Solucionario Libro Merle Potter Cap1 .Mecanica De Fluidos

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 a) density = mass/volume = M / L3 b) pressure = force/area = F / L2 = ML / T 2 L2 = M / LT 2 c) power = force × velocity = F × L / T = ML / T 2 × L / T = ML2 / T 3 d) energy = force × distance = ML / T 2 × L = ML2 /T 2 e) mass flux = ρAV = M/L 3 × L2 × L/T = M/T f) flow rate = AV = L2 × L/T = L3 /T M FT 2 / L a) density = 3 = FT 2 / L4 3 L L b) pressure = F/L 2 c) power = F × velocity = F × L/T = FL/T d) energy = F × L = FL M FT 2 / L e) mass flux = = = FT / L T T f) flow rate = AV = L2 × L/T = L3 /T (C) (B) m = F/a or kg = N/m/s2 = N.s2 /m. [µ] = [τ/du/dy] = (F/L2 )/(L/T)/L = F.T/L2 .

1.2

1.3

1.41.5 1.6

a) L = [C] T2 . ∴[C] = L/T 2 b) F = [C]M. ∴[C] = F/M = ML/T 2 M = L/T 2 c) L3 /T = [C] L2 L2/3 . ∴[C] = L3 / T ⋅ L2 ⋅ L2 / 3 = L1 / 3 T Note: the slope S0 has no dimensions. a) m = [C] s2 . ∴[C] = m/s2 b) N = [C] kg. ∴[C] = N/kg = kg ⋅ m/s2 ⋅ kg = m/s2 3 2 2/3 c) m /s = [C] m m . ∴[C] = m3 /s⋅m2 ⋅ m2/3 = m1/3 /s a) pressure: N/m2 = kg ⋅ m/s2 /m2 = kg/m⋅ s2 b) energy: N⋅ m = kg ⋅ m/s2 × m= kg⋅ m2 /s2 c) power: N⋅ m/s = kg ⋅ m2 /s3 kg ⋅ m 1 d) viscosity: N⋅ s/m2 = 2 ⋅ s 2 = kg / m ⋅ s s m

1.7

1.8

1

N ⋅ m kg ⋅ m m = ⋅ = kg ⋅ m 2 / s 3 s s2 s J N ⋅ m kg ⋅ m m f) specific heat: = = 2 ⋅ = m 2 / K ⋅s 2 kg ⋅ K kg ⋅ K s kg ⋅ K e) heat flux: J/s = 1.9 m m + c + km = f. Since all terms must have the same dimensions (units) we 2 s s require: [c] = kg/s, [k] = kg/s2 = N ⋅ s 2 /m ⋅ s 2 = N / m, [f] = kg ⋅ m / s 2 = N. Note: we could express the units on c as [c] = kg / s = N ⋅ s 2 / m ⋅ s = N ⋅ s / m kg a) 250 kN e) 1.2 cm2 a) 1.25 × 108 N d) 5.6 × 10−12 m3 (A) b) 572 GPa f) 76 mm3 c) 42 nPa d) 17.6 cm3 c) 6.7 × 108 Pa f) 7.8 × 109 m3

1.10

1.11

b) 3.21 × 10−5 s e) 5.2 × 10−2 m2

1.12 1.13

2.36 ×10 −8 = 23.6 ×10 −9 = 23.6 nPa. 0.06854m
2 2

0.00194 ρ ×3.281 d ρd 2 where m is in slugs, ρ in slug/ft3 and d in feet. We used the conversions in the front cover. 20 20 /3600 = 5.555 ×10 −5 m/s /3600 = 5.555 ×10 −5 m/s 100 100 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 m3 /s e) 2000 kN/cm2 = 2 × 106 N/cm2 × 1002 cm2 /m2 = 2 × 1010 N/m2 f) 4 slug/min = 4 × 14.59/60 = 0.9727kg/s g) 500 g/L = 500 × 10−3 kg/10−3 m3 = 500 kg/m3 h) 500 kWh = 500 × 1000 × 3600 = 1.8 × 109 J a) 20 cm/hr = 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. (C) The mass is the same on earth and the moon: τ = µ du = µ[4(8 r )] = 32µr. dr

λ = 0.225

= 0.738

m

1.14

1.15

1.16 1.17

The massis the same on the earth and the moon:

2

m= 1.18 (C)

60 = 1.863. ∴ Wmoon = 1.863× 5.4 = 10.06 lb 32.2

1.19

Fshear = F sinθ = 4200sin30o = 2100 N. F 2100 τ= shear = = 84 kPa A 250 × 10−4 m 4.8 × 10 −26 a) λ =.225 2 =.225 =.43 × 10 −6 m or 0.00043 mm −10 2 ρd .184 × (3.7 × 10 ) m 4.8 × 10 − 26 b) λ =.225 2 =.225 = 7.7 × 10 −5 m or 0.077 mm −10 2 ρd .00103 × (3.7 × 10 ) c) λ = .225 mρd 2 = .225 4.8 ×10−26 .00002 × (3.7 ×10−10 ) 2 = .0039m 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). d) 52.3 + 26.49 = 78.8 kPa. e) 52.3 + 1.196 = 53.5 kPa. a) 101 − 31 = 70 kPa abs. c) 14.7 − 31 × 760 = 527 mm of Hg abs. 101 31 d) 34 − × 34 = 23.6 ftof H2 O abs. 101 b) 760 −

1.21

31 × 14.7 = 10.2 psia. 101 31 e) 30 − × 30 = 20.8 in. of Hg abs. 101 1.22

p = po e−gz/RT = 101 e−9.81 × 4000/287 × (15 + 273) = 62.8 kPa From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is 62.8 − 61.6 % error = × 100 = 1.95 %. 61.6 22,560 − 20,000 (785 - 973) = 877 psf 25,000 − 20,000 22,560 − 20,000 T = −12.3 + (−30.1 + 12.3) = −21.4°F 25,000 −...