# Primer capitulo mott

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• Publicado : 4 de agosto de 2010

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CHAPTER ONE THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS
Conversion factors 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1250 mm(1 m/103 mm) = 1.25 m 1600 mm2[1 m2/(103 mm)2] = 1.6 × 10−3 m2 3.65 × 103 mm3[1 m3/(103 mm)3] = 3.65 × 10−6 m3 2.05 m2[(103 mm)2/m2] = 2.05 × 106 mm2 0.391 m3[(103 mm)3/m3] = 391 × 106 mm3 55.0 gal(0.00379 m3/gal) = 0.208 m3
80 km 103m 1h × × = 22.2 m/s h km 3600 s

25.3 ft(0.3048 m/ft) = 7.71 m 1.86 mi(1.609 km/mi)(103 m/km) = 2993 m 8.65 in(25.4 mm/in) = 220 mm 2580 ft(0.3048 m/ft) = 786 m 480 ft3(0.0283 m3/ft3) = 13.6 m3 7390 cm3[1 m3/(100 cm)3] = 7.39 × 10−3 m3 6.35 L(1 m3/1000 L) = 6.35 × 10−3 m3 6.0 ft/s(0.3048 m/ft) = 1.83 m/s
2500 ft 3 0.0283 m3 1 min × × = 1.18 m3/s min ft 3 60 s

Consistent units in an equation1.17

υ= =

s t

0.50 km 103 m × = 47.2 m/s 10.6 s km

The Nature of Fluids

1

1.18

υ= = υ= = υ= =
2s t
t=
2

s t

1.50 km 3600 s × = 1038 km/h 5.2 s h 1000 ft 1 mi 3600 s × × = 48.7 mi/h 14 s 5280 ft h 1.0 mi 3600 s × = 632 mi/h 5.7 s h (2)(3.2 km) 103 m 1 min 2 = 8.05 × 10−2 m/s2 × × 2 2 km (4.7 min ) (60 s )

1.19

s t

1.20

s t

1.21

a=

=

1.22

2s(2)(13 m) = = 1.63 s a 9.81 m/ s 2 2s t
2

1.23

a=

=

(2)(3.2 km) 103 m 1 ft 1 min 2 ft × × × = 0.264 2 2 2 s km 0.3048 m (60 s ) (4.7 min )

1.24

t=

2s (2)(53 in) 1 ft = × = 0.524 s 32.2 ft/ s 2 12 in a mυ 2 (15 kg)(1.2 m/s ) kg ⋅ m 2 = = 10.8 = 10.8 N ⋅ m 2 2 2 s
2

1.25

KE =

1.26

1 h2 kg ⋅ m 2 mυ 2 (3600 kg) ⎛ 16 km ⎞ (103 m ) ×⎜ × = 35.6 × 103 KE = = ⎟ × 2 2 s22 2 (3600 s ) ⎝ h ⎠ km KE = 35.6 kN ⋅ m
2

2

1.27

mυ 2 75 kg ⎛ 6.85 m ⎞ kg ⋅ m 2 = 1.76 kN ⋅ m KE = = ×⎜ = 1.76 × 103 ⎟ 2 2 2 ⎝ s ⎠ s m= m= 2( KE ) = (2)(38.6 N ⋅ m) ⎛ h 1 km 2 ⎞ 1 kg ⋅ m (3600 s ) ×⎜ × 2 × × ⎟ 2 1 (103 m)2 ⎝ 31.5 km ⎠ s ⋅N h
2 2

2

1.28

υ2

(2)(38.6)(3600) 2 kg = 1.008 kg 2 (31.5) (103 )2 2( KE ) = (2)(94.6 m N ⋅ m) 10−3 N 1 kg ⋅ m 103 g = 37.4 g × × 2 × 2 mNs ⋅N kg (2.25 m/s )

1.29

m=

υ2

1.30

υ=

2( KE ) 2(15 N ⋅ m) 1 kg ⋅ m/s 2 = 1.58 m/s = × m 12 kg N

2

Chapter 1

1.31

υ=

2( KE ) 2(212 m N ⋅ m) 10 −3 N 103 g 1 kg ⋅ m = 1.56 m/s × × × 2 = 175 g mN kg s ⋅N m mυ 2 (1 slug)(4 ft/s ) 2 1 lb ⋅ s 2 /ft = 8.00 lb ⋅ ft = × 2 2 slug 1 h2 (5280 ft ) 2 mυ 2 wυ 2 (8000 lb)(10 mi ) 2 × × = = 2 2 2 g (2)(32.2 ft/s 2 )(h ) 2 (3600 s) 2 mi

1.32

KE =

1.33

KE =

(8000)(10) 2(5280) 2 KE = lb ⋅ ft = 26700 lb ⋅ ft 2 (2)(32.2)(3600)

1.34

KE =

mυ 2 wυ 2 (150 lb)(20 ft/s ) 2 = 932 lb ⋅ ft = = 2 2g (2)(32.2 ft/ s 2 ) = 2(15 lb ⋅ ft) lb ⋅ s 2 = 6.20 slugs = 6.20 2 ft (2.2 ft/ s 2 ) 2(32.2 ft)(38.6 lb ⋅ ft)(h 2 ) 1 mi 2 (3600 s ) 2 × × 2 2 2 2 (5280 ft ) h s (19.5 mi )

1.35

m=

2( KE )

υ2

1.36

w=w=

2 g ( KE )

υ2

=

(2)(32.2)(38.6)(3600) 2 lb = 3.04 lb 2 2 (19.5) (5280) 2 g ( KE ) 2(32.2 ft/ s 2 )(10 lb ⋅ ft) = 4.63 ft/s = 30 lb w 2 g ( KE ) 2(32.2 ft/ s 2 )(30 oz ⋅ in) 1 ft = 5.18 ft/s × = 6.0 oz 12 in w 39 runs 9 innings = 2.49 runs/game × 141 innings game

1.37

υ=

1.38

υ=

1.39

ERA =

1.40

3.12 runs 1 game × × 150 innings = 52 runs game 9 innings 40 runs× 1 game 9 innings = 129 innings × 2.79 runs game

1.41

1.42

ERA =

49 runs 9 innings = 3.59 runs/game × 123 innings game

The Nature of Fluids

3

The definition of pressure 1.43 1.44 1.45 p = F/A = 2500 lb/[π(3.00 in)2/4] = 354 lb/in2 = 354 psi p = F/A = 8700 lb/[π(1.50 in)2/4] = 4923 psi
p= 12.0 kN 103 N (103 mm ) 2 N F = × × = 2.72 × 106 2 = 2.72 MPa 2 2 A π (75 mm ) / 4 kNm m F 38.8 × 103 N (103 mm ) 2 N = × = 30.9 × 106 2 = 30.9 MPa 2 2 A π (40 mm ) / 4 m m F 6000 lb = 119 psi = A π (8.0 in ) 2 / 4 F 18000 lb = 3667 psi = A π (2.50 in ) 2 / 4 20.5 × 106 N π (50 mm ) 2 1 m2 = 40.25 kN × × 2 4 (103 mm) 2 m

1.46

p=

1.47

p=

1.48

p=

1.49

F = pA =

1.50

F = pA = (6000 lb/ in 2 ) π [2.00 in ]2 / 4 p= D=

(

)

= 18850 lb

1.51

F...