Fluid mechanic - white (cap1)

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Chapter 1 • Introduction
1.1 A gas at 20°C may be rarefied if it contains less than 1012 molecules per mm3. If Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent? Solution: The mass of one molecule of air may be computed as m= Molecular weight 28.97 mol −1 = = 4.81E−23 g Avogadro’s number 6.023E23 molecules/g ⋅ mol

Then the density of air containing 1012molecules per mm3 is, in SI units,
molecules öæ g æ ö ρ = ç 1012 ÷ç 4.81E−23 ÷ 3 molecule ø mm è øè g kg = 4.81E−11 = 4.81E−5 3 3 mm m

Finally, from the perfect gas law, Eq. (1.13), at 20°C = 293 K, we obtain the pressure:

kg ö æ m2 ö æ p = ρ RT = ç 4.81E−5 3 ÷ ç 287 2 ÷ (293 K) = 4.0 Pa Αns. m øè s ⋅K ø è

1.2 The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20km and average density 0.6 kg/m3 (see Table A-6). Use these values to estimate the total mass and total number of molecules of air in the entire atmosphere of the earth. Solution: Let Re be the earth’s radius ≈ 6377 km. Then the total mass of air in the atmosphere is
m t = ò ρ dVol = ρavg (Air Vol) ≈ ρavg 4π R 2 (Air thickness) e = (0.6 kg/m 3 )4π (6.377E6 m)2 (20E3 m) ≈ 6.1E18 kg Ans.Dividing by the mass of one molecule ≈ 4.8E−23 g (see Prob. 1.1 above), we obtain the total number of molecules in the earth’s atmosphere: N molecules = m(atmosphere) 6.1E21 grams = ≈ 1.3E44 molecules m(one molecule) 4.8E −23 gm/molecule Ans.

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Solutions Manual • Fluid Mechanics, Fifth Edition

1.3 For the triangular element in Fig. P1.3, show that a tilted free liquid surface, in contact withan atmosphere at pressure pa, must undergo shear stress and hence begin to flow. 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, dueto the shaded excess-pressure triangle on the right side BC. Thus hydrostatic pressures cannot keep the element in balance, and shear and flow result.
Fig. P1.3

1.4 The quantities viscosity µ, velocity V, and surface tension Y may be combined into a dimensionless group. Find the combination which is proportional to µ. This group has a customary name, which begins with C. Can you guess itsname? Solution: The dimensions of these variables are {µ} = {M/LT}, {V} = {L/T}, and {Y} = {M/T2}. We must divide µ by Y to cancel mass {M}, then work the velocity into the group:
ì µ ü ì M / LT ü ì T ü ìLü í ý=í ý = í ý , hence multiply by {V } = í ý ; 2 î Y þ î M /T þ î L þ îT þ finally obtain

µV = dimensionless. Ans. Y

This dimensionless parameter is commonly called the Capillary Number.1.5 A formula for estimating the mean free path of a perfect gas is:
l = 1.26

µ µ = 1.26 √ (RT) p ρ √ (RT)

(1)

Chapter 1 • Introduction

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where the latter form follows from the ideal-gas law, ρ = p/RT. What are the dimensions of the constant “1.26”? Estimate the mean free path of air at 20°C and 7 kPa. Is air rarefied at this condition? Solution: We know the dimensions of everyterm except “1.26”: ì L2 ü ìMü ìMü {l} = {L} {µ} = í ý {ρ} = í 3 ý {R} = í 2 ý {T} = {Θ} î LT þ îL þ îT Θþ Therefore the above formula (first form) may be written dimensionally as
{L} = {1.26?} {M/L⋅T} = {1.26?}{L} {M/L } √ [{L2 /T 2 ⋅ Θ}{Θ}]
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Since we have {L} on both sides, {1.26} = {unity}, that is, the constant is dimensionless. The formula is therefore dimensionally homogeneous and shouldhold for any unit system. For air at 20°C = 293 K and 7000 Pa, the density is ρ = p/RT = (7000)/[(287)(293)] = 0.0832 kg/m3. From Table A-2, its viscosity is 1.80E−5 N ⋅ s/m2. Then the formula predict a mean free path of
l = 1.26

1.80E−5 ≈ 9.4E−7 m (0.0832)[(287)(293)]1/2

Ans.

This is quite small. We would judge this gas to approximate a continuum if the physical scales in the flow...
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