Chapter01

Mechanics of Fluids

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1 CHAPTER 1 Basic Considerations 1.1 Conservation of mass — Mass — density Newton’s second law — Momentum — velocity The first law of thermodynamics — internal energy — temperature 1.2 a) density = mass/volume = M L / 3 b) pressure = force/area = F L ML T L M LT / / / 2 2 2 2 = = c) power = force × velocity = F L T ML T L T ML T × = × = / / / / 2 2 3 d) energy = force × distance = ML T L ML T / / 2 2 2 × = e) mass flux = r AV = M/L 3 × L 2 × L/T = M/T f) flow rate = AV = L 2 × L/T = L 3 /T 1.3 a) density = M L FT L L FT L 3 2 3 2 4 / / = b) pressure = F/L 2 c) power = F × velocity = F × L/T = FL/T d) energy = F × L = FL e) mass flux = M T FT L T FT L = = 2 / / f) flow rate = AV = L 2 × L/T = L 3 1.4 ( C ) m = F/a or kg = N/m/s 2 = N . s 2 /m. 1.5 ( B ) [ m ] = [ t/ du/dy ] = ( F / L 2 )/( L / T )/ L = F . T / L 2 . 1.6 a) L = [ C ] T 2 . [ C ] = L/T 2 b) F = [ C ] M. [ C ] = F/M = ML/T 2 M = L/T 2 c) L 3 /T = [ C ] L 2 L 2/3 . [ C ] = L T L L L T 3 2 2 3 1 3 / / / = Note: the slope S 0 has no dimensions. 1.7 a) m = [ C ] s 2 . [ C ] = m/s 2 b) N = [ C ] kg. [ C ] = N/kg = kg m/s 2 kg = m/s 2 c) m 3 /s = [ C ] m 2 m 2/3 . [ C ] = m 3 /s m 2 m 2/3 = m 1/3 1.8 a) pressure: N/m 2 = kg m/s 2 /m 2 = kg/m s 2 b) energy: N m = kg m/s 2 × m = kg m 2 c) power: N m/s = kg m 2 3 d) viscosity: N s/m 2 = kg m s s 1 m kg / m s 2 2 =
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2 e) heat flux: J/s = N m s kg m s m s kg m / s 2 2 3 = = f) specific heat: J kg K N m kg K kg m s m kg K m / K s 2 2 2 = = = 1.9 kg m s m s m 2 + + = c k f . Since all terms must have the same dimensions (units) we require: [ c ] = kg/s, [ k ] = kg/s 2 = N s / m s N / m, 2 2 = [ f ] = kg m / s N. 2 = Note: we could express the units on c as [ c ] = kg / s N s / m s N s / m 2 = ⋅ = 1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm 3 e) 1.2 cm 2 f) 76 mm 3 1.11 a) 1.25 × 10 8 N b) 3.21 × 10 - 5 s c) 6.7 × 10 8 Pa d) 5.6 × 10 -12 m 3 e) 5.2 × 10 - 2 m 2 f) 7.8 × 10 9 m 3 1.12 ( A ) 89 2.3 6 1 0 23. 6 1 0 23.6 nPa. -- ×=×= 1.13 2 22 0.06854 0.22 5 0.738 0.0019 4 3.281 mm dd l rr == × where m is in slugs, r in slug/ft 3 and d in feet. We used the conversions in the front cover. 1.14 a) 20 cm/hr = 5 20 /360 0 5.55 5 1 0 m/s 100 - 5 20 0 5.55 5 1 0 100 - b) 2000 rev/min = 2000 2 × π /60 = 209.4 rad/s c) 50 Hp = 50 × 745.7 = 37 285 W d) 100 ft 3 /min = 100 × 0.02832/60 = 0.0472 m 3 /s e) 2000 kN/cm 2 = 2 × 10 6 N/cm 2 × 100 2 cm 2 /m 2 = 2 × 10 10 N/m 2 f) 4 slug/min = 4 × 14.59/60 = 0.9727 kg/s g) 500 g/L = 500 × 10 - 3 kg/10 -3 m 3 = 500 kg/m 3 h) 500 kWh = 500 × 1000 × 3600 = 1.8 × 10 9 J 1.15 a) F = ma = 10 × 40 = 400 N . b) F - W = ma. F = 10 × 40 + 10 × 9.81 = 498.1 N . c) F - W sin 30 ° = ma. F = 10 × 40 + 9.81 × 0.5 = 449 N . 1.16 ( C ) The mass is the same on earth and the moon: [4( 8 ) ] 3 2. du dr t m = 1.17 The mass is the same on the earth and the moon:
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3 m = 60 32 2 1863 . . . = W moon = 1.863 × 5.4 = 10.06 lb 1.18 ( C ) shear si n 4200sin3 0 2100 N. FF q = == o shear 4 2100 = 84 kPa 25 0 10 F A t - × 1.19 a) l r = = × × × = × - - - . . . . ( . ) .43 225 225 4 8 10 184 3 7 10 10 2 26 10 2 6 m d m or 0.00043 mm b) l r = = × × × = × - - - . . . . ( . ) . 225 225 4 8 10 00103 3 7 10 7 7 10 2 2 5 m d m or 0.077 mm c) 26 2 1 02 4. 8 10 .22 5 .22 5 .0039m .0000 2 (3. 7 1 0) m d l r - - × = ×× or 3.9 mm 1.20 Use the values from Table B.3 in the Appendix. a) 52.3 + 101.3 = 153.6 kPa . b) 52.3 + 89.85 = 142.2 kPa . c) 52.3 + 54.4 = 106.7 kPa (use a straight-line interpolation).
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Chapter01 - CHAPTER 1 Basic Considerations 1.1 Conservation...

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