# Supp. Fluids Answers f11 - (2 gh ) 1/2 (b) 2.21 m/s (c)...

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Fluid Mechanics Homework Answers – ME 2124 – Fall 2011 Problem Answer(s) FL1 1000 N/m 2 FL2 (a) diagram (b) F = mg sin θ (c) sketch (d) F = A μ v/h (e) v = mgh sin / A (f) 31.7 cm/s, lower FL3 (a) 265 kPa (b) 196.2 Pa 1.67 3.43 × 10 –4 lbf 1.70(a) i o i R R l R - = μω π τ 3 2 FL4 (a) 0.6079 kg/m 3 (b) 1.2756 kg/m 3 (c) -98.3 Pa FL5 (a) 1/2 in (b) From top: 1 1/8 in 2-3 3/16 in 4-5 ¼ in 6-7 5/16 in 8-9 3/8 in 10-12 ½ in 2.16 (a) 1240 lbf/ft 2 (b) 1040 lbf/ft 2 (c) 1270 lbf/ft 2 2.30 -3.32 kPa 2.34 h = ( p 1 p 2 )/ g ( ρ 2 1 ) 2.52 889 N FL6 B and C FL7 (a) 250 ft/s (b) 3.39 lbm/s FL8 140.7 ft 3.21 -76.0 lbf/ft 2 gage, +88.0 lbf/ft 2 gage 3.36 Fig. (a) 3.44 43.0 psig FL9 11.47 m, 112.3 kPa, momentum change

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FL10 (a) v=
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Unformatted text preview: (2 gh ) 1/2 (b) 2.21 m/s (c) 43.4 ml/s (d) Decreases FL11 (a) 18 m/s (b) 52.9 m/s (c) -1399 Pa gage (d) Yes FL12 (a) 1.098 kg/s (b) No 3.23 1.40 mph 3.49 2.45 × 10 5 kPa, 5.50 × 10-6 m 3 /s FL13 13.89 ft/s FL14 Water FL15 1,610,000 3.47 11.4 m 3 /s 7.41 0.0652 ft/s 7.45 3760 mph, not realistic FL16 18.38 ft FL17 (a) D V πμ ρ & 4 Re = (b) 15,530 (c) Turbulent FL18 (a) 43.0 psi (b) -13 psi (c) 0.312 psf FL19 (a) 3.18 m/s, laminar (b) 30.6 kPa gage FL20 (a) 50.0 N/m 2 (b) 133.3 kPa 8.9 ~ 3 m, 8.83 N/m 2 8.55 f = 0.292, ε = 0.0132 ft FL21 (a) Turbulent (b) 24.5 psi FL22 (a) 2.234 m/s, 1,113,000 (b) 77.7 kPa/km, 77.700 kPa (c) 682 kW...
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## This note was uploaded on 10/05/2011 for the course ME 2124 taught by Professor Uzgoren during the Spring '08 term at Virginia Tech.

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Supp. Fluids Answers f11 - (2 gh ) 1/2 (b) 2.21 m/s (c)...

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