Capitulo 2 Solucionado Balance De Masa Y Energia Felder

CHAPTER TWO
3 wk

7d

24 h 3600 s 1000 ms

= 18144 × 10 9 ms
.
1 wk 1 d
1h 1
s
38.1 ft / s 0.0006214 mi 3600 s
(b)
= 25.98 mi / h ⇒ 26.0 mi / h
3.2808
ft
1h

2.1 (a)

(c)

2.2 (a)

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

1
m
1h
760 mi
= 340 m / s
h 0.0006214 mi 3600 s
1 m3
= 57.5 lb m / ft 3
35.3145 ft 3(b)

921 kg 2.20462 lb m
m3
1
kg

(c)

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

1.34 × 10 -3 hp
= 119.93 hp ⇒ 120 hp
1
J/s

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
Theestimate could vary by an order of magnitude or more, depending on the assumptions made.
2.4 4.3 light yr 365 d 24 h
1 yr

3600 s 1.86 × 10 5 mi

1d

1h

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

0.0006214 mi

1 report
0.001 m

= 4 × 1011 reports

2.6

19 km 1000 m 0.0006214 mi1000 L
= 44.7 mi / gal
1L
1 km
1
m 264.17 gal
Calculate the total cost to travel x miles.
Total Cost

Total Cost

American

European

= $14,500 +

= $21,700 +

$1.25 1 gal
gal 28 mi

x (mi)

= 14,500 + 0.04464 x

$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

106 cm3

14 h 365 dplane ⋅ h

1d

1 yr

0.965 g

220.83 imp. gal

1

cm

1 kg
3

1 tonne

1000 g

1000 kg

tonne kerosene
plane ⋅ yr

= 1.188 × 105

4.02 × 109 tonne crude oil 1 tonne kerosene

plane ⋅ yr

7 tonne crude oil 1.188 × 10 tonne kerosene
5

yr

= 4834 planes ⇒ 5000 planes

2.8 (a)
(b)
(c)

2.9

2.10
2.11

32.1714 ft / s 2

25.0 lb m

1

lb f32.1714 lb m ⋅ ft / s 2
25 N

1 kg ⋅ m/s 2

1
9.8066 m/s 2

10 ton

1N

1 lb m
5 × 10

-4

50 × 15 × 2 m 3

= 2.5493 kg ⇒ 2.5 kg

980.66 cm / s 2

1000 g
ton

85.3 lb m
1 ft 3

1

1 m3

kg

2.20462 lb m

1 dyne
1 g ⋅ cm / s 2

2.20462 lb m

35.3145 ft 3
1 m3

500 lb m

= 25.0 lb f

11.5 kg

= 9 × 10 9 dynes

32.174 ft
1 lb f
= 4.5 × 10 6 lb f
22
1s
32.174 lb m / ft ⋅ s

≈ 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

ρc

ρfh

(30 cm − 14.1 cm)(100 g / cm 3 )
.
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

2.12

Vs =

πR 2 H
3

⇒ Vf =

; Vf =

3

−πr 2 h
3

;

πh Rh

2

2

f

H
H−

h3
H2

3

2

= ρs

H3
= ρs
H 3 − h3

h

3

r

2

2

ρf
h

Rr
R
= ⇒r = h
Hh
H

FG IJ = πR FG H − h IJ
−
3
3 H HK
3H
HK
πR F
hI
⇒ρ
GH H − H JK = ρ πR3 H
3

πR H
2

ρ f V f = ρ sVs
⇒ ρ f = ρs

πR 2 H

H

H

2

ρs

s

R

1

1−

FG h IJ
H HK

2-2

3

ρf

2.13

Say h( m) =depth of liquid

y
y= 1
dA
y–
–1+
y==1 h h
xx

⇒
A(m 2 )

1− y

dA = dy ⋅

∫

h

x = 1 y2
–
y= –1

dA

2

−1+ h

( )=2 ∫

dx = 2 1 − y dy ⇒ A m
2

2

1 − y 2 dy

−1

− 1− y 2

⇓

1m

Table of integrals or trigonometric substitution

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

()

bg

WN =4 m × A( m 2 ) 0.879 g 10 6 cm 2
cm

3

1m

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

2.14

3

1 kg

9.81 N

3

kg

10 g

= 3.45 × 10 4 A

g g0

2

b g π OPQ
2

+ sin −1 h − 1 +

1 lb f = 1 slug ⋅ ft / s 2 = 32.174 lb m ⋅ ft / s 2 ⇒ 1 slug = 32.174 lb m
1
1 poundal = 1 lb m ⋅ ft / s 2 =
lb f
32.174
(a) (i) On the earth:
175...