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FluidMechWhite5eCh01.pdf

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Department
Engineering
Course
ENGG 2230
Professor
Bahram Gharabaghi
Semester
Winter

Description
Chapter 1Introduction 12311 A gas at 20C may be rarefied if it contains less than 10 molecules per mm If Avogadros number is 6023E23 molecules per mole what air pressure does this represent Solution The mass of one molecule of air may be computed as 1Molecular weight2897 molm481E23 gAvogadros number6023E23 moleculesgmol123 molecules per mm is in SI units Then the density of air containing 10moleculesg 1210 481E233moleculemmgk g481E11 481E5 33mmmFinally from the perfect gas law Eq 113 at 20C293 K we obtain the pressure 2kgmpRT481E5 287 293Kns40Pa32ms K 12 The earths atmosphere can be modeled as a uniform layer of air of thickness 20 km 3and average density 06 kgm see Table A6 Use these values to estimate the total mass and total number of molecules of air in the entire atmosphere of the earth Solution Let R be the earths radius6377 km Then the total mass of air in the eatmosphere is 2m dVolAir Vol4RAir thickness ta v g a v g e 3206 kgm46377E6 m20E3 mAns61E18 kgDividing by the mass of one molecule48E23 g see Prob 11 above we obtain the total number of molecules in the earths atmosphere matmosphere61E21 gramsAns13E44 molecules N moleculesmone molecule48E23 gmmolecule 2 Solutions ManualFluid Mechanics Fifth Edition13 For the triangular element in Fig P13 show that a tilted free liquid surface in contact with an atmosphere at pressure p amust undergo shear stress and hence begin to flowFig P13 Solution Assume zero shear Due to element weight the pressure along the lower and right sides must vary linearly as shown to a higher value at point C Vertical forces are presumably in balance with element weight included But horizontal forces are out of balance with the unbalanced force being to the left due to the shaded excesspressure triangle on the right sideBCThus hydrostatic pressures cannot keep the element in balance and shear and flow result14 The quantities viscosityvelocity V and surface tension Y may be combined into a dimensionless group Find the combination which is proportional toThis group has a customary name which begins with C Can you guess its name Solution The dimensions of these variables are MLT VLT and Y2MT We must divideby Y to cancel mass M then work the velocity into the group MLTTLhence multiply byV 2LTYMT Vfinally obtain AnsdimensionlessYThis dimensionless parameter is commonly called the Capillary Number15 A formula for estimating the mean free path of a perfect gas is126126RT l 1 pRT Chapter 1Introduction 3where the latter form follows from the idealgas law pRT What are the dimensions of the constant 126 Estimate the mean free path of air at 20C and 7 kPa Is air rarefied at this condition Solution We know the dimensions of every term except 126 2MMLlLRT32LTLTTherefore the above formula first form may be written dimensionally as MLT L126126L32 2MLLTSince we have L on both sides 126unity that is the constant is dimensionless The formula is therefore dimensionally homogeneous and should hold for any unit system For air at 20C293 K and 7000 Pa the density is pRT70002872933200832 kgm From Table A2 its viscosity is 180E5 Nsm Then the formula predict a mean free path of 180E5Ansl94E7 m 1261200832287293This is quite small We would judge this gas to approximate a continuum if the physical scales in the flow are greater than about 100 l that is greater than about 94 m16 If p is pressure and y is a coordinate state in the MLT system the dimensions of 22the quantities a py bp dy c py d p 2223222Solution a MLT b MT c MLT d MLT17 A small village draws 15 acrefoot of water per day from its reservoir Convert this water usage into a gallons per minute and b liters per second 222Solution One acre1 mi6405280 ft64043560 ft Therefore 15 acreft333365340 ft1850 m Meanwhile 1 gallon231 in2311728 ft Then 15 acreft of water per day is equivalent to 3ft1728gal1daygal Q65340 aAns3403day2311440minminft
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