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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