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Unformatted text preview: Chemistry 132, Winter 2008 Solutions to Homework No. 6 Problem 1. See Levine problem 16.6. (a) False. By integrating Newton’s viscosity law (Levine eq. 16.15) it can be shown that for a layer with radius s the veloc ity v y along the pipe is a parabolic function of s (Levine eq. 16.16). Therefore, as shown in Levine Fig. 16.10a, the flow speed is not the same at all points in a plane perpendicular to the axis of the pipe. (b) True. Setting the derivative of v y ( s ) (Levine eq. 16.16) equal to zero and solving for s gives dv y ds = s 2 η dP dy = 0 ⇒ s = 0 . Since the pressure P decreases in the direction of flow, then dP/dy < 0 and d 2 v y /ds 2 = (1 / 2 η ) dP/dy < 0, so that v y has a maximum at s = 0, which is the center of the pipe. (c) True. Viscosity is a measure of the friction between adja cent fluid layers in laminar flow (Levine p. 493). The viscos ity of liquids decreases with increasing T (Levine fig. 16.7a) because at higher T more molecules have sufficient energy to overcome the relatively strong intermolecular attractions that characterize the liquid phase. The increased probability of overcoming intermolecular attractions between adjacent layers results in lower friction between the layers and hence in lower viscosity. According to kinetic theory, the viscos ity of a hardsphere gas is proportional to T 1 / 2 (Levine eq. 16.25). Experimentally, it is found that the viscosity of gases increases with T more rapidly than predicted by kinetic the ory (Levine fig. 16.7b). (d) True. Newton’s law of viscosity (Levine eq. 16.13) is experimentally found to be a good model of fluid flow when the flow rate is not too high (Levine p. 493). Problem 4. See Levine problem 16.14. The following equations and data are needed to solve the problem: v = 2( ρ ρ fl ) gr 2 9 η (Levine eq. 16.22) η water (25 ◦ , 1 atm) = 0 . 89 cP = 0 . 0089 P (Levine p. 494) η glycerol (25 ◦ , 1 atm) = 954 cP = 9 . 54 P (Levine p. 494) 1 P ≡ . 1 N m 2 s (Levine eq. 16.14) ρ water (25 ◦ , 1 atm) = 0 . 997044 g / cm 3 = 997 . 044 kg / m 3 (Levine problem 1.51) ρ glycerol = 1 . 25 g / cm 3 = 1250 kg / m 3 ρ steel = 7 . 8 g / cm 3 = 7800 kg / m 3 d = 2 . 00 mm = 2 . 00 × 10 3 m g = 9 . 80665 m / s 2 (Levine back cover) Figure 1: Excel spreadsheet for problem 7 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C D D ∞ sucrose,H2O / (cm 2 /s) 5.20E06 t unit t / s (Δ x ) rms / cm 0.5 min 30 0.017664 0.5 hour 1800 0.136821 0.5 day 43200 0.670284 1 week 604800 2.507971 cell formula C3 =A3*60 C4 =A4*3600 C5 =A5*24*3600 C6 =A6*3600*24*7 D3 =SQRT(2*$D$1*C3) Then, for a steel ball falling in water: v = 2( ρ steel ρ water ) gr 2 9 η water = 2 (7800 kg / m 3 ) (997 . 044 kg / m 3 ) × (9 . 80665 m / s 2 )(2 . 00 × 10 3 m / 2) 2 9(0 . 0089 P)(0 . 1 N m 2 s / P) = 16 . 6577 m / s ....
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 Winter '08
 Lindenberg
 Chemistry, Fluid Dynamics, Mole, Levine, Mass flow rate, Levine eq

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