solução   princípios elementares dos processos químicos 3ed   felder & rousseau
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solução princípios elementares dos processos químicos 3ed felder & rousseau


DisciplinaPrincípios de Processos Químicos201 materiais2.057 seguidores
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2- 1
CHAPTER TWO 
 
2.1 (a) 
3 24 3600
1
18144 109
 wk 7 d h s 1000 ms
1 wk 1 d h 1 s
 ms= ×. 
 (b) 
38 3600
25 98 26 0
.
. .
1 ft / s 0.0006214 mi s
 3.2808 ft 1 h
 mi / h mi / h= \u21d2 
(c) 
554 1 1
1000 g
3
 m 1 d h kg 10 cm
 d kg 24 h 60 min 1 m
85 10 cm g
4 8 4
4
4 4
\u22c5 = × \u22c5. / min 
 
2.2 (a) 
760 mi
3600
340
 1 m 1 h
 h 0.0006214 mi s
 m / s= 
 (b) 
921 kg
35.3145 ft
57 5
2.20462 lb 1 m
 m 1 kg
 lb / ftm
3
3 3 m
3= . 
 (c) 
5 37 10 1000 J 1
1
119 93 120
3.
.
× × = \u21d2 kJ 1 min .34 10 hp
 min 60 s 1 kJ J / s
 hp hp
-3
 
 
2.3 Assume that a golf ball occupies the space equivalent to a 2 2 2in in in × × cube. For a 
classroom with dimensions 40 40 15 ft ft ft× × : 
 nballs
3 3
6 ft (12) in 1 ball
ft in
10 5 million balls= × × = × \u224840 40 15
2
518
3
3 3 3 . 
The estimate could vary by an order of magnitude or more, depending on the assumptions made. 
 
2.4 4 3 24 3600 s
1 0 0006214
.
.
 light yr 365 d h 1.86 10 mi 3.2808 ft 1 step
 1 yr 1 d h 1 s mi 2 ft
7 10 steps5 16× = × 
 
2.5 Distance from the earth to the moon = 238857 miles 
 
238857 mi 1
4 1011
 1 m report
0.0006214 mi 0.001 m
 reports= × 
 
2.6 
 
19 0 0006214 1000
264 17
44 7
500
25 1
14 500 0 04464
700
25 1
21 700 0 02796
 km 1000 m mi L
1 L 1 km 1 m gal
 mi / gal
Calculate the total cost to travel miles.
Total Cost 
 gal (mi)
 gal 28 mi
Total Cost 
 gal (mi)
 gal 44.7 mi
 
Equate the two costs 4.3 10 miles 
American
European
5
.
.
.
$14,
$1.
, .
$21,
$1.
, .
=
= + = +
= + = +
\u21d2 = ×
x
x
x
x
x
x
 
 
2- 2
2.7 
 
6 3
3
5
5320 imp. gal 14 h 365 d 10 cm 0.965 g 1 kg 1 tonne
 plane h 1 d 1 yr 220.83 imp. gal 1 cm 1000 g 1000 kg
tonne kerosene 1.188 10 
plane yr
\u22c5
= × \u22c5
 
 
9
5
4.02 10 tonne crude oil 1 tonne kerosene plane yr
 yr 7 tonne crude oil 1.188 10 tonne kerosene
 4834 planes 5000 planes
× \u22c5
×
= \u21d2
 
 
2.8 (a) 
25 0
25 0
.
.
 lb 32.1714 ft / s 1 lb 
32.1714 lb ft / s
 lbm
2
f
m
2 f\u22c5 = 
 (b) 
2
2
25 N 1 1 kg m/s
2.5493 kg 2.5 kg
9.8066 m/s 1 N 
\u22c5 = \u21d2 
 (c) 
10 1000 g 980.66 cm 1
9 109
 ton 1 lb / s dyne
5 10 ton 2.20462 lb 1 g cm / s
 dynesm
2
-4
m
2× \u22c5 = × 
 
2.9 
50 15 2 85 3 32 174
1
4 5 106
× ×
\u22c5 = ×
 m 35.3145 ft lb ft 1 lb
 1 m 1 ft s 32.174 lb ft s
 lb
3 3
m f
3 3 2
m
2 f
. .
/
. 
 
2.10 
500 lb
5 10 1
2
1
10
252m
3
m
3 1 kg 1 m
2.20462 lb 11.5 kg
 m\u2248 × FHG
I
KJ
F
HG
I
KJ \u2248 
 
2.11 
 
(a) 
m m V V h r H r
h
H
f f c c f c
c
f
displaced fluid cylinder
3
3 cm cm g / cm
30 cm
 g / cm
= \u21d2 = \u21d2 =
= = \u2212 =
\u3c1 \u3c1 \u3c1 \u3c0 \u3c1 \u3c0
\u3c1 \u3c1
2 2
30 14 1 100 0 53( . )( . ) .
 
(b) \u3c1 \u3c1f c Hh= = =
( )( . ) .30 0 53 171 cm g / cm
(30 cm - 20.7 cm)
 g / cm
3
3 
H
h
\u3c1f
\u3c1c
 
 
2.12 
 V
R H V R H r h R
H
r
h
r R
H
h
V R H h Rh
H
R H h
H
V V R H h
H
R H
H
H h
H
H
H h h
H
s f
f
f f s s f s
f s s s
= = \u2212 = \u21d2 =
\u21d2 = \u2212 FHG
I
KJ = \u2212
F
HG
I
KJ
= \u21d2 \u2212FHG
I
KJ =
\u21d2 =
\u2212
= \u2212 = \u2212 FHG
I
KJ
\u3c0 \u3c0 \u3c0
\u3c0 \u3c0 \u3c0
\u3c1 \u3c1 \u3c1 \u3c0 \u3c1 \u3c0
\u3c1 \u3c1 \u3c1 \u3c1
2 2 2
2 2 2 3
2
2 3
2
2
3
2
3
3 3 3
3 3 3
3 3 3
3 3
1
1
; ; 
 
 
\u3c1f\u3c1s
R
r
h
H
 
2- 3
 
2.13 Say h m( ) = depth of liquid 
 
A ( ) m 2 h 
1 m 
\u21d2 
y 
x 
y = 1 
y = 1 \u2013 h 
x = 1 \u2013 y 2 
d A 
 
 
( )
( ) ( ) ( ) ( )
2
2
1 1
2 2 2
11
1 22 2 1 1
1
Table of integrals or trigonometric substitution
2 1 2 1
m 1 sin 1 1 1 sin 1
2
 
\u3c0
\u2212 \u2212 +
\u2212\u2212 \u2212
\u2212\u2212 \u2212
\u2212
= \u22c5 = \u2212 \u21d2 = \u2212
\u23a4= \u2212 + = \u2212 \u2212 \u2212 + \u2212 +\u23a5\u23a6
\u21d3
\u222b \u222b
y h
y
h
dA dy dx y dy A m y dy
A y y y h h h
 
W N
A
A
A
g g
b g =
×
= ×
E
4 0 879 1
1 10 3 45 10
2
3 4
0
m m g 10 cm kg 9.81 N
cm m g kg
 Substitute for 
2 6
3 3
( ) .
.N 
W h h hNb g b g b g b g= × \u2212 \u2212 \u2212 + \u2212 +LNM
O
QP
\u22123 45 10 1 1 1 1
2
4 2 1. sin \u3c0 
 
2.14 1 1 32 174 1
1 1
32 174
 lb slug ft / s lb ft / s slug = 32.174 lb
 poundal = 1 lb ft / s lb
f
2
m
2
m
m
2
f
= \u22c5 = \u22c5 \u21d2
\u22c5 =
.
.
 
 (a) (i) On the earth: 
 
M
W
= =
= \u22c5 = ×
175 lb 1
5 44
175 1
1
m
m
m
2
m
2
3
 slug
32.174 lb
 slugs
 lb 32.174 ft poundal
 s lb ft / s
5.63 10 poundals
.
 
 (ii) On the moon 
 
M
W
= =
= \u22c5 =
175 lb 1
5 44
175 1
1
m
m
m
2
m
2
 slug
32.174 lb
 slugs
 lb 32.174 ft poundal
 6 s lb ft / s
938 poundals
.
 
 
 
( ) /b F ma a F m= \u21d2 = = \u22c5
=
355 pound 1 1als lb ft / s 1 slug m
 25.0 slugs 1 poundal 32.174 lb 3.2808 ft
 0.135 m / s
m
2
m
2
 
 
y= \u20131 
y= \u20131+h 
x
dA
2- 4
2.15 (a) F ma= \u21d2 FHG
I
KJ = \u22c5
\u21d2 \u22c5
1 1
6
5 3623
1
 fern = (1 bung)(32.174 ft / s bung ft / s
 fern
5.3623 bung ft / s
2 2
2
) .
 
 (b) On the moon: 
3 bung 32.174 ft 1 fern
 6 s 5.3623 bung ft / s
 fern
On the earth: = 18 fern
2 2W
W
= \u22c5 =
=
3
3 32 174 5 3623( )( . ) / .
 
 
2.16 (a) \u2248 =
=
( )( )
( . )( . )
3 9 27
2 7 8 632 23
 (b) 
4
5
4 6
4.0 10 1 10
40
(3.600 10 ) / 45 8.0 10
\u2212
\u2212
\u2212 \u2212
×\u2248 \u2248 ×
× = ×
 
 (c) \u2248 + =
+ =
2 125 127
2 365 125 2 127 5. . .
 (d) \u2248 × \u2212 × \u2248 × \u2248 ×
× \u2212 × = ×
50 10 1 10 49 10 5 10
4 753 10 9 10 5 10
3 3 3 4
4 2 4.
 
2.17 
1 5 4
2 3
6
3
(7 10 )(3 10 )(6)(5 10 ) 42 10 4 10
(3)(5 10 )
3812.5 3810 3.81 10exact
R
R
\u2212× × ×\u2248 \u2248 × \u2248 ××
= \u21d2 \u21d2 ×
 (Any digit in range 2-6 is acceptable) 
 
2.18 (a) 
 A: C
C
 C
o
o
o
R
X
s
= \u2212 =
= + + + + =
= \u2212 + \u2212 + \u2212 + \u2212 + \u2212\u2212
=
731 72 4 0 7
72 4 731 72 6 72 8 73 0
5
72 8
72 4 72 8 731 72 8 72 6 72 8 72 8 72 8 73 0 72 8
5 1
0 3
2 2 2 2 2
. . .
. . . . . .
( . . ) ( . . ) ( . . ) ( . . ) ( . . )
.
 
 B: C
C
 C
o
o
o
R
X
s
= \u2212 =
= + + + + =
= \u2212 + \u2212 + \u2212 + \u2212 + \u2212\u2212
=
1031 97 3 58
97 3 1014 987 1031 1004
5
1002
97 3 1002 1014 1002 987 1002 1031 1002 1004 1002
5 1
2 3
2 2 2 2 2
. . .
. . . . . .
( . . ) ( . . ) ( . . ) ( . . ) ( . . )
.
 
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate. 
 
2- 5
 
2.19 (a) 
 
X
X
s
X
X s
X s
i
i i= = =
\u2212
\u2212 =
= \u2212 = \u2212 =
= + = + =
= =
\u2211 \u2211
1
12
2
1
12
12
735
735
12 1
12
2 735 2 12 711
2 735 2 12 75 9
.
( . )
.
. ( . ) .
. ( . ) .
C
C
min=
max=
 
(b) Joanne is more likely to be the statistician, because she wants to make the control limits 
stricter. 
(c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor 
temperature (failure of reactor control system), problems with the color measurement 
system, operator carelessness 
2.20 (a), (b) 
 (c) Beginning with Run 11, the process has been near or well over the upper quality assurance
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