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Get Full Access to Fundamentals Of Fluid Mechanics - 7 Edition - Chapter 6.108 - Problem 108
Get Full Access to Fundamentals Of Fluid Mechanics - 7 Edition - Chapter 6.108 - Problem 108

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# An incompressible Newtonian fluid flows steadily between

ISBN: 9781118116135 135

## Solution for problem 108 Chapter 6.108

Fundamentals of Fluid Mechanics | 7th Edition

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Problem 108

An incompressible Newtonian fluid flows steadily between two infinitely long, concentric cylinders as shown in Fig. P6.108. The outer cylinder is fixed, but the inner cylinder moves with a longitudinal velocity V0 as shown. The pressure gradient in the axial direction is . For what value of V0 will the drag on the inner cylinder be zero? Assume that the flow is laminar, axisymmetric, and fully developed.

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4/4/16 EXAMAPRIL14 th Ø 4/12NickTreichishostingareviewsession@415pminFulmer125 Ø 4/12Finneganishostingareviewsessioninthepit@7 Ø 4/11@515theChemclubishostingareviewsession AXE • • n=thenumberofbondinggroupsonthecentralatom • m=thenumberoflonepairsonthecentralatom • ANYtypeofbond(single,double,triple)isonebondinggroup • n+m=thenumberofelectrongroupsonthecentralatom ElectronGeometry • Thenumberofelectrongroupsdeterminesthis • Thereareonly5,seetable10.1 • Determinestheidealbondangles-elementstryandarrangethemselvesasfar awayfromeachotheraspossiblebecausetheelectronsrepulsethemselves MolecularGeometry(shape)-electrongeometryandthenumberoflonepairs determinethemoleculargeometry • Themolecularshape,thebonpolarities,andtheformalchargedistribution determinethemolecularpolarity 4/6/16 BondAngles • ElectronGeometrydeterminestheidealbondangles(theyareapproximate) Linear:180degreeangles Triganolplanar:120degreeangles Tetrahedral:109.5degreeangles Trigonalbipyramidal:90,120degreeangles Octahedral:90degreeangles • Lonepairstakeupmorespacethanbondingpairs • Doublebondstakeupmorespacethanasinglebonds • ImportantNote:Whendealingwithresonancestructures,theamountof spacethe“rotating”doublen=bondtakesupisnegligiblebecauseitis technicallyalldoublebonds ▯ • ElectronGeometry:TriganolBipyramidal • ℎ:Linearduetothe180degreeangle HowdowedothebondanglesforthismoleculeByusingtheideathatdouble bondstakeupmorespacethanasingle,andlonepairstakemorespacethanbonds • OctahedralElectronGeometry: ▯ ▯ • Molecularshape:squareplanar,duetothe90degreeangle • BondanglesforF-Xe-Fis90degrees • ElectronGeometry:Tetrahedral • MolecularShape:Triganolpyramidal • BondAnglesforH-N-Hislessthan109.5becausethelonepairwillmakethe anglesmallerthanthatofanormaltetrahedralmolecule NoticehowitisTetrahedralwiththelonepair,fortheshapethelonepairisnot drawn PolarandNon-PolarMolecules • Somethingispolarifithasanegativesideandanonpolarside • Inmoleculesthisisdeterminedbysymmetryaroundthecentralatom • If a molecule is symmetric (by polar bonds)around the central atom, the charges cancel eachother out • If the molecule is not semetric(by polar bonds) than the molecule is polar • A numeric measure of polarity is the Dipole moment(μ) • The more polar a molecule the larger the dipole moment • Is ▯ polar or non-polar Non-polar • Is ▯polar or non polar Non polar • Is ▯olar or non polar Polar • Look at the bonds and the way the bonds are arranged to see if they are polar or not! ValenceBondTheory • Acovalentbondisproducedbytheoverlapoforbitalsintheregion betweenthetwoatoms • Thegreatertheoverlap,thestrongerthebond. • But,theatomicorbitalsdonotalwayspointintherightdirectionsto producetheshapesthatthemoleculesaresupposedtohave Toexplainthis,itisassumedthattheatomicorbitalsaremixedtoproduce‘hybrid’ orbitalsthatpointinthecorrectdirection. • Hybridorbitalsarenamedfortheorbitalsthatcontributetothem.For Example isacombinationofonesorbitalandtwoporbitals • ImportantNote:Allorbitalsthataremixedmustcomefromthesame shell.ForExample3pand3scanbehybrids,but2pand4scannotbe combinedtogether. • The▯totalnumberoforbitalsisunchanged(Therewillbe2orbitals,three orbitalsandsoon) EachhybridhasitsownGeometryassociatedwithit,asshownbelow NoticehowforthegroundstateofCarbonthe2sorbitalisfull,butthe2porbitalis not▯evenhalffull,thismakescarbonunstablesothetwoorbitalscombinetoform ,makingcarbonstable • Singlebondsareformedbyoverlappingorbitalsinbetweenthe atoms.Thesebondsarecalledsigma(σ)bonds. • Doubleandtriplebondsalsocontainaσ-bondbutthesecond(andthird bond)cannotformdirectlybetweentheatoms(becausethereisalreadya pairofelectronsthere) • Theadditionalbondsareformedbyunhybridizedp-orbitalsthat overlapintheareastoeithersideofthesigmabond.Thesebondsare calledpi(π)bonds. • Adoublebondconsistsofaσ-bondandaπ-bond. • Atriplebondconsistsofaσ-bondandtwoπ-bonds. ThisDiagramshouldhelpyouvisualizetheprincipalofsigmaandpibonds

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##### ISBN: 9781118116135

This textbook survival guide was created for the textbook: Fundamentals of Fluid Mechanics, edition: 7. This full solution covers the following key subjects: . This expansive textbook survival guide covers 1484 chapters, and 1484 solutions. The full step-by-step solution to problem: 108 from chapter: 6.108 was answered by , our top Engineering and Tech solution expert on 11/10/17, 06:08PM. The answer to “An incompressible Newtonian fluid flows steadily between two infinitely long, concentric cylinders as shown in Fig. P6.108. The outer cylinder is fixed, but the inner cylinder moves with a longitudinal velocity V0 as shown. The pressure gradient in the axial direction is . For what value of V0 will the drag on the inner cylinder be zero? Assume that the flow is laminar, axisymmetric, and fully developed.” is broken down into a number of easy to follow steps, and 67 words. Since the solution to 108 from 6.108 chapter was answered, more than 329 students have viewed the full step-by-step answer. Fundamentals of Fluid Mechanics was written by and is associated to the ISBN: 9781118116135.

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