Fluid Mechanics-Frank M White Solution Ch1

소개글

Fluid Mechanics-Frank M White Solution Ch1에 대한 보고서 자료입니다.

본문내용

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

Then the density of air containing 1012 molecules per mm3 is, in SI units,

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


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 ρ versus altitude z in the atmosphere, from Table A.6:
1.2255 kg/m3
ρ




0 z 30,000 m
This writer’s approximation: The curve is approximately an exponential, ρ ≈ ρ o exp(-b z), 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:

Density in the Atmosphere
2

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:

This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere.

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.

키워드

인하대,   유체역학,   솔루션