# Answers Selected Problems - Introduction to Fluid Mechanics...

This preview shows pages 1–5. Sign up to view the full content.

Introduction to Fluid Mechanics 7 th Edition Fox, Pritchard, & McDonald Answers to Selected Problems, Chapter 1 1.5 M = 5913 kg 1.7 L = 27.25 in. D = 13.75 in. 1.9 2 2 05 . 2 t g W y = 1.11 d = 0.109 mm 1.15 y = 0.922 mm 1.17 a) N·m/s, lbf·ft/s b) N/m 2 , lbf/ft 2 c) N/m 2 , lbf/ft 2 , d) 1/s, 1/s e) N·m, lbf·ft f) N·s, lbf·s g) N/m 2 , lbf/ft 2 h) m 2 /s 2 ·K, ft 2 /s 2 ·R i) 1/K, 1/R j) N·m·s, lbf·ft·s 1.19 a) 6.89 kPa b) 0.264 gal 47.9 N·s/m 2 1.21 a) 0.0472 m 3 /s b) 0.0189 m 3 c) 29.1 m/s d) 2.19 x 10 4 m 2 1.23 101 gpm 1.25 SG = 13.6 v = 7.37 x 10 5 m 3 /kg γ E = 847 lbf/ft 3 , γ M = 144 lbf/ft 3 1.27 2.25 kgf/cm 2 1.29 c = 0.04 K 1/2 ·s/m 0 0 max 36 . 2 T p A m t = & (ft 2 , psi, R) 1.31 C D is dimensionless 1.33 c : N·s/m, lbf·s/ft k : N/m, lbf/ft f : N, lbf 1.35 H (m) = 0.457 3450·( Q (m 3 /s)) 2 1.37 ρ = 0.0765 ± 2.66 x 10 4 lbm/ft 3 (± 0.348%) 1.39 = 1130 ± 21.4 kg/m 3 SG = 1.13 ± 0.0214 1.41 = 930 ± 27.2 kg/m 3 1.43 t = 1, 5, 5 s Flow rate uncertainty = ± 5.0, 1.0, 1.0% 1.45 μ = 1.01 x 10 3 N·s/m 2 ± 0.609% 1.47 δ x = ± 0.158 mm 1.49 H = 57.7 ± 0.548 ft θ min = 31.4 o

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Introduction to Fluid Mechanics 7 th Edition Fox, Pritchard, & McDonald Answers to Selected Problems, Chapter 2 2.1 1) 1D, Unsteady 2) 1D, Steady 3) 2D, Unsteady 4) 2D, Unsteady 5) 1D, Unsteady 6) 3D, Steady 7) 2D, Unsteady 8) 3D, Steady 2.3 Streamlines: x c y = 2.5 Streamlines: t a b x c y = 2.7 Streamlines: + = c x a b y 2 1 2.9 Streamlines: y = 3 x Δ t = 0.75 s 2.11 Streamlines: x 2 + y 2 = c 2.13 Pathlines: y = 2/ x Streamlines: y = 2/ x 2.15 a K π ω 2 = 2.17 Pathlines: t C e y y 0 = , () 2 2 1 0 At t B e x x + = Streamlines: t x y 5 . 0 1 1 + = 2.19 Pathlines: t b e y y = 0 , 2 2 1 0 at e x x = Streamlines: t a b Cx y = 2.21 Pathlines: 1 4 + = t y , 2 05 . 0 3 t e x = Streamlines: + = 3 ln 40 1 x t y 2.23 Streamlines: () ( ) ( ) 0 0 0 sin t t t v t y = , ( ) ( ) 0 0 0 t t u t x = 2.25 Streaklines: τ = t e y , 2 2 1 . 0 + = t t e x Streamlines: t x y 2 . 0 1 1 + = 2.29 Streamlines: 4 4 2 + = x y (4 m, 8 m) (5 m, 10.25 m) 2.31 (2.8 m, 5 m) (3 m, 3 m) 2.33 T S T b + = 1 2 3 ν b ´ = 4.13 x 10 9 m 2 /s·K 3/2 S ´ = 110.4 K 2.35 yx = 4.56 N/m 2 2.39 a = 0.491 ft/s 2 2.41 L = 2.5 ft 2.43 t = 1.93 s 2.45 = t Md A e A Mgd V μ θ 1 sin V = 0.404 m/s = 1.08 N·s/m 2 2.47 F = 2.83 N 2.49 F = 0.0254 N 2.51 = 8.07 x 10 4 N·s/m 2 2.53 H R V a Mgr m 3 2 2 = = 0.0651 N·s/m 2
2.55 t = 4.00 s 2.57 = t C B e B A 1 ω max = 25.1 rpm t = 0.671 s 2.59 h r z μω τ θ = h R T 2 4 πμω = 2.61 Dilatant k = 0.0499 n = 1.21 μ = 0.191 N·s/m 2 , 0.195 N·s/m 2 2.63 Bingham plastic p = 0.652 N·s/m 2 2.65 a H R T 3 2 = b R T 2 4 = 2.69 () cos 1 sin + = R a R max = 79.2 N/m 2 2.75 a = 0 b = 2U c = U 2.77 V = 229 mph 2.79 M = 2.5 x trans = 0.327 m 2.81 SG = 0.9 γ = 8830 N/m3 Laminar flow 2.83 V = 667 km/hr

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Introduction to Fluid Mechanics 7 th Edition Fox, Pritchard, & McDonald Answers to Selected Problems, Chapter 3 3.1 M = 62.4 kg t = 22.3 mm 3.3 z = 9303 ft Δ z = 5337 ft 3.5 F = 45.6 N 3.9 Δ p = 972 Pa ρ = 991 kg/m 3 3.11 D = 0.477 in 3.13 Δ / 0 = 4.34% Δ p / p 0 = 2.15% 3.15 p = 0.217 psig 3.17 p = 6.39 kPa (gage) h = 39.3 mm 3.19 p = 128 kPa (gage) 3.21 Δ p = 59.5 Pa 3.23 H = 30 mm 3.25 SG = 0.900 3.27 Δ p = 1.64 psi 3.29 () [] 2 oil 1 D d p L + Δ = L = 27.2 mm 3.31 h = 1.11 in 3.33 θ = 11.1 o S = 5/SG 3.35 p atm = 14.4 psi Shorter column at higher temperature 3.37 Δ h = 38.1 mm Δ h = 67.8 mm 3.39 Δ h = 0.389 cm 3.43 Δ z = 587 ft Δ z = 3062 ft 3.45 p = 57.5 kPa p = 60.2 kPa 3.49 p A = 1.96 kPa p B = 8.64 kPa p C = 21.9 kPa p air = 11.3 kPa p air = 1.99 kPa 3.51 F A = 79,600 lbf 3.53 W = 68 kN 3.55 F R = 0.407 lbf 3.57 F R = 8.63 MN R = (8.34 MN, 14.4 MN) 3.59 F = 600 lbf 3.63 D = 8.66 ft 3.65 d = 2.66 m 3.67 SG = 0.542 3.69 F = 137 kN 3.71 F V = 7.62 kN x ´ F V = 3.76 kN·m F A = 5.71 kN 3.73 F V = gwR 2 π /4 x ´ = 4 R /3 3.75 F V = 1.05 x 10 6 N x ´ = 1.61 m 3.77
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .

### Page1 / 23

Answers Selected Problems - Introduction to Fluid Mechanics...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online