Mechanics of Fluids Solutions Manual Mechanics of Fluids Eighth edition Solutions manual Bernard Massey ReaderEmeritusinMechanicalEngineering UniversityCollege,London Revised by John Ward-Smith FormerlySeniorLecturerinMechanicalEngineering BrunelUniversity SeventheditionpublishedbyStanleyThornes(Publishers)Ltdin1998 Eightheditionpublished2006 byTaylor&Francis 2ParkSquare,MiltonPark,Abingdon,OxonOX144RN SimultaneouslypublishedintheUSAandCanada byTaylor&Francis 270MadisonAve,NewYork,NY10016 Taylor&FrancisisanimprintoftheTaylor&FrancisGroup This edition published in the Taylor & Francis e-Library, 2005. “To purchase your own copy of this or any of Taylor & Francis or Routledge’s collection of thousands of eBooks please go to www.eBookstore.tandf.co.uk.” ©2006BernardMasseyandJohnWard-Smith TherightofB.S.MasseyandJ.Ward-Smithtobeidentifiedasauthorsof thisworkhasbeenassertedbytheminaccordancewiththeCopyright DesignsandPatentsAct1988. 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BritishLibraryCataloguinginPublicationData AcataloguerecordforthisbookisavailablefromtheBritishLibrary LibraryofCongressCataloginginPublicationData Acatalogrecordforthisbookhasbeenrequested ISBN 0-203-01231-3 Master e-book ISBN ISBN0–415–36204–0(Print Edition) Chapter 1 1.1 V T p SincepV =mRT, 1 = 1 2 V T p 1 2 1 π 288.15 1.1 ∴V = (20m)3 =56.2m3 1 6 233.15101.3 1.2 (cid:4)= p = 1.4×105 N·m−2 =1.51kg·m−3 RT 287J·kg−1·K−1×323.15K ∂p p−p 1.3 K =(cid:4) AssumeK constant.Thenln((cid:4)/(cid:4) )= 0 ∂(cid:4) (cid:1) (cid:2) 0 (cid:3) K (cid:4) ∴(cid:4)=(cid:4) exp p−p0 =1025kg·m−3exp 81.7×106 0 K 2.34×109 =1061kg·m−3 1.4 (cid:4)= µ = 2×10−5 N·s·m−2 =1.333kg·m−3 ν 15×10−6 m2·s−1 R= p = 1.013×105 N·m−2 =259.2J·kg−1·K−1 (cid:4)T 1.333kg·m−3×293.15K 8310 ∴M= =32.06 259.2 1.5 µ=ν(cid:4)=400×10−6 m2·s−1×850kg·m−3 =0.34Pa·s Velocitygradient= 0.12m·s−1 =1200s−1 0.1×10−3 m Area=π0.2×1.2m2 =0.754m2 Force=0.754m2×0.34Pa·s×1200s−1 =307.6N 2 Solutionsmanual (cid:5)(cid:1) (cid:2) (cid:1) (cid:2) (cid:6) ∂u ∂u 1.6 Totalforceonplate=Area×µ + ∂y ∂y sideA sideB (cid:7) (cid:8) 0.15m·s−1 0.15m·s−1 =(0.25m)2×0.7Pa·s + 0.006m 0.019m =1.439N 1.7 Forannulus,radiusr,widthδr Velocity ωr Force=Area×µ× =2πrδrµ Clearance c µω ∴Torque=Force×r=2πr3δr c (cid:9) R µω πR4µω Totaltorque= 2πr3 dr= c 2c 0 π(0.1m)40.14Pa·s×2π ×7rad·s−1 = =7.44N·m 2×0.00013m 1.8 2γ 2×0.073N·m−1 p= = =36.5Pa d 0.004m 1.9 4γ cosθ 4×0.073N·m−1×1 h= = (cid:4)gd 1000kg·m−3×9.81N·kg−1×0.005m =0.00595m=5.95mm 1.10 4×0.377N·m−1×cos140◦ h= (13.56−1)1000kg·m−3×9.81N·kg−1×0.006m =−1.563mm 1.11 ud(cid:4) 4Q(cid:4) 4×0.0025m3·s−1×900kg·m−3 Re= = = =1508 µ πdµ π0.05m×0.038N·s·m−2 u= 2000µ = 2000×0.038N·s·m−2 =1.689m·s−1 d(cid:4) 0.05m×900kg·m−3 1.12 4Q(cid:4) 4×0.01m3·s−1 Re= = =430 ∴Laminar πdµ π0.08m×370×10−6 m2·s−1 Chapter 2 p 200×103 N·m−2 2.1 h= = =12.82m (cid:4)g 1590kg·m−3×9.81N·kg−1 2.2 Pressuredependsonlyondepthbelowfreesurface. (a) p=(cid:4)gh=(820kg·m−3×9.81N·kg−1)(3−0.15)m =22930N·m−2 =22.93kPa (b) p=820×9.81N·m−3×(3+2)m=40.2kPa (c) p=820×9.81N·m−3×{3+2−(1.2sin30◦+0.6)}m =820×9.81×3.8N·m−2 =30.57kPa (d) Load=Pressure×Area =820×9.81×3N·m−2×(3.5×2.5)m2 =211.2kN p (cid:4) gh (cid:4) 2.3 h = = water water = waterh air (cid:4) g (cid:4) g (cid:4) water air air air 1000kg·m−3×287J·kg−1·K−1×288.15K = 0.075m 1.013×105 N·m−2 =61.2m 2.4 pV =constant (cid:1) (cid:2) d 3 101.3×103 Pa+1000kg·m−3×9.81N·kg−1×9m ∴ = 4mm 101.3×103 Pa whenced =4.93mm 2.5 (cid:13)p=820kg·m−3×9.81N·kg−1×2m+(13.56−0.82) ×1000kg·m−3×9.81N·kg−1×0.225m=44.2kPa 4 Solutionsmanual (cid:13)p∗ 0.225m(13.56−0.82)1000kg·m−3×9.81N·kg−1 =(cid:13)h= (cid:4)g 820kg·m−3×9.81N·kg−1 =3.496m 44200N·m−2=820×9.81×2N·m−2+x(0.82−0.74)1000 ×9.81N·m−3 whencex=35.83m 2.6 New levels Movementoffluidin A x y B C =60mm×70mm2 =(500mm2)x X X =(800mm2)y Initial C ∴x=8.4mm; surface of separation y=5.25mm MeasuringaboveXX:Initially0.8h =0.9h A B Later:800×9.81(Oldh −60+8.4)10−3 Pa A =p+900×9.81(Oldh −60+5.25)10−3 Pa B ∴ p=9.81×10−3(−800×51.6+900×65.25)Pa=171.1Pa (cid:1) (cid:2) λz g/Rλ 2.7 Fromeqn2.7 p=p 1− 0 T (cid:1)0 (cid:2) 0.0065×7500 9.81/287×0.0065 =101.5Pa 1− 288.15 =38.3kPa (cid:1) (cid:2) (cid:1) (cid:2) 2.8 p = T0−λz g/Rλ = Ttop g/Rλ p T T +λz 0 0 top (cid:7)(cid:1) (cid:2) (cid:8) ∴z= Ttop p0 Rλ/g−1 λ p (cid:7)(cid:1) (cid:2) (cid:8) 268.15 749 287×0.0065/9.81 = m −1 0.0065 566 =2257m Chapter2 5 2.9 F =(1.2×1.8)m2×1000kg·m−3×9.81N·kg−1 ×(x+0.9sin30◦)m (a) 2160×9.81N·m−1×0.45m=9.54kN (b) 2160×9.81N·m−1×0.95m=20.13kN (c) 2160×9.81N·m−1×30.45m=645kN (bd3/12)+bd(2x+0.9)2 Centreofpressureisatslantdepth bd(2x+0.9) d2 = +2x+0.9(metres) 12(2x+0.9) (1.8m)2 = +2x+0.9m 12(2x+0.9) (cid:7) (cid:8) 1.82 thatis +0.9 mfromupperedge 12(2x+0.9) =(a)1.2m; (b)1.042m; (c)0.904mfromupperedge 2.10 Bysymmetry,centre X X ofpressureison x θ r verticalcentre-line 2ndmomentaboutXX Depth= 1stmomentaboutXX (cid:10) rx22(r2−x2)1/2dx = (cid:10)0 rx2(r2−x2)1/2dx 0 (cid:10) 0 (rcosθ)22rsinθ(−rsinθdθ) = π(cid:10)/2 0 rcosθ2rsinθ(−rsinθdθ) π/2 (cid:10) r π/2cos2θsin2θdθ = (cid:10)0 π/2sin2θcosθdθ 0 (cid:10) r π 1sin22θd(2θ) = 0(cid:11)8 (cid:12) π/2 1sin3θ 3 0 r/8[2θ/2−(1/4)sin4θ]2θ=π = 0 1/3 3 π 3πd = r = 8 2 32