# Capitulo 2 felder principio de los procesos quimicos

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CHAPTER TWO
2.1 (a)
= 18144 × 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms

(c)

554 m 4 1d 1h d ⋅ kg 24 h 60 min

1 kg 108 cm 4 = 3.85 × 10 4 cm 4 / min⋅ g 1000 g 1 m 4

2.2 (a) (b) (c)

1 m 1 h 760 mi = 340 m / s h 0.0006214 mi 3600 s 921 kg 2.20462 lb m m3 1 kg 1 m3 = 57.5 lb m / ft 3 35.3145ft 3 1.34 × 10 -3 hp = 119.93 hp ⇒ 120 hp 1 J/s

5.37 × 10 3 kJ 1 min 1000 J min 60 s 1 kJ

2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12) 3 in 3 1 ball n balls = . = 518 × 10 6 ≈ 5 million balls ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, dependingon the assumptions made. 2.4 4.3 light yr 365 d 24 h
1 yr 1d 3600 s 1.86 × 10 5 mi 1 h 1 s 3.2808 ft 0.0006214 mi 1 step = 7 × 1016 steps 2 ft

2.5 Distance from the earth to the moon = 238857 miles
238857 mi 1 m 1 report 0.001 m 0.0006214 mi = 4 × 1011 reports

2.6

19 km 1000 m 0.0006214 mi 1000 L = 44.7 mi / gal 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. TotalCost
American

= \$14,500 +

\$1.25 1 gal gal 28 mi

x (mi)

= 14,500 + 0.04464 x

Total Cost

European

= \$21,700 +

\$1.25 1 gal x (mi) = 21,700 + 0.02796 x gal 44.7 mi

Equate the two costs ⇒ x = 4.3 × 10 5 miles

2-1

2.7
5320 imp. gal plane ⋅ h = 1.188 × 105 14 h 365 d 1 d 1 yr 106 cm3 220.83 imp. gal 0.965 g 1 cm
3

1 kg 1000 g

1 tonne 1000 kg

tonne keroseneplane ⋅ yr

4.02 × 109 tonne crude oil 1 tonne kerosene yr
5

plane ⋅ yr

7 tonne crude oil 1.188 × 10 tonne kerosene

= 4834 planes ⇒ 5000 planes

2.8 (a) (b) (c)

25.0 lb m 25 N

32.1714 ft / s 2 1

1

lb f

32.1714 lb m ⋅ ft / s 2 1 kg ⋅ m/s 2 1N 9.8066 m/s 2

= 25.0 lb f

= 2.5493 kg ⇒ 2.5 kg

10 ton

1 lb m 5 × 10
-4

1000 g ton 2.20462 lb m 85.3 lb m 1 ft 3 1 m311.5 kg

980.66 cm / s 2

1 dyne 1 g ⋅ cm / s 2

= 9 × 10 9 dynes

2.9

50 × 15 × 2 m 3

35.3145 ft 3 1 m3 1 kg

32.174 ft 1 lb f = 4.5 × 10 6 lb f 2 2 32.174 lb m / ft ⋅ s 1 s

2.10 2.11

500 lb m

2.20462 lb m

≈ 5 × 10 2

FG 1 IJ FG 1 IJ ≈ 25 m H 2 K H 10K

3

(a) mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2

(30 cm − 14.1 cm)(100 g / cm3 ) . ρc = = = 0.53 g / cm 3 H 30 cm ρ H (30 cm)(0.53 g / cm 3 ) (b) ρ f = c = = 171 g / cm 3 . (30 cm - 20.7 cm) h

ρfh

ρc H ρf h

2.12

Vs =

πR 2 H
3

; Vf =

πR 2 H
3

πr 2 h
3
2

;

⇒ Vf =

πR H
2

ρ f V f = ρ sVs
⇒ ρ f = ρs

FG IJ = πR FG H − h IJ − 3 3 H HK 3 H H K πR F h I ⇒ρ GH H − H JK = ρ πR3 H 3
πh Rh
2 2 3 2 3 2 f 2 s

R r R = ⇒r = h H h H
Hh r

ρs R
3

ρf

H H− h3 H2

= ρs

H3 = ρs H 3 − h3

1−

FG h IJ H HK

1

2-2

2.13

Say h( m) = depth of liquid

y y= 1 ⇒ A(m 2 ) h 1m y= –1
1 − y 2 dy

dA y= – –1+h y= 1 h xx x = 1 y2 – dA
−1

1− y

2

dA = dy ⋅

− 1− y 2

dx = 2 1 − y dy ⇒ A m
2

( )=2 ∫
2

−1+ h

Table of integrals or trigonometric substitution
h −1

π 2 A m 2 = y1 − y 2 + sin −1 y ⎤ = ( h − 1) 1 − ( h − 1) + sin −1 ( h − 1) + ⎥ −1 ⎦ 2

( )

W N =

b g

4 m × A( m 2 ) 0.879 g 10 6 cm 2 cm
3

1 kg 10 g
3

9.81 N kg
g g0

1m

3

= 3.45 × 10 4 A

E Substitute for A L W b N g = 3.45 × 10 Mbh − 1g 1 − bh − 1g N
4

2

+ sin −1 h − 1 +

b g π OPQ 2

2.14

1 lb f = 1 slug ⋅ ft / s 2 = 32.174 lb m ⋅ ft / s 2 ⇒ 1 slug = 32.174 lb m1 1 poundal = 1 lb m ⋅ ft / s 2 = lb f 32.174 (a) (i) On the earth: 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 5.63 × 10 3 poundals 2 s 1 lb m ⋅ ft / s 2 (ii) On the moon 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 938 poundals 2 6 s 1 lb m ⋅ ft / s 2 (b) F = ma ⇒ a = F / m = = 0.135 m / s 2 355 poundals 25.0 slugs 1 lb m ⋅ ft...