Chapter1_SM

# Chapter1_SM - Chapter 1 Introduction 1.1 A gas at 20C may...

This preview shows pages 1–3. Sign up to view the full content.

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 ⎛⎞ =− ⎜⎟ 12 molecules g 10 4.81E 23 ⎝⎠ 3 33 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 = 32 kg m p RT 4.81E 5 287 (293 K) . ms K ns 4.0 Pa 2 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 Re 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 Solutions Manual Fluid Mechanics, Sixth 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 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 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 P1.4 The Saybolt Universal Viscometer, now obsolete but still sold in scientific catalogs, measures the kinematic viscosity of lubricants [Mott, p. 40]. A container, held at constant temperature, is filled with 60 ml of fluid. Measure the time t for the fluid to drain from a small hole or short tube in the bottom. This time unit, called Saybolt universal seconds , or SUS, is correlated with kinematic viscosity ν , in centistokes (1 stoke = 1 cm 2 /s), by the following curve- fit formula: SUS 100 40 for 145 215 . 0 < < = t t t ( a ) Comment on the dimensionality of this equation. ( b ) Is the formula physically correct? ( c ) Since ν varies strongly with temperature, how does temperature enter into the formula? ( d ) Can we easily convert from centistokes to mm 2 /s?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 70

Chapter1_SM - Chapter 1 Introduction 1.1 A gas at 20C may...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online