Fluid Mechanics , 6th Edition Solution Manual

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.1.Many centuries ago, a mariner poured 100 cm3of water into the ocean. As timepassed, the action of currents, tides, and weather mixed the liquid uniformly throughout theearth’s oceans, lakes, and rivers. Ignoring salinity, estimate the probability that the next sip (5ml) of water you drink will contain at least one water molecule that was dumped by the mariner.Assess your chances of ever drinking truly pristine water. (Consider the following facts:Mwforwater is 18.0 kg per kg-mole, the radius of the earth is 6370 km, the mean depth of the oceans isapproximately 3.8 km and they cover 71% of the surface of the earth.)Solution 1.1.To get started, first list or determine the volumes involved:υd= volume of water dumped = 100 cm3,υc= volume of a sip = 5 cm3, andV= volume of water in the oceans =€4πR2Dγ,where,Ris the radius of the earth,Dis the mean depth of the oceans, andγis the oceans'coverage fraction. Here we've ignored the ocean volume occupied by salt and have assumed thatthe oceans' depth is small compared to the earth's diameter. Putting in the numbers produces:€V=4π(6.37×106m)2(3.8×103m)(0.71)=1.376×1018m3.For well-mixed oceans, the probabilityPothat any water molecule in the ocean came from thedumped water is:€Po=(100 cm3of water)(oceans' volume)=υdV=1.0×10−4m31.376×1018m3=7.27×10−23,Denote the probability that at least one molecule from the dumped water is part of your next sipasP1(this is the answer to the question). Without a lot of combinatorial analysis,P1is not easyto calculate directly. It is easier to proceed by determining the probabilityP2that all themolecules in your cup are not from the dumped water. With these definitions,P1can bedetermined from:P1= 1 –P2. Here, we can calculateP2from:P2= (the probability that a molecule was not in the dumped water)[number of molecules in a sip].The number of molecules,Nc, in one sip of water is (approximately)Nc=5cm3×1.00gcm3×gmole18.0g×6.023×1023moleculesgmole=1.673×1023moleculesThus,P2=(1−Po)Nc=(1−7.27×10−23)1.673×1023. Unfortunately, electronic calculators and moderncomputer math programs cannot evaluate this expression, so analytical techniques are required.First, take the natural log of both sides, i.e.ln(P2)=Ncln(1−Po)=1.673×1023ln(1−7.27×10−23)then expand the natural logarithm using ln(1–ε)≈–ε(the first term of a standard Taylor seriesfor€ε →0)ln(P2)≅ −Nc⋅Po=−1.673×1023⋅7.27×10−23=−12.16,and exponentiate to find:P2≅e−12.16≅5×10−6... (!)Therefore,P1= 1 –P2is very close to unity, so there is a virtual certainty that the next sip ofwater you drink will have at least one molecule in it from the 100 cm3of water dumped manyyears ago. So, if one considers the rate at which they themselves and everyone else on the planetuses water it is essentially impossible to enjoy a truly fresh sip.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.2.An adult human expels approximately 500 ml of air with each breath duringordinary breathing. Imagining that two people exchanged greetings (one breath each) manycenturies ago, and that their breath subsequently has been mixed uniformly throughout theatmosphere, estimate the probability that the next breath you take will contain at least one airmolecule from that age-old verbal exchange. Assess your chances of ever getting a truly freshbreath of air. For this problem, assume that air is composed of identical molecules havingMw=29.0kgperkg-moleand that the average atmospheric pressure on the surface of the earth is 100kPa. Use 6370kmfor the radius of the earth and 1.20 kg/m3for the density of air at roomtemperature and pressure.Solution 1.2.To get started, first determine the masses involved.m= mass of air in one breath = densityxvolume =€1.20kg/m3()0.5×10−3m3()=€0.60×10−3kgM= mass of air in the atmosphere =€4πR2ρ(z)dzz=0∞∫Here,Ris the radius of the earth,zis the elevation above the surface of the earth, andρ(z) is theair density as function of elevation. From the law for static pressure in a gravitational field,€dP dz=−ρg, the surface pressure,Ps, on the earth is determined from€Ps−P∞=ρ(z)gdzz=0z= +∞∫sothat:M=4πR2Ps−P∞g=4π(6.37×106m)2(105Pa)9.81ms−2=5.2×1018kg.where the pressure (vacuum) in outer space =P∞= 0, andgis assumed constant throughout theatmosphere. For a well-mixed atmosphere, the probabilityPothat any molecule in theatmosphere came from the age-old verbal exchange is€Po=2×(mass of one breath)(mass of the whole atmosphere)=2mM=1.2×10−3kg5.2×1018kg=2.31×10−22,where the factor of two comes from one breath for each person. Denote the probability that atleast one molecule from the age-old verbal exchange is part of your next breath asP1(this is theanswer to the question). Without a lot of combinatorial analysis,P1is not easy to calculatedirectly. It is easier to proceed by determining the probabilityP2that all the molecules in yournext breath are not from the age-old verbal exchange. With these definitions,P1can bedetermined from:P1= 1 –P2. Here, we can calculateP2from:P2= (the probability that a molecule was not in the verbal exchange)[number of molecules in one breath].The number of molecules,Nb, involved in one breath is€Nb=0.6×10−3kg29.0g/gmole×103gkg×6.023×1023moleculesgmole=1.25×1022moleculesThus,€P2=(1−Po)Nb=(1−2.31×10−22)1.25×1022. Unfortunately, electronic calculators and moderncomputer math programs cannot evaluate this expression, so analytical techniques are required.First, take the natural log of both sides, i.e.€ln(P2)=Nbln(1−Po)=1.25×1022ln(1−2.31×10−22)then expand the natural logarithm using ln(1–ε)≈–ε(the first term of a standard Taylor seriesfor€ε →0)€ln(P2)≅ −Nb⋅Po=−1.25×1022⋅2.31×10−22=−2.89,and exponentiate to find:

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Fluid Mechanics, 6thEd.Kundu, Cohen, and Dowling€P2≅e−2.89=0.056.Therefore,P1= 1 –P2= 0.944 so there is a better than 94% chance that the next breath you takewill have at least one molecule in it from the age-old verbal exchange. So, if one considers howoften they themselves and everyone else breathes, it is essentially impossible to get a breath oftruly fresh air.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.3. The Maxwell probability distribution,f(v) =f(v1,v2,v3), of molecular velocities in agas flow at a point in space with average velocityuis given by (1.1).a) Verify thatuis the average molecular velocity, and determine the standard deviations (σ1,σ2,σ3) of each component ofuusingσi=1n(vi−ui)2allv∫∫∫f(v)d3v#$%&'(1 2fori= 1, 2, and 3.b) Using (1.27) or (1.28), determinen=N/Vat room temperatureT= 295 K and atmosphericpressurep= 101.3 kPa.c) DetermineN=nV= number of molecules in volumesV= (10μm)3, 1μm3, and (0.1μm)3.d) For theithvelocity component, the standard deviation of the average,σa,i, overNmoleculesisσa,i=σiNwhenN>> 1. For an airflow atu= (1.0ms–1, 0, 0), compute the relativeuncertainty,2σa,1u1, at the 95% confidence level for the average velocity for the three volumeslisted in part c).e) For the conditions specified in parts b) and d), what is the smallest volume of gas that ensuresa relative uncertainty inUof one percent or less?Solution 1.3.a) Use the given distribution, and the definition of an average:(v)ave=1nvallu∫∫∫f(v)d3v=m2πkBT"#$%&'3 2v−∞+∞∫−∞+∞∫−∞+∞∫exp−m2kBTv−u2*+,-./d3v.Consider the first component ofv, and separate out the integrations in the "2" and "3" directions.(v1)ave=m2πkBT!"#$%&3 2v1−∞+∞∫−∞+∞∫−∞+∞∫exp−m2kBT(v1−u1)2+(v2−u2)2+(v3−u3)2*+,-./0123dv1dv2dv3=m2πkBT!"#$%&3 2v1−∞+∞∫exp−m(v1−u1)22kBT*+,-./dv1exp−m(v2−u2)22kBT*+,-./dv2−∞+∞∫exp−m(v3−u3)22kBT*+,-./−∞+∞∫dv3The integrations in the "2" and "3" directions are equal to:€2πkBT m()1 2, so(v1)ave=m2πkBT!"#$%&1 2v1−∞+∞∫exp−m(v1−u1)22kBT*+,-./dv1The change of integration variable toβ=(v1−u1)m2kBT()1 2changes this integral to:(v1)ave=1πβ2kBTm!"#$%&1 2+u1!"##$%&&−∞+∞∫exp−β2{}dβ=0+1πu1π=u1,where the first term of the integrand is an odd function integrated on an even interval so itscontribution is zero. This procedure is readily repeated for the other directions to find (v2)ave=u2,and (v3)ave=u3. Thus,u= (u1,u2,u3) is the average molecular velocity.Using the same simplifications and change of integration variables produces:σ12=m2πkBT!"#$%&3 2(v1−u1)2−∞+∞∫−∞+∞∫−∞+∞∫exp−m2kBT(v1−u1)2+(v2−u2)2+(v3−u3)2*+,-./0123dv1dv2dv3=m2πkBT!"#$%&1 2(v1−u1)2±∞+∞∫exp−m(v1−u1)22kBT*+,-./dv1=1π2kBTm!"#$%&β2±∞+∞∫exp−β2{}dβ.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingThe final integral overβis:€π2, so the standard deviations of molecular speed are€σ1=kBT m()1 2=σ2=σ3,where the second two equalities follow from repeating this calculation for the second and thirddirections.b) From (1.27),n V=p kBT=(101.3kPa) [1.381×10−23J/K⋅295K]=2.487×1025m−3c) Fromn/Vfrom part b):€n=2.487×1010forV= 103μm3= 10–15m3€n=2.487×107forV= 1.0μm3= 10–18m3€n=2.487×104forV= 0.001μm3= 10–21m3d) From (1.29), the gas constant isR= (kB/m), andR= 287m2/s2Kfor air. Compute:2σa,1u1=2kBT m n()1 21m/s[] =2RT n()1 21m/s=2 287⋅295n()1 2=582n. Thus,forV= 10–15m3:2σa,1u1= 0.00369,V= 10–18m3:2σa,1u1= 0.117, andV= 10–21m3:2σa,1u1= 3.69.e) To achieve a relative uncertainty of 1% we needn≈(582/0.01)2= 3.39€×109, and thiscorresponds to a volume of 1.36€×10-16m3which is a cube with side dimension≈5μm.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.4. Using the Maxwell molecular speed distribution given by (1.4),a) determine the most probable molecular speed,b) show that the average molecular speed is as given in (1.5),c) determine the root-mean square molecular speed =vrms=1nv2f(v)0∞∫dv#$%&'(1 2,d) and compare the results from parts a), b) and c) withc= speed of sound in a perfect gasunder the same conditions.Solution 1.4.a) The most probable speed,vmp, occurs wheref(v) is maximum. Thus, differentiate(1.4) with respectv, set this derivative equal to zero, and solve forvmp. Start from:f(v)=4πnm2πkBT!"#$%&3 2v2exp−mv22kBT()*+,-, and differentiatedfdv=4πnm2πkBT!"#$%&3 22vmpexp−mvmp22kBT()*+*,-*.* −mvmp3kBTexp−mvmp22kBT()*+*,-*.*/0112344=0Divide out common factors to find:2−mvmp2kBT=0orvmp=2kBTm.b) From (1.5), the average molecular speedvis given by:v=1nv0∞∫f(v)dv=4πm2πkBT#$%&'(3 2v30∞∫exp−mv22kBT*+,-./dv.Change the integration variable toβ=mv22kBTto simplify the integral:v=4m2πkBT!"#$%&1 2kBTmβ0∞∫exp−β{}dβ=8kBTπm!"#$%&1 2−βe−β−e−β()0∞=8kBTπm!"#$%&1 2,and this matches the result provided in (1.5).c) The root-mean-square molecular speedvrmsis given by:vrms2=1nv20∞∫f(v)dv=4πm2πkBT#$%&'(3 2v40∞∫exp−mv22kBT*+,-./dv.Change the integration variable toβ=v m2kBT()1 2to simplify the integral:vrms2=4π2kBTm!"#$%&1 2β40∞∫exp−β2{}dβ=4π2kBTm!"#$%&3π8=3kBTm.Thus,vrms= (3kBT/m)1/2.d) From (1.28),R= (kB/m) sovmp=2RT,v=(8 /π)RT, andvrms=3RT. All three speedshave the same temperature dependence the speed of sound in a perfect gas:€c=γRT, but arefactors of2γ,8πγand€3γ, respectively, larger thanc.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.5.By considering the volume swept out by a moving molecule, estimate how themean-free path,l, depends on the average molecular cross section dimensiondand themolecular number densitynfor nominally spherical molecules. Find a formula for(the ratioof the mean-free path to the mean intermolecular spacing) in terms of the nominalmolecularvolume() and the availablevolume per molecule(1/n). Is this ratio typically bigger or smallerthan one?Solution 1.5.The combined collision cross section for two spherical molecules having diameter€dis€πd2. The mean free pathlis the average distance traveled by a molecule betweencollisions. Thus, the average molecule should experience onecollision when sweeping a volume equal to€πd2l. If the molecularnumber density isn, then the volume per molecule isn–1, and themean intermolecular spacing isn–1/3. Assuming that the swept volumenecessary to produce one collision is proportional to the volume permolecule produces:πd2l=C norl=Cnπd2(),whereCis a dimensionless constant presumed to be of order unity. The dimensionless version ofthis equation is:mean free pathmean intermolecular spacing=ln−1 3=l n1 3=Cn2 3πd2=Cnd3()2 3=Cn−1d3"#$%&'2 3=Cvolume per moleculemolecular volume"#$%&'2 3,where all numerical constants likeπhave been combined intoC. Under ordinary conditions ingases, the molecules are not tightly packed sol>>n−1 3. In liquids, the molecules are tightlypacked sol~n−1 3.ln1 3d3€d

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.6.Compute the averagerelativespeed,vr, between molecules in a gas using theMaxwell speed distributionfgiven by (1.4) via the following steps.a) Ifuandvare the velocities of two molecules then their relative velocity is:vr=u–v. If theangle betweenuandvisθ, show that the relative speed is:vr= |vr|=u2+v2−2uvcosθwhereu= |u|, andv= |v|.b) The averaging ofvrnecessary to determinevrmust include all possible values of the twospeeds (uandv) and all possible anglesθ. Therefore, start from:vr=12n2vrf(u)allu,v,θ∫f(v)sinθdθdvdu,and note thatvris unchanged by exchange ofuandv, to reach:vr=1n2u2+v2−2uvcosθθ=0π∫v=u∞∫u=0∞∫sinθf(u)f(v)dθdvduc) Note thatvrmust always be positive and perform the integrations, starting with the angularone, to find:vr=13n22u3+6uv2uvv=u∞∫u=0∞∫f(u)f(v)dvdu=16kBTπ#$%&'(1 2=2v.Solution 1.6.a) Compute the dot produce ofvrwith itself:vr2=vr⋅vr=(u−v)⋅(u−v)=u⋅u−2u⋅v+v⋅v=u2−2uvcosθ+v2.Take the square root to find: |vr|=u2+v2−2uvcosθ.b) The average relative speed must account for all possible molecular speeds and all possibleangles between the two molecules. [The coefficient 1/2 appears in the first equality belowbecause the probability density function of for the angleθin the interval 0≤θ≤πis (1/2)sinθ.]vr=12n2vrf(u)allu,v,θ∫f(v)sinθdθdvdu=12n2u2+v2−2uvcosθf(u)allu,v,θ∫f(v)sinθdθdvdu=12n2u2+v2−2uvcosθθ=0π∫v=0∞∫u=0∞∫sinθf(u)f(v)dθdvdu.Inu-vcoordinates, the integration domain covers the firstquadrant, and the integrand is unchanged whenuandvareswapped. Thus, theu-vintegration can be completed above thelineu=vif the final result is doubled. Thus,vr=1n2u2+v2−2uvcosθθ=0π∫v=u∞∫u=0∞∫sinθf(u)f(v)dθdvdu.Now tackle the angular integration, by settingβ=u2+v2−2uvcosθso thatdβ= +2uvsinθdθ. This leads tovr=1n2β1 2β=(v−u)2(v+u)2∫v=u∞∫u=0∞∫dβ2uv f(u)f(v)dvdu,u!v!dv!du!u=v!

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Fluid Mechanics, 6thEd.Kundu, Cohen, and Dowlingand theβ-integration can be performed:vr=12n223β3 2()(v−u)2(v+u)2v=u∞∫u=0∞∫f(u)uf(v)vdvdu=13n2(v+u)3−(v−u)3()v=u∞∫u=0∞∫f(u)uf(v)vdvdu.Expand the cubic terms, simplify the integrand, and prepare to evaluate thev-integration:vr=13n22u3+6uv2()v=u∞∫u=0∞∫f(v)vdv f(u)udu=13n2u3+6uv2()v=u∞∫u=0∞∫4πvm2πkBT#$%&'(3 2v2exp−mv22kBT*+,-./dv f(u)udu.Use the variable substitution:α=mv2/2kBTso thatdα=mvdv/kBT, which reduces thev-integration to:vr=13n2u3+6u kBTmα!"#$%&mu2kBT∞∫u=0∞∫4πm2πkBT!"#$%&3 2exp−α{}kBTm dαf(u)udu=23nm2πkBT!"#$%&1 22u3+6u kBTmα!"#$%&mu2kBT∞∫u=0∞∫e−αdαf(u)udu=23nm2πkBT!"#$%&1 2−2u3e−α+6u kBTm−αe−α−e−α()!"#$%&mu2kBT∞u=0∞∫f(u)udu=23nm2πkBT!"#$%&1 28u3+12u kBTm!"#$%&u=0∞∫exp−mu22kBT!"#$%&f(u)udu.The finalu-integration may be completed by substituting in forf(u) and using the variablesubstitutionγ=u m kBT()1 2.vr=132mπkBT!"#$%&1 28kBTm!"#$%&3 2γ3+12kBTm!"#$%&3 2γ!"##$%&&0∞∫4πm2πkBT!"#$%&3 2kBTm!"#$%&γexp−γ2()dγ=23πkBTm!"#$%&1 28γ4+12γ2()0∞∫exp−γ2()dγ=23πkBTm!"#$%&1 28 38π+1214π!"#$%&=4πkBTm!"#$%&1 2=16kBTπm!"#$%&1 2=2vHere,vis the mean molecular speed from (1.5).

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.7.In a gas, the molecular momentum flux (MFij) in thej-coordinate direction thatcrosses a flat surface of unit area with coordinate normal directioniis:MFij=1Vmvivjf(v)d3vallv∫∫∫wheref(v) is the Maxwell velocity distribution (1.1). For a perfectgas that is not moving on average (i.e.,u= 0), show thatMFij=p(the pressure), wheni=j, andthatMFij= 0, wheni≠j.Solution 1.7.Start from the given equation using the Maxwell distribution:MFij=1Vmvivjf(v)d3vallu∫∫∫=nmVm2πkBT"#$%&'3 2vivj−∞+∞∫−∞+∞∫−∞+∞∫exp−m2kBTv12+v22+v32()*+,-./dv1dv2dv3and first consideri=j= 1, and recognizeρ=nm/Vas the gas density, as in (1.28).MF11=ρm2πkBT!"#$%&3 2u12−∞+∞∫−∞+∞∫−∞+∞∫exp−m2kBTv12+v22+v32()*+,-./dv1dv2dv3=ρm2πkBT!"#$%&3 2v12exp−mv122kBT()*+,-dv1−∞+∞∫exp−mv222kBT()*+,-dv2−∞+∞∫exp−mv322kBT()*+,-−∞+∞∫dv3The first integral is equal to2kBT m()3 2π2()while the second two integrals are each equal to2πkBT m()1 2. Thus:MF11=ρm2πkBT!"#$%&3 22kBTm!"#$%&3 2π22πkBTm!"#$%&1 22πkBTm!"#$%&1 2=ρkBTm=ρRT=pwherekB/m=Rfrom (1.28). This analysis may be repeated withi=j= 2, andi=j= 3 to find:MF22=MF33=p, as well.Now consider the casei≠j. First note thatMFij=MFjibecause the velocity product underthe triple integral may be written in either ordervivj=vjvi, so there are only three cases ofinterest. Start withi= 1, andj= 2 to find:MF12=ρm2πkBT!"#$%&3 2v1v2−∞+∞∫−∞+∞∫−∞+∞∫exp−m2kBTv12+v22+v32()*+,-./dv1dv2dv3=ρm2πkBT!"#$%&3 2v1exp−mv122kBT()*+,-dv1−∞+∞∫v2exp−mv222kBT()*+,-dv2−∞+∞∫exp−mv322kBT()*+,-−∞+∞∫dv3Here we need only consider the first integral. The integrand of this integral is an odd functionbecause it is product of an odd function,v1, and an even function,exp−mv122kBT{}. Theintegral of an odd function on an even interval [–∞,+∞] is zero, soMF12= 0. And, this analysismay be repeated fori= 1 andj= 3, andi= 2 andj= 3 to findMF13=MF23= 0.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.8.Consider the viscous flow in a channel of width 2b. The channel is aligned in thex-direction, and the velocityuin thex-direction at a distanceyfrom the channel centerline isgiven by the parabolic distribution€u(y) =U01−y b()2[].Calculate the shear stressτas afunctiony,μ,b, andUo. What is the shear stress aty= 0?Solution 1.8.Start from (1.3):€τ=μdudy=μddy Uo1−yb$%&'()2*+,-./=–2μUoyb2. Aty= 0 (the location ofmaximum velocity)τ= 0. At Aty= ±b(the locations of zero velocity),€τ=2μUob.

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.9.Hydroplaning occurs on wet roadways when sudden braking causes a movingvehicle’s tires to stop turning when the tires are separated from the road surface by a thin film ofwater. When hydroplaning occurs the vehicle may slide a significant distance before the filmbreaks down and the tires again contact the road. For simplicity, consider a hypothetical versionof this scenario where the water film is somehow maintained until the vehicle comes to rest.a) Develop a formula for the friction force delivered to a vehicle of massMand tire-contact areaAthat is moving at speeduon a water film with constant thicknesshand viscosityμ.b) Using Newton’s second law, derive a formula for the hypothetical sliding distanceDtraveledby a vehicle that started hydroplaning at speedUoc) Evaluate this hypothetical distance forM= 1200 kg,A= 0.1 m2,Uo= 20 m/s,h= 0.1 mm, andμ= 0.001 kgm–1s–1. Compare this to the dry-pavement stopping distance assuming a tire-roadcoefficient of kinetic friction of 0.8.Solution 1.9.a)Assume that viscous friction from the water layer transmitted to the tires is theonly force on the sliding vehicle. Here viscous shear stress at any time will beμu(t)/h, whereu(t)is the vehicle's speed. Thus, the friction force will beAμu(t)/h.b) The friction force will oppose the motion so Newton’s second law implies:M dudt=−Aμuh.This equation is readily integrated to find an exponential solution:u(t)=Uoexp−Aμt Mh(),where the initial condition,u(0) =Uo, has been used to evaluate the constant of integration. Thedistance traveled at timetcan be found from integrating the velocity:x(t)=u(!t)d!tot∫=Uoexp−Aμ!tMh()d!tot∫=UoMh Aμ()1−exp−Aμt Mh()$%&'.The total sliding distance occurs for large times where the exponential term will be negligible so:D=UoMh Aμc) ForM= 1200 kg,A= 0.1 m2,Uo= 20 m/s,h= 0.1 mm, andμ= 0.001 kgm–1s–1, the stoppingdistance is:D= (20)(1200)(10–4)/(0.1)(0.001) = 24 km! This is an impressively long distance andhighlights the dangers of driving quickly on water covered roads.For comparison, the friction force on dry pavement will be –0.8Mg, which leads to avehicle velocity of:u(t)=Uo−0.8gt, and a distance traveled ofx(t)=Uot−0.4gt2. The vehiclestops whenu= 0, and this occurs att=Uo/(0.8g), so the stopping distance isD=UoUo0.8g!"#$%& −0.4gUo0.8g!"#$%&2=Uo21.6g,which is equal to 25.5 m for the conditions given. (This is nearly three orders of magnitude lessthan the estimated stopping distance for hydroplaning.)

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.10. Estimate the height to which water at 20°C will rise in a capillary glass tube 3mm in diameter that is exposed to the atmosphere. For water in contact with glass the contactangle is nearly 0°. At 20°C, the surface tension of a water-air interface isσ= 0.073 N/m.Solution 1.10.Start from the result of Example 1.4.h=2σcosαρgR=2(0.073N/m)cos(0°)(103kg/m3)(9.81m/s2)(1.5×10−3m)=9.92mm

Fluid Mechanics , 6th Edition Solution Manual - Page 15 preview image

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.11.Amanometeris a U-shaped tube containing mercury of densityρm. Manometersare used as pressure-measuring devices. If the fluid in tank A has a pressurepand densityρ, thenshow that the gauge pressure in the tank is:p−patm=ρmgh−ρga. Note that the last term on theright side is negligible ifρ«ρm. (Hint: Equate the pressures atXandY.)Solution 1.9.Start by equating the pressures atXandY.pX=p+ρga=patm+ρmgh=pY.Rearrange to find:p–patm=ρmgh–ρga.

Fluid Mechanics , 6th Edition Solution Manual - Page 16 preview image

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Fluid Mechanics, 6thEd.Kundu, Cohen, and DowlingExercise 1.12.Prove that ife(T,υ) =e(T) only and ifh(T,p) =h(T) only, then the (thermal)equation of state is (1.28) orpυ=kT, wherekis constant.Solution 1.12. Start with the first equation of (1.24):de=Tds–pdυ, and rearrange it:€ds=1T de+pT dυ=∂s∂e$%&'()υde+∂s∂υ$%&'()edυ,where the second equality holds assuming the entropy depends oneandυ. Here we see that:€1T=∂s∂e#$%&'(υ, and€pT=∂s∂υ$%&'()e.Equality of the crossed second derivatives ofs,€∂∂υ∂s∂e$%&'()υ$%&'()e=∂∂e∂s∂υ$%&'()e$%&'()υ, implies:€∂1T()∂υ$%&'()e=∂p T()∂e$%&'()υ.However, ifedepends only onT, then (∂/∂υ)e= (∂/∂υ)T, thus€∂1T()∂υ$%&'()e=∂1T()∂υ$%&'()T=0, so€∂p T()∂e#$%&'(υ=0, which can be integrated to find:p/T=f1(υ), wheref1is an undetermined function.Now repeat this procedure using the second equation of (1.24),dh=Tds+υdp.€ds=1T dh−υT dp=∂s∂h%&'()*pdh+∂s∂p%&'()*hdp.Here equality of the coefficients of the differentials implies:€1T=∂s∂h#$%&'(p, and€−υT=∂s∂p%&'()*h.So, equality of the crossed second derivatives implies:€∂1T()∂p#$%&'(h=−∂ υT()∂h#$%&'(p. Yet, ifhdependsonly onT, then (∂/∂p)h= (∂/∂p)T, thus€∂1T()∂p#$%&'(h=∂1T()∂p#$%&'(T=0, so€−∂ υT()∂h%&'()*p=0, which canbe integrated to find:υ/T=f2(p), wheref2is an undetermined function.Collecting the two results involvingf1andf2, and solving forTproduces:€pf1(υ)=T=υf2(p)or€pf2(p)=υf1(υ)=k,wherekmust be is a constant sincepandυare independent thermodynamic variables.Eliminatingf1orf2from either equation on the left, producespυ=kT.And finally, using both versions of (1.24) we can write:dh–de=υdp+pdυ=d(pυ).Wheneandhonly depend onT, thendh=cpdTandde=cvdT, sodh–de= (cp–cv)dT=d(pυ) =kdT, thusk=cp–cv=R,whereRis the gas constant. Thus, the final result is the perfect gas law:p=kT/υ=ρRT.

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