Elementary Principles of Chemical Processes [Solutions Manual]

497 Pages·2004·3.56 MB·English

by  Richard M. Felder, Ronald W. Rousseau

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
2598
26 0
.
.
.
1 ft / s
0.0006214 mi
s
3.2808 ft
1 h
mi / h
mi / h
=
⇒

(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
⋅
=
×
⋅
.
/ 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
/ ft
m
3

3
3
m
3
=
.

(c) 537
10
1000 J
1
1
119 93
120

3
.
.
×
×
=
⇒
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
2
in
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
=
×
×
=
×
≈
40
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
steps
5
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
26417
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.
,
.

=

=
+
=
+

=
+
=
+

⇒
=
×

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

⋅

=
×
⋅

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

×
⋅

×

=
⇒

2.8 (a) 250
250
.
.
lb
32.1714 ft / s
1 lb

32.1714 lb
ft / s
lb
m
2
f

m
2
f
⋅
=

(b)

2

2
25 N
1
1 kg m/s
2.5493 kg
2.5 kg
9.8066 m/s
1 N

⋅
=
⇒

(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
dynes
m
2

-4
m
2
×
⋅
=
×

2.9
50
15
2
853
32174
1
4 5
106
×
×
⋅
=
×
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
25
2
m
3

m

3
1 kg
1 m

2.20462 lb
11.5 kg
m
≈
×
FHG IKJFHG IKJ ≈

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

=
⇒
=
⇒
=

=
=
−
=

ρ
ρ
ρ
π
ρ
π

ρ
ρ

2
2

30
141
100
053
(
.
)( .
)
.

(b) ρ
ρ
f
cH
h
=
=
=
(
)( .
)
.
30
053
171
cm
g / cm
(30 cm - 20.7 cm)
g / cm

3
3

H

h

ρf

ρc

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

=
=
−
=
⇒
=

⇒
=
− FHG IKJ
=
−
FHG
IKJ

=
⇒
−
FHG
IKJ =

⇒
=
−
=
−
=

−FHG IKJ

π
π
π

π
π
π

ρ
ρ
ρ
π
ρ π

ρ
ρ
ρ
ρ

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

;
;

ρf
ρs

R

r

h

H

2- 3

2.13
Say h m
( ) = depth of liquid

A ( )
m 2
h 1 m
⇒

y

x

y = 1

y = 1 – h

x =
1 – y 2

d A

(
)

(
)
(
)
(
)
(
)

2

2

1
1
2
2
2

1
1

1
2
2
2
1
1

1

Table of integrals or trigonometric substitution

2 1
2
1

m
1
sin
1
1
1
sin
1
2

π

−
− +

−
−
−

−
−
−

−

=
⋅
=
−
⇒
=
−

⎤
=
−
+
=
−
−
−
+
−
+
⎥⎦

⇓

∫
∫

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
0879
1

1
10
345
10

2

3
4

0

m
m
g
10 cm
kg
9.81 N

cm
m
g
kg

Substitute for

2
6

3
3
(
)
.
.
�

W
h
h
h
Nb g
b
g
b
g
b
g
=
×
−
−
−
+
−
+
LNM
OQP
−
345
10
1
1
1
1
2

4
2
1
.
sin
π

2.14
1
1
32174
1

1
1
32174

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

=
⋅
=
⋅
⇒

⋅
=

.

.

(a) (i) On the earth:

M

W

=
=

=
⋅
=
×

175 lb
1
544

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

=
=

=
⋅
=

175 lb
1
544

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
=
⇒
=
=
⋅

=

355 pound
1
1
als
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= –1

y= –1+h
x

dA

2- 4

2.15 (a) F
ma
=
⇒
FHG IKJ =
⋅

⇒
⋅

1
1
6
53623

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
2
W

W

=
⋅
=

=

3

3 32174
53623
( )(
.
) / .

2.16 (a) ≈
=

=

( )( )

( . )( .
)

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

−
−

−
−

×
≈
≈ ×

×
=
×

(c) ≈
+
=

+
=

2
125
127

2 365
1252
127 5
.
.
.

(d)
≈
×
− ×
≈
×
≈
×

×
−
×
=
×

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 10
exact

R

R

−
×
×
×
≈
≈
×
≈
×
×

=
⇒
⇒
×

(Any digit in range 2-6 is acceptable)

2.18 (a)

A:
C

C

C

o

o

o

R

X

s

=
−
=

=
+
+
+
+
=

=
−
+
−
+
−
+
−
+
−
−

=

731
72 4
0 7

72 4
731
72 6
72 8
730
5
72 8

72 4
72 8
731
72 8
72 6
72 8
72 8
72 8
730
72 8
5
1

0 3

2
2
2
2
2

.
.
.

.
.
.
.
.
.

(
.
. )
(
.
. )
(
.
. )
(
.
. )
(
.
. )

.

B:
C

C

C

o

o

o

R

X

s

=
−
=

=
+
+
+
+
=

=
−
+
−
+
−
+
−
+
−
−

=

1031
973
58

973 1014
987
1031 1004
5
1002

973 1002
1014
1002
987
1002
1031 1002
1004
1002
5 1

23

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
=
=
=

−

−
=

=
−
=
−
=

=
+
=
+
=

=
=
∑
∑
1

12
2

1

12

12
735

735

12
1
12

2
735
2 12
711

2
735
2 12
759

.

(
. )

.

.
( . )
.

.
( . )
.

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
limit. An overhaul would have been reasonable after Run 12.

2.21 (a)

4
2
2
2

2
2.36 10
kg m
2.20462 lb
3.2808 ft
1 h
'
h
kg
m
3600 s
Q

−
×
⋅
=

(b)

4
( 4 3)
6
2
approximate
3

6
2
2
exact

(2 10
)(2)(9)
'
12 10
1.2 10
lb ft /s
3 10

'
=1.56 10
lb ft /s
0.00000156 lb ft /s

Q

Q

−
− −
−

−

×
≈
≈
×
≈
×
⋅
×

×
⋅
=
⋅

(a) Run
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
X
134
131
129
133
135 131 134 130 131 136 129 130 133 130 133
Mean(X)
131.9
Stdev(X)
2.2
Min
127.5
Max
136.4

(b) Run
X
Min Mean
Max
1
128 127.5 131.9 136.4
2
131 127.5 131.9 136.4
3
133 127.5 131.9 136.4
4
130 127.5 131.9 136.4
5
133 127.5 131.9 136.4
6
129 127.5 131.9 136.4
7
133 127.5 131.9 136.4
8
135 127.5 131.9 136.4
9
137 127.5 131.9 136.4
10
133 127.5 131.9 136.4
11
136 127.5 131.9 136.4
12
138 127.5 131.9 136.4
13
135 127.5 131.9 136.4
14
139 127.5 131.9 136.4

126

128

130

132

134

136

138

140

0
5
10
15

2- 6

2.22 N
C

k
C

C

N

p
o

o
Pr

Pr

.
.

.

(
)(
)(
)
(
)(
)( )
.
.
.

=
=
⋅

⋅
⋅

≈
×
×
×
×
×
≈
×
≈
×
×

−

−

μ
0583
1936
32808

0 286

6
10
2
10
3
10
3
10
4
10
2
3
10
2
15
10
163
10

1
3
3

1
3

3
3
3

J / g
lb
1 h
ft
1000 g

W / m
ft h
3600 s
m
2.20462 lb

The calculator solution is

m

m

2.23

Re
.
.
.

.

Re
(
)( )(
)(
)
( )(
)(
)(
)

(

=
=
×
⋅

≈
×
×
×
×
≈
×
≈
×
⇒

−

−
−

−

− −

Duρ
μ
0 48
2 067
0805

0 43
10

5
10
2 8
10
10
3 4
10 10
4
10
5
10
3
2
10

3

1
1
6

3
4

1
3)
4

ft
1 m
in
1 m
g
1 kg
10 cm

s
3.2808 ft
kg / m s
39.37 in
cm
1000 g
1 m

the flow is turbulent

6
3

3
3

2.24

1/ 2
1/3

1/3
1/ 2
5
2
3

3
5
2
5
2

(a)
2.00
0.600

1.00 10
N s/m
(0.00500 m)(10.0 m/s)(1.00 kg/m )

2.00
0.600
(1.00 kg/m )(1.00 10
m /s)
(1.00 10
N s/m )

(0.00500

44.426

ρ
μ
ρ
μ

−

−
−

⎛
⎞
⎛
⎞
=
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠

⎡
⎤
⎡
⎤
×
⋅
=
+
⎢
⎥
⎢
⎥
×
×
⋅
⎣
⎦
⎣
⎦

=
⇒

g
p
p

g

k d y
d u

D
D

k

5
2
m)(0.100)
44.426
0.888 m/s
1.00 10
m /s
−
=
⇒
=
×
g
k

(b) The diameter of the particles is not uniform, the conditions of the system used to model the
equation may differ significantly from the conditions in the reactor (out of the range of
empirical data), all of the other variables are subject to measurement or estimation error.

(c)
dp (m) y
D (m2/s) μ (N-s/m2) ρ (kg/m3) u (m/s)
kg

0.005 0.1 1.00E-05 1.00E-05
1
10
0.889

0.010 0.1 1.00E-05 1.00E-05
1
10
0.620

0.005 0.1 2.00E-05 1.00E-05
1
10
1.427

0.005 0.1 1.00E-05 2.00E-05
1
10
0.796

0.005 0.1 1.00E-05 1.00E-05
1
20
1.240

2.25 (a) 200 crystals / min mm; 10 crystals / min mm2
⋅
⋅

(b) r =
⋅
−
⋅

=
⇒
=

200
10

4 0

crystals
0.050 in
25.4 mm
min mm
in
crystals
0.050 in
(25.4) mm
min mm
in

238 crystals / min
238 crystals
1 min
60 s
crystals / s

2
2
2
2

2
2

min
.

(c) D
D
D
mm
in
mm
1 in
b g
b g
=
′
=
′
254
254
.
.
; r
r
r
crystals
min

crystals
60 s
s
1 min
FHG
IKJ = ′
=
′
60

⇒
′ =
′ −
′
⇒ ′ =
′ −
′
60
200 254
10 254
84 7
108
2
2
r
D
D
r
D
D
.
.
.
b
g
b
g
b g

2- 7

2.26 (a) 705.
/
;
lb
ft 8.27
10
in / lb
m
3
-7
2
f
×

(b)

7
2
6
2
f
3
m
2
5
2
f

3
3
m

3
3
6
3
m

8.27 10
in
9 10 N
14.696 lb /in
(70.5 lb /ft )exp
lb
m
1.01325 10 N/m

70.57 lb
35.3145 ft
1 m
1000 g
1.13 g
ft
m
10 cm
2.20462 lb

ρ

−
⎡
⎤
×
×
⎢
⎥
=
×
⎢
⎥
⎣
⎦

=
=
3
/cm

(c) ρ
ρ
ρ
lb

ft

g
lb
cm

cm
g
1 ft

m
3
m
3

3
3
FHG
IKJ =
′
=
′
1
28 317

453593
62 43
,

.
.

P
P
P
lb

in

N
.2248 lb

m

m
N
39.37 in

f
2
f
2

2
2
2
FHG
IKJ =
=
×
−
'
.
'
0
1

1
145
10

2
4

⇒
′ =
×
×
⇒
′ =
×
−
−
−
62 43
705
8 27
10
145
10
113
120
10
7
4
10
.
. exp
.
.
'
.
exp .
'
ρ
ρ
d
id
i
d
i
P
P

P'
.
'
.
exp[( .
)( .
)]
.
=
×
⇒
=
×
×
=
−
9 00
10
113
120
10
9 00
10
113
6
10
6
N / m
g / cm
2
3
ρ

2.27 (a) V
V
V
cm
in
28,317 cm

in

3
3
3

3
d
i
d i
=
=
'
.
'
1728
16 39
; t
t
s
hr
b g
b g
=
′
3600

⇒
=
′ ⇒
=
′
16 39
3600
0 06102
3600
.
'
exp
'
.
exp
V
t
V
t
b
g
b
g

(b) The t in the exponent has a coefficient of s-1.

2.28 (a) 300
.
mol / L, 2.00 min-1

(b) t
C
C
=
⇒
=
⇒
=
0
300
300
.
.
exp[(-2.00)(0)] = 3.00 mol / L
t =1

exp[(-2.00)(1)] = 0.406 mol / L

For t=0.6 min: C

C

int
.
.
( .
)
.
.

.

=
−
−
−
+
=

=

0 406
300
1
0
0 6
0
300
14

300

mol / L

exp[(-2.00)(0.6)] = 0.9 mol / L
exact

For C=0.10 mol/L: t

t

int

exact

min

= -
1
2.00 ln C
3.00 = - 1
2 ln 0.10
3.00 =1.70 min

=
−

−
−
+
=
1
0
0 406
3 010
300
0
112
.
( .
.
)
.

(c)

0

0.5

1

1.5

2

2.5

3

3.5

0
1
2

t (min)

C (mol/L)

(t=0.6, C=1.4)

(t=1.12, C=0.10)

Cexact vs. t

2- 8

2.29 (a)
p*
.
. (
. )
=
−
−
−
+
=
60
20
199 8
166 2 185
166 2
20
42 mm Hg

(b)
c
MAIN PROGRAM FOR PROBLEM 2.29
IMPLICIT REAL*4(A–H, 0–Z)
DIMENSION TD(6), PD(6)
DO 1 I = 1, 6

READ (5, *) TD(I), PD(I)

1
CONTINUE
WRITE (5, 902)

902
FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X,

*
‘ (C) (MM HG)’/)

DO 2 I = 0, 115, 5
T = 100 + I
CALL VAP (T, P, TD, PD)
WRITE (6, 903) T, P

903
FORMAT (10X, F5.1, 10X, F5.1)

2
CONTINUE
END
SUBROUTINE VAP (T, P, TD, PD)
DIMENSION TD(6), PD(6)
I = 1

1
IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2
I = I + 1
IF (I.EQ.6) STOP
GO TO 1

2
P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I))
RETURN
END

DATA

OUTPUT

98.5
1.0

TEMPERATURE
VAPOR PRESSURE

131.8
5.0

(C)
(MM HG)

�
�

100.0
1.2

215.5
100.0

105.0
1.8

�
�

215.0
98.7

2.30 (b) ln
ln

(ln
ln
) / (
)
(ln
ln ) / (
)
.

ln
ln
ln
.
( )
.
.

y
a
bx
y
ae

b
y
y
x
x

a
y
bx
a
y
e

bx

x

=
+
⇒
=

=
−
−
=
−
−
= −

=
−
=
+
⇒
=
⇒
=
−

2
1
2
1

0.693
2
1
1
2
0 693

2
0 63 1
4 00
4 00

(c) ln
ln
ln
(ln
ln
) / (ln
ln
)
(ln
ln ) / (ln
ln )
ln
ln
ln
ln
(
) ln( )
/

y
a
b
x
y
ax
b
y
y
x
x
a
y
b
x
a
y
x

b
=
+
⇒
=
=
−
−
=
−
−
= −
=
−
=
− −
⇒
=
⇒
=

2
1
2
1
2
1
1
2
1
2
1
1
2
2

(d)
/
/

2
1
2
1

3 /

ln(
)
ln
( / )
( / )
[can't get
( )]

[ln(
)
ln(
) ]/[( / )
( / ) ]
(ln807.0
ln 40.2)/(2.0
1.0)
3

ln
ln(
)
( / )
ln807.0
3ln(2.0)
2
2

[can't solve explicitly for

by x
by x

y x

xy
a
b y x
xy
ae
y
a x e
y
f x

b
xy
xy
y x
y x

a
xy
b y x
a
xy
e

=
+
⇒
=
⇒
=
=

=
−
−
=
−
−
=

=
−
=
−
⇒
=
⇒
=

( )]
y x

2- 9

2.30 (cont’d)
(e) ln(
/ )
ln
ln(
)
/
(
)
[
(
) ]

[ln(
/ )
ln(
/ ) ] / [ln(
)
ln(
) ]
(ln
.
ln
. ) / (ln .
ln . )
.

ln
ln(
/ )
(
)
ln
.
.
ln( . )
.

/
. (
)
.
(
)

/

.33
/
.

y
x
a
b
x
y
x
a x
y
ax x

b
y
x
y
x
x
x

a
y
x
b x
a

y
x
x
y
x
x

b
b
2
2
1 2

2
2
2
1
2
1

2

2
4
1 2
2 165

2
2
2

2
2
807 0
40 2
2 0
10
4 33

2
807 0
4 33
2 0
40 2

40 2
2
6 34
2

=
+
−
⇒
=
−
⇒
=
−

=
−
−
−
−
=
−
−
=

=
−
−
=
−
⇒
=

⇒
=
−
⇒
=
−

2.31 (b) Plot
vs.
on rectangular axes. Slope
Intcpt
2
3
y
x
m
n
=
= −
,

(c)
1
1
1
a
1
Plot
vs.
[rect. axes], slope =
, intercept =
ln(
3)
ln(
3)
b
b
a
x
x
y
b
b
y
=
+
⇒
−
−

(d)

1
1
3
1
1
3
2
3
2
3
(
)
(
)
(
)
(
)
,
,
y
a x
y
x
a
+
=
−
⇒
+
−
Plot
vs.
[rect. axes] slope =
intercept = 0

OR

2
1
3
3
1
3

2

ln(
)
ln
ln(
)
ln(
)
ln(
)

ln

y
a
x
y
x

a

+
= −
−
−
+
−

⇒
−
−

Plot
vs.
[rect.] or (y +1) vs. (x - 3) [log]

slope =
3
2 , intercept =

(e) ln

ln

y
a x
b

y
x
y
x

=
+

Plot
vs.
[rect.] or vs.
[semilog ], slope = a, intercept = b

(f)

Plot
vs.
[rect.]
slope = a, intercept = b

log (
)
(
)

log (
)
(
)

10
2
2

10
2
2
xy
a x
y
b

xy
x
y

=
+
+

+
⇒

(g)
Plot
vs.
[rect.] slope = , intercept =

OR
b
Plot 1 vs. 1 [rect.] , slope =
intercept =

1

1
1

2
2

2
2

y
ax
b
x
x
y
ax
b
x
y
x
a
b

y
ax
b
x
xy
a
x
xy
x
b
a

=
+
⇒
=
+
⇒

=
+
⇒
=
+
⇒

,

,

2- 10

2.32 (a) A plot of y vs. R is a line through ( R = 5, y = 0 011
.
) and ( R = 80 , y = 0169
.
).

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0
20
40
60
80
100

R

y

y
a R
b
a

b
y
R
=
+
=
−
−
=
×

=
−
×
=
×
UV|
W|
⇒
=
×
+
×

−

−
−

−
−

0169
0 011
80
5
211
10

0 011
211
10
5
4 50
10
211
10
4 50
10

3

3
4

3
4
.
.
.

.
.
.
.
.
d
ib g

(b) R
y
=
⇒
=
×
+
×
=
−
−
43
211
10
43
4 50
10
0 092
3
4
.
.
.
d
ib g
kg H O kg
2

1200
0 092
110 kg
kg h
kg H O kg
H O h
2
2
b
gb
g
.
=

2.33 (a) ln
ln
ln
(ln
ln
) / (ln
ln
)
(ln
ln
) / (ln
ln
)
.

ln
ln
ln
ln
(
.
)ln(
)
.
.
.

T
a
b
T
a
b
T
T

a
T
b
a
T

b
=
+
⇒
=
=
−
−
=
−
−
= −

=
−
=
− −
⇒
=
⇒
=
−

φ
φ
φ
φ

φ
φ

2
1
2
1

119
120
210
40
25
119

210
119
25
9677 6
9677 6

(b) T
T

T
C

T
C

T
C

=
⇒
=

=
⇒
=
=

=
⇒
=
=

=
⇒
=
=

−
9677 6
9677 6

85
9677 6 85
535

175
9677 6 175
291

290
9677 6 290
19 0

119
0.8403

0.8403

0.8403

0.8403

.
. /

. /
.

. /
.

. /
.

.
φ
φ

φ

φ

φ

b
g
b
g
b
g
b
g

o

o

o

L / s

L / s

L / s

(c) The estimate for T=175°C is probably closest to the real value, because the value of
temperature is in the range of the data originally taken to fit the line. The value of T=290°C
is probably the least likely to be correct, because it is farthest away from the date range.

2- 11

2.34 (a) Yes, because when ln[(
) / (
)]
C
C
C
C
A
Ae
A
Ae
−
−
0
is plotted vs. t in rectangular coordinates,
the plot is a straight line.

-2

-1.5

-1

-0.5

0
0
50
100
150
200

t (min)

ln ((CA-CAe)/(CA0-CAe))

Slope = -0.0093
k = 9.3
10 min
-3
⇒
×
−1

(b)

3
0
0

(9.3 10
)(120)
-2

-2

ln[(
)/(
)]
(
)

(0.1823
0.0495)
0.0495
9.300 10 g/L

9.300 10 g
30.5 gal
28.317 L
=
/
=
10.7 g
L
7.4805 gal

−

−

−
×
−
−
= −
⇒
=
−
+

=
−
+
=
×

×
⇒
=
=

kt
A
Ae
A
Ae
A
A
Ae
Ae

A

C
C
C
C
kt
C
C
C
e
C

C
e

C m V
m CV

2.35 (a) ft and h , respectively
3
-2

(b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln( .
)
353
10 2
×
−
; or

V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353
10 2
.
×
−

(c) V (
)
.
exp( .
)
m
t
3
2
=
×
×
−
−
100
10
15
10
3
7

2.36 PV
C
P
C V
P
C
k
V
k
k
=
⇒
=
⇒
=
−
/
ln
ln
ln

lnP = -1.573(lnV ) + 12.736

6

6.5

7

7.5

8

8.5

2.5
3
3.5
4

lnV

lnP

slope
(dimensionless)

Intercept = ln
mm Hg cm4.719
k

C
C
e

= −
= − −
=

=
⇒
=
=
×
⋅

(
.
)
.

.
.
.736
1573
1573

12 736
340
10
12
5

2.37 (a) G
G
G
G
K C
G
G
G
G
K C
G
G
G
G
K
m
C
L

L
m
L
L
m

L
L
−
−
=
⇒
−
−
=
⇒
−
−
=
+

0

0
0
1
ln
ln
ln

ln(G 0-G)/(G-G L)= 2.4835lnC - 10.045

-1

0

1

2

3

3.5
4
4.5
5
5.5

lnC

ln(G0-G)/(G-G L)

2- 12

2.37 (cont’d)

m

K
K
L
L

=
=

= −
⇒
=
×
−
slope
(dimensionless)

Intercept = ln
ppm-2.483
2 483

10 045
4 340
10 5
.

.
.

(b) C
G

G

G
=
⇒
−
×
×
−
=
×
⇒
=
×

−

−
−
−
475
180
10
300
10
4 340
10
475
1806
10

3

3
5
2
3
.
.
.
(
)
.
.483

C=475 ppm is well beyond the range of the data.

2.38 (a) For runs 2, 3 and 4:

Z
aV p
Z
a
b
V
c
p
a
b
c

a
b
c

a
b
c

b
c
=
⇒
=
+
+
=
+
+

=
+
+

=
+
+

�
ln
ln
ln �
ln
ln( . )
ln
ln( .
)
ln( . )

ln( .
)
ln
ln( .
)
ln(
. )

ln( .
)
ln
ln( .
)
ln(
. )

35
102
91

2 58
102
112

372
175
112

b

c

=

⇒
= −

⋅

0 68

146

.

.

a = 86.7 volts kPa
/ (L / s)
1.46
0.678

(b) When P is constant (runs 1 to 4), plot ln
�
Z vs. lnV . Slope=b, Intercept= ln
ln
a
c
p
+

lnZ = 0.5199lnV + 1.0035

0

0.5

1

1.5

2

-1
-0.5
0
0.5
1
1.5

lnV

lnZ

b

a
c
P

=
=

+
=

slope

Intercept = ln

052

10035

.

ln
.

When �V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept=ln
ln �
a
c
V
+

lnZ = -0.9972lnP + 3.4551

0

0.5

1

1.5

2

1.5
1.7
1.9
2.1
2.3

lnP

lnZ

c
slope

a
b
V

=
= −
⇒

+
=

0 997
10

34551

.
.

ln �
.

Intercept = ln

Plot Z vs �V P
b
c . Slope=a (no intercept)

Z = 31.096VbPc

1
2
3
4
5
6
7

0.05
0.1
0.15
0.2

VbPc

Z

a
slope
=
=
⋅
311. volt kPa / (L / s)

.52

The results in part (b) are more reliable, because more data were used to obtain them.

2- 13

2.39 (a)

s
n
x y

s
n
x

s
n
x
s
n
y

a
s
s s

s
s

xy
i
i
i

n

xx
i
i

n

x
i
i

n

y
i
i

n

xy
x
y

xx
x

=
=
+
+
=

=
=
+
+
=

=
=
+
+
=
=
=
+
+
=

=
−

−
=
−

=

=

=
=

∑

∑

∑
∑

1
0 4 0 3
21 19
31 32
3
4 677

1
0 3
19
32
3
4 647

1
0 3
19
32
3
18
1
0 4
21
31
3
1867

4 677
18 1

1

2

1

2
2
2

1
1

2

[( . )( . )
( . )( . )
( . )( . )] /
.

( .
.
.
) /
.

( .
.
. ) /
. ;
( .
.
. ) /
.

.
( . )( .
b g

867
4 647
18
0 936

4 647 1867
4 677 18
4 647
18
0182

0 936
0182

2

2
2

)
.
( . )
.

( .
)( .
)
( .
)( . )
.
( . )
.

.
.

−
=

=
−

−
=
−
−
=

=
+

b
s s
s s

s
s

y
x

xx
y
xy x

xx
xb g

(b) a
s

s
y
x
xy

xx
=
=
=
⇒
=
4 677
4 647
10065
10065
.
.
.
.

y = 1.0065x

y = 0.936x + 0.182

0

1

2

3

4

0
1
2
3
4

x

y

2.40 (a) 1/C vs. t. Slope= b, intercept=a

(b) b
a
=
⋅
=
slope = 0.477 L / g h
Intercept = 0.082 L / g
;

1/C = 0.4771t + 0.0823

0
0.5
1
1.5
2
2.5
3

0
1
2
3
4
5
6

t

1/C

0

0.5

1

1.5

2

1
2
3
4
5

t

C

C
C-fitted

(c) C
a
bt

t
C
a
b

=
+
⇒
+
=

=
−
=
−
=

1
1
0 082
0 477 0
12 2

1
1 0 01
0 082
0 477
209 5

/ (
)
/[ .
.
( )]
.

( /
) /
( / .
.
) / .
.

g / L

h

(d) t=0 and C=0.01 are out of the range of the experimental data.

(e) The concentration of the hazardous substance could be enough to cause damage to the
biotic resources in the river; the treatment requires an extremely large period of time; some
of the hazardous substances might remain in the tank instead of being converted; the
decomposition products might not be harmless.

2- 14

2.41 (a) and (c)

1

10

0.1
1
10
100

x

y

(b) y
ax
y
a
b
x
a
b
=
⇒
=
+
ln
ln
ln ; Slope = b, Intercept = ln

ln y = 0.1684ln x + 1.1258

0

0.5

1

1.5

2

-1
0
1
2
3
4
5
ln x

ln y

b

a
a

=
=

=
⇒
=

slope

Intercept = ln

0168

11258
308

.

.
.

2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0

(b)

Lab 1
ln(1-Cp/Cao) = -0.0062t
-4
-3
-2
-1
0
0
200
400
600
800

t

ln(1-Cp/Cao)

Lab 2
ln(1-Cp/Cao) = -0.0111t
-6

-4

-2

0
0
100
200
300
400
500
600

t

ln(1-Cp/Cao)

k = 0 0062
.
s-1

k = 0 0111
.
s-1

Lab 3
ln(1-Cp/Cao) = -0.0063t
-6

-4

-2

0
0
200
400
600
800

t

ln(1-Cp/Cao)

Lab 4

ln(1-Cp/Cao)= -0.0064t
-6

-4

-2

0
0
200
400
600
800

t

ln(1-Cp/Cao)

k = 0 0063
.
s-1

k = 0 0064
.
s-1

(c) Disregarding the value of k that is very different from the other three, k is estimated with
the average of the calculated k’s. k = 0 0063
.
s-1

(d) Errors in measurement of concentration, poor temperature control, errors in time
measurements, delays in taking the samples, impure reactants, impurities acting as
catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty
reactor.

2- 15

2.43

y
ax
a
d
y
ax
d
da
y
ax x
y x
a
x

a
y x
x

i
i
i
i

n

i
i
i

n

i
i
i

n

i
i
i
i

n

i
i

n

i
i
i

n

i
i

n

=
⇒
=
=
−
⇒
=
=
−
⇒
−
=

⇒
=

=
=
=
=
=

=
=

∑
∑
∑
∑
∑

∑
∑

φ
φ
( )

/

2

1

2

1
1
1

2

1

1

2

1

0
2
0
b
g
b
g

2.44
DIMENSION X(100), Y(100)
READ (5, 1) N

C
N = NUMBER OF DATA POINTS

1FORMAT (I10)
READ (5, 2) (X(J), Y(J), J = 1, N

2FORMAT (8F 10.2)
SX = 0.0
SY = 0.0
SXX = 0.0
SXY = 0.0
DO 100J = 1, N
SX = SX + X(J)
SY = SY + Y(J)
SXX = SXX + X(J) ** 2

100SXY = SXY + X(J) * Y(J)
AN = N
SX = SX/AN
SY = SY/AN
SXX = SXX/AN
SXY = SXY/AN
CALCULATE SLOPE AND INTERCEPT
A = (SXY - SX * SY)/(SXX - SX ** 2)
B = SY - A * SX
WRITE (6, 3)

3FORMAT (1H1, 20X 'PROBLEM 2-39'/)
WRITE (6, 4) A, B

4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/)

C
CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF
RESIDUALS
SSQ = 0.0
DO 200J = 1, N
YC = A * X(J) + B
RES = Y(J) - YC
WRITE (6, 5) X(J), Y(J), YC, RES

5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X

* 'RESIDUALb =', F6.3)

200SSQ = SSQ + RES ** 2
WRITE (6, 6) SSQ

6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3)
STOP
END

$DATA

5

1.0
2.35
1.5
5.53
2.0
8.92
2.5
12.15

3.0
15.38

SOLUTION: a
b
=
= −
6536
4 206
.
,
.