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ENGG 2230 (29)
Lecture

# Ch 1 Solutions.pdf

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School
Department
Engineering
Course
ENGG 2230
Professor
Bahram Gharabaghi
Semester
Winter

Description
Chapter 1 Introduction 123 1.1 A gas at 20C may be rarefied if it contains less than 10 molecules per mm. If Avogadros 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 weight28.97 mol 4.81E23 g m Avogadros number6.023E23 moleculesgmol 123 Then the density of air containing 10 molecules per mm is, in SI units, moleculesg 12 10 4.81E23 3 olecule m mm gkg 4.81E11 4.81E5 33 mmm Finally, from the perfect gas law, Eq. (1.13), at 20C 293 K, we obtain the pressure: 2 kgm pRT4.81E5 287 (293 K).ns4.0Pa 32 ms K 2 Solutions Manual Fluid Mechanics, Sixth Edition P1.2 Table A.6 lists the density of the standard atmosphere as a function of altitude. Use these values to estimate, crudely, say, within a factor of 2, the number of molecules of air in the entire atmosphere of the earth. Solution: Make a plot of density versus altitude z in the atmosphere, from Table A.6: 3 1.2255 kgm Density in the Atmosphere z 30,000 m 0 This writers approximation: The curve is approximately an exponential, exp(b z), o with b approximately equal to 0.00011 per meter. Integrate this over the entire atmosphere, with the radius of the earth equal to 6377 km: 2bz ()[] ( 4)mdvolR dze atmosphereoearth 0 2 32 4R (1.2255)4(6.3776)kgmEm oe a r t h g 5.718Ek 0.00011bm Dividing by the mass of one molecule 4.8E23 g (see Prob. 1.1 above), we obtain the total number of molecules in the earths atmosphere: m(atmosphere)5.7E21 grams N moleculesAns. 1.244 molecules m(one molecule)4.8E23 gmmolecule This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere. Chapter 1 Introduction 3 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. Fig. P1.3 Solution: Assume zero shear. Due to elem ent 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 excesspressure triangle on the right side BC. Thus hydrostatic pressures cannot keep the element in balance, and shear and flow result. P1.4 Sand, and other granular materials, definitely flow, that is, you can pour them from a container or a hopper. There are whole textbooks on the transport of granular materials [54]. Therefore, is sand a fluid? Explain. Solution: Granular materials do indeed flow, at a rate that can be measured by flowmeters. But they are not true fluids, because they can support a small shear stress without flowing. They may rest at a finite angle without flowing, which is not possible for liquids (see Prob. P1.3). The maximum such angle, above which sand begins to flow, is called the angle of repose. A familiar example is sugar, which pours easily but forms a significant angle of repose on a heaping spoonful. The physics of granular materials are complicated by effects such as particle cohesion, clumping, vibration, and size segregation. See Ref. 54 to learn more. ________________________________________________________________________ 1.5 A formula for estimating the mean free path of a perfect gas is: 1.261.26RT (1) p RT
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