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Student Solutions Manual Third Edition Fundamentals of Fluid Mechanics BRUCE R. MUNSON DONALD F. YOUNG Department of Aerospace Engineering and Engineering Mechanics THEODORE H. OKIISHI Department of Mechanical Engineering Iowa State University Ames, Iowa, USA John Wiley & Sons. Inc. New York Chichester Weinheim Brisbane Singapore Toronto

COVER PHOTOS Whirlpool: Francoise Sauzc/Socncc Photo Library /Photo Researchers, Inc. Computer Simulation of Vortex: Dr. Fred Espenak/Science Photo Library/Photo Researchers, Inc. ACQUISITIONS EDITOR Chanty Robey MARKETING MANAGER Harper Mooy PRODUCTION EDITOR Tony VenGraitis COVER DESIGNER Madelyn Lesure This book was printed and bound by Malloy Lithograph. The cover was printed by Phoenix Color. This book is primed on acid-free paper. @ The paper in this book was manufactured by a mill whose forest management programs include sustained yield harvesting of its timberlands. Sustained yield harvesting principles ensure that the number of trees cut each year does not exceed the amount of new growth. Copyright C 1998 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmined in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherw ise, except as permitted under Sections 107 or 108 of the 1976 United Stales Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive. Danvers. MA 01923. (508) 750-8400. fax (508) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department. John Wiley & Sons. Inc.. 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008. E-Mail PERMREQWWILEY.COM ISBN 0-471-24011-7 Primed in the United States of America 10 987654321

PREFACE This Student Solutions Manual has been developed as a supplement to Fundamentals of Fluid Mechanics, by Munson, Young, and Okiishi. At the end of each chapter, the e- Book contains a section called Review Problems. These review problems are representative of the types of problems that students should be able to solve after completing the chapter, and this Student Solutions Manual contains the detailed solutions to these review problems. We believe that as students prepare for an examination, or feel the need for some additional work on a particular topic, it will be helpful to have available such a set of review problems with their corresponding solutions. Each review problem is preceded by a brief phrase which gives an indication of the main topics to be used in solving the problem. Thus, the student can conveniently select those topics, and the corresponding review problems, of interest. This information is also presented in the table of contents. The solutions contained in this manual are worked in a logical, systematic way with sufficient detail so that they can be readily followed. Except where a greater accuracy is warranted, all intermediate calculations and answers are given to three significant figures. Unless otherwise indicated in the problem statement, values of fluid properties used in the solutions are those given in the properties tables found on the inside text cover. The authors hope that this supplement to our text will be a useful tool to help the student gain a better understanding of basic fluid mechanics. We believe that practice through solving a variety of problems, with immediately available feedback by way of the Student Solutions Manual, can be a valuable component in the spectrum of teaching tools needed in the study of fluid mechanics. Any suggestions and comments from you, the user, are certainly welcome and appreciated. Bruce R. Munson Donald F. Young Theodore H. Okiishi

CONTENTS PROBLEM CHAPTER TOPIC NUMBER 1 INTRODUCTION Dimensions 1.1R, 1.2R Units 1.3R, I.4R Specific gravity 1.5R Specific weight 1.6R Ideal gas law 1.7R, 1.8R Viscosity 1.9R, 1.I0R, 1.11R Bulk modulus 1.12R, 1.13R Speed of sound 1.14R Vapor pressure 1.15R 2 FLUID STATICS Pressure head 2.1R Pressure depth 2.2R • relationship Gage/absolute 2.3R pressure Manometer 2.4R, 2.5R, 2.6R Force on plane 2.7R, 2.8R, 2.10R, surface 2.11R Center of pressure 2.9R Force on curved 2.12R, 2.13R, 2.14R surface Buoyancy 2.15R, 2.16R Buoyancy, force on 2.17R plane surface Rigid body motion 2.18R, 2.19R vii

viii Contents 3 ELEMENTARY FLUID DYNAMICS—THE BERNOULLI EQUATION F = ma along 3.1R streamline F = ma normal to 3.2R streamline Stagnation pressure 3.3R Bernoulli equation 3.4R Heads 3.5R Free jet 3.6R Bemoulli/continuity 3.7R, 3.8R, 3.9R, 3.10R, 3.12R Bemoulli/continuity/ 3.11R Pitot tube Cavitation 3.13R • Flowrate 3.14R Channel flow 3.15R, 3.16R Energy line/ 3.17R hydraulic grade line Restrictions on 3.18R, 3.19R Bernoulli equation 4 FLUID KINEMATICS Streamlines 4.1R, 4.2R Material derivative 4.3R Acceleration 4.4R, 4.5R, 4.6R, 4.7R Reynolds transport 4.8R theorem Flowrate 4.9R, 4.1 OR Control volume/ 4.11R system

Contents ix 5 FINITE CONTROL VOLUME ANALYSIS Continuity equation 5.1R, 5.2R, 5.3R Average velocity 5.4R Linear momentum 5.5R, 5.6R, 5.7R, 5.8R, 5.9R, 5.10R, 5.11R, 5.12R Moment-of- 5.13R, 5.14R, 5.15R, momentum 5.16R Energy equation 5.17R, 5.18R, 5.20R, 5.21R, 5.22R Linear momentum/ energy 5.19R 6 DIFFERENTIAL ANALYSIS OF FLUID FLOW Acceleration 6.1R Vorticity 6.2R Conservation of mass 6.3R, 6.4R Stream function 6.5R Velocity potential 6.6R Potential flow 6.7R, 6.10R, 6.11R Inviscid flow 6.8R, 6.9R Viscous flow 6.12R, 6.13R, 6.14R, 6.15R. 6.16R

X Contents 7 SIMILITUDE, DIMENSIONAL ANALYSIS, AND MODELING Common Pi terms 7.1R Dimensionless 7.2R variables Determination of Pi 7.3R, 7.4R, 7.5R, terms 7.6R Model ing/similarity 7.7R, 7.8R, 7.9R Correlation of 7.10R, 7.11R experimental data Dimensionless 7.12R, 7.13R governing equations 8 VISCOUS FLOW IN PIPES Laminar flow 8.1R, 8.2R • Velocity profile 8.3R Turbulent velocity 8.4R profile Moody chart 8.5R, 8.6R Minor losses 8.7R Noncircular conduits 8.8R Single pipe— 8.9R, 8.10R determine pressure drop Single pipe— 8.11R determine flowrate Single pipe— 8.12R determine diameter Single pipe with 8.13R, 8.14R pump Single pipe with 8.15R turbine Multiple pipes 8.16R Flow meters 8.17R, 8.18R

Contents xi 9 FLOW OVER IMMERSED BODIES Lift/drag calculation 9.1R External flow 9.2R, 9.3R character Boundary layer flow 9.4R, 9.5R Friction drag 9.6R Momentum integral 9.7R equation Drag—low Reynolds 9.8R number Drag 9.9R, 9.10R, 9.HR, Drag—composite 9.12R, 9.13R 9.14R body Lift 9.15R, 9.16R 10 OPEN-CHANNEL FLOW Surface waves 10.1R, 10.2R Froude number 10.3R, 10.4R Specific energy 10.5R, 10.6R Wall shear stress 10.7R Manning equation 10.8R, 10.9R, Best hydraulic cross 10.10R, 10.11R 10.12R section Hydraulic jump 10.13R, 10.14R Sharp-crested weir 10.15R Broad-crested weir 10.16R Underflow gate 10.17R

xii Contents 11 COMPRESSIBLE FLOW Speed of sound Sound waves Mach number Isentropic flow Fanno flow Rayleigh flow Normal shock waves 11.1 R, 1I.2R 11.3 R 11.4 R 11.5 R. 11.6R 11.7R 11.8 R 11.9 R 12 TURBO- MACHINES Angular momentum 12.1R Velocity triangles 12.2R Centrifugal pump 12.3R, 12.4R Similarity laws 12.5R Specific speed 12.6R Turbine 12.7R. I2.8R

TABLE 1 .3 Conversion Factors from BG and EE Units to SI Units'* To Convert from to Multiply by Acceleration ft/s2 m/s2 3.048 E - 1 Area ft2 m2 9.290 E - 2 Density Ibm/ft3 kg/m3 1.602 E + 1 slugs/ft’ kg/m3 5.154 E + 2 Energy Btu J 1.055 E + 3 ftlb J 1.356 Force lb N 4.448 Length ft m 3.048 E - 1 in. m 2.540 В - 2 mile in 1.609 E + 3 Mass Ibm kg 4.536 E - 1 slug kg 1.459 E + 1 Power ft-lb/s W 1356 hp W 7.457 E + 2 Pressure in. Hg (60 T) N/m2 3.377 E + 3 lb/ft2 (psf) N/m2 4.788 E + 1 lb/in.2 (psi) N/m2 6.895 E + 3 Specific weight lb/ft3 N/m3 1.571 E + 2 Temperature °F °C Tc = (5/9)(7> - 32°) °R К 5.556 E - 1 Velocity ft/s m/s 3.048 E - 1 mi/hr (mph) m/s 4.470 E - 1 Viscosity (dynamic) Ibs/ft2 N-s/m2 4.788 E + 1 Viscosity (kinematic) ft2/s m2/s 9.290 E - 2 Volume flowrate ft3/s m3/s 2.832 E - 2 gal/min (gpm) m3/s 6.309 E - 5 It more Hum four place accuracy u desired, refer Io Appendix A.

TABLE 1.4 Conversion Factors from SI Units to BG and EE Units* To Convert from to Multiply by Acceleration m/s2 ft/s2 3.281 Area m2 ft2 1.076 E + 1 Density kg/m’ Ibm/ft3 6.243 E - 2 kg/mJ slugs/fr' 1.940 E - 3 Energy J Btu 9.478 E - 4 J ft lb 7.376 E - 1 Force N lb 2.248 E - 1 Length in ft 3.281 in in. 3.937 E + 1 m mile 6.214 E - 4 Mass kg Ibm 2.205 kg slug 6.852 E - 2 Power W ftlb/s 7.376 E - 1 w hp 1.341 E - 3 Pressure N/m2 in. Hg (60 °F) 2.961 E - 4 N/m2 lb/ft2 (psf) 2.089 E - 2 N/m2 lb/in? (psi) 1.450 E - 4 Specific weight N/m' lb/ft’ 6.366 E - 3 Temperature *C •F TF = 1.8 Tc + 32* К *R 1.800 Velocity m/s ft/s 3.281 m/s mi/hr (mph) 2.237 Viscosity (dynamic) N-s/m2 lbs/ft2 2.089 E - 2 Viscosity (kinematic) m2/s ft2/s 1.076 E + 1 Volume flowrate m’/s ft'/s 3.531 E + 1 mJ/s gal/min (gpm) 1.585 E + 4 •Jf more than four-place accuracy is desired, refer to Appendix A.

TABLE 1.5 Approximate Physical Properties of Some Common Liquids (BG Units) Liquid Temperature (*F) Density. P (slugs/ft3) Specific Weight. У (Ib/fU) Dynamic Viscosity, M (Ib-s/fl2) Kinematic Viscosity, V (ft2/») Surface Tension,* V (Ib/ft) Vapor Pressure. Pv [Ib/bi.2 (abs)] Bulk Modulus,1* B, (Ib/in.2) Carbon tetrachloride 68 3.09 99.5 2.00 E - 5 6.47 E - 6 1.84 E - 3 1.9 E + 0 1.91 E + 5 Ethyl alcohol 68 1.53 49.3 2.49 E - 5 1.63 E - 5 1.56 E - 3 8.5 E - 1 1.54 E + 5 Gasoline^ 60 1.32 42.5 6.5 E - 6 4.9 E - 6 1.5 E - 3 8.0 E + 0 1.9 E + 5 Glycerin 68 2.44 78.6 3.13 E - 2 1.28 E - 2 4.34 E - 3 2.0 E - 6 6.56 E + 5 Mercury 68 26.3 847 3.28 E - 5 1.25 E - 6 3.19 E - 2 2.3 E - 5 4.14 E + 6 SAE 30 oil1 60 1.77 57.0 8.0 E - 3 4.5 E - 3 2.5 E - 3 — 2.2 E + 5 Seawater 60 1.99 64.0 2.51 E - 5 1.26 E - 5 5 .03 E - 3 2.56 E - 1 3.39 E + 5 Water 60 1.94 62.4 2.34 E - 5 1.21 E - 5 5.03 E - 3 2.56 E - 1 3.12 E + 5 •In contact with air. bl«enlrop«c bulk modulus calculated from speed of sound ‘Typical values- Property* of petroleum products vary. TABLE 1 .6 Approximate Physical Properties of Some Common Liquids (SI Units) Liquid Temperature (*C) Density, P (kg/mJ) Specific Weight, У (kN/m‘) Dynamic Viscosity, P (N-s/m2) Kinematic Viscosity, V (m2/s) Surface Tension,* fT (N/m) Vapor Pressure, P„ [N/m2 tabs)) Bulk Modulus,1* E. (N/m2) Carbon tetrachloride 20 1.590 15.6 9.58 E - 4 6.03 E - 7 2.69 E - 2 1.3 E + 4 1.31 E + 9 Ethyl alcohol 20 789 7.74 1.19 E - 3 1.51 E - 6 2.28 E - 2 5.9 E + 3 1.06 E + 9 Gasoline1' 15.6 680 6.67 3.1 E - 4 4.6 E - 7 2.2 E - 2 5.5 E + 4 1.3 E + 9 Glycerin 20 1.260 12.4 I.50E + 0 1.19 E - 3 6.33 E - 2 1.4 E - 2 4.52 E + 9 Mercury 20 13,600 133 1.57 E - 3 1.15 E - 7 4.66 E - 1 1.6 E - 1 2.85 E + 10 SAE 30 oil" 15.6 912 8.95 3.8 E - 1 4.2 E - 4 3.6 E - 2 — 1.5 E + 9 Seawater 15.6 1,030 10.1 I.20E — 3 1.17 E - 6 7.34 E - 2 1.77 E + 3 2.34 E + 9 Water 15.6 999 9.80 1.12 E - 3 1.12 E - 6 7.34E - 2 1.77 E + 3 2.15 E + 9 •In contact with air. isentropic bulk modulus calculated from speed of sound ‘Typical values. Properties of petroleum products vary.

TABLE 1.7 Approximate Physical Properties of Some Common Gut» at Standard Atmospheric Premure (BG Unit») Gas Temperature CF) Density, P (slugs/ft*) Specific Weight, V (lb/ft3) Dynamic Viscosity, M (Ibs/ft1) Kinematic Viscosity, (ft1/») Gas Constant," К (ft lb/slug ‘R) Specific Heat Ratio,1* к Air (standard) 59 2.38 E - 3 7.65 E - 2 3.74 E - 7 1.57 E - 4 1.716 E + 3 1.40 Carbon dioxide 68 3.55 E - 3 1.14 E - 1 3.07 E - 7 8.65 E - 5 1.130 E + 3 1.30 Helium 68 3.23 E - 4 1.04 E - 2 4.09 E - 7 1.27 E - 4 1.242 E + 4 1.66 Hydrogen 68 1.63 E - 4 5.25 E - 3 1.85 E - 7 1.13 E - 4 2.466 E + 4 1.41 Methane (natural gas) 68 1.29 E - 3 4.15 E - 2 2.29 E - 7 1.78 E - 4 3.099 E + 3 1.31 Nitrogen 68 2.26 E - 3 7.28 E - 2 3.68 E - 7 1.63 E - 4 1.775 E + 3 1.40 Oxygen 68 2.58 E - 3 8.31 E - 2 4.25 E - 7 1.65 E - 4 1.554 E + 3 1.40 •Values of the gas constant arc independent of temperature ^Values of the specific beat ratio depend only slightly on temperature TABLE 1.8 Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (SI Units) Gas Temperature CC) Density, P (kg/nr*) Specific Weight, У (N/m3) Dynamic Viscosity, M (N-s/m1) Kinematic Viscosity, V (m2/s) lias Constant," R (J/kg-K) Specific Heat Ratio,1* к Air (standard) 15 1.23 E + 0 1.20 E + 1 1.79 E - 5 1.46 E - 5 2.869 E + 2 1.40 Carbon dioxide 20 1.83 E + 0 1.80 E + 1 1.47 E - 5 8.03 E - 6 1.889 E + 2 1.30 Helium 20 1.66 E - 1 1.63 E + 0 1.94 E - 5 1.15 E - 4 2.077 E + 3 1.66 Hydrogen 20 8.38 E - 2 8.22 E - 1 8.84 E - 6 1.05 E - 4 4.124 E + 3 1.41 Methane (natural gas) 20 6.67 E - 1 6.54 E + 0 1.10 E - 5 1.65 E - 5 5.183 E + 2 1.31 Nitrogen 20 1.16 E + 0 1.14 E + 1 1.76 E - 5 1.52 E - 5 2.968 E + 2 1.40 Oxygen 20 1.33 E + 0 1.30 E + 1 2.04 E - 5 1.53 E - 5 2.598 E + 2 1.40 •Values of the gas constant are independent of temperature. hValoes <rf the specific heat ratio depend only slightly on temperature.

1Ш LIR (Diihensioas) During a study of a certain flow system the following equation relating the pressures p, and p3 at two points was developed: Pi Pi + /£V Dg In this equation V is a velocity, t the distance between the two points, I) a diameter, g the acceleration of gravity, and/a di- mensionless coefficient. Is the equation dimensionally consis- tent? (ANS: No) ^ince each htrm in Hie equation does nof Иан "Hie $amc dimension^ +he e^uahion is dimensionally Censislrn-h. No. I.2R | 1.2R (Dunensions) If V is a velocity, t a length. IF a weight, and д a fluid property having dimensions of FL~2T, determine the dimensions of: (a) VCW/p., (b) ИдС, (c) Уд/f, and (d) УС2д/ТУ (ANS: L4T ~ F;L 'T; Fl. J: I.) (i) Wfi = i F*L-'T (cl) V/У ± W (F) l-IR

MR (Units) Make use of Table 1.4 to express the follow- ing quantities in BG units: (a) 465 W, (b) 92.1 J, (c) 536 N/m1. (d) 85.9 mm1, (e) 386 kg/m2. (ANS: 3.43 x IO? ft Ib/s: 67 9 ft lb; 11.2 Ib/ft2: 3.03 x 10 6 ft1; 2.46 slugs/ft2) fa.) ftSh/sfafV) /"z 374x/o"Z Qr ) - Г-V <?2./ J r ^2.U^) (1- 374 X/d-' ) = 47, f f£-U> (C) 53L^- » (S3L fl.omio'2 Ц.2 -Гг /т*" />и4 * • .. / ___ (d) S5.4 = (rs.4 л J) [fe l^3^3.^ fts /.¥/? | ' "* MR (Units) A person weighs 165 lb at the earth’s surface. Determine the person's mass in slugs, kilograms, and pounds mass. (ANS: 5.12 slugs; 74.8 kg; 165 Ibm) /ГНА3 5 /ГИ OSS’ /Ь5 /4 'KCfS />»a ss ?/ zw /sz Ibrmn fmass /&>5 Из/уи I-2.R

I.5 R Specific gravity > Макс use of Fig. 1.1 to determine the specific gravity of water al 22 and 89 °C. What is the specific volume of water at these two temperatures? (ANS: 0.998: 0.966: 1.002 X 10 'm'/kg; 1.035 X 10 1 m’/kf) /> - Thus, nee a not M'c 56 - S6 = ^с,9ГС I ООО //n * specific 1/0/ume. -------= 6. f<?g @ Л2*С /ОРО---^== /in 3 €>. Ш @ 8<ГС , . 3 />71 -/.ooztio -jr: Lo35 x/o"’7 /-ЗЯ

I.6R 1.6 К (Specific weight) A 1-ft-diametercylindrical tank (hat is 5 ft long weighs 125 lb and is filled with a liquid having a specific weight of 69.6 lb/ft’. Determine the vertical force re- quired to give the tank an upward acceleration of 9 ft/sJ. (ANS: 509 )b up) /•7g | I.7 K (Ideal gas law) Calculate the density and specific weight of air at a gage pressure of 100 psi and a temperature of 100 °F. Assume standard atmospheric pressure. (ANS: 1.72 X 10 4 slugs/ft’; 0.554 lb/ft1) У-rom^he. ideal gts lav/i x *T (1.7/4 ) [(,„ ^/1?] = 1.12 x ID1 _________________ Pl * //-72 z/»’2 )(зга o.ssh- I-4H

TeR~ I.MR iIdeal gas law) A large dirigible having a volume of 90.000 m3 contains helium under standard atmospheric condi- tions (pressure = 101 kPa (abs) and temperature ~ 15 °C). Determine the density and total weight of the helium. (ANS: 0.169 kg/m3; 149 X iO'N) From 1 he ideal qas law, volume ~ Чо,оооm3 W?~| I.9R (Viscosity) A Newtonian fluid having a specific grav- ity of 0.92 and a kinematic viscosity of 4 X 10”° m2/s flows past a fixed surface. The velocity profile near the surface is shown in Fig. PI ,9R. Determine the magnitude and direction of the shearing stress developed on the plate. Express your answer in terms of V and 6. with U and 3 expressed in units of meters per second and meters, respectively. (ANS: 0.578 U/ 8 N/m: acting to right on plate) where P • VP - 0.51? У о М/тР acPthq /ъ 6* plahe. I-SR

ПйП I. IOR <Viscosity i A large movable plale is located between two large fixed plates as shown in Fig. PI.IOR. Two Newtonian fluids having the viscosities indicated arc contained between the plates. Determine the magnitude and direction of the shearing stresses that act on the fixed walls when the moving plate has a velocity of 4 m/s as shown. Assume that the velocity distri- bution between the plates is linear. (ANS: 15.3 N/m’ in direction of moving platet FIGURE P1 .1 OR >>». Stresses ad on ffxcd co a. I Is fn diredion of- moving I-6R

TiiR~] 1.11R (Viscosity» Determine the torque required to rotate a 50-mm-diameter vertical cylinder at a constant angular veloc- ity of 30 rad/s inside a fixed outer cylinder that has a diameter of 50.2 mm. The gap between the cylinders is filled with SAE 10 oil at 20 "C. The length of the inner cylinder is 200 mm. Neglect bottom effects and assume the velocity distribution in the gap is linear. If the temperature of the oil increases to 80 *C, what will be the percentage change in the torque? (ANS: 0.589 N nr. 92.0 percent) /л rhe 1.0 * ID 0 J(/0 я ZOO mm SOitnm (co niined) /-7Я

l.flR continued] Thus at to* C y= ITT У) 0.050 2 /»м - O. 0500 nn^ - O. 584 At !o*C} ^T= (o.514 JV.nn ) Ja Au/ reduction in 0J0 Hduthon iu L = tvwfrut is €ju*l h> : 0.584 ~ &П1 ~^o. t. 100 <o. 584 I^2£3]xioo = 7ЛОЙ/ 1.0 i IO'1 J — 1.12 R 1.I2R iBulk modulus) Estimaie the increase in pressure (in psi) required to decrease a unit volume of mercury by 0.1%. (ANS. 4.14 X 10’psi) s ~ ^he^etr^im Table. EYc -fc/yxjd* Thus, Др x - ЁуД^ _ _ o.ooT) 3 Ap z Ч-./Ч it TSL /-8R

l.flR continued] Thus at to* C y= ITT У) 0.050 2 /»м - O. 0500 nn^ - O. 584 At !o*C} ^T= (o.514 JV.nn ) Ja Au/ reduction in 0J0 Hduthon iu L = tvwfrut is €ju*l h> : 0.584 ~ &П1 ~^o. t. 100 <o. 584 I^2£3]xioo = 7ЛОЙ/ 1.0 i IO'1 J — 1.12 R 1.I2R iBulk modulus) Estimaie the increase in pressure (in psi) required to decrease a unit volume of mercury by 0.1%. (ANS. 4.14 X 10’psi) s ~ ^he^etr^im Table. EYc -fc/yxjd* Thus, Др x - ЁуД^ _ _ o.ooT) 3 Ap z Ч-./Ч it TSL /-8R

1.13 R I 1.13 R i Bulk modulus) What is the isothermal bulk mod- ulus of nitrogen al a temperature of 90 °F and an absolute pres- sure of 5600 lb/ft2? (ANS'. 5600 lb/ft21 for iioi'hfrmol bulk modulus t So ly^ti E* - ft ° ° Tmr~\ 1.I4 K (Speed of sound) Compare the speed of sound in mercury and oxygen at 20°C. (ANS:cM?/cw = 4.45) For mercuryf _______________ io dh«b . / * C-- \ s r /,34>x/z>* s dM s for ot-gyen c = (£t ,2o) c(or ye e») 3zu-^- (See Tables 1.6 and 1.3 for ^aIucs of kta.nd R.} 1-ЯК

TT5R] I.IS R (Vapor pressure! Al a certain altitude it was found that water boils at 90 °C. What is the atmospheric pressure al this altitude? (ANS: 70.1 kPa (abs)l The l/apor pressure о/ iva.itr ai ?D*C zj 1. ОI X Ю* Pa Рабл) (Table B.2- in Appt л dir fl) . Thu^ if uait> boil л a.i Phis hem pt ra.hu ret i~he a-hmaspheric pressure musl be fflual /о 7.D! t /0*Pa. 7p. / -4Л. (aba} l-IOR

2,lR 1 2.1R (Pressure bead» Compare the column heights or water, carbon tetrachloride, and mercury corresponding to a pressure of 50 kPa. Express your answer in meters. (ANS: 5.10 m; 3.21 m; 0.376 m) e ' J, = То *‘°3 £* с/л S far : -*. = j- — s: /0 г* a. 1”*" 4 У 3JL. Per car ben bebrech/oriJc, -Z = 5~<7х/< '*11 = 3. Д/ rm rWT t 3 For mercury .* * Jl> =г 0.311* rm 1ЪЗ*1о' s 1 J ) t 1 2.г Я 2.2 R (Pressure-depth relationship) A closed tank is par- tially filled with glycerin. If the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft, what is the pressure in lb/ft2 at the bottom of the tank? (ANS: 1650 lb/ft2) qlycerin- hn ft) (l = ait ) ♦ • 6 psi 10ft J 2~lR

2.3R (Gage-absolute pressure) On the inlet side of a pump a Bourdon pressure gage reads 600 Ib/ft2 vacuum. What is the corresponding absolute pressure if the local atmospheric pres- sure is 14.7 psia? (ANS: 10.5 psia) «10.£ psia 2.4R 2.-IR t Manometer) A tank is constructed of a scries of cyl- inders having diameters of 0.30. 0.25, and 0.15 m as shown in Fig. P2.4R. The tank contains oil, water, and glycerin and a mercury manometer is attached to the bottom as illustrated. Cal- culate the manometer reading, h. (ANS: 0.0327 m) FIGURE P2.4R -7L “ -------—------------------------------------------------ /35 ЛИ * й. 0327 аи 2-2%

2.5R | 2.5R i Manometer) A mercury manometer is used to mea- sure the pressure difference in the two pipelines of Fig. P2 5R. Fuel oil (specific weight = 53.0 lb/ft*) is flowing in A and SAE 30 lube oil (specific weight = 57.0 lb/ft’) is flowing in B. An air pocket has become entrapped in the lube oil as indi- cated. Determine the pressure in pipe В if the pressure in A is 15.3 psi. (ANS: 18.2 psi) Thu$} 2.6R 2.6R (Manometer) Determine the angle 0 of the inclined tube shown in Fig. P2.6R if the pressure at A is 1 psi greater than that at B. (ANS: 19.3 deg» £ , О.оМьг.ч*,) ~ 2-3R

2.7R | 2.7 R (Force on plane surface) A swimming pool is 18 m long and 7 m wide. Determine the magnitude and location of the resultant force of the water on the vertical end of the pool where the depth is 2Л m. (ANS: 214 kN on centerline. 1.67 m below surface) b = vidlh - I'm Fa* Ж A = Л.ГО % ‘ **-) - *» /m — 7”4« farce of- 214 1.Ы be tout surface ver-had сы/егЬ*е »f 2-W

2.8 R (Force nn plane surface) The vertical cross section of a 7-m-long dosed storage tank is shown in Fig. P2.8R. The tank contains ethyl alcohol and the air pressure is 40 kPa. De- termine the magnitude of the resultant fluid force acting on one end of the tank. (ANS: 847 kN) FIGURE P2.8R Л' (From Table 1.6 о ethyl . alcohol + IT - (yo (пл*%£)(-?-)( = k n F'ar are* з: ^3 s Ai r ^3 - + x***) = 2-*3 kt/ ~ ^ki Г ^г. r ^3 = /6>O ku г t- w kN G Л. hl 2-5R

2.4R | 2.9 R (Center of pressu re) A 3-ft-diameter circular plate is located in the vertical side of an open tank containing gasoline. The resultant force that the gasoline exerts on the plate acts 3.1 in. below the centroid of the plate. What is the depth of the liquid above the centroid? (ANS: 2.18 fll 5o fe' 3.13 fj 2.-6R

2.IOR | 2.I0 R (Force on plane surface) A gate having the trian- gular shape shown in Fig. P2. IOR is located in the vertical side of an open tank. The gate is hinged about the horizontal axis AB. The force of the water on the gate creates a moment with respect to the axis AB. Determine the magnitude of this mo- ment. (ANS: .1890 kN mi A whtrt ₽ Thittj * 2/ZoA.H л** ' t], ~;-rh & 1 4 а» Л./ ZZ 4 е z О fa = ~Thu5/ hr de£trrnnH. fa abtui. A-В - (A1 Z.° = WfO JlN-m ----------- - — ' =c Ulnae. ДВ 2-7/?

2.IIR | 2.11R (hire: <>n plane surfacei The rectangular gale CD of Fig P2.11R is 1.8 tn wide and 2.0 m long. Assuming the material of the gate to be homogeneous and neglecting friction at the hinge C, determine the weight of the gate necessary to keep it shut until the water level rises to 2.0 m above the hinge. IANS: 180 kN) 2'8R

2.I2R1 2.I2R (Force on curved surface) A gate in the form of a partial cylindrical surface (called a Tainter gate) holds back water on top of a dam as shown in Fig. P2.12R. The radius of the surface is 22 ft. and its length is 36 ft. The gate can pivot about point A. and the pivot point is 10 ft above the seat. C. Determine the magnitude of the resultant water force on the gate. Will the resultant pass through the pivot? Explain. (ANS; 118.000 lb) Lei F^ be -force o-f- jo-ie on flmd and t? - V / Д ' - J [2^00 Л A-l^ 4= ’ (Lz'1 = 53^00 lb С в A fl» A7.0* Л = 22-J2a. De ec Ae « 21- а г cos© ЧПИ x VrHt£; 4A<.« K 3L^ и here : A CDg ' A0CDe ~ A(jc£ “ " Aecf з 2. ч о fi As«'\« * \es = " ('•Л-Г” с««.3й) г /U о -Pt1 Thus) ^CD£ ~ (lD/Ьм Z*o-f±) - iLo-Pt* » ?-oo/e (iZf k} = l!jCOI> IL continued 2-qR

2.l2Rconlinued\ For QQui ЛЬнит = o F, - F, - IL Ahoj ZFj = o or F^cy = F^-'W - 53,1£>o IL-j^odo il = 35)^00 H = H?,ooi> H> The. direction of all differenfal Forces ach»^ a» i~he ale. is f>erf>ertdiCiflar -h> The <Jale surface t artU There!*» -f~he rfSulhjri/ tn us! pass "through 1"he )nlersec.-hort of ell these -forces и hick pomt A, yes. 2- / OF

2.l2Rconlinued\ For QQui ЛЬнит = o F, - F, - IL Ahoj ZFj = o or F^cy = F^-'W - 53,1£>o IL-j^odo il = 35)^00 H = H?,ooi> H> The. direction of all differenfal Forces ach»^ a» i~he ale. is f>erf>ertdiCiflar -h> The <Jale surface t artU There!*» -f~he rfSulhjri/ tn us! pass "through 1"he )nlersec.-hort of ell these -forces и hick pomt A, yes. 2- / OF

2.13 R | 2.13R (Force on curved surface) A conical plug is located in the side of a tank as shown in Fig. 2.13R. (a) Show that the horizontal component of the force of the water on the plug does not depend on h. (b) For the depth indicated, what is the mag- nitude of this component? (ANS: 735 lb) (a) Consider < cylinder of fluid of diameler)Oi and uith The pH* j removed fat figure J. F*« pressure cUsfribuhdoi over The. right and left sur faces are shown. Ld< role that the pressures cancel ercepi Tor the center area, of diamehetj d. The pressure di s-f-nb uTor aver this center area. (**/// yield < ttSul/dni UhtcA is independent of X and depends on /he ftm'd specific, weight , the fluid dtplhj and the hole diamekr^ 4. (b) For <* circular area, of dianncfert al91 ft, F^- A = 7^ lb 2'HR

Z.l+R | 2.14R (Force on curved surface) The 9-ft-long cylinder of Fig. P2.I4R floats in oil and rests against a wall. Determine the horizontal force the cylinder exerts on the wall at the point of contact, A. (ANS: 2300 lb) The hori^onia I -/erces ac.Ttn^ on "Hie -free - bady-d>a ъкои>н <s>*i "The . Гъг e^ui h'kriumj F = F-F A ' i F^ n The hon'jOfi/eJ Ta F, -- *4, A = И» & Ui/iere Si нее. "flt ui«l| exerts on "foe Cylinder s 8 ЛЗо lb F^? й Чех Ai = (51.0 - (рЯЗо /Ь 4^еи R = ^ХЗо 11» ~ Зо 11 “ 2 Зоо > ox "^e umU A - 2-/2 R

2./5R\ 2.I5R (Buoyancy) A hot-air balloon weighs 500 lb, in- cluding the weight of the balloon, the basket, and one person. The air outside the balloon has a temperature of 80 °F, and the heated air inside the balloon has a temperature of 150 °F. As- sume the inside and outside air to be at standard atmospheric pressure of 14.7 psia. Determine the required volume of the balloon to support the weight. If the balloon had a spherical shape, what would be the required diameter? (ANS: 59.200 ft3: 48.5 ft I F»r equilibrium } ^vertical ~ ° So P's 5 * 4 Ulbere •’ /Г - bueytni /erct. s •/ air imide btlloo* Utecht bosAei and lead Thus, +4 From Hie < idee I J43 ^,ur У-- ° R-T cn «т or For ou-fatde air until 7”s 8b*F-t 5'¥o*R t _ а.О7 3Г4 £, (l7/L -gjb )($*>*) Similarltj /or inside air unth T - is"o*F +*/bo * faio*R } ~Thu5 -i-rom Fed) c /L . = 500 _______________________________________________ "V" ~ у у • .j Л 6135 L 7^» ~ ^bS)l •>_____ q »</!•.Jt - cirticie u --------- ••r ttr for spherical shape, (luiih dianueder^ ТГ Н3= S^UX) -ft2 So iht-1 /.= /8.3 -Pi ~r 4 - 2-I3R

2./6R | 2.16R (Buoyancy) An irregularly shaped piece of a solid material weighs 8.05 lb in air and 5.26 lb when completely submerged in water. Determine the density of the material. t ANS: 5 60 slugs/ft‘i / W (in ^л1ег} S’. Zb !b ?. OS lb S.Lo 2-/VR

2.I7R | 2.17R i Buoyancy, force on plane surface) A cube. 4 ft on a side, weighs 3000 lb and floats half-submerged in an open tank as shown in Fig. P2.17R. For a liquid depth of 10 ft, de- termine the force of the liquid on the inclined section AB of the tank wall. The width of the wall is 8 ft. Show the magnitude, direction, and location of the force on a sketch. (ANS: 75.000 lb on centerline. 13.33 ft along wall from free surface) ъМ к = 3^,00 lb Де At, frhcA Uktrt h^s S’ J» (п.зз л) - 7St00otL _ z-£ CS/tJl'zo/i) * 1й& = /3.33 fl £ (/» 2-ISR

2.I8R | 2. IKK (Kigid motion) A container that is partially filled with water is pulled with a constant acceleration along a plane horizontal surface. With this acceleration the water sur- face slopes downward at an angle of 40° with respect to the horizontal. Determine the acceleration. Express your answer in m/s2. IANS: 8.2.3 m/s'i and , 'Thus, » $ tan Vo‘ = (j, H = 8.21 ±1 For Ду «8.23-^ 2-/6R

2.I4R I 2.I9R I Rigid-body motion) An open, 2-ft-diameter tank contains waler to a depth of 3 ft when at rest. If the tank is rotated about its vertical axis with an angular velocity of 160 rev/min. what is the minimum height of the tank walls to pre- vent water from spilling over the sides? (ANS: 5 IX ft) The 1/о1ц/пс of fluid м lhe. rotzlin^ ftnk is ушч by 2V tf. : \ ХП-Л--Й c Jo _ irco'R * тгЛ R- rfr * ZTT Md 'TUI ft? ♦ . + 7T^e f/A) = 7Г (2dt + ft* ottlb fe in ft} Slhct ibe. Im fa I volvmt. V-TTR'i - = 1Г (iflYfe/t)- ^ft* 4* and th*. final vo/фпе tnuii hr e^unl^ S ТГ (2J2 1-f.} ft*- 37Г ft3 -fos a 920 ft Thus, fvom 7fc« first1 -Кг * fi Sri0 И . , = s./г ft tr',n Л. (52.2 2-I7R

3.IR (F = ma along streamline) What pressure gradient along the streamline, dp!ds, is required to accelerate air at stan- dard temperature and pressure in a horizontal pipe at a rate of 300 ft/sJ? (ANS: -0.714 lb/ft3) where 6*0 and. Thus, = -O.7/¥ V 3-lR

3.2R 3.2R (F = ma normal to streamline) An incompressible, inviscid fluid flows steadily with circular streamlines around a horizontal bend as shown in Fig. P3.2R. The radial variation of the velocity profile is given by rV = roVo. where Vo is the velocity at the inside of the bend which has radius r = ru. Determine the pressure variation across the bend in terms of Vq. Го. p. r. and Po. where p0 is the pressure at r = r0. Plot the pressure distribution, p = p(r), if r0 = 1.2 m, r, = 1.8 m, Vo = 12 m/s, p0 = 20 kN/m2, and the fluid is water. Neglect gravity. (ANS:p„ + 0.5pV-|l (r^r)2]» X FIGURE P3.2R and Г'Х'Я* Г,-П^г fa Thu? Hencet f-f.- -=^===^ж==^==== For -the given data • /> =20^ % '(¥)*] = 2.0 + 7/.?p - , vhere Г"m 3-2.R

З.ЗК (Stagnation pressure) A hang glider soars through standard sea level air with an airspeed of 10 m/s. What is the gage pressure at a stagnation point on the structure? (ANS 61.5 Pa) (!) (*) f,s0 = Ф +'м *2л Wi^ Z, -Zz j fl -OJ ' vt */оф and V,*0 Thus, 10 f/ = 61.5 ^/m* s 6/.5 Pg ЗЛР | 3.4R (Bernoulli equation) The pressure in domestic water pipes is typically 60 psi above atmospheric. If viscous effects are neglected, determine the height reached by a jet of water through a small hole in the top of the pipe. (ANS: 138 fl) Thu^ n 6z.¥ Jh. = /33ft ___ c*) z,*A FT t у but f=6o psi fi=° V^O V, *0 if the hole diameter is much smaller than the pipe diameter Note: Because of viscous effects between the water and the p'Pe and the water and, the air, the actual value of would be less than 139 ft. 3~3R

3.5R | 33R i Heads i A 4-in.diameter pipe carries 300 gal/min of waler al a pressure of 30 psi. Determine (a) the pressure head in feet of water, <b) the velocity head, and <c) the total head with reference to a datum plane 20 ft below the pipe. (ANS: 69.2 ft: 0.909 ft: 90.1 ft) 30 psi 20H (.C) s 69.2 +0.909+20 - 40.! f/ 3-¥R

T6R~T 3.6R (Free jet t Water flows from a nozzle of triangular cross section as shown tn Fig. P3.6R. After it has fallen a dis- tance of 2.7 ft. its cross section is circular (because of surface tension effects) with a diameter I) = 0.11 ft. Determine the flowrate. Q. (ANS: 0.158 ft'/s) /f>! . J. » — a. > 7 where *fz *0 t 2z*o,2t •Z.ifi FIGURE P3.6R AV*?*, //-(A/4)1 = = O.OOfSO Щ and ^1 = i(o.iUi)(o./i^H) = 0.01*6 ft2 O.l^n Thus, _______________* n = (Q.O09S0ff) /2 (32.1&(Z.7H) Ч Г. (0.009SQ (0.0!s& m3- J = OJS8^- *0 3-Sf{

3.7R I 5.7R i Bernoulli i ontinuit) Water flows into a large tank at a rate of 0.011 m’/s as shown in Fig. P3.7R. The water leaves the lank through 20 holes in the bottom of the lank, each of which produces a stream of 10-mm diameter. Determine the equilibrium height, h. for steady state operation. (ANS 2.50 ml «(»<Ь but where 0.01/ anol QZ'Z0/f2^ ^2.of0^Vz + jr + ay 4 whet -r-L and / nvs, so that 0- OH-!?- = 2of (о. Dim/ ^2(4.8!h or 3-6R

Ж! 3.XR < Bernoulli continuity I Gasoline flows from a 0.3-m- diamcter pipe in which the pressure is 300 kPa into a 0.15-m- diameter pipe in which the pressure is 120 kPa. If the pipes are horizontal and viscous effects are negligible, determine the flowrate. (ANS: 0.420 m‘/s) 0.3 m 0.15 m jO. =300kPa r‘ f^lZOkPa T Thus, Q V, = %(o.3mf(S.4* t) = O.vzo^ ЗЛЯ~[ 3.9R i Bernoulli continuity ) Water flows steadily through the pipe shown in Fig. P3.9R such that the pressures at sections (I) and (2) are 300 kPa and 100 kPa. respectively. Determine the diameter of the pipe at section (2), D2. if the velocity at section I is 20 m/s and viscous effects are negligible. IANS: 0.06XX in) 300 kPa 100 kPa FIGURE P3.9R 300 х/o'3= lOQ^o N/*\ + VzZ f.ftfX/oW/»3 U4.9lm/sK) 9. Sox Ю*Mb3 2(9Sln/s3) Thusj Vz=^2.zm/s so ihd since V,At sl/z/)l or Vi^b, , then Ч-2.3. m/sJ 3-7R

3.10 R I ЛЮК (Bernoulli continuity) Water flows steadily through a diverging tube as shown in Fig. P3.10R. Determine the velocity, V, al the exit of the tube if frictional effects arc negligible. (ANS: 1.04 ft/s) FIGURE P3.10R where AM eAVx , or °rf. +*Г«>-»-г(вЧ = -Л+'2*‘в& ,a By combining (!\M. end(3}- * -даТРАГ) s -Пл^- +I(5Z2^ 0Г Vi- 2./3 ff/4 and V, Thus, since Уз Ал * К At., iben 3-8 R

Т7Щ 3.11R (Bernoulli continuity Pilot tubel Two Pitot tubes and two static pressure taps are placed in the pipe contraction shown in Fig. P3.11R. The flowing fluid is water, and viscous effects are negligible. Determine the two manometer readings. h and H. (ANS: 0; 0 252 fl) S ф + with and 1^14*0 Г a/ r r Thl/S) f>t -fl so that manometer considerations j>t + r (i^h)s + + biofht or with f, this fives h~0 Also, +57 +Jj s *’* w^ere 2»~ j z* ? 14 s2$, Aj Ц = Af 14 or 4 - * - 4^-15 = 4 u4 Ш, - (k- A)«* ’ °-“,n (l) B"f A-Wi - ГН * ft - f(г^-г> °Г *Н-(г,-Ъ) Ггог» and (s~) м/e obtain О.149Я* Hft) or H- 0-252. {I 3-qR

3.UR | 3.12R (Bernoulli continuity i Water collects in the bottom of a rectangular oil tank as shown in Fig. P3.12R How long will it take for the water to drain from the tank through a 0.02- m-diameter drain hole in the bottom of the tank? Assume quasi- steady flow. (ANS: 2 45 hr) Rectangular tank: 2.6 m x 9.5 m FIGURE P3.12R *г‘ where TL & Thus, since -SG*O.87t and XO -4^4 h = or 4 = (j 0 v or _________________________t 14 = h + 0.87(1.9 m))‘ = ’jh+/.6S3 ' %-, where h^m A/so, Q « /)2V2 » -£44 « т(а02”?[*¥31Ь +1Л531 -?] (I) anal = /.3<?x/o'3-{h^6ss where fl, » 2.6m (9.Sm) = 24(7/»a Hence, Combine E^$. (0 and (ti io give -^’.63x/o's h +/.653 or । n h -o Г db J +/.653‘ Л-0.7 or _________ 2 ^h+MS3 = -X^3K/(? where if B lime io drain ihe waier = -5.b3x/dsi{ ТЬ VS; if - 8.83 у io3 s = Z.v-rhn 3-/0R

эд 3.13 R (Cavitation) Water flows past the hydrofoil shown in Fig. P3.13R with an upstream velocity of Vo. A more ad- vanced analysis indicates that the maximum velocity of the wa- ter in the entire flow field occurs at point 8 and is equal to 1 1 Vo. Calculate the velocity, V», at which cavitation will begin if the atmospheric pressure is 101 kPa (abs) and the vapor pressure of the water is 3.2 kPa (abs). (ANS’31.4 m/s) (I) Ви/ 14 = II Vo and fl - 101 кРл (abs) 3<? /ha/ fy (ft fives ,о.бт or Vo = 3/.¥ £ 3-/IR

з.т | 3.14 R ( Flowrate) Water flows through the pipe contraction shown in Fig. P3.14R. For the given 0.2-m difference in ma- nometer level, determine the flowrate as a function of the di- ameter of the small pipe, D (ANS: 0.015ft m'/s) FIGURE P3.14R where Z/^Zj. ™d Thus, But , fl, * rh, and flz « trhi , so -that flz -fl, - * °-2 * Thus, ( or ________________, J =^(o./u>fy2.(9.e/^)(o.2m) = 3-I2R

ЗЛИ 3.15 R (Channel How > Water flows down the ramp shown in the channel of Fig. P3.15R. The channel width decreases from 15 ft at section (1) to 9 ft at section (2). For the conditions shown, determine the flowrate. i ANS: 509 ti'/s) /l/зо, HzVz Z Лк v - kklikil и ~ (зН)(№ 1 (I) where > fax' ® > 22.' 3 and = 3.3314 Thus, Ey. (!) becomes [з.ззх-/]14Я= 2(32*2 jL)(8-3)H , or V, = S-d-S’ Hence, Q = A,/, = (6 H) (IS M $) = fotzr 3-/3R

3.I6R | 3.16 R (Channel flowi Water flows over the spillway shown in Fig. P3.16R. If the velocity is uniform at sections (1) and (2) and viscous effects are negligible, determine the flow- rate per unit width of the spillway. (ANS: 7.44 m7s) + Za where, if points (') and are located on the free surface, *n<i A/so, /?, И ' Ла 14 » ho M. \ f 14 = '0778 *4 Thus, (!) becomes ZH^l' -(0.273)1] = 3.Sm-7.0 m, or Hence, f= V2Z2 =(7.^)f/.On>) = 7.W&- V2 = 7.^ 2- о 3-HR

3.I7 R | < (Energy line hydraulic grade line) Draw the en- ergy line and hydraulic grade line for the flow shown in Problem 3.43. 3.43 A smooth plastic, 10-m-long garden hose with an in- side diameter of 20 mm is used to drain a wading pool as is shown in Fig. P3.43. If viscous effects are neglected, what is the flowrate from the pool? FIGURE P3.43 5/лсе * 2, e t wHh ft, *f)x *0, V, =oj ?z *o and 2/ • O.Z/я +О.13/Я * O.VJm, if follows that *0.¥3m *9 HGL For inviscid flow with nt pumps or IurLines, the energy Uno (EL} is horizontal t at an elevation of the free surface. The hydraulic grade line (HGL} is one velocity head lower-even wild the pipe outlet. Since the fluid velocity is constant through wt the pipe with ^O.uJnfhe above diagram is obtained. 3-/SR

3.10 R I 3.I8R I Restrictions un Bernoulli equation I A 0.3-m- diameter soccer ball, pressurized lo 20 kPa. develops a small leak with an area equivalent to 0.006 mm:. If viscous effects are neglected and the air is assumed to be incompressible, de- termine the flowrate through the hole. Would the ball become noticeably softer during a 1-hr soccer game? Explain. Is it rea- sonable to assume incompressible flow for this situation? Ex- plain. (ANS: O.rip X 10“ ’ m'/s; yes: no, Ma > 0.3) v/here -2,9 z о and % ‘° Thus, or ^here ? = Assume 7/ 9 st that (I0l^20)^o3^_________ Г (286j9 ^(Z 73 K ThUZ,____________, ’ v J 2 tutf-k) _ n_ so thad <? ' A H « (0.004 = r.rox/o-7-^ If this fIon/rate continued for one hoorj fhe vtb/me of air leaving the ball would be V = Qi ^(9.46X/o'7-^)(3/oos) = 3.S9mo3m3 Since the volume of the ball is approximately ~r _ 3.sfx/o m* _ 0 2.55^2S.s% of the ain would escape. ybaii I*J x/о т3 Ihe ball would become /loficably softer A/ote that Маг= where ^^kRT1 or if fz^/s*C cz = [Zy<7 (^6.9 ^,)(z7J^/s/')^3' 3¥0 so that the Mffh flu/nber is Mn = 1*4 Jt- s 0 ¥80 >0.3 T//US; should assume compressible ’ ai 3*0£ f/ow. ------------ 3-/6R

3.I4R | Л. I9R (Restrictions on Bernoulli equation) Niagara Falls is approximately 167 ft high. If the water flows over the crest of the falls with a velocity of 8 ft/s and viscous effects are neglected, with what velocity does the water strike the rocks at the bottom of the falls? What is the maximum pressure of the waler on the rocks? Repeat the calculations for the 1430-ft-high Upper Yosemite Falls in Yosemite National Park. Is it reason- able to neglect viscous effects for these falls? Explain. (ANS: 104 ft/s. 72.8 psi; 304 ft/s. 620 psi; no) Thus and А. V/ 4/7/ -h а) И//7Л h*M7fit f fa* td2.¥-^(U7fl) =/0,500^ = 72.9 psi b) With h-/^3of/l aftd (Mf « ^зао Ж 620psi Aerodynamic drag on the wafer would reduce ~fhe values о/ 14 and As (especially hr the h C ° ff case). 3.17 R

*77П 4.1 R t Streamlines I The velocity field in a flow is given by V = rM + jftj- () Plot the streamline through the origin at times t = 0, t = 1. and t “ 2. (b) Do the streamlines plotted in part (a) coincide with the path of particles through the origin? Explain. (ANS; lx + C: no) а) и.e X2y and nr » Xх I where the streamlines are obtained from ~^c'Wsx*y or ydy-idx whicht for a given time can be integrated to give •£yz -ix *C where C is a constant. For streamlines through the origin (**0, y*o)^ C*0. Thus’ yz= 2tx or y=0 for 1*0 y- fzx for i*!? and y-zVx for l“2. These streamlines are plotted below. b) Since this is an unsteady flow (i.e. ^^xz^o) and since *y (i.e. of a given location the slope is a function of Итв)) streamlines and palhlines do not coincide.

±2Rj 4.2 R । Sireainlint-s । A velocity field is given by и = у — I and и = у - 2. where и and v are in m/s and x and у are in meters. Plot the streamline that passes through the point (x, y) (4. 3). Compare this streamline with the streakline through the point (x, y) = (4. 3). (ANS: л - у + ln(y 2) ♦ I) U- у-I } v*y~2. where the streamlines are obtained from dx u y-/ °r (#3* = fa °r - f(A =х+г' J (У J J y J'' /s a constant. From integral tables : = У"2 +21п(У'2^ s 1л(У'2) Thus, the streamlines are given by у -2 +2In(y-2)-/n(y-2) = X/G у +ln(y-2) = X + C , where C is a constant (I) For the streamline that passes through and у s3f the Value of 0 is found from Fj.(0 as: 3+1п(э-2)*Ч+С or C-l Thus, Xs y+ln(y~2)+l This streamline is plotted be low: Mole •’ fts X—-~oo} y—*2 , Also} since U^y^l^ilh У^ anywhere on this streamlinet it follows that U>0. The flow is from left to right. Since the flow is steady, streamlines are the same as streak lines. 4-2R

4.3 R 4.3R (Material derivative) The pressure in the pipe near the discharge of a reciprocating pump fluctuates according to p = [200 + 40 sin(8r)] kPa, where I is in seconds. If the fluid speed in the pipe is 5 m/s, determine the maximum rate of change of pressure experienced by a fluid particle. (ANS: 320 kPa/ч) Since U = , wc0 d follows Ma/ jr/'M p- p(1) or $ • 40(8)cos(8t) Thu3) = 40(8)^=320^- mex /Vote •’ Since =0 the value of U. is not important. 4.4 R 4.4 R (Acceleration) A shock wave is a very thin layer (thickness “ () in a high-speed (supersonic) gas flow across which the flow properties (velocity, density, pressure, etc.) change from state (I) to state (2) as shown in Fig. P4.4R. If V( = 1800 fps, V2 = 700 fps. and ( = IO'4 in., estimate the average deceleration of the gas as it flows across the shock wave. How many g’s deceleration does this represent? (ANS: 1.65 x 10" tt/s’; 5.12 X IO9) FIGURE P4.4R д = + V-W so with V • !/(*)< f а ж — U> jy t Without knowing the actual velocity distribution f W=U.(x)) the iu. ~ (Vi+Vi) ( Vi -16) _ (1800+700) fps (~roo~ 1800) fps <^x9 UTX~ ~ 2 "T 2. / JOj*) Г1 Qx = -/.65^/0 _f . Ox -1.65 X/O 3 This is 3i.i = -5.12 x/o9 4.3R

4.5 R i Acceleration i Air flows through a pipe with a uni- form velocity of V » 5 ri ft/s, where / is in seconds. Determine the acceleration at time I = —1.0, and I s. (ANS; 10 1 ft/s*:O: 101 fi/sj a*-^r+7*vV With u = j v *0,^*0 /his becomes sb.ctg-0. Thus, a • -Wt at ts~/s a = 0= at t = 0 and a = at /•! s 4-4R

32П 4.6 R * federation) A fluid flows steadily along the streamline as shown in Fig. P4.6R. Determine the acceleration at point A. At point A w hat is the angle between the acceleration and the x axis? At point A what is the angle between the accel- eration and the streamline? (ANS: 10 n 301 ft s' 4K.5 dec: 1X5 deg) a = n ♦ Os n + 3 = (l°0^ n +(/о#)(з or a - /On +3o$ Q, In terms of unit vectors i'and j n *~sin3rfi +cos3(fj and S - cos, 30*t + sin 30*f л f. а =ю(-о.5$ +-O.B66J) +-3o(O.S66f +0.Sj) = 2l.0i +23.7J Hence, 0 • tan ‘ж ten ' or 0= ¥Q.5‘ and , a o(=e-30 -/8.5° ¥-5R.

4.7R 4.7 R (Acceleration) in the conical nozzle shown in Fig. P4.7R the streamlines are essentially radial lines emanating from point A and the fluid velocity is given approximately by V “ C/t2, where C is a constant. The fluid velocity is 2 m/s along the centerline at the beginning of the nozzle (x “ 0). Determine the acceleration along the nozzle centerline as a func- tion of x. What is the value of the acceleration at x = 0 and x = 0.3 m? ANS: 1.037/(0.6 x)' i m/s-; 13.3 i m/s2; 427 i m/s j Atony the nozzle centerline. 0.6 m FIGURE P4.7R + becomes a = U $ ? where U = Thus, with F+X or F-0.6-X. U. = —1 U (0.6-X) Hence, a [(0.6-x)3-] S/дсе Ul-2.^ at X~o it follows that s (0.6m) 1. » _ s2- m* 2.C * Д С л = 2(0.72) -ух Л (o.6-x/ ~ (o.6-x)s ~ (o.i^xfm3 _ /.037 (o.6'X)s /Н X'O , d 22 /3.3? . at Х^о.зт ^a= ^27^

¥gR| 4.8 R (Reynolds transport theorem) A sanding operation injects IO5 particles/s into the air in a nx>m as shown in Fig. P4.8R. The amount of dust in the room is maintained at a con- stant level by a ventilating fan that draws clean air into the room at section (1) and expels dusty air at section (2). Consider a control volume whose surface is the interior surface of the room (excluding the sander) and a system consisting of the material within the control volume at time I = 0. (a) If N is the number of particles, discuss the physical meaning of and evaluate the terms [>Ntyt/Dt and dNnJdt. (b) Use the Reynolds transport theorem to determine the concentration of particles (parti - cles/m’l in the exhaust air for steady state conditions. (ANS: 5 X 10'perticles/m'j FIGURE P4.8R -0^ = time rate of change of the member of particles in the system, di time 1*0 [he system consists of Nparticles. ftssuminq that the particles do not get glued together, /V remains constant. Thus,^O = time rate of change of the number of particles in the control volume. Depending on the rale at which the sander creates particles and ejects them into the room compared to the rate al which the fan draws them from the poomt we could have 1 O. b) +ne] rate of flow of particles out of control volume or for steady state ^r=0 so that: flow of particles indo control volume (from sander, none enter fit) - flow of particles out of control volume (thrush fan exhaust, ) Thus, JO5 Р—И - \^Агпх where nz- particle concentration НеПСе' .„s particles ... Л ’ „ - 10 Petioles. ' fifiM - SXI° 4--7R

таи 4.ЧК (Flo» rate) Water flows through the rectangular chan- nel shown in Fig. P4.9R with a uniform velocity. Directly in- tegrate Eqs. 4.16 and 4.17 with b I to determine the mass flowrate (kg/s) across and Л-B of the control volume. Repeat for C-D. (ANS: 18.000 kg/s; 18.000 kg/s) Equation ¥.171 orf with p = and bsi this qives B- = -W-fi [(зЧЩГМЛ - /Л *** j */6'000 & or B.„t -18000^

Ч.ЮН | 4.1 OR (How rale Air blows through two windows as in- dicated in Fig. P4. IOR. Use Eq. 4.16 with b I Ip to determine the volume flowrate (ft’/sl through each window. Explain the relationship between the two results you obtained. (ANS: SO ft1/»; 160 fl7») Top V«w Wind Wind V-20ft/s V= 20 ft/s • f LU* m FIGURE P4.10R BarjpbV-Л^ For the left window (? *= У’л • ^V'fidA ° Q = ^20sin 30*dfi *(20sin 30* &)(Ч- ft)(2 Ц) s 8^ s' For the right window Q * J V'h dA °Г = (20^)(gf/)(Zft) - 160 V-h s V sin30* V V-h=V

4-.//R | 4.1 IK (Control volume system) Air flows over a flat plate with a velocity profile given by V = u(y)i, where и 2y ft/s for 0 S у S 0.5 ft and и « 1 ft/s for у > 0.5 ft as shown in Fig. P4.11R. The fixed rectangular control volume ABCD co- incides with the system at time t = 0. Make a sketch to indicate (a) the boundary of the system at time t = 0.1 s. (b) the fluid that moved out of the control volume in the interval 0 S / £ 0.1 s. and (c) the fluid that moved into the control volume during that time interval. Since V sU(y)i, each fluid particle -travels only in the x~diredionf with the distance of travel AXSU it г where bi-О. I s. Thus, iXja = <$ХЛ = 0 since U-0 at y-O. Also • ixc - (t^)(O.ls) S0.l ft. The fluid originally atom; tines A~B and C-b move to positions A-E-B' and b-F-C' as shown it the figure below. The location of the system at ic0.!s and the fluid that moved into or out of the control volume is indicated. .control Volume .-------system at t^O.la Jt-IOR

5.IR | 5.1 R (Continuity equation) Water flows steadily through a 2-in.-inside-diameter pipe at the rate of 200 gal/min. The 2- in. pipe branches into two l-in.-insidc-diameter pipes. If the average velocity in one of the 1 -in. pipes is 30 ft/s. what is the average velocity in the other 1-in. pipe? (ANS: 51.7 ft/s) 5'1 К

5.2R | 5.2R (Continuity equation) Air (assumed incompressible) flows steadily into the square inlet of an air scoop with the nonuniform velocity profile indicated in Fig. P5.2R. The air exits as a uniform flow through a round pipe I ft in diameter, (a) Determine the average velocity at the exit plane, (b) In one minute, how many pounds of air pass through the scoop? (ANS: 191 ft/s; 688 Ib/tnin) b) Weig/Н flowrate = (150&) = - &89JL £-2R

T3R~[ 5.ЛН (Continuity equation i WateratO.l m’/s and alcohol (SG = 0.8) al 0.3 mJ/s arc mixed in a v-duct as shown in Fig. P5.3R. What is the average density of the mixture of alcohol and water? (ANS: 849kg/m'l Alcohol (SG = 0.81 Q 0.3 m’/$ FIGURE P5.3R For steady -flow §df) = 0 t0F Fo у w = m /A3 AlsOj stnee Hie wafer arid alcohol may he considered m com pre $5/ hie Я ^2 Я ^3 / Z i4€ gef - 5-3R

5.4R 5.4R (Average vetodly) The flow in an open channel has a velocity distribution V - ft/s where U = frec-surface velocity, у = perpendicular diMance from the channel bottom in feet, and Л = depth of the channel in feet. Determine the average velocity of the channel stream as a fraction of U. (ANS: 0.833» For aoy F/ovS Cross section m = pAu • Г f v. A dA '4 5-W?

5.5R | S.5R 11 invar momentum) Water flows through a right an- gle valve at the rate of 1000 Ibm/s as is shown in Fig. P5.5R. The pressure just upstream of the valve is 90 psi and the pres- sure drop across the valve is 50 psi. The inside diameters of the valve inlet and exit pipes are 12 and 24 in. If the flow through the valve occurs in a horizontal plane, determine the л and у components of the force exerted by the valve on the water. (ANS: 18.2001b; 10.800 lb> a) of ihe momentum equation is X Vi fl 4 ^VALV£ Ax cs V or F = */*» fii f^x but U^O, АГ, = - Ц t , and ~ fzftz * di 14 } where m = f4 A -30 FAx = (9o-so)^(^(2^/}=/8,200/Z where we have Used И * -dp- or \J = 30 S-SR (continued)

5.5 R continued] Similarly, in the у - direction £nrf> V-n d/l t or ex f6»/ or *0 = -fi^i^ny t or Улу^ Thus, since s/4'7^ we Mn s (^dzief} +30 ^-(/9.7^} « 10,900 !b 5-6R

5.5 R continued] Similarly, in the у - direction £nrf> V-n d/l t or ex f6»/ or *0 = -fi^i^ny t or Улу^ Thus, since s/4'7^ we Mn s (^dzief} +30 ^-(/9.7^} « 10,900 !b 5-6R

5.6R | 5.6R I Linear momentum* A horizontal circular jet of air strikes a stationary flat plate as indicated in Fig. P5.6R, The jet velocity is 40 m/s and the jet diameter is 30 mm. If the air velocity magnitude remains constant as the air flows over the plate surface in the directions shown, determine: (a) the mag- nitude of FA, the anchoring force required to hold the plate sta- tionary. (b) the fraction of mass flow along the plate surface in each of the two directions shown, (c) the magnitude of FA. the anchoring force required to allow the plate to move to the right at a constant speed of 10 m/s. (ANS: 0.696 N; 0 933 and 0.0670: 0.391 N) The. toon - deform/y control w/име. shown in the sketch above is usd. (a) To determine the magnitude of F* &e apply the component of the hhear- momentum eyua-fion (Eg. 5.22 J atony the direction of FA. Thus, Scs/\T - £ГУ , or F • /я V-sin 3o* » pA, V1/ sin io* ~ pity Sm3oe A J I * J J аГ Д 7? TO OSO'”')' 3o‘)f'dd-\ « p.SUU л /»’/ fy) *9‘%i.t ------- 6t>9 To determine -the fraction of mats Flaw along the plate surface m eack of the 2- directions shown /л the sketch above, we apply the component of the linear momentum eguafion parallel to the surface of the PJakJ d/le£rx, h obinin <» Surfoct (continued) S-7R

5.6 R continued the value vf R /2) Съ) (V Since the. air veto Pty моуп) facte /г**мз constant it yen. Thus fn»n fy./ cue obfam "i'j4e"3a' Since V3^^X'S J £f- 2 becomes n>2 « /» - to - Cos 3o * рпм conserva/ion of- mass we conclude that n,3 Crmbimnj Eujf- 3 n~id У- *mc fet /П, • to; -to - tocos 3o‘ 3 J J J • c >4. (l-COS9»J - to {O. 0170} у 2 J dud to s to-ft-o.017^ = i-frto) 2 J ' ✓ Лим, to involves 93-33, $ to. л„</ m-involves 6-73, of fa. . • •*—*______________________________________ J (C)To tetevnine. +ht v^ayniiudo of P^ n^g/riol-fp allch* fld p/qfc to nove to toe riaht af я esnstont spot a of 70 за, we use . , ~ o’ 4 no* - cterwM/oy Contop/ tVo/име /lice the One /Л /<♦& SieJtP above -that Moves to the n'yht with о sfeed of /qm . Z^e 'h'wts/oh'nj control vo/unc l/neor мопеп/им /ец. 5-2n) /easts fn F = f7'' /vj’ ,0y)^'m^0 л f F • (,гз (to a. tonЗо^-/^Л л v I hf. at J ол/ and F ‘ o. 3f/ H 'л -== * Since И Vi. ~ a/>^ ft - Л *fs an^ 2<s S^J the Bernoulli equation is valid fro/n /-*2 and i-*3. Thus, ihere are no viscous e(f'ecis(8ennoul/i eqi/aiicn is Valid only (or inviscid Hod') SO thad -0. HenceJ fghny^aie £-8R

5.6 R continued the value vf R /2) Съ) (V Since the. air veto Pty моуп) facte /г**мз constant it yen. Thus fn»n fy./ cue obfam "i'j4e"3a' Since V3^^X'S J £f- 2 becomes n>2 « /» - to - Cos 3o * рпм conserva/ion of- mass we conclude that n,3 Crmbimnj Eujf- 3 n~id У- *mc fet /П, • to; -to - tocos 3o‘ 3 J J J • c >4. (l-COS9»J - to {O. 0170} у 2 J dud to s to-ft-o.017^ = i-frto) 2 J ' ✓ Лим, to involves 93-33, $ to. л„</ m-involves 6-73, of fa. . • •*—*______________________________________ J (C)To tetevnine. +ht v^ayniiudo of P^ n^g/riol-fp allch* fld p/qfc to nove to toe riaht af я esnstont spot a of 70 за, we use . , ~ o’ 4 no* - cterwM/oy Contop/ tVo/име /lice the One /Л /<♦& SieJtP above -that Moves to the n'yht with о sfeed of /qm . Z^e 'h'wts/oh'nj control vo/unc l/neor мопеп/им /ец. 5-2n) /easts fn F = f7'' /vj’ ,0y)^'m^0 л f F • (,гз (to a. tonЗо^-/^Л л v I hf. at J ол/ and F ‘ o. 3f/ H 'л -== * Since И Vi. ~ a/>^ ft - Л *fs an^ 2<s S^J the Bernoulli equation is valid fro/n /-*2 and i-*3. Thus, ihere are no viscous e(f'ecis(8ennoul/i eqi/aiicn is Valid only (or inviscid Hod') SO thad -0. HenceJ fghny^aie £-8R

5.7R t 1 invar топи ntuni i An axisymmetric device it used to partially "plug" the end of the round pipe shown in Fig. P5.7R. The air leaves in a radial direction with a speed of 50 ft/s as indicated. Gravity and viscous forces are negligible. De- termine the () flowrate through the pipe, (b) gage pressure at point (1). (c) gage pressure at the tip of the plug, point (2). tdl force, F, needed to hold the plug in place. (ANS: 23.6 ft’/s: 1.90 Ib/ft-. 2.97 Ib/ft2; 3.IK lb) For part (a) we dekrmine +be volume -flowndc -through +he P'P^ calculating the volume -f-lownate ot the dir leaving pi-f+er being turned by Hie ак\$дттеЬх pl<^. “Bias 3 - V3A3 = ₽ 22^ v For pari (b') we determine gage pressure at- (0 by applying -the. Bernoull i egua+ion +© +he tiow between 0") and +Ье nacLlol -Row leaving the plug., StzvKrn C*>). TVtus (continued} S-9K

5.7R continued | Fof pe\r+ <jc} ууе determine foe. cpqe pressure aVHt+ip of Hie plu^ point by applying the Bernoulli equation between poirtk (l} and <a). Thus, since \£-0, 4. Vi - fk e T " f For рдг+ Ш rfc. apply Hie linear «мотеи-кил 4o Hie CLon-tenH of ft®. cenVrbl volume s^ck-^ej atoue 4® get- Cu^V'hd/) or ^cs Thus, f upV' df\ since ^3-o J(l) -4?V,A, ’ f>A,-F o/-F«^A+4<Q- Thus, F - (b,fc }^‘ F - З.в lb

5.7R continued | Fof pe\r+ <jc} ууе determine foe. cpqe pressure aVHt+ip of Hie plu^ point by applying the Bernoulli equation between poirtk (l} and <a). Thus, since \£-0, 4. Vi - fk e T " f For рдг+ Ш rfc. apply Hie linear «мотеи-кил 4o Hie CLon-tenH of ft®. cenVrbl volume s^ck-^ej atoue 4® get- Cu^V'hd/) or ^cs Thus, f upV' df\ since ^3-o J(l) -4?V,A, ’ f>A,-F o/-F«^A+4<Q- Thus, F - (b,fc }^‘ F - З.в lb

5.QR | 5.XR (I.incur momentum I A nozzle is attached to an 80- mm inside-diameter flexible hose. The nozzle area is 500 mtn2. The Control volume shown in the sketch is used. Me assume that /he vertical co^pontd of /he anchoring ferrety / i$ exerted by the hose material. Ms further assume that the horiyontal component of the anchor/by do ret. F must be eteded by the hands holding the hose and noyyle stationary. Application of the horizontal ar X direction component of the. linear momtnteim equation leads do = F II) Application of Bernoulli's equation between seofions(l)am<^.)yields P a 2. or r. = К 6 Л, Combining fyS- * And 3 I/'. 2(p-pt) г ' ,> 1 ' and becomes A,X (continued) 5'11 R

XZ!-S

XZ!-S

5Т₽~Т 5.4R ( Linear momentum ( A horizontal air jet having a velocity of 50 m/s and a diameter of 20 mm strikes the inside surface of a hollow hemisphere as indicated in Fig. P5.9R. How large is the horizontal anchoring force needed to hold the hem- isphere in place? The magnitude of velocity of the air remains constant. (ANS: 1.93 N) FIGURE P5.9R The con/го! Volume shown in /he sketch is used. The x-component of /he momentum equation q/i/es JU.f> V-n dfl - or where for conservation of mass (J, 4 И - pa-ftiVz 53 m, /he mass flow rati. Thus, « rh (V, *14 ) - 2 mV since V, = V2^SoT //ote /hat V^Vi (i.e./he speed is constant\ but Vt = Sot * Vz * -SOtnr (i.e. /he veloci/y changes). rh = i.^ (^(0.020п/-)(50?) = 0.0193 If we obtain R ^(0.0143^(50^(2^/.93/V

5.IOR | 5.IOR (Linear momentum» Determine the magnitude of the horizontal component of the anchoring force required to hold in place the 10-foot-wide sluice gate shown in Fig. P5. IOR. Compare this result with the size of the horizontal component of the anchoring force required to hold in place the sluice gate when it is closed and the depth of waler upstream is 6 ft. (ANS: 5’10 lb; 11.200 lb. When the gate is dosed the wafer is stationary and the resultant wafer force on /he gate is ^closed ~' fc A Xhcf] = ft-?; b where H~ 6 ft and b iOff * width Thus, R,dosei = i («- //>00 Zi From /he control volume diagram shown, the X- component of the momentum equation becomes: $ UpV-tid/) or (!) open or - д л, v^ -irH'k-irUb + ₽lf M -M'bb where Ц-*# and Vz • ‘ V,4 ‘ £ Tfo, (¥^-ГМ)(ИЮ((-Я-)1 Miliatl) or 5-WR

5.IIR | 5.IIK il.incur momentum» Two jets of liquid, one with specific gravity 1.0 and the ocher with specific gravity 1.3. col- lide and form one homogeneous jet as shown in Fig. PS. 11R. Determine the speed. V. and the direction, в, of the combined jet. Gravity is negligible. (ANS: 6.97 ft/s: 70.3 deg) For the control volume shown, -the X-component of the momentum equalion hecom&s cs or (-VfCosSo^^Fv,)/), -(Vcose)f(v)A *o since there /к no force acting on the control Volume. Similarly j in the у-direction FIGURE P5.11 R $ЛГ?У-п *£fyt or as (-V/S/пЗо')^ ('Vjdi + Vz (-VZ)AZ +(Vsine)f> (V)A ' 0 a) By combining Eqs. (B and» (x) we oblain (divido (/) by (*)): cote* --------- Thus, cof 6 = so that 6 ~ У0,3е _______fi CDs 30* £ (о.2Н)г____________ X (0.2flF - e, (8^(o.zf/)xs/n3^ = 0,3S7 Also, for conservation of mass • f И4 = (?, * fi 14 Az (з) By combining Eqs. (B and L3) ^Vtcasdcf^, - Vcos6[q,V//}/ -O or since /^=A e, Vх соаЗО* = ^(8^ сосЗов_______________ и V cos6(^Vi^M (cos 70.3°)[^ (8^)4 !.з^ (И g)] - * £~ISR

5.12 R | 5.12K 11 .incur momentum । Waler flows vertically upward in a circular cross-sectional pipe as shown in Fig. P5.12R. At section (1). the velocity profile over the cross-sectional area is uniform. At section (2), the velocity profile is where V = local velocity vector, wc = centerline velocity in the axial direction. R = pipe radius, and г - radius from pipe axis. Develop an expression for the fluid pressure drop that oc- curs between sections (I) and (2). (ANS: P| Pi ~ RjirR' + 0.50 ptq + gp/», where R - friction force) FIGURE P5.12R = ох/о/ force o-f pipe wall on Me fluid weight of Waler For I he control volume shown in the figure the Z'componeni of lhe momentum equation is V-n dH or p>pr, ft, j (2-irrdr) * ft 4 -/?* fil-Rz-'Ww r*° where dA^zirrdr Thus, with Й this becomes R * prf. ‘ % rJr (l> But with X2 %, **' Л rdo I x-0 (continued) 5-/6R

S.IZR coniinued j Thus / since Eq. (!) becomes ~ (тк ^)m-oz (z) k/e can determine urc in berms of by using ihe con fin oily egucdion- ^V-nd/) *Ot or since * cons+anb “ r rR / ]/<£ d/J jurc[l--fc](zirr<T) = rz\> (x-xz)dx FeO XeO Trib^t = 2ТГДГС (/) Thus, jurc = Зиг, and Eq. (i) becomes fl-fl.* / or +ie^t2 or with - У/lh ж flrpl S-I7R

S.IZR coniinued j Thus / since Eq. (!) becomes ~ (тк ^)m-oz (z) k/e can determine urc in berms of by using ihe con fin oily egucdion- ^V-nd/) *Ot or since * cons+anb “ r rR / ]/<£ d/J jurc[l--fc](zirr<T) = rz\> (x-xz)dx FeO XeO Trib^t = 2ТГДГС (/) Thus, jurc = Зиг, and Eq. (i) becomes fl-fl.* / or +ie^t2 or with - У/lh ж flrpl S-I7R

5.13 R | 5.13R (Moinent-of-motnentum) A lawn sprinkler is con- structed from pipe with {-in. inside diameter as indicated in Fig. P5.I3R. Each arm is 6 in. in length. Water flows through the sprinkler at the rate of 1.5 Ib/s. A force of 3 lb positioned halfway along one arm holds the sprinkler stationary. Compute the angle. 0. which the exiting water stream makes with the tangential direction. The flow leaves the nozzles in the horizon- tal plane. (ANS: 23.9 deg) The Studionanj , non' deftrrtwnj condn>l volume m He. зкеМ h used. 4fpHca/fo^ of dhe axJal Сотри'* d of /иолле^Т- of' тотеи/иъ e^pa/iav\ 5-5o) feeds du T = fh Г V =. m г V COS О she//- * 0tz 2 * (0 Since. cohere V =. rh t zTa----------------- Г Hoyjk epOiiy a , У Dntjjk hods do з. cos e = sial io nary con-1 Го I volume rhzr2 tdhere IT) - &Q = >-5 !b/s c ' v 32.zH7s* -3 z . , m-' (з U- (ft n) (o.o»66 I'hus 0 m ' 5-/8R

S.WR | 5.MR (MomciM-of-mmnentum) A water turbine with ra- dial flow has the dimensions shown in Fig. P5.14R. The abso- lute entering velocity is 15 m/s. and it makes an angle of 30° with the tangent to the rotor. The absolute exit velocity is di- rected radially inward The angular speed of the rotor is 30 rpm. Find the power delivered to the shaft of the turbine. (ANS: - 7.68 MW) The. stationary and non- deform/ту control Volume shewn in the sketch above is used. We- use €a. 5-S3 to determine /he shaft power involved Ihus ^sh9f! * “ (I) The mass /faverate may be obtained fronn (г} h, ° - p\,2n>-.k where radial component of velocity at $echbn(i) The blade velocity at section (!) is V . r b, » (2 " )( 30 g)' = «.»J The values of z and V , may be Obtained with the heJp of a velocity triat1yit ftn -Me. flow at section(/} as sketchedLdov. (continued) 5-I9R

И6/4 /he teloci/y /yianq/с vse conclude. -that V = V ii*3o9 . I/ cos io9 = (lSf»ysM3o9)^ 7,^22? *,t 1 • 4 s ' * The* fror* Eq z M = ( 9Э9 ksjfrJ 2v(2rr\)(i^) dty/oo h_ ftho, w'dh the. /rievf^e we $ec that I/ ^\Еса$3»*^ V sthso9 ^65 Casio* B 13.0 &,/ ' 7 1 s ' s Thet vhH*. £<j- I ^e obt*/* 1?4,2 >= -7.48MW sha-fi —— 5-20R

И6/4 /he teloci/y /yianq/с vse conclude. -that V = V ii*3o9 . I/ cos io9 = (lSf»ysM3o9)^ 7,^22? *,t 1 • 4 s ' * The* fror* Eq z M = ( 9Э9 ksjfrJ 2v(2rr\)(i^) dty/oo h_ ftho, w'dh the. /rievf^e we $ec that I/ ^\Еса$3»*^ V sthso9 ^65 Casio* B 13.0 &,/ ' 7 1 s ' s Thet vhH*. £<j- I ^e obt*/* 1?4,2 >= -7.48MW sha-fi —— 5-20R

5./SR I 5.15 R tMomcnt-of-momentuni) The single stage, axial- flow turbomachinc shown in Fig. P5.I5R involves water flow at a volumetric flowrate of 11 m’/s. The rotor revolves at 600 rpm. The inner and outer radii of the annular flow path through the stage are 0.46 and 0.61 m. and 0? = 30°. The flow entering the rotor row and leaving the stator row is axial viewed from the stationary casing. Is this device a turbine or a pump? Esti- mate the amount of power transferred to or from the fluid. (ANS: pump: 7760 kW) device is а, римр Ье.саще the lift force acting on each rotor hlade is opposite tn dtrec/>*m to the blade мо/ion. The. direction of the blade /if/ forte is ascertained by oot/'np hod the, blade turns the f/a-r past the blade . The pomer transfdrrtd trtrm the ritr blades to the- rva/er may be era tinted the Moment- of - Momentum pome-r ejuAfion (equation (5.S3)with rh V И A 33-6 t Inhere. к At A m de obtain tf irith Ihus 7 760 ft W 'O.Wn, r- 0.3/m 2 rotor exit fiom velocity triony/e U/e = id.8 dudt 7.76xга No/e. •' Since Wshaft >0 ibis device is a pump net In S-2IR

5.I6R | 5.16 R (Moment-of-momentum) A small water turbine is designed as shown in Fig. P5.I6R. If the flowrate through the turbine is 0.0030 slugs/s, and the rotor speed is 300 rpm, esti- mate the shaft torque and shaft power involved. Each nozzle exit cross-sectional area is 3.5 X 10"’ft2. (ANS: — 0.0107 ft lb; -0.336 ft Ib/sj FIGURE P5.16R For shaft -torque use. can use the axial of the monte* / - of- £-$0). Thuij u/ifh T Shafi (ftnstdera-hon at ike absolute and relative velocities о/ the out of each rto^le. (see. sketch bol**') leads -to (continued) 5-22R

5-23R

5-23R

f>./7R | 5.17 k (Energy equation) Water flows steadily from one location to another in the inclined pipe shown in Fig. P5.I7R. At one section, the static pressure is 12 psi. Al the other section, the static pressure is 5 psi. Which way is the waler flowing? Explain. I ANS: from A to B) FIGURE P5.17R To dektwii dinc.-Hor\ of- wakr {low we арй1у fht (^.B.gi) for Мм ^)fo(e) fa fa* sechons(B)^)- The Ion obtained u>ifc Gf-S-П d ponti*- feril* cowtd- fa/dmefen, siuj For Flew fan fech’o^s (в) ft> (A), fy. fo I OH - +- 1Ki fhtA/ Secfafa) {q iecfi^pb) {i.t. uphill} 5-2W?

5./8R | 5.I8R (Energy equationi The pump shown in Fig. P5.18R adds 20 kW of power to the flowing water. The only loss is that which occurs across the filter at the inlet of the pump. Determine the head loss for this filter. (ANS: 7.69 in) FIGURE P5.18R The energy equation for Ibis flow can be wrillen as *hp = + 3^ **2 *4 where (I) Z!s^Zj J fl ~ ~20 kPa J and 20*/03 filsO} XQ (?.8x/o3$3)(0.05^) " Thus, Eq. (!) becomes (-zo*lo3-/£T) + (6.37%}z (9.8*К?-g3) 2(9.1! &T) + ¥0.8m (2S.5 &) 2(9.81-^) ”L or hfs ^69 m S-25R

SJ4R S.19К (Linear momentum energy) Eleven equally spaced turning vanes are used in the horizontal plane 90° bend as in- dicated in Fig. P5.I9R. The depth of the rectangular cross- sectional bend remains constant at 3 in. The velocity distribu- tions upstream and downstream of the vanes may be considered uniform. The loss in available energy across the vanes is O.2V?/2. The required velocity and pressure downstream of the vanes, section (2). are 180 ft/s and 15 psia. What is the average magnitude of the force exerted by the air flow on each vane? Assume the force of the air on the duct walls is equivalent to the force of the air on one vane. (ANS: 4.61 lb) To e.s/)'rna.k. the average °f Ae. -force exerted ky ike ЛП~ -flow On eack vane, tve /‘Ac rnoymhrde of -fke. resutfont farce exerted by fac air on the vanes Ord ike duct and divide. that resu/f by U- assume fas! tte. duct walls act as ore. additional sane.- 7Лс tiheav momentum equaJiinn (e<i r.2-L) ii used * ike X andу of the fCSulhint forest. exerted by the varies jma dud **Hs on ike air- between Sections Q) andCzj. Thus, F • ,or cs (l) and , . *y = РЛ + VpA.V, {г) from ike Cbnservuhtn of enatS principle чме ha^e l/= k s (MO ff)C/2irjF3ih.) . ^2? ' sJ(2¥ihjO;n.) (continued) S-26R

5.IQR continued | With the ЦиН* (fy. £f3) o№i„ As su^ested •* Section 7.2.2, W prestos *t^cAd,s tiJa^J. Thus, fv**> Ja D *. -36. / H> nx Рпнч ^ 2 k^e obtain D = flS.21 ps;a-l4.7fb*)(2<fin )(3<\.) tM)(^x^^)(2^)(3^fLf^ Г (1ЧЧ \ F or ( /7> ] о *<К.Ч- lb Пу Then Я Qnd ^(-30.! lb)t(H.4-lb)' - 55.3 lb R Vdni IZ 57.3 lb ж ЧЛ! lb IZ 5-27R

5.IQR continued | With the ЦиН* (fy. £f3) o№i„ As su^ested •* Section 7.2.2, W prestos *t^cAd,s tiJa^J. Thus, fv**> Ja D *. -36. / H> nx Рпнч ^ 2 k^e obtain D = flS.21 ps;a-l4.7fb*)(2<fin )(3<\.) tM)(^x^^)(2^)(3^fLf^ Г (1ЧЧ \ F or ( /7> ] о *<К.Ч- lb Пу Then Я Qnd ^(-30.! lb)t(H.4-lb)' - 55.3 lb R Vdni IZ 57.3 lb ж ЧЛ! lb IZ 5-27R

5.20R | 5.20R (Energy equation * A hydroelectric power plant op- erates under the conditions illustrated in Fig. P5.2OR. The head loss associated with flow from the waler level upstream of the dam. section (I), to the turbine discharge at atmospheric pres- sure. section (2k is 20 tn. How much power is transferred from the water to the turbine blades? (ANS: 23.5 MW) Zir -fl»* fnm icctianM Л’ гесН»»Сг), * u).t. - У<Ч-А?- A" thefr x Out since. end net ntt For pother, multiply by tbe mtSS bpi net ou^ But m =. у» Q. 5-28R

S.2IR I 5.21R (Energy equation I A pump transfers water from one large reservoir to another as shown in Fig. P5.2IRa The dif- ference in elevation between the two reservoirs is 100 ft. The friction head loss in the piping is given by where V is the average fluid velocity in the pipe and K, is the loss co- efficient. which is considered constant. The relation between the total head rise. H, across the pump and the flowrate. Q. through the pump is given in Fig. 5.21 R/>. If KL = 40. and the pipe diameter is 4 in., what is the flowrate through the pump? ANS: 0.653 ft’/s) fry flow Xc/ioH (l) /p (2 E$. 5.14 /eads -h> \ since (П From Fiy. PF.//7 b we. conclude Hirt L . ZOO - IDO Q £ Fro^t Me problem s/a/emeni (2) or Since. . 8/Л d & (i) QotnLinmy Fjs. /, Z 3 we o4/ai~ 81.6 й + /00 GL - /40 * О (4) root of 4 thd maket pkysicn/ sense is d = 0.653 5'2</R

S.22R | • f2) 5.22 R t Energy equation) The pump shown in Fig. P5.22R adds 1.6 horsepower to the water when the flowrate is 0.6 ft’/s. Determine the head loss bewteen the free surface in the large, open tank and the top of the fountain (where the velocity is zero). (ANS: 7.50 fit FIGURE P5.22R The energy equation for this flow can be written as where (I) Л - A = A hl so, . и а. s ™ (62.* fa) (0.6$) = 23 5 ft Thus^ £q.(D becomes г, +hs ~?z s 8ft + 235ff-2*H = 7.soft Some of this head loss may occur in the pipe and some in the wafer jet as it in!er ads wtth /he surrounding air. 5-loR

6.1k I 6.IK (Acceleration) The velocity in a certain flow field is given by the equation V - 3yz4 + xJ + yfc Determine the expressions for the three rectangular components of acceleration. (ANS: At.-’ + 6yJz: 3v.*’ + From /»r vtlocrhj; us 3yg-z } iWt, ЛПЛ Ur = . Since. CL ж * к + ir in f ur * Ji r Л fa*a^- o + (?углУ°) + Ь*)(з&) - ЗхУ + S/milerly^ anz and a, = o f Сз^Уе) i- (хъУо) Л/з0> /7 г 4ffT + и. viar-r ur№ Ui Mr 3*^ Ъ 50 = О + (3^tz}(e>) -f- (b)(o) = хг 6-/R

6.2R (Vorticity I Determine an expression for the vorticity of the How field described by V = x’yi - xy2j Is the flow irrotational? (ANS: — (x* + y')k;no> The vorticity is 1 vice the roiafioh vector •’ 5 = v*V ( E?. L.n) ( Ej. C./3) (E^_ k. ifj (El. ^/hce /л п°4 Vго ev-eryidlieref -f^e flout /5 boi. Irrirhuiio/ieJhip. 6-2 к

63R (Conservation of mass I For a certain incompressible, two-dimensional flow field the velocity component in the у di- rection is given by the equation v = x1 + 2xy Determine the velocity component in the x direction so that the continuity equation is satisfied. (ANS: -x- + ft t)) To ts-h) “the c*,fLinuthj du c><r J* + Ti, <O Since. do- ~ > a Then from d U. = ~2.n l2> djt. t2) Can be i/l+tyedrd iniih ret peel *. К о He \ du. s -/zx dx i- £(<;) 0Г U = - )cz -t- f(>, ) cohere 4 '(•]) сл an undetermined Tunebon of ij ,

б.чк | 6.4R (Согьег» ation of mass) For a certain incompressible flow field it is suggested that the velocity components are given by the equations и = хгу v = 4y’z н = 2z Is this a physically possible flow field? Explain. (ANS: No) Any phyUCally poSStUt. intfimpreSSiUe. field mu$i iainf] C4>*.sera.Fe>» at muss as expressed by /Ле relu.F<pMsh>'p For the 1/eJoc.i-hj clistribu.ii£>t « ZXtj &JT s- /£4*2 and = Z 9A • 2>4 J Sabsiiikhbu /ЛХе (!) shouis 1bti -rlifi+b ф О for dI X,^E. Th"this is not л. phijsiculltf po^siUe fiou fizld. Uо. 6-4-R

T5R~[ 6.5 R (Stream hinctinn I The velocity potential for a certain flow field is ф = 4xy Determine the corresponding stream function. (ANS; 2tyJ - jt) + О Ike. Veloci-hj pe/beo-hil > и Л*** , $ c. U.^ From the de /ini-hoM о the shbv/n -funohon} U - - *y tn -Z/ifejrele £$.(>) LJitb rested. J* t fdp = / Ч-y dy or + where, fd*) is ан arbiirary(t) Similtrlfj 1 th fu.Hc.tior of*‘ = ¥-X anti J' d l/J * “ J ' 4-tdx. or id! = - XX 2 * A. ) v^ere fy)/s en <s ) 7ё> 5лЬ»$ * arbitrary function ofy. s. tz} anti (3) fit*) * (/y) for al/X andy. Thvst i * fj. ecenstant. + c Inhere. C ds Л COM3 boot . 4--5R.

6.6 R (Velocity potential! A two-dimensional flow field is formed by adding a source at the origin of the coordinate system to the velocity potential Ф r2 cos 20 Locate any stagnation points in the upper half of the coordinate plane (Os Ss ar). (ANS: 0, = w/2; i\ “ Thus, a*ti V,- -21-^20 it 2trr It CoS 2£> Лауль&еи pointe cuill occur 1^-°, Vj.*ot-far- Thuij о - ~2$ s/') 2 $s о ~ +Zfs CasZ^ <*> ЁчцаЬоп U) te s*tes/v*cJ at er Q %jIT • Fn>m Cos Z&4 " Jar the pomkle Valuer »f €*lty 9 Satis ftj .(3). Recall {hat tm>o for a source. T/lus/ p. a \T^~ ! Sr *rTT e3) bji/l 6'6R

6.7R | 6.7 R (Potential fl»») The stream function for a two-di- mensional. incompressible flow field is given by the equation Ф - 2л - 2y where the stream function has the units of ft2/s with x and у in feel, (a) Sketch the streamlines for this flow held. Indicate the direction of flow along the streamlines, (b) Is this an irrotational flow field? (c) Determine the acceleration of a fluid particle at the point x = 1 ft, у “ 2 ft. (ANS: yes; no acceleration) OV Lines Canstani f are s/ream/roas .ThrSj и Uh ZX-Zfy The. of A Jihn a/ream hnt. J Ц J (where /3 some Con she nt) is of The form if a Zt-Zy straight lines as illasharlerf m fte fry urn a/reamhaes. Thus> streamlined are - 2 Jx " the direction of flow is as shown -o field is irrotahona/ lOhtre = for the stream -function given dm ~ Ju _ д So that and The. lC) 2/nce the velocity ij csnstani throu^houi The f/ou> fhM, the acceleration of a/J fluid pa hides / 5 lero. Д sihce V--2.l-'2/ 6-7R

6.XR (Inviscid flow » In a certain steady, incompressible, inviscid, two-dimensional flow field (w = 0. and all variables independent of z) the x component of velocity is given by the equation: и = x1 — у Will the corresponding pressure gradient in the horizontal jt di- rection be a function only of x. only of y, or of both x and y? Justify your answer. (ANS: only of a) Smcc the flotJ -field rnai-L Satisfy lhe Cfifrt/nuitjj е^чл.Еюл and rt АНшл tfixf and there Иге v = -2.x ex } For steadyt htuo-di/nfunoHtl -f/ouj of an inviscid -fluid (/th "the. X-uxLs heri'^oniol ло thrb п-o') the. ^~CMyPonent of the momentum equalion It Thusf f»r dhe. и and v fi'uen alw. The. pressure (jnacheni is ч -Tune-Lion pnly dT X. 6-BR

6.9R (Ins tscld How) The stream function for the flow of a nonviscous, incompressible fluid in the vicinity of a comer (Fig. P6.9R) is ф « 2r*/s sin $0 Determine an expression for the pressure gradient along the boundary в = 3rr/4, (ANS: -64 p/27 / ,/’i At on or О -ЭТТ/Ч boundarywhich ti s tornloni. P * 4 4/глтline <и=о For the stream /цнсЪон ytven vr Г fa 3 • a rd a lent) Лг е^ЗТГ/V- btun^ri/, r г «э ~ О t -5 0 Since if Ч от 6-ЯК

&IDR I 6.I0R (Potential flow) A certain body has the shape of a half-body with a thickness of 0.5 m. If this body is to be placed in an airstream moving at 20 m/s, what source strength is re- quired to simulate flow around the body? (ANS: 100 m'A) ‘he. uidth jo •Д<£ From Ey. of the body is ж (.О. S'on J IT thA distance, between the source and the nose b = — 2jrU ^here m й The source sirenjth} a„d there Jbre /m = 27rLrb * Л7Г fan™ 64DR

77шГ~1 6.1 IK 'Potential Hou) A source and a sink are located along the x axis with the source at » = - 1 ft and the sink at x 1 ft. Both the source and the sink have a strength of 10 ft2/s. Determine the location of the stagnation points along die x axis when this source-sink pair is combined with a uniform velocity of 20 ft/s in the positive x direction. (ANS: ± 1.08 ft) Fko/7) . L. !oL Thus (or <Z s I -ft rrn ~ у . 9 net V ® 24? / ) J I * = ± I. 08 -ft 6-UK

6.I2R | 6.I2K (Viscous How i In a certain viscous, incompressible flow field with zero body forces the velocity components arc и « ay — b(cy — y2) v = w = 0 where a. b. and c are constant, (a) Use the Navier-Stokes equa- tions to determine an expression for the pressure gradient in the x direction, (bi For what combination of the constants a. b. and c (if any) will the shearing stress, r„, be zero at у = 0 where the velocity is zero? (ANS: 2h/i',a - M l The. к-component of-the Navier 'Stokes equation is (£^.6./27ц} Far the specified 6-/27A above reacts ba 0 s + о + Сгк) so that: For the qiven velocity distribution = A -be + zby and Note’- With s0 then twhere if a-heathen U~b^ "thus, $ =2 by sQ when y-0. 6'I2R

~6J3R~] 6.1 JR (Viscous flow l Л viscous fluid is contained between two infinite, horizontal parallel plates that arc spaced 0.5 in. apart. The bottom plate is fixed, and Ibe upper plate moves with a constant velocity. U. The fluid motion is caused by the move- ment of the upper plate, and there is no pressure gradient in the direction of flow. The flow is laminar. If the velocity of the upper plate is 2 ft/s and the fluid has a viscosity of 0.03 Ibs/ft2 and a specific weight of 70 lb/ft1, what is the required horizontal force per square foot on the upper plate to maintain the 2 ft/s velocity? What is the pressure differential in the fluid between the top and bottom plates? <ANS: 1.44 lb/ft-; 2 02 lb/ft) У и th /о Ileus 4* That (о. оз /2 !!L S/лсе 'there n no f’luid mokio* /j, Th? y- dlrec.be>* > "/he pressure i/urtabe» lit The y-direabo* ц hydro ita-hc ja 6-I3R

6J4-R | 6.I4R (> iscous Howl A viscous liquid (д = 0.016 lb s/ft2, p = 1.79 slugs/ft3) flows through the annular space between two horizontal, fixed, concentric cylinders. If the radius of the inner cylinder is 1.5 in. and the radius of the outer cylinder is 25 in., what is the volume flowrate when the pressure drop along the axis of the annulus is 100 lb/ft2 per ft? (ANS: 0 317 ft3/») Гог laminar flow in an annulus (see £<{. 6,16 6)- «?- & so lhal (№fl}z-№rftt * - в(^) - 0.317 £ о k/e must check to see if the flow is laminar. If nuf ihe above result is not correct. NovJ where bh = з(г,-Г^ ОГ if . ef3 £______ 0.3/7^ ,, (гг-гг)^ 17(1.5^/^^^ fllsoJ = 67.8 (!2^)(ЗЛЗ. & * O.9Ii Since P.e - 67.8 <<^!oo -Ihe flow is laminar and 6'7^R

6.15 R | ft.lSR (Viscous How) Consider the steady, laminar flow of an incompressible fluid through the horizontal rectangular chan- nel of Fig. P6.15R. Assume that the velocity components in the x and у directions are zero and the only body force is the weight. Start with the Navier-Stokes equations, (a) Determine the ap- propriate set of differential equations and boundary conditions for this problem. You need not solve the equations, (b) Show that the pressure distribution is hydrostatic at any particular cross section. (ANS: *р/гП - 0; - -pg, Лр/dl ~ pidw/Ax' + with w - 0 for v = ±b/2 and у = ±<t/2> FIGURE P6.15R (a.) Fre/n dhe desertpbon of ihe prollem) ir~oj 2t~ai and "fbe Corrbnuiitj P^uaFeu /rdicif/ej dhet ’-<5, -these Ceadihdns the На<лег-5Ше$ epit/o^ (ej. ь.м) reduce, do ( jC-dirfcbo'» ) »o <4 f y. tlirecii» *> ) г>4 ' i *) ( i - direction ) g-Xg-Ж? Be>undarij Cendi-f-юлл ere; Ur^O -for Xs- 4 ! *1s ~Z Xn/e^ra-hoh of Ej.fi.) yields fa -- - fa Jt !>* -eis* Ъ ЦоикиеГ' from ** *** * f»"sdioa * 30 becomes J3*-/?!) * f.(^ Thust ad °- jw* Cron seefaou ( Z - constant'') ~^g er> cJJier^. C is o. cettsian'i. E^uafioit indicates -th»i "the f>Yess и кв chsfn buboa /л hydrosi^a-bc af 6 ytffn Cress sechoei. 6-/£R

6.I6R | 6.16R (Viscous Howl A viscous liquid, having a viscosity of IO4 Ibs/ft2 and a specific weight of 50 lb/ft3, flows steadily ---► £ through the 2-in.-diamcter, horizontal, smooth pipe shown in Fig. P6.I6R. The mean velocity in the pipe is 0.5 ft/s. Deter- mine the differential reading. Ah, on the inclined-tube manom- eter. (ANS: 0.0640 ft) у =65 Ibrtt3 FIGURE P6.16R C-htck. Retfnohli nurttbtr to Мхгпняе Sf flout Is •’ /2 fa < L ЮО !о±И- Thus Мои) й So 'tkai. к _ F (to* Uh. 12-!^ Thus ft ft» D. 5П 3 3 6-/6R

7.IR | 7.I R (Common Pi terms i Standard air with velocity V flows past an airfoil having a chord length, b. of 6 ft. (a) De- termine the Reynolds number. pVb/p. for V = 150 mph. (b) If this airfoil were attached to an airplane flying at the same speed in a standard atmosphere at an altitude of 10,000 ft, what would he the value of the Reynolds number’’ (ANS: X.40 X !</’: 6 50 X IO6) 35 34-Ji lb 7 — 3. ГЗ У- X )b * It) 3.7V- x/o Wo X lb b = 6ft 7~/R

72R | 7.2 R (Dimensionless variables) Some common variables in fluid mechanics include: volume flowrate, Q. acceleration of gravity, g, viscosity, p, density, p and a length, t. Which of the following combinations of these variables arc dimensionless? (•) C7l?<'2- (b) pQ/pt. (c) gt'/Q2. <d) pQ(/p. (ANS: (b); (cl) = £ 3 not JinttAuofiJezs 7-2 R

7.3 R | 7.3 R (Determination of Pi terms A fluid flows at a ve- locity V through a horizontal pipe of diameter D. An orifice plate containing a hole of diameter d is placed in the pipe. It is desired to investigate the pressure drop. Др, across the plate. Assume that Др = /(D. d. p, V) where p is the fluid density. Determine a suitable set of pi terms. IANS: Др/рУ7 = d»d/l)b FL'2 L dL'*L p± FL*T* y=LT"' from the pi theorem ST-3 = 2 pi terms rofrurtd. Use as rtpfebh$ variables . Thus, 7Г,- ОУу (Ft^fL^fLT'^ (FL^T')C • fW 1-eC-o (forFj ^3-f-ath (forL) D, V) and p and so th»t — b+Zc.=O (tor T) -Pollotos that o.-o) bs~2j C~-l t and therefore Check dimensions usinj MLi sheer* дA = \ = M°L6T° of. V2/° (кт-) (мгь) C = o 1+4. —ЧС so — i +zc~o th follows "thei a.~ b~ ot c^Oj and therefore _ d 7TX~ D Lohich is obi/iouslij dimension less- 7-3R

7VR | 7.4K (Determination of I’i terms) The flowrate. Q. in an open canal or channel can be measured by placing a plate with a V-notch across the channel as illustrated in Fig. P7.4R. This type of device is called a V-notch weir. The height. H, of the liquid above the crest can be used to determine Q Assume that Q - fW. g. в) where g is the acceleration of gravity. What are the significant dimensionless parameters for this problem? (ANS: 2 - FIGURE P7.4R 0= L3T~‘ LT'2 ©- F4T‘ From "the pi. Theorem, Ч--2 = 2 pi terms required • *0f7Tz) By inspection, /or TTt (containing Q) : 7Г = ——— = L*T__________________= L* Г * Since, "the anglef 9j is dimensimitts О So 1ho± . Ф (ei 7-4R

7.SR | 7.5R <Determination of Pi terms» In a fuel injection sys- tem. small droplets are formed due to the breakup of the liquid jet. Assume the droplet diameter, d. is a function of the liquid density, p, viscosity, д. and surface tension. <r. and the jet ve- locity. V. and diameter. D. Form an appropriate set of dimen- sionless parameters using д, V. and D as repeating variables. (ANS: d/D n ddpVD/p.. <г/дР»> d_= L FL*T* u.-FL~xT (TiFL4 V±LT~' D=L From the pi Theorem, £>-3=3 pi termi repaired. U^e l/ ana D as repealing variables. Thus, 77J = (L^L-‘Tt(LT-t(l.r-rLT A*O t^F) I-la-rh + g = о (tor L ) <Z - b - O (tor T'i I/- tv Hou/s Thai 0.-0j b 3 <s>, С = -I, a*<f Therefore. tr- "l D Ljhich /3 obv/ouslt^ dimensionless. 74--/z 'V‘o- (ргЧ'Хрс-'т) *(l)1 = РЧТ l-t a, =0 ( p) -tf-latb-rG -o (tor L) A i-a.- b - о (t»r T) Uh -/ollaius Thai a. - -I, b-11 C "/? ana therefore. ~rr FVD z " i7~ (continued) 7-£R

7.5R continued | Check dimMuhns usinj MLT itjikm' /OVD fHL'^fi-T^fLt ok Hi- fifor F) —/-Д< th tc -o Q.~b = a Z-J- fvf/ows 'that 4 - b* , £ - &; 4*d Therefore. 3 " /<-1/ Check cJ/mtH^nns MLT : 3 , M ° iff6 :.Ok 7-6 R

7.5R continued | Check dimertuhns usinj MLT itjikm' /OVD fHL'^fi-T^fLt ok Hi- fifor F) —/-Д< th-tc -o CL-b = a Z-J- fvf/ows 'that 4 - b* , £ - &; 4*d Therefore. 3 " /<-1/ Check cJ/mtH^nns MLT : 3 , M ° iff6 :.Ok 7-6 R

7.6R (Determination of Pi terms) The thrust. J. devel- oped by a propeller of a given shape depends on its diameter. D. the fluid density, p. and viscosity, д. the angular speed of rotation, ш. and the advance velocity. V. Develop a suitable set of pi terms, one of which should be р&ы/р.. Form the pi terms by inspection. <ANS: T/pt'/>' — tbipVD/p. рГ> ш/p)i *т-г v= From the pi theorem } (p~3- 3 pi terms required-, 77J 'z f>Y\ Check, usm^ MLT : Л/Z 7 М°С.еТ* - CJhich is the. Reynolds number Л fc^oum 3 " Check u&m^ MLT 1 CopDx ML /<- ML.-* Ok US

7.7R 7.7R (Modeling similarity) The water velocity at a certain point along a 1:10 scale model of a dam spillway is 5 m/s. What is the corresponding prototype velocity if the model and prototype operate in accordance with Froude number similarity? (ANS 15.8 m/s) For Fro и de. number similori/цf Frfn ~ 'F~ Jo , v* l/Ж) - 5/*/: t then /io (f^ = 7'8R

7.SR (Modeling similarity) The pressure drop per unit length in a 0.25-tn.-diameter gasoline fuel line Ls to be deter- mined from a laboratory test using the same tubing but with water as the fluid. The pressure drop at a gasoline velocity of 1.0 ft/s is of interest, (a) What water velocity is required? (b) At the properly scaled velocity from part (a), the pressure drop per unit length (using water) was found to be 0.45 psf/ft. What is the predicted pressure drop per unit length for the gasoline line? (ANS: 2.45 ft/s: 0.0510 lb/ft7 per ft) For flou> /л о dosed condurtl Dependent pi ierm * $ (jT* p 1 For this ptrhcular problem the dependent variable. /3 the. pressure drop per unit length t 3o 1h*t Dependent pi he rm - * b. [d/d0 A-hof the. dtaradtridic- length /or pipe fiou /$ the pip* diameter, D. Thus, n) ~7^ r I о > /-! (л) 72 теш 4a/п dynamic Si rm hr Ль, and uith f -t !h‘S v (. shy ) -- fr.sfx* п-зг ( 5-X ll)-b l-k^ ) ( 1.ЧЧ- ) 3 ~ — * (b) file seme Reynolds num^ and wi& ^/b (same |ul»i>• $ ) ike* /гот 5o _ 11.61^»/(IXZ (|)(o.4!F Ft R

7.4R I 7.9R (Modeling, similarity) A thin layer of an incom- pressible fluid flows steadily over a horizontal smooth plate as shown in Fig. P7.9R. The fluid surface is open to the atmos- phere, and an obstruction having a square cross section is placed on the plate as shown. A model with a length scale of } and a fluid density scale of 1.0 is to be designed to predict the depth of fluid, y, along the plate. Assume that inertial, gravitational, surface tension, and viscous effects arc all important. What are the required viscosity and surface tension scales? (ANS: O.I25.O.O625) Free surface FIGURE P7.9R. Л //utd dynamics problem /or uhich men hit I, surface kftsien' Qhd VISCOU3 e&eck, are all import»** requires Fronde, Reynold*, a*d U/ebtr number similwhs (jee Table 7./). Thue, ( Freudt number similwfy} /Л -fallows' (with 14и _ Y ' У7 For V</eWr number зГт|)4н4ь> V(/e = \Fe. or flr*Va* ----—--- (Г ^ce, (Г„ Л" vJ = $£. d /cHo/d ^(d/ 7-IOR,

7.I0R | 7.I0R (Correlation of experimental datai The drag on a 30-ft long, vertical. 1.25-ft diameter pole subjected to a 30 mph wind is to be determined with a model study. It is expected that the drag is a function of the pole length and diameter, the fluid density and viscosity, and the fluid velocity. Laboratory model tests were performed in a high-speed water tunnel using a nxxlel pole having a length of 2 ft and a diameter of 1 in. Some model drag data are shown in Fig. P7.I0R. Based on these data, predict the drag on the full-sized pole. (ANS: 52.2 lb) FIGURE P7.10R 0 350 300 250 100 50 #200 o 1 150 Model velocity, ft/s uhere: •О'» draj = F, Я ~ Pole lenf/h* L } D~ p>!e diameter <-1 fluid dfnn’hj = Vfftoirfy -FL. *Tj l/Z'v delocrhj - LT [ fro,» He pi -theorem L>-3* 3 pi -firms required. A diiwsiwl anahjsis ('for etarn pie usm$ as repeadinj variable* fields <£._ = a> (1,^] I \ D 1 J Geometric similan-hj ye^uirtj a -A and -this tondi-hdn is Sa&'s-fied since, = 3,^ , 2.4.*Ls. Req/iol^s number jimilarrh/ requires -fya.-i Д, " Л" «Зо и г \ / .i \ у = -b \/= l2-^¥xlt -&A/o.00231 ^§Yj.25-(t | у /* A [ аду x j»’’ Д j, Sbjs J\l/n44: j - UsV •X' V=(3emfi.X^(£^T^tf,f^ iP /oJitwi ibst required model delocrfy Л A* If <fnj -frffeGre -7-HR.

7. HR I 7.11R (Correlation of experimental datai A liquid is con- tained in a U-tube as is shown in Fig. P7.11R. When the liquid is displaced from its equilibrium position and released, it oscil- lates with a period t. Assume that т is a function of the accel- eration of gravity, g. and the column length, (. Some laboratory measurements made by varying t and measuring r, with g - 32.2 ft/s2, are given in the following table. FIGURE P7.1 1 R T(s) 0.548 0.783 0.939 1.174 <(ft) 0.49 1.00 1.44 2.25 Based on these data, determine a general equation for the period. (ANS: т = 4.441 f/g»' 2) r= ТС;,я) Ti T }*Lr' ЯИ From -the pi Theorem, 3-2* ! pi term required'. 1Г, cconstant. Since -there is onhj I -pt berm, /7 tollous 1ht± where C/5 * Constant. F»n the data. yn/en: 0+4 1.00 l,W 2.25 f.44 «W Wh ThusС - and 7-T2R

7.12 R | 7.12R (Dimensionless governing equations) An incom- pressible fluid is contained between two large parallel plates as shown in Fig. P7.I2R. The upper plate is fixed. If the fluid is initially at rest and the bottom plate suddenly starts to move with a constant velocity, U, the governing differential equation describing the fluid motion is du d2u where u is the velocity in the x direction, and p and д are the fluid density and viscosity, respectively. Rewrite the equation and the initial and boundary conditions in dimensionless form using h and U as reference parameters for length and velocity, and t^p/fx as a reference parameter for time. (ANS: = d^u’/dv*' with u* = 0 at r* 0. u* I at y* « 0. and и* - 0 at y* = I) —► (' FIGURE P7.12R z«t % -Lfi), w •• , Ju . MVu*) jtf _ V Jjf fl ) . u йц? it' jtf >t ' >t*l с/ f du . . г, in*- Zj.), ay ’ iy* du U O* 77>i/s, 1be опум! d:ffemh*/ etutiitn and <53 at 77?« ini-hiJ Cs>ndi^oM k3O (jnd boundary c&ndf/wdi became. !.o at i-c bec&ffts u,Kp &t Lc=lT at ^=o at yzi.) and a*'- о at 1-I3R

7.13R i Dimensionless governing equations! The flow be- tween two concentric cylinders (see Fig. P7.I3R) is governed by the differential equation where ut is the tangential velocity at any radial location, r. The inner cylinder to fixed and the outer cylinder rotates with an angular velocity ш. Express the equation in dimensionless form using R„ and w as reference parameters. (ANS: * </(v;/r*)rfr» = 0) FIGURE P7.13R let and vf * So -hta-t dh*' ' dr*/ dr*2' ihe origin*! dif/erenhal becomes T’-'W

8.1R (Laminar How) Asphalt at 120 °F, considered to be a Newtonian fluid with a viscosity 80,000 times that of water and a specific gravity of 1.09. flows through a pipe of diameter 2.0 in. If the pressure gradient is 1.6 psi/ft determine the flowrate assuming the pipe is (a) horizontal; (b) vertical with flow up. (ANS: 4 69 x 10 ’ fp/s; 3.50 x 10 ’ ft'/’s) If the flow is laminar, then Q - a) where and from Table В J for A10 /204 /L= 60,000^ Qt/o* (J./Wx/o's jp, IV>‘F Q al Гог horizontal flow, 0*0 y » t Thus, from b) For vertical flow up, 0*90 T 2 m. lu(M3iS*)(IH)кЗМ’и> T Note: We must check to see if our assumption of laminar flow is correct. $/»« v~ Я - ' 0,2/5 ‘ il Mna 1M 0.93/ The flow is laminar when в *0. Ti is laminar for 6е 90е also. в-IR

&2R~| X.2R (Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2f. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.44 times greater than it is for the second pipe. If the diameter of the first pipe is D. deter- mine the diameter of the second pipe. (ANS: 1.095 D) for lani/nar flow} <? • , where Qt = Q3 and Thus, 9-2 R

8.3R | H.3R (Velocity profile I A fluid flows through a pipe of ra- dius R with a Reynolds number of 100,000. At what location, r/R. does the fluid velocity equal the average velocity? Repeat if the Reynolds number is 1000. (ANS O.75X; 0.707) For Re• 10s, • (/"£) t where n *7.1 f see Fty. 8.17) The relationship between the average velocity, V, and /he centerline velocity, Vc, can be ob/ained from QsRVs^udA or r.R irf^V^ й (rvrdr) = 27Г vc^ (I'^rJr Bed, with r X.s l~-fj- so Ihai dxs-£dr and r-R(hx),we have (l'^)nrdr (l-x)dx - r‘° 7 n I 2 n - К [pttl)Л Л2Л?/) |fl л (nil)(Znti) Thus, from (i) and (a.)’. TR V = 27Г Vc R foHifrjHi) V ~ (пнМгп+О (i) Hence, U when i/ t. r\fi i/ /1. JZxfi HsVc(IrT or —^1 V(l-R) For Re =l°3 the flow Is laminar wi/h U* Vc\j~f%) J and Thus, U*V when iVc . .or 0707 d~3R

9^П X.4R (Turbulent velocity profile) Water at 80 °C flows through a 120-mm-diamcter pipe with an average velocity of 2 m/s. If the pipe wall roughness is small enough so that it does not protrude through the laminar sublayer, the pipe can be con- sidered as smooth. Approximately what is the largest roughness allowed to classify this pipe as smooth? (ANS: 231 x 10 'ml Let h= roughness. ThuSj hsLs , where 4 е у? Ъ* or With Re- ~ 6.S8K ios and a smooth pipe we obtain f = 0.0/25 (see Fig. Э.20Х Hence, and 0 07>/^ = 2.3IAI0~Sm 8 *// *3.65*\07^ (see Table B.2) f^vb^"60mm eo*c « laminar sublayer Sickness. If the reuyhnesr element js snal/er than o.oizi mm it lies within the laminar Sublayer 8-^R

8.5R | 8.5K (Moody chart i Water flows in a smooth plastic pipe of 200-mm diameter at a rate of 0.10 m’/s- Determine the fric- tion factor for this flow. (ANS: 0.0128) plasiic pipeJ ^-0 (see Table в.I) \/_ Q GJ0 sT - 3 iqID. У~Ъ %(0Л& 9 3JQ^ « £.68* 10s f so from Fiy. 9.20 obtain For a D *0.200m Also, Re s where Hence, _ n (3./8ISr)(0'Liri /./ZVff'taF" <=0.0/28 8.6R (Moody chart I After a number of years of use. it is noted that to obtain a given flowrate, the head loss is increased to 1.6 limes its value for the originally smooth pipe. If the Reyn- olds number is IO*, determine the relative roughness of the old pipe. tANS: 0.00070» Let ( )0 denote the old pipe and ( )л the smooth pipe. Thus, (i\ ‘O t , Vo - Vs and hL=L6 hLj . Hence, with 0о-й, t ' fs , or Н0*!’Ь where from wHhRe-!0* f5*O.O//S Thust f0*/.6 (o.OHS^ 0.0/8* which with Re*/o‘implies 9-5R

6.7R | 8.7R i Minor losses) Air flows through the fine mesh gauze shown in Fig P8.7R with an average velocity of 1.50 m/s in the pipe. Determine the loss coefficient for the gauze. <ANS: 56.7) G»urt wer end al pipe where 2^2, } И, where fie '° °"d fi = Bww vet'? . _2/Z8J^2_ = 56.7 ff-6R

OR~|------------------------------------------------------ K.XR (Nuncircular conduits) A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round shape. The amount of material in each straw is to be the same. That is. the length of the perimeter of the cross section of each shape is the same. For a given pressure drop, what is the ratio of the flowrates through the straws? Assume the dnnk is viscous enough to en- sure laminar flow and neglect gravity. <ANS:C>„_, - 1.83 0^) D -4 — a — Af, = v/here &f> '6nL Since hL/ -hLl , f, -o^ - > vl,ere l' 'Dr ‘"J Thus, 4K = 4^ Since the perimeters are equal, f}sPz, or D a TrD-Va. Непса, For laminar Flow j where Ct964 and Czs 56.9 (Table 0.3) , Яъ- Thus, from Eq. ID 6? wx 36.9 v2 ^.1 = or K=/-**/i4 Also, Q,~ A/V, s^1)гУ1 and Q3.9Л14 * & 14 so that round ~h93 Qsqtere 9-7R

8.9R (Single pipe—determine pressure drop) Determine the pressure drop per ЗОО-m length of new 0.20-m-diamctcr horizontal cast iron water pipe when the average velocity is 1.7 m/s. (ANS: 47.6 kN/nr) . O.Zm (I) h7m/j 1 (X) V . — ,------- i I • — 300 m-3----H or M I4S14S^-/-7’3L > b*0.2m this yVes e + f 4 ip]/1 = f 4 г lhe PW* is horizontal. Bui fron, Table 8.11 г L3xl^ -з^‘ SO Ы frm fy. 8.10, 0.022 ^Af =0.022^')ЦЧЧЧ^(1-7^^‘Чо'^ 8-8R

d.lOR I X.10R (Single pipe—determine pressure drop) A tire protection association code requires a minimum pressure of 65 psi at the outlet end of a 250-ft-long, 4-in.-diameter hose when the flowrate is 500 gal/min. What is the minimum pressure allowed at the pumper truck that supplies water to the hose? Assume a roughness of e = 0.03 in. (ANS: *1.0 psi) f/) D=*in. €=0.03 in. W -**» • ------и Q=500<pl//nin /*250 ft and Q^oo^fz» Pl = рг + f /з?уХwhere V= % • = 12 8s From Fig. 8.20 will) * . fifty- - 3-SJtl°smJ i" °-°°7s twe obtain 0.035 Hencef from E^. (6 fi, or S 65 psi 4- 24.0 psi = ЯФ.0 psi (>) 8-9R

И~[ x.liR (Siqjle pipe—deteratine flowrate) An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 100 ft and diameter 5/8 in. If the pressure at the faucet to which the hose is attached remains at 55 psi. how long will it take to till the pool? The water exits the hose as a free jet 6 ft above the faucet. (ANS: 32.0 hr) =* v+1 *ktn ’ss&> г,‘°‘ fVZ~ ^-°6, where V~ff АЦ Re'^‘> °r *'¥3°°V'wher’ V~" and from F7y. &20 w/'lh % *0 (I) (•) Re Trial and error soh/hon of Efs- (2), and (3) for f,V, and Re’. Assume f^O.02- from Ef.W V=/¥.2&; from Ef.fr) Re from E^.fr) f= 0.0/96 + 0.01 CbiA closely Assume f •0.0/96; fromEf.U) from Ef.fr) Re-6.i4x/o^so fn» Ej.(s) f* 0.0/96, which agrees with the assumed value. Thm, V=/^9-, « Al», ¥=0t, where ^fHhhfhwe ee/V-P^^o. (SH) = <• / / »• \ - 32.0 hr Q 0.0307^ \ 36DOS J . 8-IOR

а/2/Г[ 8.I2R «Single pipe—determine pipe diameieri Water is to flow at a rate of 1.0 m3/s through a rough concrete pipe (e 3 mm) that connects two ponds. Determine the pipe diameter if the elevation difference between the two ponds is 10 m and the pipe length is 1000 tn. Neglect minor losses. (ANS: 0.74X ml (I) Thust f •£ = tOm with Q Q /Г»*Л _ 1.17 m , n v ~£ —537 м b rn r,(/,000m) (/.27 /D3-^)2 _ . T ~~D — Ofq.81^) or Ds inhere D~m • 0./2/ (V (/./ZX/0~6 «s (F^.e.V>) Re (3) (^) //.27 ) p₽ L vd _ (-gj ле or < Re =/./¥*!<>'/[) M'. f Trial and error solution • equations and Moody char!) and ¥ unknowns ( Re j Djfj and &/b) Assume f в 0.02. so that from Eq. (ft, D * /.53(o.o2.)& * o. 7oom. from EqM, = l-бзхю6 and from Eq. (¥\ £ * ^¥i2Jx/o~3 Thusj from the Moody chard (Fiy 8.20) > f = O.O19 *0.021 fhe assumed Value. Try ayain. Assume f-0.029 which qives D- o.7£¥mj Ее */.s/x/o\ and 75 * з.зиг/о^, Цепов; from F/'y. 8,20^ f= 0.02.8 Ф0.029. A no!her fry wt'fh f=0.028 к/ill show ihai fhis isihe correct value. ThUsj D = LS3 (0.028)^ * O.7VS m 8-HR

8J37H 8.I3R (Sinelv pipe with pump) Without the pump shown in Fig. P8.13R it is determined that the flowrate is too small. Determine the horsepower added to the fluid if the pump causes the flowrate to be doubled. Assume the friction factor remains at 0.020 in either case. IANS: I 51 hp) V /V Water FIGURE P8.13R hp+ +7/ * * V * +({& , where pf sp^0, И Zz = 0, and . Assume a sharp edyed entrance: 0.5 (Fff. 8.22.) Wi/houl !he pump, h/>-0 Th as, 6.28 & Wdh the pump |/= 2(6.28^) =^/2.Л# Thus, hp = (I * f/ *o.s) hp-- s f< *« + ‘-sfuzofl'/sff Hence, from E^s.es Tif= fQhp -(£2.¥^)(0М)'(11.^)(^Ю -831 or W=/.5/hp 8-/2R

8.I4R 8.14R (Singh- pipe uifh pump) The pump shown in Fig. P8.14R adds a 15-ft head to the water being pumped when the flowrate is 1.5 ft3/s. Determine the friction factor for the pipe. (ANS: Far f/ow from the upper tank fo /he lower /ank! where V,‘O, з, - *>e ft, z^iKft, V^of, aed +ZK. + К/ *0.6+2(0.3)Ц*гл (see Ffy, 8.25) Ъ V %/^и, T nzex// Thin; £e. ft) becomes .z „ П . ,QC ff+fl (Z°Oft) +22) У M 2-00 ft + /5 ft = 62.*$ +(9£H+(j (osW ^z'z)z(32.zSj w bvi wHh #3 v= 4 = = Z&g, 9/m f=0^306 Nole ‘ 1-f !he flow wes from lhe lower lank (2) Io /he upper lank ft)t ihen ОГ И • 1 »i 2. AgWpi) „ w „ t /^/ - zoefi .z.2]^^ which yules f = 0.0412 Bolh„ lhe assumed dov/nflow (f = 0.030s) and lhe upflow (f'0.0412) yjve reasonable friction factor values. 8-I3R

8ЖТ 8.15R (Single pipe with turbine) Water drains from a pressurized tank through a pipe system as shown in Fig. P8.15R. The head of the turbine is equal to 116 m. If entrance effects are negligible, determine the flow rate. (ANS: 3.71 X 10 ' m /s) where 2,-200m, fix * V/ -?2.-0, and hs - (hs <0 since il is a turbine) or vz= I7¥S (8+zooo-F) where V~Js’ Also, Re- A l.IZrii3^ or Re = 8-92*10* V (I) W The final equation is the Moody chart (Eq. 8.2») with £ s o.oooem s Q 00Q ° o.l m r \ \ (s) £*0.008 Trial and error solution •* ---------- (3 equations' 3 unknowns-Ке4 ( И) e Assume f- 0.038 so that fron>Tq.(Df y-y.S6-^-. {rom Re - *f.Q7xю£' and from {he Moody chartj (-0.03S which does not equal the assumed value. Try aqain. Assume f*0.03f which qives /- ^.73and Re *¥.22x10*. Hencet from Me Moody chart f- 0.03S which agrees Wi th th» assumed value. Thus, 1/-¥.тз-^ and Q- AV- f£(0.!(¥.73%) * 3.7/xio2^ 8'/¥R

S.I6R (Multiple pipes) The three tanks shown in Fig. P8.16R are connected by pipes with friction factors of 0.03 for each pipe. Determine the water velocity in each pipe. Neglect minor losses. (ANS: (A) 4.73 ft/s. (B) 8.35 ft/». (C) 10.3 ft/s) FIGURE P8.16R /issume the flow from both banks /) and В is info bonk C, or Q3 * Qt tQ^ Thus, /.zxK = Z^*/.l‘l4 Hence, 14 s 0.694 V, ♦ 0.840™ For the flow from Й Io C, with Уч ~0, we obtain Z* = у 838 ft- 80S И +2(^2л^-в^У^ V* ] or 5 33 — 0.373 V* + 0.233 Vj2 w Similarly for the flow from BboC, with = we obtain 2, -Л Ч& $ 4 £ «»« ОГу5-0.2«ЦХ + 0.233 ^2 Thus, 3 equations and 13)) for Vt ,7^ andV3. Solve as follows- Subtract (z) from (з) to obtain 12 = 0.296 141 - 0.373 V,2 W From (i): V3 */141.6 -16 V,2, or when combined with (to JlW.6-UVr's 0.694 Vi ♦& WO Vx )Or Vt = JZOO -2.21V,X -0.826 V, Combine E^s .(9)and to obtain1 - \]zoo -2.27V,2 -0.826 У,! - t *hich can be simplified Io x ]200-2.27Vi2' - 96.S -/.725 V* By squaring Ibis epaalien we obtain (after simplification) : -----—-----? -j0, V, -101.3 V,* H77¥ =0 Hence • И1»-------------- 27y Thus, 7^8.89^ orVrWg Note- The 7, -8 89 solution is an extra root introduced by squaring It is nut a solution of the original Eqs. (DtM,(3), For this Value, Eq. becomes S^i/ioo-2.21 (8.89xF = 96.S-!. 7X5 (8.89)* or ¥0* ' Thus У^У.73^ , from Eq. (2) V3 =[ 33g , and from Eq.d) 7^ 8-ISR

6.17 R | 8.I7R (Flow meters) Water flows in a 0.10-m-diameter pipe at a rate of 0.02 mJ/s. If the pressure difference across the orifice meter in the pipe is to be 28 kPa, what diameter orifice is needed? (ANS 0.070 nit V b~O.I0m d and p •<?<?<? %з- Hence, - »a ~3 Ood 3.¥0* 0 = I---------7 M*°- Re-ty, , ' Thus, D (2.zs^)(0.lm') Trial and error solution1 fl-иите a value of d; calculate = 77 ; lookup C„ in fig-9.¥ I and calculate З.¥0х/О~3 (see Eg.O)}. where Co ~Ct( Re ,£) fro/» fig. 9.¥l. (0 where Vе Д E fto.nrf "*-55& = 2.29t/0s ; compare wiih Jl~()od)¥ d d, m P c. 7/ -(iod)¥ 0.070 0.70 0.61 3. ¥3xi0~3 0.069 0.69 0.6I 3.30x)0~3 0.07) 0.71 0.61 3.56* Hf3 Thus, d - 0-070 m Q-I6R

8.18R |Но» meters) A 2.5-in.-diameter flow nozzle is in- stalled in a 3.8-in.-diamctcr pipe that carries water at 160 °F. If the flowrate is 0.78 cfs. determine the reading on the inverted air-water U-tube manometer used to measure the pressure dif- ference across the meter. • ANS: 6.75 ft) (I) where ₽”о’1^"лл58 = 9.90^ and fnm НЫвВ.Г- р‘П8Чб TL il ‘ /U=Q.3ZX/0'* Thus, wtin D pW> (1996 ^)(9.9ф9)(т(Н) .I . r o„,. ле = *тг~ =---°—/J.-3-------7./¥хЮ we obtain from Fn.9.¥3- 3.31ХЮ Cn - 0.989 Thus, from Ey, (!) (0.989)^/ Ofl-h ‘ M Hewer, so Ы h cn 8-I7R

V.lRd.ift drag calculation) Determine the lift and drag coefficients (based on frontal area) for the triangular two-di- mensional object shown in Fig. P9.IR. Neglect shear forces. (ANS: 0; I 70) <£/= [p cos 6 dA + sin & dA , where %r~0 Thus, oft' * $ f cosG dA p cos6dA pcos6dfi e dA ~2 \ P Ч~5'M I Hence, ') 2b ~2.(-/i°(£pV*)cos where . , n c о b/*_ / = length of object Z jr /.7(iev')ib “ tpu'ib Because of symmetry of the object, T=O °Г £ = /,70 Note • Г - /7 + fG * 6)cos ° _5—c = -id fV^lb .z (i-2>dfv‘)^^ i.7Hev‘)/b which checks wilb fhe above answer. G G ?-/£

9.2ft I 9.2R 11 sternal How character> A 0.23-m-diameter soccer ball moves through the air with a speed of 10 m/s. Would the flow around the ball be classified as low. moderate, or large Reynolds number flow? Explain. (ANS. Luigi* Reynolds number flow) Яе =3T , where D-0.23m } 4*10%, and Vs WW*%- for standard air. Thus, pfi - (1оФ)(О-*3^ = i.sgxitf This is a large Key golds ™ 'MW % number flow. (See Tig.1. 6 g) Я.ЗК | 9.5R i External flow character) A small 15-mm-long fish swims with a speed of 20 mm/s. Would a boundary layer type flow be developed along the sides of the fish? Explain. rANS: No) /?e = , or with t~ /^x/o~3m j U*zox/o SL and l/= (t'.e., /S.S'C water) Re" -268 This Reynolds number js not large eqwgh Io have true boundary layer type ^oW. (Re ^/ООО « often assumed to be the lower limit.) V-2O <?~2R

9.4R ilhnindan laser Но» i Air flows over a flat plate of length i = 2 ft such that the Reynolds number based on the plate length is Re 2 X IO5. Plot the boundary layer thickness. 8, for 0 s r £ /S.7^ •Ш?, where l*2fi and y* I.S7*lo-*& * T, (И/оЧОМ/М?) ThuSj u=--------yti— For /?ey~-Sx/^5 the boundary layer flow as lowrinar and r r IW ~ c lf/.57x/o'v)x or 6= 0.0/50 where X~H.<f~ft 6~siu —тп * This resuH is plotted below for О * X *2 ft. h/ofe the X~axis The boundary layer drawn to scale for оsX-0.5 ft is shewn below, h/ote" If is "thin' (t.e^ V*I5.7 £ x-o X = 0.25 ft sl = 0.0112 ft X*0.5ft i X <0.5 ft Я'ЗК

4.5R | 9.5 R iltoundars later How At a given location along a flat plate the boundary layer thickness is 6 “ 45 mm At this location, what would be the boundary layer thickness if it were delined as the distance from the plate where the velocity is 97% of the upstream velocity rather than the standard 99% ? Assume laminar flow. (ANS: 38 5 mm» The standard definition of <f is the value of у where U “О.ЯЯ fdThis occurs af a Value of tja у “ 5-° that Д » mm when )7 *5-0 For qiven fJ, Vt andX or CD From Table fd we interpolate to find the value of when tf-O.47 Thusj * У-.О +0.ц(03700^O9SSS\ = 2g f 7 'r-. 7S9 -msss / so -ihat £9.U) (jivesi 3s. 5 mm

тал 9.6 R (I- ridion drag) A laminar boundary layer formed on one side of a plate of length € produces a drag 3. How much must the plate be shortened if the drag on the new plate is to be 3/4? Assume the upstream velocity remains die same. Ex- plain your answer physically. (ANS: (пл - f/16) U - t -1 where ° /yj? and f y/Re^ R* bl where b" plate width. Thus, J)( - ipV2™2- ь/ = a^pt^bfr/r (/) Consider huo f/ows with , Ц = Uz , b, *b2 t and I or/z so from Ey. (I) =^~^~ so ^“t w'^ °fyzs^ Ai ) Or Since the boundary layer is thinner on the front portion of the plate than on the rear portion) the wall shear stress is greater near the front. To reduce the drag by a factor of four} more than three-fourths of the plate must be removed (i.a^ from length Я to length ji). 4-5R

PREFACE This Student Solutions Manual has been developed as a supplement to the third edition of Fundamentals of Fluid Mechanics by Munson. Young, and Okiishi. In this third edition a new section has been added at the end of each chapter which contains a series of review problems. These problems are representative of the types of problems that students should be able to solve after completing the chapter, and this Solutions Manual contains the detailed solutions to these review problems. We believe that as students prepare for an examination, or feel the need for some additional work on a particular topic, it will be helpful to have available such a set of review problems with their corresponding solu- tions. Each review problem is preceded by brief phrases which give an indication of the main topics to be used in solving the problem. Thus, the student can conveniently select those topics, and the corresponding review problems, of interest. This information is also presented in the table of contents. The solutions contained in this manual are worked in a logical, systematic way with sufficient detail so that they can be readily followed. Except where a greater accuracy is warranted, all intermediate calculations and answers are given to three significant figures. For con- venience, some tables of properties and tables of unit conversion factors taken from Fundamentals of Fluid Mechanics are included on the inside of the front and back covers of this manual. Unless otherwise indicated in the problem statement, values of fluid properties used in the solutions are those given in the properties tables found on the inside of the front cover. Some occasional references to equations made in the solutions refer back to Fundamentals of Fluid Mechanics. Although this Student Solutions Manual has been developed as a supplement to a particular text, the solutions are sufficiently detailed, using conventional notation and basic equations, so that the manual could also be a useful supplement to other standard fluid mechanics texts. Problem statements and figures are included with each solution. v

9.7R 9.7 R I Momentum integral equation) As is indicated in Table 9.2. the laminar boundary layer results obtained from the momentum integral equation are relatively insensitive to the shape of the assumed velocity profile. Consider the profile given by и = U far у > 6, and и = I/{1 — [(у — S)/S12)''2 for у £ f> as shown in Fig. P9.7R. Note that this satisfies the con- ditions и = 0 at у = 0 and и = U at у — 6. However, show that such a profile produces meaningless results when used with the momentum integral equation. Explain. From the momentum integral equation, б=7^^- > tt• (»-p t (,) У I/ v L «* /Vote"' -frsO of Y~0 and * I and Y*l } as required. Also, C, x y(l~g) dY which can be evaluated for the given g(Y). о However, * , ^re % - -(У-Л f-хХУ-П- [T^ Y*0 , Thus, as Y—ot ft-°0 so that Cz-°°, which from Eq. (I) gives 6 = 00 This profile cannot be used since it gives due to the physically unrealistic ^•‘oa at the surface (y*o). See the figure below.

9.8ft | 9.KR (Drag—low Reynolds number) How fast do small water droplets of 0.06-pm (6 X IO"’ m) diameter fall through the air under standard sea-level conditions? Assume the drops do not evaporate. Repeat the problem for standard conditions at 5000-rn altitude (ANS: 1 .10 X 10 m/s; I 20 x 10 7 m/s) die. =D-6*/6 m R'-ty fv •*ir For steady conditions, d)+Fe sld, where ti{ Re-^^t, dJ-dray = 37TDU/x Also, ^weiyht an<* s Kir s twoyar! force Since Кну, <<^1d j con neyted the bvoyoni force. That is, f or orV=^~ fit sea level /s = /.7духю^ (see Table C.2) so that u= llotlf^ (I) Ai/e M Re - ““ °{ л the low Re drag elation is l/alid. fit an аНЫе of SooOm , and from Ef.lD = /.20x/o'7^- u= 4-7R

тал 9.9R iDragi A 12-mm-diamcter cable is strung between a series of poles that are 40 m apart. Determine the horizontal force this cable puts on each pole if the wind velocity is 30 m/s. (ANS: 372 N) ( D - /2 etm Fp = force on one pole = of) where A Since Be - ж *2.+7*Ю* if follows front F/j.9-2-1 lhaiCD . Hence, R * ^)(3o^(Wn,)(o.oi2n>)^377^ 4-8R

Я. IDR | 4.1(IR i llrayi How much less power is required to pedal a racing-style bicycle at 20 mph with a 10-mph tail wind than at the same speed with a 10-mph head wind? (See Fig. 9.30.) (ANS: 0.375 hp) V “(20-io) mph s Ю mph fail wind U* (2.0*10) mph » 30 mph head wind "P ~ power io overcome aerodynamic dray " 17B c&, where Ug* bike speed and , wilh V - wind speed relative io bike. from Fig. for a racing bike Cb = 0.08 and А'З.ЯИ2 WUh а Ш vie J U, =(«»?) f^-) • 24.3 # «"</ Ц-(з.0-ю)пр1 ‘ /*./£ 7»«, PM = (2t3tf)l(o.40Z3B ^)(№7#Г(з.4Я‘)(<>ев) ‘ O.vm 4» \Tdh a head wind Vs (Ю^Ю) mph * ^Sj ~^ead ж №^)л^-оогзв ° 08) s Hence, ~ УЗ , *(0.42-2 - 0.0¥7/} hp - 0.375 hp 9-9/?

таи 9.11R < Drag t A rectangular car-top carrier of 1.6-ft height. 5.0-ft length (front to hack), and a 4.2-ft width is attached to the top of a car. Estimate the additional power required to drive the car with the carrier at 60 mph through still air compared with the power required to drive only the car at 60 mph. <ANS: 12.9 hp) "P = power = Uofi where d? * dray я -% PU*/I (!) From Ey. 4.18 with £ = -fr = 3./3 we attain CD =/-3 (Ntte • Re - = ^57мо~¥ = ^^7X/°S, whereas the Value from Fiy. 4.1-8 is iha] for Re ~/os, Howevert for a ttuni oi\jett life (his boxt (he Value of is esseniially independent of Re.) Thust iron Ey. (!) /3 (i)(0.0O3.3Q *-%&)(/.бЮ(¥.1Н) (88-^)2 = eo.zlh So fhtt ‘(88^) (90.5lb) = /2.^bp 550 Г 7-/0R

9.12К (Dragt Estimate the wind velocity necessary to blow over the 250-kN boxcar shown in Fig. P9.12R. (ANS: approximately 32.6 m/s io 35.1 m/s) 7Г the boxcar is about у Io tip around point 0t then -0j or 2.55^ = O.7SVJ = 7.3*40% or U- If we consider the boxcar as a flat plale y then from Fig. 9J9 Cb */.9 so that JI Г 7. (7.3¥ХЮ¥/^) -^1 /Vote '• The Cb value used above is for a thin plate normal to the flow (i.e. I-0 in /he above figure). From Fig. 9.2S with 2/D *! (ti e. assume the width equals the height) we have Cb *2.2 which (gives Us 32.^ ^L. The two Values are not too different.

тати 9.I3R (Dragi A 200-N rock (roughly spherical in shape) of specific gravity SG = 1.93 falls at a constant speed U. De- termine U if the rock falls through (a) air: (b) water. (ANS: 176 m/s: 5.28 m/si <Nw Jap L-’ For steady fall orFg'^'d л lj) where J^eV^D1 and fU Нысе, w=2OqN = /лз(ч.90*к?-H*) or & = °-272 Also, FB = = p(%8/&) = 0.103 f> Nj where Thus, Eg.(D becomes О. юз p f CD i (o.mf 2oo f or о. юз p+0.02.4/pCbUz =2oot where U~* « a) For falling through air, and Eg. (2) becomes 0.0359 CD Uz«/94.4 or CbUl = sse° Ю , Л< ’ > • - ! Н»? U. „her. V- ? (в e„J Vi,. 9.^, <i е,ьт Note •' * Ena/ and error solu!ion' Assume CQ obtain U from Eo.(3^) He from check Cb from the graph. Assume Сй = 0.5-* V= /об g —Re* /У7*/о‘ — CB*o.!64o.s Assume Cqk 0.16—* {^=/37^ —* Re*З.Ч-7*/о*—* CB* 0.18 eo.!6 Assume Cb -O.fQ— U* ПЬ — Re •8Л7 */<>‘ ^CD*0. IB (checks) Thus} U /76 jf) air. (b ) For falling through water} p s 944 -^s and Eg. (2) becomes CbJJZ^3.3^ Trial and error solution with Eys. (t)tll)fand(&). Assume Cb ~ 0.2 - U* ¥.o9 ^—Re= W¥*K>& “* ^>= O./O Ф 0.2 Assume Cb»0J2-^ U-5.29^- —^Re */.7^*/о* —* Cb-o./2 (checks) Thus, U~5.2&in water. <H 2R

чж I 4.I4R (Drag—composite bodsi A shortwave radio an- tenna is constructed from circular tubing, as is illustrated in Fig. P9.I4R. Estimate the wind force on the antenna in a 100 km/hr wind. IANS: ISON) The qnienna is a composife body consisting of one main pole, one horiiontal bar, and four vertical rods. Thus, - /.У и'*'ге Ohio in CD(- from Tig. 9.2! for the given Re, s 1 Thus, R , (27.8^)(0.04m) = 4 c Пв' /.46X10'^ ^ZX/V Rtr ° CAt* Z* so -that -^(1.13 ^(zzeS-fa.^SmyOMm) +(l.sn,)(0.02.f>i)+4(i,n)(o.oim)^ or <H3R

Ч.15К (I ift> Show that for level flight the drag on a given airplane is independent of altitude if the lift and drag coefficients remain constant. Note that with CL constant the airplane must fly faster at a higher altitude. у For level flight <£= where airplane weigh! -constant and Alto, Thus, if CL f /)} and X are constant (independent of altitude), it follows from Hg.O) that pU is also constant. Hence, for constant CD it follows from Eg.(i) that X) is constant (independent of altitude). (Since p decreases with increasing attitude the idea that pU is constant implies that F increases with altitude.) R (D

9.16R (Lift) The wing area of a small airplane weighing 6.22 kN is 10.2 m2. (a) If the cruising speed of the plane is 210 km/hr, determine the lift coefficient of the wing, (b) If the en- gine delivers 150 kW at this speed, and if 60% of this power represents propeller kiss and body resistance, what is the drag coefficient of the wing. (ANS: 0.292; 0.0483» U=210 km/hr^S-L-J , /о. 2*? W=6.22kk For equilibrium, • Й/ = 6.2 2 kN t where Xs CL£f>Vy). Th", „М 58.3 9- Г Ж * - 6.22* ________ 4 if 17 /? i (1.23 ^)(58.3^-)Z( 10.2 Ш*) ' ft I so} = Tbody+TL * tin, J vhere = 0.6 fa _. pnp ^=0.6(150 к») Thas> = <?okW fa s,S0 kW-90kW = 60 kb/ where ^^D-c,ieuMv °C = . г ^>'°J = 0.0*83 ° pz/i/? 6/.z3^X5s.3^-)3f/a2/»x) ==

IQ.IR | IO.IK (Surface waves) If the water depth in a pond is 15 ft. determine the speed of small amplitude, long wavelength (X > y) waves on the surface. (ANS: 22.0 ft/s) For Л , tank so that (see Eg /o.¥) -[(32.2 £)(15W)]* = 22-0# 10.2 R I I0.2K (Surface waves) A small amphtude wave travels across a pond with a speed of 9.6 ft/s. Determine the water depth. (ANS: v ? 286 ft) For long wavelength waves C- '/$/' or у g ° 32.аЧ. s 2-06 ft In general, from Eg. IQ.V. Hence lonlj Pick various Л and calculate y. (I) we do know that у £ 2.86 ft /0-1R

|O.3R | I D.3R I Froude number) The average velocity and average depth of a river from its beginning in the mountains to its dis- charge into the ocean arc given in the table below. Plot a graph of the Froude number as a function of distance along the river. Distance (mi) Average Velocity (ft/s) Average Depth (ft) 0 13 1.5 5 10 2.0 10 9 2.3 30 5 3.7 50 4 5.4 W) 4 6.0 90 3 6.2 (ANS: Fr I.H7 at the beginning. 0.212 at the discharge) f where anJ values of Vend у are given in /he /able. The following resultsare obtained. Distance , mi Fr 0 1.87 5 /.23 Ю l.os 30 0.4-38 so 0.303 80 о. гев 90 0.212 Distance, mi No/e that the flow « supercritical for the first Ю miles or so. I0-2R

тал I0.4R (Froude nemberi Water flows in a rectangular channel at a depth of 4 ft and a flowrate of Q = 200 cfs. De- termine the minimum channel width if the flow is to be sub- critical. t ANS: 4.41 ft) Q*200cf.s у И e , where b - witllh~ ft <»rd И* ft/s. Thus, Note'- /)$ b decrnseSj frincreases. Sei Fr*l for minimum w^ih hr subrcriifcal {low. Hencet | =____50/i}___ or b= [(32.2)(4)]^ — ID-3R

Ю.5R | 1D.5R i Specific energy > Plot the specific energy diagram for a wide channel carrying q = 50 fr/s. Determine (at the critical depth, (b) the minimum specific energy, (c) the alternate depth corresponding to a depth of 2.5 ft. and Id) the possible flow velocities if E = 10 ft. tANS: 4.27 ft; 6.41 ft. X 12 ft; 5 22 ft/sor 22.3 ft/s) Thus. L Mote- r'“r —, where E~ ft, y~ ft Fguation (l) is plotted belw -]* = ¥.27 ft c) If then - 8.71 Thus, solve for the positive reel roofs of 8.7/ = y+ , or y‘- 8.7/ У* 138.8 *0 One roof is the given у *2.5. Thus, divide бу-2.^ info Ед.Ы to obtain y^-i.Vy ~/i.S3*0 which hes reels Thus, the alternate depth corresponding to у *25 fl /s у *8.12 ft d) For 5*/Off Еъ(!) is IO=y + l~pfor y3-/oyx 1-38.8*0 which has roofs • у *9.57, 2.2¥,Md-/,8/. Thus, y*9.57fl»ry*2.7.¥fl <i) (3) The specific energy diagram is as shown above. Ю-flR

IQ.6R III.6R (Specific cnei-д?» Water flows at a rate of 1000 ft3/s in a horizontal rectangular channel 30 ft wide with a 2-ft depth. Determine the depth if the channel contracts to a width of 25 ft. Explain. (ANS: 2.57 ft» lop view side view ^ + ^‘*7,= + Z,*Y/*2H, (”'?9)2.2fl (%? 2(32.2$) 2(32.2$) or У2 - 6'23y* #• 2¥.8 = 0 which has roots y^S.5)^ 2.571 -/.75 ft. Note ’ FrtB^s a =-----л = 2. OB >1 [(32.2$(2f/)]i If there is no relative minima/» area between (Dand(2) where critical flow can occur, it follows that frt >l also. Thus, the y2sS.Sl (subcrihca! flow root) is not valid. Thvsf yz‘Z2B 2.57 ft Note: If уг = 5.5/ ft anol Уг- £si = 7.26^-, then r_ - JL В 7.2-6 ft* _n Fr’ W pziW.ys.wW]14 -aS*5 v/hereas if u0 „ yt = 2.57ft and Vi ~ 2^7 ->5Л then VriS[(3252)(2.S7^ ~ /,7/ >Z Ю-SR /

I0.7R | I0.7R i W all shear Stress! Water flows in a 10-ft-wide rec- tangular channel with a flowrate of 150 cfs and a depth of 3 ft. If the slope is 0.005. determine the Manning coefficient, n. and the average shear stress at the sides and bottom of the channel. (ANS: 0.0320; 0.585 lb/ft-1 y-.3fr •I- — ’ bc /Off = widih QslS0 cfs • ^>>>>/7 ^0.005 =3. t where X*/.¥9t /) = by «= (ЮЮОЮ - Soft* P= b *2y « /Off *2fyff) hQ7£ ff Also, X = SO = (^¥^)(/.d7SH)(O.OOS) = 0.S85 I0-6R

IO.BR | 10.8R (Manning equation) The triangular flume shown in Rg. PI0.8R is built to cany its design flowrate. Qo. at a depth of 0.90 m as is indicated. It the flume is to be able to carry up to twice its design flowrate. Q = IQq. determine the freeboard, f. needed. (ANS: 0.378 in) FIGURE P10.8R Ze/ ( )0 denote the design conditions and ( \ denote conditions with Q 9 2-Qo Thus, Q^^R^S" > where K*t, /),• (o.9m)2= 0.81m* Po= 2l2 (0.9m) » 2.55 m Thus, = = °-3,8 m Hence, QB = (O.eio) (0.318 j* Sof hlso, Q2 « /)2 R* SO3 , where ftx- У2 , ''''ХА// Hence, with n0sn0 and t 91= J4a(o wii)4S.? or from Eft. (I) and (2) with = 2Q we obtain 0.500y^* = 2 (0.377) or y3* U67 m Ho we /er, уг-0.9т s iainf** so that v . ^-О.Чт _ 1Л7т-О.9т _0^78m T sin 95* sin 95е - I0-7R

IQ.^R | 10.ЧК iMannini, equation i Water flows in a rectangular channel of width b at a depth of b/3. Determine the diameter of a circular channel (in terms of b) that carries the same flow- rate when it is half-full. Both channels have the same Manning coefficient, n. and slope. (ANS. 0.889 b) ln For the rectangle, ft* b(^) and P*bt?(4)“'fb For the semi-circle f /)” and PX?D Thus, from fyd) with equal Q, X,n, and s0 for either case'- or D= o. eeq Ь Io-br

IO.IOR | IO.IOR (Manning equation) A weedy irrigation canal of trapezoidal cross section is to carry 20 m’/s when built on a slope of 0.56 m/km. If the sides are at a 45° angle and the bottom is 8 m wide, determine the width of the waterline at the free surface. (ANS: 12.0 tn) 8 m Q ш % f) Rh$ , where from Table 10.11 ng0.°3 Л1«, (0 and ° Sti-8JX so lhal D -A = nf> P (8t2.83y) Thus, E^d) becomes 20*о^У^У[Ъ °Г (y* + Byfo 2S.¥ л/ГЛ Дх 0Г 0.4 y* +8y-6.96(8 +2.93 y) s0sf(y) Solve for F(y)=O From lhe graph, Fs О when у = 1.98 m (D Thus, t*2y+8m Or /« 2(l-98m') iStn = 12.0m

IO.IIR | HI.HR (Manning equation* Determine the maximum flowrate possible for the creek shown in Fig. P10.11R if it is not to overflow onto the floodplain. The creek bed drops an average of 5 ft/half mile of length. Determine the flowrate dur- ing a flood if the depth is 8 ft. (ANS: 18: ft’/s 1517 ft’/s) ^•(sx+4*)^ -6.40ft (П /в2-Н-Л (a) Mau mum Q without overflowing- whir. X-I-Vt and from Table 10.1 п=О.озс Also, and P* 6.¥0H + 6fi + 8.O6H*2 so that Rfi=-ps^Hs2’3¥fi Th us, from Q = аозо (^в)(2.3¥)^(0.00101) (d) A f a flood depth of 8 ft: QsQi +01 ♦ Qs, where 9.- X*/*f, S0=aoo/e<1, and from Table /«•/* п^0.озо}Пя=о.оз5 ^о.о5о From part M* S.vofj and /^S.oeff while Jta-(sox^)4KSo.2fl and ^(tf+4^-So. )ft Thus, /}, • (¥H)(l8H) * i(6ft ★I6H)(4H) = /20 X?2= i (fH)(8Oft)*/doft* ond fa •i(¥fl)(doft) - /ooftx an^ Pl»(6.W^6^8.06)ft*2o.sf{ fpx= 80.1 H, and P3 = so ^Ff Hence, since we obtain fa Чц** * or icf (izo)*4 П60)*Ь (юо^л П Q - 1.ЧЧ(.0.0019% (iat?4(0.03) + (в0.1)^(о.03£) + (so.2)^(o.os)] =(842+4-70 4.20s) £ or Q= 1517 10' IOR

I0.I2R | 10.I2R (Best hydraulic cross section* Show that the tri- angular channel with the best hydraulic cross section (i.e.. min- imum area, A. for a given flowrate) is a right triangle as is shown in Rg. Е10.8Л. (? = •$/) Rh^ or K, n, s, constant Thus, fora given flowrale and for the mini num area Eg.(i) gives ^-0 Also, t where Ps2% and A*£b(! cosO) °r -it, corf “> However, since sin6* $ or i* zsine follows IM Thus, Eg. W becomes • Rh = (21^ ~ (sinG смв)* •i(sMcoseiiiHi^ tifA аХмвекв^соге-ап’-е) Thus, with ^‘0 (i.e- minimum area), ^--0 when cos^ff-sin^Q^O Q =¥5* (i.e. (he besd hydraulic cross -section occurs with a right triangle) 10-HR

I0.I3R | I0.I5R (Hydraulic jump) At the bottom of a water ride in an amusement park, the water in the rectangular channel has a depth of 1.2 ft and a velocity of 15.6 ft/s. Determine the height of the "standing wave" (a hydraulic jump) that the boat passes through for its final "splash." (ANS: 2.50 ft) The depth ratio for a hydraulic jump is & = i f-l +J| ♦8Fr,xl where fr, • --—к = ь -2 51 * L * ' J ' еГ ' [(ЗГ.2?ХЛ2П^ -ZS‘ Hence, £ =^[-1 +Jt+ 9(2.51)*] = 3.08 so that У2 = (3 °8)(l.2fi) = 3.70ft Thus, ya -y » (3.7O-/.2o)H » 2.50 ft Ш] I0.I4R (Hsdraulic jump) Water flows in a rectangular channel with velocity V 6 m/s. A gate at the end of the channel is suddenly closed so that a wave (a moving hydraulic jump) travels upstream with velocity V, = 2 m/s as is indicated in Fig. P10.I4R. Determine the depths ahead of and behind the wave. Note that this is an unsteady problem for a stationary observer. However, for an observer moving to the left with ve- locity V„. the flow appears as a steady hydraulic jump. iANS: 0 652 nr, 2 61 nit FIGURE P10.14R For an observer moving to the left with speed » 2^ the flow appears as shoe* below. v-V^= в® „ Thus, treat as a jump with ' y, --------y* V,s вф f Vis2.& ssrss/ssslsss 'V Since /), И 14 or 4 it follows that ^ = 1[-| +^l+8Fr^] = ¥ Hence, Fr, = 3.16 However, So that and = s^(0-6S2m)^2.6lm I0-I2R

I0.L5R | 10.15K i Sharp-crested weir) Determine the head. H. re- quired to allow a flowrale of 600 m'/hr over a sharp-crested triangular weir with в = 60°. (ANS: 0536 m) , where Q = -g- and 6 *6°* Thus, Ы61 or .’л 0 122 = Н* (I) The value of is attained from Fig. 10.25 as a function of H. о '6°' Trial and error solution of Egs.(l),(2): p^jo.xs (t) Assume Cwi = 0.60, or 0.122 = 0.60 so that HK0.S24m 0Г H‘(0.524m)(3.28£)*l.73 ft From Fig. /0.25 f/)/s gives ~0.5B ° ЙН 1 (extrapolate for Assume C^s0.S8t or 0.122 =0.S8 so that (0.536rn)(3.2в = /.76 ft which checks with C^~0.SB, the assumed value. Thus, 0.536 m I0-I3R

IO./6R\ 1U.16R (Broad-crested weir) The top of a broad-crested weir block is at an elevation of 724.5 ft. which is 4 ft above the channel bottom. If the weir is 20-ft wide and the flowrate is 400 cfs, determine the elevation of the reservoir upstream of the weir. (ANS: 730 86 ft) <?- (i^ . « ь'хой- , where H~H Thus, H3-2*.8H-49.2=0 Ths roots of this cubic equation аге H- 6,36 and two complex roots. Thus, Hs 6.36 ft so that 1,= 72 УStt + 6. 36 ft • 730. 86 ft I0-/9R

10.17 К 10.17R (Underflow gate» Water flows under a sluice gate in a 60-ft-wide finished concrete channel as is shown in Fig. P10.17R. Determine the flowrate. If the slope of the channel is 2.5 ft/100 ft, will the water depth increase or decrease down- stream of the gate? Assume Cc = y2/a = 0.65. Explain. (ANS: 1670 it'1 A: decrease) ¥ .. P 1 0 . 1 7 R 0.55 Q = ba- = b CA а where, b = 60 # . a = 2 ft and from fy.10.21 " “ _ zfree outflow since e =S, /7 follows that C^= 0.55 Hence, * (боЮ(о.££)(2Ю[г(з2.2^)(юН}] = /67O-g- Determine the slope needed io maintain uniform flow downstream of the gate- Q^ARh6^, where К-1Я9 and from Table WJ 0.012 (» A,so> cca -0.65(2f^=l-3 ft /.3ff~fe so that *— j Я = - 7Q P= (60+2(l.3))ft = &2.£H °nd R - A - 7efjZ - 1 245 U 6of\ 1610 ' >0Г 0.0222 Hence, the required slope for uniform flow is Sts0.0222, but lhe actual slope is So= -O.ozsOj more than required. The fluid will speedup and the depth decrease. IO-I5R

IIAR | 11.1 К (Speed of sound) Determine the speed of sound in air for a hot summer day when the temperature is 100 °F; for a cold winter day when the temperature is — 20 °F. (ANS: 1160 ft/s; 1028 ft/s) For a perfect gas C * ykRTt where for air к” ТЧ- and R= / 7/6 iff ja (see Tab/e/.7). •J* U4 I' Hence, _____________ ________________ ,—__________________________- Cootf: = /z f ) (^o +/oo) 'R = or c/w*f ~ U60 = 77/ mph Similarly, .----------------------- e /oxg У = 701 mph 7E2R1 11.2R (Speed of sound) Compare values of the speed of sound in ft/s in the following liquids at 68 °F: (a) ethyl alcohol, (b) glycerin, (c) mercury. (ANS: 3810 ft/s; 6220 ft/s; 4760 ft/s) For a liquid с where the values o{ Fv (bulk modulus') and. P (density) are qiven in Table /£. Thus, for el/iyl alcohol /.53 = 39Ю\ or fl С=38Ю al " Г (t.sitio^)//* and for mercury 220 Q * #270 mph J //-IR

7W] I UR (Sound waves) A stationary point source emits weak pressure pulses in a flow that moves uniformly from left to right with a Mach number of 0.5. Sketch the instantaneous outline at time = 10s of pressure waves emitted earlier at time = 5s and time = 8s. Assume the speed of sound is 1000 ft/s. Since Ma- 'c' if follow that I he fluid speed is ]/ - c Ma = /000^ (0.5) *SOoQ. Assume that the sound source is located at **y-0. Al time !0sj the pulse that leas emitted af h&s has traveled a distance R9a c Ai - I ooo Q (/0s-8s) *2.oooft relative lo the moving -fluid that has moved 4 distance Xg «= V*t = 500 Q (/Os -8s) - 1000 ft. 5/т/1аг1ул for the pulse that was emitted at fc-SSj Rs - /ooo (Ms -5s)* ^оооМ and Xs < 500& (/Os -5s) « 2 soo H These pulses at ts/0s are shown below. They are circles. II-2.R

7Ш~Г 11.4R (Mach number» An airplane moves forward in air with a speed of 500 mph al an altitude of 40,000 ft. Determine the Mach number involved if the air is considered as U.S. stan- dard atmosphere (see Table C.l). t ANS: 0.757» Ma " "o' j where C* tfkRT Fro/n Table C.lj the le/vperalure а! Чоооо f4 is -69.7'F. Thusj c « [/.?(/7/j£fa)(* 968 Q so -that , ,, t. л= O,7S7 II-3P

«Ей 11.SR । Isentropic flow) At section (I) in the isentropic flow of carbon dioxide, p, = 40 kPa(abs). Tt = 60 ®C. and Vi = 350 m/s. Determine the flow velocity, V2. in m/s. at another section, section (2). where the Mach number is 2.0. Also calculate the section area ratio. Л2/Л,. (ANS: 500 m/s; 1.71) fl = V^3S0/n/s C°* Vz = = 2C^2 ^kRT2' (I) where far C02j k*!’30 afldR*i88;9 (see fable/д') Also, from Ец. I/.56 ‘ ' wh‘re f’r is‘"jrl,f'c floW^ foj-To.z. T/p^ by dividitli) these equations we obtain r, s 7WW " However м - И 350 lyia,w 7>—» = . = 1,22 V к 8T, ]//и J (1,3) f273 * £0)K Thus from Eq. Tz « (273^0)К (2)1^ 2S5K/S0 from Ея.(!) so fhal with . /42 =wr .з.ез, we obtain 3 83 A , 1Л2 f 1+0.15(1? J ' _ . . 2 [//0,/5(/.22)*] “ L= //-V₽

11,6R i Isentropic How I An ideal gas in a large storage tank at 100 psia and 60 *F flows isen tropically through a converging duct to the atmosphere. The throat area of the duct is 0.1 ft . Determine the pressure, temperature, velocity, and mass flow- rate of the gas at the duct throat if the gas is (a) air, (b) carbon dioxide, (c) helium. (ANS: (a) 5X8 psia; 433 °R. 1020 ft/s; l.(M slugs/s; (bi 54 6 psia; 452 "R; 815 ft/s; 125 slugs/s; (ci 48.8 psia; 391 R; 2840 ft/s; 0 411 slugs/s) ^•lOO/uia ------ Тв‘60Г paim If I he nozzle is choked t then the fluid speed of the exit Zr equal Io the speed of sound there and the pressure is the critical pressure p' where from Eq. II.61 r z WM If p9< fltim , then the nozzle is not choked and sfide. If P4> P»i/»j then the nozzle is choked and p*. A shown below, the nozzle is choked in each case considered- a) For air /</¥ » /М р*~ (юоpsia) (yjppi) = 52.0psia >* Pt.7psia Thus, pik* p* = 52.5 psia From Eq. U.63, since me have Th * T'- T. (-&) or s So that FT'I(>AV-(^a-\A M m ?HV (RTtk)^ ^Жи^) or ! S n rri q (continued) 11-SR

II.6R continue^ b) Similarly for carbon di ok ide к *1.3 so ibaf « z . X/ 2. f = (1°°Р31^{7ТП) = £¥.4psia > fam so fa*fl Also '7ih “ r‘ “ To (f^) * (¥60 tjo) (7^77) * ¥ S2*R so that Vth-fkn, -[>-3(M0^i)(WR)]* ~ etstt and (£U sfe. _L^= c) F/'nallyj for helium so Ihal , , x i.u/(l.u-D а P • (toopsia) (j^i) = ¥8.8psia > ра^ so lhal fa у Also, s T** To (f^i) s(¥6O/£o) (yjiTf) * 3<?Гя so that (39/ 9R)]^28¥C^ (¥8.9&)(№^) (O.lWl&H) 9) . (12^20^. )(3ffR) 0. ¥// dPSL s II-6R

II.6R continue^ b) Similarly for carbon di ok ide к *1.3 so ibaf « z . X/ 2. f = (1°°Р31^{7ТП) = S¥. 6psia >fa,„ so fa*fl Also '7ih “ r‘ “ To * (¥60 /60) (7^77) * ¥ 52*R so that Vth-fkn, -[>-3(M0^i)(WR)]* ~ etstt and (£U sfe. -L^= c) F/'nallyj for helium so Ihal , , x i.u/(l.u-D а p • (lOOpsia) (j^i) = ¥8.8 psi a > ра^ so lhal fa у Also, s T** To (f^i) -(¥60/60) (y^7f) * 3<t/'fi so that (39/ •R)]^28¥O^ (¥8.9&)(/9*&) (О./^)(29¥О . (12^20^. )(3ffR) 0. ¥// dPSL s II-6R

II.7R I I I.7R ll anno Нои) A long, smooth wall pipe (f = 0.01) is to deliver 8000 ft3/min of air at 60 °F and 14.7 psia. The inside diameter of the pipe is 0.5 ft and the length of the pipe is 100 ft. Determine the static temperature and pressure required at the pipe entrance if the flow through the pipe is adiabatic. (ANS: 539 T<; 214 psia) 1*100" T^iO'F = 0.608 Th US' from Fit], b.2. (Fann о flow) wrlh /Ьа^О.бОв wa obtain г ~8г тй 7Г(0.5")г so that Ma2”- = 0.4£ so that from f (kJ,) _ (о.оЫЮоН) ,2 00 a fU4.) f(f-k) ' D as fl ' ' 5 d v/e о bl a in 1(H) = 2.00+0.4-5 * 2.¥£ D Hence, from Fit]. D.2 we find Ма^О.ЗЯ, -Д = U6 / and ^s2.7S Thus, wilh^ 4 TtaT2^)(and />, A we can obtain Tandpt. From F/f. D2 with /Vtqx = 0.608 we have &e/./2 and T Thus, . T ^боноУН - Wfi - l¥.7Psia Z3.¥psia II-7R

7IgR~[ 11 .HR (Ray leigh flow» Air enters a constant-area duct that may be considered frictionless with Г, = 300 К and V, = 300 m/s. Determine the amount of heat transfer in kJ/kg required to choke the Rayleigh flow involved. (ANS: 5O2OJ/kgi Т^зоо к f For Rayleigh flow the energy equation (Eq. 5.69) is m[h2-h, ♦ £ (Vi1-Vrx)] «=Qnei . in fynet - ж ho ~h0 where h0' h*£V ' Cfi70 is in m * ' v г /he stagnation enthalpy. Thus] fynet - Cy>(To2.~'f)i) 0) ht section (D, Ria, a ~v/,#?-• = r— c°-86¥ i ikKT, [/у (2S6.f Thust from Eq. (H.-rt), I, ' 'orT°'° (3°oK^" <^(as^l =3¥s x Since the flow is choked (<‘.e. /У1аг-1) J ^here from Eq, (l/J3i)t T,. (iTkM^r • 7 - (зч-sK)_________(!*/.^ (o.869) )__________ - зso К '°* " ( K) 2a9H)lH(^fa869)^.86g)- ” 35OK r.z i T,a • 350 К se iM wHh Eq. (D gives s/ooq^-K <in»t =(!00^j^-K) (3SoK-39SK) = Е020Я- //-8R

7Ш7 I I.9R (Nornul shock waved Standard atmospheric airen- ters subsonically and accelerates isentropically lo supersonic flow in a duct. If the ratio of duct exit to throat cross-section areas is 3.0. determine the ratio of back pressure to inlet stag- nation pressure that will result in a standing normal shock at the duct exit. Determine also the stagnation pressure loss across the normal shock in kPa. (ANS 0.375: 56.1 kPal fl* Щ/ к Pt (0) Vo*O For isentropic flow with * 3.0 y/e obtain /Y)q* from Fiq.D.t or £^.<//-7/;: MaK*2-64 Then from Eq. (Н-ЗЧ) we obtain r - k/(k~i) /.№*-/) = 0.047! white k-1 T77 j where from E<p (Ihiso) (orFi (/.4+1) (1.4+1) ' ™ c (7^ (О.0Ч7Г) • 0.375 , ОГ 101 kP(O.3!S)* 37.4 kPa Finatty from Eq. (//•!5^)(orFiq^‘f) 7 / Ira/ Y ЛЛ*-')/ \ . .*/?!-*) = 0.4У5 Thi>Si A,-fi«y = Arp - $У ж "" kR> M[' - °- ^SJ ‘ 1 *p‘ IMR

7ГЩ 12.1 R i Annular momentum) Water is supplied to a dish- washer through the manifold shown in Fig. PI2.1R. Determine the rotational speed of the manifold if bearing friction and air resistance are neglected. The total flowrate of 2.3 gpm is divided evenly among the six outlets, each of which produces a 5/16- in. -diameter stream. (ANS 0.378 rev/s) section FIGURE P12.1R Wi/h points and (3) located as in /he divjra/n above, ~ rinVpin) where I4,n=0. Thus, 7s 2.П)/Г/ V6l + 2/пя гг +2т3г3 Vg3 * 0 since {here is no friction. But m^rh^-rhi so /ha/ the above becomes r, 14/ Ife +r3V93 *0 (/) Bid «1^3/ ,4-42,3 ft) ' Q = 6AM =б[?(-^Я)2]м- ~0.00320id(- ^г,ш wi/h , .» <?= -O.OOSllg Thus, I Ao so /hat from Vg; *Wtcos3(f-l/i or l/e; s(i-^сохЗс/"-rt OJ. ^i/h Г, - n.f/, rts ^f/,and Eq. (!) becomes 4(7.306 -4^) +&(1.зы-&ш) +^(/.396 or 2¥.9 "Ю.5Ш 71m, = =o.37i^ 12-IR

I2.2R | I2.2R (Velocity triangles) An axial-flow turbomachine ro- tor involves the upstream (1) and downstream (2) velocity tri- angles shown in Fig. P12.2R. Is this turbomachine a turbine or a fan? Sketch an appropriate blade section and determine the energy transferred per unit mass of fluid. IANS: turbine; 36.9 ft'/s1) from the qiven in lei conJ J ions *(2O$) * 36-04 Q Thwith -fhe M velocity triangle is as shown below, or sin60*-30# Th^ since T* oi (fiVoi~ F/Vqi) w/th 14/ an J /7 follows ihat T<0, i.e.j the machine is a turbine. from the fiyire it follows that or 0-S6.3* Hence., the blaJe wool J be I2-2R.

I2.3R (< entrifiigal pump I Shown in Fig. P12.3R are front and side views of a centrifugal pump rotor or impeller. If the pump delivers 200 liters/s of water and the blade exit angle is 35* from the tangential direction, determine the power require- ment associated with flow leaving at the blade angle. The flow entering the rotor blade row is essentially radial as viewed from a stationary frame. (ANS: 34K kW) Fro/w /2.// with ~0 we have Xwf a) where и2'г1ш= (°‘,£(3000^)(2irt^v){6o 3TJ - % To deter/nine ^e2 we use {he exit velocity hrianqle shown below. Thus, v I/ -11 --/г— И?2 - 4? f.ar)3S* wilh n г liters./—\ I/ _ Q _ Ь200 ~~S ) (i°Q° hiers) _ у 07fn, - 27Trzb2 ~ ITT(0.15m)(0.03m} ~ Hence, I/ ts. 1 - -7 Q—т x 370 И02 «= 4- M л /^35 * J'.V s From Hq. (I): ^1чч ^°o (37CI = 3.?8xios = 348 kW о , _ /2.-3R

I2.4-R | 12.4R (Centrifugal pump) The velocity mangles for water flow through a radial pump rotor are as indicated in Fig. PI2.4R. (a) Determine the energy added to each unit mass (kg) of water as it flows through the rotor. (h> Sketch an appropriate blade section. (ANS: 404 N m/kg) FIGURE P1 2. a) From Eq. 12.. 12 y I s ^2 ^2'^1 14/ i where since the relative (Л velocity al the exit is in the radial direction (see the figure below), 16$ Also, from the inlet conditions, - Vn fan 30*, where the minus siqn means that 14, is in ex/7 the opposite direction of U, From conservation of mass (with p, -pi) we have ThuS, Л И-I ‘ 2vq bt = S,nce A,SOt r- H I/, = rt(JJ and or S°‘s Th vs, Ur, 9(I6^)/(O.S) = 32# SO that, V9I —yri )0n3O--(32^)fan3O,=-/8.S$- From Eq. (I): x « (I6^)(I6^) -(ff^X-ie.s^) 40¥ An appropriate blade section wovld be fanqent to IV, and h/i al the inlet and exit. Thus, al the exit the blade would be radial. At the inlet, *1 --O) blade 12-¥Я

/2.5 R | I2.5R (Similarity laws) When the shaft horsepower sup- plied to a certain centrifugal pump is 25 hp. the pump discharges 700 gpm of waler while operating at 1800 rpm with a head rise of 90 ft. (a) If the pump speed is reduced to 1200 rpm, determine the new head rise and shaft horsepower. Assume the efficiency remains the same, (b) What is the specific speed, for this pump? (ANS: 40 ft. 7.41 hp; 1630) a) From Eg. (12.37} t for a given pump operating at different speeds hOi a)f . V " so '"°' h<n• m = mojt WshaHZ VshaflZ "I . \3 , a /^2 | i't /200 Грт\. , < . ( UJ/ J ^sha/-// ( igoo rpfri) - 7.¥/ hp kj W(rpm) YQ (gpm) so that M =. (1800 грт}У 700<jpm (№)’" ' = Note. that for conditions of state (z}j from Eg. (12.36} or e ’ (7ёоГ7рт~) 700 s^7 7^ Ehust s = /63Q *dz (¥o.o/f)^ That isNsj( s Nsjz . The specific speed is constant the pump regardless of its speed when operating at a constant efficiency. I1-3R

7Z6R~[ 12.6K (Specific speed* An axial-flow turbine develops 10,000 hp when operating with a head of 40 ft. Determine the rotational speed if the efficiency is 88%. (ANS: 65.4 rpm) * 10'000 hp t t ty*o.88 Vrom Th^ /2.32, -65 = 12'6 R

12 7Ц | I2.7R (Turbine) A water turbine with radial flow has the dimensions shown in Fig. P12.7R. The absolute entering veloc- ity is 50 ft/s, and it makes an angle of 30° with the tangent to the rotor. The absolute exit velocity is directed radially inward. The angular speed of the rotor is 120 rpm. Find the power de- livered to the shaft of the turbine. (ANS: - 1200 hp) FIGURE P12.7R * PQ(UiVn' where V9Z*0 and M V9,.(SO frees Also, , f,» Q = 2irr,bV. Sin30* = 27Г (2H)(IH)(sefrtiii30 » 3/¥-+ 1 у ^7 end -'"Ту/W Thus, £<{.(!) (fives <Sr/WTT Г» ortyheH '(6.61ГЮЛ ssohk&) s'l,200 12-7R

/2.8Р| I2.8R (Turbine) Water enters an axial-flow turbine rotor with an absolute velocity tangential component. of 30 ft/s. fhe corresponding blade velocity. U, is 100 ft/s. The water leaves the rotor blade row with no angular momentum. If the stagnation pressure drop across the turbine is 45 psi. determine the efficiency of the turbine. (ANS: 0.8981 The power removed from the fluid is *r • where . wh'rt Ф it the sfatjrefion pressure head drop. Thus, The power removed by fhe turbine is t where Ц • (>QU(V91-V,i) *-?QUV9, since Vez*0 Hencet * = I2-8R