Fluid Dynamics Sol ch1

Fluid Dynamics Sol ch1 - Chapter 1 Introduction 1.1 A gas...

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Chapter 1 Introduction 1.1 A gas at 20 ° C may be rarefied if it contains less than 10 12 molecules per mm 3 . 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 1 Molecular weight 28.97 mol m 4.81E 23 g Avogadro’s number 6.023E23 molecules/g mol == = Then the density of air containing 10 12 molecules per mm 3 is, in SI units, ρ ±² ± ² =− ³´ ³ ´ µ¶ µ 12 3 33 molecules g 10 4.81E 23 molecule mm gk g 4.81E 11 4.81E 5 mm m Finally, from the perfect gas law, Eq. (1.13), at 20 ° C = 293 K, we obtain the pressure: Α = 2 32 kg m p RT 4.81E 5 287 (293 K) . ms K ns 4.0 Pa 1.2 The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20 km and average density 0.6 kg/m 3 (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 R e be the earth’s radius 6377 km. Then the total mass of air in the atmosphere is 2 ta v g a v g e m dVol (Air Vol) 4 R (Air thickness) (0.6 kg/m )4 (6.377E6 m) (20E3 m) . Ans ρρ π =≈ ± 6.1E18 kg 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: molecules m(atmosphere) 6.1E21 grams N m(one molecule) 4.8E 23 gm/molecule Ans. 1.3E44 molecules
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2 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 with an atmosphere at pressure p a , 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 ele- ment weight included. But horizontal forces are out of balance, with the unbalanced force being to the left, due to 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 . This group has a customary name, which begins with C . Can you guess its name? Solution: The dimensions of these variables are { } = {M/LT}, {V} = {L/T}, and {Y} = {M/T 2 }. We must divide by Y to cancel mass {M}, then work the velocity into the group: 2 / , { } ; Y / . MLT T L hence multiply by V LT MT finally obtain Ans ±² == = ³´³ ´³´ ³´ µµ µ ¶¶ µ V dimensionless. 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: 1.26 1.26 (RT) p (RT) ρ ± (1)
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Chapter 1 Introduction 3 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?
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This note was uploaded on 05/07/2008 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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Fluid Dynamics Sol ch1 - Chapter 1 Introduction 1.1 A gas...

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