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444 Solutions Manual x Fluid Mechanics, Fifth Edition Solution: (a) Assume no pressure drop and neglect velocity heads. The energy equation reduces to: p1 V12 z U g 2g 1 0 0 (L l ) For laminar flow, h f p2 V22 z h U g 2g 2 f 128P LQ SUgd 4 Solve for 't 0 0 0 h f , or: h f | L l and, for uniform draining, Q 128 P LX SU gd 4 ( L l ) X 't Ans. (a) (b) Apply to 't 6 s. For water, take U 998 kg/m3 and P 0.001 kg/ms. Formula (a) predicts: 't 128(0.001 kg/ms)(0.12 m)(8E6 m3 ) , S (998 kg/m3 )(9.81 m/s2 )d 4 (0.12 0.02 m) 6s Solve for d | 0.0015 m Ans. (b) 6.18 To determine the viscosity of a liquid of specific gravity 0.95, you fill, to a depth of 12 cm, a large container which drains through a 30-cm-long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of draining is found to be 1.9 cm3s. What is your estimate of the fluid viscosity? Is the tube flow laminar? Fig. P6.18 Solution: The known flow rate and diameter enable us to find the velocity in the tube: Q A V Evaluate U liquid 0.95(998) and the tube exit: 0.605 m s 948 kgm3. Write the energy equation between the top surface pa Ug or: 0.42 1.9 E6 m 3 /s (S /4)(0.002 m)2 2 Vtop 2g ztop V 2 32P LV 2g Ugd 2 pa V 2 0 hf , Ug 2 g (0.605)2 32P (0.3)(0.605) 2(9.81) 948(9.81)(0.002)2