fluid force on vertical side of tank the weight density of water is 62.4 trapezoid

In fact, B is not a constant, it hides one of the variables in pipe flow. The parameter B must have dimensions of inverse length. Can you guess its name? Solution: The dimensions of these variables are Ans. This group has a customary name, which begins with C. Find the combination which is proportional to µ. Thus hydrostatic pressures cannot keep the element in balance, and shear and flow result.ġ.4 The quantities viscosity µ, velocity V, and surface tension Y may be combined into a dimensionless group. 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. Vertical forces are presumably in balance with element weight included. Due to element weight, the pressure along the lower and right sides must vary linearly as shown, to a higher value at point C. 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. M(atmosphere) 6.1E21 grams = ≈ 1.3E44 molecules m(one molecule) 4.8E −23 gm/moleculeġ.3 For the triangular element in Fig. 1.1 above), we obtain the total number of molecules in the earth’s atmosphere: N molecules =

Then the total mass of air in the atmosphere is m t = ò ρ dVol = ρavg (Air Vol) ≈ ρavg 4π R 2e (Air thickness) = (0.6 kg/m 3 )4π (6.377E6 m)2 (20E3 m) ≈ 6.1E18 kgĭividing by the mass of one molecule ≈ 4.8E−23 g (see Prob. Solution: Let Re be the earth’s radius ≈ 6377 km.

Use these values to estimate the total mass and total number of molecules of air in the entire atmosphere of the earth. m øè s ⋅K ø èġ.2 The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20 km and average density 0.6 kg/m3 (see Table A-6). Then the density of air containing 1012 molecules 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 mįinally, from the perfect gas law, Eq. Molecular weight 28.97 mol −1 = 4.81E−23 g Avogadro’s number 6.023E23 molecules/g ⋅ mol The mass of one molecule of air may be computed as m=

If Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent? Solution: Introduction 1.1 A gas at 20☌ may be rarefied if it contains less than 1012 molecules per mm3.

Fluid force on vertical side of tank the weight density of water is 62.4 trapezoid