Mechanics of FluidsSolutions ManualMechanics of FluidsEighth editionSolutions manualBernard MasseyReader Emeritus in Mechanical EngineeringUniversity College, LondonRevised byJohn Ward-SmithFormerly Senior Lecturer in Mechanical EngineeringBrunel UniversitySeventh edition published by Stanley Thornes (Publishers) Ltd in 1998Eighth edition published 2006by Taylor & Francis2 Park Square, Milton Park, Abingdon, Oxon OX14 4RNSimultaneously published in the USA and Canadaby Taylor & Francis270 Madison Ave, New York, NY 10016Taylor & Francis is an imprint of the Taylor & Francis Group 2006 Bernard Massey and John Ward-SmithThe right of B. S. Massey and J. Ward-Smith to be identied as authors ofthis work has been asserted by them in accordance with the CopyrightDesigns and Patents Act 1988.All rights reserved. No part of this book may be reprinted orreproduced or utilised in any form or by any electronic, mechanical, orother means, now known or hereafter invented, including photocopyingand recording, or in any information storage or retrieval system,without permission in writing from the publishers.The publisher makes no representation, express or implied, with regardto the accuracy of the information contained in this book and cannotaccept any legal responsibility or liability for any efforts oromissions that may be made.British Library Cataloguing in Publication DataA catalogue record for this book is available from the British LibraryLibrary of Congress Cataloging in Publication DataA catalog record for this book has been requestedISBN 0415362040This edition published in the Taylor & Francis e-Library, 2005.To purchase your own copy of this or any of Taylor & Francis or Routledgescollection of thousands of eBooks please go to www.eBookstore.tandf.co.uk.(Print Edition)I SBN 0-203-01231-3 Master e-book I SBNChapter 11.1Since pV = mRT, V1V1= T1p2T2p1 V1 = 6(20 m)3288.15233.151.1101.3 = 56.2 m31.2 = pRT = 1.4 105N m2287 J kg1 K1323.15 K= 1.51 kg m31.3 K = pAssume K constant. Then ln(/0) = p p0K = 0exp_p p0K_= 1025 kg m3exp_81.7 1062.34 109_= 1061 kg m31.4 = = 2 105N s m215 106m2 s1 = 1.333 kg m3R = pT = 1.013 105N m21.333 kg m3293.15 K = 259.2 J kg1 K1 M = 8310259.2 = 32.061.5 = = 400 106m2 s1850 kg m3= 0.34 Pa sVelocity gradient = 0.12 m s10.1 103m = 1200 s1Area = 0.2 1.2 m2= 0.754 m2Force = 0.754 m20.34 Pa s 1200 s1= 307.6 N2 Solutions manual1.6Total force on plate = Area __uy_side A+_uy_side B_= (0.25 m)20.7 Pa s_0.15 m s10.006 m + 0.15 m s10.019 m_= 1.439 N1.7 For annulus, radius r, width rForce = Area VelocityClearance = 2rrrc Torque = Force r = 2r3rcTotal torque =_ R02r3c dr = R42c= (0.1 m)40.14 Pa s 2 7 rad s12 0.00013 m = 7.44 N m1.8p = 2d = 2 0.073 N m10.004 m = 36.5 Pa1.9h = 4 cos gd = 4 0.073 N m111000 kg m39.81 N kg10.005 m= 0.00595 m = 5.95 mm1.10h = 4 0.377 N m1cos 140(13.56 1)1000 kg m39.81 N kg10.006 m= 1.563 mm1.11Re = ud= 4Qd= 4 0.0025 m3 s1900 kg m30.05 m0.038 N s m2 = 1508u = 2000d= 2000 0.038 N s m20.05 m900 kg m3 = 1.689 m s11.12Re = 4Qd= 4 0.01 m3 s10.08 m370 106m2 s1 = 430 LaminarChapter 22.1 h = pg = 200 103N m21590 kg m39.81 N kg1 = 12.82 m2.2 Pressure depends only on depth below free surface.(a) p = gh = (820 kg m39.81 N kg1)(3 0.15) m= 22930 N m2= 22.93 kPa(b) p = 820 9.81 N m3(3 +2) m = 40.2 kPa(c) p = 820 9.81 N m3{3 +2 (1.2sin 30+0.6)} m= 820 9.81 3.8 N m2= 30.57 kPa(d) Load = Pressure Area= 820 9.81 3 N m2(3.5 2.5) m2= 211.2 kN2.3hair = p

airg = waterghwater

airg = water

airhwater= 1000 kg m3287 J kg1 K1288.15 K1.013 105N m2 0.075 m= 61.2 m2.4 pV = constant_ d4 mm_3= 101.3 103Pa +1000 kg m39.81 N kg19 m101.3 103Pawhence d = 4.93 mm2.5 p = 820 kg m39.81 N kg12 m+(13.56 0.82)1000 kg m39.81 N kg10.225 m = 44.2 kPa4 Solutions manualpg = h = 0.225 m(13.56 0.82)1000 kg m39.81 N kg1820 kg m39.81 N kg1= 3.496 m44200 N m2=820 9.81 2 N m2+x(0.82 0.74)10009.81 N m3whence x = 35.83 m2.6 New levelsAx y BX XC Initialsurface ofseparationMovement of uid inC = 60 mm70 mm2= (500 mm2)x= (800 mm2)y x = 8.4 mm;y = 5.25 mmMeasuring above XX: Initially 0.8hA = 0.9hBLater: 800 9.81(Old hA60 +8.4)103Pa= p +900 9.81(Old hB60 +5.25)103Pa p = 9.81 103(800 51.6 +900 65.25)Pa = 171.1 Pa2.7 From eqn 2.7 p = p0_1 zT0_g/R= 101.5 Pa_1 0.0065 7500288.15_9.81/2870.0065= 38.3 kPa2.8 pp0=_T0zT0_g/R=_ TtopTtop+z_g/R z = Ttop__p0p_R/g1_= 268.150.0065 m__749566_2870.0065/9.811_= 2257 mChapter 2 52.9 F = (1.2 1.8) m21000 kg m39.81 N kg1(x +0.9sin 30) m(a) 2160 9.81 N m10.45 m = 9.54 kN(b) 2160 9.81 N m10.95 m = 20.13 kN(c) 2160 9.81 N m130.45 m = 645 kNCentre of pressure is at slant depth (bd3/12) +bd(2x +0.9)2bd(2x +0.9)= d212(2x +0.9)+2x +0.9(metres)= (1.8 m)212(2x +0.9)+2x +0.9 mthat is_ 1.8212(2x +0.9)+0.9_m from upper edge= (a) 1.2 m; (b) 1.042 m; (c) 0.904 m from upper edge2.10 By symmetry, centreof pressure is onvertical centre-lineX XxrDepth = 2nd moment about XX1st moment about XX=_r0 x22(r2x2)1/2dx_r0 x2(r2x2)1/2dx=_0/2(r cos )22r sin (r sin d)_0/2r cos 2r sin (r sin d)= r_/20 cos2 sin2d_/20 sin2 cos d= r_018 sin22d(2)_13 sin3_/20= r/8[2/2 (1/4) sin 4]2=01/3= 38r2 = 3d326 Solutions manual2.11 X Xx60 60 Full depth = (2.5 m) sin 60Breadth of strip= 2.5 m_(2.5 m) sin 60x(2.5 m) sin 60_= 2.5 mx cosec 60 Second moment of area about XX=_ (2.5 m) sin 600(2.5 mx cosec 60)x2dx = 2.5412 sin360 m4First moment = Area Depth3= 122.5 2.5sin 60 2.5sin 603 m3= 2.536 sin260 m3 Depth of C.P. = 2.52 sin 60 m = Depth2 Thrust is equally divided between XX and bottom.Thrust = Area Pressure at centroid= 122.52sin 601000 9.81 2.5sin 603 N = 19160 N Load at bottom = 9580 N; at each upper corner 4790 N2.12 Let shaft be at depth h below free surface. Then force on disc= R2gh.By parallel axes theorem, 2nd moment of area about freesurface = R4/4 +R2h2.1st moment of area about free surface = R2h Depth of C.P. = R24h +h below free surface, that is, R2/4hbelow shaft Turning moment on shaft= R2gh R24h = R4g4 [independent of h]= (0.6 m)41000 kg m39.81 N kg14 = 999 N mChapter 2 72.13l1.5 m0.5 m2 mCForce on plate= 1150 kg m39.81 N kg11.5 m(2 m)2= 33.84 kN(Ak2)c, plate = Al26 (Ak2)c, diagonal = Al212since diagonals are perpendicular Depth of C.P. below free surface= (Al2/12) +Ay2Ay = y + l212y =_1.5 + (2)212 1.5_m= 1.611 m, that is, 1.111 m from top of aperture Total moment about hinge = 33.84 kN1.111/2 m= 26.59 kN m2.14 Width of gates = (3 m) sec 30 = 3.464 mThrust on deep side of gate= (1000 9.81 4.5)(9 3.464)N = 1.376 MNTrust on shallow side of gate= (1000 9.81 1.35)(2.7 3.464)N= 0.124 MNNet thrust = (1.376 0.124) MN = 1.252 MN Force between gates = 1.252 MN2sin 30 = 1.252 MNResultant force F acts at height y given byF1h13 F2h23 = Fy, since F1, F2 act at 23h1, 23h2 below free surfaces y = 1.376 9/3 0.124 2.7/31.252 m = 3.208 mTotal hinge reaction R also acts at this height.8 Solutions manualIf top hinge is distance x above bottom hinge,Rtopx = R(3.208 0.6) m x = RRtop2.608 m = 3 2.608 m = 7.82 m,that is, 8.42 m above base2.15ABCVRHHorizontal component H= Thrust on verticalprojection AC divided by width= 121000 9.81 272N m1= 3.576 MN m1acting at 23 27 m= 18 m below BCVertical component V = Weight of water ABCArea ABC =_ 270xdy =18_ 270y1/2dy = 2318(27)3/2m2= 396.8 m2 Vertical component = 1000 9.81 396.8 N m1= 3.893 MN m1It acts through centroid of ABC. Moments of area about AC:396.8x =_ 270xdyx2 =_ 2709ydy = 9 2722 m3whence x = 8.27 m = arctan 3.5763.893 = 42.57Resultant =_(3.5762+3.8932) MN m1= 5.29 MN m1It intersects free surface at (18tan 8.27) m from C= 8.27 m from CChapter 2 9that is {8.27 +_(18 27)} m= 30.31 m from B2.160.8 m1.2 mABCxyOilWaterRelevant forces are only those onvertical plane 0.5 m wide.Total force on AB = F1=_ 0.8 m0

oilgxbdx =_12

oilgbx2_0.8m0= 12850 kg m39.81 N kg10.5 m(0.8 m)2= 1334 NTotal force on BC =_ 1.2 m0(oilg0.8 m+watergy)b dy= bwaterg_ 1.2 m0(0.85 0.8 m+y)dy= bwaterg_(0.68 m)y + y22_1.2m0= 0.5 1000 9.81(0.816 +0.72) N= 7535 NTotal force F = (7535 +1334)N = 8869 NLet total force act at height z above base of tank. Then momentsabout axis through C:Fz = F1_1.2 + 0.83_m+_ 1.2m0bwaterg(0.68 m+y)(1.2 my)dy= 1334 1.467 N m+0.5 1000 9.81 N m2_(0.816 m2)y + 0.52 m2 y2 y33_1.2m0= 1957 N m+500 9.81[0.9792 +0.3744 0.576] N m= 5771 N m z = 57718869 m = 0.651 m10 Solutions manual2.17bacxyOilWater WaterForce = gba2Force = g_c a4_ a22+_gc +ga4_ a22= gca2+ ga38(1 )For zero net forceb = c + a8(1 )Total moment about centre-line for forces on left= _ a/20g(b x)ax dx +_ a/20g(b +y)ay dy= ga_ba28 + a324 + ba28 + a324_= 112ga4 NetForce acts at 112ga4gba2= a2/12b below centre-line.Total moment about centre-line for forces on right= _ a/20g(c x)axdx +_ a/20(gc +gy)ay dy= ga_ca28 +a324 +ca28 + a324_= 124ga4(1 +) Netforce acts at 124ga4(1 +) gba2= a2(1 +)/24b below centre-line. Axis of couple is 12_ a212b + a2(1 +)24b_= a248b(3 + ) below centre-line2.18 Pressure at centroid = (15000 +900 9.81 1) Pa = 23829 Pa Total force = 23829 Pa 0.24 m2= 5719 NChapter 2 11This acts on vertical centre-line Force on lock = 12 5719 N = 2859 NAir pressure is equivalent to 15000900 9.81 m = 1.699 m of oil Equivalent free (atmospheric) surface is at 2.699 m abovecentre-line Depth of C.P. below centre-line = (Ak2)c/Ay= (0.4 m)212_2.699 m = 0.00494 mMoments about horizontal axis through upper hinge:5719(0.125 +0.00494) N m = 2859 0.125 N m+FL(0.25 m) Force on lower hinge = FL = 1543 Nand force on upper hinge = (2859 1543) N = 1317 N2.19 2.7 kg of iron occupy 2.7 kg7500 kg m3 = 0.00036 m3 Buoyancy force = 0.00036 m31000 kg m39.81 N kg1= 0.36 9.81 N Spring balance reads (2.7 0.36) kgf = 2.34 kgfParcel balance reads (5 +0.36) kgf = 5.36 kgf2.20x3d0.9lx9 sl/ 20(l-s-l/ 20)Archimedes for case II:0.9l = 1 s +0.8_1920l s_whence s = 0.7lVolume of water is constant 4(3d)2x +0.9l_4(3d)2 4d2_= 4(3d)2x

+0.7l_4(3d)2 4d2_ 9x +0.9l{9 1} = 9x

+0.7l{9 1} x

x = 0.1778l12 Solutions manual2.21HdhxPressure pp0Archimedes 4d2(x h)g= 27 9.81 NAt base of cylinder, pressure= p0+gx = p +gh p p0 = g(x h)= 27 9.81 N(/4)(0.3 m)2= 3747 PaFor isothermal compression pV = constant p(H h) = p0H h = p p0p H = 3747105047 450 mm = 16.05 mmx h = 27 9.81 N(/4)(0.3 m)21000 kg m39.81 N kg1 = 0.382 m x = 398 mm2.22 From eqn 2.7, p at 6000 m is p0_1 zT0_g/R= 101 kPa_1 0.0065 6000288.15_9.81/2870.0065= 47.01 kPa at 6000 m is 47 010287(288.15 0.0065 6000)kg m3= 0.6574 kg m3which must be same as effective density of balloon. Total mass of balloon = 0.6574 60.83kg = 0.17625 kg Mass of helium = (176.25 160)g = 16.25 g2.23 BM= Ak2/V = 64d4__4d20.6l_= d2/9.6lB is at 0.3l above base.Chapter 2 13 When M and G coincide, BM= 0.2l d2= 0.2 9.6l2 d/l =1.92 = 1.3862.24 Weight of pontoon= (6 3 0.9) m31000 kg m39.81 N kg1= 158.9 kNBM= Ak2/V = [(6 33/12)/(6 3 0.9)] m = 0.833 mGM=_0.833 + 0.92 0.7_ m = 0.583 m7600 N m = W(GM) sin sin = 7600158.9 1030.583 = 4.702.25 If relative density = , depth of immersion h = 150 mm Height of B = 75 mmBM= Ak2/V = 64d4_4d2h = d216h = 75216 150mm= 7532mmFor stability BM > BG 7532>1502 75that is 132> 1 32232 +1 > 0 >16 +256 3232 = 0.9677or 17 0.013000= 173 105[m5] d > 1.415 101m = 141.5 mm, say 150 mm7.6 10 m water 1000 9.81 10 Pa. Now hf = KQ2l,where K = constant. (4 +9.81)104Pa = K(0.049)2900 m7 s2New p = K(0.067)2450 m7 s2+K(0.049)2450 m7 s2= 13.81 1042__0.0670.049_2+1_ Pa = 19.81 104Pa Inlet pressure = (19.81 9.81)104Pa = 100 kPa40 Solutions manual7.7 For steady ow, head difference = lossesEntry to pipe (1): 0.5u21/2gFriction in pipes (1) and (3): 2 4f l1d1u212gwhere f = 0.005_1 + 2575_= 0.00667 Loss = 2 4 0.00667 7.50.075u212g = 5.33u21/2gEnlargement, (1) (2):Loss = (u1u2)22g = u212g_1 116_2= 0.88u21/2g_since u2 = A1A2u1 =_d1d2_2u1_Friction in pipe(2): 4f l2d2u222gwhere f = 0.005_1 + 25300_= 0.00542 Loss = 4 0.00542 600.3_u116_2 12g = 0.017u21/2gLoss at entry to (3) is 0.5u23/2g = 0.5u21/2gExit loss = u23/2g =Total=1.0u21/2g8.23u21/2g 1.5 m = 8.23u21/2g whence u1 = 1.891 m s1 Q = 4(0.075)21.891 m3 s1= 0.00835 m3 s1= 8.35 L s17.8Inlet head = 370 103820 9.81 m = 46.0 m; Static lift = 20 m hf = 26 m = 4f 185d19.62_ 0.040(/4)d2_2m6 d5= 0.003763f m5Re = 0.040(/4)d2d2.3 106 [m] = 2.214 104d [m]Try f = 0.006 Then d = 0.1177m; Re = 1.886 105;kd = 0.15 1030.1177 = 0.001274 f = 0.0057Chapter 7 41Then d = 0.1165 m; Re = 1.901 105; kd = 0.001288 f = 0.0057 d = 116.5 mm, say 120 mm7.9Q 2d5/2_2ghfl_1/2log_ k3.71d + 2.51d3/2_2ghf/l_1/2_= 2(0.075)5/2_19.62 26185_1/2log_ 0.15 1033.71 0.075+ 2.51 2.3 106(0.075)3/2(19.62 26/185)1/2_ m3 s1= 0.0127 m3 s1= 12.7 L s17.1050 mmX X(1)(2)AB Let head differenceabove level XX be hmetres of liquid. Then(h +0.05)1200 9.81= 0.05 13600 9.81 h = 0.517Manometer measures_p2g + u222g_ p1gthat is, 12g_u21(u1u2)2_= 12g_2u1u2u22_= Q22g_ 2A1A2 1A22_= h m Q2= 2ghA2(2/A1) (1/A2)= 19.62 0.517 (/4)(0.1)2_2/(/4)(0.05)2__1/(/4)(0.1)2_ m6 s2= 8.93 105m6 s2 Q = 9.45 103m3 s1 Mass ow rate = 9.45 103m3 s11200 kg m3= 11.34 kg s142 Solutions manual7.11A:B:123(hf)A = 4f ldu22g(hf)B = 4f l/2du212g + 4f l/2du222g= 2f ldu212g_1 + 14_since u2 = u3 = u12_hf_B =_hf_A u1u =_85 = 1.265 Increase = 26.5%7.12A1.8 mh(m)4 m1.2 m4.6 mXLength to X = x =_7.52+h2[m]4.6 m =_1.5 + 4 0.01 360.1_ u22g u22g = 0.2893 mEnergy/weight at X = 40 1031000 9.81 m+ u22g +_h 1.8_m= Energy/weight at AHead losses between A and X= 101.3 1031000 9.81 m_0.5 + 4fxd_ u22g 0.2893_1 +0.5 + 4 0.010.1_7.52+h2_+h= 61.3 1031000 9.81 +1.8 whence h = 6.47 [m]7.13Velocity at C = 0.0425 m3 s1(/4)(0.1 m)20.62 = 8.73 m s1Velocity at B = 0.0425(/4)(0.2)2 m s1= 1.353 m s1Chapter 7 43Head lost C B is(8.73 1.353)219.62 m = 2.772 m= Loss between A and BpBpA g = u2Au2B2g h1 = {(200/150)41}1.353219.62 m2.772 m= 2.571 mPower = 0.0425 m3 s11000 kg m39.81 N kg12.772 m= 1156 W7.14Re = ud= 4Qd= 3.20.6 106d (m)which is >106for all reasonable value of d.For given Q and , power hf fd5f depends on relative roughness; f _kd_1/3 Power d16/3 If power halved, d multiplied by 23/16= 1.139,that is, d = 569 mmFor n pipes in parallel Re = 4(Q/n)d2 d2 = d1/nTotal power = ngQn hf = gQ32f l(Q/n)22gd52= 32Q3l2d52n2 const_ kd2_1/3_d1n_16/3n2 Power for n pipesPower for 1 pipe = n10/3. This is >1 No advantage.7.15Pipe 1: hf + u212g[which is lost] = u212g_4f l1d1+1_= 16Q212d412g_4f l1d1+1_44 Solutions manual= 16Q212(0.1)419.62_4 0.008 450.1 +1_ s2 m5= 1.272 104Q21 s2 m5Pipe 2: hf + u222g = 16Q222(0.15)419.62_4 0.008 600.15 +1_ s2 m5= 2252Q22 s2 m5 Total head at pump outlet (inlet as datum)= 6.6 m+12 720Q21 s2 m5= 8.6m+2252Q22 s2 m5 12720(0.037 Q2)2= 2 +2252Q22whence Q2 = 0.02153 m3 s1 Head at outlet =_8.6 +2252(0.02153)2_m = 9.64 m7.16hAD = 4 0.01 16 1030.3 19.62 u2A = 108.7u2A [s2 m1] = 125 mhwhere h = head at D above C.hDB = 4 0.01 95000.2 19.62 u2B = 96.8u2B [s2 m1] = h 30 mhDC = 4 0.01 80000.15 19.62 u2C = 108.7u2C [s2 m1] = hContinuty: 4 (0.3)2uA = 4 (0.2)2uB+ 4 (0.15)2uCHence h = 89.2 m. uB =_59.296.8 m s1= 0.781 m s1; uC =_ 89.2108.7 m s1= 0.906 m s1 QB = 4(0.2)20.781 m3 s1= 0.02456 m3 s1QC = 4(0.15)20.906 m3 s1= 0.0160 m3 s17.17Q from A is 0.48 m3 s1 uA = 0.48(/4)(0.75)2 m s1= 1.086 m s1_hf_A = 4 0.006 1040.751.086219.62 m = 19.25 mChapter 7 45Head at junction above level of pipe=_3 +(2.2 103)10419.25_ m = 5.75 m_hf_B =_5.75 +_2.75 103_5500 3_ m = 17.875 m= 4 0.006 5500dB19.62_ 0.24(/4)d2B_2m6whence dB = 0.512 m_hf_C =_5.75 +_3.2 103_3000 3_ m = 12.35 m= 4 0.006 3000dC19.62_ 0.24(/4)d2C_2m6whence dC = 0.488 m7.1830 m = 4 0.008 10000.6u219.62 s2 m1 u = 3.322 m s1 Q = 4(0.6)23.322 m3 s1= 0.939 m3 s10 = 12u2f = 121000 3.32220.008 Pa = 44.1 PaIn rst half of pipe ux = uin uinx1500at x metres from inlet since ()dQdx = Q/3500 mhf1 = 4f2gd_ 500m0u2xdx = 4f2gd_ 500m0_uin uinx1500_2dx= 0.03219.62 0.6u2in_13_1 x1500_3(1500)_5000= 1.359u2in_13_23_3_= 1.359u2in1927hf2 = 4 0.008 5000.6 19.62_23uin_2= 1.359u2in49 30 m = hf1+hf2 = 1.359u2in_1927 + 49_whence uin = 4.385 m s146 Solutions manual uout = 23 4.385 m s1= 2.923 m s1 Qout = 2.9234(0.6)2m3 s1= 0.826 m3 s17.191 m1 mhFor upper, rectangular,portion _18 9 m2_dh= 0.94 (0.15 m)22_2gh dt t1 = 18 90.9(/4) 0.152219.62_ 1m2mh1/2dh = 952 sFor lower, triangular, portion A = h118 9 m2= 162h m2t2 = 10.9(/4) 0.152219.62_ 01mAh1/2dh= 7.10_ 01m162h1/2dh = 767 sTotal time = t1+t2 = 1719 s7.20h = (zAzB) = zA_1.57.5zA_= 65zA zA = 56h(1.5 m2)56h = 0.64(0.025 m)2_19.62 m s2h1/2t t = 1.5 (5/6)0.6(/4) 0.025219.62_ 01.2mh1/2dh = 958 21.2 s= 2100 s7.21 Let surface areas of tanks be A, B. Then A(zA) = BzB= Qth = (zAzB) = zA_1 + AB_= zA_1 + 2.521.52_= 349 zA Qt = 934Ah = (1.30 m2)hh =_4f ld +1.5_ u22g =_4 0.01 750.05 +1.5_ u22g = 61.5u22gChapter 7 47 u=_0.565 m1/2 s1_h1/2and Q=_0.00111 m5/2 s1_h1/2t = _ x1m1.300.00111h1/2 dh = 2.600.00111_1 x1/2_= 1200 swhence x = 0.238 m h = 0.762 m zA = 934 0.762 m = 0.2017 m7.220.6 m1.5 m0.45 m1.5 mh(m)A = r2= _0.45 +(h 1.5) 0.151.5_2m2= _0.3 +0.1h_2m2t = _ h2h1Aaudh =_ h2h1_0.3 +0.1h_2d2/4_4f l2gdh1/2dh= 4d2_4f l2gd_ 2.81.7(0.09h1/2+0.06h1/2+0.01h3/2) dh= 4d2_4f l2gd_0.18h1/2+0.04h3/2+0.004h5/2_2.81.7= 4d2_4f l2gd 0.2027 = 600 s d =_424 0.008 36 0.2027219.62 6002_1/5m= 0.0404 m, say 40 mm7.23Initially u = 0.017 m3 s1(/4)(0.075 m)2 = 3.848 m s11.5 m+h1 = u22g_1.0 + 4 0.01 30.075_= 2.6u22g48 Solutions manualwhence h1 = 0.462 m (depth in tank)A = 4_4.25 + 1.751.5 h_2m2When Q = 0.034 m3 s1,_0.034 m3 s1_ dt = Adh +au dtthat is, dt_0.034 4(0.075)2_19.62(1.5 +h)2.6_1/2_= 4_4.25 + 76h_2dh_metre, second units_ t = 4_ 1.50.462(4.25 +(7/6)h)2dh0.034 0.01214(1.5 +h)1/2 secondsAn analytical solution is possible. Put denominator =x,then_0.034 x0.01214_= 1.5 +hand dh = 2(0.034 x)0.012142 dxThen t = 420.012142_ 0.012980.017_4.25+76__0.034x0.01214_21.5__2(0.034 x)dxx= 1.067 104_ 0.0170.01298_11.66 539x +7921x2_2_0.034x 1_ dx= 1.067 104_ 0.0170.01298_4.62x 563 +2.870 104x 7.65 105x2+1.067 107x36.27 107x4_ dx= 1.067 104_4.62ln x 563x +1.435 104x2 2.55 105x3+2.667 106x41.255 107x5_0.0170.01298= 1626 s (This depends on relatively small differences.)Chapter 7 497.24P3 mhAt any time, head required= h +hf = h + 32f lQ22gd5= 500 1031000 9.81 m = 50.97 m Q2=_50.97 mh_ 2gd532f l=_50.97 mh_2.955 106m5 s2Also Q = Adh/dt t =_ 5.5m3.2m5dh1032.95550.97 h= 50002.955_2_50.97 h_1/2_5.53.2 s= 1042.955_47.77 45.47_s = 980 sPower required = 1Qg(h +hf) = 1Q(500 kPa) Electricity used in time t is 1Q(500 kPa)t Total used = 1(500 kPa)_ Qdt= 1(500 kPa)(Volume transferred)= 10.52500 1035 2.3 N m=5.75 1060.52 103 kW s= 5.75 1030.52 3600 kW h = 3.072 kW h7.25At any instant, head required = h +hf = h + 32f lQ22gd5= 300 1031000 9.81 m = 30.58 m Q2=_30.58 mh_ 2gd532f l50 Solutions manualAlso Q = Adh/dt t =_ 5.6m3.0mAdh(30.58 h)1/2_32f lgd5= A_32f lgd5_2(30.58 h)1/2_5.63.0= 2A_32f lgd5 (27.581/224.981/2) d5/2= 2At_32f lg 0.2537 1215 60_32 0.007 359.81 0.2537= 9.624 104_m5/2_d 0.0621 m that is, d > 62.1 mm7.26 Final difference of levels = (40 2 15) mm = 10 mmA_dz1dt_= Q = Adz2dt dhdt = ddt (z1z2) = 2dz1dt = 2QA = 2gd4hA128l t = 64lAgd4_ 0.01m0.04mdhh = 64lAgd4 ln4 = gd4t64lA ln 4 = 840 9.81 101247864 0.4 14(0.02)2ln 4 Pa s= 1.110 mPa s7.27 p = (7 1054.62 105) Pa 1000 kg m39.81 N kg1_30sin 45_ m = 29900 Pa0 = R2dpdx = 0.0125 m_29900 Pa30 m_= ()12.46 Pa = r2dpdx = 0.005 m_29900 Pa30 m_= ()4.98 PaChapter 7 517.28 (p1p2) R2F = Rate of increase of momentum=_ R02r dru2R2V2= 24V2_ R0r_1 r2R2_2dr R2V2= 43R2V2R2V2= 13R2V2 F = R2(p1 p2 V2/3)Chapter 88.1=_ 10_1 uum_d =_ 10_1 sin 2_d =_ + 2cos 2_10= 1 2= 0.3634=_ 10uum_1 uum_d =_ 10_sin 2 sin2 2_d=_2cos 2 2 + 12sin _10= 12 + 2= 0.13668.2 From Blasiuss solution, 4.91x/(Rex)1/2Rex = 6 m s12.4 m14.9 106m2 s1 = 9.66 105 4.91 2.4 m(9.66 105)1/2= 0.01199 m0 = 0.332 u2m(Rex/2)1/2[Table 8.1]= 0.332 1.21 kg m3_6 m s1_2_9.66 1052_1/2= 0.0208 PaTotal drag on one side = 0.9 m0.664 1.21 kg m3(6 m s1)2 2.4 m(9.66 105)1/2 = 0.0636 N Power = 2 0.0636 N6 m s1= 0.763 WFor turbulent layerCF = 0.074(Rel)1/5= 0.074(9.66)1/510 = 0.0470Chapter 8 53 Power = 2 12u2m(Area)CFum= 1.21 kg m3(6 m s1)3(2.4 m0.9 m)0.00470= 2.654 W8.3 Since U = constant, eqn 8.9 gives0 = x_ 0(U u)udy = U2 x_ 101/7(1 1/7)d[where = y/]= U2 x_788/7 799/7_10= 772U2ddx= 0.023U2_ U_1/4 7721/4d = 0.023_U_1/4dx 772 455/4+const = 0.023_U_1/4xbut const. = 0 since = 0 when x = 0 = 0.3773x4/5(/U)1/5Total drag = 2b_ x00 dx = 2b 772U2= 736bU20.3773x4/5(/U)1/5= 7360.8 m(1000 kg m3)(3 m s1)20.3773(2.5 m)4/5_106m2 s13 m s1_1/5= 55.69 NPower = 55.69 N3 m s1= 167.1 W8.4 (a)Rel = 10.5 m s13 m15 106m2 s1 = 2.1 106Ret = 5 105 xt/l = 5/21xtx0xt= 36.9(5 105)3/8 = 0.269 [eqn 8.29] x0/xt = 0.73154 Solutions manual x0l = x0xtxtl = 0.731 521 = 0.1740CF = 0.074(2.1 106)1/5[1 0.174]4/5= 0.003454 Drag = 12u2m(Area)0.003454(b) Rel = 10.5 0.315 106 = 2.1 105 Wholly laminarCF = 2x = 1.328(Rel)1/2[Table 8.1]= 0.00290 Ratio = 0.0034540.00290 = 1.1928.5um = 160 1033600 m s1= 44.44 m s1 Rel = 44.44 110 1.221.79 105 = 3.332 108If transition occurs at Re = 5 105, length of laminar layer= 5 1053.332 108 110 m = 0.165 m Assume layer turbulent throughout.From eqn 8.26, CF = 0.455(8.5227)2.58 = 0.001808Power = 110 m8.25 m 12 1.22 kg m3(44.44 m s1)20.001808 44.44 m s1= 87.9 kW8.6Increasein lam.xx=0.5 m3 motTurb.At transition Rext= 29 m s10.5 m14.5 106m2 s1= 106(t)lam = 0.664 0.5 m(106)1/2= 0.332 mmChapter 8 55So for turbulent layer with virtual origin at x = x01.15 0.332 103m = 0.037(xtx0){U(xtx0)/}1/5= 0.037(xtx0)4/5x1/5t(Uxt/)1/5whence xtx0 =_106/51.15 0.332 103m0.037(0.5 m)1/5_5/4= 0.1237 m x0 = 0.3763 mAt x = 3 m = 0.037(3 0.3763) m{29(3 0.3763)/14.5 106}1/5 = 0.004397 mCD = Drag/width12U2length = U2trailing edge12U2length = 2t.e.length= 2 0.004397 m3 m = 0.002938.7 If main-stream velocity remained unchanged,Rel = 1.8 0.1501.21 106 = 2.231 105 Laminar boundary layer = 1.721 0.150 m(2.231 105)1/2 [Table8.1] = 0.546 mmOutlet velocity = u1b2(b 2)2 = 1.8 m s1 252(25 1.092)2= 1.968 m s1Pressure drop = 121000 kg m3(1.96821.82) m2 s2= 317.2 Pa8.8Re = 22 0.00315 106 = 4400Eqn 8.33 : 0.003 m22 m s1 = 0.198_1 19.74400_ = 1445 Hz56 Solutions manual8.9 p = const 12(2Usin )2if boundary layer thin.If separation occurs at = , total component of pressure drag whereboundary layer is attached= 2_ 0(const 2U2sin2)rd cos per unit length= 2r_Csin 2U213 sin3_0= 2r_Csin 23U2sin3_p at separation = C 2U2sin2 Drag divided by length, due to pressure in wake= 2r(C 2U2sin2) sin Net pressure drag/length= 4rU2sin3 43rU2sin3 = 83rU2sin3CD =83rU2sin312U22r = 83 sin3 = 1.24 sin = 0.7747 = 50.78 or 129.22But separation occurs only where pressure gradient adverse. = 129.2, skin friction neglected.8.10 um = 2Usin xr dumdx = 2rUcos xrAt x = x1, 2= 0.452Usin6(x1/r)_ x10sin5 xr dx= 0.45r2Usin6(x1/r)_ x10_1 cos2 xr_2d_cos xr_= 0.45r2Usin6(x1/r)_cos xr 23 cos3 xr + 15 cos5 xr_x10= 0.45r2Usin6(x1/r)_ 815 cos x1r + 23 cos3 x1r 15 cos5 x1r_Chapter 8 57 = 2dumdx= 0.45r2Usin6(x1/r)_ 815 cos x1r + 23 cos3 x1r 15 cos5 x1r_ 2rUcos x1rwhich is satised by = 0.09 and x1r 103.18.11 Re = 60 0.15015 106 = 6 105(a) From Fig. 8.14, CD 0.08 Drag 12 1.21 kg m3(60 m s1)24(0.15 m)20.08= 3.08 N(b) From Fig. 8.14, CD 1.10 Drag 12 1.21 60240.1521.10 N = 42.0 N8.12 Drag = 90 9.81 N when u = 6 m s1Assume CD = 1.32. Then90 9.81 N = 12 1.22 kg m3(6 m s1)21.32 4d2 d = 6.23 mCheck: Re = 6 6.2315 106 exceeds 1038.13(4Re/3CD)1/3= uT_

2/g()_1/3= 10 m s1__1.21 kg m3_2__9.81 N kg118.0 106N s m22798.79 kg m3__1/3= 14.36 From Fig. 8.15, Re = 1148 d = 1148 18 106N s m210 m s11.21 kg m3 = 1.708 mm58 Solutions manualFor water_34CD(Re)2_1/3= d_()g/2_1/3= 0.001708 m_1000 kg m31800 kg m39.81 N kg1(103N s m2)2_1/3= 44.48 From Fig. 8.15, Re = 443 uT = 443 103N s m20.001708 m1000 kg m3 = 0.259 m s18.14u = 0.0185 m3 s1(/4)(0.1 m)2 = 2.355 m s1f = 19.62 m s21.89 m0.1 m4 25 m(2.355 m s1)2 = 0.00668Re = 2.355 m s10.1 m1.2 106m2 s1 = 1.963 105Eqn 8.56:(0.00668)1/2= 4log_ k3.71d + 1.261.963 105(0.00668)1/2_whence k/d = 0.002955(umu)R2=_ R0(umu)2r dr= 2A_0

_1/2_ 0R_ln Ry_(R y)dy [from eqn 8.50]= 2 2.5u_f2_1/2_3R24_ umu = 3.75u(f /2)1/2 um = u_1 +3.75(f /2)1/2_= 2.866 m s10 = 12u2f = 12 1000 kg m3(2.355 m s1)20.00668= 18.54 Pa8.15 hfl = 1000 kg m30.7 kg m3 50 103m1000 m = 0.07143Q = 4d2u = 4d2_2ghfd4f l_1/2= 4d5/2_2ghf4l_1/2 _4log10(Re f1/2) 0.4_Chapter 8 59= 4d5/2_ghf2l_1/2 _2log10(Re2f ) 0.4_= 4d5/2_ghf2l_1/2_2log10_2gd3hf42l_0.4_= 2(0.25 m)5/2_9.81 m s20.071432_1/2_log10_9.81 0.2530.071432 1821012_0.2_= 0.204 m3 s10 = pld4 = ghfld4 = 0.7 kg m39.81 N kg10.07143 116 m= 0.03066 PaChapter 99.1 ux + vy = 0 Continuity is satisedBut vx uy = 0 Flow is not irrotational9.2 Only (a) and (e) satisfy eqn 9.99.3u = y = 4y = x = 4xy +f (y)v = x = 1 +4x = y = y 4xy +f (x) = y 4xy + const. = vx uy = 4 4 = 0 Flow irrotationalAt (1, 2) q2= u2+v2= (64 +25) m2 s2= 89 m2 s2 p0 =_4800 + 12 1.12 89_PaAt (9, 6) q2= u2+v2= (576 +1369) m2 s2= 1945 m2 s2 p = p0 12q2=_4800 + 12 1.12(89 1945)_Pa= 3761 Pa9.4qr = r = Ar = 1r = A and = A +f (r)qt = r= 0 = r = f () = A + const.Chapter 9 619.5r = (1.52+22)1/2m = 2.5 m qr = 3/222.5 m s1= 0.3 m s1Along radial streamlines p+ 12q2r = constant pr = qrqrr = qr_ m2r2_= 800 kg m30.3 m s1 3/22(2.5 m)2 m2 s2= 28.8 Pa m1Acceleration = qrqrr = 0.3 3/222.52 m s2= 0.036 m s29.6 dhdr = u2gr and ur = constant = C dhdr = C2gr3 h = C2g_ r2r1drr3 = C22g_1r21 1r22_Q =_ r2r1s udr = sC_ r2r1drr = sCln_r2r1_= s_2g(h)r21r22r22r21_1/2ln_r2r1_= sr1r2_2g(h)r22r21_1/2ln_r2r1_= s_r + s2_ _r s2____2g(h)_r + s2_2_r s2_2___1/2ln_r +s/2r s/2_=_r2 s24_(sgh/r)1/2ln

2r +s2r s

9.7 p/r = 2r. Here = const., p = 122r2+const.At free surface p = 0 gz = 122r2+C(which is eqn of paraboloid)62 Solutions manualTake z = 0 at vertex. Then C = 0 z = 2r22gVolume of air =_ r2dz =_ R0r222g2r dr = 2R44g= R2_150 mm3_ 2= 4g(0.05 m)R2 = (0.2 9.81)1/20.05 rad s1= 28.01 rad s19.8 Pressure at circumference = 122R2= 12900kg m3_2 15rad s1_2(0.125 m)2= 62.5 kPaForce on elemental annulus = 122r22r dr Total force = 2_ R0r3dr = 2R4_4=_900(2 15)2(0.125)4_4_N = 1533 N9.9 In forced vortex p2p1 = 122_r22r21_= 12

q21r21_r22r21_In free vortex p2p1 = 12

_q21q22_= 12q21_1 q22q21_= 12q21_1 r21r22_If p2 p1 for forced vortex = 2(p2 p1 for free vortex) and q1 issame for both,then r22r21r21= 2_1 r21r22_ whence r423r21r22+2r41 = 0 r22 =3r21_9r418r412 = 2r21 or r21Reject r22 = r21 because then p2p1 = 0. r2 = r12Chapter 9 639.10 For forced vortex q = r and p1p0 = 122r2For free vortex qr = C and p = K 12q2At interface q = C_R = R C = R2p = K 122R2= 122R2+p0 K = p0+2R2At r = , q = 0 p = K Difference in surface levels = pp0g = 2R2/g9.11In forced vortex p = 122r2= 12q2= 121000 62Pa= 18 kPa where q = 6 m s1For free vortex qr = C = R2=_30 rad s1_(0.1 m)2= 0.3 m2 s1p = D 12

C2r2At R = 0.1 m pforced = pfreethat is, 121000kg m3_30 rad s1_2(0.1 m)2= D 121000kg m3(0.3 m2 s1)2(0.1 m)2whence D = 90002PaIn free vortex q = 6 m s1whenr = Cq = 0.36 m = 20 mThen p =_90002 121000(0.3)2(/20)2_Pa= (9000218000)Pa= 70.83 kPa p = (70.83 18) kPa = 52.83 kPa9.12 For forced vortex = 8 2 rad s1= 50.27 rad s1and p = 122r2+A. Let p = 0 when r = 0. A = 0.64 Solutions manualTotal force on top of drum caused by forced vortex=_ 0.15m0122r22rdr = 42(0.15 m)4= 4900kg m3_50.27 rad s1_2(0.15 m)4= 904 NFor free vortex qr = C and p = D 12q2At r = 0.15 m, p = 122(0.15 m)2and q = (0.15 m) as forforced vortex 122(0.15 m)2= D 122(0.15 m)2whence D = 2(0.15 m)2= 900 50.2720.152Pa= 51.2 kPaC = qr = (50.27 0.15)0.15 m2 s1= 1.131 m2 s1 Total force for free vortex =_ 0.3m0.15m_D 12

C2r2_2r dr= 2_Dr22 12C2ln rr0_0.3m0.15m(where ln r0 = integration constant)= 2_512002 (0.320.152) 12900 1.1312ln 2_N= 8343 N Grand total = 904 N+8343 N 9250 N9.13 For a sink = 2m 1 = 312m2 s1; 2 = 22m2 s1 = 1_321+22_m2 s1qr = 1r= 34m s1for (1) when r = 2 mand 12 m s1for (2) when r = 22mChapter 9 65 u = 12 cos 135m s1= 12m s1v =_ 34 12 cos 45_m s1= 54m s1___ q = (u2+v2)1/2= 1429 m s1= 0.429 m s1at 360arctan 2.5= 291.8By symmetry, stagnation point is on x-axis and at (s, 0) where 32s = 42(2 ms) s = 67 mthat is, stagnation point is at_67 m, 0_9.14Uyx 1 2Rankine ovalSource at (b, 0),sink at (b, 0), each ofstrength |m|.Then = UY + m2(21)= UY + m2arctan 2byx2b2+y2 Eqn of surface is: the above = 0For the point (0, Y) on line = 0, 21 = 2UYm= 21 by symmetry UYm = 2 1 = arctan bY that is, bY = tan_UYm_Also (X, 0) is stagnation point. 0 = U m2(X b)+ m2(X +b)whence mU = X2b2bMax velocity, at (0, Y), isU + m2(b2+Y2)1/2 cos 1+ m2(b2+Y2)1/2 cos( 2)= U + m2(b2+Y2)1/2_ b(b2+Y2)1/2 + b(b2+Y2)1/2_66 Solutions manual= U + mb(b2+Y2)= U + (X2b2)Ub2+Y2 = U_X2+Y2Y2+b2_ Max pressure difference = 12U2_X2+Y2Y2+b2_29.15 Eliminating m from the two simultaneous eqns in Problem 9.14 givesbY = tan bYX2b2Then with X = 0.1 m and Y = 0.05 m, b = 0.0781m (by trial) Distance between source and sink = 2b = 156.2 mm|m| = U_X2b2b_= _3 m s1_ 0.120.078120.0781 m= 0.471 m2 s1Max velocity = U_X2+Y2Y2+b2_[as in Problem 9.14]=_3m s1__ 0.12+0.0520.052+0.07812_= 4.36 m s19.16 ymax = 0.05 = m2 15(metre, second units) m = 1.5 m2 s1Stagnation where r = m2U = 1.52 15 m = 15.92 mmEqn of surface is UY m2= m2 that is, = _1 2UYm_ yx = tan = tan_ 2UYm_= tan 2UYm x = y cot 2UYm = ycot(20y) (metre units)When y = 0.025 m, x = 0 that is, = 2 qt = Usin = 15 m s1qr = Ucos + m2r = 0 + 1.52 0.025 m s1= 30m s1Chapter 9 67p = 12q2= 12 1.23kg m3_152+ 3022_m2 s2= 194.5 Pa9.17 = Uy m2(1+2); tan 1 = y ax ; tan 2 = y +ax tan(1+2) =_(y a)_x_+_(y +a)_x_1 _y2a2_ _x2 = 2xyx2y2+a2 = Uy m2arctan_ 2xyx2y2+a2_v = x = m2_1 + 4x2y2_x2y2+a2_2_1_x2y2+a2_2y 4x2y_x2y2+a2_2= m22y_a2x2y2__x2y2+a2_2+4x2y2= 0 when (i) y = 0 or (ii) a2= x2+y2u = y =U + m2_1 + 4x2y2_x2y2+a2_2_1_x2y2+a2_2x +4xy2_x2y2+a2_2= U+ m22x_x2+y2+a2__x2y2+a2_2+4x2y2For case (i): u = U + mx_x2+a2_= 0 when Ux2+mx +Ua2= 0that is, when x = m2U(+)_ m242U2 a2For case (ii): u = U + m2x = 0 when x = m2U andy2= a2 m242U2 which is < 0 Impossible68 Solutions manual Reject case(ii)Stagnation point is at__ m2U_ m242U2a2, 0__At stagnation point y = 0 and 1+2 = 2 = m Eqn of contour is Uy m = m2(1+2)When x = 0, Uy = m m2( or 3) that is, y = m2UThen u = U and v = m2_mU__a2 m24U2_1= 2Um2_4U2a2m2_ q = U_1+ 4m42(m24U2a2)2_1/29.18Fx_divided by depth_=_ 3/2/2pa cos d andp p = 12(U2q2) = 12U2(1 4sin2) Fx = 12U2a_ 3/2/2(4sin2 1) cos d= 12U2a_43 sin3 sin _3/2/2= 13U2a Total |Fx| = 13 1000kg m3_1.2m s1_20.9 m3 m= 1296 N9.19 Mass of cylinder= 7800kg m3_4(0.320.2882)4 +240.28820.006_m3Added mass = 1000kg m3_40.324_m3= 90 kg Ratio= 7800_ 0.588 0.012 + 0.28820.003_+907800_ 0.588 0.012 + 0.28820.003_= 2.580Chapter 9 699.20sin =32 = 4aU |Lift| = U32 4aU CL = _3/2_4aU212U22a = 239.216 ms114 ms1UUPropulsive force (SE)/Length of cylinder= U cos = _14 m s1_80 103N2 9 m = 1.225 kg m3 14 m s1 = 259.2 m2 s1But = 0.5 2r r = r2 = 259.21.52 rad s1= 36.67 rad s1(clockwise)sin = 4aU = 259.241.5_142+62whence = 9025.49 = arctan 6/14 = 23.20Propulsion line comes from 4523.20 = 21.8W of N. Stagnation points are 3.69 E of N and 47.29 W of N.Max pressure at stagnation point. Min pressure at max velocity,which occurs at (r = a, = 3/2).(qt)r=a = 2Usin 32 + 2a =_2_142+62+ 259.23_m s1= 57.96 m s1Pressure difference = 12q2= 12 1.225kg m3_57.96m s1_2= 2058 Pa9.22 Flow pattern is same as right-hand half of that formed by sources at(a, 0) and (a, 0) = m2(1+2) ; tan 1 = yx +a; tan 2 = yx a70 Solutions manual tan (1+2) = [y/ (x +a)] +[y/ (x a)]1 y2/_x2a2_ = 2xyx2y2a2 = m2arctan 2xyx2y2a29.23 Irrotational vortex + sink: = 2ln rr0+ m2qt = 2r = 2rqt = 2(1.2 m)(20sin 70m s1)= 48 sin 70 m2 s1 Inner qt = 48 sin 7020.9 m s1= 803 sin 70 m s1qr = m2r m = 2rqr = 21.2 20cos 70m2 s1= 48 cos 70m2 s1 Inner qr = 48 cos 7020.9 m s1= 803 cos 70m s1 q = (qt+qr)1/2= 803 m s1 Pressure drop = 12 1000kg m3__803_2202_m2 s2= 155.6 kPa9.24CL12U2S = T cos 7 CL = 102cos 712 1.23 13.521.8 0.9= 0.558CD = 102sin 712 1.23 13.521.8 0.9 = 0.06859.25CL = 32.812 1.23 3020.75 0.1 = 0.790CD = 1.68same = 0.0405AR = 7.5 CDi = 0.79027.5 = 0.02650 CD = 0.0405 0.0265 = 0.0140Chapter 9 71Effective is less nominal by arctan 0.7907.5 = 1.92. Effective = 5.08For AR=5, reduction of is arctan 0.7905 = 2.88 Corresponding nominal = 5.08+2.88 = 7.96CL = 0.790cos 2.88cos 1.92 = 0.790;CD = 0.0140 + 0.79025 = 0.0537Chapter 1010.10.9 m1.8 m0.6 m0.6 mxyx0.6 m = 0.60.9 x = 0.4 my =_0.42+0.62m=0.52 mA = 12{(0.6 +2 0.4) +0.6}0.6 m2= 0.6 m2P = 0.6 m+2y = 2.042 mm = A P = 0.2938 mu = 60 m1/2 s1_0.2938 m 12600 = 0.638 m s1Q = 0.638 m s10.6 m2= 0.3827 m3 s110.2 Q2Q1= A2m2/32 i1/2A1m2/31 i1/2 =_A2A1_5/3_P1P2_2/3 Q2 = 1.24 m3 s1_ 2.5 0.5 +0.52tan 302.5 0.35 +0.352tan 30_5/3_2.5 +2 0.35cosec 602.5 +2 0.5cosec 60_2/3= 2.216 m3 s110.3n = 0.018; A = (1 m)_b + (1 m)1.5_;P = b +2_12+_ 11.5_2m = b + 2313 mChapter 10 73Eqn 10.9 : 2.8b + 11.5=_ b + 11.5b + 2313_2/3 _ 17000_1/2_0.018,whence (by trial) b = 4.456 m10.4n = 0.015. For trapezium mmax = h2 = bh +h2/3b +2 2h/3 b = 2h30.3 [m3 s1] =_bh + h23__h2_2/3_ 11800_1/2_0.015=_2h23 + h23__h2_2/3_ 11800_1/2_0.015 h = 0.520 m and b = 0.601 m10.5arcos = ar = 0.20.5 = 66.42 2 = 132.84Area = 4(1 m)2360 132.84360+0.2 0.5sin m2= 0.587 m2Perimeter = (1 m)227.16360 = 1.982 m m = 0.2962 mu = 0.3250.587 m s1= K(0.2962 m)2/3_ 11500_1/2 K = 48.2 m1/3 s1When conduit is full i = 4.53.6 103 = 1800 andm = d/4 = 0.25 mQ = 4(1 m)248.2 m1/3 s1(0.25 m)2/3_ 1800_1/2= 0.532 m2 s174 Solutions manual10.660d/4Full : 0.13 m3 s1= 4d2(58 m1/2 s1)_d4i (1)At depth d/4 : A = 1203604d2 d2 sin 60d4= 0.1535d2P = 120360d = 1.047d m = 0.1466d 0.6 m s1= (58 m1/2 s1)0.1466di (2)Divide eqn (1) by eqn (2): 0.130.6 m2= d280.1466Whence d = 0.460 mThen, from eqn (2), i = 0.00158810.7Max. Q requires maxAm2/3= A5/3/P2/3A = 12r2 +r sin 2r cos_ 2_= r2_2 sin 2 cos 2_= r22 ( sin )P = rMax ( sin )5/3_2/3requires2/3 53( sin )2/3(1 cos ) ( sin )5/3 231/3= 0i.e. either = sin [which corresponds to minimum at = 0]or 3 5 cos +2sin = 0, whence (by trial) = 5.278 [in radian measure] 302.4Chapter 10 75Then A = (r2/2){5.278 (0.844)} = 3.061r2andP = 5.278r2.8 m3 s1= 3.061r2(80 m1/3 s1)_3.061r5.278_2/31102,whence r = 1.205 m d = 2.41 m10.8A = bh +h2/3 A1 =_1 0.5 + 0.523_m2= 0.644 m2;A2 =_1 0.65 + 0.6523_m2= 0.894 m2u1 = 0.850.644 m s1= 1.319 m s1By continuity, A1u1 = A2u2+(A2A1)c 0.85 m3 s1= (0.894 m2)u2+(0.250 m2)cMomentum eqn: gA1x1gA2x2 = (u1+c)A1(u2u1)where x = depth of centroidthat is, gA1(A1x1A2x2) = (u1+c)(u2u1)= (u1+c)2A1A2A2Ax = bh h2 + h23 h3 = bh22 + h333 A1x1 =_1 0.522 + 0.5333_m3= 0.1491 m3;A2x2 =_1 0.6522 + 0.65333_m3= 0.2641 m3 (u1+c)2= 9.810.644(0.2641 0.1491)0.8940.894 0.644 m2 s2= 6.27 m2 s2 c = (6.27 1.319) m s1= 1.185 m s1Q2 = (0.85 0.250 1.185) m3 s1= 0.554 m3 s1Surface width = 1 m+2hcot 60 B1 = 1.577 m; B2 = 1.751 m76 Solutions manual Fr1 = u1_gA1/B1= 1.319_9.81 0.644/1.577 = 0.659Fr2 = u2_gA2/B2= 0.5540.894_ 1.7519.81 0.894 = 0.27710.9h + u22g = 1.8 m and 12 m3 s1= u(3 m)hwhence h + 12g_4h_2= 1.8 in metre, second units h31.8h2+0.815 = 0 h = 1.027 m or 1.357 mFr = u_gh = 4_gh3 = 1.226 or 0.808hc = (q2/g)1/3For critical ow_ghc = m2/3i1/2/n i = n2ghcm4/3 = n2ghc_b +2hcbhc_4/3= n2(gq)2/3b4/3_bg1/3q2/3 +2_4/3= 0.0142(9.81 4)2/334/3_3 9.811/342/3 +2_4/3= 0.0039410.10 For parabola x2= 4ay. Area =_ y02xdy = 4a1/2_ y0y1/2dy= 83a1/2y3/2B = 2x = 4a1/2y1/2 A/B = 23y uc =_gA/B =_23ghcE = h + u22g = hc+ u2c2g = hc+ 23ghc2g = 43hc hc = 34E uc =_23g34E =_gE210.11 For min upstream depth, ow between piers is critical. E = 32_q2g_1/3= 32_ 45022529.81_1/3m = 4.81 m= h1+ Q2b21h212g = h1+ 4502(m3)30219.62h21= h1+ 11.47(m3)h21whence h1 = 4.15 mChapter 10 7710.12 Wave stops when ow is critical, that is, when Q2B = gA3A = bh +h2cot 60 = (1 0.15 +0.152/3) m2= 0.1630 m2B = b +2hcot 60 = (1 +0.3/3) m = 1.173 mP = b +2hcosec 60 =_1 +0.3 23_m = 1.346 m m = AP = 0.1211 mgAB = u2c = (Km2/3i1/2)2 K2= gABm4/3i = 9.81 0.16301.173(0.1211)4/30.004 m2/3 s2 K = 75.4 m1/3 s1For uniform ow 0 = mgi = 0.1211 1000 9.810.004 Pa = 4.75 Pa10.13u1 = 5.43.5 0.38 m s1= 4.06 m s1 From eqn 10.25h2 =__0.19 +_0.192+ 2 0.38 4.0629.81__m = 0.956 mHead lost = (h2h1)34h1h2= 0.57634 0.38 0.956 m = 0.1315 m Power dissipated = 5.4 m3 s11000 kg m39.81 N kg10.1315 m= 6970 W10.14 For the jump h1+h22 = q2gh1h2 1.75h1 = (8.5/2.5)29.81 2.5h21[m3]whence h1 = 0.646 m and h2 = 1.615 mDownstream of jump A = 2.5 1.615 m2= 4.04 m2P = (2.5 +2 1.615) m = 5.73 m_78 Solutions manual m = 0.75 mu = 8.54.04[m s1] = (0.705)2/3(0.002)1/2n whence n = 0.016810.15h1+ 12g_ qh1_2= h2+ 12g_ qh2_2whence q = h1h2_ 2gh1+h2_1/2= 6 1.2_19.627.2_1/2m2 s1= 11.89 m2 s1 Q = 6 m11.89 m2 s1= 71.3 m3 s1Critical depth = (q2/g)1/3= (11.892/9.81)1/3m = 2.433 mDownstream depth > critical Jump must have occurred.Steady-ow momentum eqn:121000 kg m39.81 N kg1(1.223.12) m2F= 1000 kg m3(11.89 m2 s1)2_ 13.1 11.2_m1whence F = force on blocks divided by width = 32080 N m1 Total force = 6 32080 N = 192.5 kN10.16 h1 = (1.2 0.2) m = 1.0 m; h2 = 0.85 m; u1 = q1.2 m;u2 = q0.85 m 1.0 + q21.2219.62 = 0.85 + q20.85 19.62[metre, second units]whence q = 2.066 m2 s1and Q = 5.58 m3 s110.17u1 =_ 2.5 0.92.5 +2 0.9_2/3_ 11200_1/2_0.015 m s1= 1.250 m s1q = 1.250 0.9 m2 s1= 1.705 m1/2 s1_0.9 y + 1.250219.62_3/2m3/2whence y = 0.222 mChapter 10 7910.18(a) With metre, second units 0.6 + 119.62_ Q1.2 0.6_2= 0.56 + 119.62_ Q0.6 0.56_2whence Q = 0.337 m3 s1(b) h10.2 + 119.62_ Q1.2h1_2= 32h2where h2 = (q2/g)1/3=_ 0.33720.629.81_1/3m = 0.318 mwhence h1 = 0.6674 m Increase = 0.0674 m = 67.4 mm10.19E = 0.43 m+_ 0.1400.7 0.43_2119.62 m = 0.441 m Critical depth = 23 0.441m = 0.294 mSince for rectangular section, hc =_Q2b2g_1/3,b =_Q2gh3c_1/2= 0.2804 m Flow in throat of ume is critical and depth there = 0.294 mE = 0.441 m = h3+_ Qb3h3_212g = h3+ 0.002039 m3h23i.e. h330.441h23+0.002039 = 0 [metre units]But h = 0.43 m must t this eqn also. h 0.43 is a factor of LHS.(h 0.43)(h20.011h 0.00474) = 0whence h3 = (0.0055 +_0.00552+0.00474) m = 0.0746 m12gh1b1h1 12gh3b3h3F = Q(u3u1)= Q2_ 1b3h3 1b1h1_ F = (h1h3)_gb12 (h1+h3) Q2b1h1h3_= 305.5 N80 Solutions manual10.20Near outlet h = (q2/g)1/3=_ 1.2521.529.81_1/3m = 0.414 mMetre, second units:h Average h A P m u u2/gh 1 u2/gh i 104(i s)104dl/dh h l0.4140.526 0.470 0.705 2.440 0.2889 1.773 0.6818 0.3182 37.03 30.78 103.4 0.112 11.580.5260.638 0.582 0.873 2.664 0.3277 1.432 0.3591 0.6409 20.42 14.17 452 0.112 50.60.6380.750 0.694 1.041 2.888 0.3605 1.201 0.2118 0.7882 12.65 6.40 1232 0.112 138.0200.18i = u2n2/m4/3Add about 3.5(q2/g)1/3= 1.45 m; Total = (200.18 +1.45) m= 201.63 m, say 202 m10.21q = 28050 m2 s1= 5.6 m2 s1. For uniform ow downstream5.6h = h2/30.00040.0151/2 h2 = 2.366 mh1 =__2.3662 +_2.36624 + 2 5.629.81 2.366__m = 0.842 mMetre, second units:h Average h (=m) u = q/h u2/gh 1 i 104(i s)104dl/dh h l0.60.681 0.6405 8.74 11.17 311.5 307.5 363.1 0.081 29.410.6810.762 0.7215 7.76 7.51 209.5 205.5 365.6 0.081 29.610.7620.842 0.802 6.98 5.20 147.2 143.2 362.9 0.080 29.0388.05 mi u2n2h4/3 = q2n2h10/310.22c2= gm tanh mh = 2T2 = 42m2T2 m = 42gT2tanh mhAs rst approximation, assume deep water tanh mh 1 m = 0.0497 m1Iteration then gives m = 0.0657 m1 = 95.6 mPressure is max under crest, that is, whensin mx = 1, cos mx = 0Then (steady velocity)2= [c +Amcosh{m(h +z)}]2Chapter 10 81 pmax = 12[A2m2sinh2mh +c2]12[c +Amcosh{m(h +z)}]2gz= 12A2m2sinh2mh cAmcosh{m(h +z)}12A2m2cosh2{m(h +z)} gz= 12c2a2m2+ c2amsinh mh cosh{m(h +z)}12c2a2m2cosh2{m(h +z)}sinh2mhgzNeglect terms in a2. Thenpmax+gz = amsinh mhgm tanh mhcosh{m(h +z)}= agcosh{m(h +z)}cosh mh a =_pmaxg +z_ cosh mhcosh{m(h +z)}=_ 145 1031025 9.81 14_ cosh(0.0657 15)cosh(0.0657 1)m = 0.6404 m Height = 2a = 1.281 m10.23 c = f = 25 s10.0085 m = 0.2125 m s1 (0.2125 m s1)2=_2+ g2_tanh 2h=_ 2875 kg m30.0085 m+9.81 m s20.0085 m2_tanh_20.0040.0085_whence = 0.038 N m1Rate of energy transmission = c2(1 +2mhcosech 2mh)= 0.2125 m s12_1 +2 20.00850.004cosech40.0040.0085_= 0.1096 m s1Group velocity = c2_3 m2+g m2+g + 2mhsinh 2mh_= m2c m2+g +Rate of energy transmission82 Solutions manual= 0.038(2/0.0085)20.21250.038(2/0.0085)2+875 9.81 m s1+0.1096 m s1= 0.2600 m s110.24 Assume Airy waves a = 0.5 m2 = 0.25 m ah = 0.254 = 116 : OK. = cT c2=_2cT + gcT2_tanh 2hcTAs rst approximation, assume tanh(2h/cT) 1(deep water waves) and effect of negligible.Then c = gT/2 = (9.81 5/2) m s1= 7.807 m s1 = 7.807 m s15s = 39.09 m 0.017 m indeed negligibleSecond approx. : c =_9.81 52m s1_tanh_24c5 m s1_=_7.807 m s1_tanh_1.6c m s1_By iteration c = 5.589 m s1 = 5.589 m s15 s = 27.95 m. a/ < 0.01 : OK.m = 2/ = 0.2248 m1For negligible , cg = c2_1 + 2mhsinh 2mh_= 5.5892_1 + 2 0.2248 4sinh 1.7984_m s1= 3.407 m s1Energy/width for one wavelength= 12a2g =12(0.25 m)2(27.95 m)1000 kg m39.81 N kg1= 8570 J m1 Total power = 8570 J m1 4.5 m5 s = 7.71 kWMidway between crest and trough x = 0 or /2.At half still-water depth z = 2 mA = ca cosech mh = 5.589 m s10.25 m cosech(0.2248 4) = 1.363 m2 s1Chapter 10 83(Absolute velocity)2= A2m2cosh2{m(h +z)} sin2{m(x ct)}+A2m2sinh2{m(h +z)}cos2{m(x ct)}= A2m2sinh2{m(h +z)} in this case |Absolute velocity| = Amsinh{m(h +z)}= 1.363 0.2248sinh(0.2248 2) m s1= 0.142 m s1(Steady velocity)2= A2m2sinh2{m(h +z)} cos2mx+[c Amcosh{m(h +z)} sin mx]2 (Steady velocity at = 0)2= A2m2sinh2mh +c2By Bernoulli for x = 0,p=0 + 12(A2m2sinh2mh +c2)12[A2m2sinh2{m(h +z)} +c2] gz=12A2m2[sinh2mh sinh2{m(h +z)}] +g(z)=121000 1.36320.22482[sinh2(0.2248 4)sinh2(0.2248 2)] +1000 9.81 2 (Pa)=19.66 kPaHorizontal semi-axis =Ac cosh{m(h + z)}= 1.3635.589 cosh(0.2248 2) m = 0.2688 mVertical semi-axis =Ac sinh{m(h + z)}= 0.1136 m10.25 Mean rate of energy transfer/width= 14ca2g(1 +2mhcosech 2mh) = 14ca2g for deep water.It remains constant as waves approach shore.2T2 = c2= gm = g2for deep water. = gT22and c = gT284 Solutions manual Power/width = a2g2T8= 0.621025 9.81288W m1= 11300 W m1 Total power produced = 11300 W m180 m0.5= 452 kW2= 42m2 = c2T2= gmT2tanh mh m = 42gT2 coth mh When h = 5 m m = 429.81 82 coth(m5)[metre units]whence m = 0.1184 m111300 W m1= 14ca2g(1 +2 mhcosech 2mh)= 14_ 2mT_a2g(1 +2mhcosech 2mh)Hence a = 0.614 mChapter 1111.1 T2288.15 K = 11 +( 1)/2 = 11.2 T2 = 240.125 K u2 = a =_(1.4 287 J kg1 K1240.125 K)= 310.6 m s1

2 = 1_T2T1_1/( 1)= p1RT1_ 11.2_2.5= 101.3 103287 288.15 1.22.5 kg m3= 0.777 kg m3Max u when p = 0. 1p0

0= u22 +0 umax =_2 RT0 1 =_2 1.4 287 288.150.4 m s1= 761 m s111.2 T2288.15 K = 1 +0.2 021 +0.2 1.62 T2 = 190.6 K a2 =_1.4 287 190.6 m s1= 276.7 m s1 u2 = 1.6 276.7 m s1= 443 m s1_or u22 = 2cp(T0T2) = 2010 J kg1 K1(288.15 190.6) Kwhence u2 = 443 m s1__ p2101.5 kPa_0.4/1.4= T2T0= 11.512 p2 = 23.88 kPa

2 = p2RT2= 23.88 103287 190.6 kg m3= 0.437 kg m386 Solutions manual m = 0.437 kg m3(0.6 m)2443 m s1= 69.6 kg s1Min area is where M = 1. From eqn 11.59At = m_p0

0 [2/( +1)]( +1)/( 1)_1/2= 69.6{[(101.5 103)2/(287 288.15)]1.4(2/2.4)6}1/2 m2= 0.288 m211.3 M = cosec40. a =_1.4 287 283.15 m s1= 337.3 m s1Temperature rise = u22cp= a2M22cp= 337.32cosec2402010 K= 137 K Temperature = 147 C11.4 Using eqns 11.32, 11.24 and 11.293 = (p0)2p1= (p0)2p2p2p1=_1 +0.2M22_3.5 1 +1.4M211 +1.4M22=_1 +0.2_ 1 +0.2M211.4M210.2__3.5 1 +1.4M211 +1.4__1 +0.2M21_/_1.4M210.2__whence M1 = 1.386p2p1=_1 +1.4M21_ _1.4M210.2_1.68M21+1.2 = 1.4M210.21.2 = 7M2116From eqn 11.30: 2

1= 0.4 +2.4p2/p12.4 +0.4p2/p1= 6M21M21+5 = 1.665Eqn 11.26: T2283.15 K = 1 +0.2M211 +0.2M22M22 = 1 +0.2M211.4M210.2 M2 = 0.746 T2 = 352.7 K a2 =_1.4 287 352.7 m s1= 376.4 m s1 u2 = a2M2 = 280.7 m s1Chapter 11 87For subsequent expansion T3 = T2(p1/p2)0.4/1.4= 352.7_ 67M211_1/3.5K = 286.3 K = 13.2 C11.5 1 = 40, 2 = 30.From eqn 11.37 tan 10 =2cot 40_M21 sin2401_M21_1.4 +cos 80_+2whence M1 = 1.967Normal component M1sin 1 = 1.264p2p1= 7 (M1sin 1)216 [as in Problem 11.4 above] p2p1 = p1_p2p11_= 76p1_(M1sin 1)21_= 76 35 kPa(1.26421) = 24.43 kPa_M2sin 30_2= 1 +0.2_M1sin 40_21.4_M1sin 40_20.2= 0.648 M2 = 1.610303.15 KT2= 1 +0.2M22 = 1.518 T2 = 199.7 K u2 = M2_ RT2 = 1.610_1.4 287 199.7 m s1= 456 m s111.6M34040From eqn 11.37tan 10 = 2cot (1.6102sin2 1)1.6102(1.4 +cos 2) +2 = 50.69 that is, 40.69to wallM23 sin240.69 = 1 +0.2 1.6102sin250.691.4 +1.6102sin250.690.2 [eqn 11.29] M3 = 1.25088 Solutions manual11.7 p2p1= 0.5 =_1 +0.2 1.521 +0.2M22_1.4/0.4 M22 = 3.838From eqn 11.42,2 =6arctan_2.8386_1/2arcsec(3.858)1/2= 25.251 =6arctan_1.256_1/2arcsec1.5 = 11.91 21 = 13.34 [or from Table A3.2]11.8 p0p > 1.893 Flow supersonic.From eqn 11.47 or Table A3.1, M1 = 1.20T = T0(1 +[( 1)/2]M2)1= 363.15 K(1.288)1= 281.9 K a =_1.4 287 281.9 m s1= 336.6 m s1u = 336.6 1.20 m s1= 404 m s111.9 a at sea level =_ RT =_1.4 287 288.15 m s1= 340.3 m s1 Assumed Mach no. = 740 10003600_340.3 = 0.604This corresponds to p2/p1 at sea level of_1 +[( 1)/2] M2_/( 1)=_1 +0.2(0.604)2_3.5= 1.280 p2p1 = 0.280 101.3 kPa = 28.33 kPa Actual p2 = 63.83 kPa63.8335.5 = (1 +0.2M21)3.5whence M1 = 0.955a at 8000 m altitude is_1.4 287 236.15 m s1= 308.0 m s1 u = 0.955 308.0 m s1= 294.2 m s1= 1059 km h1T0T = u22cp= 294.222 1005 K = 43.07 K T0 = 6.07 CChapter 11 8911.10 Critical ambient pressure = 700 kPa 0.528 = 369.6 kPa In (a) ow does not become sonic Exit pressure =400 kPaT = T0_ pp0_( 1)/= 313.15_47_0.4/1.4K = 266.9 K 6 Cu2= 2cp(T0T) = 2010 46.25 m2 s2 u = 305 m s1 = pRT = 400 103287 266.9 kg m3= 5.22 kg m3 m = Au = 5.22 650 106305 kg s1= 1.035 kg s1(b) Exit pressure = critical = 369.6 kPaT = T0_1 + 12 12_1= 2 +1T0= 313.151.2 K = 260.96 K 12.2 Cu =_ RT =_1.4 287 260.96 m s1= 323.8 m s1 = 369.6 103287 260.96 kg m3= 4.94 kg m3 m = 4.94 650 106323.8 kg s1= 1.039 kg s111.11p0 = p_1 +_ 12_M2_/( 1)= 14 kPa_1 +0.2 2.82_3.5= 380 kPaPressure in throat = 0.528 380 kPa = 200.6 kPaEqn 11.59: m2= A2tp0

0_ 2 +1_( +1)/( 1)= A2tp20RT0_ 2 +1_( +1)/( 1) At = m_RT0p0__ +12_( +1)/( 1)_1/2= 1.2287 293.15380 1031.4 1.23m2= 1.338 103m290 Solutions manualEqn 11.59:_AtAe_2=_ +12_( +1)/( 1) 2 1__ pp0_2/_ pp0_( +1)/_= 1.265__ 12.5683.5_2/1.4_ 12.5683.5_2.4/1.4_= 1.265_ 12.5685 12.5686_= 2.822.146 Ae = At 2.1432.8 = 4.68 103m2T = T0_1 + 12 M2_1= 293.15 K(2.568)1= 114.15 K 159 Cu = Ma = 2.8_1.4 287 114.15 m s1= 600 m s111.12 At design conditions_AtA_2=_ +12_( +1)/( 1)M2_1 +[( 1)/2] M2_( +1)/( 1)= 1.26 1.82_1 +0.2 1.82_6 = 1.261.821.6486With argon_AtA_2=_43_4M21_1 + 13M21_4_43_2M1_1 + 13M21_2 = 1.231.81.6483 whence M1 = 1.902p01p1=_1 +[( 1) /2] M21_/( 1)=_1 + 13M21_2.5= 7.23p2p1= 2 +1_M21 12_[from eqns 11.24 and 11.29]= 5M2114 = 4.27 p01p2= p01p1p1p2= 7.234.27 = 1.692Chapter 11 9111.13 Provided that steam remains superheated and that Boyles law isvalid then by eqn 11.58:p2p0=_ 2 +1_/( 1)= 0.5457 p2 = 109.14 kPaInitial specify entropy = 7.280 kJkg1K1(fromtables) assumedto remain constant. Steamis not superheated under these conditions.Limitation is that steamdoes not become wet at throat. Limitingp2 for saturation (with s = 7.280 kJ kg1 K1) is 126.7 kPa.Then v2 = 1

2= 1.359 m3 kg1 m2= 2A22

22 1_p0

0 p2

2_= 2_4 (0.05 m)2_2 11.3592 kg2 m61.30.3(200000 0.9602 126700 1.359) N m kg1 m = 0.600 kg s1Alternatively: 12u22 = hs = (2770 2685) kJ kg1= 85 103m2 s2 m = u2A2/v2 = 0.596 kg s111.14 From eqn 11.58: pc = 106Pa _ 22.3_1.3/0.3= 5.46 105PaSince nozzle is choked, use eqn 11.59. Apply eqn 11.60 down-stream of shock and substitute for m2:_ pp02_2/_ pp02_( +1)/= 1_AtA_2p01p02

01

02_ 2 +1_( +1)/( 1)By Boyles law 01

02= p01p02(T02)(T01)= p01p02(h02)(h01)= p01p02assuming that h depends on T only (implied by constant cp). Alsoh0 unchanged across shock._ pp02_2/1.3_ pp02_2.3/1.3= 12_ 2 +1_( +1)/( 1)_AtA_2_p01p02_292 Solutions manualFor exit conditionsx2/1.3x2.3/1.3= 0.15_ 22.3_2.3/0.3(0.5)2_ 106x700 103_2= 0.0262x2where x = pexit/p02 and pexit = 700 kPai.e. 1 x3/13= 0.0262x6/13 x3/13=1 +()1 +0.10480.0524 = 0.975 x = 0.8964 p02 = 7000.8964 kPa_p0p_( 1)/= 1 + 12 M2 M2exit = 2 1__ 10.8964_0.3/1.31_= 0.1704 Mexit = 0.413p02p01= 700 1030.8964_106= 0.781 By eqn 11.33 M1 = 1.840Upstream of shock p0p =_1 +_ 12_M2_/ 1=_1 +0.15 1.8402_1.3/0.3= 5.92Eqn 11.61_AtA_2=_ +12_( +1)/( 1)_ 2 1___ 15.92_2/_ 15.92_( +1)/_= 0.424 AAt=_ 10.424_1/2= 1.535and for exit conditions,0.52=_2.32_2.3/0.3_ 20.3___pexitp0_2/1.3_pexitp0_2.3/1.3_ pexit/p0 0.940 or 0.1063 pexit = 940 kPa or 106.3 kPa (without shocks)Chapter 11 9311.15 a1 =_1.4 287 323.15 m s1= 360.3 m s1

1 = 180 103287 323.15 kg m3= 1.941 kg m3u1 = 2.7 kg s11.941 kg m3/4 (0.1 m)2 = 177.1 m s1 M1 = 177.1360.3 = 0.492From Table A3.3 or eqn 11.78: f lmaxP/A = 4f lmax/d = 1.140 lmax = 1.140 0.1 m4 0.006 = 4.75 mT2 = Tc = T1T1/Tc=_1 +[( 1)/2] M21( +1)/2_T1 = 282.3 K 9.2 Cp2 = pc = p1(pc/p1) = p1M1_1 +[( 1)/2] M21( +1)/2_1/2= 82.7 kPaFor mid-point f lmaxP/A = 1.140/2 = 0.570 M = 0.581T = Tc(T/Tc) = 282.3 K 1.124 = 317.3 K 44.2 Cp = pc(pc/p) = 82.7 kPa 0.548 = 150.9 kPa11.16(a) 1 = 730 103287 303.15 kg m3= 8.39 kg m3u1 = 2.3 kg s18.39 kg m3/4 (0.15 m)2 = 15.51 m s1 M1 = 15.511.4 287 303.15 = 0.0444From eqn 11.78 _f lmaxPA_1 = 355.7[Interpolation in Table A3.3 not accurate for small M.]f lPA = 0.006 2000 m0.15 m 4 = 320.0 (f lmaxP/A)2 = 35.7 M2 = 0.1342 from Table A3.3 or iteration of eqn 11.78Eqn 11.76: p2 = 730 kPa 0.04440.1342_0.4 0.04442+20.4 0.13422+2_1/2= 241.4 kPa94 Solutions manual(b) Eqn 11.84 : (/4)20.154_7302p22_1062 2.32287 303.15 ln_730p2_= 2 0.006 20000.15_p2 in kPa_whence p2 = 240.8 kPa11.17 Eqn 11.84:(/4)2d4(8262)10102 0.322287 288.15 ln 86 = 2 0.006 140d[d in metres]i.e. 1.02 107d40.2877 = 1.68/dAs rst approx. neglect log term. Then d5= 1.681.02 107 m5 d = 0.044 mThen 1.02 107d5= 1.68 +0.2877 0.044 = 1.693,whence d = 0.0441 m11.18 (/4)2(0.075)4(60022402)1062m2287 288.15 ln_600240_= 2 0.08 450.075whence m = 1.842 kg s1

1 = 600 103287 288.15 kg m3= 7.26 kg m3 u1 = 1.842 kg s17.26 kg m3/4 (0.075 m)2 = 57.5 m s1q = 12_u22u21_= u212__u2u1_21_= u212__

1

2_21_= u212__p1p2_21_= 57.522 (6.25 1) m2 s2= 8670 m2 s2Q = mq = 1.842 kg s18670 m2 s2= 15970 W11.19 p1/p2 = 300/120 = 2.5f lP/A = 4f l/d = 4 0.006 85/0.05 = 40.8 From Table A3.3: M1 0.12; M2 0.030Chapter 11 95Put inlet conditions in eqn 11.79:300 103Pa = m(/4)(0.05 m)20.12_287 288.15 m2 s21.4_1 +0.2 0.122__1/2whence m = 0.2913 kg s1or put in outlet conditions:120 103Pa = m(/4)(0.05 m)20.30_287 288.15 m2 s21.4_1 +0.2 0.32__1/2whence m = 0.2934 kg s1 say m = 0.292 kg s1Increase by 50% gives m = 0.438 kg s1From eqn 11.80:M22 = 10.4__1 +__1 + 0.81.40.4382287 288.151202106(/4)2(0.05 m)4___= 0.1964 M2 = 0.4332 _4f lmax/d_2 = 1.653_4f lmax/d_1 = 40.8 +1.653 = 42.453 M1 = 0.1245 p1 = 0.438(/4)(0.05)20.1245_ 287 288.151.4_1 +0.2 0.12452__1/2Pa= 435 kPaChapter 1212.1 Values of lgdudt are trivial.12.2u0 = 2.5(/4)0.12 m s1= 3.183 m s1dudt = Kt5/4 u = 49Kt9/4+u0 K = u0(4/9) 89/4 = 0.0665 [m s13/4] When t = 6 su =_3.183 490.0665 69/4_m s1= 1.517 m s1hi = lgdudt = 16009.81 0.0665t5/4= 101.9 m when t = 6 sAlso hf is reduced by4 0.005 16001 19.62 (3.18321.5172) m = 12.77 m H = (300 +101.9 +12.8) m = 414.7 m12.3k = 4 0.005 6001 +10 +2 = 2423 m = 24u2max/2g umax = 4.34 m s1t = 12 s = 600 m24 4.34 m s1 ln_4.34 +u4.34 u_whence u = 3.37 m s112.4 l1gdu1dt + l2gdu2dt + 4f1l1d1u212g + 4f2l2d2u222g = 12 mChapter 12 97 309.81dQdt1(/4)(0.225)2 + 1009.81dQdt1(/4)(0.15)2+ 4 0.007 300.225 19.62Q2_(/4)(0.225)2_2+ 4 0.008 1000.15 19.62Q2_(/4)(0.15)2_2= 12 [meter, second units] 654dQdt +3602Q2= 12that is, 0.1815dQdt = Q20.003331 t =_ Q0.1150.1815dQQ2+0.003331= 0.18150.003331_arctan Q0.003331_0.115Q= 3.144[arctan 1.992 arctan(17.33Q)] When t = 2 s 23.144 = 1.106 arctan(17.33Q)whence Q = 0.02929 m3 s112.5 Load to be accelerated = (1000 +225 +180) kg = 1405 kgIf plunger diam. = D m, acceleration of water in pipe= 1.5_ D0.04_2m s2= 937.5D2m s2Pressure drop due to this acceleration= 1000 kg m360 m937.5D2m s2= 5.625 107D2PaForce required at plunger= {(1000 +225 180)9.81 +1130 +1405 1.5} N = 13490 N 4D2(2.75 1065.625 107D2) N = 13490 N98 Solutions manual D40.0489D2+0.0003053 = 0whence D2= 0.02444 0.01709and D =0.00735 = 0.0857 is the smaller value of D Plunger diameter = 85.7 mmAt steady speed, force at plunger= {(1000 +225 180)9.81 +1130} N = 11380 NPressure in cylinder = 11380 N(/4)(0.0857 m)2 = 1.971 106PaPressure loss in pipe = (2.75 1061.971 106) Pa= 0.778 MPa hf = 0.778 1061000 9.81 m = 4 0.006 600.04u219.62 m s2 u = 6.58 m s1 Lift speed = 6.58_ 4085.7_2m s1= 1.432 m s112.6 Eqn 12.8: 1K

=_ 11.035 109 + 0.60.0125 207 109_ m2 N1 K

= 8.35 108Pa c =_8.35 108N m2880 kg m3 = 974 m s1 2lc = 900974 s = 0.924 s12.7 Hoop stress = (p)d/2t p = 2d/th = 40 106d/t PaAlso (p)2= 2c2u20 = 2u20K

/ = u20_ EE/K +d/t__40 106Pad/t_2= 1000 kg m3(2 m s1)2_ 200 109Pa(200 109)/(2 109) +d/t_ (d/t)22d/t 200 = 0whence d/t = 1 +()201 = 15.18Chapter 12 99c = (p)/u0 =_40 10615.18_(1000 2)_m s1= 1318 m s1 2l/c = (2 500/1318) s = 0.759 s12.8 Put: 3 m s1= B0_700 103N m2 B0 = 3.586 103m2 N1/2 s1300 103Pa = u_1.24 109N m2850 kg m3whence u = 0.2922 m s1Then 3.2922 m s1= B_400 103N m2 New B = 5.205 103m2 N1/2 s1 Increase in area =_ BB01_100%= 45.2%For rapid movement, time