562 psi

10.64 !ip = sg(Q/Cv)2 = 0.495(5000/4230)2 = 0.692 psi

10.65 Sp = sg(QICv)2 = 1.590(60/34)2 = 4.952 psi

10.66 Sp = sg(QICv)2 = 0.68(300/160)2 = 2.391 psi

10.67 Sp = sg(QICv)2 = 0.495(1500/700)2 = 2.273 psi

10.68 !ip = sg(QICv)2 = 1.030(18/25)2 = 0.534 psi

10.69 Sp = sg(QICv)2 = 0.823(300/330)2 = 0.680 psi

10.70 Sp = sg(QICv)2 = 1.258(3500/2300)2 = 2.913 psi

For Problems 10.62 to 10.70, values of sg are found from Appendix B.

10.60 Sp = sg(QICv)2 = 0.989(600/640)2 = 0.869 psi sg = 61.7 lb/ft3/62.4 lb/ft3 = 0.989

10.61 Sp = sg(Q!Cv)2 = 0.997(15/25)2 = 0.359 psi sg = 62.2 lb/ft3/62.4 lb/ft3 = 0.997

141 SERIES PIPE UNE SYSTEMS

38 he = l.Ohu + lOOfrhu + 150frhu + 30frhu + l.Ohu + f 0.0525 hu = hu[7.32 + 724/]

Ent. Chk. valve Ang. V. Elbow Exit Friction

h = !!___ (in i e): u= Q = 435 L/min x 1 m3/s = 3.344 mis u 2g pp A 2.168xl0-3m2 60000L/min hu= (3.344)2/(2)(9.81) = 0.570 m NR = vDp = (3.344)(0.0525)(820) = 8.47 x 104; D = 0.0525 = 114l;f = 0.0222

Õ l. 70 X 10-3 e 4.6X10-S PA = y[4.5 m + hL] = (0.82)(9.81kN/m3)[4.5m+0.570 m(7.32 + 724(0.0222)] = 143.5 kPa

p v2 P v2 Class IĿ _d.+ z +_A.__ h =----Û- + z8 +---Û--: VA= Vi3 =O; PB =O

, r A 2g L r 2g

PA = y[zn -ZA+ hL] = y[4.5 m + hL]Jr= 0.019

11.2

Entr. 3 Elbows Friction

h =Velocityheadinpipe= v~ =-1-(Ü)2= (L99mls): =0.202m ull 2g 2g A 2(9.81 mis ) NR = vD/v = (1.99)(0.098)/(1.30 x 10-6) = 1.50 x 105 Die= 0.098/1.5 x 10-6 = 65300 -e> f =O.O 165; fr-;::; 0.010 in zone of complete turbulence

pB= 9Ŀ81~N[12m 0.202-0.202-(3)(0.010)(30)(0.202)-(0.0165) 80Ŀ5 (0.202)] m 0.098

PB = 85.1kN/m2=85.1 kPa

p v2 P v2 Class I; Pt. A at tank surface: _A_+ z +_A_ -hL =----Û- + z8 +---Û--: PA =O, VA= O

r A 2g r 2g

r [ZA -z8 =V~ hL]=r [12m-hu -1.0hu -3f.r(30)hu f L hu, J PB = w . 2g \V ll il B D l

11.1

Class I systems

CHAPTER ELEVEN

SERIES PIPE LINE SYSTEMS

Chapter 11

Similarly: At 210ÜF, v = 7.85 x 10-5; NR = 8.49 x 104 ~ f= 0.0205 hL = 26.2 ft; PA = 509.6 psig

At lOOÜF, v0 = 7.21x10-4 ft2/s; NR = vADA = (20.9)(0.3l~8) = 9.25 x 103 A V 7.21X10

D/c=0.3188/1.5 X 10-4=2125 ~f=0.032 ¶á = [1.30 + 125(0.032)](6.79 ft) = 36.0 ft

PA = 500 psig + (0.895)(~2.4 lb) [ 4.0 + (1.32 - 6. 79) + 36.0] ft(l ~t2j = 513.4 psig ft lMm

2 40 J'1 2 2 2 hl = 2(30)/r VA + j 1ĉ VA + 0.28 VA = (1.30+125 j) VA

2g 0.3188 ft 2g 2g 2g Elbows Friction Enlarge

Where Di/D1=0.4801/0.3188=1.51~K=0.28 (Table 10.1)

. 1 ft3 Is Q = 750 gal/mm x = 1.67 ft3 Is

449 gal/min

= JL = 1.67 ft3 Is = 20.9 ft/s VA AA 0.07986ft2

= JL = 1.67 ft3 Is= 9.23 ft/s Va A8 0.181 ft2

fr=0.017

Class 1

p v2 p u2 ; +zA + 2~ =b: =-;+zB + 2; PA =pa +rÜ[(za-zA)+v~2-gv~ +hL]

v~ = (2ÜĿ9)2 ft = 6.79 ft; v~ = 1.32 ft 2g 2(32.2) 2g

p v2 p v2 Classl; ~+zA +_1'_-hL =-Û-+z8 +-Û-: r, = Pa +y[(z8 -zA)+hL] r 2g r 2g

v= Q = 60 gal/min x 1 ft3/s = 5.73 ft/s: hu=~= (5Ŀ73)2 = 0.509 ft A 0.02333 ft2 449 gal/min 2g 2(32.2)

50 b: = 6.5hu+ 2(30)frhu+ J hu= (6.5 + 60Jr+ 290.f)hu: Jr= 0.019

0.1723 . NR = vDp = (5.73)(0.1723)(0.90)(1.94) = 2.87 x 104: D = 0.1723 = 1150

rá 6.0X10-s e 1.5X10-4

f= 0.0260: Then hL = [6.5 + 60(0.019) + 290(0.0260)](0.509 ft) = 7.73 ft

200 . (0.90)(62.4 Ib) [25 f 7 73 f] 1 ft2 212 8 . p A = ps1g + 3 t + . t . 2 = . ps1g ft 144m

142

11.4

11.3


hr = 1.0 vi+á; 100 u{ +2(30)/n u{ +160/n vi +0.43 vi+ Ir, 300 u~ 2g 0.0843 2g 2g 2g 2g 0.1560 2g Entr. Friction 3-in Elbows 3-in Gate valve Enl. Friction 6-in

v2 v2 + 2(30)fT6 -6 + 1.0-6

2g 2g Elbows 6-in Exit

Zá Zz = _LhL = 1.47 + 38.4 + 1.90 + 5.06 + 0.63 + 4.97 + 0.140 + 0.126 z1-z2=52.7m

11.6 Class 1 Pts. 1 and 2 at reservoir surfaces: Pi = p2 =O; Vá= Vz =O p v2 p v2 _1_ + z + -1 - h = -2 + z + -2 : z1 - z2 = hr = sum of 8 losses r 1 2g L r z 2g

Water at lOÜC; v = 1.30 x 10-6 m2ls

3-inpipe: V3= S2_ =5.37mls: vi =1.47m: NR =v3D3 =3.48xl05 ~ 2g 3 V

D3/c = 0.084311.2 x 10-4 = 703;/J = 0.022; F13 = 0.0215 Q u2 vD 6-inpipe: v6= -=1.57mls: -6 =O.l26m;NR =-6-6=1.88x105 A6 2g 6 V

D61e = 0.156011.2 X 10-4 = 1300;á;, = 0.0205;fT6 = 0.0185

PA = 12.74 MPa

P v2 P v2 Class 1 Ŀ _A_ + z + __!!__ - h = ---Û- + z + _!L ' r A 2g L r B 2g '

ThenpA=pB+ ro[(zA-zB)+v~2gvl ".

u = g = O.O 15 m3 Is = 0.892 misĿ u~ = 0.8922 m = 0.0405 m A AA 1.682xl0-2m2 '2g 2(9.81)

Vá3 = g = 0.015 = 7.87 mis; u~ = (7.87)2 m = 3.16 m A8 1.905xl03 2g 2(9.81)

N =u ADA= (0.892)(0.l463) = 6.15 x 103Å Die= 0Ŀ1463 = 3180Ŀá; = 0.035 RA v 2.12xl0-5 ' 4.6x10-5 ' A

N = vaDa = (7.87)(0.0493) = 1.83 x 104Ŀ Die= 0.0493 = 1072;fB = 0.028 Rn V 2.12xl0-5 ' 4.6xl0-5

fm = 0.019

hr = j~ 0.\~~3 (0.0405) + 0.37(3.16) + 2(20)Cfm)(3.16) + /B 0.0:93 (3.16)

Friction 6-in Contr. Elbows Friction 2-in Where D1ID2 = 0.146310.0493 = 2.97; K= 0.37, Table 10.3 hi. = 19.68 m

8.80 kN PA = 12.5 MPa + 2 [4.5 + 3.16 - 0.0405 + 19.68]m = 12.5 MPa + 240 kPa m

11.5

Chapter 11 144

(See Computer solution on next page)

PA - PB = (1.25)(9.81kN/m3)[1.2m+0.577 m] = 21.79 kPa

11.7 Class 1 p vz p o: _A+ ZA+ _A__hL =-B +zB + _IL: PA - Ps + r[(zB -zA)+hd; VA= Us r 2g r 2g

VA = Us Q = l. 70 L/min x 1 m3 Is = 0.346 mis: v2 = 6.1 O x 10-J m A 8.189x10-s m" 60000 L/min 2g

NR = vDp = (0.346)(0.01021)(1250) = 1.47 x 104: D = 6807;/= 0.0282; fr= 0.013 Õ 3.0xl0-4 e 30 v2 v2 v2 v2

hL =f -+150fr-+340fr-+8(50)fr-=0.577m 0.010212g 2g 2g 2g Friction Ball chk. Globe v. 8 Ret. bends


*NOTE: Pressure at point 1 set arbitrarily to 100 kPa. Pressure at point 2 is computed for il/ustration of effect of pressure drop.

Resu/ts:

Element 2: K2 = Element 3: K3 - Element 4: K4 = Element 5: Ks =

Element 6: K6 = Element 7: K7 -

Element 8: Ka

0.00 Energy losses-Pipe 2:

Pipe: K1 = f(UD) =

Energy loss hL1 = 0.5064 rn Friction Energy loss hL2 = 0.0119 m (h = 0.013) Energy loss hL3 = 0.0270 m (h = 0.013) Energy loss hL4 = 0.0317 rn (h = 0.013) Energy loss hLs = 0.00 rn Energy loss hL6 = 0.00 m Energy loss hL7 = 0.00 rn Ener loss n.ç = 0.00 m

Energy loss hL1 = 0.00 m Friction Energy loss hL2 = 0.00 m Energy loss hL3 = 0.00 rn Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hL6 = 0.00 m Energy loss hu = 0.00 m Ener loss hLB = 0.00 m

Total ener loss huot = 0.5770 m

-21.79 kPa

K Qty.

Energy losses-Pipe 1: K Qty. Pipe: K1 = f(UD) = 82.96 1 Check Va/ve: K2 Globe Va/ve: K3 Return Bends: K4 Element 5: ç, Element 6: K6 Element 7: K1

Element 8: Ka -

Area: A = 7.54E-03 m2 [A= nD214] DIE= 65313 Relative roughness LID= O

Flow Velocity = 0.0038 mis [v = QIAJ Velocity head = 0.0000 rn Reynolds No. = 1.53E+03 [NR = vDlv] Friction factor: f= 0.0563 Usin E . 8-7

Area: A = 8.19E-05 m2 DIE= 6807 LID= 2938

Flow Veloc²ty = 0.346 mis Velocity head = 0.006104 m Reynolds No. = 1.47E+04 Friction factor: f = 0.0282

CLASS 1 SERIES SYSTEMS I Pressure SI:

Chapter 11 146

Spreadsheet solutions to Problems 11.8, 11.9, and 11.10 are on next page.

Class n .... & 1..r-_-:_-:_-_-30_m_====-::i.~[ ..QJ

Method ne Iteration h = P1-P2= 68kN/m2 =7.70m=f~~Āu=J2ghLD L r ((0.90)(9.8lkN/ ) D2g' JL u= 2(9.81)(7.70)(0.04658) = J0.235: D = 0.04658 = 31053

j (30) j e 1.5X10-6

Tryf= 0.03; u= .Jo.23510.03 = 2.80 mis Náá = uDp = u(0.04658)(900) = l .40 x l04(u) = 3.92 x 104

rá 3.0xl0-3

New f = 0.022; l) = 3.27 mis; Náá = 4.58 x 104; f= 0.0210 u= 3.345 mis; Náá = 4.69 x 10\f= 0.0210 No change

Pi V12 P2 u2 z1 = Zz -+z +--h =-+z +-2 : y á 2g L y 2 2g Uá = U2

1 2

Class n [Repeated using computational approach - Sec. 11.5] hr, = 30.0 ft; Die= 2120;~se Eq. 11-3. l

2~gDhL lo 1 + 1.784v Q = -2.22D L g 3.7D!& D.jgDhá/L

(32.2)(0.3188)(30) ) [ 1 (l. 784 )(7.3 7 X 10-6 )] Q=-2.22(0.3188)2 25 og 3.7(2120) + (0.3188).Jl2.32

L u2 Class Il há = 30.0 ft =f~- Method ne Iteration , D2g

Then u= J2gh1,D = ~2-(3-2-.2-)(-30-)(-0-.3-18_8_) =J24.64 jL !(25) f

Náá= uD = u(0.3188) =4.33x104(u): D = 0.3188 =2120 V 7.37X10-6 e l.5X10-4

Tryf= 0.02; then u= .}24.64/ f = 35.1 ft/s Nu=4.33x104(35.1)=1.52 x 106; Newf=0.0165 u= .J24.64/0.0165 = 38.6 ft/s; Náá = 1.67 x 106;/= 0.0165 OK Q =Av= (0.07986 ft2)(38.6 ft/s) = 3.08 ft31s

11.9

11.8

Class II systems


Using Eq. 11-3 1 .4500ft3Js 7.04ft/s

Resu/ts: Maximum va/ues Volume flow rate: Q =

Velocit : v =

Method 11-A: No minor losses CLASS 11 SERIES SYSTEMS 11-A & 11-B US:

Results: Maximum values Volume flow rate: Q = 0.0056m3/s Using Eq. 11-3

Velocit : v = 3.31m/s Area: A = 0.001704 m2

DIE = 31053.33

Method 11-A: No minor losses Uses Equation 11-3 to find maximum allowable volume flow rate

CLASS 11 SERIES SYSTEMS 11-A & 11-B US: PPLIED FLUID MECHANICS

Using Eq. 11-3 3.0546ft3/s 38.27ft/s

esults: Maximum va/ues Volume flow rate: Q =

Velocit : v =

Method 11-A: No minor losses CLASS 11 SERIES SYSTEMS 11-A & 11-B US: PPLIED FLUID MECHANICS

Chapter 11 148

Vel. head at point 1 = Vel. head at oint 2 =

Qty. Energy loss hu = 2.45 m Friction Energy loss hL2 = 1.48 m fr = 0.0107 Energy loss hL3 = 0.00 m Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hLa = 0.00 m Energy loss hL7 = O.OOm

loss hLB = 0.00 m

loss huot = 3.93m

K

dditional Pipe Data: LID= 272

Flow Velocity = 3.01 mis Velocity head = 0.462 m Reynolds No. =7.21E+04 Friction factor: f = 0.0194

esu/ts: Maximum values Volume flow rate: Q = 0.0023 m3/s Using Eq. 11-3

Velocit : v = 3.91 m/s Area: A= 0.000595m2

DIE= 18353.33


CLASS 11 SERIES SYSTEMS 11-A & 11-B SI: PPLIED FLUID MECHANICS


Energy loss hL 1 = 16.90 ft Friction Energy loss hL2 = 2.97 ft fr = 0.0215 Energy loss hL3 = 2.45 ft Energy loss hL4 = 0.00 ft Energy loss hL5 = 0.00 ft Energy loss hLs = 0.00 ft

Energy loss hL7 = 0.00 ft Ener loss hLB = 0.00 ft

Total energ loss huot = 19.86 ft

K Qty. 3.45 ft


dd²tiona/ Pipe Data: LID= 217

Flow Velocity = 14.91 ft/s Velocity head = 3.452 ft Reynolds No. = 2.43E+05

0.0226

Results: Maximum values Volume flow rate: Q = 0.9782 ft3/s Using Eq. 11-3

16.23 ft/s Velocit : v =


Chapter 11 150

*Nozzle K3 restated in terms of ve/ocity head in pipe rather than out/et ve/ocity head. *K3 = 0.15(v:/v1)2 = 0.15(4)2 = 2.40 Nozzle Velocity: vN = vp(A,IAN) = vp(D,IDNJ2 = 4.0vp = 4.0(8.11ft!s)=32.44 ft!s

Energy loss hu = 10.36 ft Friction

Energy loss hL2 = 0.35 ft fT = 0.01 Assumed Energy loss hL3 = 2.45 ft

Energy loss hL4 = 0.00 ft

Energy loss hLs = 0.00 ft Energy loss hL6 = 0.00 ft


Ener loss hLB = 0.00 ft

loss huot = 13.15 ft

K Qty. 16.32 ft


Flow Velocity = Velocity head = Reynolds No. = Friction factor: f =

8.11 ft/s 1.020 ft

4.59E+04 0.0212


esults: Maximum values Volume flow rate: Q = 0.0194 ft3/s Using Eq. 11-3

Velocit : v = 14.23 fUs


CLASS 11 SERIES SYSTEMS 11-A & 11-B US:


*Nozzle K3 restated in terms of ve/ocity head in pipe rather than out/et velocity head. *K3 = 0.15(vfv1)2 = 0.15(4)2 = 2.40 Nozzle Ve/ocity: vN = vp(A/AN) = vp(D/DNJ2 = 4.0vp = 4.0(20.36 ft/s) = 32.44 ftls

loss hu01 = 71.21 ft

53.57 ft Friction 2.19ft fy=0.01 Assumed 15.45 ft 0.00 ft 0.00 ft 0.00 ft 0.00 ft

0.00 ft

Energy loss hL 1 = Energy loss hL2 = Energy loss hL3 = Energy loss hL4 = Energy loss hL5 = Energy loss hL6 = Energy loss hL7 = Ener loss hLB =

Energy losses in Pipe: Pipe: K1 = f(UD) =

K Qty. 8.32 1



20.36 ft/s 6.439 ft

1.15E+05 0.0173



Velocit : v = 38.20 ft/s

Method ll~A: No minor losses CLASS 11 SERIES SYSTEMS ll~A & ll~B US:

Chapter 11 152

12.62 m loss huot =

10.33 m Friction 1.47 m fT = 0.0095 0.48 m 0.33 m 0.00 m 0.00 m 0.00 m

o.oo m

Energy loss hL1 = Energy loss hL2 = Energy loss hL3 = Energy loss hL4 = Energy loss hLs = Energy loss hL6 = Energy loss hL7 = Ener loss hLs =

Energy losses in Pipe: Pi e: K1 = f(UD) =

K Qty. 0.967 ft Vel. head at oint 2 =

4.36 mis 0.967 m

109E+05 0.0177

Flow Velocity = Velocity head = Reynolds No. =

Friction factor: f =


5.05 mis Velocit : v = Using Eq. 11-3

esu/ts: Maximum va/ues Volume flow rate: Q = 0.0098 m31s Area: A= 0.001948 m2

DIE= 33200

Method llĿA: No minor losses Uses Equation 11-3 to find maximum allowable volume flow rate

CLASS 11 SERIES SYSTEMS JI-A & 11-B SI:


2 2 h1, = (2. 77 + 652J3)!!.1_ + (2.43 + 192 f 6)~

2g 2g

_ u; u; 5 5 v; v; } há - 1.0- + 2(30) ĉJr - + J; + 0.45- Based on 3-in pipe;J3r = 0.022 , 2g 2g ' 0.08432 2g 2g

D 0.156 Entrance Elbows Friction Enlarge. -2 = = l.85 D1 0.0843

3 O v2 u2 u2 u2 } + j6 ----6 + 30j67 -6 + 45ii;r -6 + 1.0-6 Based on 6-in pipe;.f6r= 0.019 0.156 2g 2g 2g 2g Friction Elbow Valve Exit

11.15 Class 11 with 2 pipes: Pts. A and B at tank surfaces. Method IIC p u2 p v2 P A = Pa = O rA +ZA +

2gA -hL = ya +Za +

2ga : ZA - Za= hi = ] 0 m:

VA =Va =0

Tryf= 0.02 u= 266.6 = 4.15 m/s; NJ? = uD p = ( 4.15)(0.0498)(823) = l.04 x 105

3.46 + 602(0.02) Õ 1.64 X 10-3 Newj= 0.018; v= 4.32 m/s; NR = 1.08 x 105; Newf= 0.018 No change Q =A u= (1.945 x 10Ŀ3 m2)( 4.32 mis)= 8.40 x 10Ŀ3 m' /s

In Eq. I: 2 2 2

13.59 m = !:2_ + (2.46 + 602j)!:2_ = (3.46 + 602j)1 2g 2g 2g

2g(13.59 m) 2(9.81)(13.59) 266.6 u= =

3.46 + 602f 3.46 + 602f 3.46 + 602f

Die= 0.0498 m = 33200 1.5 X 10-6 m

fr= 0.010 (Approx.)

11.14 Class 11 Pt. l at surface oftank A; Pt. 2 in stream outside pipe. u1 = O,p2 =O P u2 P u2 p u2 __!_+z +-1 =h =-2 +z +-2 : _!_+(z -z )=-2 +h r 1 2g L r z 2g r ' z 2g L

150kN/m2 -5m= 13.59m=u; +hz CD MethodIIC 8.07 kN/m3 2g ,

2 2 2 30 2 2 he= 0.50!:2_+160J;.!:2_+2(18)fr!:2_+ f Ŀ mm Vz =(2.46+602j)!:2_

2g 2g 2g 0.0498 m 2g 2g Entrance Valve Bends Friction

Lr/D= 300mm =6.02~ Le =18(Fig.10.23) 49.8mm D

Chapter 11 154

U3 = 3.30u6 = 3.51 mis NR = 1.29 x 105(3.50) = 4.20 x 105-> f3 = 0.0195 No change

3

NR = 2.38 x 105(1.02) = 2.43 x 105-> fe= 0.020 No change 6

Å-----1-9-6Ŀ-2---~ = 1.02 mis 34.9+7646(0.0195) + 192(0.02)

NR = 2.38x105 (1.012) = 2.41x105 -> New fe= 0.020 6

U3 = 3.42u6 = 3.46 mis NR = 1.29x105(3.46) = 4.46x105 -> New 13 = 0.0195

3

Try 13 =fe= 0.02

u = 196.2 = 1.012 mis 6 34.9 + 7646(0.02) + 192(0.02)

Iterate for both f andfe: D3 = 0.0843 m = 703: N = u3D3 = u3(0.0843)=1.29x105 (u) e 1.2X10-4 ID R, V 6.56X10-7

3

D6 = 0.1560 =1300: NR = u6D6 = u6(0.156) =2.38xl05(u6) e 1.2 X 6 V 6.56X10-7

34.6+ 7646.f; + 192J;;

(34.9+7646.f;+192J;;)

196.2

34.9+7646.f;+192J;; 2(9.81)(10)

Solve for v6

11 73u2 u2 u2 hL=(2.77+65213) . 6 +(2.43+192j6)-6 =(34.9+7646J;+192j~)-6 2g 2g 2g


New 12 = 0.0255,./4 = 0.0305 288.1

Vi3 = -5.60+9030(0.0255)+1020(0.0305) = 1.06 mis; VA= 3'87 Vi3 = 4Ŀ10 mis

N11A = 2.01x104 =rf: = 0.0255 No change; N11fl = 1.02x104 ~ J4 = 0.0305 No change Q = AaVB = (7.538 x 10-3m2)(1.06 mis)= 7.99 x 10-3 m31s

Try 12 = J4 = 0.02 288.1

Uá3 = = 1.21 misĀ vA = 3.87 Vi3 = 4.69 mis -5.60+9030(0.02)+1020(0.02) '

NR = 4.88 X 103(4.69) = 2.29 X 104; N; = 9.59 X 103(1.21) = 1.16 X 104 A H

both 12 and 14Å -5.60+903012+102014 -5.60+903012+102014 = VJ3=

Iterate for

15 2 2 2 ~+ VB [8.40+903012 +102014]= VB (-5.60+9030/z +1020.fi] 2g 2g 2g

2g(l4.68) 288.1

2

14.68 = ~ 2g

In Eq.(D

v2 v2 h1, = (0.52 + 60212) ~ + (0.6+102014)-Û-

2g 2g

But VA= Vi3 AB = v8 [DB )2 = v8(1.97)2 = 3.87 VJ3: v~ = 15.0v~ AA DA

Then: hL=(0.52+60212) l5.0v~ +(0.6+102014) v~ = v~ [8.40+903012+ 1020}4] 2g 2g 2g

L D ID = 0.098 = 1.97 2 1 0.0498

_ f 30 v~ v~ + 100 v~ ) I' v~ . +. _ bi=: 2 -+0.52-+ J4---+2(30 J,.4-. JJ;á-0.0lOapprox. 0.0498 2g 2g 0.098 2g 2g Friction(2-in) Enlarge. Friction ( 4-in) Elbows

z = 175 kNlmz -4.5m = 14.68m = v~ - v~ +h (!) 8 0.93(9.81 kNlm3) 2g 2g L

Method IIC 11.16 Class 11 with two pipes Pt. B in stream outside pipe. Pè = O v2 v2 p v2 -v2

PA + z + ~ - h = PB + z +-Û-: -1_ + z - z = B A + h A2 L B2 A B 2 L

Yo g Yo g Yo g

Chapter 11 156

L M²nimum Specify 4-in type K copper tube; D = 97.97 mm

New 1= 0.0105 D

D = 0.0901 m; NR = 2.35 x 106; - = 60100 -+ 1= 0.0105 No change e

Tryj'> 0.02 D = [5.67 x 10-4(0.02)]0Ŀ2 = 0.103 m;

NR= 2.12x105 =2.07x106; D = 0.103 =68400 0.103 e l.5xl0-6

NR = 4Q = 4(0.06) = 2.12X105 1CVD 1C(3.60X10-1)D D

150 kN/m2 = 15.74m 9.53 kN/m3 r

11.18 Class 111 Same method as Prob. 11.17.

D = [ 8LQ2 1]0.2 = [ 8(30)(0.06)2 1]0.2=[5.67X10-4 !]Ü2 1C2ghl 1²2(9.81)(15.74)

New 1= 0.0175 D = 0.342 ft; NR = 4.25 x 105; Die= 2279-+ I = 0.0175 No change

L M²nimum Specify: 5-in Sch. 80 pipe, D = 0.4011 ft

Tryl= 0.02 D = [0.267(0.02)]Á"2 = 0.351 ft;

NR= 1.453xl05 =4.14x105; D = 0.351 =2330 0.351 e 1.5X10-4

NR = uD = 4QD = 4(0.5) = 1.453x105 v 1CD2v 1C(4.38x106)D D

p1-p2 =h =10.0lbĿft3144in2 =23_61ft=I~~= jLQ2I6 r L in ' (61.0 lb) ft2 D 2g D2g1²2D4

D = [ 8LQ2 1]0.2 = [ 8(1000)(0.5)2 1]0.2=(0.2671)2 1C2ghl 1²2(32.2)(23.61)

b !' 1 1000

ft 2'1 6 z,-~.u,-~

11.1 7 Class 111

P1 + z + u12 - h = P2 + z + u~ r 1 2g L r 2 2g

Class III systems

157

Specify 5-in Sch. 80 steel pipe; D = 0.4100 ft

Final Minimum Diameter: M²nimum diameter: D = 0.3479 ft

1.11 E-07 Ar ument in bracket:

lntermediate Results in Eq. 11-8: L/ghl = 1.315562

Method 111-A: Uses Equation 11-8 to compute the inimum size of pipe of a given /ength that will flow a given volume flow rate of fluid with a /imited pressure drop. (No minar losses)

111-A & 111-B US: CLASS 111 SERIES SYSTEMS

Final Minimum Oiameter: Minimum diameter: D = 1.9556 ft

SERIES PIPE UNE SYSTEMS

6.22E-11 Ar ument in bracket:


ethod 111-A: Uses Equation 11-8 to compute the inimum slze of pipe of a given /ength that w²ll flow a given volume flow rate of fluid


Specify 4-in type K copper tube: D = 97.97 mm

Final Minimum Diameter: Minimum diameter: D = 0.0908 m

lntermediate Results in Eq. 11-13: L/ghL = 0.194292

Ar ument in bracket: 2.89E-22

ethod 111-A: Uses Equation 11-13 to compute the inimum size of pipe of a given length hat w²/J flow a given vo/ume flow rate of fluid with a l²m²ted pressure drop. (No m²nor /osses)

111-A & 111-B SI: CLASS 111 SERIES SYSTEMS

PPLIED FLUID MECHANICS

PPLIED FLUID MECHANICS

Chapter 11 158



Energy loss hL3 = 3.47 ft fr = 0.016 Energy loss hL4 = 0.31 ft fr=0.016 Energy loss hL5 = 0.00 ft

Energy loss hL6 = 0.00 ft Energy loss hL7 = 0.00 ft

En erg loss hLB = 0.00 ft

Total energ loss huot = 6.09 ft

Energy losses in Pipe: Pipe Frict²on: K1 = f(LID) =

K Qty.

3.12 1

are actual with desired ressure at

o psig O psig

2.28 psig

Results: G²ven pressure at po²nt 1 = Desired pressure at point 2 = ctual pressure at point 2 =

0.000 ft 0.638 ft


Flow area: A = Relative roughness: Die =

LID= Flow Velocity = Velocity head = Reynolds No. = Friction factor: f=

0.13894 ft2 2804 178 6.41 ft/s 0.638 ft

2.95E+05 0.0175

Final Minimum Diameter: Minimum diameter: D = 0.3007ft

2.91 E-09 Ar ument in bracket:


ethod 111-A: Uses Equation 11-8 to compute the ²n²mum size of pipe of a given /ength hat w²ll flow a given volume flow rafe of fluid



11.22 Class 1 Pt. 1 at collector tank surface. Pt. 2 at um inlet. 1 = O, v1 = O

El+ z + Vá2 - h = P2 + z + u; : p = r[(z - z ) - h - u; ] <D r 1 2g L r 2 2g 2 1 2 L 2g

. 1 ft3 /s Q = 30 gal/mm x 1 = 0.0668 ft3 /s h = 0.019 for 2-in pipe 449 gal/min

= g = 0.0668 ft3 Is= 2_86 ft/s; u; = (2.86)2 = 0.127 ft lJ:à Ćz 0.02333 ft2 2 g 2(32.2)

2 2 lOOft 2 2 2 hL = o.51+t.851+ á Ā v2 + fi.(8)1=1(2.50+58.0f)

2g 2g 0.1723 ft 2g 2g 2g Entrance Filter Friction Valve

Náá= vDp =(2.86)(0.1723)(0.92)(1.94)=2.45xl04: D = 0.1723 = 1149 Õ 3.6X10-5 8 1.5X10-4

f= 0.0265 hL = (0.127 ft)(2.50 + 58.0(0.0265)] = 0.513 ft In Eq.(ĉ):

)(62.4lb) 513 o 2 ] ft ¶" . p2 = (0.92 ft3 [3.0- O. - . l 7 144 in2 = 0.94 ps1g

But h = z1 - z2 - 0.5 m = 3.85 m 0.5 m = 3.35 m

v2 z1-z2=2~ +hL =0.782+3.07= 3.85 m

In Eq. (D

2 2 2 2 - V2 V2 -r ) Vz Vz J, o ni= 0.5-+ fi.(160)-+ Jr(30 -=-[0.5+190 rJ= .782[0.5+190(0.018)]

2g 2g 2g 2g = 3.07 m

11.21 Class 1 Pt. 1 at tank surface, Pt. 2 in stream outside i e: pi = Pi =O; v1 =O Pi V12 h - P2 u; . u; + h É -+z1 +-- L --+z2 +-. Zá -z2 - L r 2g r 2g 2g

1J:2=.fl= 1500L/min lm3/s =3.92m/s: u; =(3.92)2 =0.782m Aá 6.38x10-3m2 60000 L/min 2g 2(9.81)

h=0.018

Practice problems f or any class

Chapter 11 160

Final M²nimum Diameter: Minimum diameter: D = 0.1059ft

1.37E-09 Ar ument in bracket:

lntermediate Results in Eq. 11-8:

L/ghl = 0.134576

Method 111-A: Uses Equation 11-8 to compute the inimum size of pipe of a given length that will f/ow a g²ven vo/ume f/ow rate of fluid

111-A & 111-B US: CLASS 111 SERIES SYSTEMS PPLIED FLUID MECHANICS

11.23 Class 1 Pt. 1 at collector tank surface; Pt. 3 at upper tank surface. Pi = ps =O, Vá = 1J.3 =O

Pi +z1 + v12 =b: +hA = p3 +z3 +u:: hA =(z3 -zi)+hL =19.0ft+hL r 2g r 2g

hL = h +h =0.513ft+ f (100) v,7 + f lSft u~ +l.O v,7 " l,." LJ;"h dT 2g d OJ 15 ft 2g 2g

~m Prob. 11.22] Ch. valve Friction Exit

vd = g_ = 0.0668 ft3 /s = 6.43 ft/s: v?t = ( 6.43)2 = 0.642 ft Ad 0.01039 ft2 2g 2(32.2)

NR = vdDp = (6.43)(0.115)(0.92)(1.94) =3_6?xl04: " Õ 3.6x 10-5

D 0.115 = 4 = 767 ~ f= 0.0265; fdr= 0.022 e l.5X10- he = 0.513 ft + (0.022)(100)(0.642) + (0.0265)(157)(0.642) + 1.0(0.642) = 5.24 ft hA = 19.0 ft + he = 19.0 + 5.24 = 24.24 ft

Power = PA = hAyQ = (24.24 ft)(0.92)(62.4/b)(ÜĿ0668 ft3) 1 hp 2 = 0.169 hp ft3 s 550 ft-lb/s


Energy loss hL 1 = 5.68 ft Energy loss hL2 = 0.64 ft Energy loss hL3 = 0.85 ft fT = 0.022 Energy loss hL4 = 0.28 ft fT = 0.022 Energy loss hL5 = 0.00 ft Energy loss hL6 = 0.00 ft Energy loss hu = 0.00 ft Ener loss hLB = 0.00 ft

loss huot = 7.45 ft

Energy losses in Pipe: K Qty. Pi e Friction: K1 = f(UD) =

O psig O psig

0.36 psig ressure at

Flow Velocity = 6.43 ft/s Velocity head = 0.643 ft Reynolds No.= 3.67E+04 Friction factor: f = 0.0261

Results: Given pressure at point 1 = Desired pressure at point 2 = ctual pressure at point 2 = Com are actual with desired

Flow area: A = Relativa roughness: Die=

LID=

0.000 ft 0.643 ft


0.01039 ft2 767 339

162 Chapter 11




CLASS 11 SERIES SYSTEMS 11-A & 11-B US: PPLIED FLUID MECHANICS

Q = 0.0333 ft3/s v = Q = 0.0333 ft3 /s = 6.68 ft/s

A 0.00499 ft2

2 (32.2)(0.07975)(23.08) lo [ 1 + (l.7841)(1.2lxl0-5)]

= -2.22(0.07975) 100 g 3.7(532) (0.07975)-J0.5927

11.25 Class 11 Computational approach, Eq. 11.12. hà = P1 Pz = 10.0 lb ft3 144inz = 23.08 ft

y in 2 62.4 lb ft2 1-in Schedule 80 steel pipe; D = 0.07975 ft; A= 0.00499 ft2 Die= 0.07975/1.5 x 10-4 = 532 Water at 60ÜF; v = 1.21 x 10-5 ft2/s Q=-2.22D2


Energy loss hL 1 = 19.75 ft Energy loss hL2 = 0.00 ft Energy loss hL3 = 0.00 ft Energy loss hL4 = 0.00 ft Energy loss hLS = 0.00 ft Energy loss hL6 = 0.00 ft Energy loss hL7 = 0.00 ft Ener loss hLB = 0.00 ft

Total energ loss huot = 19.75ft

Energy /osses in Pipe: K Qty. Pipe Friction: K1 = f(UD =

Results: Given pressure at point 1 = 108 psig Desired pressure at point 2 = 100 ps²g ctua/ pressure at point 2 = 102.18 psig Com are actual with desired ressure at

1.417ft 1.417ft


Flow area: A= 0.02332 tt2 Relative roughness: Die= 1149

LID= 696 Flow Velocity = 9.55 ft/s Velocity head = 1 .417 ft Reynolds No.= 3.62E+05 Friction factor: f = 0.0200

Final Minimum Diameter: Minimum diameter: D = 0.1655 ft

lntermediate Results in Eq. 11 Ŀ8: L/ghl = 0.137164


ethod lllĿA: Uses Equation 11-8 to compute the inimum size of pipe of a given length hat will f/ow a given volume f/ow rate of fluid with a limited pressure drop. No minar losses)

11/ĀA & 111Ā8 US: CLASS 111 SERIES SYSTEMS

Chapter 11

2

Va =45.7ft 2g

= L = 64.0 lb/ft3 = 1.988 slugs p g 32.2 ft/s2 ft3

u _ g_ _ 0.50 ft3 /s = 9. 743 ft/s A - A - 0.05132 ft2

A

u~ = (9.743)2 ft = 1.474 ft 2g 2(32.2)

u = 2__ = 0Ŀ50 = 54.2 ft/s B AB Jr(l .3/12)2

4

164

In 2 1/2-in discharge line:

Q 0.50 Ud=-= =]5.03ft/s Ad 0.03326

u; = (15.03)2 =3.509 ft 2g 2(32.2)

11.29 Class 1 Pt. B in stream outside nozzle. pa = O

P u2 P u2 ~+z +~-h +h =-Û-+z +---Û-- y A 2g I A y B 2g

o: -u2 p h = (z Z ) + B A - ~ + h A B A 2g y L

11.28 Design problem with variable solutions: Pressure at pump inlet can be increased by: lowering the pump, raising the reservoir, reducing the :tlow velocity in the pipe by using a larger pipe, reducing the h: in the valve by using a less restrictive valve (gate, butter:tly), eliminating elbows or using long-radius elbows, using a well-rounded entrance, and shortening the suction line.

L = 11.5 m + 1.40 m = 12.90 m z1 -z2 = 0.75 m - 1.40 m = 0.65 m

In Eq.Q) 9.53 kN P2 = 3 [-0.65 m - 0.335 m - 4.09 m] = -48.4 kPa m

fr= 0.018

11.27 Class 1 Q = 475 L/min (1 m3/s/60000 L/min) = 7.917 x 10-3m3/s Pt. 1 at reservoir surface; Pt. 2 at pump inlet. p1 = O, u1 = O P u2 p u2 [ u2 l __!_ + z + _1 - h = __!_ + z + _2 : P2 =y (z, - zz)- 2g2 h1, (V y 1 2g L y 2 2g

= Q = 7.917x10-JmJ /s = 2.56m/s: u; = (2.56)2 = 0.335 m ~ A 3.090x 10-3 m2 2g 2(9.81)

be= u; +(0.0195) 12Ŀ90m u; +(0.018)(2)(30)u; +(0.018)(340)u; =4.09m 2g 0.0627 m 2g 2g 2g Entrance Friction 2 Elbows Valve

N11 = uD - (2.56)(0.0627) = 4.46x1 os: D = 0.0627 1363 -+ f = 0.0195 V 3.60X10-7 e 4.6X10-5


See spreadsheet on page 167 for solution to Problem 11.31 using the spreadsheet I Pressure SI.

Then hL = [(0.0235)(672) + (0.021)(370)](0.328 m) = 7.73 m 9.81kN PA = [25 m + 7.73 m] = 321.1 kPa

11.31 Class 1 Pt. B outside pipe. PB = O, Vu = VA. 2 2

P A U A h _ Pa Va . _ [( ) h ] -+ZA+-- L --+zB +-.pA -r ZB -ZA + L r 2g r 2g

hi. = fr(340)!!_ + f 27Ŀ5 m u2 + fr(30)!!_ = [f(672) + fr(370) ]!!___ 2g 0.0409 m 2g 2g 2g

Val ve Friction Elbow u= Q = 200 L/min 1 m3 /s = 2.54 mis: u2 = (2.54)2 = 0.328 m

A 1.31x10-3m260000L/min 2g 2(9.81)

NR = uD = (2.54)(0.0409) = 7.98x104: D = 0.0409 = 889: f = 0.0235 V 1.30X10-6 s 4.6 X 10-5

fr= 0.021

11.30 Class 1 See Problem 11.29. 3-in Sch. 40 discharge line. fr = 0.018 u3

Ud= VA= 9.743 ft/s: zs: = 1.474 ft 2g

N = udDp = (9.743)(0.2557)(1.988) =1.24xl05: R,á Õ 4.0 X 10-5

D = 0Ŀ2557 = 1705; f = 0.0205 e 1.5X10-4

hL = (0.0205) 82 (1.474) + 0.018(30)(1.474) + 32.6(1.474) = 58.5 ft 0.2557

hè = 80 + 45.7 - 1.474 + 7.875 + 58.5 = 190.6 ft PA = hAyQ = (256.8)(64.0)(0.50)/550 = 11.1 hp Input power = P1 = PA!eM = (11.1 hp)/0.76 = 14.6 hp = P1

Then hA = 80 ft + 45.7 ft- 1.474 ft - (-3Ŀ5Ü lb)ft3 l44in2 + 144.8 ft = 276.9 ft in ' 64.0 lb ft2

Power = PA = hAyQ = (276.9 ft)(64.0 lb/ft3)(0.50 ft3/s)l hp/550 ft-Ib/s = 16.1 hp Input power = P1 = PA/eM = (16.1 hp)/0.76 = 21.2 hp = P1

Nozzle Elbow Friction

he = (0.0205) 82 ft (3.509) + 0.018(30)(3.509) + 32.6(3.51ft)=144.8 ft 0.2058 ft

N = udDp = (15.03)(0.2058)(1.988) = 1.54x105 RJ Õ 4.0X10-5 Die= 0.2058/1.5 X 10Ŀ4 = 1372 ......,. f = 0.0205; fr = 0.018

Chapter 11 166

In Problem 11.33, the globe valve is replaced with a fully open gate valve having much less energy loss. The result is that Q = 0.003394 m3/s (204 L/min) can be delivered with the same pressure at point A. This is approximately a 20% increase in flow. Also, it is only about 3% less than the flow that could be delivered through the pipe with no minor losses at all, as shown in the results of Method II-A at the top of the spreadsheet.

Problem 11.32 shows that Q = 0.00283 m' /s (170 L/min) can be delivered through the given system with a fully open globe valve installed and with a pressure of 300 kPa at point A at the water main.

Recall that Method 11-A is set up and solved first in the spreadsheet. This ignores the minor losses and gives an upper limit for the volume flow rate that can be delivered through the system with a given pressure drop or head loss. Then, Method II-B is used iteratively to hone in on the maximum volume flow rate that can be carried with the minor losses considered. The designer enters a series of estimates for the value of Q and observes the resulting pressure at the outlet from the piping system. Obviously this pressure for this problem should be exactly zero because the pipe discharges into the free atmosphere. Each trial requires only a few seconds to complete and the designer should continue making estimates until the pressure at the outlet is at or close to zero.

Problems 11.32 and 11.33 are sol ved using the spreadsheet on the following two pages. These problems are of the Class II type and are solved using the procedure described in Section 11.5. Method 11-B is used because the system contains significant minor losses. In fact, one objective of these two problems is to compare the performance of two design approaches for the same system, using a different, more efficient valve in Problem 11.33 as compared with Problem 11.32.

SERIES PIPE LINE SYSTEMS

Energy losses-Pipe 2: K Qty. Pipe: K1 = f(UD) = 0.00 Element 2: K2 Element 3: K3 Element 4: K4 Element 5: Ks Element 6: K6 Element 7: K7

Element 8: Ka

Energy loss hu = 5.13 m Friction Energy loss hL2 = 0.21 m (fr = 0.021) Energy loss hL3 = 2.34 m (fr = 0.021) Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hL6 = O.OOm Energy loss hL7 = 0.00 m En erg loss hLa = 0.00 m

Energy loss b; 1 = O.OOm Friction Energy loss hL2 = O.OOm Energy loss hL3 = 0.00 m Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hL6 = 0.00 m Energy loss hL7 = 0.00 m Energy loss hLB = 0.00 m

Total ener loss huot = 7.68 m Total chan e in pressure = -320.58 kPa

320.58 kPa

167

Energy losses-Pipe 1: K Qty. Pipe: K1 = f(UD) = 15.64

Elbow: K2 Globe Va/ve: K3 Element 4: K4 Element 5: Ks Element 6: K6 Element 7: K7 Element 8: Ka

Area: A= 7.54E-03 m2 [A= nD2!4] Die= 65313 Relative roughnes LID= O

Flow Velocity = 0.4422 m/s [v = Q/A] Velocity head = 0.0100 m Reynolds No. =3.33E+04 [NR = vD/v] Friction factor: f= 0.0228 Usin E . 8-7

Area: A = 1.31 E-03 m2 DIE= 889 LID= 672

Flow Velocity = 2.537 m/s Velocity head = 0.328084 m Reynolds No. =7.98E+04 Friction factor: f = 0.0233

CLASS 1 SERIES SYSTEMS I Pressure SI: PPLIED FLUID MECHANICS

Chapter 11 168

Vel. head at po²nt 1 = Vel. head at oint 2 = 0.236 m

Qty. Energy loss hL1 = 3.75 m Friction Energy loss hL2 = 0.15 m Energy loss hL3 = 1.68 m Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hL6 = 0.00 m Energy loss hL7 = 0.00 m Ener loss hLa = 0.00 m

Total energ loss huot = 5.58 m

K


Flow Velocity = 2.15 m/s Velocity head = 0.236 m Reynolds No. =6.76E+04

0.0237

Results: Maximum values Volume flow rate: Q = 0.0035 m3/s

2.66 m/s Velocit : v =

Uses Equation 11-3 to estimate the allowable volume flow rate

LASS 11 SERIES SYSTEMS

SERIES PIPE UNE SYSTEMS 169

Vel. head at point 1 = Vel. head at oint 2 = 0.340 m

Energy loss hu = 5.31 m Friction Energy loss hL2 = 0.21 m Energy loss hL3 = 0.06 m Energy loss hL4 = 0.00 m Energy loss hL5 = 0.00 m Energy loss hL6 = 0.00 m Energy loss hL7 = 0.00 m Ener loss hLB = 0.00 m

loss huot = 5.58 m

K Qty.

dditiona/ Pipe Data: LID= 672

Flow Velocity = 2.58 m/s Velocity head = 0.340 m Reynolds No. =8.13E+04 Friction factor: f= 0.0232

Resu/ts: Maximum values Volume flow rate: Q = 0.0035 m3/s

2.66 m/s Velocit : v =

Method ll~A: No minor losses Uses Equation 11-3 to estimate the allowable volume flow rate

CLASS 11 SERIES SYSTEMS PPLIED FLUID MECHANICS

Chapter 11 170

11.34 Class 1 Pt. 1 at surface of tank A; Pt. 2 outside pipe in tank B. u1 = O

Pi U12 h Pi u; - + [< ) u; h ] (i'\ -+Zá +-- 1. =-+Zz +-: P1 - P2 Y _Z2 -Zá +-+ J, \Y y 2g y 2g 2g

Vz = Q = 250 gal/min. 1 ft3 /s = 23.87 ft/s: v; = (23.87)2 = 8.844 ft A 0.02333 ft2 449 gal/min 2g 2(32.2)

N = uDp = (23.87)(0.1723)(1.53) =3.00xlOs: 11 Õ 2.lOxl0-5

D = O.l723 =1149~/=0.0205 e 1.5X10-4

_ u; J;( u; J;(30)u; á( llOft )u; Āji-O ¶i= 0.5-+ T 160)-+2 T -+ -. r= .019 2g 2g 2g 0.1723 ft 2g Entr. Valve 2 Elbows Friction 2

he = ~~ [4.68 + 6381] = 8.844 ft[4.68 + 638(0.0205)] = 157.1 ft

. 49.0llb lft2 p1 = 40.0 psig + 3 [20 ft + 8.844ft:+157.1 ft] . 2 = 103.3 psig ft: 144 111

171 SERIES PIPE LINE SYSTEMS

Energy loss hu = 159.05 ft Friction Energy loss hL2 = 14.18 ft fT = 0.019 Energy loss hL3 = 37.81 ft fT = 0.019 Energy loss hL4 = 6.22 ft Energy loss hLs = 0.00 ft Energy loss hL6 = 0.00 ft Energy loss hu = 0.00 ft Ener loss hLB = 0.00 ft


K Qty. 12.79 1

Energy losses in Pipe: Pipe: K1 = f(UD) =

12.44 ft Vel. head at point 1 = Vel. head at olnt 2 =

Results: Maximum va/ues Volume flow rate: Q = 0.7981 ft3/s Using Eq. 11-3


28.30 ft/s 12.438 ft

3.56E+05 0.0200




CLASS 11 SERIES SYSTEMS 11-A & 11-B US:

Chapter 11 172

K Qty. Energy losses in Pipe: Pi e: K1 = f(UD) = Energy loss hL1 = 188.44 ft Friction

Energy loss hL2 = 16.86 ft fT = 0.019 Energy loss hL3 = 2.25 ft fT = 0.019 Energy loss hL4 = 7.40 ft Energy loss hLs = 0.00 ft Energy loss hLs = 0.00 ft Energy loss hL7 = 0.00 ft Ener loss hLB = 0.00 ft


14.79 Ft Vel. head at point 1 = Vel. head at oint 2 =

30.86 ft/s 14.790 ft

3.88E+05 0.0200



Resu/ts: Maximum va/ues Volume flow rate: Q = 0.7981 ft3/s Us²ng Eq. 11-3

Veloc²t : v = 34.23 ft/s

Uses Equat²on 11-3 to find max²mum allowable volume flow rate

CLASS 11 SERIES SYSTEMS JI-A & 11-B US:


K Qty. Energy /osses in Pipe: Pipe: K = f(UD = Energy loss hu = 193.14 ft Friction

Energy loss hL2 = 11.53 ft fr = 0.019 Energy loss hL3 = 2.31 ft fr = 0.019 Energy loss hL4 = 7.58 ft



Energy loss hl7 = 0.00 ft

Ener loss hLB = 0.00 ft


15.17 Ft Vel. head at point 1 = Vel. head at oint 2 =

31.25 ft/s 15.167 ft

3.93E+05 0.0199

Flow Velocity = Velocity head = Reynolds No.=

Friction factor: f =


Results: Maximum va/ues Volume flow rate: Q = 0.7981 ft3/s Using Eq. 11-3

Velocit : v = 34.23 ft/s 1148.67


Chapter 11

21.00 m Total energy loss huot =

15.75 m Friction 5.25 m fT = 0.017 0.00 m 0.00 m 0.00 m 0.00 m 0.00 m 0.00 m

Energy loss hu = Energy loss hL2 = Energy loss hL3 = Energy loss hL4 = Energy loss hLs = Energy loss hL6 = Energy loss hL1 = Ener loss hLB =


174

K Qty.


Flow Velocity = 4.22 mis Velocity head = 0.908 m Reynolds No. =3.75E+05 Friction factor: f = 0.0178

5.02 m/s Using Eq. 11-3

Velocit : v =

Resu/ts: Maximum values Volume flow rate: Q = 0.0412 m3/s Area: A = 0.008219 m2

01&= 2224

Method 11-A: No minor losses Uses Equation 11-3 to estimate the allowable volume flow rate

CLASS 11 SERIES SYSTEMS 11-A & 11-B SI:

Energy loss hL2 = Energy loss hL3 = Energy loss hL4 = Energy loss hL5 = Energy loss hL6 = Energy loss hL7 = Ener loss hLa =

SERIES PIPELINE SYSTEMS 175

Total energy loss huot = 21.00 m

Energy loss hu= 15.75 m Friction 5.25 m fT = 0.017 O.OOm 0.00 m 0.00 m 0.00 m 0.00 m 0.00 m


Energy losses in Pipe: K Qty. Pipe: K1 = f(UD) = 17.35


Flow Velocity = 4.22 mis Velocity head = 0.908 m Reynolds No. =3.75E+05 Friction factor: f= 0.0178

Using Eq. 11-3 Results: Maximum values

Volume flow rate: Q = Velocit : v =

Area: A = 0.008219 m2 Die= 2223.913

0.0412 m31s 5.02 mis

CLASS 11 SERIES SYSTEMS Method 11-A: No minor losses Uses Equation 11-3 to estimate the allowable volume flow rate

11-A & 11-B SI: PPLIED FLUID MECHANICS

176 Chapter 11


K Qty. Energy loss hu= 14.69 m Friction Energy loss hL2 = 6.10 m fT = 0.016 Energy loss hL3 = 0.00 m Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hLB = 0.00 m Energy loss hL7 = 0.00 m Ener loss hLB = 0.00 m



Flow Velocity = 4.69 mis Velocity head = 1.122 m Reynolds No. =5.23E+05

0.0168

0.0746 m31s Using Eq. 11-3 5.78 mis

Results: Maximum values Volume flow rate: Q =

Velocit : v = Area: A =0.012908 m2 DI& =2786.957


CLASS 11 SERIES SYSTEMS JI-A & 11-8 SI:

SERIES PIPE UNE SYSTEMS 177


K Qty. Energy loss hu= 19.35 m Friction Energy loss hL2 = 1.07 m fr = 0.016 Energy loss hL3 = 0.00 m Energy loss hL4 = 0.00 m Energy loss hLs = 0.00 m Energy loss hLB = 0.00 m Energy loss hL7 = 0.00 m

Energ loss hLa = 0.00 m



Flow Velocity = 5.41 m/s Velocity head = 1.491 m Reynolds No. =6.03E+05 Friction factor: f = 0.0166

0.07 46 m3/s Using Eq. 11-3 5.78 m/s

Results: Maximum values Volume flow rate: Q =

Velocit : v = Area: A =0.012908 m2

01& =2786.957



Chapter 11 178

550 ft-lb/s Then PA = eP1= (0.75)(0.20 hp) = 82.5 ft-lb/s hp

hA = PA = 82.5 ft-lb/s = 65_27 ft rQ (62á;1b)(ÜĿÜ2~3ft3)

u= Q = 0Ŀ0203 ft3 /s = 6.60 ft/s: !t._= (6Ŀ60)2 = 0.677 ft A 7r(0.0625 ft)2 / 4 2g 2(32.2)

hà = f !:_.!!._ = (0.0225) IOO (0.677 ft) = 24.36 ft D 2g 0.0625

NR = vD = (6Ŀ60)(0.0625) = 3.41 x 104 ~ f= 0.0225 (Smooth curve) v 1.21xlo-s

Z2 - Zá = 65.27 ft - 24.36 ft - 0.677 ft = 40.2 ft

o-fl:J Ŀ o.0825 tt

z ĿZy

11.42 Class 1 Note: Pump delivers 1.0 in3/rev Q = 1.0 in3 x 2100 rev x ft3 x mm

rev min 1728 in ' 60 s = 0.0203 ft3/s P vz p v2 _!_+z +-1 -h +h =_!_+z +-2 r 1 2g L A r z 2g


Problem 11.44 answer is the result of the bottom portian for Method 111-B solution. Pipe size is set to 0.3355 ft for the 4-in Schedule 40 steel pipe. With minar losses considered, the pressure at point 2 is 60.80 psig, greater than the desired 60.0 psig. Therefore, it is satisfactory.

NOTES: Problem 11.43 answer is the result of the top portian for Method 111-A so/ution. Dmin = 0.3259 ft. Standard 4-in Schedule 40 steel pipe is closest standard size; D = 0.3355 ft.

0.07 ft fT = 0.017 0.38 ft fT = 0.017 0.76ft fT=0.017

0.27 ft fr = 0.017 0.84ft fT=0.017

0.00 ft 0.00 ft

Energy loss hL2 = Energy loss hL3 = Energy loss hL4 = Energy loss hLs = Energy loss hL6 = Energy loss hL1 =

Energy loss hu = 16.98 ft K Qty. 34.19

Energy losses in Pipe: Pipe Friction: K1 = f(UD) =

Results: Given pressure at point 1 = 80 psig Desired pressure at point 2 = 60 psig ctual pressure at point 2 = 60.80 psig Com are actual with desired ressure at

Vel. head at point 1 = 0.497 ft Vel. head at oint 2 = 0.497 ft

Flow area: A = 0.08840 tt2 Relative roughness: DIE = 2237

LID= 1788 Flow Velocity = 5.66 ft/s Velocity head = 0.497 ft Reynolds No. = 1.57E+05 Friction factor: f = 0.0191


lntermediate Results in Eq. 11-8: L/ghl =0.880858

ethod 111-A: Uses Equation 11-8 to compute the inimum size of pipe of a given /ength hat will flow a given volume flow rafe of fluid


Chapter 11

7.31 ft

180

Total ener loss huot =

4.90 ft 0.92 ft (fT = 0.01 est) 1.48 ft (fT = 0.01 est) O.DO ft 0.27 ft 0.84 ft O.DO ft O.DO ft

Energy loss hu = Energy loss hL2 = Energy loss hu = Energy loss hL4 = Energy loss hL5 = Energy loss hL6 = Energy loss hL7 = Ener loss hLB =

Energy losses in Pipe: K Qty.

Flow area: A= 0.01414 ft2 Relative roughness: Die = 134200

LID= 412 Flow Velocity = 6.30 ft/s Velocity head = 0.616 ft Reynolds No. =6.99E+04 Friction factor: f = 0.0193

Results: Given pressure at point 1 = 15 psig Desired pressure at point 2 = O psig ctua/ pressure at point 2 = 2. 13 psig Com are actual with desired ressure at oint 2

Vel. head at point 1 = 0.616 ft Vel. head at oint 2 = 0.616 ft


lntermediate Results in Eq. 11-8: L/ghL =O .140592

ethod 111-A: Uses Equat²on 11-8 to compute the inimum s²ze of pipe of a given fength that will flow a given vofume f/ow rafe of fluid

111-A & 111-B US: CLASS 111 SERIES SYSTEMS PPLIED FLUID MECHANICS


Area: A= 0.01414 tt2 Area: A= 0.78540ft2 [A= n02/4] Dle = 134200 Die= 6667 Relative roughness LID= 412 LID= o

Flow Velocity = 6.30 fUs Flow Velocity = 0.11ft/s [v = Q/A] Velocity head = 0.616 ft Velocity head = O.OOOft [v2/2g] Reynolds No. =6.99E+04 Reynolds No. = 9.37E+03 [NR = vD!v]

Friction factor: f = 0.0193 Friction factor: f = 0.0318 U sin E . 8-7 Energy /osses-Pipe 1: K Qty.

Pipe: K1 = 7.96 1 Energy loss hu = 4.90ft

Ball check value: K2 - Energy loss hL2 = 0.92 ft [fT = 0.01 O assumed]

Std. e/bows: K3 = Energy loss hL3 = 1.48 ft [fT = 0.010 assumed] Element 4: K4 - Energy loss hL4 = O.OOft

Element 5: Ks = Energy loss hLs = O.OOft

Element 6: K6 = Energy loss hL6 = O.OOft

Element 7: K1 Energy loss hL7 = O.OOft

E/ement 8: Ka Ener loss hLa = O.OOft Energy /osses-Pipe 2: K Qty.

Pipe: K1 = 8.16 1 Energy loss hL 1 = 0.00 ft Elbow: K2 = Energy loss hL2 = 0.00 ft Nozzle: K3 - Energy loss hL3 = 0.00 ft

Element 4: K4 Energy loss hL4 = 0.00 ft Element 5: Ks Energy loss hLs = 0.00 ft Element 6: K6 - Energy loss hLB = 0.00 ft Element 7: K1 Energy loss hL7 = 0.00 ft Element 8: Ka Ener loss hLB = 0.00 ft

Total ener loss huot = 7.31 ft

Results: Total head on pump: hA = 30.3 ft

CLASS 1 SERIES SYSTEMS I PowerUS:

Chapter 11

.0025 m3/s 150 L/min

182

Actual flow rate will be greater than design value - See next page Oesign Volume flow rate: Q = Oesign Volume flow rate: Q =

14.27 m Total energy loss hLtot =

4.83 m 7.64m 0.90m 0.90 m 0.00 m 0.00 m O.OOm

O.OOm

Energy loss hu = Energy loss hL2 = Energy loss hL3 = Energy loss hL4 = Energy loss hLs = Energy loss hL6 = Energy loss hl7 = Ener loss hLs =

Results: Given pressure at point 1 = O kPa Desired pressure at point 2 = O kPa Actual pressure at point 2 = 5.74 kPa

Actual 2 should be > desired ressure)

0.000 m 0.000 m


Energy losses in Pipe: K Pi e Friction: K1 = f(UD =

dditional Pipe Data: Flow area: A = 0.000595 m2

Relative roughness: DIE= 18353 LID= 254

Flow Velocity = 4.20 mis Velocity head = 0.899 m Reynolds No. = 4.84E+04 Friction factor: f= 0.0211

Final Minimum Diameter: M²nimum diameter: D = 0.0220 m



ethod 111-A: Uses Equation 11-13 to compute the inimum size of pipe of a given length that w²l/ flow a given vol u me f/ow rafe of fluid

111-A & 111-8 SI: CLASS 111 SERIES SYSTEMS


Actual volume flow rate when p2 = 0.00 kPa Actual volume flow rate: Q = 0.002566 m3/s Actual volume flow rate: Q = 153.93 L/min


Energy loss hu = 5.06 m Friction Energy loss hL2 = 8.05 m Energy loss hL3 = 0.95 m Energy loss hL4 = 0.95 m Energy loss hL5 = O.OOm Energy loss hL6 = O.OOm Energy loss hL7 = O.OOm Ener loss hLB = O.OOm

K Qty. 0.000 m



Flow Velocity = 4.31 m/s Velocity head = 0.947 m Reynolds No. = 4.96E+04 Friction factor: f= 0.0210


Results: Maximum values Volume flow rate: Q = 0.0047 m3/s Area: A =0.0005953 m2

DI&= 18353.333

Method ll~A: No minor losses CLASS 11 SERIES SYSTEMS ll~A & ll~B SI: PPLIED FLUID MECHANICS

Chapter 11 184

Actual volume flow rate when p2 = O.DO kPa Actual volume flow rate: Q = 0.002121 m3/s Actual volume flow rate: Q = 127.26 L/min Change from Problem 11.47A: -26.67 L/min

Percent change: -17.3 %



Energy loss hu = 3.61 m Friction Energy loss hL2 = 5.50 m Energy loss hL3 = 0.65 m Energy loss hL4 = 0.65 m Energy loss hLs = O.OOm Energy loss hu; = O.DO m Energy loss hL7 = O.DO m Ener loss hLB = O.DO m

Energy /osses in Pipe: K Qty. Pipe: K1 = f(UO) = 5.57 1


Flow Velocity = 3.56 m/s Velocity head = 0.647 m Reynolds No. = 4.1 OE+04 Friction factor: f= 0.0219



O!&= 18353.333

Method 11-A: No minor losses CLASS 11 SERIES SYSTEMS 11-A & 11-B SI: PPLIED FLUID MECHANICS


Actual volume flow rate when p2 = 0.00 kPa Actual volume flow rate: Q = 0.002170 rn3/s Actual volume flow rate: Q = 130.218 L/rnin Change from Problern 11.47A: -23.712 L/min



Energy /osses in Pipe: K Pi e: K1 = f(UO) =


Qty. Energy loss hu = 3.76 m Friction Energy loss hL2 = 5.76 m Energy loss hL3 = 0.68 m Energy loss hL4 = 0.68 m Energy loss hLS = O.OOm Energy loss hLB = 0.00 m Energy loss hL7 = 0.00 m Energ loss hLB = 0.00 m


Flow Velocity = 3.65 m/s Velocity head = 0.678 m Reynolds No. = 4.20E+04 Friction factor: f= 0.0218

6.58 m/s Velocit : v = Using Eq. 11-3

Resu/ts: Maximum values Volume flow rate: Q = 0.0039 m3/s Area: A =0.0005953 m2

Ole= 18353.333

Method 11-A: No minor losses 11-A & 11-B SI: CLASS 11 SERIES SYSTEMS PPLIED FLUID MECHANICS

Chapter 11 186

Actual volume flow rate when p2 = 0.00 kPa Actual volume flow rate: Q = 0.002430 m3/s Actual volume flow rate: Q = 145.77 L/min Changefrom Problem 11.47A: -8.16 L/min



Energy losses in Pipe: K Pi e: K1 = f(UO) =


Qty. Energy loss hu = 4.59 m Friction Energy loss hL2 = 7.22 m Energy loss hL3 = 0.85 m Energy loss hL4 = 0.85 m Energy loss hLs = 1.49 m fT = 0.011 Energy loss hLs = 0.00 m Energy loss hL7 = 0.00 m Ener loss hLa = O.OOm


Flow Velocity = 4.08 m/s Velocity head = 0.849 m Reynolds No.= 4.70E+04 Friction factor: f = 0.0213



01& = 18353.333

Method 11-A: No minor losses Uses Equation 11-3 to est²mate the allowable volume flow rate


187 PARALLEL PIPE UNE SYSTEMS

ThenQ. =A.u. =(8.213xl0-3 m2)(5.08 mis)= 4.17 xl0-2m3/s} _2 3 QT =6.02x10 m Is Qb =Abvb =(4.768xl0-3 m2)(3.87 mis)= 1.85 xl0-2 )m3/s

= 3.87 mis; NR = 6.28 x 104---)- fb = 0.022 No change b

Recompute ub: 2(9.81)(17.05)

Vá,= [770 á;, + 5.4]

Assume la = Jb = 0.02: hl =17.05m=(587f.+1.02]v; ;v.= r9Ŀ81)(l7.05) =5.08m/s a 2g 587f.+1.02]

NR =v.D. =(5.08)(0.1023)=1.08xl05: D= 0.1023 =2224-)-f.=0.02 n V 4.8X10-6 e 4.6X10-s

No change

hL =17.05m=[770/b +5.4] v~ ;vb = [(9Ŀ81)(l7.oy =4.01 mis b 2g 770/b + 5.4

NR =vbDb =(4.01)(0.0779)=6_51xl04: D= 0.0779 =l693-)-fb=0.022 b V 4.8X10-6 e 4.6X10-5

Elbows Valve Friction fbr=0.018

Lower branch b:

Friction Elbows 60 2 2 2 2

h = á; vb + I; (240)!Ñ.. + 2á; (30)!Ñ.. = [770á; + 5.4]!Ñ.. Lb b 0.0779 2g bT 2g bT 2g b 2g

f.r = 0.017

p v2 p v2 _A_+ z +~-h =_.!!..+za+_..!!..: ZA= Zs, VA= Un r A 2g L r 2g

PA Pa =h =h =h =(700-550)kN/m2 =17.05m r L L., Lá, 8.80 kN/m3

60 v2 v2 v2 Upper branch a: hl =f. a + 2 á;IT (30)-Û = [ 5 87 f. + 1.02 ]-Û a 0.1023 2g 2g 2g

12.1

Systems with two branches

CHAPTER TWELVE

PARALLEL PIPELINE SYSTEMS

Chapter 12 188

h = + 30 v; =(571/) v; and h = h : fo7. = 0.019 L, Ja 0.0525 2g a 2g L, Lb

Friction 60 v2 v2 v2 v2 há = fb b +3J;,T(30)--b + hr(l50)-b =[l ]42fb +4.56]-b-

-b 0.0525 2g 2g 2g 2g (i) Assume fa= Ji,= 0.02 and set hL, = hL.

2 2 571(0.02)~ = [1142(0.02) + 4.56]!:i.

2g 2g ll.42v; =27.4v~ Va= ._/27.4 /11.42 1Jb = 1.551.Jb

12.3

(D/s)b = 768; Try Ji,= 0.023 2 2

h, = (174(0.023)+8.80]5-=12.805- -h 2g 2g

v2 vz Let hL, = h1,,,: 10.69-Û = 12.8-b : Va= 1.0941.Jb

2g 2g Qá = ĆaVa + Ćbl.Jb = Aa(l.0941.Jb) + Ćbl.Jb = !Jb[l.094Aa + Ab]

Q 0.223 ViJ= 1 = =6.21 ft/s; va= l.094ViJ=6.79ft/s 1.094A. + Ab 1.094(0.0233) + 0.01039

N = vbDb = 6.21(0.115) =5.90xl04Ŀ á; =0.025 Rb V 1.21 X 10-5 ' b

v2 Repeat: h; =13.15-b; va= 1.1091.Jb; ViJ=6.15 ft/s; Va=6.83 ft/s

-b 2g

N11 = 5.85 x 104;Ji, = 0.025 No change b

Qa = AaVa = (0.0233)(6.83) = 0.159 ft3/s; Qb = Ćbl.Jb = 0.01039(6.15) = 0.0639 ft3/s

h = h = 10.69 v; = (1 ÁĿ69)(6Ŀ83)2 = 7. 74 ft L L,. 2g 2(32.2)

pá p2 = yhL ( 62~; lb )(7.74ft{1~4~:2) = 3.36 psi

Friction 2 Elbows Valve

12.2 DatafromexampleProblem 12.1 Q1 =0.223 ft3/s=Aava+Ćb!Jb; l.r =0.019; fbT =0.022

v2 vz v2 vz ( v2 ) hL = /.1(160)-Û + /.r(8)-Û + 7.5-Û = (168far + 7.5)-Û = 10.69 -Û " 2g 2g 2g 2g 2g Valve 1 Valve 2 Heat exch.

2 2 20 2 2 2 - vb + vb á; vb_ + vb vb h1 -2J;,r(30)-+ Jbr(340)-+ b----[174Jb +400J;,r]-[174jb +8.80]- -b 2g 2g 0.115 2g 2g 2g


Ll = 2902(0.02) + 1.14 = 1.59 o Vi 989(0.016) + 7.5 Vi Qs = Q6 + Q2 = A6Uá, + A2Vi = A6(1.59Ui) + A2Vi = Ui[1.59A6 + A2] = 3.01 ft31s

l²D) = 0.505~4 =3369: Tryfi,=0.016: (D) = O.l72~4 =1149; 8 6 1.5 X 10 8 2 1.5X10 Tryh = 0.02

lft3/s Qs-in = 1350 gallmin = 3.01 ft31s: Benzene: p = 0.87(1.94) = 1.69 slugs/fr'

449 gal/min f6r=0.015; !21'=0.019 rá=8x 10-61b-slft2

500 u2 u2 u2 u2 u2 h1, =fà 6 +J;,r(340)-6 +J;,r(l00)-6 +2f6r(30)-6 =[989j6+7.5]-b 6 0.5054 2g 2g 2g 2g 2g

500 u2 u2 u2 há = !2 2 + 2fzr(30)-2 + [2902/z + 1.14]-2 -z 0.1723 2g 2g 2g

u2 u2 But hL =hL: (989j6+7.5]-6 =[2902j2+1.l4]-2

6 2 2g 2g

u6 =Vi 2902j2+1.14 989 J;, + 7.5

12.4

Q. =A.u.= (2.168 x 10-3m2)(3.98 mls)[60000 L/minll m31s] = 518 L/min Qb = ĆbUá, = (2.168 X 1o-3)(2.55)(60000)=332 Llmin

9.8lkN[ (3.98)2 ] p.-pb=yh1 = rhz = 3 571(0.021) m =95.0kPa , -u m 2(9.81)

Recompute from CD: 2 2

571(0.021)~ = [1142(0.0215) + 4.56] !:1_ 2g 2g

11.91 u~ = 29.11 u~ Va= 1.56Uá, Q 850160000 Uá,= A = = 2.55 mis: u = 1.56Uá, = 3.98 mis 2.56Ab (2.56)(2.168x10-3) Û

No change infa orfi,

N11 = u.D. = (3 .97)(0.0525) = 1.60x105 : D = 0.0525 = 1141 ~la= 0_021 a V 1.30X10-6 8 4.6X10-5

N = ubDb = (2.56)(0.0525) = 1.03 x 105: D = 1141 ~ fi, = 0.0215 11Å V 1.30 X 10-6 8

Uá,= QA = 850 L/min 1 m3 Is = 2_56 mis 2.55Ab 2.55(2.168x10-3m2)(60000 L/min)

Va= l.55Uá, = 3.97 mis

Chapter 12 190

Equate hL, = hL" v2 u2 16v2 l 66v2

(6.18 + K)-Û =10.38-b = 10.38--Û =--Û ; then 6.18 + K= 166 2g 2g 2g 2g

K= 166- 6.18=160

Q. = Qb =A.u.= AbUt,; Vi,= v._1__ = v (DÛ )2 = 4v Ā vb2 = l6v.2 A Û D Û' b b

2 2

hl =(600(0.0163)+0.60)5_=10.385_ b 2g 2g

30 2 2 2 2

h = I. m v. + 2J; (30)!:!.e__ + K !!.e_= (300J; + 0.60 + K)!:!.e__ L, Û O.lOOm 2g aT 2g 2g Û 2g

Friction 2 Elbows Valve 30 2 2 2

h =J; _5_+2/, (30)5_=(600J; +0.60)5_ Lb b 0.05 2g bT 2g b 2g

Friction 2 Elbows 2 2

hl = [300(0.0186) + 0.60 + K]!:!.e__ = (6.18 + K)!:!.e__ a 2g 2g

12.5 For illustration, use water at lOÜC; Qa = Qb = 500 L/min = 8.33 x 10-3 m3/s.

u= Jl=8.33x10-3m3/s=l.061m/s: (D) = 0.100 =66667 Û AÛ JT(0.100)2 rrr' /4 E: a 1.5X10-6

NR u.D. = (1.06l)(O.l_?O) = 8.18 x 104 ~fa= 0.0186; f.T = (O.OlO)(approx.) a V 1.30X10

Vi,= Jl = 8.33x10-3 m3 Is= 4.24 mis: (D) = 0.05 = 33333 Ab Jr(0.05)2 m214 e b 1.5x10-6

N ubDb = (4.24)(0.05) = 1.63 x 105 ~ 1: = 0.0163Ŀ á; O 010 ( ) Rb V 1.30X10-6 Jb ' bT = . approx.

No change inf6 orj; Q6 = A6u6 = (0.2006 ft2)(13.98 ft/s) = 2.80 ft3/s [1258 gal/min] Q2 = A2LJ:á = (0.02333)(8.79) = 0.205 ft3/s [92 gal/min]

NR = u6D6p = (13.98)(0.5054)(1.69) = 1.49 x 106 ~ f6 = 0_016 6 Õ 8X10-6

N = u2D2p = (8.79)(0.1723)(1.69) = 3_20 x 105 ~ r = 0_020 R2 Õ 8X10-6 }2

3.01 ft3/s U;á= 1.59(0.2006) + (0.02333) ft2 = 8.79 ft/s; u6 = l.59LJ:á = 13.98 ft/s


12.7 Hardy Cross technique -Data from Prob. 12.4 vz

Qs = 3.01 ft3/s; hà = [989.f6 + 7.5]-6 2g

vz h2 = [2902.fi + 1.14]-2

2g

Restate h6 and h2 in terms of Q6 and Q2

hà = [989.f6 + 7.5] Ü: 2 = [989 á;, + 7.5] Ü: 2 2gA6 2(32.2)(0.2006)

hà = [989.16 + 7.5](0.3859)Q~ = [381.6.f6 + 2.894] Ü~ h2 = [2902.fi + 1.14] Ü~ 2 = [2902/i + 1.14] Ü~ 2

2g~ 2(32.2)(0.02333)

h2 = [290~á; + 1.14](28.53) Ü~ = [82791.fi + 32.52] Ü~ N = v6D6p = Q6D6p R" lJ A6Õ

N = Q6(0.5054)(1.69) = (5.322 X 105)Q R, (0.2006)(8X10-6)

6

N = V2D2P = Q2D2P R, lJ AzÕ N = Qz(0.1723)(1.69) = (1.560 x 106)Q R, (0.2333)(8x10-6)

2

c. Valve in branch 1 open: Q = Q1 = 0.456 ft3 /s

b. Valve in branch 2 open: Q = Qz = 1.385 ft3/s

a. Both valves open: u = ~2ghL = ,.......2(-3-2.-2)-(4-6-.2-) = 20.9 ft/s: A = ;ir(2 in)2 ft2 = 0.0218 ft2

1 6.8 6.8 '

1 4 144in2

Q1 = A1 u1 = (0.0218 ft2)(20.9 ft/s) = 0.456 ft3/s

lh = ~2ghL = 2(32.2)(46.2) = 15.87 ft/s: Ćz = ;ir(4 in)2 ft2 = 0.0873 ft2 11.8 11.8 4 144 in '

Qz = A2!h = (0.0873 ft2)(15.87 ft/s) = 1.385 ft3/s Qtotal = Q1 + Qz = 0.456 + 1.385 = 1.841 ft3/s

2 2 2 2 2 2

h, = 2(0.9)-'i_ + 5 _li_ = 6.81: h2 = 2(0.9)!:2+10!:2=11.8!:2 2g 2g 2g 2g 2g 2g

P vz P v2 ___A_+ z + .:s: - hL = -1L + Zg + -1L: VA= Vs, PB =O, ZA= ZB r A 2g r 2g

h =pA =20lbft3 144in2 =46.2ft=h =h L r in2(62.4 lb)ft2 I 2

12.6

Chapter 12

-0.00037 (Negligible)

-0.01253

-0.07635

-0.21575

192

h = [1142+- + 4.56) u; = [1142J;, + 4.56]Q~ b :tb 2g 2gA2

h - [l l42A + 4Ŀ56)Q~ = [l 142Jb + 4.56)(1.084 X 104) Q~ b- 2(9.81)(2.168x 10-3)2

(J)h = kQ2 @2kQ 56.25 45.00 -428.4 1679.94 -372.15 1724.94

66.39 48.89 -145.9 992.5 -79.51 1041.4

70.16 50.26 -80.00 734.95 -9.84 785.21

70.81 50.49 -71.05 692.65 -0.24 642.16

ha= 571fa u; = 571faQ; 2g 2gA;

h = 57lfaQ; =(6.192xl06)1ÁQ2 Û 2(9.81)(2.168xl0-3)2 Ja Û

12.8 Data from Prob. 12.3 - Hardy Cross technique Ća = Ab = 2.168 x 10-3 m2; v= 1.30 x 10-6 m2/s

. 1 m3/s QA = 850 L/mm x = 1.417 x 10-2 m3/s 60000 L/min

Q6 = 2.805 ft3/s; Q2 = 0.205 ft3/s

6 2

4

6 2

2.792 J.49 X 106 0.016 9.000 -0.2177 3.40 X 105 0.020 1688

2.805 J.49 X 106 0.016 9.000 -0.205 3.20 X 105 0.020 1688

3

2 -0.294 2.716 6 2

2 -0.51 2.50 6

NR f k 1.33 X 106 0.0160 9.000 7.96 X 105 0.0195 1647

1.45 X 106 0.016 9.000 4.59 X 105 0.020 1688

TRIAL PIPE Q

hà = [381.6(0.016) + 2.894) Qà = 9.000Qà = k6Qà

h2 = [82791 (0.0195) + 32.52) Q; = l 647Q; = k2Q;

Friction factors found from Moody's diagram.

Try Q6 = 2.50 ft3/s; Q2 = 3.01 - Q6 = 0.51 ft3/s

Náá. = (5.322 X 105)(2.50) = 1.33 X 106;_1<; = 0.0160 Náá = (1.560 X 106)(0.51) = 7.96 X 10\/2 = 0.0195

2

( ~ l = 3369; ( ~ l = 1149 (From Prob. 12.4)


Ji,= [ [ 1 0.25 5.74 )]2 = 0.0225 log 3.7(1141) + (7.77x104)Á"9

Computefusing Eq. (9.9):

la= [1 [ 1 0.25 5.74 )]2 = 0.0208 og + 09 3.7(1141) (1.86x105) Ŀ

Try Q. = 0.0100 m3/s Qb = QA - Q. = 0.01417 - 0.0100 = 0.00417 m3/s

Then NR,, = (1.863 x 107)(0.010) = 1.86 x 105

NR =(1.863x107)(0.00417)=7.77x104 b

(D/c)a = (D/c)b = (0.0525/4.6 x 10-5) = 1141

Tria) 1

N = v.D. = Q.D. = Q. (0.0525) R,, V A. V (2.168 X 10-3)(1.30X10-6) N R. = (1.863 X 107)Q.; N Rb = (1.863 X 107)Qb

Chapter 12

<')

~ ~ ..J < a: ::::; ::::; J- o (.) 2 o a: tl) {'\j u. <O IO (/) co ll) w o o 1- o o ~ o ·

s JI Å ..J 00 u, a: a; .o o w w u, o.. e, (/) áá: áá: w ::::> ~ > ..J < z u:

u n 00 a; .o ww o.. 2: áá: o..

e: e: ee :::J :::J

w (!J o ,... "' ~ tl) ~ o (5 8 8 :e eo; C\I q o o o u - ~ -s. o (") <O co

o 9 9 < w w w ... o ll) O) ..J IO o (O

w ~ o o o - ~ ~

o ll) <O o o - - - o - ,... (') .... ll) (') CX) ll) (') <X) -"' ll) ,... (') C\I ll) ...... C\I tl) ...... C\I C\I t\l ll) C\I <") ll) C\I (') tl)

C\I (') ~ <X) (') tl) C\I ,... co o

< ,... <O (") ..,. - <") (') o o CIO ,... - ,... ...... o ,... ...... o -"' N "? ....: o² ~ q o) ~ o 11 - .t:

Å 11 11 ll) ll) o ll) ll) o ll) tl) o o o .>C o o -"' o o ~ + + C\I + + t\l + + C\I

.>C w w o w w o w w o O) <X) o o o o "! C\I z ~ t\l z ~ C\I z - cr) < - cr) < - eo; < s: ..e: .t:

u. u. LL á....; o o o o <X) ll) 2 o CX) 2 o co 2 < o C\I - - - - u. {'\j C\I ::::> C\I C\I ::::> C\I C\I ::::> o o (/J o o (./) o o en a: o o o o o o u.

o ll) ..,. ll) ll) ll) ll) o o o o o o z + + + + + +

~ w w w w w w <O ,... - (') - (') w <X! ,... ~ ~ ~ ~ a: - ,... - - - - {'\j (') (') (') C") (') o 9 o o o o t.Đ w w w t.Đ uJ o o o o o C\I co o ,... tl) C\I tl) - ~ - <O ll) <O tl) - ~ c· Ič ex) Ič . .

w 2: ca .Q ca .e ca .Q a. ..... 5 o - - - g; o ..J < C\I (") a: - ...

194

2 w ..J al o a: w e,

::::> 2

ÜÜ - a: ~ u. o -e w ... ... < (./Jo en ...., o <X) a: N (.) - b2 ~~ :e o (!J a: z a, C¶ ::::> ~ en >- ..J (./)

~~ ~~ ~<

~~ z~


Qb + Qc = Qe + 0.30 ft3/s 0.70 + 0.10 = 0.50 + 0.30 (check)

Qe + Qr= 0.60 ft3/s Try Qf= 0.10; o, = 0.50

Qd - Qf= 0.30 ft3/s Try Qd = 0.40, Qr= 0.10

Qc + Qd = Qa = 0.50 Try Qc = 0.10; Qct = 0.40

QÛ + Qb = 1.2 ft3/s Try QÛ = 0.50; Qb = 0.70

For Trial 1: Flow equations at nodes:

N = vD = QD = Q(0.2058) = (S.l14 x 10s)Q R v Av (0.03326)(1.21x10-5) Use Eq. (9.9) to compute! f= 0.25 2

[log( 3. 7(11372) + (~ :);')] f= 0.25 2

[log(l970~10~ + (~~7);' )]

Values of/to be computed. For all pipes DIE= 0.2058/1.5 x 10-4 = 1372 Use water at 60ÜF; v= 1.21 x 10-5 ft2/s

a+bÅ 1.2 e+ f Å 0.8 c+dÅa d-fÅo.3

b+cÅÅ+0.3

k = k = k. = k: = f (50) a b d e (0.2058)(64.4)(0.03326)2

= 3410/

k k f(30) = 2046/ e= r= (0.2058)(64.4)(0.03326)2

0.8

Å 1.2 -

d o. General forro for k h = kQ2 = fi, !!__ = fi,Q2

D 2g D2gA2

:. k= JL D2gA2

12.9 2 1/2-in Sch. 40 pipes: D = 0.2058 ft; A= 0.03326 ft2

Chapter 12 196

PIPE f k Eq. for k a 0.0197 67.18 ka= 3410fa b 0.0194 66.00 kb = 3410fb e 0.0233 47.65 k; = 2046.fc d 0.0200 68.26 kd = 3410.fd e 0.0197 67.18 kà = 3410.fe f 0.0233 47.65 k.= 2046.fr

Similarly:

ka= (3410)(0.0197) = 67.18

ĉa = 0.25 2 = 0.0197

[}og((l.970X10-4) + 5Ŀ74 5 09)] (2.557X10 ) .

Computefvalues for Eq. 9.9:

NR = 5.114 X 104; NR = 2.045 X 105 e d

NR e = 5.114 X 104

Similarly, NR = 3.580 X 105;

b

NR, = 2.557 X 105;

NR, = (5.114 X 105)(Qa) = (5.114 X 105)(0.50) = 2.557 X 105

Compute NR values for Trial 1:


1 O) - ,... o IO <O o C\I CD co co C\I C\I - o Å (") Å - ,..... ,.... CD ~ Å ~ ,.... co ~ '(! ~ - Å - Å ,,. ,..: ~

o (1) <O ~ .... ,,. - t\i - cO - 9 o ~ - ..,. e? ci 9 ci 9 ci 9 - - ,,. - ..,. ,,. . - . . -s. o C\I C\I C') (") Å Å q q q q q q < w w w w w w 1- ,.... o - C\I Å o ...1 CD co Å ,.... CD IO w co - C\I O) C\I - Q a; "! ~ ~ ~ ~

Å O) <O o Å o ..,. C\I o <O o CD <O ,... C\I co (O (O Å c:o ,... co C\I <O s C\I (1) (') - C') <O C\I C') O) ..,. "! (1) O) C') co ~ (O o CI) <O o ,... o ,... C! ,,. C')

,..: N 11) a) 11) ;,; ,..: IO o cO M c:\ĉ M a) C\ĉ cO cO o M t\i M ci M cO

C\I a) <O a) a) ..,. - ,... - ..,. ,.... - ..,. co O) - co - ,.... (O - - - U) <O - - ,... co - - ,... <O <O - - N o ,.... <O - <O - o <O O) - C') c:o 1..,. co o <O (') N in O) IO O> IO <O - co - < .... C') ,... IO ,... C\I - ,... (O o - ,... (') ,... C') O) - (') C\I ,... ..,. o ..,. (") (O O) O)

o <XI C') ..,. q ..,. O> ca ..,. (O - ~ O> ,... O) C\I ... O) q Å "". O) - O) <O ,... O> o ~ <.c:i c:\ĉ ci IO 9 ci <.c:i ci ~ M ..,. ci 9 9 C') .. ci - t"i Å ci 9 9 t"i t"i ci ci 11 - ~ - - - N ~ - - . C\I ~ - - . s: . . . '

u 11 11 11 11 11

o o o s o o ..,. O) ~ ~ ~ U) Å ~ ~ O) <O ,.... ~ ,... co C\I co ,... eo ee ~ co .... O> - .Jo: C\I ~ C\I C\I "! C\I IO Å - N - ,... <O (') N "'! Å C'l.I C\I C\I ,... <O - C\I ~ ,..: ,..: Q ,..: cO ,..: e <.c:i .o .o e .o ,..: ,..: .n e .o .n e .o ,..: ,..: .o o IO ,... <O

<O <O Å z ..,. <O <O ..,. z <O <O Å z Å <O <O Å z <O <O Å z ..,. <O <O Å z < -e < -c < -c ..c: .e: .e: ..e: .e: .e: u. u. u. u. u. u.

...,: o o o o o o c.) ,.... Å (') ::e (') o ,... <') ::e IO ll) - :e - O) co C\I ::e IO IO - ::e - O) co o ::e < a> O) (') (') o O> C') O) O> C'l.I C\I O) O) C\I O) O) C'l.I C\I O) CI) C\I

u.. ...... ...... C'l.I :::> C\I C\I .... C'l.I :::> - - C'l.I :::::> C\I - - C\I :::::> - - C\I :::> C\I - - C\I :::> o o o CI) o o o o CI) o o o CI) o o o o CI) o o o (J) o o o o (/) a: o o o o ci ci ci ci ci ci ci c::i ci ci c::i c::i c::i ci ci o ci u..

· U') 'Á ..,. Å ll) 'Á Å 'Á 'Á ..,. Å IO IO Å IO IO Å ..,. IO ll) Å o o o o o o o o o o o o o o o o o o o o o z + + + + + + + + + + + + + + + + + + + + + >= w w w w w w w w w w w w w w w w w w w w w

<O CIO .,... - IO <O .... .... C\I C') (') <O Å IO (') o O> O) o - - w l.() IO - - o 11) - o - IO IO C\I C') C\I C! - ~ (") (') (') <O a: N C".ĉ .o ll) C\ĉ c:\ĉ ll) C".ĉ C') ,..: ,..: C\ĉ C\i ,..: (') C') ,... ,..: c:\ĉ C\i ,..:

o o o o o o o o o C'l.I C\I co C\I co C\I co ..,. ..,. co C\I co o o 8 o o o 8 O> - ,... ,... - co - C') <O Å Å co - co o o o o o o co - Å Å Å IO Å O> o Å ..,. Å I,(') Å l.()

,... - - "". IO - 11) <O - - ..,. .,. - ll) <O - - .,. .,. - ci 9 ci 9 o 9 ci ci 9 c::i 9 ci 9 ci ci 9 ci 9 o 9 ci

w !:!:: <12 ..o o o 'C qJ - <12 ..o o o 'C Gt - a:J ..o o o 'C Gt - Q.

1- 5 c.) - C\I - C\I - C\I a: (3

...J < C\I C') a: - 1-

Chapter 12

o CD 11) - - CD ~ _g .Q ~ o .s ..,. <O ..,.

O) o O') 11)

..,. lt) ..,. s ...,. ..,. 111 ci ci d .... ~ 11 11 11 ,g 00 o -o Qi - < w w w - e, a. a. ..,. a: a: a: ....l < a: 1- :E CX) C\I ll) o C') <O ..,. a:: O) o ..,.

lt) <O - u, ci ci ci en 1- JI 11 11 ....l 00 o :::) (J) a; .o ¿ w a: w w w u. a.. Cl. 9: o a: a: e, >- a: -c :E :E :::) <()

w

~ - - ..,. tO C\I C\I lt) o o o o o o o 9 ci 9 ci q ci 9 u -s. o lt) lt)

q o -e w w ... - lt) ....l U) IO w C\I lt)

e ~ ,..: . IO o CX) ..,. co lt) - ,.... - o o <O o ,.... o CD q ..,. ...,.

~ o; ci ("') C\i ("') ci ("') a) ('\I ,.... - ..,. ,.... CX) - - - U) <O - - C\I ...,. O) lt) - ll) (') lt) U) - o ,.... C\I ..,. - ...,. ,.... ..,. o - ..,. ~ O) o O') (D ,.... q o ~ (") ..,. ci 9 9 (") ("') .... q a C\I C\I .... .... .J: . .

JI 11 o o

<O O) co ~ co o O) O) ~ lt) ~ C\I C\I C\I ..... <O o C\I ~ c® ~ e ~ ,..: ,..: ~ e <O <O <O ..,. z ..,. U) U) ..,. z -e <

.:::: .:::: u. u.

~ o o u lt) l.(.) .... :E - O> O) o :E -e O) O) C\I C\I O> O') C\I u. - - C\I :::) C\I - - C\I :::) o o o cn q o o o cn a: ci ci ci o ci ci ci u.

o l.O U) ..,. ..,. ID U) ..,. o o o o o o o z + + + + + + + > w w w w w w w ..,. o O) O') o o ..,. w q - C') (') (") C') <O a: C') ("') ,.... ,.... C\i C\I ,....

CX) C\I lt) ll) ..,. <O ..,. (") <O ..,. ..,. O> o a> o O') o ..,. ...,. ..,. lt) ..,. l.O <O - - ..,. ..,. - ci 9 ci ci ci 9 ci .

w e, 111 .Q o o "O G) - a: 1- 5 u - C\I a: (3 ....l -e ..,. a: 1-

198


o, = 2.0 X 10"2 È Qh = 0.5 X 10-2 @

Qd = 4.0 X 10"2 É Qe = J.0 X 10Ŀ2@ Qr= 2.0 x 10Ŀ2 (Assume)

For Trial 1:

Qa = 2.0 x 10Ŀ2 (Assume) Qb = 0.5 X 10Ŀ2 È Qc = 4.0 x 10Ŀ2 (Assume)

For continuity atjoints:

o-. -o.1oom3/t

d \ f - e

1 ,,

1 0~ ( 2i

' ~ ' - \ (1 .. ..

0.025 0.025 0.025

f = 0.25 = 0.25

[lo ( 1 + 5.74 )]2 [io (5.49x10-6 + 5.74 )]2

g 3.7DI e (N11)09 g (N11)09

Use Eq. 9.9 to compute/

For ali tubes: Water at 15ÜC; v= 1.15 x 10Ŀ6 m21s N = uD = QD = Q(0.07384) = (1.500 x 101)Q 11 v Av ( 4.282 x 10-3 )(1.15 x 10-6)

DIE= (0.07384)1(1.5 x 10Ŀ6 m) = 49227

For tubes e, f: L = 6.0 m; k= (2.259 x 105)! For tubes d, b, e, h: L = 15 m; k = (5.647 x 105)! For tubes a, g: L = 18 m (Ignore minor losses); k = (6.776 x 105)!

Ü. = 6000 L/min x l.O m3ls = 0.100 m31s = 1.0 x 10-1 m31s m 60000 L/min Q0u1 = 1500 L/min = 0.025 m3 /s (Each) = 2.5 x 10-2 m3 Is h= á~!!_-á!:___S!__kQ2Å k= fL

D 2g - D 2gA2 - ' D(2)gA2

3-in Type K copper tubes: D = 0.07384 m; A = 4.282 x 10Ŀ3 m2

k = jL = jL = (3.765 X 104)/L D(2)gA2 (0.07384)(2)(9.81)( 4.282X10-3)2

12.10

Chapter 12

w (!J

O) co co co - co :; ~ <O o IO C') ..,. - ,... ~ co IO IO C\I C\I - IO ~ <O <O IO ...,. ,...

<O ,...

N IO - ..,. iri a:i ll) C\I (") ,... a:i

~ r...: M <O cD C\i a:i ..,. q; cD ~ C\i M ~ cD ~ o; C\I - - ~ (') - - ":" IO . - u . . . '$.

a (') (') C') C') (") (')

o 9 o 9 o 9 -e w w w w w w t- O) .... co .... <O <O ...1 ,... - N lt') O> co w ~ 1() ,... o C! ~ o - ce; M ~ - - . . . . o - O) - O) O) ll) C\I 1() C\I o - C') C\I ll) ,... U) ll) CX) o ,... o - 8 o a C! c:o """. C') Å <O ..,. o <O o o c:o ca ..,. o ca O) co CX) ~ ,... C\I ,... C'! C)

r...: o; iri a:i ,..,: C\i M C\i u:i ,..,: in a:i ,... ci C\i ci cD cD cD o a:i a:i

~ <O lt) ..,. M - co ..,. en o C') C') (') co co (') .... ('t) O) o C') - ca <O <O co :5 ca .... - - ('t) <O .... C') - C\I ,... C\I ll) - ... .... - (") - <O ..,. C\I ,... C\I C\I - - - (") - <O

N o o o (1) o ..,. .... o ,... o s o o (') C\I <O lt) CD ,... ,... 1() co 1() O) (') - < <O ,... - .... - ,... (') C\I ..,. C\I ,... ,.... ,... IO ca co c:o co C\I o ,... ~ - (") co a O) C\I ,... C! "": "": O> ~ Å ~ O) C\I (') ..,. - c:o ~ c:o Å C') C! ~ ,... ,... (1)

~ cri 9 ~ - Å ..... ci - ,..,: - <1 ci ~ ~ ci ~ - 'Á a? C\i - - - ~ ci 9 11 . ... . . . . . . ..e: .

11 11 11 11 a ti

a a 9 a o o o ..... ..,. ~ ..,. O> (") o ~ o o - co ,... o ~ o ao <O O> ...: O) (1) ..,. ...: o c:o ..,. C\I ..,. IO ,... o C\I o o c:o C\I CX) IO O> C\I O> <O (') ee C\I co <O C\I C\I ~ O) ,... O> e O> C') C') C') e C') O) ,... e ,... ll) ca e co lt') lt') (') e <") C\I <O e O) o C\I z C\I ,... O> (') z (") O) o z O> ..... C\I z C\I ,... co (') z C') o O) z - .... - .....

-c -c -c < < < ..c: ..c: ..c: ..c: s: ..c: u.

~ u. u. lL u.

t3 o o o o o <O - o 2 o o <O <O 2 <O <O - 2 Å lt') co 2 co Å - o 2 o C\I o 2 < .. O) (") (") (') <O ..,. ...,. ..,. O) ..,. o C\I C\I (W) ll) IO lt') IO ,...

u. .... - .... ::::> .... - - .... ::::> - - .... ::::> - C\I - ::::> ,.. - .... - :::::> - - - ::::> C! o o C/l o o o o C/l o o o (/J o o o rn o o o C! <n o o o rn c® o ci ci ci ci ci ci ci c:i ci ci ci ci ci c:i ci o ci ci ci lL

g U') ..,. ll) IO IO ll) ll) ll) IO ...,. ll) Å ll) l.() ll) ll) IO "' ll) ll)

o o o o o o o o o o o o o o o o o o o o + + + + + + + + + + + + + + + + + + + +

> w w w w w w w w w w w w w w w w w w w w 8 o o g o o o o o o .... (") ,... .... C\I co co co Å - w ll) o C! ~ C! C! C! ~ C\I ..,. ,... .... o <lf'. ll) IO ..,. (')

a: cri ,.... <O <O <O - (W) (") (') ,... M iri <O <č iri N C\i C\i C\i - o o o o o o o o o o ..,. <O - - lt') IO C\I C\I (") ,... o ll) o o o o o o o ll) - (") ll) lt') (W) <O

,... ,... <O co a C\I o '<I" ..,. ..,. - C\I C\I C\I o C\I o '<I" ;g (W) - ... - ... o o o o o o o o o o o o o o o o o o o o ci 9 9 ci ci ci 9 ci 9 ci ci 9 9 ci 9 ci 9 ci 9 ci .

w a. ca ..o (.J (.J "O Q) - - Á' .e tU .o (.J (.J "O CD - - OI .S: ii: t: ::::> u - C\I (") - N (')

g;; u ....1 < - C\I a: 1-

200


~ o <O C\I O) (f) C') ... ~ O) <O co C') ,... .... ... ,... C\I IO "' CX)

C') ~ ~ "". o ,... ,.... C\I C\I ~ ~ IO "" co ~ "": <X) ~ ~ ~ ..,. - ... <'i c.; c.; - <'i e; ... ..,. ci 9 o - .... 9 - - o . (") . . ' ' C\I . .

s. o ""

..,. ...,. ..,. "" ..,. o o o o o q

e( w w w w w w ... co IO <O C\I - o ...J ,.... ,... <D ...,. IO ,... w (") IO C') U) ~ "". Q ~ ~ e; <'i - - . ' .

N IO - IO o IO C\I - (\1 - a) - 8 ...,. - ..,. o ..,. Å - .... .... .... C\I .... <O ,.... ... o "". q - <O - O> - "": ..n ..... (f) - O> - ,... O) ,... ~ ~ <D ..n co ~ C\I ...,. .io: - ...,. cD - cD C\i ,...; O> <O Q ,...; C\i a) g C\i <O a) <O .... <O ,... <O

,...; - ,...; <O C\I IO IO IO IO O> O) - - o O) - lt) ..,. IO CX) o .... - o O> - ...,. -e- C\I ,... C\I ...,. C\I ... - - (f) ... <O Å (") C\I ,.... C\I ..,. C") ... - - C') - ce - - C\I CIO l.C) co ..,. CXJ (') <O ,... <O ,... ,.... - CD C') ,.... <O ~ <O co

,.... ("') ,... C') :; o - Û <X) C") l.O o IO <C - l.C) <O IO IO O> o - - a) c:o <C O> "' C\I C\I IO O> (\1 o <O ..,. <O O> <O ~ ,... q N O> C\I IO o <O - ce <O ,... ~ C\I ~ - ~ o .io: ..n ci uj ci ..n ,...; <'i - ci - <'i ci 9 ..n ci uj ci ..n ,...; <'i - ci - C\i - 9 11 . ' ' ' . . ' . ' Å Å .

.e 11 11 u 11 11 11

o o o o a 9 ('I) O> O> .X. O> CX) co C\I .X. C\I C\I ,.... .io: <O CX) c:o .io: co C') C") (f) .io: ('I)

,... o C") C\I O> C\I O> o (') ,.... C\I ,... <O - C\I O> CX) O> C\I (7) C') CX) CD C\I CD o <O C\I ~ <O o a:> o CX) <O .... C') o C') Å C') o IO IO CX) o <X) <O (') C') o (') IO C\I e .... o IO o O> ..... C\I z C\I ,.... co C') z C') - O> z O> ... "' z C\I ,.... co {") z C') ... O> z

-c e( -e -e e( < .e .e .e s: .e .e u.. l5 u.. u. u. u,

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u.. - "' - :::::) - - - ..... :::::) - ..... - :J - C\I - :J - ..... - ... :J .... - - :J o o o en q o o o C/J o o o C/J o o o C/J o o o o C/J o o o C/J a: ci ci ci o ci ci ci ci o ci ci ci ci ci ci ci ci o ci o u..

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Chapter 12 202

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PARALLEL PIPE UNE SYSTEMS

Chapter 12 204

Reynolds numbers: Assume water at 60ÜF; v = 1.21 x 10-5 ft2/s 16-in pipes: (j), (i),É,È

@ k = f (l 500) = 290.51/ (0.6651)(64.4)(0.3472)2

8-in pipe

fil) k = f ( 4000) = 77 4.69/ (0.6651)(64.4)(0.3472)2

8-in pipe

È k = /(4000) = 103.39/ (0.9948)( 64.4)(0. 7771)2

12-in pipe

È k = f ( 4000) = 64.31/ (1.094)(64.4)(0.9396)2

14-in pipe

(J) o k= /(1500) = 38.77/ ' (0.9948)(64.4)(0.7771)2

12-in pipe

È k = f (2000) = 16.50/ (1.25)(64.4)(1.227)2

16-in pipe

@ k= /(2000) = 51.70/ (0.9948)(64.4)(0. 7771)2

12-in pipe

(3) k= !<2000) -9.158/ (1.406)(64.4)(1.553)2

18-in pipe

CD <UÈk= f(15oo) 1=12.38/ ' (1.25)(64.4)(1.227)2

16-in pipe

h = kQ2 = fl !!__ = fl,Q2 . k = fl . D 2g D2gA2 Å D2gA2

12.11 Hardy Cross technique =Data preparation


Initial estimates for flow rates for Tria! 1: See spreadsheet.

É+CD= 15.5 n31s @+É=CD @= l.5 ft3ls +@ È=È+È @+È=È+È+ 1 ft31s

To satisfy continuity atjoints:

Similarly: 18-in pipe: È: N11 = (7.482 x 104)Q; DIE= 93 73 14-in pipe: È: N11 = (9 .623 x 104)Q; DIE = 7293 12-in pipes:É (VÈ@: N11 = (1.058 x 105)Q; DIE= 6632 8-in pipes: @),@, N11 = (1.583 x 105)Q; DIE= 4434

N11 = uD = QD = Q(l.25) = (8.419 x 104)Q v Av (1.227)(1.21x10-5)

DIE= 1.2511.5 X 10Ŀ4 = 8333

Chapter 12 206

continued

Trlal 1 Circuit Pipe a (ft;j/s) NR f k h= kQà 2kQ ~a %Chg

1 1 7.0000 5.89E+05 0.0144 0.1788 8.7592 2.5026 2.12 4 2.5000 2.64E+05 0.0162 0.8347 5.2171 4.1737 5.94 6 -4.0000 3.37E+05 0.0154 0.1904 -3.0472 1.5236 -3.71 3 -8.5000 6.36E+05 0.0142 0.1299 -9.3847 2.2082 -1.75

Summations: 1.5445 10.4081 0.14839 2 2 4.5000 3.79E+05 0.0152 0.1877 3.8008 1.6892 -3.60

5 3.0000 2.53E+05 0.0160 0.2639 2.3748 1.5832 -5.39 7 -2.0000 2.12E+05 0.0166 0.6451 -2.5806 2.5806 8.09 4 -2.5000 2.64E+05 0.0162 0.8347 -5.2171 4.1737 6.47

Summations: -1.6221 10.0267 -0.16178 3 6 4.0000 3.37E+05 0.0154 0.1904 3.0472 1.5236 1.60

9 3.5000 3.70E+05 0.0155 1.6029 19.6360 11.2206 1.83 11 -1.5000 1.59E+05 0.0174 0.6730 -1.5142 2.0190 -4.28 8 -4.5000 4.33E+05 0.0151 0.9710 -19.6619 8.7386 -1.43

Summations: 1.5071 23.5017 0.06413 4 7 2.0000 2.12E+05 0.0166 0.6451 2.5806 2.5806 -18.43

10 1.0000 1.58E+05 0.0179 13.8120 13.8120 27.6240 -36.87 12 -2.0000 3.16E+05 0.0164 4.7639 -19.0556 19.0556 18.43 9 -3.5000 3.70E+05 0.0155 1.6029 -19.6360 11.2206 10.53

Summations: -22.2991 60.4808 -0.3687

NETWORK ANAL YSIS USING THE HARDY CROSS TECHNIQUE U.S. Cust Units Prob# 12.11 4 Circuits 12 Pipes Fluid: Water at 60 deo F Fluid kinematic viscositv: 1.21E-05 ffls

Pipe# ID (ft} Length (ft} Roughness Die C1 C2 E (ft} (k=C1*f) (NR=C2*Q)

1 1.2500 1500 1.50E-04 8333 12.37 8.42E+04 2 1.2500 1500 1.50E-04 8333 12.37 8.42E+04 3 1.4060 2000 1.50E-04 9373 9.16 7.48E+04 4 0.9948 2000 1.50E-04 6632 51.68 1.06E+05 5 1.2500 2000 1.50E-04 8333 16.50 8.42E+04 6 1.2500 1500 1.50E-04 8333 12.37 8.42E+04 7 0.9948 1500 1.50E-04 6632 38.76 1.06E+05 8 1.0940 4000 1.50E-04 7293 64.26 9.62E+04 9 0.9948 4000 1.50E-04 6632 103.35 1.06E+05 10 0.6651 4000 1.50E-04 4434 773.68 1.58E+05 11 0.9948 1500 1.50E-04 6632 38.76 1.06E+05 12 0.6651 1500 1.50E-04 4434 290.13 1.58E+05

207 PARALLEL PIPE LINE SYSTEMS

continued

Trial 2 Circuit Pipe a (fe/s) NR f k h = kQ" 2kQ AQ %Chg

1 1 6.8516 5.77E+05 0.0145 0.1791 8.4099 2.4549 -0.56 4 2.1898 2.32E+05 0.0164 0.8495 4.0737 3.7205 -1.76 6 -4.0843 3.44E+05 0.0154 0.1899 -3.1686 1.5516 0.94 3 -8.6484 6.47E+05 0.0142 0.1297 -9.6983 2.2428 0.44

Summations: -0.3833 9.9698 -0.03845 2 2 4.6618 3.92E+05 0.0151 0.1869 4.0617 1.7426 -1.07

5 3.1618 2.66E+05 0.0159 0.2619 2.6186 1.6564 -1.57 7 -2.2069 2.33E+05 0.0164 0.6365 -3.0999 2.8092 2.25 4 -2.1898 2.32E+05 0.0164 0.8495 -4.0737 3.7205 2.27

Summations: -0.4932 9.9287 -0.04967 3 6 4.0843 3.44E+05 0.0154 0.1899 3.1686 1.5516 -3.65

9 3.0672 3.24E+05 0.0157 1.6276 15.3121 9.9845 -4.86 11 -1.5641 1.65E+05 0.0173 0.6687 -1.6360 2.0919 9.53 8 -4.5641 4.39E+05 0.0151 0.9695 -20.1952 8.8495 3.27

Summations: -3.3504 22.4775 -0.14906 4 7 2.2069 2.33E+05 0.0164 0.6365 3.0999 2.8092 -0.23

10 1.3687 2.17E+05 0.0171 13.2579 24.8364 36.2921 -0.38 12 -1.6313 2.58E+05 0.0168 4.8700 -12.9597 15.8888 0.32 9 -3.0672 3.24E+05 0.0157 1.6276 -15.3121 9.9845 0.17

Summations: -0.3356 64.9746 -0.00517 Trial3 Circuit Pipe a (ft~/s) NR f k h = kQ" 2kQ AQ %Chg

1 1 6.8901 5.80E+05 0.0145 0.1790 8.4997 2.4672 -0.60 4 2.1786 2.30E+05 0.0165 0.8501 4.0348 3.7041 -1.90 6 -4.1949 3.53E+05 0.0153 0.1893 -3.3314 1.5883 0.99 3 -8.6099 6.44E+05 0.0142 0.1297 -9.6165 2.2338 0.48

Summations: -0.4134 9.9934 -0.04136 2 2 4.7115 3.97E+05 0.0151 0.1867 4.1435 1.7589 -0.39

5 3.2115 2.70E+05 0.0158 0.2614 2.6958 1.6788 -0.57 7 -2.1624 2.29E+05 0.0165 0.6382 -2.9843 2.7602 0.84 4 -2.1786 2.30E+05 0.0165 0.8501 -4.0348 3.7041 0.83

Summations: -0.1799 9.9020 -0.01817 3 6 4.1949 3.53E+05 0.0153 0.1893 3.3314 1.5883 -0.32

9 3.2111 3.40E+05 0.0157 1.6189 16.6920 10.3965 -0.42 11 -1.4151 1.50E+05 0.0175 0.6791 -1.3599 1.9220 0.95 8 -4.4151 4.25E+05 0.0151 0.9730 -18.9662 8.5916 0.30

Summations: -0.3027 22.4984 -0.01345 4 7 2.1624 2.29E+05 0.0165 0.6382 2.9843 2.7602 -1"12

10 1.3739 2.17E+05 0.0171 13.2518 25.0127 36.4123 -1.76 12 -1.6261 2.57E+05 0.0168 4.8717 -12.8824 15.8442 1.48 9 -3.2111 3.40E+05 0.0157 1.6189 -16.6920 10.3965 0.75

Summations: -1.5774 65.4132 -0.02411

12.11 (continued)

Chapter 12 208

Trial4 Circuit Pipe a (ft;s/s) NR f k h = kQà 2kQ ·Q %Chg

1 1 6.9314 5.83E+05 0.0145 0.1789 8.5969 2.4806 -0.15 4 2.2018 2.33E+05 0.0164 0.8489 4.1153 3.7381 -0.48 6 -4.1670 3.51E+05 0.0153 0.1895 -3.2899 1.5791 0.26 3 -8.5686 6.41E+05 0.0142 0.1298 -9.5290 2.2242 0.12

Summations: -0.1067 10.0219 -0.01065 2 2 4.7296 3.98E+05 0.0151 0.1866 4.1736 1.7649 -0.46

5 3.2296 2.72E+05 0.0158 0.2612 2.7242 1.6870 -0.68 7 -2.1684 2.29E+05 0.0165 0.6380 -2.9996 2.7667 1.01 4 -2.2018 2.33E+05 0.0164 0.8489 -4.1153 3.7381 0.99

Summations: -0.2171 9.9568 -0.02180 3 6 4.1670 3.51E+05 0.0153 0.1895 3.2899 1.5791 -0.34

9 3.2004 3.39E+05 0.0157 1.6195 16.5878 10.3661 -0.44 11 -1.4016 1.48E+05 0.0175 0.6801 -1.3361 1.9065 1.00 8 -4.4016 4.23E+05 0.0151 0.9733 -18.8571 8.5683 0.32

Summations: -0.3155 22.4199 -0.01407 4 7 2.1684 2.29E+05 0.0165 0.6380 2.9996 2.7667 -0.19

10 1.3980 2.21E+05 0.0171 13.2238 25.8437 36.9730 -0.29 12 -1.6020 2.53E+05 0.0168 4.8800 -12.5245 15.6358 0.26 9 -3.2004 3.39E+05 0.0157 1.6195 -16.5878 10.3661 0.13

Summations: -0.2689 65.7416 -0.00409 Trial 5 Circuit Pipe a (ft;s/s) NR f k h = kQà 2kQ ·Q %Chg

1 1 6.9421 5.84E+05 0.0145 0.1789 8.6220 2.4840 -0.15 4 2.1906 2.32E+05 0.0164 0.8495 4.0765 3.7217 -0.47 6 -4.1704 3.51E+05 0.0153 0.1895 -3.2950 1.5802 0.25 3 -8.5579 6.40E+05 0.0142 0.1298 -9.5065 2.2217 0.12

Summations: -0.1030 10.0076 -0.01029 2 2 4.7514 4.00E+05 0.0151 0.1865 4.2099 1.7721 -0.13

5 3.2514 2.74E+05 0.0158 0.2609 2.7586 1.6969 -0.19 7 -2.1506 2.27E+05 0.0165 0.6387 -2.9541 2.7472 0.29 4 -2.1906 2.32E+05 0.0164 0.8495 -4.0765 3.7217 0.29

Summations: -0.0621 9.9379 -0.00625 3 6 4.1704 3.51E+05 0.0153 0.1895 3.2950 1.5802 -0.08

9 3.2104 3.40E+05 0.0157 1.6189 16.6854 10.3946 -0.10 11 -1.3875 1.47E+05 0.0176 0.6812 -1.3115 1.8904 0.24 8 -4.3875 4.22E+05 0.0152 0.9737 -18.7434 8.5439 0.08

Summations: -0.0745 22.4091 -0.00332 4 7 2.1506 2.27E+05 0.0165 0.6387 2.9541 2.7472 -0.15

10 1.4021 2.22E+05 0.0171 13.2191 25.9860 37.0681 -0.23 12 -1.5979 2.53E+05 0.0168 4.8814 -12.4642 15.6004 0.20 9 -3.2104 3.40E+05 0.0157 1.6189 -16.6854 10.3946 0.10

Summations: -0.2095 65.8103 -0.00318

12.11 (continued)


Q1 = 350 L/min = 5.833 x 10Ŀ3 m3ls +-- Initial assumption Q2 = Q1 - 115 L/min = 235 L/min = 3.917 x 10Ŀ3 m31s .J, Continuity cond.@ Q3 = 880 L/min - Q1 = 530 L/min = 8.833 x 10-3 m31s .J, Cont. cond(D Q4 = 60 L/min = 1.000 < 10Ŀ3 m ' Is +-- Initial assumption Q5 = Q2 + Q4 - 200 L/min = 95 L/min = 1.583 x 10Ŀ3 m31s .J, Cont. cond.@ Q6 = Q3 - Q4 - 375 L/min = 95 L/min = 1.583 x 10Ŀ3 m31s .J, Cont. cond.@ Q7 = 115 L/min - Q6 = 20 L/min = 3.333 x 10Ŀ4 m31s ~ Cont. cond.@

Trial 1: Assumptions

Q2 + Q4 = Qs + 200 L/min = Qs + 3.333 x 10Ŀ3 m31s

V Q6 + Q1 = 115 L/min = 1.917 x 10Ŀ3 m31s

Y Qs = Q1 + 75 L/min = Q1 + 1.250 x 10-3 m31s

Computefusing Eq. 9.9. For continuity at the joints: G) Qá + Q3 = 880 L/min = } .467 X } 0"2 m3 Is @ Q1 = Q2 + 115 L/min

= Q2 + 1.917 x 10Ŀ3 m31s @Q3 = Q4 + Q6 + 375 L/min

= Q4 + Q6 + 6.250 X 10"3

Pi es 4 5 6 7: N = QDp = Q(0.022l)(920) = (2.651 x 107)Q p ' ' ' R Ará (3.835X10-4)(2.00X10-3)

Die= 0.0221/1.5 x 10-6 = 14733

Pi e 2: N = QDp = Q(0.0348)(920) =(1.683xl07)Q p R Ará (9.510xl0-4)(2.00xl0-3)

Die = 0.0348/1.5 X 1 o-6 = 23200

Fluid is a coolant: sg == 0.92, rá == 2.00 x 10-3 Pa-s Reynolds numbers and relative roughness Pi es l 3: N = uDp = QDp = Q(0.0475)(920) p ' R rá Ará (1. 772X10-3)(2.00X10-3)

M1 = (1.233 X 107)Q Die= 0.047511.5 x 10-6 = 31667

880

É@@(i)

L u2 JLQ2 jL 12.12 h=kQ2= !--= ;k=

D 2g D(2g)A2 D(2g}A2

k= !(7.5) 2=(2.56xl06)f ! (0.0475)(2)(9.81)(1.772X10-3)2

k= f(7.5) = (1.2} X 107)/ á (0.0348)(2)(9.81)(9.510 X 10-4)2

k= !(7.5) =(1.18xl08)f (0.0221)(2)(9.81)(3.835X10-4)2 "'--~~

Chapter 12 210

continued

Trial 1 Circuit Pipe a (m;j/s) NR f k h = kQà 2kQ ~a %Chg

1 1 -5.833E-03 7.19E+04 0.0193 4.95E+04 -1.6845 577.6 -1.99 2 -3.917E-03 6.59E+04 0.0197 2.40E+05 -3.6759 1876.9 -2.97 3 8.833E-03 1.09E+05 0.0177 4.55E+04 3.5492 803.6 1.32 4 1.000E-03 2.65E+04 0.0243 2.85E+06 2.8539 5707.8 11.63

Summations: 1.0427 8965.9 1.163E-04 2 4 -1.000E-03 2.65E+04 0.0243 2.85E+06 -2.8539 5707.8 13.36

5 -1.583E-03 4.20E+04 0.0219 2.57E+06 -6.4387 8134.8 8.44 6 1.583E-03 4.20E+04 0.0219 2.57E+06 6.4387 8134.8 -8.44 7 -3.333E-04 8.83E+03 0.0322 3.78E+06 -0.4201 2520.9 40.10

Summations: -3.2740 24498.2 -0.000134

NETWORK ANAL YSIS USING THE HARDY CROSS TECHNIQUE U.S. Cust Units Prob# 12.12 2 Circuits 7 Pipes...<Jrawn steel tubing Fluid: Coolant Fluid kinematic vlscositv: 2.17E-06 m2/s Pipe# D (m) Length (m) Roughness DIE C1 C2 lnitial a

E (rn) (k=C1*f) (NR=C2*Q) Umin 1 0.0475 7.5 1.50E-06 31667 2.56E+06 1.23E+07 350 2 0.0348 7.5 1.50E-06 23200 1.21E+07 1.68E+07 235 3 0.0475 7.5 1.50E-06 31667 2.56E+06 1.23E+07 530 4 0.0221 7.5 1.50E-06 14733 1.18E+08 2.65E+07 60 5 0.0221 7.5 1.50E-06 14733 1.18E+08 2.65E+07 95 6 0.0221 7.5 1.50E-06 14733 1.18E+08 2.65E+07 95 7 0.0221 7.5 1.50E-06 14733 1.18E+08 2.65E+07 20

211 PARALLEL PIPE LINE SYSTEMS

continued

Trial 2 Circuit Pipe Q (m"'/s) NR f k h=kQ;.: 2kQ L\Q %Chg

1 1 -5.949E-03 7.34E+04 0.0192 4.93E+04 -1.7451 586.67 0.92 2 -4.033E-03 6.79E+04 0.0196 2.38E+05 -3.8736 1920.79 1.36 3 8.717E-03 1.07E+05 0.0178 4.56E+04 3.4655 795.13 -0.63 4 7.501E-04 1.99E+04 0.0260 3.06E+06 1.7210 4588.97 -7.30

summanons: -0.4322 7891.55 -5.477E-05 2 4 -7.501E-04 1.988E+04 0.0260 s.ose-oe -1.7210 4588.97 -0.22

5 -1.449E-03 3.841 E+04 0.0223 2.62E+06 -5.5043 7595.56 -0.12 6 1.717E-03 4.549E+04 0.0215 2.52E+06 7.4382 8665.94 0.10 7 -1.997E-04 5.291E+03 0.0373 4.38E+06 -0.1747 1750.47 -0.84

Summations: 0.0381 22600.94 1.684E-06

Trial 3 Circuit Pipe Q (m"'/s) NR f k h=kQ;.: 2kQ L\Q % Chg

1 1 -5.895E-03 7.27E+04 0.0193 4.94E+04 -1.7165 582.39 0.08 2 -3.979E-03 6.70E+04 0.0197 2.39E+05 -3.7799 1900.15 0.11 3 8.771E-03 1.08E+05 0.0178 4.56E+04 3.5048 799.13 -0.05 4 8.065E-04 2.14E+04 0.0256 3.01 E+06 1.9547 4847.36 -0.56

Summations: -0.0368 8129.03 -4.531E-06 2 4 -8.065E-04 2.137E+04 0.0256 3.01 E+06 -1.9547 4847.36 1.21

5 -1.451 E-03 3.845E+04 0.0223 2.62E+06 -5.5157 7602.41 0.67 6 1.715E-03 4.545E+04 0.0215 2.52E+06 7.4252 8659.29 -0.57 7 -2.013E-04 5.336E+03 0.0372 4.37E+06 -0.1773 1760.80 4.83

Summat²ons: -0.2225 22869.86 -9.731E-06

12.12 (continued)

Chapter 12 212

Trial4 Circuit Pipe Q (m~/s) NR f k h=kQ;à 2kQ AQ %Chg

1 1 -5.890E-03 7.26E+04 0.0193 4.94E+04 -1.7141 582.04 0.10 2 -3.974E-03 6.69E+04 0.0197 2.39E+05 -3.7722 1898.44 0.14 3 8.776E-03 1.08E+05 0.0178 4.55E+04 3.5080 799.46 -0.06 4 8.013E-04 2.12E+04 0.0256 3.01 E+06 1.9327 4823.72 -0.70

Summations: -0.0456 8103.65 -5.625E-06 2 4 -8.013E-04 2.124E+04 0.0256 3.01 E+06 -1.9327 4823.72 0.25

5 -1.441 E-03 3.820E+04 0.0223 2.62E+06 -5.4502 7562.81 0.14 6 1.725E-03 4.571E+04 0.0215 2.52E+06 7.5004 8697.68 -0.12 7 -1.916E-04 5.078E+03 0.0378 4.44E+06 -0.1629 1700.83 1.04

Summations: -0.0454 22785.04 -1.992E-06

Trial 5 Circuit Pipe Q (m~/s) NR f k h = kQ;à 2kQ AQ %Chg

1 1 -5.884E-03 7.26E+04 0.0193 4.94E+04 -1.7112 581.60 0.03 2 -3.968E-03 6.68E+04 0.0197 2.39E+05 -3.7626 1896.31 0.04 3 8.782E-03 1.08E+05 0.0178 4.55E+04 3.5121 799.87 -0.02 4 8.050E-04 2.13E+04 0.0256 3.01 E+06 1.9481 4840.24 -0.21

Summations: -0.0136 8118.03 -1.680E-06 2 4 -8.050E-04 2.133E+04 0.0256 3.01E+06 -1.9481 4840.24 0.16

5 -1.439E-03 3.814E+04 0.0223 2.62E+06 -5.4368 7554.70 0.09 6 1.727E-03 4.576E+04 0.0214 2.52E+06 7.5158 8705.54 -0.07 7 -1.896E-04 5.025E+03 0.0379 4.45E+06 -0.1601 1688.47 0.67

Summations: -0.0291 22788.95 -1.278E-06 Final flows in Umin

Circuit Pipe Q 1 1 -353.1 Umin

2 -238.1 Umin 3 526.9 Umin 4 48.3 Umin

2 4 -48.3 Umin 5 -86.4 Umin 6 103.6 Umin 7 -11.4 Umin

12.12 (continued)

213 PUMP SELECTION AND APPLICATION

13 .25 Q = 280 gal/min; P = 26 hp; e= 53%; NPSHR = 10.9 ft

1 1-x3-6 2

1 1-x3-10 2

1 1 x3-6 2

l l L 6 in casing-size of largest impeller 3 in nominal suction connection size

1 1 in nominal discharge connection size

2

h = h (D2 )2 = h (ÜĿ75D, )2 = 0.5625h, : 44% reduction. Û2 Û1 DI Û1 D1 (á

13.24

13.23

13.22

13.21

13.20

13.19 Q2=Q1 D2 =Q, o.7SD1 =0.75Q1: 25%reduction. D, D,

h = h ( N2 )2 = h ( 0.5N, )2 = 0.25h ha divided by 4. Û2 Û1 N Û1 N Û1

1 1

13.17

13 .15 Affinity laws relate the manner in which capacity, head and power vary with either speed or impeller diameter.

13 .1 to 13 .14: Answers to questions in text.

CHAPTER THIRTEEN

PUMP SELECTION ANO APPLICATION

Chapter 13 214

N __ N.jQ. N-- N,H314 __ (5000)(40)¿.75 13.37 = 795 rpm s H314 Å .jQ .JlOOOO

13 .36 Q = 2750 gal/min; H = 200 ft; D = 15 in; N = 1780 rpm N = N .jQ = 1780.Ji750 = 1755. D, = DH114 = 15(200)0.zs = 1.0S s H314 (200)ÜĿ15 ' .JQ .J2150 Point in Figure 13.34 lies in radial flow centrifuga} region.

13.35 Q = 390 gal/min; H= 550 ft; D = 12 in; N= 3560 rpm _ N .JQ _ 3560J390 _ . _ DH114 _ (12)(550)ÜĿ25 _

Ns-~- 075 -619,Ds- In - r;:;;:;;:; -2.94 H (550) Ŀ .,áQ vt390

Point in Fig. 13.48 lies in radial flow centrifugal region.

13.34 a. Rotary or 3500 rpm centrifuga} b. Rotary c. Rotary d. Reciprocating e. Rotary or high speed centrifuga} f. 1750 rpm centrifuga! g. 1750 rpm centrifugal or mixed flow h. Axial flow

13.33 The same capacity is delivered but the head capability increases to the sum of the ratings of the two pumps.

13.32 Total capacity doubles.

13 .31 As fluid viscosity increases, capacity and efficiency decrease, power required in creases.

13 .30 Throttling val ves dissipate energy from fluid that was delivered by pump. When a lower speed is used to obtain a lower capacity, power required to drive pump decreases as the cube of the speed. Variable speed control is often more precise and it can be automatically controlled.

13.28 6 in 7 in 8 in 9 in 10 in

ha 120 ft 190 ft 250 ft 320 ft 390 ft

Q 145 gal/min 187 gal/min 220 gal/min 260 gal/min 290 gal/min

emax 51% 54% 56% 57.3% 58%(Est)

13.29 NPSHR increases.

Then Q2 = 125 gal/min; P2 = 19 hp; e2 = 45%; NPSHR = 5.5 ft (approximate values) 13 .27 From Problem 13 .26, h0 = 250 ft: Let h0 = 1.15 hÛ = 288 ft

1 2 1

13 .26 At ha = 250 ft, emax = 56%: Q = 220 gal/min; P = 24.0 hp; NPSHR = 8.0 ft

215 PUMP SELECTION ANO APPLICATION

13.53 NPSHa = hsp b, - há- hvp = 34.48 - 4.8 - 2.2 - 6.78 = 20.70 ft h = 14.7 lb X ft3 X 144in2 = Patm = 34.48 ft sp in ' 61.4 lb ft2 r hs = 4.8 ft; há= 2.2 ft; hvp = 6.78 ft (Table 13.2) Water at 140ÜF

Water at 180ÜF, hvp = 17.55 ft; r= 60.6 Ib/ft3 hsp = (14.4)(144)/(60.6) = 34.22 ft NPSH0 34.22 - 10 - 6 - 17.55 = 0.67 ft Cavitation will likely occur!

b. a. h = Patrn = 14.4lb ~ 144in2 = 33.34 ft sp r in 2 62.2 lb ft2 h, = 10.0 ft; há= 6.0 ft; hvp= 1.17 (Table 13.2) NPSH0 = 33.34-10- 6- 1.17=16.17 ft

13.52 NPSH0=hsp-hs-há-hvp: SomedatafromProb. 7.14.

13.51 (NPSHR)2 = (NPSHR)1 (N2 )2 = 7.50 ft(2850)2 = 4.97 ft N1 3500

13.50 Air pockets will not form in an eccentric reducer as they will in a concentric reducer.

13.49 Large pipe sizes reduce flow velocity and reduce energy losses, thus increasing NPSH0Å

13 .48 Elevating reservoir raises pressure at pump inlet and increases NPSH0Å

13.47 At inlet to pump. Pressure at this point and fluid properties affect pump operation, particularly the onset of cavitation. Pump manufacturer's NPSHR rating related to pump inlet.

13.43 to 13.46 See text.

b. Ns = 260 too low d. Ns = 18.5 too low f. Ns= 2943 radial h. Ns = 24277 axial

13.42 Same method as Problems 13.38 to 13.41. a. Ns = 1463 radial c. Ns= 3870 radial or mixed e. Ns= 104 too low g. Ns = 7260 mixed

_ N JQ (3500)"'5¦¿ _ . . Ns - ~= 075 -2475, Twice Ns from 13.40. H (100) Ŀ

13.41

13.40 N = NJQ = 0750)"'5¦¿ = 1237 s H314 (lOO)o.15

13.39 N = NjQ = 0750).Jl2000 = 2659 s H314 (300)o.15

13.38 N = NjQ = (l750)v'S¿o0 = 3913 s H314 (lOO)o.75

216 Chapter 13

13.57 Find NPSH0: Carbon tetrachloride at 65ÜC; sg = 1.48; Patm == 100.2 k:Pa; h, == -1.2 m; há= 0.72 m hvp = 4.8 m Open tank: hsp == Psply= Pa1mlr= (100.2 kN/m2)/[l.48(9.81kN/m3)]=6.90 m NPSHa = 6.90 - 1.2 - 0.72-4.8 == 0.18 m (Very low)

13.56 Find NPSH0: Carbon tetrachloride at 150ÜF; sg = l.48;pa1m == 14.55 psia; h, == -3.6 ft; há= 1.84 ft hvp == 16.3 ft Open tank: hsp = Psply= Patmlr== (14.55 lb/in2)(144 in2/ft2)/[l.48(62.4 lb/ft3)] = 22.69 ft NPSHÛ = 22.69 - 3.6 - 1.84 - 16.3 = 0.95 ft (Low)

For all problems 13.56 - 13.65: NPSHa = hsp Ñ h, - há- hvp See Section 13 .11, Equation 13-14. See Figure 13 .3 7 for vapor pressure head hvpĀ

NPSHa = 10.68 - 2.0- 0.280 - 4.97 = 3.43 m

há= 0.280 m

há= (0.0195) 2Ŀ0 m (0.056 m) + 1.35(0.056) + (0.018)(20)(0.056)

0.0779m

+ (0.0202) I.5 (0.271) 0.0525

Vz = __Q_ = 300160000 = 2.31 m; u~ = (2.31)2 = 0.271 m Az 2.168xl0-3 s 2g 2(9.81)

N = U2D2 = (2.31)(0.0525) = 3.4 X 105Ŀ D2 = 1141 Ŀ á: = 0.0202 R, v 3.60x10-1 ' e '12

u3 = 2_ = 300 L/min 1 m3 /s = 1.05 m; u; = (1.05)2 = 0.0560 m Ći 4.768xl0-3m260000L/min s 2g 2(9.81)

N = V3D3 = (1.05)(0.0779) =2.3x10s. D3 = 0.0779 1=1694Ŀ.f; = 0.0195 e, v 3.60xl0-7 ' e 4.6x10-5 ' 3

NOTE: hr= 0.018

K1 = 75/Jr= 75(0.018) = 1.35

há= J; (~) u; + K1 u; + hr (20) u; + /2 (~) u~ D 3 2g 2g 2g D 2 2g

Friction 3 in Foot valve Elbow Friction 2 in

13.55 NPSHa = hsp - h, - há- hvp

h - Paun _lOl.8.kN m2 -1068 i ¶ -20 Ŀh -48 sp - - - 2 - Å m, s - Å m, vp - Å m r m 9.53kN

13.54 NPSHa = hsp + h, - há- hvp = 11.47 + 2.6 - 0.80 - 1.55 = 11.72 m p 98.5 kN m' h = --3!!!!.. = = 11.47 m: h = 2.6 m: h = 0.80 m sp r m" 8.59 kN ' s ' f

h = Pup = 13.3kN m3 = 1.55 m op r m ' 8.59 kN

217 PUMP SELECTION ANO APPLICATION

13.65 Find required pressure above the fluid in a closed, pressurized tank so that NPSH0 2: 150.0 m. Propane at 45ÜC; sg = 0.48; Patm = 9.47 kPa absolute; hs = -1.84 m; há= 0.92 m hvp = 340 m Solve Eq. 13-14 for required hsp = NPSH0 - h, + há+ hvp = 1.50 + 1.84 + 0.92 + 340 = 344.3 m Psp = rhsp = (0.48)(9.81 kN/m3)(344.3 m) = 1621kN/m2=1621 kPa absolute Gage pressure: Ptank = Psp - Patm = 1621 kPa - 98.4 kPa = 1523 kPa gage

13.64 Find required pressure above the fluid in a closed, pressurized tank so that NPSH0 2: 4.0 ft. Propane at 1 lOÜF; sg = 0.48; Patm = 14.32 psia; h, = +2.50 ft; há= 0.73 ft hvp= 1080ft Solve Eq. 13-14 for required hsp = NPSH0 =h, + há+ hvp = 4.0 -2.5 + 0.73 + 1080 = 1082.2 ft Psp = rhsp = (0.48)(62.4 lb/ft3)(1082.2 ft)(l ft2/144 irr') = 225.1lb/in2=225.l psia Gage pressure: Ptank = Psp - Patm = 225.1 psia - 14.32 psia = 210.8 psig

13.63 FindNPSH0: Gasoline at40ÜC; sg = 0.65;p01m = 99.2 kPa; h, = +0.65 m; há= 1.18 m hvp = 14.0 m Open tank: hsp = Ps/r= Pa1,,/r= (99.2 kN/m2)/[0.65(9.81 kN/m3)) = 15.55 m NPSH0 = 15.55 + 0.65 - 1.18 - 14.0 = 1.02 m

13.62 Find NPSH0: Carbon tetrachloride at 65ÜC; sg = 1.48; Patm = 100.2 kPa; h, = + 1.2 m; há= 0.72 m hvp =4.8 m Open tank: hsp = Ps/r= Pa1mlr= (100.2 kN/m2)/[1.48(9.81kN/m3))=6.90 m NPSH0 = 6.90 + 1.2 - O. 72 - 4.8 = 2.58 m

13.61 Find NPSH0: Gasoline at 1 lOÜF; sg = 0.65;p01m = 14.28 psia; hs = -2.25 ft; há= 0.87 ft hvp = 51.0 ft Open tank: hsp = Ps/r= Paimlr= (14.28 lb/in2)(144 in2/ft2)/[0.65(62.4 lb/ft3)] = 50.70 ft NPSH0 = 50.70 - 2.25 - 0.87 - 51.0 = -3.42 ft (Cavitation expected)

13.60 Find NPSH0: Carbon tetrachloride at 150ÜF; sg = I.48;p01m = 14.55 psia; h, = +3.66 ft; há= 1.84 ft hvp = 16.3 ft Open tank: hsp =ps/r= Pa1mlr= (14.55 lb/in2)(144 in2/ft2)/[1.48(62.4 lb/ft2)] = 22.69 ft NPSH0 = 22.69 + 3.67 1.84 - 16.3 = 8.22 ft

13.59 Find NPSH0: Gasoline at 1 lOÜF; sg = 0.65;p01m = 14.28 psia; h, = +4.8 ft; há= 0.87 ft hvp = 51.0 ft Open tank: hsp = Ps/r= Paimlr= (14.28 lblin2)(144 in2/ft2)/[0.65(62.4 lb/ft3)] = 50.70 ft NPSH0 = 50.70 + 4.8 - 0.87 - 51.0 = 3.63 ft

13.58 Find NPSH0: Gasoline at 40ÜC; sg = 0.65;p01m = 99.2 kPa; hs = -2.7 m; há= 1.18 m hvp = 14.0 m Open tank: hsp = Ps/r= Paimlr= (99.2 kN/m2)/[0.65(9.81 kN/m3)] = 15.55 m NPSH0 = 15.55 - 2.7 - 1.18 - 14.0 = -2.33 m (Cavitation expected)

Chapter 14 218

14.7 DatafromProb.14.6. d=d2=3.50in

A= (4)(2) + _!_ (2)(2) + (6)(3.50-2.00) = 19.0 in2 2

WP = 3.50 + 4 + 2.828 + 1.50 = 11.828 in R = A/WP = 19.0 in2/11.828 in= 1.606 in

IÅ e .. á G __ -_________ - ilr-4~-L=~12

14.6 d = d, = 2.0 in; L = ,J22 + 22 = 2.828 in

A = ( 4)(2) + _!_ (2)(2) = 1 O in2 2

WP = 4 + 2 + 2.828 = 8.828 in R = AIWP = 1.133 in

14.5 W= 150 mm; d= 62 mm;X= l.5d= 1.5(62) = 93 mm L = ,J X2 + d2 = 111.8 mnr' R = _.i_ = Wd + Xd = (150)(62) + (93)(62) = 15066 mm2 ""40Å3 mm

WP W + 2L 150 + 2(111.8) 373.5 mm

14.4 Data from Prob. 14.3, but e= 45Á. X= dtan 45Ü = d= 1.50 ft L = d/cos 45Á = 1.50/cos 45Á = 2.121 ft A= (3.50)(1.50) + (1.50)(1.50) = 7.50 ft2 WP=3.50+2(2.121)=7.743 ft R = AIWP = 7.50 ft2/7.743 ft = 0.969 ft

WP = W + 2L = 3.50 + 2(1.732) = 6.964 ft R = AIWP = 6.549 ft2/6.964 ft = 0.940 ft

d = 1.50 ft; W = 3 .50 ft, x = d tan 30Ü = 0.866 ft L = d/cos 30Á = 1.732 ft

14.3

A Wd (2.75 m)(0.05 m) R = = = = 0.367 m WP W + 2d [2.75 + 2(0.50)]m

14.2

n_ti ~D/2 t

R _.i_=rcD2 x~=D =300mm =7Smm WP 8 çt: 4 4

14.1

CHAPTER FOURTEEN

OPEN CHANNEL FLOW

OPEN CHANNEL FLOW 219

14.13 A= (0.205)(0.250) = 0.05125 m'; WP = 0.205 + 2(0.250) = 0.705 m R = AIWP = 0.0727 m; n = 0.012; Q = 1.00 1 AR213S112

n

S= [~]2 =[ (0.012)(0.0833) ]2 = O.Ol25 AR213 (0.05125)(0.0727)213

. 1 m3/s where Q = 5000 L/mm x = 0.0833 m3/s

60000 L/min

1rD2 7r(6)2 S= 1ft/500ft=0.002; n = 0.024; A=--=--= 14.14 ft2 8 8 WP = nD/2 = n(6)/2 = 9.425 ft; R = AIWP = 1.50 ft

Q = I.49 AR213 Su2 = I.49 (14.14)(1.50)213 (0.002)112 = 51.4 ft3/s n 0.024

14.12

See Prob. 14.7. d2 = 3.50 in; A= 19.0 in2; WP = 11.828 in; R = 1.606 in A= 19.0 irr'(I ft2/144 in') = 0.132 ft2

WP = 11.828 in x _!!!__ = 0.986 ft 12in

R = l. 606 in x 1 ft = O .134 ft 12 in

h 4.0 in ft O 00 6 O 01 . S= -=--x--= . 55 ;n= . 3g1ven L 60 ft 12 in

Q = i.49 AR213 S112 = I.49 (0.132)(0.134)213 (0.00556)112 = 0.295 ft3/s n 0.013

14.11

1-- T 12.o l--3.50--j

14.10 A= (3.50)(2.0) = 7.00 m2; WP = 3.50 + 2(2.0) = 7.50 m R = A/WP = 0.933 m; n = 0.017 u= 1.00 R213 8112 = 1.00 (0.933)213(0.001)112 = l. 777 mis

n 0.017 Q =Av= (7.00 m2)(1.777 mis)= 12.44 m3/s Yh = AIT= 7.00 m2/3.50 m = 2.00 m

NF = u/ .ág;;; = l. 777 mis = 0.401 ~(9.81 mls2)(2.00 m)

14.9 A= (1.0)(2.50) + 2[~(1.9)(3.8) J = 9.72 m2 WP = 2(4.249) + 2(0.60) + 1.0 = 10.697 m

R = AIWP = 0.909 m

14.8 A= (1.0)(0.5) = 0.50 m2; WP = 1.0 + 2(0.5) + 2.0 m R = AIWP = 0.50/2.0 = 0.25 m

Chapter 14 220

By trial and error, y = 3.10 ft;

[ l>/3

AR213 = 10(3.1) l0(3.1) - = 47.78 10 + 2(3.1)

[ ]

2/3

AR213 = lOy lOy Findy such that AR213 = 47.75 10+2y

ICCb** 1 ~ r-1ott--1

L2 Å .á2f"+'22 Å 2.828

L,Å./41+ .. Å5.867 ~

J,~l j ==-: ~ĉT 2 10 4 l--1~ Å l-10~ á_ 2

14.16 AR213 = nQ = (0.015)(150) = 47.75 1.49S112 (1.49)(0.001)112

A = 101r WP = 10 + Zv; R = __::!__ = l O y .r s :.n WP 10 + 2y

b. Depth = 6.0 ft:

A= (4)(12) + 2[1(4)(4)]

+ (2)( 40) + 2 [ l (2)(2)]

A= 148 ft2 WP = 2(2.828) + 2(10) + 2(5.657)

+ 12 = 48.97 ft R = AIWP = 148/48.97 = 3.022 ft Q= l.49(148)(3.022)213(0.00015)112=141.1 ft3/s

0.04

WP = 12 + 2(4.243) = 20.485 ft R = AIWP = 45/20.485 = 2.197 ft

Q= t.49 (45)(2.197)213(0.00015)112 =34.7ft3/s 0.04

J 3 ,_ .. __ 12_ft __ ., 3 r_ Ć0 -:.- t/a.ott YL Å./32+31Å4.243ft

Depth = 3.0 ft:

A= (3)(12) + 2[1(3)(3)] = 45 ft2 14.15 a.

14.14 See Prob. 14.8 for d = 0.50 m; Prob. 14.9 for d = 2.50 m S = 0.50% = 0.005, n = 0.017 a. d= 0.50 m; A= 0.50 m2; R = 0.25 m

Q = LOO AR213 S112 = 1.00 (0.50)(0.25)213(0.005)112 = 0.825 m3/s n 0.017

b. d = 2.50 m; A= 9.72 m2; R = 0.909 m Q = 1.00 (9.72)(0.909)213(0.005t2 = 37.9 m3/s

0.017

221 OPEN CHANNEL FLOW

Find critica! depth for Q = 15 .89 m3 Is: Let NF = 1.0 = ~ (!) -iss,

A = 3y + Xy = 3y + (l .5y)(y) = 3y + 1.5y2 T= 3 + 2X= 3 + 2(1.5)y = 3 + 3y

From Eq.Ci) Q = ág;i or Q2 = gA or _!_ = g = 9Ŀ81 = 0.0388 ' A T A2 T A3 Q2 (15.89)2

B T _ 3+3y . T ut -3 - 3Å Findy such that -3 = 0.0388 A [3y + 1.5yz] A

By trial, y= 1.16 m =Ye= Critica! depth

Q 15.89m3/s 2018 / T idth u= - = . ms; opw1 t =T=3.0+2X=7.50m A 7.875 m2

A 7.875 m ' = l.05 m Hydraulic Depth = Yh = - T 7.50m

V 2.018 Fraude No. = NF = -- = = 0.629 ágy; ~(9.81)(1.05)

Q = 1.00 (7.875)(0.937)213(0.001)112 = 15.89 m3/s 0.015

WP = 3.0 + 2(2.704) = 8.408 m R = AIWP = 7.875/8.408 = 0.937 m

For y= 1.50 m,X= 2.25 m L = 1.803( 1.5) = 2. 704 m ._:I XÅ 1.5y l--3.0 m~ XÅ 1.5y 1/ A~(J)(l.5)+ 2[~(1.5)(2.25)] ~7.875 m2 ~ !/? r

"-;tLÅ./yZ+(1:5ij2 Å 1.803y s- ~Ja .W Ŀ o.oct1

14.19

_l'_-=-=_- __ w ., OAm AR213 = nQ = (0.013)(2.0) = 0_671 l.OOS112 (0.0015)112

A= 0.4W; WP = W + 0.8

R =~ = 0.4W : Then AR213 = 0.4W[ 0.4W ]213: By trial, W== 3.55 m WP W +0.8 W +0.8

14.18 S=0.075/50=0.0015

[

2 ]2/3 AR213 = (3y + y2) 3 y+ y : By trial,

3 + 2.828y for y= 1.69 m, AR213 = 8.02

AR213 = nQ = (0.017)(15) = 8_064 l .OOS112 (0.001)112

A= 3.0(y) + 2[l(y)(y) J = 3y + y2 WP=3.0+2L=3.0+ 2J2y

14.17

222 Chapter 14

14.22 A= l.OO(y); WP = 1.00 + 2y; R = f-1.00'"-1 1+2y -

1

-== ly l. y(m) A(m) R(m) y A R 1 g

~ 11:: A 0.10 0.10 0.0833 0.80 0.80 0.3077 ~r 0.20 0.20 0.1429 1.00 1.00 0.3333 1.1 .4

1.2 0.30 0.30 0.1875 1.50 1.50 0.3750 .a .2 0.40 0.40 0.2222 2.00 2.00 0.4000 .4 o

o o OA 0.8 1.2 1.8 2.0 0.50 0.50 0.250 y(m) 0.60 0.60 0.2727

QJ_ J '37p . 1 ft3 /s Q = 500 gal/mm x = 1.114 ft3/s

449 gal/min

AR213 = nQ = (0.013)(1.114) = 0307 1.498112 (l.49)(0.001)112

2 A nD2 2 D A=nD/8Ā WP=nD/2ĀR= -=----=- ' ' WP 8 JrD 4

AR213 = 7rD2 [D]213 = lrD2(D)213 = 7r(D)s13 = 0.156(D)813 = 0.307 8 4 8( 4)213 8(2.52)

[o 307]318 Then D = ziz:.': = 1.29 ft 0.156

14.21

By trial, h = 0.458 ft

Find h such thatAR213 = 0.0635

á--1.0tt-i 14.20 Each trough: . 1 ft3/s Q = 250 gal/mm = 0.557 ft3/s 449 gal/min

AR213 = nQ = (0.017)(0.557) = 0_0635 1.498112 (1.49)(0.01)112

1 A= (l.O)(h)l = h/2

WP = h + L = h + .J 1 + h2 R= A = h/2

WP h+.JI;h2

AR213 = !!_[ h/2 ]213 2 h + .J1 + h2


E o.28 a: 0.28 ,,; ::2 J 0.24 ..Sl 0..22 l 0.2

o

1~~-r\--+-~__,-t-~~+--t-~O.D18 .~-t-'IM-__,t---t---1-~~-r0.o18

el)

.75--+lf.--+-"""-'~_._--l--+--+-+ 0.014 ² a 1.5Q---l--Jo\.c---l----F=!9......C:::--:::::;~;....+-~~012

Prob. 14.25

R=AIWP s 0.2105 m 0.0162 0.2637 0.0120 0.2857 0.0108 0.2878 0.0107 0.2791 0.0111 0.2654 0.0119 0.2500 0.0129

T Prob. 14.26

0.667 m1 0.667 0.667 0.667 0.667 0.667 0.667

A=Wy W(m) y= 2/3W WP= W+2y 0.50 1.333 m 3.167 m 0.75 0.889 2.528 1.00 0.667 2.333 1.25 0.533 2.317 1.50 0.444 2.389 1.75 0.381 2.512 2.00 0.333 2.667

1 --- 1 y

r--w--1

14.25 and 14.26 Q 2.00 Q=Au= Wyu;y= -=-- Wu W(3.0)

WP = W + 2yĀ R = ___:!_ ' WP

s = [_!!!2_]2 = [(0.015)(2.00)]2 AR213 AR213

14.24 Q= 1.00 AR2138112.s=[_!!!2_]2 =[ (0.015)(2.00) ]2 =O.Ol16m3 n ' AR213 (0.666)(0.270)213

Data from Problem 14.23.

14.23 Q=Au= (0.80 m)(y)(u)

y = _2_ = 2.00 m3 Is = 0.833 m O.Su (0.80 m)(3.0 mis)

WP = 0.8 + 2y = 0.8 + 2(0.833) = 2.467 m; A= 0.8(y) = 0.8(0.833) = 0.666 m2

R = A = 0.666m2 = 0.270 m WP 2.467m

Chapter 14

6 10 14 18 22 26 Oepth, y (In)

o

224

; 10 50 ®

o 81------ ........ ---.----140 Å

~8 30 1 14 ~ 21--4-7"-;..q..--+---+---+--~10

- 1.0 e. ix= ,,; 0.8

~ 0.6 ,g ::s 0.4

i :e 0.2 o

4.325 11.147 21.476 35.745 54.380 65.466

0.3616 0.5412 0.7049 0.8605 1.0114 1.0856

3.803 5.005 6.206 7.408 8.610 9.211

1.375 2.708 4.375 6.375 8.708 10.000

0.500 0.833 1.167 1.500 1.833 2.000

6.00 10.00 14.00 18.00 22.00 24.00

R(ft) WP(ft) y(ft) y(in)

14.30 and 14.31 b = 2.00 ft; z = 1.50; S = 0.005; n = 0.017

A= (b + zy)y; WP = b + 2y.JJ + z2 ; R = AIWP; Q = I.49 AR213S112 n

14.29 Same as 14.28 except n = 0.010 - plastic:

Q = i.49 (7.5)(0.936)213(0.005)112 = 75.63 ft3/s 0.010

zÅ1.SO f Å 20.Q in X ²2t Å 1.667 ft

14.28 Data from Prob. 14.27 n = 0.017 - formed unfinished concrete.

Q = l .49 AR213 S112 = l .49 (7 .5)(0.936)213(0.005)112 = 44.49 ft3 Is n 0.017

"-......, /

~-:.. à_I 1 2.0ft 1

b

14.27 From Table 14.2: A = (b + zy)(y) = (2.00 + 1.5(1.667)](1.667) = 7.50 ft2 WP = b + 2y .J 1 + z2 = 2.0 + 2(1.667)..f~l +-1-.5-2 = 8.009 ft

R = AIWP = 7.50/8.009 = 0.936 ft


R = y/2 = 0.4767 ft/2 = 0.2384 ft

S= [ nQ ]2 -[ (0.015)(1.25) ]2 -[0.02768]2 -[ 0.02768 ]2 1.49 AR213 - (1.49)(0.4545)R213 - R213 - (0.2384)213

= 0.00519 A 2y2 u 2.75

yh= =- =y=0.4767ft:NF= --=-;::::==== T 2y JiY: ~(32.2)(0.4767)

= 0.702 < 1.0 Subcritical Triangle: y= JA= .Jo.4545 = 0.674 ft; R = 0.354y = 0.2387 ft

s = [ 0.02768 ]2 = 0.00518 (0.2387)213 A y2 y 2.75

Yh = T = 2y = 2 = 0.337 ft: NF = ~(32.2)(0.337) = 0.835 < 1.0 Subcritical

Q = 1.25 ft3 Is; u= 2. 75 ft/s; A = Q/ u= 0.4545 ft2 (All)

~0.4545 Rectangle: y= = = 0.4767 ft; b = 2y = 0.9535 ft 2

14.36, 14.37, 14.38

14.35 S = 0.0012; n = 0.013; Data from Prob. 14.33: Ü = 1.00 AR213 Sl/2 = 1.00 (0.0358)(0.0742)213(0.0012)112 = 1.68 X 10-2 m3/s 0.013 0.013

14.34 S = 0.0012; n = 0.013; Data from Prob. 14.32: Q = 1.00 AR213 A112 = 1.00 (0.06919)(0.104)213(0.0012)112 = 4.08 x 10-2 m3!s

0.013 0.013

14.33 D = 0.375 m; y= .135 m < D/2---+ See Table 14.2 .

. -1 . -1[1- 2(0.135)] O= n - 2 sm [1 - 2y/D] = n - 2 sm 0.375 = 2.574 rad

A= (B-sin B)D2 = (2.574-sin 2.574)(0.375)2 = 0.0358 m2

8 8 WP = OD/2 = (2.574)(0.375 m)/2 = 0.482 m R= B-sin D =[2.574-sin2.574]0.375m =0.0742m

B 4 2.574 4

14.32 D = 0.375 m; y= 0.225 m > D/2---+ See Table 14.2. O= n + 2 sin-1[(2y/D) - l] = 3.544 rad A= (B- sin B)D2 = (3.544-sin 3.544)(0.375)2 m2 = 0.06919 m2

8 8 WP = OD/2 = (3.544)(0.375 m)/2 = 0.665 m R= B-sinB D =[3.544-sin3.544]0.375m =O.l04m

B 4 3.544 4

Chapter 14 226

u= Q = Q =~Ŀ N =_!!__ A by 2.0y' F fiY

For y= 0.5 m, u= 5.50 mis; NF = 2.48 For y= 1.934 m, v = 1.418 m/s; NF = 1.325

e.

d.

See spreadsheet and graph for values of y versus E. 5 52 For y= 0.50 m: E= 0.50 + ---Ŀ---

' 2(9.81)(2.0)2(0.5)2

=2.042 m From spreadsheet, altemate depth = 1.934 m

c.

Emin = 1.375 m

b. MinimumE occurs wheny =Ye; From Eq. 14.18: E . =y + _Jĉ_ _ + Q2 _ + Q2 mm e 2 A2 - Y e 2 (b )2 - Ye 2 b2 2 g g '.Ye g Ye

= 0.917 + 5Ŀ52 2(9.81)(2.0)2(.917)2

14.39 a. Wheny=yc,NF=l.O= ~= Üe= Üágy= Jf Ā -isè, A-vgy (by) gy b g y312

Theny= [ Q ]213 =[ 5Ŀ5 ]213 =0.917m=y bjgNF 2.0"'9.81(1.0) e

A n y' n y Yh = -=--=- = 0.4225 ft

T 2(2y) 4

N - 2Ŀ75 = 0.746 < 1.0 Subcritical F - ~(32.2)(0.4225)

S . . A ny2 á:A 2(0.4545) O 5379 f em²c²rcle: =-;y=-= =. t 2 re rc

[

1

0.02768 " R = y/2 = 0.269 ft: S = 213] = 0.00441 (0.269)

T . Jd: ~0.4545 12 o c, rapezo²d: y= - = = 0.5126 ft; R =y = .2563 tt 1.73 1.73

s = [ 0.02768 ]2 = 0.00471 Ŀ (0.2563)213 '

A A 0.4545 = 0.3841 ft Yh = T = 2.309y 2.309(0.5126)

Nè= ~ 2Ŀ75 = 0.782 < 1.0 Subcritical (32.2)(0.3841)

227

5 4 3

E(m) 2

Y= 1.934 m Alternate depth Y= 0.05 m

2

4

5

t)

Equation for E (in m) as a function of y: A = by; A2 = b2y2 Given: b = 2.0 m; Q = 5.5 m3/s; v = Q/A

E= y+ v2/2g =y+ Q212gA2 = Q2/2gb2y2 E= y= (5.5)2 / [3(9.81)(2.0)2y2] =y+ 0.3854/y2

S = [_5L_2] ĀA= b11 = 2.011 AR213 ' .r .r

WP = b + 2y = 2 + 2y R = ___:!_ = 2.0 y = -2:'._

WP 2+2y 1+ y

e)

NF R S 2.483 0.333 0.0378 Giveny 0.325 0.660 0.0010 Alternate depth for given depth

by iteration on y to make E the same as for the given y

Velocity 5.500 1.418

Critical depth-See Solution Manual

Q = 5.5 n = 0.017 Only for this problem

OPEN CHANNEL FLOW

E 154.2 38.6 9.835 2.809 1.671 1.402 1.385 1.468 1.597 1.751 1.919 2.096 2.280 2.467 2.657 2.849 3.043 3.238 3.433 3.630 3.827 4.024 4.222 4.420 4.618

y 0.5 0.1 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.20 3.40 3.60 3.80 4.00 4.20 4.40 4.60

d), e), f) Giveny = E A 0.50 2.04160 1.000 1.9391 2.04160 3.878

Rectangular channel b=2

E= y+ 0.3854/y2 Y Emin= 0.0917 1.375

14.39 (continued)

Chapter 14 228

E(m) e) Velocity = v = Q/A, NF= vi JiY: = Q!( AJii:). See spreadsheet

For y= 0.50 m: v = 3.251 mis, NF = 1.690. Supercritical For y= 0.913 m: v = 1.571 mis, NF = 0.528. Subcritical.

0.5 1.0 1.5 2.0 2.5

-.1.0 E ~ 0.5

Specific energy for y= 0.50 m: E= 1.039 m from spreadsheet. Iterate on y to find altemate depth for which E= 0.1039 m. See spreadsheet: Yait = 0.913 m.

d)

1.5 Specific energy versus y: See spreadsheet using equation in b).

e)

a) For given Q, D, andy: Compute B, A, Tusing equations in Table 14.2. Compute NF = v / JiY: = Q( AJii:). Iterate values of y until NF = 1.000. See spreadsheet: Ye= 0.658 m.

b) M²nimum specific energy: E= y+ v2/2g =y+ Q2/(2gA2) From spreadsheet, withy =Ye= 0.658 m: Emin = 0.924 m.

14.40 Circular Channel Q= 1.45 m3/s n= 0.015 Finished concrete D= 1.20m

y(m) ~rad) A(m2) T(m) y,,(m) NF E(m) Velocity y less than D (mis) 0.10 1.171 0.0450 0.6633 0.068 39.478 52.982 32.211 0.20 1.682 0.1239 0.8944 0.139 10.039 7.181 11.703 0.25 1.896 0.1707 0.9747 0.175 6.481 3.928 8.494 0.30 2.094 0.2211 1.0392 0.213 4.539 2.492 6.558 0.40 2.462 0.3300 1.1314 0.292 2.597 1.384 4.394 0.50 2.807 0.4460 1.1832 0.377 1.690 1.039 3.251 Giveny 0.60 3.142 0.5655 1.2000 0.471 1.193 0.935 2.564 y greater than D 0.658 3.335 0.6349 1.1944 0.532 1.000 0.924 2.284 Critical depth 0.70 3.476 0.6849 1.1832 0.579 0.888 0.928 2.117 0.80 3.821 0.8010 1.1314 0.708 0.687 0.967 1.810 0.90 4.189 0.9099 1.0392 0.876 0.544 1.029 1.594 0.913 4.238 0.9231 1.0240 0.901 0.528 1.039 1.571 Alt depth for given y 1.00 4.601 1.0071 0.8944 1.126 0.433 1.106 1.440 1.199 6.168 1.1309 0.0693 16.330 0.101 1.283 1.282 Nearly full pipe depth 1.20 6.283 1.1310 0.0000 #DIV/O! #DIV/O! 1.284 1.282 Full pipe

(yh and NF undefined)

Part f of problem: Slopes for given y and alternate depth

y(m) R(m) s 0.50 0.2649 0.0140 S for giveny 0.913 0.3630 0.0021 S for altemate depth

Problem 14.40 Procedure: Refer to Table 14.2 for geometry of a partially full circular pipe.


e) Specific energy versus y: See spreadsheet using equation in b ). E(ft)

d) Specific energy for y= 0.25 ft: E= 1.067 ft from spreadsheet. Itera te on y to find alterna te depth for which E = O .1067 ft. See spreadsheet: Yau = 1.065 ft.

0.5 1.0 1.5 2.0 2.5

- .::= -s.:: 0.5

1

:/r= o.zs tt

1.0 Iterate values of y until NF = 1.000. See spreadsheet: Ye= 0.418 ft.

b) M²nimum specific energy: E= y+ v2/2g =y+ Q2/(2gA2)

From spreadsheet, withy =Ye= 0.418 ft: Emin = 0.523 ft.

Problem 14.41 Procedure: Refer to Table 14.2 for geometry of a triangular channel.

a) For given Q, z, and y: Compute A, Tusing equations in Table 14.2. Compute NF = v / JiY: = Q I ( AJiY:) . 1.5

14.41 Triangular channel z= 1.5 n = 0.022 Q= 0.68 ft3/s

y(ft) A(ft2) V(ft/s) T(ft) Yh(ft) NF E(ft) 0.20 0.060 0.60 0.100 6.316 2.194 0.25 0.094 7.253 0.75 0.125 3.615 1.067 Giveny 0.30 0.135 0.90 0.150 2.292 0.694 0.40 0.240 1.20 0.200 1.116 0.525 0.418 0.262 2.594 1.254 0.209 1.000 0.523 Critical depth 0.50 0.375 1.50 0.250 0.639 0.551 0.60 0.540 1.80 0.300 0.405 0.625 0.70 0.735 2.10 0.350 0.276 0.713 0.80 0.960 2.40 0.400 0.197 0.808 0.90 1.215 2.70 0.450 0.147 0.905 1.00 1.500 3.00 0.500 0.113 1.003 1.065 1.701 0.400 3.20 0.533 0.097 1.067 Alternate depth 1.10 1.815 3.30 0.550 0.089 1.102 1.20 2.160 3.60 0.600 0.072 1.202 1.30 2.535 3.90 0.650 0.059 1.301 1.40 2.940 4.20 0.700 0.049 1.401 1.50 3.375 4.50 0.750 0.041 1.501

Slopes at given depth and alternate depth y(ft) R(ft) s 0.25 0.10401 0.521 Slope for given depth 1.065 0.44307 0.000229 Slope for altemate depth

f) Compute WP = BD/2 (See Table 14.2). Compute R = AIWP.

[ nQ ]2 Compute S: AR213

See spreadsheet: For y== 0.50 m, S =O.O 140. For y= 0.913 m, S = 0.0021.

Chapter 14 230

14.42 Trapezoidal channel z= 0.75 n = 0.013 Q= 0.80 ft3/s b = 3.000 ft

y(ft) A(ft2) V(ft/s) T(ft) Yh(ft) NF E(ft) 0.05 0.152 5.267 3.08 0.049 4.177 0.481 Giveny 0.1 0.308 2.602 3.15 0.098 1.467 0.205 0.1288 0.399 2.006 3.19 0.125 1.000 0.191 Critical depth 0.20 0.630 1.270 3.30 0.191 0.512 0.225 0.25 0.797 1.004 3.38 0.236 0.364 0.266 0.30 0.968 0.827 3.45 0.280 0.275 0.311 0.40 1.320 0.606 3.60 0.367 0.176 0.406 0.4770 1.602 0.499 3.72 0.431 0.134 0.481 Alternate depth 0.50 1.688 0.474 3.75 0.450 0.125 0.503 0.60 2.070 0.386 3.90 0.531 0.093 0.602 0.70 2.468 0.324 4.05 0.609 0.073 0.702 0.80 2.880 0.278 4.20 0.686 0.059 0.801 0.90 3.308 0.242 4.35 0.760 0.049 0.901 1.00 3.750 0.213 4.50 0.833 0.041 1.001 1.065 4.046 0.198 4.60 0.880 0.037 1.066 1.10 4.208 0.190 4.65 0.905 0.035 1.101 1.20 4.680 0.171 4.80 0.975 0.031 1.200 1.30 5.168 0.155 4.95 1.044 0.027 1.300 1.40 5.670 0.141 5.10 1.112 0.024 1.400 1.50 6.188 0.129 5.25 1.179 0.021 1.500

Slopes at given depth and alternate depth y(ft) R(ft) s 0.05 0.0486 0.264 Slope for given depth 0.477 0.3820 0.000152 Slope for altemate depth

Problem 14.42 Procedure: Refer to Table 14.2 for geometry of a trapezoidal channel.

a) For given Q, b, z, andy: Compute A, Tusing equations in Table 14.2. Compute NF = v / ágy;, = Q I ( Aágy;,) .

Iterate values of y until NF = 1.000. See spreadsheet: Ye= 0.1288 ft.

b) M²nimum specific energy: E= y+ v2/2g =y+ Q21(2gA2)

From spreadsheet, withy =Ye= 0.1288 ft: Emin = 0.191 ft.

e) Specific energy versus y: See spreadsheet using equation in b ).

[ ]

2 . nQ

Compute S. AR213

See spreadsheet: For y= 0.25 ft, S = 0.0521. For y= 1.065 ft, S = 0.000229.

e) Velocity = v = QI A, Np = v I ágy;, = Q I ( Aágy;,) . See spreadsheet

For y= 0.25 ft: v = 7.253 ft/s, NF = 3.615. Supercritical For y= 1.065 ft: v = 0.400 ft/s, NF = 0.097. Subcritical.

f) Compute WP = 2y.áá-;-;z (See Table 14.2). Compute R = AIWP.


14.47 (a) Q = (3.27 + 0.4HIHc)LH312 = [3.27 + 0.4(1.5/4)](3)(1.5)312 = 18.8 ft3/sec (b) Q = (3.27 + 0.4HIHc)(L - 0.2H)H312 = (3.42)(2.70)(1.5)312 = 16.95 ft3/sec (e) Q = 2.48H512 = (2.48)(1.5)512 = 6.84 ft3/sec

2 4 8 8 10 12 H(lnches)

J. 1 14.44 H = 1.5 ft; He= 3 ft; Q = 15 ft3/sec; use Eq. (14.15)

Q = [3.27 + 0.40(1.5/3.0)][L - 0.2(1.5)](1.5)312

= (3.47)(L - 0.3)(1.838)

L-0.3 = Q 15

= 2.36 ft = (3.47)(1.838) (3.47)(1.838)

L = 2.36 ft + 0.30 ft = 2.66 ft

14.45 (Q = 3.27 + 0.40HIHc)LH312

He = 2 ft; L = 6 ft

Prob. (15.10) (15.11) H(in) H(ft) Q(ft3/sec) Q i o o 0.00 0.00 "' 1 Pi""

2 .167 1.35 1.34 s. o

4 .333 3.84 3.80 6 .500 7.14 7.03 8 .667 11.10 10.90 10 .833 15.70 15.48 12 1.000 20.80 20.40

14.46 Q = (3.27 + 0.40HIHc(L 0.2H)H312 J Negligible difference on graph.

14.43 H,nax = 6 in/tan 30Á = 10.4 in x ft/12 in= 0.867 ft Qmax = 1.43 H,~~ = (1.43)(0.867)512 = 1.00 ft3/sec

[ nQ ]2 Compute S: AR213

See spreadsheet: For y= 0.05 ft, S = 0.264. For y= 0.4770 ft, S = 0.000152. 1r1 ----

E(ft) 2.5 2.0 1.0 1.5 0.5

1.0

~ 0.5

1.5 d) Specific energy for y= 0.05 ft: E= 0.481 ft

from spreadsheet. Iterate on y to find alterna te depth for which E= 0.481 ft. See spreadsheet: Yau = 0.4770 ft.

e) Velocity = v = Q/ A, NF = v / ágy;, = Q / ( Aágy;,) . 0 lz:::.-::::::!;:=:::=::::;á:::==;:==;:==.... See spreadsheet For y= 0.05 ft: v = 5.267 ft/s, NF = 4.177. Supercritical For y= 0.4770 ft: v = 0.499 ft/s, NF = 0.134. Subcritical.

f) Compute WP = b + 2y~ (See Table 14.2). Compute R = AIWP.

Chapter 14 232

14.52 Trapezoidal channel-Long-throated flume - Design C: H = 0.84 ft; Q = K1(H + K2)" K1 = 16.180; K2 = 0.035; n = 1.784 Q = 16.180[0.84 + 0.035)1-784 = 12.75 ft3/s = Q

b) L=IO.Oft Q = (3.6875L + 2.5)H1.6 = 39.375H1.6

( Q )1/1.6 ( 50 )0.625

H= = = 1.155 ft 39.375 39.375

14.51 a) Q = 50 ft3/s; L = 4.0 ft; Q = 4.00 LH'; n = 1.58 H= [Q/(4.00)L]11Û = [50/(4.00)(4.0)]111.58 = [3.125]Á'633 = 2.06 ft

H(ft) Q(ft /sec) 150 0.25 3.434 - (/)

1.00 32.000 à;:á- 100 ~ -

1.50 61.469 o 50 2.00 97.681 2.25 118.077 o 2.50 139.905 0.5 1.0 1.5 2.0 2.5

H(ft)

14.50 L = 8.0 ft; Qmin = 3.5 ft3/s; Qmax = 139.5 ft3/s Q = 4.00 LH'; H = [Q/4.00)L)11n = [Q/(4.00)(8.0)]v1.61 = [Q/32]ÁĿ621 Qmin = 3.5 ft3/s; H= [3.5/32]0Ŀ621 = 0.253 ft Qmax = 139.5 ft3/s; H= [139.5/32]Á'621=2.496 ft

2 4 a a 10 12 HQnches)

14.49 Q = 3.07H153

H1.53 = Q/3.07 :. H = (Q/3.07)111.53

Min Q = 0.09 ft3/sec H = (0.09/3.07)11153 = (0.0293)ÜĿ654 = 0.10 ft

Max Q = 8.9 ft3/sec H = (8.9/3.07)111.53 = (2.90)0Ŀ654

= 2.01 ft

2.0

!1.s ... s. o 1.0

2.5 1 1 1

J V 1/ 1/

!___,+/ .5

14.48 Q = 2.48H512

H(in) H(ft) Q(ft~/sec) o o o 2 .167 .0283 4 .333 .159 6 .500 .439 8 .667 .900 10 .833 1.57 12 1.000 2.48


39.06 121.7 238.4 385.7

0.087 0.271 0.531 0.859

0.10 0.20 0.30 0.40

Q(gpm) Q(ft /s) H

14.60 Selecta long-throated flume for 30 gpm < Q < 500 gpm. Using 449 gpm = 1.0 ft3/s, 0.0668 ft3 /s < Q < 1.114 ft3/s; Select Circular channel; Design A; Q = D25 K1(H/D + K2)" H = D{[Q/D2'5)(K1)]u" - K2}; D = 1.000 ft; K1 = 3.970; K2 = 0.004; n = 1.689 For Q = 0.0668 ft3/s: H= -l.0{[0.0668/(1.02'5)(3.970)]1ii.689 - 0.004} = 0.0851 ft =H For Q = 1.114 ft3/s: H = -l.O{[l.114/(1.02'5)(3.970)t1.689 - 0.004} = 0.467 ft = H

14.59 Circular channel=-Long-throated flume - Design C: Q = 6.80 ft3/s; Find H. Q = D2Ā5 K1(HID + K2)1'; D = 3.000 ft; K1 = 3.507; K2 = 0.000; n = 1.573 Solving for H: H = D{[Q/D2Ŀ5)(K1)]11" - K2} = 3.0{[6.80/(3.02'5)(3.507)]111.573 - 0.00} = 0.797 ft =H

14.58 Rectangular channel=-Long-throated flume - Design B: Q = 1.25 ft3/s; Find H. Q =bà K1(H + K2)"; be= 1.00 ft; K1 = 3.696; K2 = 0.004; n = 1.617 Solving for H: H = [Ql(beK1)]lln - K2 = [1.25/(1.0)(3.696)]111.617 - 0.004 = 0.507 ft = H

14.57 Circular channel=-Long-throated flume - Design A: H= 0.09 ft; Q = D2Ŀ5 K1 (HID + K2)" be= 1.00 ft; K1 = 3.970; K2 = 0.004; n = 1.689 Q = (1.00)25(3.970)[0.09/1.00 + 0.004)1-689 = 0.0732 ft3/s = Q

14.56 Circular channel=-Long-throated flume - Design B: H = 0.25 ft; Q = D2'5 K1 (HID + K2)" D = 2.00 ft; K1 = 3.780; K2 = 0.000; n = 1.625 Q = (2.00)2'5(3.780)[0.25/2.00 + 0.000)1-625 = 0.729 ft3/s = Q

14.55 Rectangular channel=-Long-throated flume - Design C: H = 0.40 ft; Q = bcK1(H + K2)" be= 1.500 ft; K1 = 3.375; K2 = 0.011; n = 1.625 Q = (1.500)(3.375)[0.40 + 0.01l]i.625=1.194 ft3/s = Q

14.54 Rectangular channel=-Long-throated flume - Design A: H = 0.35 ft; Q = bcK1(H + K2)" be= 0.500 ft; K1 = 3.996; K2 = 0.000; n = 1.612 Q = (0.500)(3.996)[0.35 + 0.000]1'612 = 0.368 ft3/s = Q

14.53 Trapezoidal channel-Long-throated flume - Design B: H = 0.65 ft; Q = K1(H + K2)" K1 = 14.510; K2 = 0.053; n = 1.855 Q = 14.510[0.65 + 0.053)1-855 = 7.547 ft3/s = Q

234 Chapter 14

H(m) H(ft) Q(ft3is) Q(m3/h) 0.100 0.328 0.622 63.38 0.125 0.410 0.888 90.49 0.150 0.492 1.190 121.3 0.175 0.524 1.524 155.3

14.61 Given 50 m3/h < Q, 180 m3/h; Convert to ft3/s; 0.4907 ft3/h < Q < 1.766 ft3/h Specify Rectangular channel long-throated flume, Design B. Find H for each limiting flow rate. Q =b; K1(H + K2t; he= 1.00 ft; K1 = 3.696; K2 = 0.004; n = 1.617 Solving for H: Hmin = [Q/(beK1)]11n -K2 = (0.4907/(1.0)(3.696))111.617 - 0.004 = 0.2829 ft = Hmin Converting to m: Hmin = 0.0863 m for Q = 50 m3/h Hmax = [Q/beK1)]11n -K2 = [1.766/(1.0)(3.696)1/1.617 - 0.004 = 0.629 ft = Hmax Converting to m: Hmax = 0.1917mforQ=180 m3/h

235 FLOW MEASUREMENT

15.3 A, = 0.02333 ft'; A2 = 0.00545 ft2; AzA, = 0.234 D = 2.067 in; d/D = 1.0012.067 in = 0.484; try C = 0.605 (Fig. 15.7)

Q = CA1[2g(p, - Pz)lr]'" ~(0.60SX0.02333.)[ 2(32.2)(0.53)(144)151.2 ]112 ( A1 I A2 )2 -1 L Assumed [(0.02333)1(0.00545)]2 -1

= 0.0332 ft31sec u = _SI= 0.0332 ft3 lsec = 1 .42 ft/sec 1

A, 0.02333 ft2

N = V1D1P = (1.42)(0.1723)(1.60) = 1.14 X ] 04 R Õ 3.43 X ]0-5

Then C = 0.612 (Fig. 15.7)-Recompute Q Q = 0.0332 ft3lsec x 0.61210.605 = 0.0336 ft31sec V= Q/A = 1.44 ftlsec; Náá = 1.16 X 104; No change in e

Check C from Fig. 15.5. Q 6.96x10-2 rrr' Is u= -= = 8.87 mis A, 7.854X10-3

N = váD, = (8Ŀ87)(0.lOO) = 6.82 X 104} Then e= 0.98, OK R V 1.3 X 10-5

15.2 p1 - p2 = Ywhw = (9.81 kNlm3)(0.081 m) = 0.795 kNlm2

p1 - p2 = 0.795 kNlm2 x 103 N = 62_57 m y air 12.7 Nlm3 1 kN

Assume C = 0.98

Q = CA,á2gh[ym1Yw -1)]'" ~(0.9S)(7.854XIO~')[ 2(9.81)(0.81)1(9810/12.7-1] ]112 (A/A2)2 -1 L Assumed [(7.854xl0-3)1(1.964xl0-3)]2-I

Q = 6.96 x 10-2 m3ls

_ Q _ 2.12xl0-2 m3ls _ 2 70 1 V1 - -- - Å m S

A1 7.854x10-3 m '

N = váDá = <2-7o)(O.lO) = 7.5 X 105 Then e= 0.984Ŀ, OK R V 3.60X10-?

15.1 Q = CA1[2g(p1 - P;)lr]'" ~(0.984X7.854x!O'')[ 2(9~~1)(55)19.53 -3 z ]112 (A1IA2) -1 LAssumed [(7.854xl0 )l(l.964x10 )] -1

Q = 2.12 x 10-2 m3ls= 0.0212 m31s Now check Nu and re-evaluate C.

CHAPTER FIFTEEN

FLOW MEASUREMENT

236 Chapter 15

15.5 Find 11p across nozzle, Q = 1800 gal/rnin(l ft3/s)(449 gal/min) = 4.009 ft3/s v1 = QIA1 = (4.009 ft3is)/(0.1810 ft2) = 22.15 ft/s Water at 120ÜF; r= 61.7 lb/fr'; V= 5.94 X 10-6 ft2!s NR = VáDá/V = (22.15)(0.4801)(5.94 X 10-6) = 1.79 X 106; Then e= 0.992 From Eq. (15-4), solving for p1 - P2=11p 11p = [v1/C]2 [(A1/A2)2 - 1] [)1'2g] = [22.15/0.992]2 [(0.18/0.0668)2 - 1][61.7/(2(32.2))] Sp = 3029 lb/ft2 (1 ft2/144 in") = 21.04 psi

An orifice size between 1.0 in and 7 .O in would be preferred to give a more convenient mano meter deflection.

h = P1 - P2 = 0.0694 lb/ft2 = 0.00655 ft 10.6 10.6 lb/ft'

= o .0694 lb/ft2

h = Pi - Pz But YA = (0.83)(62.4 lb/fr') = 51.8 lb/ft3 Ammon r; YA

h = Pi - P2 = Pi -p2 = 238 lb/ft2 =22.4 ft 62.4-51.8 10.6 lb/ft3 10.6 lb/ft3

(b) For d = 7 .O in, A2 = 7i(7.0)2 in2 x ft2 2 = 0.267 fr'

4 144in A1/A2 =0.545/0.267 =2.04

_ yQ2[(Ai I Az)2 -1] _ (0.83)(62.4)(0.0557)2[(2.04)2 -1] Pi - P2 - 2g C2 A12 - 2(32.2)(0.620)2 (0.545)2

Solve for p, - p2 from Eq. (15.5)

Q = CA [2g(pá - P2)/ r]l/2; Then Q2 = c2 Aá2 [2g(pá - P2)! '. 1 (A1/Az)2-l (A1/Az)2-l _ rQ2[(A1 I Az)2 -1]

P1 - P2 - 2g e: At (a) Ford =l.Oin,A2 =0.00545ft2;A1/A2 =0.545/0.00545 =100

_ yQ2[(A; Az)2 -1] _ (0.83)(62.4)(0.0557)2[(100)2 -1] = 238 lb/ff Pi P2 - 2g C2 A12 - 2(32.2)(0.595)2(0.545)2

For Manometer P1 +y Ah -yjz = P2 P1 - P2 =yjz -yAh =h(yw-YA)

15.4 Q =25 gal/min x 1 ft3/sec/449 gal/min =0.0557 ft3/sec u = Q!A =0.0557 ft3/sec/0.545 ft2 =0.1022 ft/sec pipe NR = vD p = (0.1022)(0.833)(0.83)(1.94) = 5 _5 x 104

Õ 2.5X10-6

(a) d!D = 1.0/10.0 = 0.10; e = 0.595 (Fig. 15.7) (b) d!D =1.0110.0 =0.70; e =0.620


15.9 Flow nozzle-Design: Install in 5 1/2 inch Type K copper tube; D1 = 4.805 in; A1 = 0.1259 ft2 Specify the throat diameter d. NOTE: Mu/tiple solutions possible. Fluid: Linseed oil at 77ÜF; r= 58.0 lb/²r'; v= 3.84 X 10-4 ft2/s U se mercury manometer with scale range O - 8 inHg Range of flow rate: Q111á0 = 700 gpm = 1.559 ft3/s; Qmax = 1000 gpm = 2.227 ft3/s Velocity in pipe: V1-min = Qmá0/A1 = 12.38 ft/s; Vi-max = QmaJA1 = 17.69 ft/s Reynolds No.: NR-min = V1-minD1lv= 1.29 X 104; NR-max = V1-maxD1/v= 1.84 X 104 From Figure 15-5: Cmin = 0.955; Cmax = 0.961

Similarly, V1max = 0.730 mis; NR = 1.06 X 103; e= 0.610 hmax = 0.1754m=175.4 mm

h . = [(A1 / Ćz)2 - l]v1~in = [(6.26)2 -1](0.292 mls)2 = 0.0273 m = 27.3 m mm 2gC2[(y"'/yÛ)-1] 2(9.81mls2)(0.619)2[(132.8/7.87)-1] m

h = [(A1 / Ćz)2 l]v12 2gC2[(Ym I Ya)-1)]

V1min = Qmin/Aá = (1.0 m3/h)/(9.51 X 10-4 m2)(1 h)/(3600 s) = 0.292 mis NRI = VáDálv= (0.292)(0.0348)/(9.51 X 10-4 rrr') = 4.25 X 103; e= 0.619

15.8 Or²fice meter. Propyl alcohol at 25ÜC; r= 7.87 kN/m3; V= 2.39 X 10-6 m2/s 1 1/2 in x 0.065 in wall steel tube: D1 = 34.8 mm= 0.0348 m; A1 = 9.51 x 10-4 m2 Let f]= 0.40 = d/D; Then d = 0.40 D = 0.40(34.8 mm)= 13.92 mm= 0.01392 m A2 = TCD2/4 = n(0.01392 m)2/4 = 1.52 x 10-4 m2; A1IA2 = 6.26 From Eq. (15-6), solve for hin mercury manometer; Yrn = 132.8 kN/m3

v = 0.766 mis; lteration: New NR = 5.07 x 103; New C= 0.623 New Vá= 0.785 mis; New NR = 5.19 X 103; New e= 0.623 - Unchanged. Final value of Q = A1 Vá= (7.419 x 10-3 m2)(0.785 mis)= 5.824 x 10-3 m3/s = Q

[ ]1/2 [ ]1/2 v = C 2gh[y m / Yeg -1] = (0.608) 2(9.81)(0.095)/(132.8/10.79-1]

1 (A1 / Ćz)2 -1 [3.778]2 -1

15.7 Find Q through an or²fice meter. D1 = 97.2 mm; A1 = 7.419 x 10-3 m2; d = 50 mm= 0.050 m A2 = 1.963 X 10-3 m2; A1A2 = 3.778; d/D = 0.05/0.0972 = 0.514; Trial 1: e= 0.608 for NR = 1X105 Ethylene glycol at 25ÜC: r= 10.79 kN/m3; V= 1.47 X 10-s m2/s

15.6 Find Sp across Venturi, Q = 600 gal/min(l ft3/s)(449 gal/min) = 1.336 ft3/s v1 = Q/A1 = (1.336 ft3/s)/(0.0844 ft2) = 15.12 ft/s Kerosene at 77ÜF; r= 51.2 lb/ft3; V= 2.14 X 10-s ft2/s NR = v1D1/v = (15.12)/(0.3355)/(2.14 x 10-5) = 2.37 x 105; Then C= 0.984 From Eq. (15-4), solving for P1 - P2 = Sp !ip = [v1/C]2 [(A1/A2)2 - 1] [;12g] = [15.12/0.984]2 [(0.0884/0.01227)2 - 1][51.2/(2(32.2))] Sp = 9551 lb/ft2 (1 ft2/144 in2) = 66.3 psi

Chapter 15 238

Let h = 10.0 in= 0.833 ft when v= Vi-max = 10.90 ft/s and C = 0.610

Where B = 2g[(rmlrw) -1] = 2(32.2)[(844.9162.4) - 1] = 807.6

a) Use Eq.(15-6) and solve for A2 from which we can obtain the throat diameter d. A - A1

,- [B~r +!r

15 .1 O Or²fice Meter - Design: Install in 12 inch ductile iron pipe; D1 = 12 .24 in = 1. 020 ft; Ćá = 0.8171 ft2 Specify the or²fice diameter d. NOTE: Mu/tiple solutions possible. Fluid: Water at 60ÜF; r= 62.4 lb/²t'; v= 1.21 X 10-5 ft21s Use mercury manometer with scale range O - 12 inHg Range of flow rate: Qmin = 1500 gpm = 3 .341 ft' Is; Qrnax = 4000 gpm = 8.909 ft3 Is Velocity in pipe: Vá-n¼n = QminlA1=4.089 ftls; Vá-max = QmaJA1 = 10.90 ftls Reynolds No.: NR-min = Vá-minD1lv= 3.45 X 105; NR-max = Vá-maxD1lv= 9.19 X 105 From Figure 15-5: C,11án = 0.612; Cmax = 0.610 [Assumed f]= dlD = 0.70]

Nozzle throat diameter = d = 3. 7 41 in When Q = 1000 gpm, h = 8.00 in manometer deflection When Q = 700 gpm, h = 3.97 in manometer deflection

Summary:

h = (v1 I C)2 [(A1 I Ćz )2 -1 = [(12.38 I 0.955)2[(0.1259 I 0.07635)2 -1] = 0_3306 ft = 3Å97 in B 873.73

h = (v1 / C)2[(A1 / Az)2 -1] B

For Vá = V1-min = 12.38 ftls ande= 0.955,

b) Use Eq. (15-6) and solve for h to determine the manometer reading when Q = QminĿ

Then d = (4Ai/7r)112 = [(4)(0.07635)/7r]112 = 0.3118 ft = 3.741 in= Throat diameter

A2 = ÜĿ 1259 112 = 0.07635 ft2 = 7rd214

[(873.7)(0.667)(0.961)2 + 1]

(17.69)2

Let h = 8.0 in= 0.667 ft when v= Vá-max = 17.69 ft/s and C = 0.961

Where B = 2g[(rmln0) - 2(32.2)[844.9/58.0) - 1] = 873.7

a) Use Eq. (15-6) and solve for A2 from which we can obtain the throat diameter d


v = .j2gh(yw -yÛ)/ Ya= ~2(32.2 ft2/s)(0.02 ft)(62.4-0.0736)/0.0736 = 33.0 ft/s

15.15 Pitot-static tube carrying air at atmospheric pressure and 80ÜF. Ya== 0.0736 lb/ft3. h = 0.24 inH20; h = (0.24 in)(l .O ft/12 in)= 0.02 ft; yg = Yw = 62.4 lb/ft3 Find flow velocity v.

v = .j2gh(yw - Ya)/ Ya = ~2(9.81 m/s2)(0.0048 m)(9810-10.71)/10.71 = 9 .28 mis

15.14 Pitot-static tube carrying air at atmospheric pressure and 50ÜC. Ya= 10.71 N/m3Å h = 4.80 mmH20; h = ( 4.80 mm)(l mil 000 mm) = 0.0048 mH20; yg = Yw = 981 O N/m3 Find flow velocity v.

15.13 u=~2gh(yg -y)ly =.j2(9.8l)(0.106)(132.8-9.81)/9.81 =5.11 mis

_(_l _I._05_)_(2_5_)2- m =O .036 m x 103 mm/m = 36 mm 2(9.81)(9719)

h

(yg = 9 .73 kN/m3 = 9 .73 x 103 N/m3)

(yg-y =9.719 x 103 N/m3 =9719 N/m3)

yu2 h =----

2g(yg -y)

(Yair = 11.05 N/m3) 15.12 u=~2gh(yg-y)ly Solveforh

15.11 u=~2gh(yg -rv r . h =225 mm =0.225 m Yg = 132.8 kN/m3 (mercury); y = 7 .74 kN/m3 (methyl alcohol) u= .j2(9.81)(0.225)(132.8- 7.74)/7.74 = 8.45 mis

Nozzle throat diameter = d = 9 .219 in. When Q = 4000 gpm, h = 10.0 in manometer deflection When Q = 1500 gpm, h = 1.40 in manometer deflection

Summary:

h = (v1 /C)2[(A1 / Az)2 -1] = [(4.089/0.612)2[(0.817110.4636)2 -1) = 0.1164ft=1.40 in B 807.6

b) Use Eq. (15-6) and solve for h to determine the manometer reading when Q = QminĿ h = (v1 I C)2[(A1 I Ćz)2 -1)

B For Vi = Vi-min = 4.089 ft/s and C = 0.612

Then d = (4A2/tr)112 = [(4)(0.4636)/tr]1i2 = 0.7683 ft = 9.219 in= Throat diameter Actual f]= d/D = 9.219/12.24 = 0.753; Assumed C values OK.

A = 0.8171 =O 4636 ft2 = m12á4 2 [(807.6)(0.833)(0.610)2 + 1]112 Å

(10.90)2

240 Chapter 16

16.4 Assume all air leaves parallel to the face ofthe sign. 103 m 1 h Vá= 125 km/h X --X--= 34.7 m/s km 3600s

Rx=pQ(v2, -v1J =pQ(O-(-v1))=pQv1 =p(Av1)(v1)= pAv12

Rx= l.34lkgx(3m)(4m)x(34.7)2m2 =19.4xl03kgĿm =19.4kN nr' s2 s2

. R ' 19 .4 X 103 N 161 7 N Equivalent pressure = p = -Ŀ = 2 = 2 = 1617 Pa A 12m m

R

. 1 fr' /s Q = 150 gal/mm x = 0.334 ft3/s

449 gal/min

V1 = Uz = Q = 0.324 ft3 Is = 61.25 ft/s A ;r(l/12 ft )2 /4

R, = pQ{ V2, -v1,) = pQ(O- (-vi))= pQV1

R, = 1.94 lb-s2 x 0.334 ft3 x 61.25 ft = 39.7 lb ft" s s

R, = pQ( v2y -v1,) = pQ(v2 -0) = pQUz = pQv1 = 39.7 lb

16.3

16.2 See sketch for Problem 16.1: Rx = pQv1 = p(Av1)v1 = pAv[

v = {R: = 300 lb 144 in 2 /ft2 = 84.2 ft/s {;;A (1.94 lb- s2 /ft4)( ff(2.0 in)" /4)

Q =Av= ;r(0.075 m)" x 25 m = 0.1104m3 16.l 4 s s

R, = pQ{ v2, -v1,) = pQ(O- (-Vi))= pQv1

lOOOkg 0.1104m3 25m 2761kgĿm R = x x--=---- x m ' s s s2 = 2761N=2.76 kN

CHAPTER SIXTEEN

FORCES DUE TO FLUIDS IN MOTION

241 FORCES DUE TO FLUIDS IN MOTION

Ry = pQ[ u2, -u1J = pQ[Ui sin 60Ü - O]= (988)(0.118)(15)(sin 60Á) = 1512 N

R, = pQu1[l - cos 60Ü]

R = 988 kg 0.118 m3 15 m [O 5] x 3 X X X Å m s s

= 873 kgĿm = 873 N sz

16.7 Q=Au= 7r(0.10m)2 x15m =0.118m3/s 4 s

R; = pQ( u2y - u1y ) = pQ[ Vi sin 50Ü - O] = (1.88)(0.451 )(22 sin 50Á) = 14.3 lb

Rx

\11t _]' +x

' / /Uz

á~ 1 1

R, = pQ( u2, -u1,) = pQ[Ui cos 50Á - (-u1)]

R, = pQ[ Vi cos 50Ü + uá]; but Vi = u1 R, = pQu1[cos 50Á + 1] R, = 1.88 l~ Ŀ s2 x 0.451 ft3 x 22.0 ft x l .643

ft s s = 30.6 lb

16.6 Q =A u= (2.95 in2)(22.0 ft/s)(l ft2/144 in2) = 0.451 ft3/s

16.5 Q=Au= 7T(l.75in)2 X ft2 X 25ft 4 144in2 s

= 0.418 ft3/s R, = pQ( u2, - Uáx) = pQ[-Ui sin 30Ü - (O)]

R; = -pQIJi sin 30Á = -(1.94)(0.418)(25) sin 30Á R, = -10.13lb=10.13 lb to right R; = pQ( u2, - u1y) = pQ[ Vi cos 30Ü - (-u1)]

R; = pQ[Vi cos 30Á +uá]= (1.94)(0.418)[25 cos 30Á + 25] = 37.79 lb up

Chapter 16

Q = 100 gal/min x 1 ft3 /s = 0.223 ft3/s: u1 = Q = 0Ŀ223 ft3 /s = 37. l ft/s 449 gal/min A 0.0060 ft2

Assume all fluid strikes vane and is deflected perpendicular to incoming stream.

Rà = pQ( U2, -u1,) = pQ[O - (-Uá))= pQUá

R = 1.94lbĿs2 x 0.223ft3 x 37.lft = 16_0 lb X ft" S S

R, is force exerted by vane on water Rx' is force exerted by water on vane L.MA =O = Fs(0.5 in) - Rx'(l .O in)

Fs = Rx' ~ = 2Rx' = 2(16.0 lb)= 32.0 lb 0.5

A

- +x

+y r

RyÅWeight

R = 550 N = 550 kg-rn/s'' = pQ(u -u ) y 2, 1,

Rx, =pQ(u2, -u1J =pAu1[0-(-uá)]= pA1u12

A= .!_[1l"(2.0in)2 x ft2 ] =0.0109ft2 1 2 4 144in2

R = 1.94lbĿs2 x0.0109ft2x(40.0ft)2 x, ft4 s = 33.9 lb

Rx, = pQ(u2, u1J = pQ(Ui-0) = pQU;!

Assume LJi= u1,A2=A1 Rx, = pAu12 = 33.9 lb M1A = Rx, (4.00-0.424)in = (33.9 lb)(3.576in) = 121 lb-in

Moment due to R.rz is small-depends on shape of leaving stream.

= pA1 u1(0- (-u1)) = PA1uf

u1 = f!i, = (900 kg:~~)~:.~~xs;o-4 m2) = 25.2 mis Ćá = 1l"(0.035 m)2 = 9.62 X 10-4 m2

4

Q = Ćá Uá= JZ"(l.00 in)2 X ft2 X 30 ft = 0.1636 ft3 4 144in2 s s

Q 0.1636 ft3 144 in2/ft2 19.23ft LJi= -= =--- Ć2 SJZ"(4.002 -3.802)in2/4 S

Ry = pQ ( u2, - u1, )

Ry= l.88l~-s2 x 0.1636 ft3 [19.23-(-30){t = 15.14lb ft s s

242

16.11

16.10

16.9

16.8


pQv1 = (1.03)(1000 kg/m3)(0.025 m3)(3.32 mis)= 85.5 kg-rn/s" = 85.5 N pQv2 = (1.03)(1000)(0.025)(12.85)N = 331 N

_ [ vá - vi ] [ vá vi v12 ] [ vi v12 ] pz-pá+ Y -hL =pá+y - --3.5- =pá+y ---2.5- 2g 2g 2g 2g 2g 2g

= 825 kPa + (1.03)(9.81 kN/m3) [[-12.852 -2.5(3.32)2]m2/s2] Pz 2(9.81mls2)

= 825 kPa - 99 .2 kPa = 726 kPa

= 3.32 mis Q 0.025 m3/s Vi= A1 = 7.538x10~3m2

Q 0.025 m3/s Uz = A2 = 1.945x10~3 m '

16.14

= 12.85 mis

= 368 lb

= 20 ft/sec

_ [vi v12 h ]- [vi vá 0.12vi]- [1.12vi-v12] Pi-r ---+ L -r --- + -r ---- 2g 2g 2g 2g 2g 2g

= 62.4 lb [ 1.12(80)2 - (20)2] ft = 6550 lb/ft2 p, ft3 64.4

Rx-P1A1=pQ(v2, -v1J =pQ(-Uz-(-v1))=pQ(v1-Uz) R, = pQ(v1 Uz) + P1Ć1 = (1.94)(0.0218)(80)(20- 80) + (6550)(0.0873) = -203 + 571

L Av _J

2 2 _8_ + !:!L _ h = Pz + !:!L r 2g L r 2g V =V . Ćz =80Ŀ 0.0218

1 2 A1 0.0873

16.13

R

+x +-- 16.12 Rx=pQ[v2, -v1J =pAUz[Uz-0]= pAvi

A = ;r( 4.0 in)z x ftz = 0.0873 ft2 4 144in2

Rx = I.94f:~. s2 x 0.0873 ft2 x (60 ft/s)"

= 609 lb acting on water jet Force on boat is reaction to R, acting toward left. +....

Chapter 16 244

u2 (21 7)2 FromSection 10.8,hL= K-2 =0.043 Ŀ =0.314ft 2g 64.4

P u2 p u2 [ uz u2 ] __!_+_1 -hl=-2+_2 :.p2=P1+Y I 2 -0.314 y 2g y 2g 2g

p2=125+ 62.4lb[(5.55)2-(21.7)2 -0.314]ft lft2 =125-3.10=121.9psig ft3 64.4 144in2

LFx = Rx + P2Ć2 - páĆá = pQ( u2, -u1J = pQ(-Uz - (-vi))= pQ(u1 - Uz)

Rx = pQ(u1 - Uz)- P2Ć2 + P1Ć1 = (1.94)(1.11)(5.55 -21.7)- (121.9)(7.39) + (125)(28.89)

R; = -34.9 - 900 + 3611 = 2676 lb

- Q_l.l14ft3/sec =21.7ft/sec Uz - A,_ - 0.05132 ft2

A1 = 0.2006 ft2 = 28.89 in2

A2 = 0.05132 ft2 = 7.39 in2

D11D2 = 0.5054/0.2557 = 1.98; K = 0.043

_ Q _ 1.114 ft3/sec _ 5 55 fil u1 - - - - Å t sec A1 0.2006 ft2

16.15 Q = 500 gal/min x 1 ft3/sec 449 gal/min

= 1.114 ft3/sec

y-direction: R; - P2A2 = pQ ( u2v - u1,) = pQ( Uz - O)= pQUz

Ry=pQUz+p2A2=331N+1412N=1743N

R, = pQu1 + p1A1 = 85.5 N + 6219 N = 6305 N

x-direction: Rx-P1Ć1 =pQ(u -u.) =pQ(O-(-u1))=pQu1 2); 1\'.

p1A1 = (825 kN/m2)(7.538 x 10-3 m') = 6.219 kN = 6219 N p2A2 = (726 kN/m2)(1.945 x 10-3 rrr') = 1.412 kN = 1412 N

245 FORCES DUETO FLUIDS IN MOTION

Rx=29.95 kN

16.18 Q = 2000 Lfmin X 16.67 X 10-6 m' /s/1 L/min = 0.0333 rrr' /s u= QIA = 0.0333 m3/s/7.419 x 10-3 m2 = 4.49 mis Z:Fx=pQ(u2, -u1J =pQ(Uz-(-u1))=2pQu Z:Fx = Rx- p-A, -p2A2 = Rx-2P1Ć1 Then Rx- 2p1A1 = 2pQu

Rx = 2pQu+ 2p1A1 = 2(0.89) (1000 kg) x 0.0333m3 x 4.49 m + 2x 2.0x\06N x7.4l9x10_3m2

m3 s s m

= 267 ~gĿm + 29.68 x 103 N = 267 N + 29.68 x 103 N = 0.267 kN + 29.68 kN s

R =~R2 +r2 X U

R = 28.9 kN @ 45Á

L N _J y-direction: R, - P2Ć2 = pQ ( V2y - V1y ) = pQ( Uz - O) Ry = pQVz + p2A2 = 0.838 kN + 19.57 kN = 20.41 kN

x-direction: Rà - p1A1 = pQ(u2, -u1J = pQ(O- (-u1)) ~

Rà = pQu1 + p1A1 P1 1

lOOOkg 0.1253 6.71m/s 1050kN 1864 10-2 2 3 X X + 2 X. X m m s s m

s, = 838 kg-rn/s" + 19.57 kN = 838 N x l ~ + 19.57 kN = 20.41 kN 10 N

+yt +L-]

16.17 u= Q= 0Ŀ125m3/s =6.7lm/s A 1.864x10-2 m2

y-direction: R; = pQ( V2y -u1y) = pQ(Uz cos 45Á - O)= (1.93)(6.5)(20.45)(0.707) = 182 lb

x-direction: Rx-P1A1=pQ(u2, u1J =pQ(-usin45Ü-(-u1))=pQv(l-sin45Ü) R, = pQv(l - sin 45Ü) + p1A 1 = (1.93)(6.5)(20.45)(0.293) + (90.5)(0.3174) = 103.9 lb

+yt ~

Find Pi (p2 =O, Vá = Uz, z1 = z2) p L u2 -:-hL =O;p1 =rb, =rt- Jy 2g u= Q/A = 6.5/0.3174 = 20.45 ft/sec NR = uD = (20.45)(0.6354) = l .42 x 106 r 9.15xl0-6

fr= 0.014 (Table 10.5); L)D = 16 (Table 10.4) pá = yhL = (62.2)(0.014)(16)(20.45)2/64.4 = 90.5 lb/ft'

16.16

Chapter 16 246

y-direction: Ry=pQ(v2y -v1J =pQ(O-(-v1sin15Ü))=pQvsi1115Á

R; = (1000)(0.943)(30)(0.259) = 7.32 x l 03 kg-m/s' = 7.32 kN =Force on water ,J.. F~rce 011 car= 7.32 kN t .

R, = 55.6 x 103 kg-rn/s' = 55.6 kN =Force of car on water.x-- Force 011 car= 55.6 kN---)

.+x r fY

R R' -~Y~-----------' R = R; + R: = 11.0 kN R

() = tan' ...2.. = 22.3Ü Rx

y-direction: Ry - p2A2 sin 45Ü = pQ( v2y - v1y) = pQ( t'2 sin 45Ü - O)= pQv sin 45Ü

Ry = (pQv + p2A2) sin 45Ü = (1.37 kN + 4.59 kN)(0.707) = 4.18 kN

16.20 Q =Av= (3.142 x 10-2 m2)(30 mis)= 0.943 m31s a. x-direction:

R; = pQ( v2, v1J = pQ( t'2 - (-v, cos 15Ü))

R; = pQu(l + cos 15Á) = l .966pQv

s. = (1.966)(1Ü:3kg )( 0.94: m3 )(30s m)

x-direction: s. - p1A 1 p2A2 cos 45Ü= pQ ( V2x - v1, )

Rx -p1A1(l + cos 45Ü) = pQ(l..'2 cos 45Á - (-vi)) R; - páA1(1 + cos 45Ü) = pQu(l + cos 45Á) R; = pQu(l + cos 45Ü) + p1A1(1 + cos 45Ü) pQv= 1590kg x 0.12m3x7.19m=1.37xl03 kgĿm = 1.37 kN

m ' s s s2

R; = 1.37 kN(l.707) + 4.59 kN(l.707) = 10.17 kN

16.19 Vá=Ui=V

v= Q = 0.12 m3 Is = 7.19 mis A 1.670x10-2 m2

p1A1 = p2A2 275 kN X 1.670 X 10-2 m2

2 m = 4.59 kN


16.23 Let Vie = Vá - Vá, 1 L Velocity ofBlade L Velocity of Air

2.60x102-5 kgĿ m = 2_6 x 10_5

N s

R = 1.20 kg x Jr(0.015 m)2 x (0.35 m/s)2

X m ' 4 1

v1=

A= n(0.010 m)214 = 7.85 x 10-5 m2

0.196 kgm/s" ---------- = 45.6 mis (l.20kglm3)(7.85xl0-5 m")

.A

l á--1sm -i 1 ,,, Å

715""" 1 ' 1 ... -l-

16.21 R, = pQ( v2, -v1J = pĆVá(O - (-v1)) = pAv12 (D

W . h f 0.10kgĿ9.8lm 0981N eig to. carton = Wc = mg = 2

Ŀ = Ŀ s

2.MA = Rx(75 mm) - Wc(15 mm)= O Impending tipping 15 15 Rx=Wc-=0.981N- =0.196N 75 75

From(D, v1 = {R; VPA

b. Because the inlet jet acts atan angle to the x-y directions, we compute its components: v1x = v1 cos (15Á) = (30 mis )(0.966) = 28.98 mis: V1y = v1 sin (15Ü) = 30 mis )(0.259) = 7.76 mis Only v1x is affected by the moving vane. Then Ve1x = v1x - 12 mis= 16.98 mis. ve1y= v1y = 7.76 mis. The magnitude of the resultant effective velocity is:

lvetyl = ~(16.98)2 +(7.76)2 = 18.67 mis The total eff ective mass flow rate into the vane, Me, is, Me= pQ,e = pAve1 = (1000 kglm3)(3.142 x 10-2 m2)(18.67 mis)= 586.6 kg/s The velocity, Ve 1, acts atan angle a, with respect to the horizontal, where a= Tan-1(7.76)/16.98) = 24.58Ü Only the component of Vei acting parallel to the van e is maintained as the jet travels around the vane. This component is computed using fJ, the difference between a and the angle ofthe vane inlet. fJ= 24.58Á - 15Á = 9.58Á Then, VeJ(par) = (ve1)cos(9.58Ü) = (18.67 mls)(0.986) = 18.41 mis This velocity remains undiminished as the jet travels around the vane. Then ve2 = 18.41 mis to the left. Force in x-direction: R, = Me(L1vex) = Me(ve2x - Ve1x) = (586.6 kg/s)[18.41 - (-16.98)]mls =20.86 kN Force in y-direction: Ry = Me(L1vey) = Me(Ve2y - Ve1y) = (586.6 kgls)[O - (-7.76)]mls =4.55 kN

Chapter 16

80 20 40 60 Angle e (Degrees)

o o

0.08

0.24

0.32 r-, r-,

' -, 1 ' ~ 1

' j\._ -, " \ 1\ \ t\

\

0.16

248

() Rx(lb) M(lbĿin)

10 0.141 0.352

20 0.134 0.336

30 0.124 0.310 - 40 0.110 0.274 .5

50 0.0920 0.230 g_ 60 0.0715 0.179

~ s 70 0.0489 0.122 2" o 80 0.0248 0.062

1-

90 O.O O.O

See analysis=-Problem 16.24: R, = pA Uá2 sin(90 - 8) = pA Uá2 cos 8 R, = (2.06 X 10-3)(0.417)(10.0)2 cos 8 = 0.143 cos 8; Moment = Rx(2.5 in)

16.26

16.25 -è>

See Problem 16.24: V u1x R, = pA u12 sin 70Ü = (2.06 x 10-3)(0.694)(10.0)2 sin 70Á = 0.1345 lb Moment = Rx(2.5) = (0.1345 lb)(2.5 in)= 0.336 lb-in

R; = 2Ŀ06x l04-3 lbĿs2 x 0.694 ft2 x (10.0 ft/s)2 x sin 45Ü = 0.101 lb ft

Assume R, acts at middle of louver, 2.50 in from pivot Moment = Rx(2.5) = 0.1O1 lb(2.5 in) = 0.253 lb-in

u1

16.24

_ R _ (O 075 ) 40 rev 2TC rad 1 min ub- w- . m x --x-Ŀ-x-- min rev 60s

= 0.314 mis Ve= 0.35 mis - 0.314 mis= 0.0358 mis R, = pA u: = (1.20) 1C(O.Ol5)2 (0.0358)2 = 2.72 X 10-7 N

4

R, = pQ( u2, - u1,) Rx = pQ(O - (-u1 sin 45Ü)) Rx = pQu1 sin 45Ü = p(A Uá) Uá sin 45Ü

R, = pAu12 sin 45Ü Compute R, far a 20.0 in length oflouver, as shown in Fig. 16.19. A= (5 in)(20.0 in)(l ft2/144 irr') = 0.694 ft2


. u 1 O.O m rev 60 s Rotational speed = OJ = _!_ = x x -- = 4 77 rpm r s(0.20 m) 21l" rad min

VRI = 15.25 mis !16.54Ü ; VR2 = 15.25 mis 53.46Ü\ u Vmx = 14.62 m/s -e-; Vmy = 4.34 mis t; VR2x = 9.077 m/s e-; VRzy = 12.25 m/s? Me= pQe = pĆjVR = (1000 kg/m3)(.n(0.0075 m)2/4)(15.25 mis)= 0.674 kg/s R, = Me(VR2x - VR1x) = (0.674 kg/s)(9.077 - (-14.62)) mis= 15.97 kg m/s'' = 15.97 N *- R; = Me(VR2y - VR1y) = (0.674)(12.25 - 4.34) = 5.33 Nt

Results:

16.30 Compute the force on one blade when the turbine wheel is rotating and has a tangential velocity of 10 mis. Method: See vector diagram on next page. Law of sines and law of cosines used. 1. Compute the velocity relative to the blade vRI for the inlet. 2. The magnitude of this velocity remains undiminished as the jet traverses the blade. 3. The relative velocity rotates 11 OÜ as it traverses the blade, 4. Resol ve Vm and VRZ into x and y components. 5. Compute the effective mass flow rate Me= pQe = pAjvR. 6. Compute reaction forces: R, = Me(vR2x - Vmx) and Ry = Me(VR2y VR1y)

Force in the x-direction: R; = M(Llvx) = M(v2x - v1x) = (1.105 kg/s)[12.5 - (-24.62)]mls = 41.0 N Force in the y-direction: Ry = M(Llvy) = M(v2y - v1y) = (1.105 kg/s)[21.65 - 4.34]mls = 19.1 N

16.29 R, = M(L1v2) = M(v2x - v1x): Where M = pAv A= 1l"D214 = .n(0.0075 m)214 = 4.418 x 10-5 m2 M= pAv = (1000 kg/m3)(4.418 x 10-s m2)(25 mis)= 1.105 kg/s v1x = v1 cos(lOÜ) = (25 mls)(0.985) = 24.62 mis; v1y = v1 sin(lOÜ) = (25 mls)(0.174) = 4.34 mis Vz =V= 25 mis; v2x = v2 cos(60Ü) = (25 mls)(0.5) = 12.5 mis; v2y = v2 sin(60Ü) = (25 mls)(0.0866) = 21.65 mis

16.28 See Prob. 16.27: W= pAu12=1.94lbĿs2 x JZ"(0.75in)2 x (25.0ft/s)2 x 1 =6_20 lb

0.6 ft" 4 144in2/ft2 0.6

lw '

Rr 1 - 1

16.27 Maximum E.= ÕW= 0.60W Sliding is impending when R, = Fá Rx=rQ(u2, -u1J =pAu1(0-(-u1))=pAu12

R. pAu2 Then W= _x =--1 0.6 0.6

W= 2.40xl0-3 lbĿs2 x JZ"(l.5in)2 x ft2 x (25.0ft/s)2 =0_03071b ft" 4 144in2 0.6

Chapter 16 250

R; is the reaction force exerted by the blade on the water; positive R; acts to the left Then the force exerted by the water on the blade acts to the right, accelerating the blade Positive Ry acts radially outward Then the force exerted by the water on the blade acts radially inward toward the center of rotation When Ry is negative, the net radial force on the blade is outward.

Problem 16.31 Forces on rotating turbine wheel vBlade VRI <P VR2 a Me Rx Ry (mis) (mis) (Ü) (mis) (Ü) (kg/s) (N) (N) o 25.00 10.00 25.00 60.00 1.105 41.00 19.12 5 20.09 12.48 20.09 57.52 0.888 27.00 11.20 10 15.25 16.54 15.25 53.46 0.674 15.97 5.33 15 10.55 24.29 10.55 45.71 0.466 7.92 1.50 20 6.34 43.22 6.34 26.78 0.280 2.88 -0.42 25 4.36 95.00 4.36 -25.00 0.193 0.69 -1.19

Assume lvRJI =lvR21 and VR vector is rotated 110Á from inlet to outlet

Vector diagram for Problem 16.30 Vá= Velocity of water jet vR = Velocity relative to blade Vá,= Velocity ofblade

-Ass, llEl·CIT'I oF je-( Le~VINC. V/l/ltà

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Applied Fluid Mechanics-Mott-Sol 6