ALL 9 CHAPTERS COVERE
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1. Mass and Energy Transfer.
2. Environmental Chemistry.
3. Mathematics for Growth.
4. Risk Assessment.
5. Water Pollution.
6. Water Quality Control.
7. Air Pollution.
8. Global Atmospheric Change.
9. Solid Waste Management and Resource Recovery.
,1.1-1.7 The solutions for these problems are the solutions for problems 1.1-1.7 in
the 2nd edition Solutions Manual.
1.8 The washing machine is a batch reactor in which a first order decay of grease
on the clothes is occurring. The integrated form of the mass balance
equation is:
C C0 e - kt
First, find k:
1 C 1 1 1
k ln ln ln 0.128 min-1
C0
t C0 1 0.88C 1 0.88
min min
0
Next, calculate the grease remaining on the clothes after 5 minutes:
0.128 min -1
m m0 e - 0.500 g e- 0.264 g
kt
5.00 min 1
The grease that is not on the clothes must be in the water, so
0.500 g - 0.264
Cw 0.00472 g/L
g
50.1 L
,1. C.V.
9 Plateau Creek
Q0, C0 Qb, C0 Qs, Cs
Qf, C0 Qf/2, Cf
farm
a. A mass balance around control volume (C.V.) at the downstream
junction yields
Q /2 C QC 0.5 m 3 /s 1.00 mg/L 4.0 m 3 /s 0.0015 mg/L
f f b 0
Cs 0.112
Qs 4.5 m 3 mg/L
/s
b. Noting that Qb = Q0 – Qf and Qs = Q0 – Qf/2 the mass balance
becomes (Qf/2)Cf + (Q0 – Qf)C0 = (Q0 –Qf/2)Cs
and solving for the maximum Qf yields
Q C -C 5.0 m 3 /s 0.04 - 0.0015 mg/L
Qf 0 s 0
0.04 0.371 m3 /s
⎛ fC Cs ⎛ - 0.0015
⎞ 1.0
⎜ - C0 ⎟ ⎜ ⎞
mg/L
⎝ 2 2 ⎠ ⎝ 2 2 ⎠
,1.10 Write a mass balance on a second order reaction in a batch reactor:
Accumulation =
Reaction
where r(m) - km2
dm
V
Vr(m)
dt
mt t
dm 1 1
-k so, -
m m - kt
m2
dt
m0 0 0 t
1⎛ 1 1 ⎞
k t⎜
-m
m calculate mt=1 based on the equation’s stoichiometry
⎟⎟ that
⎝t 0
⎠ 1 mole of methanol yields 1 mole of carbon monoxide
⎛ mole CO ⎞⎛ 1 mole CH 3 OH ⎞⎛ 32 g CH 3 OH
⎞100 g CO
⎜⎜ 28114.3 ⎟⎟
g CH ⎜
OH ⎟⎜⎜ ⎟ 3
g CO 1
COmole mole CH OH ⎟
⎝ ⎠⎝ ⎠⎝ 3 ⎠
so, CCH OH, t 1 200 g -114.3 g 85.7 g
1 ⎛ 3
1 ⎞
an - ⎟⎟ 0.00667 d -1g -1
d 1
k ⎜⎜
1d ⎝ 200 g ⎠
85.7 g
,1.11-1.13 The solutions for these problems are the solutions for problems 1.8-1.10
in the 2nd edition Solutions Manual.
1.14 Calculate the pipe volume, Vr, and the first-order rate constant, k.
2
3.0 ft 3400 ln ln 2 0.0578 min-1
Vp 24,033 an k
ft ft3 d (12 min)
2
4 t1/2
For first-order decay in a steady-state
PFR
min ga ⎞⎤ mg L
⎡ 2.82
⎟
C Ce - kt
1.0 mg/L exp 0.0578min 1 ⎞
⎟⎜ ⎛ l
3
24,033 ft ⎜ 0.134 ft
3
0 ⎢
104 gal⎟
⎜ ⎦
⎣
⎝
,1.15 The stomach acts like a CSTR reactor in which a first order decay
reaction is occurring.
Stomac
Gastric Q, h
juices Ci Q, Digested
in Ce food stream
V, r(C) out
V = 1.15 L, k = 1.33 hr-1, Q = 0.012 L/min, m0 = 325 g, t = 1 hr
Accumulation = In – Out + Reaction
dC QCi - QCe where r(C) = -kC, Ci = 0, and C = Ce
V
dt Vr(C)
C t
dC
⎛ Q ⎞
so, - k⎟ dt
⎜ ⎠ 0
C0
C ⎝ V
⎛ Q
- k
⎞
t 325 g 0.012 L/min 60 min/
⎠
⎜ hr ⎟ ⎟⎟
⎜⎜ ⎜
⎟
⎟
1.15 ⎞ ⎛ ⎛ ⎞
⎜
L exp 1.33 hr -1 1 hr
C Ce ⎛ ⎞
⎝ V
⎝ ⎠ ⎝ ⎝ 1.15 L ⎠ ⎠
C = 40.0 g/L and then mt=1hr = (40.0 g/L)(1.15L) = 46.0 g
,1.16-1.20 The solutions for these problems are the solutions for problems
1.11-1.15 in the 2nd edition Solutions Manual.
1.21 a. Calculate the volume that 1 mole of an ideal gas occupies at 1 atm and 20 °C.
1 mole 0.082056 L atm mol -1 K -1
V 24.04 L then
293.15 K
nRT
P 1 atm
ppmv = (mg/m3)(24.04 L/mol)(mol wt)-1 = (60 mg/m3)(24.04 L/mol)(131 mol/g)-1
= 11.0 ppmv
b. Draw a sketch of the valley as the CSTR, non-steady state control volume.
Coal Valley
Q, Ca Q, Ce
V, r(C), kd, ks
kd = radioactive decay rate constant = (ln2)/t1/2 = (ln2)/(8.1 d) = 0.0856 d-1
ks = sedimentation rate constant = 0.02 d-1
c. The mass balance for the CSTR control volume is
dC QCa - QCe Vr(C) QCa - QCe V kd Ce ks Ce
V
dt
Assuming Ca = 0 and integrating yields
C0 1 11.
1 ln ln ppmv
t ⎛ 0 ⎞
k k C ⎛ 3 ⎞ 1 10 -5
d s
⎛
Q
V ⎞
⎛ 10 5 ⎞ 60 24 ⎝
⎜ 6 ⎟
m min
⎜
⎠
⎜⎜ ⎟ ⎠
min ⎟⎠
-1
⎝ 0.02d -1
⎟
⎜ 0.0856d
d ⎟
6 3
⎜ 2.0 10 m ⎟
⎜ ⎟
⎝ ⎠
t = 0.0322 d = 46.4 min
,1.22 a ⎛ ⎞⎛ ⎞⎛ ⎞
.
8.00 mg mmole 40mg
m 2⎜ ⎟⎜⎜ ⎟⎟⎜ ⎟ 1.00 L
6.53 mg NaOH
NaOH
⎝ L ⎠⎝ 98 mg ⎠⎝ mmole⎠
b. The reaction (P1.3) is second-order (see the rate constant’s units) so that
dC
H - kC C - - kC 2
dt H
OH H
After integration and noting that at t = t1/2, C = C0/2, the equation can be
written
1 1
t1/2 ⎛ ⎞⎛ ⎛ 4.38 10-8
⎞
kC seconds
1.4 1011 mol 8.00 mg ⎞⎛ mmole ⎞ ⎟
0
⎜ 2
⎜ ⎟⎜⎜ ⎟
L s L 98 mg ⎟ ⎟
⎜ ⎜
⎝ ⎠ ⎠⎝ ⎠⎠
⎝ ⎝
Therefore, neutralization occurs almost instantly.
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SOLUTIONS MANUAL
,TABLE OF CONTENTS
1. Mass and Energy Transfer.
2. Environmental Chemistry.
3. Mathematics for Growth.
4. Risk Assessment.
5. Water Pollution.
6. Water Quality Control.
7. Air Pollution.
8. Global Atmospheric Change.
9. Solid Waste Management and Resource Recovery.
,1.1-1.7 The solutions for these problems are the solutions for problems 1.1-1.7 in
the 2nd edition Solutions Manual.
1.8 The washing machine is a batch reactor in which a first order decay of grease
on the clothes is occurring. The integrated form of the mass balance
equation is:
C C0 e - kt
First, find k:
1 C 1 1 1
k ln ln ln 0.128 min-1
C0
t C0 1 0.88C 1 0.88
min min
0
Next, calculate the grease remaining on the clothes after 5 minutes:
0.128 min -1
m m0 e - 0.500 g e- 0.264 g
kt
5.00 min 1
The grease that is not on the clothes must be in the water, so
0.500 g - 0.264
Cw 0.00472 g/L
g
50.1 L
,1. C.V.
9 Plateau Creek
Q0, C0 Qb, C0 Qs, Cs
Qf, C0 Qf/2, Cf
farm
a. A mass balance around control volume (C.V.) at the downstream
junction yields
Q /2 C QC 0.5 m 3 /s 1.00 mg/L 4.0 m 3 /s 0.0015 mg/L
f f b 0
Cs 0.112
Qs 4.5 m 3 mg/L
/s
b. Noting that Qb = Q0 – Qf and Qs = Q0 – Qf/2 the mass balance
becomes (Qf/2)Cf + (Q0 – Qf)C0 = (Q0 –Qf/2)Cs
and solving for the maximum Qf yields
Q C -C 5.0 m 3 /s 0.04 - 0.0015 mg/L
Qf 0 s 0
0.04 0.371 m3 /s
⎛ fC Cs ⎛ - 0.0015
⎞ 1.0
⎜ - C0 ⎟ ⎜ ⎞
mg/L
⎝ 2 2 ⎠ ⎝ 2 2 ⎠
,1.10 Write a mass balance on a second order reaction in a batch reactor:
Accumulation =
Reaction
where r(m) - km2
dm
V
Vr(m)
dt
mt t
dm 1 1
-k so, -
m m - kt
m2
dt
m0 0 0 t
1⎛ 1 1 ⎞
k t⎜
-m
m calculate mt=1 based on the equation’s stoichiometry
⎟⎟ that
⎝t 0
⎠ 1 mole of methanol yields 1 mole of carbon monoxide
⎛ mole CO ⎞⎛ 1 mole CH 3 OH ⎞⎛ 32 g CH 3 OH
⎞100 g CO
⎜⎜ 28114.3 ⎟⎟
g CH ⎜
OH ⎟⎜⎜ ⎟ 3
g CO 1
COmole mole CH OH ⎟
⎝ ⎠⎝ ⎠⎝ 3 ⎠
so, CCH OH, t 1 200 g -114.3 g 85.7 g
1 ⎛ 3
1 ⎞
an - ⎟⎟ 0.00667 d -1g -1
d 1
k ⎜⎜
1d ⎝ 200 g ⎠
85.7 g
,1.11-1.13 The solutions for these problems are the solutions for problems 1.8-1.10
in the 2nd edition Solutions Manual.
1.14 Calculate the pipe volume, Vr, and the first-order rate constant, k.
2
3.0 ft 3400 ln ln 2 0.0578 min-1
Vp 24,033 an k
ft ft3 d (12 min)
2
4 t1/2
For first-order decay in a steady-state
PFR
min ga ⎞⎤ mg L
⎡ 2.82
⎟
C Ce - kt
1.0 mg/L exp 0.0578min 1 ⎞
⎟⎜ ⎛ l
3
24,033 ft ⎜ 0.134 ft
3
0 ⎢
104 gal⎟
⎜ ⎦
⎣
⎝
,1.15 The stomach acts like a CSTR reactor in which a first order decay
reaction is occurring.
Stomac
Gastric Q, h
juices Ci Q, Digested
in Ce food stream
V, r(C) out
V = 1.15 L, k = 1.33 hr-1, Q = 0.012 L/min, m0 = 325 g, t = 1 hr
Accumulation = In – Out + Reaction
dC QCi - QCe where r(C) = -kC, Ci = 0, and C = Ce
V
dt Vr(C)
C t
dC
⎛ Q ⎞
so, - k⎟ dt
⎜ ⎠ 0
C0
C ⎝ V
⎛ Q
- k
⎞
t 325 g 0.012 L/min 60 min/
⎠
⎜ hr ⎟ ⎟⎟
⎜⎜ ⎜
⎟
⎟
1.15 ⎞ ⎛ ⎛ ⎞
⎜
L exp 1.33 hr -1 1 hr
C Ce ⎛ ⎞
⎝ V
⎝ ⎠ ⎝ ⎝ 1.15 L ⎠ ⎠
C = 40.0 g/L and then mt=1hr = (40.0 g/L)(1.15L) = 46.0 g
,1.16-1.20 The solutions for these problems are the solutions for problems
1.11-1.15 in the 2nd edition Solutions Manual.
1.21 a. Calculate the volume that 1 mole of an ideal gas occupies at 1 atm and 20 °C.
1 mole 0.082056 L atm mol -1 K -1
V 24.04 L then
293.15 K
nRT
P 1 atm
ppmv = (mg/m3)(24.04 L/mol)(mol wt)-1 = (60 mg/m3)(24.04 L/mol)(131 mol/g)-1
= 11.0 ppmv
b. Draw a sketch of the valley as the CSTR, non-steady state control volume.
Coal Valley
Q, Ca Q, Ce
V, r(C), kd, ks
kd = radioactive decay rate constant = (ln2)/t1/2 = (ln2)/(8.1 d) = 0.0856 d-1
ks = sedimentation rate constant = 0.02 d-1
c. The mass balance for the CSTR control volume is
dC QCa - QCe Vr(C) QCa - QCe V kd Ce ks Ce
V
dt
Assuming Ca = 0 and integrating yields
C0 1 11.
1 ln ln ppmv
t ⎛ 0 ⎞
k k C ⎛ 3 ⎞ 1 10 -5
d s
⎛
Q
V ⎞
⎛ 10 5 ⎞ 60 24 ⎝
⎜ 6 ⎟
m min
⎜
⎠
⎜⎜ ⎟ ⎠
min ⎟⎠
-1
⎝ 0.02d -1
⎟
⎜ 0.0856d
d ⎟
6 3
⎜ 2.0 10 m ⎟
⎜ ⎟
⎝ ⎠
t = 0.0322 d = 46.4 min
,1.22 a ⎛ ⎞⎛ ⎞⎛ ⎞
.
8.00 mg mmole 40mg
m 2⎜ ⎟⎜⎜ ⎟⎟⎜ ⎟ 1.00 L
6.53 mg NaOH
NaOH
⎝ L ⎠⎝ 98 mg ⎠⎝ mmole⎠
b. The reaction (P1.3) is second-order (see the rate constant’s units) so that
dC
H - kC C - - kC 2
dt H
OH H
After integration and noting that at t = t1/2, C = C0/2, the equation can be
written
1 1
t1/2 ⎛ ⎞⎛ ⎛ 4.38 10-8
⎞
kC seconds
1.4 1011 mol 8.00 mg ⎞⎛ mmole ⎞ ⎟
0
⎜ 2
⎜ ⎟⎜⎜ ⎟
L s L 98 mg ⎟ ⎟
⎜ ⎜
⎝ ⎠ ⎠⎝ ⎠⎠
⎝ ⎝
Therefore, neutralization occurs almost instantly.
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TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
