Solution for Introduction to Environment Engineering and Science 3rd edition by Gilbert M. Masters

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:
kt
CC -
0e=
First, find k:
1-
0
0
0
min0.128
0.88
1
ln
min1
1
0.88
ln
min1
1
ln
1
====
C
C
C
C
t
k
Next, calculate the grease remaining on the clothes after 5 minutes:
( ) ( )( ) g0.264eg0.500e
1-1
min.005min128.0--
0 === kt
mm
The grease that is not on the clothes must be in the water, so
g/L0.00472
L50.0
g0.264-g0.500
w ==C

Q0, C0 Qs, CsQb, C0
farm
Plateau Creek
C.V.
Qf/2, Cf
1.9
Qf, C0
a. A mass balance around control volume (C.V.) at the downstream junction
yields
( ) ( )( ) ( )( )
( ) mg/L0.112
/sm4.5
mg/L0015.0/sm4.0mg/L00.1/sm0.5/2
3
33
s
0bff
s =
+
=
+
=
Q
CQCQ
C
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
( ) ( )( ) /sm0.371
mg/L
2
0.04
0.0015-
2
1.0
mg/L0.0015-04.0/sm5.0
2
-
2
- 3
3
s
0
f
0s0
f =
⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
C
C
C
CCQ
Q

1.10 Write a mass balance on a second order reaction in a batch reactor:
Accumulation = Reaction
2
-erewh kmr(m)Vr(m)
dt
dm
V ==
∫∫ =
tm
m
dtk
m
dmt
0
2
-
0
so, kt
mm t
-
1
-
1
0
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
0t
1
-
11
k
mmt
calculate mt=1 based on the equation’s stoichiometry that
1 mole of methanol yields 1 mole of carbon monoxide
( ) OHCHg114.3
OHCHmole
OHCHg32
COmole1
OHCHmole1
COg28
COmole
COg100 3
3
33
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
so, g85.7g114.3-g2001tOH,CH3
===C
and 1-1-
gd0.00667
g200
1
-
g85.7
1
d1
1
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=k

1.11-1.13 The solutions for these problems are the solutions for problems 1.8-1.10 in the
2nd
1.14 Calculate the pipe volume, Vr, and the first-order rate constant, k.
( ) ( ) 1-
1/2
3
2
p min0.0578
min)(12
2ln
t
2ln
andft24,033
4
ft3400ft3.0
==== kV
π
For first-order decay in a steady-state PFR
( ) ( )( ) Lmg2.82
ft0.134
gal
gal10
min
ft033,24min0578.0expmg/L1.0e 34
31-
0 /=⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
== −kt
CC

1.15 The stomach acts like a CSTR reactor in which a first order decay reaction is
occurring.
V, r(C)
Stomach
Q, Ce
Gastric
juices in
Q, Ci
Digested food
stream 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
)(- ei CVrQCQC
dt
dC
V += where r(C) = -kC, Ci = 0, and C = Ce
so, ∫∫ ⎟
⎠
⎞
⎜
⎝
⎛
+=
t
dtk
V
Q
C
dC
0
C
C
-
0
( )( ) ( )⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+−⎟
⎠
⎞
⎜
⎝
⎛
==
⎟
⎠
⎞
⎜
⎝
⎛
+
hr1hr1.33
L1.15
hrmin/60L/min0.012
exp
L1.15
g325
e 1-
- tk
V
Q
CC
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
1.21 a. Calculate the volume that 1 mole of an ideal gas occupies at 1 atm and 20 °C.
( )( )( ) L24.04
atm1
K15.293KmolatmL082056.0mole1 -1-1
=
⋅⋅⋅
==
P
nRT
V then
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.
Q, Ca
V, r(C), kd, ks
Coal Valley
Q, Ce
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
( )esedeaea -)r(-
d
d
CkCkVQCQCCVQCQC
t
C
V +−=+=
Assuming Ca = 0 and integrating yields
( )
⎟
⎠
⎞
⎜
⎝
⎛
×
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
++
×
⎟
⎠
⎞
⎜
⎝
⎛ ⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
⎟
⎠
⎞⎜
⎝
⎛ ++
= ppmv
101
11.0
ln
0.02d0.0856d
m102.0
d
min2460
min
m
106
1
ln
1
5-
1-1-
36
3
5
0
sd
C
C
kk
V
Q
t
t = 0.0322 d = 46.4 min

1.22 a.
( ) NaOHmg6.53L00.1
mmole
mg40
mg98
mmole
L
mg8.00
2NaOH =⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=m
b. The reaction (P1.3) is second-order (see the rate constant’s units) so that
2
HOHH
H
kC--
d
d
- ++
+
== CkC
t
C
After integration and noting that at t = t1/2, C = C0/2, the equation can be written
seconds104.38
mg98
mmole
L
mg00.8
2
sL
mol
101.4
11 -8
110
1/2 ×=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
×
==
kC
t
Therefore, neutralization occurs almost instantly.

1.23 – 1.37 The solutions for these problems are the solutions for problems 1.16-1.30 in
the 2nd

2.1 – 2.14 The solutions for these problems are the solutions for problems 2.1 - 2.14 in
the 2nd
2.15 EDTA (C10N2O8H16) has a molecular weight of 292 g/mole
Calcium molar concentration: (20 mg/L)(40.1 mol/g)-1
(103
mg/g)-1
= 0.000499 M
Mass of EDTA: (0.000499 mol/L)(292 g/mol)(44 gal)(3.785 L/gal) = 24.3 g EDTA

2.16 The balanced equation is: 2Al0
+ 6HCl → 2AlCl3 + 3H2 (g)
So,
2
0
2
2
0
0
2
0
Hg
Alg9
Hg2
Hmole
Almole
Alg27
Hmoles3
Almoles2
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛

2.17 - 2.21 The solutions for these problems are the solutions for problems 2.15 - 2.19 in
the 2nd
2.22 From Table 2.3, Ksp,CaSO4 is 2 x 10-5
and Ksp = [Ca2+
] [SO4
2-
] = [SO4
2-
]2
At saturation: [SO4
2-
] = Ksp
½
= (2 x 10-5
) ½
= 0.00447 mol/L
Check if saturation is exceeded: mol/L0.00368
g136
mole
L
g5.0
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
Since saturation is not exceeded the sulfate concentration is 0.00368 mol/L

2.23 Assume a closed system.
a. [Ca2+
] = Ctot = [H2CO3] + [HCO3
-
] + [CO3
2-
] eqn. 1
Charge balance: [H+
] + 2[Ca2+
] = [HCO3
-
] + 2[CO3
2-
] + [OH-
] eqn. 2
At pH 8.5: Ctot ≈ [HCO3
-
] » [CO3
2-
] (see Figure 2.5) and [H+
] « [OH-
]
So eqn. 2 simplifies to: 2[Ca2+
] = Ctot + [OH-
] eqn. 3
Using eqn. 1 in eqn. 3 yields: [Ca2+
] = [OH-
] = 10-5.5
= 3.16 x 10-6
Note this is less than the Ca2+
initially added, so CaCO3(s) has precipitated
b.
[ ] [ ][ ] ( )( )
( ) M10M102.24
104.47
1010HCOH
CO 7.65--8
7-
-5.5-8.5
1
3
(aq)2 =×=
×
==
−+
K
c.
[ ] ( )( ) g/L0.20g/L0.19968
mol
g100
M10-M102CaCO 5.5-3-
(s)3 ==⎟
⎠
⎞
⎜
⎝
⎛
×=
2.24 - 2.36 The solutions for these problems are the solutions for problems 2.20 - 2.32 in
the 2nd

the 2nd
3.18 a.
rabbits1,750
2
rabbits3,500
2
*
===
K
N
Use the midpoint of the first year population (350 rabbits) as N0
( ) ( )
( )
1-
0
0
0
yr0.635
rabbits3,500
rabbits350
-1rabbits350
yr
rabbits200
-1
d
d
=
⎥
⎦
⎤
⎢
⎣
⎡
=
⎟
⎠
⎞
⎜
⎝
⎛
=
K
N
N
t
N
r
and
( )( ) rabbits/yr556
4
rabbits500,3yr0.635
4d
d -1*
===
rK
t
N
b.
( )( ) ( )
( )
rabbits/yr501
rabbits3,500
rabbits1,200
-1rabbits200,1yr0.635-1
d
d 1-0
0
0
=⎥
⎦
⎤
⎢
⎣
⎡
=⎟
⎠
⎞
⎜
⎝
⎛
=
K
N
rN
t
N

3.19 a.
deer3,500
2
deer7,000
2
*
===
K
N
b. Use the mean current population of deer during the year (2,350 deer) as N0
( ) ( )
( )
1-
0
0
0
yr0.192
deer7,000
deer2,350
-1deer2,350
yr
deer300
-1
d
d
=
⎥
⎦
⎤
⎢
⎣
⎡
=
⎟
⎠
⎞
⎜
⎝
⎛
=
K
N
N
t
N
r
and
( )( ) deer/yr336
4
deer000,7yr0.192
4d
d -1*
===
rK
t
N
c.
( )
( ) ( )
( )( )
( )( )
yr3.55
deer3,500-000,7deer2,350
deer2350-000,7deer3,500
ln
yr0.192
1
-
-
ln
1
1-
0
0
=⎥
⎦
⎤
⎢
⎣
⎡
=⎥
⎦
⎤
⎢
⎣
⎡
=
NKN
NKN
r
t
the 2nd

the 2nd
4.11 a.
( ) 3-
101.05
rems8,000
deathcancer1
yr70
yr
rem0.12
riskDenver ×=⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
( ) 3-
100.35
rems8,000
deathcancer1
yr70
yr
rem0.04
risklevelSea ×=⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
b. Denver deaths due to cosmic radiation exposure:
( ) yr
deaths8.6
rems000,8
death1
people1057.0
yrperson
rem12.0 6
=⎟
⎠
⎞
⎜
⎝
⎛
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
( ) yr
deaths1,077
people000,100
yr
deaths189
people100.57Expected 6
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
×=
c.
( ) ⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞⎜
⎝
⎛==
rems8,000
deathcancer1
yrN
yr
rem0.04-0.1210risklIncrementa 6-
months)2suggests4.3(Tableyr0.1
0.08
108,000
N
-6
=
×
=
d.
( ) yr
deaths1,500
rems000,8
deathcancer1
yrperson
rem0.04
people10300 6
=⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×

4.12 Radon exposure of 1.5 piC/L equivalent to 400 mrem/yr (0.4 rem/yr):
a.
( ) yr
deaths15,000people10300
rems8,000
deathcancer1
yrperson
rem4.0 6
=×⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
b.
lifetime
deathcancer103.5
lifetime
yr70
rems8,000
deathcancer1
yr
rem0.4
Risk 3-
×=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=

4.13-4.14 The solutions for these problems are the solutions for problems 4.13 - 4.14
in the 2nd
4.15 Data is taken from Tables 4.9 and 4.10
a.
( )( )( )
( )( )( )( )durationfreqrateintakePF
yr
d365lifespanbody wtrisk
conc
⎟
⎠
⎞⎜
⎝
⎛
=
( )( )
34-
32-
6-
mg/m104.93
life
yr25
yr
d250
d
m20
mg
dkg102.9
yr
d365
life
yr70
kg7010
×=
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
b.
( )( )( )
( )( )( )( )durationconcrateintakePF
yr
d365lifespanbody wtrisk
freq
⎟
⎠
⎞⎜
⎝
⎛
=
( )( )
d/yr52
life
yr25
m
mg104.93
d
m20
mg
dkg102.9
yr
d365
life
yr70
kg7010
3
4-32-
7-
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=

4.16 a. Use Table 4.11 oral RfDs and Table 4.10 for other factors
HItotal = HIarsenic + HImethylene chloride
3.5
mg0.060
dkg
L
mg0.560
mg0.0003
dkg
L
mg0.070
kg70
d
L1
HI =⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
⎟
⎠
⎞
⎜
⎝
⎛
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
b. No, this is not a safe level, since the HI > 1.
c. Use Table 4.9 for PFs and Table 4.10 for other factors
risk = riskarsenic + riskmethylene chloride
( )
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
+⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
L
mg0.560
mg
dkg0075.0
L
mg0.0700
mg
dkg75.1
yr
d365
life
yr70
kg70
life
yr10
yr
d250
d
L1
risk
-4
101.8risk ×= Since this risk is » 10-6
, it would probably not be considered
acceptable

4.17 a.
( )
5-
3
3
107
yr
d365
life
yr70
kg70
life
yr25
yr
d250
d
m20
m
mg0.00002
mg
dkg50
risk ×=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
=
b. ( )( ) workers15107workers220,000 -5
=×
c. The 2006 U.S. cancer rate of 24% taken from text in Section 4.2 is the same
as the 1992 rate given in Table 4.1
( )( ) workers52,815workers1524.0workers000,220 =+
d. From Table 4.11 RfDs for both arsenic and mercury are 0.0003 mg/kg·d
( )
0.11
m
mg0.0001
m
mg0.00002
kg70
dkg
mg0.0003
d
m20
HI 33
3
=⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
Since the HI is less than 1 the non-cancer hazard is acceptable, despite
there being a cancer risk greater than 10-6
.

the 2nd
4.27 260 million people, 2L/day, 360d/yr, 30yr, find risk and incremental cancers for,
a. trichloroethylene at 0.005 mg/L:
Risk = CDI x Potency
Risk =
2L/d x 0.005 mg/L x 360 d/yr x 30 yr
70 kg x 365 d/yr x 70 yr
x
1.1x10-2
mg/kg - d
= 6.6x10−7
Δ cancer =
260x106
people x 6.6x10-7
cancer/ person − life
70 yr/lifetime
= 2.5 cancer/yr
b. benzene at 0.005 mg/L:
Risk =
2L/d x 0.005 mg/L x 360 d/yr x 30 yr
70 kg x 365 d/yr x 70 yr
x
2.9x10-2
mg/kg - d
= 1.75x10−6
Δ cancer =
260x106
people x 1.75x10-6
cancer/ person − life
70 yr/lifetime
= 6.5 cancer/yr
c. arsenic at 0.01 mg/L:
cancer/yr785
eyr/lifetim70
lifeperson/cancer2.11x10xpeople260x10
=cancer
10x11.2
d-mg/kg
1.75
x
yr70d/yr x365xkg70
yr30d/yr x360xmg/L0.01x2L/d
Risk
4-6
4
=
−
Δ
== −
d. carbon tetrachloride at 0.005 mg/L:
Risk =
2L/d x 0.005 mg/L x 360 d/yr x 30 yr
70 kg x 365 d/yr x 70 yr
x
0.13
mg/kg - d
= 7.9x10−6
Δ cancer =
260x106
people x 7.9x10-6
cancer / person − life
70 yr/lifetime
= 29 cancer/yr
e. vinyl chloride at 0.002 mg/L:
Risk =
2L/d x 0.002 mg/L x 360 d/yr x 30 yr
70 kg x 365 d/yr x 70 yr
x
2.3
mg/kg - d
= 5.6x10−5
Δ cancer =
260x106
people x 5.6x10-5
70 yr/lifetime
= 206 cancer/yr

f. PCBs at 0.0005 mg/L:
Risk =
2L/d x 0.0005 mg/L x 360 d/yr x 30 yr
70 kg x 365 d/yr x 70 yr
x
7.7
mg/kg - d
= 4.6x10−5
Δ cancer =
260x106
people x 4.6x10-5
70 yr/lifetime
=172 cancer/yr
the 2nd

the 2nd
5.34 On a molar basis the C:N:P ratio for algae is 106:16:1 and on a mass basis
it is 41:7.2:1
The mass ratio of the lake water is:
1
4
13.3
03.0
12.0
40.0
P
N
C
==
Therefore, carbon is limiting.
the 2nd

5.54 a. From Table 5.11, η = 0.34 for a sand aquifer
( ) ( )
( )( )
( ) d
m61.2
0.0005
d
m09.00.34
d
d
d
d
====
L
h
v'
L
h
v
K
η
b.
tcontaminan
R
v
v'
= so
d
m0.015
6
d
m0.09
tcontaminan ===
R
v'
v
and the distance traveled is (0.015 m/d) (365 d) = 5.48 m
c. Calculate the maximum contaminant mass dissolved based on the aqueous
solubility and density of PCE (use Table 5.14)
( )( )( )( ) g2,230m5.48m4.0m2.00.34
m
g150 3max =⎟
⎠
⎞⎜
⎝
⎛=m
Calculate the mass leaked
( ) ( ) g59,500d365g1.63
d
mL100leaked =⎟
⎠
⎞⎜
⎝
⎛=
mL
m
Since the mass leaked > soluble mass, the solubility limit is exceeded and the
guideline for the aqueous concentration in the presence of NAPLs (see page
258) is used: CPCE = 0.10 Aq SolPCE = 15 mg/L
d.
( ) ( )( )( ) d
m0.490
d
m0.090.34m42m22
3
capture === BvwQ
e.
( )
( ) ( )
yr22.2d8,100
mg
g10
m
L10
L
mg15
d
m0.490
g59,500
3
3
33
PCE
PCE
==
⎟
⎠
⎞⎜
⎝
⎛⎟
⎠
⎞⎜
⎝
⎛
==
−QC
m
t
5.55 The solution for this problem is the solution for problem 5.53 in the 2nd
edition
Solutions Manual.

6.1 a. A primary standard is an enforceable limit on the concentration of a
contaminant in water or an enforceable requirement that a particular treatment
technique be implemented. Primary standards apply only to contaminants that
impact human health. A secondary standard is a recommended limit on the
concentration of a water constituent or on the measured value of a water quality
parameter (e.g., turbidity). Secondary standards apply to factors that affect a
drinking water’s aesthetic but not human health attributes.
b. An MCL is a primary standard, whereas an MCLG is a maximum
concentration goal for a drinking water contaminant, which would be desirable
based on human health concerns and assuming all feasibility issues such as cost
and technological capability are not considered. An MCLG is not an enforceable
limit, but does provide the health-based concentration, which the MCL should
seek to approach as closely as possible within the constraints of practical
feasibility. There are many contaminants with MCLGs, but no numeric MCL.
These include acrylamide, copper and lead.
6.2 The CWA sets up a system by which the maximum concentration of contaminants
in discharges to surface waters and in the surface waters themselves are set and
enforced, while the SDWA provides the legislative mechanism necessary to set
and enforce the maximum contaminant concentrations in drinking water. The
CWA is designed to ensure that the quality of surface waters in the U.S. is, at a
minimum, appropriate for the beneficial uses for which the water is designated.
The SDWA is designed to ensure that water supplied by public water systems for
human consumption meets acceptable health standards at the point of use.

6.3 a. The hydraulic detention time is:
( ) ( )
( )
hr4.17d0.174
gal7.4805
ftgal/d102.5
ft0.10ft43.0
36
2
==
⎟
⎠
⎞⎜
⎝
⎛×
==
π
θ
Q
V
b. The critical velocity is:
( )
( ) hrft
ft
2.40
ft43.0
hr24
d
gal7.4804
ft
d
gal102.5
2
3
2
36
b
o
⋅
=
⎟
⎠
⎞⎜
⎝
⎛⎟
⎠
⎞⎜
⎝
⎛ ×
==
πA
Q
v
c. The weir loading rate is:
( )
( ) hrft
ft
51.5
ft43.02
hr24
d
gal7.4804
ft
d
gal102.5
rateloadingweir
3
36
w ⋅
=
⋅⋅
⎟
⎠
⎞⎜
⎝
⎛⎟
⎠
⎞⎜
⎝
⎛ ×
==
πL
Q

6.4 From Appendix C: At 20°C, ρ = 998.2 kg·m-3
and μ = 0.00100 kg·m-1
·s-1
a. The particle settling velocity is:
( ) ( )( )( )( )
s
m102.18
sm
kg
00100.018
m100.1kg/m2.9984.0m/s9.807
18
- 5-
2-5322
pp
s ×=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
×
==
μ
ρρ dg
v
b.
( )
( )( ) s
m102.89
m30m10
s606024
d
d/m7,500
4-
3
b
o ×=
⎟
⎠
⎞
⎜
⎝
⎛
⋅⋅
==
A
Q
v
Since vo > vs, less than 100% of the particles will be removed.
c. The settling velocity of the smallest particle which is 100% removed is equal
to vo. So,
( ) ( ) s
μm
36.5
s
m
103.65
m
kg
2.9984.0
s
m
9.807
s
m
1089.2
sm
kg
0.0010018
-
18 5-
2
1
32
4-
2
1
p
s
p =×=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
×⎟
⎠
⎞
⎜
⎝
⎛
⋅
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
=
ρρ
μ
g
v
d

6.5 From Appendix C: At 20°C, ρ = 998.2 kg·m-3
and μ = 0.00100 kg·m-1
·s-1
The length to width ratio is 5, so Ab = 5w2
Set vo = vs and solve for w:
( ) ( )
m42.7
m10
m
kg
998.2-200,1
s
m
9.8075
sm
kg
00100.0
s
m
0.10018
d-5
18
2
1
25
32
3
2
1
2
p
=
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
=
−ρρ
μ
g
Q
w

6.6 From Appendix C: At 15°C, μ = 0.00114 kg·m-1
·s-1
a.
bV
P
G
μ
=
( )
( ) ( )
1-
2
1
2
1 s50.0
m75.3m17.4
sm
kg
0.00114
W186
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=G
Similarly, G2 = 20.1 s-1
and G3 = 10.0 s-1
.
b.
( ) ( )( ) 4-
3
6-153-6
0
3
p
107.54
6
m
mL10mL108.1m100.20
6
×=
××
==Ω
ππ Nd
Q
GV
N
N
π
α b0 4
1
Ω
+=
( ) ( )( )( ) ( ) 17.9
s606024
d
d
m
16,000
m75.3m17.4s0.501054.741.0
1 3
2-1-4
0
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅⋅⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
+=
π
N
N
Therefore, there will be on average 18 singlets per aggregate.

6.7 From Appendix C: At 15°C, μ = 0.00114 kg·m-1
·s-1
a.
Accumulation = Reaction: )r(NV
dt
dN
V =
and
π
α GN
N
Ω
=
4
-)r(
so, ∫∫
Ω
=
tN
N
dt
G
N
dN
0
4
-
0
π
α
which yields:
⎟
⎠
⎞
⎜
⎝
⎛ Ω
= π
α Gt
NN
4
-
0e
b.
( )
1-
2
1
3
3
2
5-
s3.78
m004.0
sm
kg
0.00114
s
mkg
105.6
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
×
==
V
P
G
μ
c.
( )
4-
3
7
35-
0
3
p
101.64
6
m0040.0
100.1
m100.5
6
×=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
×
==Ω
π
π Nd
( ) ( )( )( )( )
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ ××
==
⎟
⎠
⎞
⎜
⎝
⎛ Ω
π
π
α
min30
min
s60s78.31064.141.0
-exp
L4.0
101.0
e
1-4-74
-
0
Gt
NN
L
particles
106.07 5
×=N
d.
aggregate
singlets4.12
L1007.6
L0.4
101.0
1-5
7
0
=
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
=
N
N

e.
( ) ( )( )( )( )
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ ××
==
⎟
⎠
⎞
⎜
⎝
⎛ Ω
π
π
α
min30
min
s60s78.31064.140.2
-exp
L4.0
101.0
e
1-4-74
-
0
Gt
NN
L
particles
101.88 6
×=N

6.8 a. ( ) hr
m
640m80
hr
m
8.0
3
2
fa =⎟
⎠
⎞
⎜
⎝
⎛
== AvQ
b. From the definition of filter efficiency: Vf – (Vb + Vr) = ηfVf (6.8.1)
and the definition of the effective filtration rate gives:
Vf – (Vb + Vr) = refAftc (6.8.2)
Combining (6.8.1) and (6.8.2) and rearranging yields:
( )( )( )
cycle)(perm33,370
0.96
hr52m80
hr
m7.7 3
2
f
cfef
f ===
η
tAr
v
From (6.8.1): ( ) ( )( ) cycle)(perm1,3350.96-1m370,33-1 33
ffrb ===+ ηVVV
Thus, the volume of water lost per cycle (Vb + Vr) is 1,340 m3
.

6.9 Note that this problem incorporates a practically realistic, yet possibly confusing,
nuance in that the total backwash time per cycle (tb) is double the total time that
backwash water is flowing through the filter (tb’). Thus, calculation of the water
volume used in backwash is based on 8 minutes of flow per cycle, whereas the
time the filter is off-line for backwashing is 16 minutes.
a.
( )( )( ) L307,200
min
s60min0.8m64
sm
L
10.0 2
2
'
bf
f
b
b =⎟
⎠
⎞
⎜
⎝
⎛
⋅
== tA
A
Q
V
( )( )( ) L316,800
min
s60min15m64
sm
L
5.50 2
2rfar =⎟
⎠
⎞
⎜
⎝
⎛
⋅
== tAvV
( ) ( ) ( ) min1,409min15-min16min6024rbcf =−⋅=−−= tttt
( )( )( ) L102.98
min
s60min,4091m64
sm
L
5.50 72
2f
f
f
f '
f
×=⎟
⎠
⎞
⎜
⎝
⎛
⋅
== tA
A
Q
V
%980.979
L102.98
L316,800-307,200-102.98
7
7
f
rbf
f ==
×
×
=
−−
=
V
VVV
η
b. The fraction of a filtration cycle that is not backwashing is:
( )
0.989
hr24
min60
hr
min16-hr24
c
brf
=
⎟
⎠
⎞
⎜
⎝
⎛
=
−+
t
ttt
so the number of filters required is:
( )
( )
( ) ( )
6.89
989.0
L1000
m
m64
sm
L
5.50
s
m2.40
989.0 3
2
2
3
fa
plant
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
==
Av
Q
n
Therefore, at least 7 filters are required.

6.10 a. For a steady state PFR, the Giardia concentration in the contactor effluent is:
θ*
-
0e k
NN =
so,
( )( ) ( ) 3
1-
3
0
*b m156000,1ln
min0.53
min
s60
s
m0.20
ln ===
N
N
k
Q
V
b. For a steady state CSTR, the mass balance is:
QN0 = QN + k*
NVb
so,
( )( )
( ) ( ) 3
1-
3
0
*b m22,6001-000,1
min0.53
min
s60
s
m0.20
1- ==⎟
⎠
⎞
⎜
⎝
⎛
=
N
N
k
Q
V
c. For a steady state series of CSTRs, the mass balance is:
( )m*
0
m
1
1
kN
N
θ+
=
so,
( )( )
( ) ( )( ) m67.51-000,1
min0.53
min
s60
s
m0.20
1- 35
1
1-
31
m
0
*b,1 ==
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
m
N
N
k
Q
V
and the total contactor volume is: mVb,1 = 5(67.50 m3
) = 337 m3
d.
minmg
L
0.265
L
mg2.0
min0.53
0.1
-1*
⋅
=
⎟
⎠
⎞⎜
⎝
⎛
== n
C
k
k

6.11 For a steady state CSTR, the mass balance is:
QN0 = QN + k*
NVb
so,
( )( )
( ) s
L0.0156
1-1,000
s80.7L2.00
1-
-1
0
*
b
==
⎟
⎠
⎞
⎜
⎝
⎛
=
N
N
kV
Q
and the volume produced during ten hours of operation would be:
( ) L562
hr
s6060
hr10
s
L
0.0156 =⎟
⎠
⎞
⎜
⎝
⎛ ⋅
⎟
⎠
⎞
⎜
⎝
⎛
=Qt

6.12 Hardness in meq/L:
meq/L7.50
mg/meq20.0
mg/L150
Ca2
==+
meq/L4.92
mg/meq12.2
mg/L60
Mg2
==+
Total hardness = 7.50 + 4.92 = 12.4 meq/L
Hardness as CaCO3:
3
3
CaCOasmg/L621
meq
CaCOasmg
50.0
L
meq
12.4hardnessTotal =×=
Table 6.4 would classify this as very hard water.

the 2nd
6.17
Component Concentration
(mg/L)
Equiv. Weight
(mg/meq)
Concentration
(meq/L)
CO2 14.5 22 0.6591
Ca2+
110.0 20 5.500
Mg2+
50.7 12.2 4.156
Na+
75 23.0 3.261
HCO3
-
350 61.0 5.738
SO4
2-
85.5 48.0 1.781
Cl-
16.2 35.5 0.4563
pH 8.2 - -
9.656
5.738
0
5.500
0
Ca2+
HCO3
-
CH
Mg2+
Ca-CH
NCH
Mg-NCH
9.656
5.738
0
5.500
0
Ca2+
HCO3
-
CH
Mg2+
Ca-CH
NCH
Mg-NCH
a. Mg-CH = CH – Ca-CH = 5.738 –5.500 = 0.238 meq/L
b. Mg-NCH = TH – CH = 9.656 – 5.738 = 3.918 meq/L
c. 1 meq/L Ca(OH)2 neutralizes 1 meq/L CO2(aq)
so, Ca(OH)2 required = 0.6591 meq/L

6.18
Component Concentration Equiv. Weight
(mg/meq)
Concentration
(meq/L)
CO2 6.0 mg/L 22 0.2727
Ca2+
50.0 mg/L 20 2.500
Mg2+
20.0 mg/L 12.2 1.639
HCO3
-
3.1 mmole/L 61.0 3.100
pH 7.6 - -
4.139
3.100
0
2.500
0
Ca2+
HCO3
-
CH
M g2+
Ca-CH
NCH
M g-NCH
4.139
3.100
0
2.500
0
Ca2+
HCO3
-
CH
M g2+
Ca-CH
NCH
M g-NCH
Table P6.18 Components, lime and soda ash dosage, and solids generated
Component (eq’n.) Concentration
(meq/L)
Lime
(meq/L)
Soda ash
(meq/L)
CaCO3(s)
(meq/L)
Mg(OH)2(s)
(meq/L)
CO2(aq) (6.37) 0.2727 0.2727 0 0.2727 0
Ca-CH (6.38) 2.500 2.500 0 5.000 0
Mg-CH (6.39) 0.600 1.200 0 1.200 0.600
Ca-NCH (6.40) 0 0 0 0 0
Mg-NCH (6.41) 1.039 1.039 1.039 1.039 1.039
Excess 0.400
Totals 5.412 1.039 7.512 1.639
a. Lime required = (5.412 meq/L)(37.07 mg/meq) = 200.5 mg/L
Soda ash required = (1.039 meq/L)(53.0 mg/meq) = 55.1 mg/L
b. Sludge generated:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
mg10
kg
gal
L785.3
d
gal1015
meq
mg
9.22
L
meq
.6391
meq
mg
0.50
L
meq
512.7 6
6
kg/d24,040=

6.19
(mg/meq)
Concentration
(meq/L)
Concentration
(mg/L CaCO3)
Ca2+
90 mg/L 20 4.500 225.0
Mg2+
30 mg/L 12.2 2.4596 123.0
HCO3
-
165 mg/L 61.0 2.705 135.3
pH 7.5 - - -
348.0
135.3
0
225.0
0
Ca2+
HCO3
-
CH
M g2+
Ca-CH
NCH
M g-NCHCa-NCH
348.0
135.3
0
225.0
0
Ca2+
HCO3
-
CH
M g2+
Ca-CH
NCH
M g-NCHCa-NCH
( )[ ] [ ][ ]
3
4-
7-
5.7
1
-
3
aq2 CaCOmg/L19.14M101.914
M104.47
L
mol
10
g61.0
mol
L
g
0.165
HHCO
CO =×=
×
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
−
+
K
(mg/L
CaCO3)
Lime
(mg/L
CaCO3)
Soda ash
(mg/L
CaCO3)
CaCO3(s)
(mg/L
CaCO3)
Mg(OH)2(s)
(mg/L
CaCO3)
CO2(aq) (6.37) 19.14 19.14 0 19.14 0
Ca-CH (6.38) 135.3 135.3 0 270.6 0
Mg-CH (6.39) 0 0 0 0 0
Ca-NCH (6.40) 90.00 0 90.00 90.00
Mg-NCH (6.41) 123.0 123.0 123.0 123.0 123.0
Excess 20
Totals 297.4 213.0
a.
L
mg220.1
CaCOmg100
Ca(OH)mg74
L
CaCOmg297.4
Lime
3
23
required =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
b.
L
mg225.8
CaCOmg100
Ca(OH)mg061
L
CaCOmg213.0
AshSoda
3
23
required =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=

edition
Solutions Manual.
6.21
(mg/meq)
Concentration
(meq/L)
Concentration
(mg/L CaCO3)
Ca2+
95 mg/L 20 4.75 238
Mg2+
26 mg/L 12.2 2.13 107
HCO3
-
160 mg/L 61.0 2.62 131
pH 7.0 - - -
6.88
2.62
0
4.75
0
Ca2+
HCO3
-
CH
Mg2+
Ca-CH
NCH
Mg-NCHCa-NCH
6.88
2.62
0
4.75
0
Ca2+
HCO3
-
CH
Mg2+
Ca-CH
NCH
Mg-NCHCa-NCH
( )[ ] [ ][ ] meq/L1.17M105.868
M104.47
L
mol
10
g61.0
mol
L
g
0.160
HHCO
CO 4-
7-
7
1
-
3
aq2 =×=
×
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
−
+
K
(meq/L)
Lime
(meq/L)
Soda ash
(meq/L)
CaCO3(s)
(meq/L)
Mg(OH)2(s)
(meq/L)
CO2(aq) (6.37) 1.17 1.17 0 1.17 0
Ca-CH (6.38) 2.62 2.62 0 5.24 0
Mg-CH (6.39) 0 0 0 0 0
Ca-NCH (6.40) 2.13 0 2.13 2.13 0
Mg-NCH (6.41) 2.13 2.13 2.13 2.13 2.13
Excess 0.400
Totals 6.32 4.26 10.67 2.13
a.
L
mg234
meq
mg7.073
L
meq6.32
Limerequired =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
b.
L
mg226
meq
mg35
L
meq4.26
AshSoda required =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=

edition
Solutions Manual.
6.23
(mg/meq)
Concentration
(meq/L)
Concentration
(mg/L CaCO3)
Ca2+
40.0 mg/L 20 2.00 100
Mg2+
10.0 mg/L 12.2 0.820 41.0
HCO3
-
110 mg/L 61.0 1.80 90.2
pH 7.0 (assumed) - - -
2.82
1.80
0
2.00
0
Ca2+
HCO3
-
CH
Mg2+
Ca-CH
NCH
Mg-NCH
Ca-NCH
2.82
1.80
0
2.00
0
Ca2+
HCO3
-
CH
Mg2+
Ca-CH
NCH
Mg-NCH
Ca-NCH
( )[ ] [ ][ ] meq/L0.807M104.034
M104.47
L
mol
10
g61.0
mol
L
g
0.110
HHCO
CO 4-
7-
7
1
-
3
aq2 =×=
×
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
−
+
K
(meq/L)
Lime
(meq/L)
Soda ash
(meq/L)
CaCO3(s)
(meq/L)
Mg(OH)2(s)
(meq/L)
CO2(aq) (6.37) 0.807 0.807 0 0.807 0
Ca-CH (6.38) 1.80 1.80 0 3.60 0
Mg-CH (6.39) 0 0 0 0 0
Ca-NCH (6.40) 0.200 0 0.200 0.200 0
Mg-NCH (6.41) 0.820 0.820 0.820 0.820 0.820
Excess 0.400
Totals 3.83 1.02 5.43 0.82
a.
L
mg142
meq
mg7.073
L
meq3.83
Limerequired =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
b.
L
mg54.1
meq
mg35
L
meq1.02
AshSoda required =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=

c. Sludge generated:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
mg10
kg
gal
L785.3
d
gal1073
meq
mg
9.22
L
meq
.820
meq
mg
0.50
L
meq
.435 6
6
kg/d41,400=

6.24
Qf = 5 ×106
L/d Qp = 3 ×106
L/d
Cf = 1,500 mg/L Cp = 75 mg/L
Qc, Cc
QfCf = QcCc + QpCp and Qf = Qc + Qp
So,
( )
( )
( )( ) ( )( )[ ]
( ) mg/L3,640
L/d103-105
mg/L57L/d103-mg/L500,1L/d105
-
-
66
66
pf
ppff
c =
××
××
==
QQ
CQCQ
C

6.25
Qf = 5 ×106
L/d Qp = 3 ×106
L/d
Cf = 1,500 mg/L Cp = 75 mg/L
Cc = 3,640 mg/L
Qc = 2 ×106
L/d
a. Water recovery:
60%or0.60
d
L105
d
L103
6
6
F
P
=
×
×
==
Q
Q
r
b. Salt rejection:
95%or0.95
L
mg1,500
L
mg75
-1-1
F
P
===
C
C
R
c. Log salt rejection:
( ) ( ) 1.300.95-1log--1log-log
=== RR

6.26 a. The main constituent of concern from the perspective of configuring the
POTW treatment train is biodegradable organic matter. It is, however, arguable
from the perspective of human health that the main constituents of concern are
human pathogens.
b. Unit operation function based on an overall aim of removing BOD:
i. The grit chamber removes the very largest and most settleable
particles, which may contain a modest fraction of organic matter.
ii. The primary sedimentation basin removes most of the
gravitationally settleable organic matter (as well as inorganic
matter). Typically about 35% of the BOD can be removed by
primary sedimentation.
iii. The bioreactor converts dissolved and fine particulate
biodegradable organic matter into microbial cell mass and energy
for microbial metabolism.
iv. The secondary clarifier physically removes the cell mass generated
in the bioreactor by gravitational settling.
v. Digestors further degrade the organic particles in the primary
clarifier sludge and/or the microbial cell mass separated into the
secondary clarifier sludge by exposing it to more prolonged
biodegradation.

6.27 Both primary and secondary wastewater treatment are designed to remove
biodegradable organic matter (BOM) and the superset of total solids. Primary treatment
only removes that BOM which can be physically separated from the raw sewage by
floatation, gravitational settling or screening. On the other hand, secondary treatment
removes BOM that may be biodegraded by microbes within a relatively short duration
(typically the hydraulic retention time is less than 1 day). Much of the BOM degraded in
secondary treatment is dissolved or colloidal and it is converted into microbial cell mass.
The cells are removed from the secondary effluent by settling in a secondary clarifier (or
by exclusion by a membrane).
the 2nd

6.30
well-mixed
aeration pond
Q, S0, X0
Q, S, X
V, S, X, rg’, rsu
Q = 30 m3
/d
S0 = 350 mg/L BOD5
S = 20 mg/L BOD5
Ks = 100 mg/L BOD5
kd = 0.10 d-1
μm = 1.6 d-1
Y = 0.60 mg VSS/mg BOD5
a. Assuming X0 = 0, the microbe mass balance yields equation 6.57:
( )
( )dm
ds
-1-
1
k
kK
S
θθμ
θ+
=
and solving for the hydraulic detention time, θ,
( )[ ]
( )
( ) ( )[ ] d6.0
d1.0BODmg/L100-d0.1-6.1BODmg/L20
BODmg/L20100
-- 1-
5
1-
5
5
sdm
s
=
+
=
+
=
dkKkS
SK
θ
μ
b. From equation 6.60 and 6.51,
( )( )
( )( ) L
VSSmg
7.07
L
BODmg
350d6.1d6.0
L
BODmg
20-350
L
BODmg
20100
BODmg
VSSmg
0.60
-
51-
55
50s
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
+
=
KS
SSSKY
X
θ
and at steady state with X0 = 0, the effluent flux of microbial mass must equal the
rate of microbial mass production in the pond:
'
grVQX =
and,
d
VSSkg
0.212
mg01
kg
L
VSSmg
07.7
m
L
000,1
d
m
30 63
3
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=QX

6.31
well-mixed
aeration lagoon
Q, S0, X0
Q, S, X
V, S, X, rg’, rsu
l = 60 m
w = 5 m
h = 2 m
Q = 30 m3
/d
S0 = 350 mg/L BOD5
S = 20 mg/L BOD5
Ks = 100 mg/L BOD5
kd = 0.10 d-1
μm = 1.6 d-1
a. The lagoon’s hydraulic detention time is:
( )( )( ) d1.50
d
m400
m2m5m60
3
===
Q
V
θ
and the steady state substrate concentration in the lagoon is:
( )
( )
( )( )( )
( )( ) ( )( )[ ] L
BODmg
161
d08.0d1.5-1-d10.1d1.5
d08.0d1.51
L
BODmg
76.0
-1-
1 5
1-1-
1-5
dm
ds
=
+⎟
⎠
⎞
⎜
⎝
⎛
=
+
=
k
kK
S
θθμ
θ
Therefore, the lagoon’s BOD5 removal efficiency is:
( ) ( ) 52.2%100
L
BODmg
336
L
BODmg
161-336
100
-
5
5
0
0
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
S
SS
Eff
c. The lagoon’s daily oxygen demand is:
( )
d
Okg
70.2
mg10
kg
m
L000,1
L
BODmg
161-336
d
m
400- 2
63
5
3
0 =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=SSQ

6.32
Q, S0 Qs Qe, Xe, S0
WAS
Qw, Sw, Xw
RAS
Qr, Sr, Xr
V, X, SP.C.
Act. Sludge Basin
S.C.
Q = 0.300 m3
/s X = 2,100 mg VSS/L Xr = 10,000 mg VSS/L
θc = 9.0 d S0 = 220 mg BOD5/L
dVSSmg
BODmg
0.52 5
⋅
=
M
F
a. Based on the definition of the food to microbe ratio,
( )
3
5
5
3
0
m5,220
L
VSSmg
100,2
dVSSmg
BODmg
0.52
d
s606024
L
BODmg
220
s
m
0.300
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
⎟
⎠
⎞
⎜
⎝
⎛ ⋅⋅
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
X
M
F
QS
V
b. Noting that Xr = Xw, the cell retention time can be used to calculate the WAS
flow rate.
( )
( ) d
m
122
L
VSSmg
000,10d9.0
L
VSSmg
100,2m5220 3
3
wc
w =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
==
X
VX
Q
θ
c. From a hydraulic flow balance around the activated sludge basin, recycle line
and secondary clarifier,
s
m
0.299
s606024
d
d
m
122-
s
m
0.300-
333
we =⎟
⎠
⎞
⎜
⎝
⎛
⋅⋅⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
== QQQ

d. First, write a microbial mass balance around the secondary clarifier (S.C.).
QsX = QeXe + (Qr + Qw)Xw
Noting that Xe ≈ 0 and Qr = Qs – Qe - Qw, simplify and solve for Qs.
( ) s
m
0.378
L
VSSmg
2,100-10,000
L
VSSmg
000,10
s
m
0.299
-
3
3
w
we
s =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==
XX
XQ
Q
and the hydraulic detention time is then:
( ) d0.160
d
s606024
s
m
0.378
m5220
3
3
s
=
⎟
⎠
⎞
⎜
⎝
⎛ ⋅⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==
Q
V
θ

6.33 Q, C0 Qe, Ce
Qs, Cs
Q = 20 MGD
Qs = 0.070 MGD
C0 = 800 mg TS/L
Ce = (1 – 0.18)C0 = 656 mg TS/L
a. A mass balance around the primary clarifier yields:
( ) ( )
( )
( ) ( ) ⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=+=
L
TSmg
8000.82
L
TSmg
8000.82-800
MGD0.070
ΜGD20
- ee0
s
s CCC
Q
Q
C
L
TSmg
41,800s =C
b. The mass of solids removed annually is:
( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅⋅⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
==
yr
s365606024
mg10
kg
L
mg
800,41
m
L1,000
MGD
s
m0.04381
MGD0.070 63
3
ss tCQm
yearpertonne4,040
yr
TSkg
104.04 6
=×=m

Pg. 7.1
SOLUTIONS FOR CHAPTER 7
7.1 From (1.9), mg / m
3
=
ppm x mol wt
24.465
(at 1 atm and 25
o
C)
a. CO2mg / m
3
=
5000ppm x 12 + 2x16( )
24.465
= 8992mg/m
3
≈ 9000mg / m
3
b. HCHO ppm =
24.465 x 3.6 mg/m3
2x1 + 12 + 16( )
= 2.94ppm
c. NO mg /m
3
=
25ppm x 14 +16( )
24.465
= 30.7mg/m
3
7.2 70% efficient scrubber, find S emission rate:
600 MWe
η = 0.38
600/0.38=1579 MWt
9000 Btu/lb coal 1% S
Input =
600, 000 kWe
0.38
x
3412 Btu
kWhr
x
lb coal
9000Btu
x
0.01 lb S
lb coal
= 5986 lb S/hr
70% efficient, says release 0.3 x 5986 lbS/hr = 1796 lb S/hr ≈1800 lbS/hr
7.3 If all S converted to SO2 and now using a 90% efficient scrubber:
SO2 = 0.1 x
5986 lbS
hr
x
(32 + 2x16) lb SO2
32 lb S
= 1197 lb SO2 / hr ≈ 1200 lb SO2 /hr
7.4 70% scrubber, 0.6 lb SO2/106 Btu in, find % S allowable:
a.
X lbs S
lbs coal
x
0.3 lbs S out
1 lb S in
x
2 lbs SO2
lb S
x
lb coal
15, 000Btu
=
0.6 lb SO2
10
6
Btu
X =
15,000x0.6
0.3x2x10
6 = 0.015 = 1.5% S fuel
b.
X lbs S
lbs coal
x
0.3 lbs S out
1 lb S in
x
2 lbs SO2
lb S
x
lb coal
9, 000Btu
=
0.6 lb SO2
10
6
Btu
X =
9, 000x0.6
0.3x2x10
6 = 0.009 = 0.9% S fuel

Pg. 7.2
7.5 Compliance coal:
1.2 lbs SO2
10
6
Btu
=
lb coal
12, 000 Btu
x
X lb S
lb coal
x
2 lb SO2
lb S
X =
1.2x12, 000
2x10
6 = 0.0072 = 0.7%S
7.6 Air Quality Index:
_________________________________________________________
Pollutant Day 1 Day 2 Day 3
_________________________________________________________
O3, 1-hr (ppm) 0.15 0.22 0.12
CO, 8-hr (ppm) 12 15 8
PM2.5, 24-hr (µg/m3
) 130 150 10
PM10, 24-hr (µg/m3
) 180 300 100
SO2, 24-hr (ppm) 0.12 0.20 0.05
NO2, 1-hr (ppm) 0.4 0.7 0.1
___________________________________________________________
Using Table 7.3:
a. Day 1: Unhealthy, AQI 151-200 triggered by PM2.5.
b. Day 2: Very Unhealthy, AQI 201-300, triggered by both O3 and NO2
c. Day 3: Moderate, AQI 51-100, triggered by CO, PM10 and SO2
7.7 8 hrs of CO at 50 ppm, from (7.6):
%COHb = 0.15% 1 − e
−0.402/ hr x 8hr
( )x50 = 7.2%
7.8 Tractor pull at 436 ppm CO:
a. 1 hr exposure: %COHb = 0.15% 1 − e
−0.402t
( ) ppm( )= 0.15% 1− e
−0.402x1
( )x436 = 21.6%
b. To reach 10% COHb,
10 = 0.15 1− e
−0.402t
( )x436 = 65.4 − 65.4e
−0.402t
e
−0.402t
=
55.4
65.4
= 0.871 so t = -
1
0.402
ln 0.871( ) = 0.41 hr
7.9 RH to produce HCHO:
RO • +O2 → HO2 • +R' CHO (7.19)
for R' CHO to be HCHO, R' must be H so that
RO • +O2 → HO2 • +HCHO
for the reaction to balance , R = CH3
which says RH in (7.16) must be CH4 (methane)

Pg. 7.3
7.10 RH = propene = CH2=CH-CH3 = C3H6 so, R = C3H5
so the sequence of reactions (7.16) to (7.19) is:
C3H6 + OH• → C3H5 • +H2O
C3H5 • +O2 → C3H5O2 •
C3H5O2 • +NO → C3H5O • +NO2
C3H5O • +O2 → HO2 • +C2H3CHO
The end product is acrolein, CH2CHCHO.
7.11 A 20-µm particle blown to 8000 m. From (7.24) its settling velocity is
v =
d2
ρg
18η
=
(20x10−6
m)2
x 1.5x106
g / m3
x 9.80m/s2
18 x 0.0172g/m - s
= 0.019m/s
Time to reach the ground =
8000m
0.019m /s x 3600s/hr x 24hr/d
= 4.87days
Horizontal distance = 4.87 days x 10 m/s x 3600s/hr x 24hr/d x 10-3km/m = 4200 km
7.12 Residence time for 10-µm particle, unit density, at 1000m:
Settling velocity = v =
d2
ρg
18η
=
(10x10−6
m)2
x 106
g / m3
x 9.80m/s2
18 x 0.0172g/m - s
= 0.00317m/s
Residence time = τ =
h
v
=
1000m
0.00317m /s x 3600s/hr
= 87.6hrs
7.13 Settling velocity and Reynolds numbers:
a. 1 µm: v =
d2
ρg
18η
=
(1x10−6
m)2
x 106
g / m3
x 9.80m/s2
18 x 0.0172g/m - s
= 3.2x10
−5
m /s
Re =
ρair dv
η
=
1.29x103
g/ m3
x 1x10-6
m x 3.17x10-5
m/s
0.0172 m/s
= 2.4x10
−6
b. 10 µm: v =
d2
ρg
18η
=
(10x10−6
m)2
x 106
g / m3
x 9.80m/s2
18 x 0.0172g/m - s
= 3.2x10
−3
m/s
Re =
ρair dv
η
=
1.29x103
g/ m3
x 10x10-6
m x 3.17x10-3
m/s
0.0172 m/s
= 2.4x10
−3
c. 20 µm: v =
d2
ρg
18η
=
(20x10−6
m)2
x 106
g / m3
x 9.80m/s2
18 x 0.0172g/m - s
= 0.0127m/s

Pg. 7.4
Re =
ρair dv
η
=
1.29x103
g/ m3
x 20x10-6
m x 0.0127m/s
0.0172 m/s
= 0.02
So for all of these particles, the Reynolds number is much less than 1, which means (7.24) is
a reasonable approximation to the settling velocity.
7.14 Finding the percentage by weight of oxygen and the fraction (by weight) of oxygenate
needed to provide 2% oxygen to the resulting blend of gasoline.
a. Ethanol, CH3CH2OH
€
Oxygen
Ethanol
=
16
2x12 + 6x1+1x16
= 0.347 = 34.7%
€
0.02 =
Xg oxygenate x %O in oxygenate
Yg fuel
€
Ethanol
Fuel blend
=
X
Y
=
2%
%O
=
2%
34.7%
= 0.058 = 5.8% by weight
b. Methyl tertiary butyl ether (MTBE), CH3OC(CH3)3
€
Oxygen
MTBE
=
16
5x12 +12x1+1x16
= 0.182 =18.2%
€
MTBE
Fuel
=
2%
%O
=
2%
18.2%
= 0.11=11%
c. Ethyl tertiary butyl ether (ETBE), CH3CH2OC(CH3)3
€
Oxygen
ETBE
=
16
6x12 +14x1+1x16
= 0.157 =15.7%
€
ETBE
Fuel
=
2%
%O
=
2%
15.7%
= 0.127 =12.7%
d. Tertiary amyl methyl ether (TAME), CH3CH2C(CH3)2OCH3
€
Oxygen
TAME
=
16
6x12 +14x1+1x16
= 0.157 =15.7%
€
TAME
Fuel
=
2%
%O
=
2%
15.7%
= 0.127 =12.7%
7.15 Ethanol fraction CH3CH2OH (sg = 0.791) in gasoline (sg = 0.739) to give 2% O2:
€
Oxygen
Ethanol
=
16
2x12 + 6x1+1x16
= 0.347 = 34.7%
€
0.02 =
X (mL eth) x 0.791 g eth /mL x 0.347 gO/g eth
X (mL eth) x 0.791 g eth/mL + Y (mL gas) x 0.739 g gas/mL
€
0.02 =
0.2745X
0.791X + 0.739Y
=
0.2745
0.791+ 0.739Y /X

Pg. 7.5
€
Y
X
=
0.2745 − 0.02x0.791
0.02x0.739
=17.50
€
ethanol
fuel blend
=
X(mL eth)
X(mL eth) +Y(mL gas)
=
1
1+ Y /X
=
1
1+17.50
= 5.4% by volume
7.16 The CAFE fuel efficiency for flex-fuel cars that get:
a. 18 mpg on gasoline and 12 mpg on ethanol.
b. 22 mpg on gasoline and 15 mpg on ethanol
c. 27 mpg on gasoline and 18 mpg on ethanol
€
a. CAFE mpg =
1 mile
0.5 mile
18 mile/gal gas
+
0.5 mile
12 mile/gal alcohol
x
0.15 gal gas
1 gal alcohol












= 29.4 mpg
€
b. CAFE mpg =
1 mile
0.5 mile
22 mile/gal gas
+
0.5 mile
15 mile/gal alcohol
x
0.15 gal gas
1 gal alcohol












= 36.1 mpg
€
c. CAFE mpg =
1 mile
0.5 mile
27 mile/gal gas
+
0.5 mile
18 mile/gal alcohol
x
0.15 gal gas
1 gal alcohol












= 44.1 mpg
7.17 At 25 miles/100 ft3
of natural gas, 0.823 gallons of gasoline equivalents per 100 ft3
,
and each equivalent gallon counting as 0.15 gallons of gasoline gives a CAFÉ rating of
€
100 ft3
ngas
25 mile
x
0.823 equiv. gal gasoline
100 ft3
ngas
x
0.15 gal gasoline
1 equiv. gal gasoline
= 0.004938 gal gasoline/mile
€
CAFE mpg =
1
0.004938 gal/mile
= 202.5 miles/gallon ≈ 203 mpg
7.18 The “break-even” price of E85 with gasoline at $3.50/gallon:
From Table 7.5, the energy ratio E85/gasoline = 81,630/115,400 = 0.70736
So E85 should cost no more than 0.70736 x $3.50 = $2.48/gallon.
7.19 On an energy-content basis, the cheapest would be:
a. E85 at $2/gallon or gasoline at $3/gallon
E85 < $3/gallon x 0.71 = $2.13. E85 is cheaper.
b. E85 at $2.50/gallon or gasoline at $3.30/gallon
E85 < $3.30/gallon x 0.71 = $2.34. Gasoline is cheaper
c. E85 at $2.75/gal or gasoline at $4/gal
E85 < $4.00/gallon x 0.71 = $2.84 E85 is cheaper

Pg. 7.6
7.20 With a 15-gallon fuel tank and 25 mpg on gasoline.
a. E10 = 0.10 x 75,670 Btu/gal + 0.90 x 115,400 Btu/gal = 111,427 Btu/gal
€
111,427
115,400
x 25 mpg = 24.14 mpg
Range = 15 gal x 24.14 mi/gal = 362 miles
b. E85 = 0.85 x 75,670 + 0.15 x 115,400 = 81,630 Btu/gal
€
81,630
115,400
x 25 mpg =17.7 mpg
c. E85/gasoline = 0.5 x 81,630 + 0.5 x 115,400 = 98,515 Btu/gal
€
98,515
115,400
x 25 mpg = 21.3 mpg
7.21 A 45-mpg PHEV, 30-mile/day on electricity at 0.25 kWh/mile; 50 mi/d, 5 days per
week, 2 days @ 25 mi /d .
a. At $3.50 per gallon and $0.08/kWh:
.
€
Electric = (30 mi/d x 5 d/wk +25 mi/d x 2 d/wk) x 0.25 kWh/mi x $0.08/kWh = $4/wk
€
Gasoline =
(20 mi/d x 5 d/wk)
45 mi/gal
x $3.50/gal = $7.78/wk
€
PHEV cost/mile =
$7.78 +$4.00
(50x5 +25x2) mi
= $0.0393 = 3.93¢/mile
Original 50 mpg
€
HEV cost/mile =
$3.50/gallon
50 mi/gal
= 7¢/mile
b. On an annual basis:
PHEV annual cost = $0.0393/mi x 52 wk/yr x 300 mi/wk = $612/yr
€
HEV annual cost =
$3.50/gallon
50 mi/gal
x52wk/yr x 300mi/wk = $1092/yr
Annual savings = $1092 - $612 = $480/yr
c. At $3000 for batteries:
€
Simple payback =
Extra 1st cost
Annual savings
=
$3000
$480/yr
= 6.25 yr
Batteries would need to last 6.25 yr x 300 mi/wk x 52 wk/yr = 97,500 miles

Pg. 7.7
7.22 From Figure 7.28 the well-to-wheels CO2/mile are:
a. 25 mpg car:
€
11.2 kg CO2/gal x 1000 g/kg
25 mi/gal
= 448 gCO2/mile
b. 50 mpg car:
€
50 mi/gal
= 224 gCO2/mile
c. PHEV:
€
gasoline :
45 mi/gal
= 248 gCO2/mile
€
electricity : 0.25 kWh/mi x 640 gCO2/mi =160 gCO2/mile
€
Half gas, half electricity : 0.5 x 248 + 0.5 x 160 = 204 gCO2/mile
d. EV:
€
electricity : 0.25 kWh/mi x 640 gCO2/mi =160 gCO2/mile
e. FCV:
€
14.4 gC/MJ
0.36 mi/MJ
x
(12 +2x16)gCO2
12gC
x
MJ CH4
0.61 MJ H2
= 240 gCO2/mile
7.23 At 0.25 kWh/mi from a 60%-efficient NGCC plant with a 96%-efficient grid, 14.4
gC/MJ of n. gas and 1 kWh = 3.6 MJ:
a. The EV carbon emissions would be
€
14.4 gC/MJ in
1 MJ CH4
x
44 gCO2
12gC
x
1 MJ in
0.6 MJ out
x
3.6 MJ out
kWh
x 0.96 grid = 304 gCO2/kWh
0.25 kWh/mi x 304 gCO2/kWh = 76 gCO2/mi
b. For the PHEV, half miles on gasoline and half on electricity:
€
gasoline :
45 mi/gal
= 248 gCO2/mile
electricity : 0.25 kWh/mi x 304 gCO2/kWh = 76 gCO2/mi
€
Half gas, half electricity : 0.5 x 248 + 0.5 x 76 = 162 gCO2/mile
7.24 With 5.5 hr/day of sun, 17%-efficient PVs, 75% dc-ac, 0.25 kWh/mile, 30 mi/day:
Electricity needed = 0.25 kWh/mile x 30 miles/day = 7.5 kWh/day
€
Area =
7.5 kWh/d
5.5 h/d x 1 kW/m2
x 0.17 x 0.75
=10.7 m2
=116 ft2

Pg. 7.8
7.25 A 50%-efficient SOFC, 50% into electricity, 20% into useful heat, compared to 30%-
efficient grid electricity and an 80% efficient boiler:
Assume 100 units of input energy to the SOFC, delivering 50 units of electricity and 20
units of heat.
For the grid to provide 50 units of electricity: Input energy = 50/0.30 = 167 units
For the boiler to provide 20 units of heat: Input energy = 20/0.80 = 25 units
Total for the separated system = 167 + 25 = 192 units
Energy savings = (192-100)/192 = 0.48 = 48%
7.26 NG CHP versus separated systems; Natural gas 14.4 gC/MJ, grid 175 gC/kWh. The
joule equivalent of one kWh of electricity is 3.6 MJ.
a. CHP with 36% electrical efficiency and 40% thermal efficiency versus an 85%-
efficient gas boiler for heat and the grid for electricity.
CHP: Assume 100 MJ input to the CHP delivering 36 MJ electricity and 40 MJ heat.
Carbon emissions would be 14.4 gC/MJ x 100 MJ = 1,440 gC
Separate: Grid electricity = 36 MJ/(3.6 MJ/kWh) = 10 kWh x 175 gC/kWh = 1750 gC
Boiler = 40 MJ/0.85 = 47 MJ x 14.4 gC/MJ = 677 gC
Total carbon = 1750 + 677 = 2427 gC
Savings: (2427 – 1440)/2427 = 0.41 = 41%
b. CHP with 50% electrical & 20% thermal efficiency vs a 280 gC/kWh, coal-fired
power plant for electricity and an 80% efficient gas-fired boiler for heat.
CHP: Assume 100 MJ input to the CHP:
Carbon emissions would be 14.4 gC/MJ x 100 MJ = 1,440 gC
Separate: Coal 50 MJ/(3.6 MJ/kWh)=13.89 kWh x 280 gC/kWh = 3889 gC
Boiler = 20 MJ/0.80 = 25 MJ x 14.4 gC/MJ = 360 gC
Total carbon = 3889 + 360 = 4249 gC
Savings: (4249 – 1440)/4249 = 0.66 = 66%

Pg. 7.9
7.27 Power plants emitting 0.39 x 1012
g particulates from 685 M tons coal with a heat content
of 10,000 Btu/kWh while generating 1400 billion kWh/yr.
heat input = 685x10
6
tons x 2000
lb
ton
x 10, 000
Btu
lb
= 1.37x10
16
Btu
efficiency =
output
input
=
1400x109
kWh x 3412Btu/kWh
1.37x10
16
Btu
= 0.349 ≈ 35%
At NSPS of 0.03 lb particulates per 106 Btu input, emissions would have been:
emissions =
0.03 lb
10
6
Btu heat input
x 1.37x10
16
Btu in x
1000g
2.2 lb
=1.87x10
11
g
For comparison,
emissions at NSPS
actual emissions
=
1.87x1011
g
0.39x10
12 = 0.48 = 48%
7.28 Derivation for the dry adiabatic lapse rate:
dQ = dU + dW where dU = Cvdt and dW = PdV
dQ = Cvdt + PdV (1)
ideal gas law says PV = nRT
so, d(PV) = PdV + VdP = nRT
or, PdV = nRT - VdP
plugged into (1) gives:
dQ = CvdT + nRdT − VdP
dQ
dT
= Cv + nR − V
dP
dT
(2)
at constant pressure :
dQ
dT
= Cv + nR = Cp
putting that into (2) gives,
€
dQ
dT
= Cp −V
dP
dT
or,
dQ = CpdT −VdP which is (7.37)

Pg. 7.10
7.29 Plotting the data, extending from ground level to crossing with ambient profile at the
adiabatic lapse rate, and extending from the stack height gives:
2120191817161514
0
200
400
600
800
Temperature (C)
Altitude(m)
plume risemixing depth
a) mixing depth (projecting from 20oC at 0-m at slope -1o/100m) = 400 m
b) plume rise (projecting from 21oC at 100m) = 500m
7.30 From Problem 7.29, projection from the ground at 22oC crosses ambient at 500m.
Need the windspeed at 250 m (halfway up) using (7.46) and Table 7.6 for Class C,
uH
ua
=
H
za






p
so,
u250
4m/s
=
250m
10m






0.20
= 1.90
u250 =1.90x4 = 7.6m /s
Ventilation coefficient = 500m x 7.6m/s = 3.8x103 m2/s
7.31 Below the knee, the plume is fanning which suggests a stable atmosphere, which could be
profile (a), (b) or (d).
Above the knee, the plume is looping, which suggests superadiabatic, which is d.
7.32 H=50m, overcast so Class D, A at 1.2km, B at 1.4km.
a. From Fig 7.52, Class D, H=50m, max concentration occurs at 1km.
Beyond 1 km, concentration decreases so the "A" will be more polluted than “B.”
b. Clear sky, wind < 5m/s: Class is now A, B or C. At 50m, Class A, B, or C, Fig 7.52
shows us that the maximum point moves closer to the stack.
c. It will still be house at site "A” that gets the higher concentration of pollution.

Pg. 7.11
7.33 Bonfire emits 20g/s CO, wind 2 m/s, H=6m, distance = 400m. Table 7.7, clear night,
stability classification = F
C x, 0( )=
Q
π uσ yσz
exp −
H2
2σz
2





 (7.49)
a. From Table 7.9 at 400m, σy = 15m, σz = 7m
C =
20x106
µg /s
π 2m/s x15m x 7m
exp −
62
2x7
2





 = 21x10
3
µg /m
3
= 21mg / m
3
b. At the maximum point, Fig. 7.52 we can get a rough estimate of the key parameter
CuH
Q






max
≈ 3.8x10
−3
/ m
2
Cmax =
Q
uH
CuH
Q






max
=
20x103
mg /s
2m /s
x
3.8x10−3
m
2 = 38 ≈ 40mg / m
3
7.34 H=100m, Q=1.2g/s per MW, uH
=4m/s, uAnemometer
=3+m/s, C<365µg/m3.
The more unstable the atmosphere, the higher the peak downwind concentration (see Fig.
7.51). From Table 7.7, with wind > 3m/s, B is the most unstable.
From Fig. 7.52, Xmax = 0.7 km;
CuH
Q






max
≈ 1.5x10
−5
/m
2
Cmax =
Q
uH
CuH
Q






max
= 365x10−6
g /s =
Q
4m /s
x
1.5x10
−5
m2
Q ≈
4x365x10−6
1.5x10−5
= 97g /s
Maximum power plant size = 97 g/s x
MW
1.2 g/s
= 80MW
7.35 Atmospheric conditions, stack height, and groundlevel concentration restrictions same as
Prob. 7.34 so that:
Emissions Q ≈ 97 g/s
97g /s =
0.6 lb SO2
10
6
Btu in
x
1 Btu in
0.35 Btu out
x
3412 Btu out
kWh
x
1hr
3600s
x
103
g
2.2 lb
xPkW
PkW =
97x106
x0.35x3600x2.2
0.6x3412x1000
= 131,000KW =130MW

Pg. 7.12
7.36 H = 100m, ua = 4 m/s, Q = 80g/s, clear summer day so Class B:
First, find the windspeed at the effective stack height using (7.46) and Table 7.7:
uH = ua
H
za






p
= 4m /s
100
10






0.15
= 5.65m /s
a. At 2 km, Table 7.9: σy = 290 m, σz = 234 m
C x, 0( )=
Q
π uσ yσz
exp −
H
2
2σz
2






C =
80x106
µg/s
π 5.65m/s x290m x 234m
exp −
1002
2x234
2





 = 61µg / m
3
b. At the maximum point, 0.7 km (Fig. 7.52),
CuH
Q






max
≈ 1.5x10
−5
/m
2
Cmax =
Q
uH
CuH
Q






max
=
80x106
µg/s
5.65m /s
1.5x10−5
m
2





 = 212µg/ m
3
c. At x = 2km, y = 0.1 km:
C x, y( )=
Q
π uσ yσz
exp −
H2
2σz
2





 exp
−y2
2σy
2






= 61µg / m
3
x exp
-1002
2x290
2





 = 57µg / m
3
7.37 For class C, notice from (7.47) and (7.48) with Table 7.9 and f =0
σy
σz
=
ax0.894
cx
d
+ f
=
104x0.894
61x
0.911 = 1.7x
−0.017
≈ fairly constant ≈ k (about 1.7)
So, assume σy = k σz , then from (7.49)
a. C x, 0( )=
Q
π uσ yσz
exp −
H2
2σz
2





 =
Q
π ukσz
2 exp −
H2
2σ z
2






To find the maximum concentration, differentiate and set equal to zero:

Pg. 7.13
dC
dσ z
=
Q
π uk
1
σz
2
−H2
2






−2
σz
3





e
−
H2
2σ z
2
+ e
−
H2
2σz
2 −2
σz
3














= 0
Multiply through by σz5 and cancel lots of terms to get,
2H2
2





 − 2σ z
2
= 0 or σz =
H
2
= 0.707H
b. Substituting the newly found value for σz,
Cmax =
Q
π uσyσz
exp −
H2
2 H
2
2( )








=
Q
π uσ yσz
e
−1
=
0.117Q
uσyσz
c. Using σy = k σz
Cmax =
0.117Q
ukσz
2 =
0.117Q
uk 0.707H( )2 =
f(Q, u)
H
2 (varies as inverse of H2)
7.38 Find the effective stack height of the Sudbury stack:
130 C 20 m/s
15.2m
380m
10 Co o
8 m/s
Using (7.52) for the buoyancy flux parameter
F = gr
2
vs 1−
Ta
Ts





 = 9.8
m
s
2 x
15.2m
2






2
x20
m
s
x 1−
10 + 273
130 + 273





 = 3370m
4
/s
3
The distance downwind to final plume rise xf is given on page 464 (with F>55),
xf = 120F
0.4
= 120x 3370( )0.4
= 3092 m
For stability classification C, use (7.54) for plume rise,
Δh =
1.6F1 3
xf
2 3
uh
=
1.6x 3370( )1 3
3092( )2 3
8
= 635m plume rise
H = effective stack height = h + Δh = 380 + 635 = 1015 m

Pg. 7.14
7.39 Repeat Problem 7.38 with a stable, isothermal atmosphere:
F = 3370 m4/s3 from Prob. 7.38. For isothermal atmosphere we can use (7.51) along a
stability parameter to estimate plume rise. The stability parameter (7.53) is
S =
g
Ta
ΔTa
Δz
+ 0.01K /m





 =
9.8m /s2
10 + 273( )K
0 + 0.01K/ m( ) = 3.46x10
−4
/s
2
Putting that into (7.51) for plume rise under these conditions gives
Δh = 2.6
F
uhS






1 3
= 2.6
3370m4
/s3
8m /s x 3.46x10
-4
/s
2






1 3
= 278m
(Notice the atmospheric stability lowered effective stack height vs Prob. 7.38)
7.40 Cloudy summer day, stability classification C (Table 7.7),
120 C 10 m/s
2m
100m
6.0 Co o
5 m/s
Using (7.52) for bouyancy flux parameter
F = gr
2
vs 1−
Ta
Ts





 = 9.8
m
s
2 x
2m
2






2
x10
m
s
x 1−
6 + 273
120 + 273





 = 28.4m
4
/s
3
and distance downwind to final plume rise xf given on page 464 (with F<55),
xf = 50F
5 8
= 50x 28.4( )5 8
= 406
For stability classification C, use (7.54) for plume rise,
Δh =
1.6F1 3
xf
2 3
uh
=
1.6x 28.4( )1 3
406( )2 3
5
= 54m plume rise

Pg. 7.15
7.41 Power plant, find groundlevel pollution 16 km away. Need first find H.
200 MWe
100m
r=2.5m
13.5 m/s
145 Co 15 Co
5m/s
Class E
lapse rate=5 C/km
o
Q=300g/s SO2
16km
First, find bouyancy flux parameter (7.52),
F = gr
2
vs 1−
Ta
Ts





 = 9.8
m
s
2 x 2.5m( )2
x13.5
m
s
x 1 −
15 + 273
145 + 273





 = 257m
4
/s
3
plume rise for stable (Class E) atmosphere needs S from (7.53),
S =
g
Ta
ΔTa
Δz
+ 0.01K /m





 =
9.8m /s2
15 + 273( )K
5o
1000m
+ 0.01K/ m





 = 5.1x10
−4
/s
2
plume rise is given by (7.51),
Δh = 2.6
F
uhS






1 3
= 2.6
257m4
/s3
5m /s x 5.1x10
-4
/s
2






1 3
= 121m
So, the effective height is H = 100m + 121m = 221 m
Concentration downwind at 16km: (Table 7.9) σy = 602m, σz = 95m
C x, 0( )=
Q
π uσ yσz
exp −
H2
2σz
2





 (7.49)
C =
300x106
µg /s
π 5 m/s x602m x 95m
exp −
2212
2x95
2





 = 22µg /m
3
7.42 A 20 g/s source, 5 m/s, H = 50 m want peak concentrations Class A, C, F..
Cmax =
Q
uH
CuH
Q





 =
20x106
µg /s
5m /s
CuH
Q





 = 4x10
6 CuH
Q





µg /s
50m
A,C,F
5 m/s
20 g/s
Using Fig. 7.52 with 50 m and varying stability classifications gives:

Pg. 7.16
"A"
CuH
Q





 ≈ 6x10
−5
at 0.25km, Cmax = 4x10
6
x 6x10
-5
= 240µg / m
3
"C"
CuH
Q





 ≈ 5.8x10
−5
6
x 5.8x10
-5
= 230µg /m
3
"F"
CuH
Q





 ≈ 2.4x10
−5
6
x 2.4x10
-5
= 96µg/ m
3
x (km)
C ( g/m )µ 3
240
230
96
0.25 0.55 3.7
7.43 H = 50m, 100m, 200m; Class C, 20 g/s, 5 m/s wind:
Cmax =
Q
uH
CuH
Q





 =
20x106
µg /s
5m /s
CuH
Q





 = 4x10
6 CuH
Q





µg /s
Using Fig. 7.52,
@50m: Cmax ≈ 4x106 x 5.7x10-5 = 228 µg/m3
@100m: Cmax ≈ 4x106 x 1.5x10-5 = 60 µg/m3
@200m: Cmax ≈ 4x106 x 3.4x10-6 = 14 µg/m3
Do they drop as (1/H)2? That is,
€
expectation is
C(2H)
C(H)
=
1
4
and
C(4H)
C(H)
=
1
16
Test them:
C(100m)
C(50m)
=
6
22.8
= 0.26
C(200m)
C(100m)
=
1.4
6
= 0.23 not bad!
Expect
C(200m)
C(50m)
=
1
16
= 0.0625
C(200m)
C(50m)
=
1.4
22.8
= 0.061 again, not bad.

Pg. 7.17
7.44 Paper mill emitting H2S, 1km away want 0.1 x odor threshold:
1 km
40 g/s
4-10m/s
Class B
0.01 mg/m3
Class B, at 1 km, (Table 7.9) σy = 156m, σz = 110m
C x, 0( )=
Q
π uσ yσz
exp −
H2
2σz
2






0.01x10
−3
g /m
3
=
40g /s
π u m/s x156m x 110m
exp −
H2
2x110
2






Rearranging: e
H2
24,200
=
40
π u 156 x 110 x 0.01x10
-3 =
74.2
u
or, H = 24, 200 ln
74.2
u










0.5
so, at each end of the wind speed range we can find the height needed:
Hu= 4 = 24, 200 ln
74.2
4


 






0.5
= 265m
Hu=10 = 24, 200 ln
74.2
10










0.5
= 220m
says to be conservative use H=265m
From Fig 7.52 at H=265, Class B, Xmax ≈1.8km.
Therefore, with the peak occurring beyond the 1 km house, the concentration will rise for
buildings located > 1km away. YES there could be a problem.
7.45 Stack under an inversion:
45m
150 g/s
5 m/s
Class C
X L
L=100m
At x = XL σz = 0.47 (L-H) = 0.47 (100 - 45) = 26 m

Pg. 7.18
For class C, σz = 26m at x = 0.4 km (Table 7.9), therefore XL = 0.4 km, and also
from Table 7.9, σy = 46m.
At x = XL : C XL,0( )=
Q
π uσ yσz
exp −
H2
2σz
2





 (7.49)
=
150x103
mg /s
π 5 m/s x 46m x 26m
exp −
452
2x26
2





 =1.8mg / m
3
At x = 2XL : σy = 85m (Table 7.9), so using (7.55) gives
C 2XL, 0( )=
Q
2π uσ yL
=
150x103
mg /s
2π x 5m/s x 85m x 100m
= 1.4mg / m
3
7.46 Stack under an inversion layer:
50m
80 g/s
5 m/s
X L
L=250m
4 m/s
x=4km
C=?
We need the stability classification: Clear summer day, 4m/s, Table 7.7 says Class B.
at x = XL , (7.56) gives us σz = 0.47 (L-H) = 0.47 (250 - 50) = 94 m
a. From Table 7.9, at σz = 94m Class B, XL ≈ 0.9km. Since our point of interest is
at 4 km, we are well past the point at which reflections first occur so we can use
(7.55). We need σy at 4km, which is given in Table 7.9 as 539m:
C 4km, 0( )=
Q
2π uσ yL
=
80x106
µg /s
2π x 5m/s x 539m x 250m
= 47µg /m
3
b. Without the inversion layer, at 4km σz = 498m, σy = 539m so,
C 4km, 0( )=
Q
π uσ yσz
exp
−H2
2σ z
2





 =
80x106
µg /s
π x 5m/s x 539m x 498m
exp
−502
2x498
2





 = 19µg / m
3

Pg. 7.19
7.47 Agricultural burn. Clear fall afternoon, winds 3 m/s, so stability class "C" (Table 7.7),
and σz = 26m (Table 7.9). Using (7.57),
C 0.4km( )=
2q
2π uσ z
=
2x300mg / m −s
2π x 3m/s x 26m
= 3.0mg / m
3
0.3g/s-m
u=3m/s
400m
7.48 A freeway modelled as a line source:
10,000
vehicles/hr u=2m/s
200m
1.5 g/mi
Clear summer, 2 m/s, Table 7.7 suggests Class A or A-B.
At 0.2 km, σz = 29m for Class A; σz = 20m for Class B. What should we use? Since it
is somewhere between Class A and Class B, but closer to A, let's use σz ≈ 26m:
To find the linear emission rate:
q =10, 000
vehicles
hr
x
1 hr
3600s
x
1.5g
mi − vehicle
x
mi
5280ft
x
ft
0.3048m
= 2.58x10
−3
g / m −s = 2.58mg /m − s
Then, using (7.57),
C 0.2km( )=
2q
2π uσ z
=
2x2.58mg /m − s
2π x 2m/s x 26m
= 0.04mg /m
3
7.49 Box model, 250,000 vehicles between 4 and 6pm, driving 40km ea, emitting 4g/km
CO.
a. qs = 250, 000veh. x
40km
veh
x
4gCO
km
x
1
2hrs
x
hr
3600s
x
1
15x80x10
6
m
2 = 4.6x10
−6
gCO/ m
2
s
b. Using (7.61) with t= 2hrs x 3600s/hr =7200s,
C t( ) =
qs L
uH
1 − e
− ut/ L
( )

Pg. 7.20
=
4.6x10−6
g /m2
s x 15, 000m
0.5m /s x 15m
1− e
−0.5m / s⋅7200s / 15000m
( )= 0.002g /m
3
= 2mg /m
3
c. With no wind, go back to (7.58) and solve the differential equation:
LWH
dC
dt
= qs LW
dC =
qsLW
LWH
dt so, C =
qs
H
t
C =
qs
H
t =
4.6x10−6
gCO/ m2
⋅ s
15m
x 2hrs x
3600s
hr
= 0.0022gCO /m
3
= 2.2mg / m
3
7.50 Box model, 105 m on a side, H=1200m, u=4m/s, SO2=20kg/s, steady state:
4m/s
Cin=0
10 m
5
5
10 m
1200m
20 k g/s
input rate = output rate
20
kg
s
x
109
µg
kg
= 4
m
s
x 10
5
m x 1200m x C
µg
m3( )
C =
20x109
4x10
5
x1200
= 41.7µg / m
3
7.51 Assume steady-state conditions were achieved by 5pm Friday so that from Problem
7.50, C(0) = 41.7 µg/m3.
With qs = 0, and Cin = 0, (7.60) gives us C(t) = C 0( )e
− ut/ L
.
a. At midnight, t=7hrs x 3600s/hr = 2.52x104 s
C(t) = C 0( )e
− ut/ L
= 41.7µg/ m
3
⋅ e
-4m/s ⋅ 2.52x104
s 105
m
=15.2µg /m
3
b. Starting up again at 8am on Monday, by 5pm (9hrs later):
first check to see concentration left from Friday at 5pm (63 hrs earlier):
C(t) = C 0( )e
− ut/ L
= 41.7µg/ m
3
⋅ e
-4m/s ⋅ 63hrx3600s / hr 105
m
= .005µg /m
3
≈ 0

Pg. 7.21
so we can ignore that and let C(0) = 0 at 8am Monday. First find the emissions per
unit area,
qs =
emission rate
area
=
20kg /s x 109
µg /kg
10
5
m x 10
5
m
= 2.0µg/ m
2
⋅ s
Then use (7.57) with Cin = 0:
C t( ) =
qs L
uH
1 − e
− ut/ L
( )
=
2.0µg/ m2
⋅ s x 105
m
4m /s x 1200m
1 − e
−(4m / s x 9hr x 3600s/hr/105
m)
( )= 30.2µg / m
3
7.52 Steady-state conditions from Prob 7.50, wind drops to 2 m/s, 2hrs later:
From Prob. 7.50, emission rate qs = 2.0 µg/m2-s, and C = 41.7µg/m3. Using (7.60),
C t( ) =
qs L
uH
1 − e
− ut/ L
( )+ C(0)e
−ut/ L
C 2hr( ) =
2.0µg /m2
⋅s x 105
m
2m/s x 1200m
1− e
−(2m / s x 2hr x 3600s/hr/105
m)
( )+ 41.7e
− 2m / s x 2hr x3600s/hr /105
m( )
= 47.3 µg/m3
7.53 Modified Prob. 7.50, now incoming air has 5 µg/m3 and there are 10 µg/m3 already
there at 8am. Find the concentration at noon:
4m/s
Cin=5 g/m3
10 m
5
5
10 m
1200m
2 g/m2-s
µ
µ
C t( ) =
qsL
uH
+ Cin





 1 − e
− ut/ L
( )+ C(0)e
−ut/ L
(7.60)
C 4hr( ) =
2.0µg /m2
⋅s x 105
m
4m/s x 1200m
+ 5µg /m
3




 1 − e
−(4 m/ s x 4hr x 3600s/hr/105
m)
( )
+ 10e
− 4m / s x 4hr x3600s/hr /105
m( )
C(4hr = noon) = 26.1 µg/m3.

Pg. 7.22
7.54 Now using conditions of Prob 7.50, but for a nonconservative pollutant with
K=0.23/hr:
Rate into box = Rate out of box + Rate of decay
S = u W H C + K C V
20
kg
s
x
109
µg
kg
= 4
m
s
x 10
5
m x 1200m x C
µg
m3( )
+
0.23
hr
x C
µg
m
3 x
1hr
3600s
x10
5
m x 10
5
m x 1200m






20x109 = 4.8x108 C + 7.6x108 C
C = 16 µg/m3
7.55 Starting with (7.64) and using the special conditions of this tracer-gas study; that is, a
conservative tracer (K=0), no tracer in the air leaking into the room (Ca=0), and the
tracer source turned off at t=0 (S=0) gives the exponential decay of tracer as:
€
C t( ) = C0e−nt
and then taking the log:
€
ln C t( )[ ]= ln C0( )− nt
which is of the form y = mx + b, where y = lnC, m = n(ach), and b = lnCo
time (hr) C (ppm) ln C
0 10.0 2.303
0.5 8.0 2.079
1.0 6.0 1.792
1.5 5.0 1.609
2.0 3.3 1.194
3210
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
time (hr)
lnC
From the graph, the slope is about: slope ≈
2.1−1.3( )
2.0 − 0.5
= 0.53
Thus, the infiltration rate is about 0.53 air changes per hour.

Pg. 7.23
7.56 Infiltration 0.5ach, 500m3 volume, 200 m2 floor space, radon 0.6pCi/m2s:
0.5 ach
V=500m3
0.6pCi/m s2
K=7.6x10 /hr- 3
Using (7.63) with K = 7.6x10-3/hr (Table 7.14),
S =
S
V( )
I + K
=
0.6pCi /m2
s x 200m2
500m
3






0.5 /hr + 7.6x10−3
/hr( )x
1hr
3600s
= 1700pCi /m
3
=1.7pCi /L
7.57 Same as Problem 7.56 but have half as much ground-floor area to let radon in, so:
S =
S
V( )
I + K
=
0.6pCi /m2
s x 100m2
500m
3






0.5 /hr + 7.6x10−3
/hr( )x
1hr
3600s
= 850pCi / m
3
= 0.85pCi / L
7.58 From Problem 7.56, the radon concentration is 1.7 pCi/L. Using a residential
exposure factor of 350 days/yr from Table 4.10
€
Exposure =
1.7pCi/L x 350
day
yr
x30yr
365day/yr x 70yr
= 0.70 pCi/L average over 70 yrs
From Problem 4.10 we are given a cancer death rate of 1 per 8000 rems exposure. From
Problem 4.12 a 1.5 pCi/L radon concentration yields an exposure of about 400 mrem/yr.
€
Risk = 0.70 pCi/L x
400mrem/yr
1.5pCi/L
x
1 cancer death
8000 rem
x
rem
103
mrem
x70yr = 0.0016 ≈ 0.16%

Pg. 7.24
7.59 A 300m3 house, 0.2ach, oven+2burners 6pm to 7pm, find CO at 7pm and and 10pm.
For these circumstances, (7.65) is appropriate:
€
C t( ) =
S
IV
1− e−n t
( ) (7.65)
From Table 7.13, the source strength S is
6 – 7 pm: Oven + 2 burners = 1900 mg/hr + 2 x 1840 mg/hr = 5580 mg/hr CO
solving for C after 1 hr:
C 1hr, 7pm( ) =
5580mg/hr
0.2
airchange
hr
x300
m
3
ac
1− e
−0.2/ hr x 1hr
( )= 16.8mg / m
3
Now turn off the burners and watch CO coast down until 10pm, 3hrs later:
€
C(10pm) = C(7pm) x e−n t
=16.8 e−0.2/hr x 3hrs
= 9.3mg/m3
7.60 n = 0.39 ach, V = 27m3, after 1-hr NO = 4.7ppm. Find source strength, S:
First convert NO in ppm to mg/m3 using (1.9) and assuming T=25oC,
mg / m
3
=
ppm x mol wt
24.465
=
4.7 x 14 +16( )
24.465
= 5.76mg / m
3
a. To find the NO source strength, rearrange (7.65)
€
S =
n V C
1− e−n t
( )
=
0.39
ac
hr
x27
m3
ac
x5.76
mg
m3
1− e−0.39/hr x 1hr
( )
=188 mgNO/hr
b. 1-hr after turning off the heater,
€
C = C0 e−n t
= 4.7ppm x e-0.39/hr x 1hr
= 3.2 ppmNO
c. In a house with 0.2 ach, 300m3,
€
C ∞( ) =
S
n V
=
188 mg/hr
0.2
ac
hr
x 300
m3
ac
= 3.1 mg/m3
NO
Using (1.9) again gives
€
ppm =
24.465
ppm x mol wt
=
24.465
3.1 x (14 +16)
= 2.6 ppm NO

Pg. 7.25
7.61 Find the settling velocity of 2.5-micron particles having density 1.5x106
g/m3
. In a
room with 2.5-meter-high ceilings, use a well-mixed box model to estimate the
residence time of these particles.
From (7.24) the settling velocity is
€
v =
d2
ρg
18η
=
2.5x10−6
m( )
2
⋅ 1.5x106
g/m3
( )⋅ 9.8m/s2
( )
18x0.0172g/m⋅ s
= 2.97x10−4
m/s
From Example 7.4, the residence time is
€
τ =
h
v
=
2.5 m
2.97 x 10−4
m/s x 3600 s/hr
= 2.3 hours
7.62 100,000 kW coal plant, 33.3% efficient, CF = 0.70,
a. Electricity generated per year,
€
100,000 kW x 24 hr/day x 365 day/yr x 0.70 = 613x106
kWh/yr
b. heat input = 613x10
6
kWh /yr out x
3 kWht in
1 kWhe out
x
3412Btu
kWh
= 6.28x10
12
Btu /yr
c. Shut it down and sell the allowances,
SO2 saved by shutting down = 6.28x10
12 Btu
yr
x
0.6 lb SO2
10
6
Btu
x
ton
2000 lb
= 1883 tons/yr
€
1883
tons
yr
x
1 allowance
ton
x
$400
allowance
= $753,200/yr

8.1 From (8.1),
δ18
O o
oo( )=
18
O 16
O( )sample
18
O 16
O( )standard
−1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
x103
=
0.0020150
0.0020052
−1
⎡
⎣⎢
⎤
⎦⎥x103
= 4.9
Since the sample has more 18
O in it, there would be more glaciation since ice
selectively accumulates 16
O, increasing the concentration of 18
O left behind in
seawater.
8.2 Plotting delta D versus temperature gives
.
As shown, δD changes by 6.19 per mil per o
C.
8.3 An ice core with (2
H/1
H) = 8.100 x 10-5
.
a. Using VSMOW 0.00015575 for deuterium in (8.1) gives
δD o
/oo( )=
2
H/1
H( )sample
2
H/1
H( )standard
−1
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
x103
=
0.00008100
0.00015575
−1
⎛
⎝
⎜
⎞
⎠
⎟x103 = −479.8
b. With δD(0
/00) = –435 per mil, and 6 per mil change in δD(0
/00) per o
C, the rise in
δD from –479.8 to – 435 (44.8 per mil) gives a temperature rise of 44.8/6 = 7.5o
C.
8.4 From the equation given for the ice core,
T o
C( )=1.5 δ18
O o
oo( )+ 20.4 = 1.5x −35( ) + 20.4 = −32.1o
C
Pg. 8.1

notice, by the way, that since this sample is for glacial ice, not ocean water or sediment,
the negative sign on δ18
O o
oo( )means colder temperatures.
8.5 Plotting the ice core data for T(o
C) and δD
So: T o
C( )= 0.1661 δD o
oo( )+ 72.45
8.6 The flat earth!
1370W/m 2 R
Eabsorbed = Eradiated
Sπ R
2
= σ T
4
A = σ T
4
2π R
2
( )
T =
S
2σ
⎛
⎝
⎜ ⎞
⎠
⎟
1
4
=
1370W /m2
2x5.67x10−8
W/ m2
K4
⎛
⎝
⎜
⎞
⎠
⎟
1
4
= 331.5K - 273.1 = 58.4
o
C
8.7 The basic relationship is S =
k
d
2 . Using d and S for Earth from Table 8.2 lets us find k:
k = S d2
= 1370W /m2
x 150x106
km x 103
m/ km( )
2
= 3.083x1025
W
a. Mercury: S =
k
d
2 =
3.083x1025
W
58x106
km x 103
m/ km( )
2 = 9163W/ m2
Pg. 8.2

b. The effective temperature (8.7) of Mercury would be:
Te =
S 1− α( )
4σ
⎡
⎣⎢
⎤
⎦⎥
1
4
=
9163W/ m2
1− 0.06( )
4x5.67x10−8
W/ m2
K4
⎡
⎣⎢
⎤
⎦⎥
1
4
= 441K (168
o
C)
c. Peak wavelength:
λmax =
2898
T K( )
=
2898
441
= 6.6μm
8.8 Solar flux variation of ± 3.3%, gives a range of S
Smax = 1370 (1+0.033) = 1415.2 W/m2
Smin = 1370 (1 -0.033) = 1324.8 W/m2
Te,max =
S 1− α( )
4σ
⎡
⎣⎢
⎤
⎦⎥
1
4
=
1415.2W/m2
1 − 0.31( )
4x5.67x10−8
W/ m2
K4
⎡
⎣⎢
⎤
⎦⎥
1
4
= 256.2K (-17o
C)
Te,min =
S 1− α( )
4σ
⎡
⎣⎢
⎤
⎦⎥
1
4
=
1324.8W /m2
1 − 0.31( )
4x5.67x10−8
W/ m2
K4
⎡
⎣⎢
⎤
⎦⎥
1
4
= 252K (-21
o
C)
The variation from –17o
C to –21o
C is a difference of about 4oC, or about ± 2o
C.
8.9 After a nuclear war:
2
a. Surface temperature,
σ Ts
4
= 240W /m2
Pg. 8.3

Ts =
240W/ m2
5.67x10−8
W/m2
K4
⎡
⎣⎢
⎤
⎦⎥
1
4
= 255K (-18
o
C)
b. X, Atmosphere to space: Balance Incoming from space = Outgoing to space
342 = 69 + X X = 273 W/m2
c. Y, Absorbed by earth: Incoming solar has to go somewhere,
342 = 69 + 257 + Y Y = 16 W/m2
d. Z, Radiation from atmosphere to surface: Balance earth's surface radiation,
Y + Z = 240 = 16 + Z Z = 224 W/m2
8.10 A 2-layer atmosphere:
342
107
X
Y
168
24 78
40
W
ZW
350
Z
T1
T2
390
a. At the surface: 168 + Z = 24 + 78 + 390 Z = 324 W/m2
b. Extraterrestrial: 342 = 107 + 40 + W W = 195 W/m2
c. Lower atmosphere:
Y + 24 + 78 + 350 + 195 = 2 x 324 Y = 1 W/m2
d. Incoming: 342 = 107 + X + 1 + 168 X = 66 W/m2
e. Temperatures T1 and T2 (assuming blackbody radiation) can be found from
σ T1
4
= W = 195 T1 =
195W/m2
5.67x10
−8
W /m
2
K
4
⎛
⎝
⎜
⎞
⎠
⎟
1 4
= 242K (-31
o
C)
Pg. 8.4

σ T2
4
= Z = 324 T2 =
324W/m2
5.67x10
−8
W/m
2
K
4
⎛
⎝
⎜
⎞
⎠
⎟
1 4
= 275K (2
o
C)
8.11 Hydrologic cycle:
evaporation =
78W/ m2
x5.1x1014
m2
x
1J/s
W
x3600
s
hr
x24
hr
d
x365
d
yr
2465kJ /kg x 10
3
kg/ m
3
x 10
3
J/ kJ
= 5.1x10
14
m
3
Averaged over the globe, with area 5.1x1014 m2, annual precipitation is very close to 1 m
8.12 Greenhouse enhanced earth:
100
342
67 24 78
30
Z
Y
X
291K
W
a. Incoming energy: 342 = 100 + 67 + W W = 175 W/m2
b. Find Z from radiation to space:
342 = 100 + 30 + Z Z = 212 W/m2
c. To find X, need the energy radiated by a 291 K surface:
surface radiated = σ T
4
= 5.67x10
−8
W/m
2
⋅ K
4
x 291K( )
4
= 406.6W/ m
2
so that, 406.6 = X + 30 X = 376.6 W/m2
d. Can find Y several ways; at the surface, or in the atmosphere,
W + Y = 406.6 + 24 + 78 = 175 + Y Y = 333.6 W/m2
or, 67 + 24 + 78 + X = Y + Z
67 + 24 + 78 + 376.6 = Y + 212 (Y = 333.6)
8.13 CO2 from 10 GtC/yr to 16 GtC/yr over 50 years, with initial 380 ppm and
A.F. = 40%. Since it is linear, the total emissions would be those at constant level
plus the area of a triangle rising by 6 Gt/yr:
Pg. 8.5

50 yrs x10 Gt/yr + 1/2 x 50 yrs x 6 GtC/yr = 575 GtC.
Using the 2.12 GtC/ppm ratio and the 0.40 A.F. gives
CO
2( )= 380 +
575 GtC x 0.40
2.12 GtC/ppm
= 380 +108 = 488 ppm
8.14 CO2 growing at 2 ppm/yr, fossil fuel and cement emissions at 9 GtC/yr, and A.F. of
38%. The remaining emissions due to land use changes are:
Cemissions =
2 ppm/yr x 2.12GtC/ppm
0.38
=11.15 GtC/yr
Land use emissions = 11.15 – 9 = 2.15 GtC/yr
8.15 With 40% oil, 23% coal, 23% gas and 14% carbon free:
a. Using LHV values from Table 8.3:
Coal 23% @ 25.8 gC/MJ
Oil 40% @ 20.0 gC/MJ
Gas 23% @ 15.3 gC/MJ
Other 14% @ 0
Avg C intensity = 0.23x25.8 + 0.40x 20.0 + 0.23x15.3 + 0.14x0 = 17.45 gC/MJ
b. Coal replaced by non-carbon emitting sources:
Avg C intensity = 0.23x0 + 0.40x20.0 + 0.23x15.3 + 0.14x0 = 11.52 gC/MJ
c. Modeled as an exponential growth function over 100 years:
C = C0ert
r =
1
t
ln
C
C0
⎛
⎝
⎜
⎞
⎠
⎟ =
1
100
ln
11.52
17.45
⎛
⎝
⎜
⎞
⎠
⎟ = −0.0042 = −0.42%/yr
8.16 With resources from Table 8.4 and LHV carbon intensities from Table 8.3, A.F. = 50%:
a. All the N. Gas: 15.3 gC/MJ x 36,100 x 1012
MJ = 552,330 x 1012
gC = 552 GtC
ΔCO2 =
552 GtC x 0.5
2.12 GtC/ppmCO2
=130 ppm CO2
b. All the Oil: 20.0 gC/MJ x 24,600 x 1012
MJ = 492 GtC
Pg. 8.6

ΔCO2 =
492 GtC x 0.5
2.12 GtC/ppmCO2
=116 ppm CO2
c. All the Coal: 25.8 gC/MJ x 125,500/2 x 1012
MJ = 1619 GtC
ΔCO2 =
1619 GtC x 0.5
2.12 GtC/ppmCO2
= 382 ppm CO2
d. All three: 130 + 116 + 382 = 628 ppm CO2. From (8.29) with ΔT2X = 2.8o
C:
ΔTe =
ΔT2X
ln 2
ln
CO2( )
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
2.8
ln 2
ln
628 +380
380
⎡
⎣⎢
⎤
⎦⎥ = 3.9o
C
8.17 Out of oil and gas, demand = 2 x 330 EJ/yr, 28%coal, 60% syn gas/oil@44gC/MJ,
a. Carbon emission rate:
Avg carbon intensity = 0.28 x 25.8 + 0.60 x 44 + 0.12 x0 = 33.6 gC/MJ
Emissions =
2 x 330x1018
J
yr
x
MJ
10
6
J
x
33.6gC
MJ
x
GtC
10
15
gC
= 22.2GtC /yr
b. Growth from 6.0 GtC/yr to 22.2 GtC/yr in 100 yrs,
r =
1
100
ln
22.2
6.0
⎛
⎝
⎜ ⎞
⎠
⎟ = 0.013 = 1.3%/ yr
c. Amount remaining with 50% airborne fraction, use (8.27):
Total emitted = Ctot =
C0
r
er T
−1( )=
6.0 GtC/yr
0.01308
e0.01308/yr x 100 yr
−1( )=1239 GtC
Amount remaining in atmosphere = 0.50 x 1239 = 619 GtC
d. Amount in atmosphere in 100 yrs = 750 + 619 = 1369 GtC
(CO2 ) =
1369GtC
2.12GtC/ ppmCO2
= 646ppm
e. Equilibrium temperature increase, with ΔT2x=3oC from (8.29):
ΔT =
ΔT2x
ln2
⋅ln
CO2
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
3.0
ln2
⋅ ln
645
356
⎛
⎝
⎜
⎞
⎠
⎟ = 2.57o
C
Pg. 8.7

8.18 Repeat of Prob. 8.17, but now conservation scenario:
a. Carbon emission rate:
Avg carbon intensity = 0.20 x 25.8 + 0.30 x 15.3 + 0.10 x20 = 11.75 gC/MJ
Emissions =
330x1018
J
yr
x
MJ
10
6
J
x
11.75gC
MJ
x
GtC
10
15
gC
= 3.88GtC /yr
b. Growth from 6.0 GtC/yr to 3.88 GtC/yr in 100 yrs,
r =
1
100
ln
3.88
6.0
⎛
⎝
⎜ ⎞
⎠
⎟ = −0.0044 = −0.44% /yr
c. Amount remaining with 50% airborne fraction, use (8.27):
Total emitted = Ctot =
C0
r
er T
−1( )=
6.0 GtC/yr
−0.0044
e−0.0044/yr x 100 yr
−1( )= 483 GtC
Amount remaining in atmosphere = 0.50 x 483 = 242 GtC
d. Amount in atmosphere in 100 yrs = 750 + 242= 992 GtC
(CO2 ) =
992GtC
2.12GtC/ ppmCO2
= 468ppm
e. Equilibrium temperature increase, with ΔT2x=3oC,
ΔT =
ΔT2x
ln2
⋅ln
CO2
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
3.0
ln2
⋅ ln
468
356
⎛
⎝
⎜
⎞
⎠
⎟ = 1.18o
C
8.19. Finding LHV efficiency of a condensing furnace with 95% HHV efficiency.
From Example 8.4, HHV = 890 kJ/mol and LHV = 802 kJ/mol. The output of a HHV
95% efficient furnace burning 1 mole of methane is 0.95 x 890 kJ = 845.5 kJ. On an LHV
basis, you still get the same output, but the efficiency is now
LHV efficiency=
845.5 kJ delivered
802 kJ LHVinput
=1.054 =105.4%
Pg. 8.8

This over 100% efficiency is one reason LHV values are sometimes avoided in the U.S.
8.20 Finding HHV carbon intensities:
a. Ethane, C2H6 :
2 x 12 gC/mol
1542 kJ/mol
x
103
kJ
MJ
=15.56 gC/MJ
b. Propane, C3H8 :
3 x 12 gC/mol
2220 kJ/mol
x
103
kJ
MJ
=16.36 gC/MJ
c. n - Butane, C4H10 :
4 x 12 gC/mol
2878 kJ/mol
x
103
kJ
MJ
=16.68 gC/MJ
8.21 Using HHV carbon intensities from Table 8.3, the four options are:
COP=3
η =0.95
100MJ
1380gC
95MJ
delivered 1380gC
95MJ
=14.5gC/MJ
η =0.70
100MJ
1380gC
70MJ
delivered 1380gC
70MJ
=19.7gC/MJ
1) pulse
2) conv gas
3) heat pump η =0.35
100MJ
2420gC
35MJ
2420gC
35MJ
=69.1gC/MJ
power plant
heat pump
70 from enviro.
105MJ del
2420gC
105MJ
=23.0gC/MJ
4)resistance η =0.35
100MJ
2420gC
35MJ
power plant
Notice the tremendous range: 14.5 to 69.1 gC/MJ, almost 5:1 !
8.22 Propane-fired water heater with 2200 kJ/mol vs Example 8.6:
a. Carbon intensity C3H8 :
3 x 12 gC/mol
2220 kJ/mol
x
103
kJ
MJ
=16.36 gC/MJ
b. Delivering heat at 85% efficiency to hot water
Pg. 8.9

c. Savings versus 32.5 gC/MJ with a n. gas electric water heater in Example 8.6:
propane
electric
=
19.25 gC/MJ
32.5 gC/MJ
= 0.59 so there is a 41% savings vs electricity
8.23 Initial CO2 = 356 ppm, 6 GtC/yr and 750 GtC; want 70 year scenario. Do it by scenario:
(A) Using r = 1.0 + 0.3 - 2.0 - 0.7 = -1.4%/yr in (8.27)
Ctot =
C0
r
er T
−1( )=
6.0 GtC/yr
−0.014
e−0.014/yr x 70 yr
−1( )= 268 GtC
CO2( )=
750GtC + Ctot x AF
2.12 GtC/ppmCO2
=
750 + 268 x 0.4 GtC
2.12 GtC/ppmCO2
= 404 ppm
ΔT =
ΔT2x
ln2
⋅ln
CO2
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
3
ln2
⋅ ln
404
356
⎛
⎝
⎜
⎞
⎠
⎟ = 0.55o
C
To find the doubling time, rearrange (8.27):
Ctot to double current 750 GtC =
750 GtC
AF = 0.4
=
C0
r
er Td
−1( )
Td =
1
r
ln
750/0.4( )r
6.0
+1
⎡
⎣
⎢
⎤
⎦
⎥ =
1
−0.014
ln
750/0.4( ) −0.014( )
6.0
+1
⎡
⎣
⎢
⎤
⎦
⎥ = never!
(B) r = 1.5 + 1.5 - 0.2 + 0.4 = 3.2%/yr
Ctot =
C0
r
er T
−1( )=
6.0GtC/yr
0.032
e0.032/yr x 70 yr
−1( )=1574 GtC
CO2( )=
750 GtC + Ctot x AF
2.12 GtC/ppmCO2
=
750 +1574 x 0.5GtC
2.12 GtC/ppmCO2
= 725 ppm
Pg. 8.10

ΔT =
ΔT2x
ln2
⋅ln
CO2
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
2
ln2
⋅ ln
725
356
⎛
⎝
⎜
⎞
⎠
⎟ = 2.05o
C
Td =
1
r
ln
750
AF( )r
6.0
+1
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
=
1
0.032
ln
750
0.5( )0.032
6.0
+1
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= 69 yrs
(C) r = 1.4 + 1.0 - 1.0 - 0.2 = 1.2%/yr
Ctot =
C0
r
er T
−1( )=
6.0 GtC/yr
0.012
e0.012/yr x 70 yr
−1( )= 658 GtC
CO2( )=
750 GtC+ Ctot xAF
2.12 GtC/ppmCO2
=
750 + 658 x 0.5 GtC
2.12 GtC/ppmCO2
= 509 ppm
ΔT =
ΔT2x
ln2
⋅ ln
CO2
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
3
ln2
⋅ ln
509
356
⎛
⎝
⎜
⎞
⎠
⎟ =1.55o
C
Td =
1
r
ln
750
AF( )r
6.0
+1
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
=
1
0.012
ln
750
0.5( )0.012
6.0
+1
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
=116yrs
8.24 With 1990 6.0 GtC/yr + land use 2.5 GtC/yr and the following growth rates to 2100
Population growth rate dP/dt = 0.8%
Per capita GDP growth rate d(GDP/P)/dt = 1.3%
Final Energy per GDP growth rate =d(FE/GDP)/dt = - 0.7%
Primary Energy to Final Energy growth rate d(PE/FE)/dt = 0.1%
Carbon per unit of Primary Energy growth rate d(TC/PE)/dt = -0.2%
Carbon Sequestration growth rate d(C/TC)/dt = 0.0%
Total growth rate = 0.8 + 1.3 – 0.7 + 0.1 – 0.2 + 0.0 = 1.3%/yr
a. The carbon emission rate in 2100
From energy C = C0ert
= 6.0e0.013x110
= 25.1 GtC/yr
Including land use: Total emission rate = 25.1 + 2.5 = 27.6 GtC/yr
b. Total carbon emissions:
Ctot (energy) =
C0
r
erT
−1( )=
6.0
0.013
e0.013x110
−1( )=1467 GtC
Ctot (land use and industry) = 110 yrs x 2.5 GtC/yr = 275 GtC
Total emissions = 1467 + 275 = 1742 GtC
Pg. 8.11

c. The increase in CO2 concentration with A.F. = 0.5:
ΔCO2 =
1742 GtC x 0.5
2.12 GtC/ppmCO2
= 410 ppm CO2
d. Estimated 2100 CO2 concentration = 360 + 410 = 770 ppm
e. With ΔT2X = 2.8o
C, the global equilibrium temperature increase 2100
ΔTe =
ΔT2X
ln 2
ln
CO2( )
CO2( )0
⎡
⎣
⎢
⎤
⎦
⎥ =
2.8
ln 2
ln
770
360
⎡
⎣⎢
⎤
⎦⎥ = 3.1o
C
8.25 Identification of the halocarbons:
a. C3HF7 is an HFC (no Cl), 3 1 7 - 90 = 227, HFC-227
b. C2H3FCl2 is an HCFC, 2 3 1 - 90 = 141, HCFC-141
c. C2F4Cl2 is a CFC, 2 0 4 - 90 = 114, CFC-114
d. CF3Br is a Halon, H-1301
8.26 a. HCFC-225, 225 + 90 = 315 (3C, 1H, 5F), 8 sites - (1+5) = 2 Cl, ∴ C3HF5Cl2
b. HFC-32, 32 + 90 = 122 (1C, 2H, 2F), 4 sites, 0 Cl, CH2F2∴
c. H-1301, (1C, 3F, 0Cl, 1Br) CF3Br∴
d. CFC-114, 114 + 90 = 204 (2C, 0H, 4F), 6 sites - 4 = 2 Cl, C2F4Cl2∴
8.27. Finding climate sensitivity λ and varying feedback factor g.
a. From (8.35) and (8.40) using ΔT2X = 2.5 o
C.
λ =
ΔT2X
4.2
=
2.5
4.2
= 0.595
o
C
W m2 g =1-
λB
λ
=1−
0.27
λ
=1−
0.27
0.595
= 0.546
If g = 0.1 + 0.546 = 0.646, then
λ =
λB
1− g
=
0.27
1− 0.646
= 0.763o
C W m2
( ), ΔT2X = 4.2λ = 4.2x0.763= 3.2o
C
Pg. 8.12

b. For ΔT2X = 3.5 o
C
λ =
ΔT2X
4.2
=
3.5
4.2
= 0.833
o
C
W m2 g =1-
λB
λ
=1−
0.27
λ
=1−
0.27
0.833
= 0.676
If g increases to 0.776, then
λ =
λB
1− g
=
0.27
1− 0.776
=1.205 o
C W m2
( ), ΔT2X = 4.2λ = 4.2x1.205 = 5.1o
C
Notice ΔT2X becomes more sensitive as the feedback factor increases (0.7o
C increase when g
changes from 0.546 to 0.646 versus 1.9o
C increase when g changes from 0.676 to 0.776).
8.28 Using Figure 8.39:
a. The AS probability that ΔT2X is less than 2.5o
C. Answer: 20%
b. The WR probability that ΔT2X is greater than 3o
C. Answer: 40%
c. The AS probability that ΔT2X is between 3o
C and 4o
C. Answer: 50%
d. The WR probability that ΔT2X is between 3o
C and 4o
C. Answer: ≈ 35%
Pg. 8.13

8.29. Radiative forcing for N2O,
ΔF = k2 C − C0( )
k2 =
ΔF
C − C0( )
=
0.14
311 − 275( )
= 0.133
If N2O reaches 417 ppb, added forcing would be:
ΔF = k2 C − C0( )= 0.133 417 − 311( )= 0.37W/m2
8.30 a. Combined radiative forcings from 1850 to 1992
ΔFCO2 = 6.3 ln
CO2( )[ ]
CO2( )0[ ]
= 6.3 ln
356
278
⎛
⎝
⎜
⎞
⎠
⎟ = 1.558 W/m
2
ΔFCH4
= 0.031 CH4 − CH4( )0( )= 0.031 1714 − 700( )= 0.463 W/m2
ΔFN2O = 0.133 N2O − N2O( )0( )= 0.133 311 − 275( )= 0.140W /m2
ΔFCFC−11 = 0.22 CFC − 11( )− CFC −11( )0[ ]= 0.22 0.268 − 0( ) = 0.059 W/ m2
ΔFCFC−12 = 0.28 CFC −12( )− CFC −12( )0[ ]= 0.28 0.503 − 0( )= 0.141 W/m2
Pg. 8.14

Combined forcing = 1.558 + 0.463 + 0.140 + 0.059 + 0.141 = 2.36 W/m2
b. From 1992 to 2100:
ΔFCO2 = 6.3 ln
CO2( )[ ]
CO2( )0[ ]
= 6.3 ln
710
356
⎛
⎝
⎜ ⎞
⎠
⎟ = 4.35 W/m
2
ΔFCH4
= 0.031 CH4 − CH4( )0( )= 0.031 3616 − 1714( )= 0.581W/m
2
ΔFN2O = 0.133 N2O − N2O( )0( )= 0.133 417 − 311( )= 0.370W/ m2
ΔFCFC−11 = 0.22 CFC − 11( )− CFC −11( )0[ ]= 0.22 0.040 − 0.268( ) = −0.050 W/ m2
ΔFCFC−12 = 0.28 CFC −12( )− CFC −12( )0[ ]= 0.28 0.207 − 0.503( ) = −0.083 W /m2
Combined forcing = 4.35 + 0.581 + 0.370 - 0.050 - 0.083 = 5.17 W/m2
c. From 1850 to 2100
ΔF = 6.3 ln
710
278
⎛
⎝
⎜ ⎞
⎠
⎟ + 0.031 3616 − 700( )+ 0.133 417 − 275( )
+ 0.22x0.040 + 0.28x0.207 = 7.53 W/m2
(alternatively: ΔF = 2.36 + 5.17 = 7.53 W/m2)
8.31 From Prob. 8.30 for 1850 to 2100:
ΔF = 6.3 ln
710
278
⎛
⎝
⎜ ⎞
⎠
⎟ + 0.031 3616 − 700( )+ 0.133 417 − 275( )
+ 0.22x0.040 + 0.28x0.207 = 7.53 W/m2
ΔTs = λ ΔF = 0.57 oC/(W/m2) x 7.53 W/m2 = 4.3 oC
8.32 Using and forcing
ratios of HFC-134a to CO
RCO2
t( )dt
0
20
∫ ≈13.2yrs; RCO2
t( )dt
0
100
∫ ≈ 43.1yrs; RCO2
t( )dt
0
500
∫ ≈138yrs
2 of (Fg/FCO2) = 4129 and τ = 14 yrs. First simplify GWP to
Pg. 8.15

GWPg =
Fg
FCO2
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟⋅
e−t /τ
dt
0
T
∫
RCO2
t( )dt
0
T
∫
=
Fg
FCO2
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟⋅
τ 1− e−T /τ
( )
RCO2
t( )dt
0
T
∫
a. GWP(20) = 4129⋅
14 1− e−20/14
( )
13.2
= 3330 (vs 3300 in Table 8.7)
b. GWP(100) = 4129⋅
14 1− e−100 /14
( )
43
=1340 (vs 1300 in table)
c. GWP(500) = 4129⋅
14 1− e−500/14
( )
138
= 420 (vs 400 in table)
8.33 For a greenhouse gas with τ = 42 years and a relative forcing of 1630 times that of CO2.
From Problem 8.32, GWP =
ΔFg
ΔFCO2
⋅
τ 1 − e
− t
τ
( )
RCO2
t( )dt∫
a. The 20-year GWP would be
GWP20 = 1630 ⋅
42 1− e
− 20
42
( )
13.2
= 1965
b. The 100-year GWP would be
GWP100 = 1630 ⋅
42 1− e
−100
42
( )
43.1
=1440
c. The 500 year GWP would be
GWP500 =1630 ⋅
42 1 − e
−500
42
( )
138
= 495
8.34 Applying GWPs from Table 8.7 to the emission rates given:
Pg. 8.16

8.35 Using 100-year GWPs from Table 8.7 with emission rates of 6,000 million metric tons
(Mt) of CO2, 26.6 MtCH4, and 1.2 Mt N2O. gives
6000 x 1(CO2) + 26.6 x 23(CH4) + 1.2 x 296(N2O) = 6967 MtCO2 = 6.967 GtCO2-eq
Adjusting for the ratio of C to CO2 gives
6.967 GtCO2 x (12gC/44gCO2) = 1.9 GtC-eq/yr
8.36 The actual ΔTrealized is estimated to be 0.6oC, which is 75% of the equilibrium ΔT
ΔTrealized = 0.6oC = 0.75 ΔTequilibrium
so, ΔTequilibrium = 0.6/0.75 = 0.8oC
but, ΔTequilibrium = λ ΔFactual = 0.57 x ΔFactual = 0.8
that is, ΔFactual = =
0.8
0.57
=1.40W/ m
2
The direct forcing is 2.45 W/m2, so aerosols etc. are 2.45 - 1.40 = 1.05 W/m2
8.37 Repeating Example 8.12 with the 100-yr GWP for CH4 = 23. With 1.5 MJ of
leakage, 15.3 gC/MJ we get
1.5 MJ x 15.3 gC/MJ x
16 gCH4
12 gC
x
23 gCO2
1 gCH4
= 703 gCO2 − eq
The actual CO2 emissions remain the same at 5525 gCO2
So, with 83.73 MJ of heat to the water, total CO2-eq emissions per MJ gives
703 gCO2 - eq + 5525 gCO2
83.73 MJ heat to water
= 74.4 gCO2−eq/MJ
8.38 Using Table 8.3 for the LHV carbon intensity of coal (25.8 gC/MJ), (3.18) to find
σ, and (3.20) to find tm, then plotting (3.17) gives for (a):
Q∞ = 200,000 EJ x
25.8 gC
MJ
x
1 GtC
1015
gC
x
1012
MJ
EJ
= 5160 GtC
σ =
Q∞
Pm 2π
=
5160 GtC
22 GtC/yr 2π
= 93.57 yr
Pg. 8.17

tm = σ 2ln
Pm
P0
= 93.57 yr 2 ln
22 GtC/yr
6.0 GtC/yr
=150.8 yr
then put these into P = Pm exp −
1
2
t − tm
σ
⎛
⎝
⎜
⎞
⎠
⎟
2⎡
⎣
⎢
⎤
⎦
⎥
Putting this into a spreadsheet so it can be plotted yields…
8.39 With a carbon tax of $20/mt of C (as CO2):
a. Assuming a capacity factor of 100% (plant operates all of the time):
Cemissions =
50 MW
0.35
x
1 MJ/s
MW
x
3600 s
hr
x
8760 hr
yr
x
24 gC
MJ
x
1 mtC
106
gC
=1.08x105
mtonC/yr
Pg. 8.18

Carbon tax =1.08 x105
mtC/yr x
$20
mton
= $2.16 million/yr
b. With carbon sequestering:
Area =
1.08x105
mtonC/yr
5000 kgC/yr ⋅ acre
x
103
kg
mton
= 21,600 acres
c. Biomass instead of paying the tax:
Forestry could cost =
$2.16 million/yr
21,600 acres
= $100/yr per acre
8.40 Landfill leaking 10 tonnes (1 tonne = 1 mt = 1000 kg) CH4 per year
a. 20-year GWP for methane = 62 (Table 8.7)
10 tonnes CH4 /yr x 62 = 620 tonnes/yr
b. Burning the methane
CH4 + 2 O2 CO2 + 2 H2O
1molCO2
1molCH4
x
12 + 2x12( )gCO2/mol
12 + 4x1( )gCH4/mol
x
10tonneCH4
yr
= 27.5tonneCO2/yr
c. Equivalent CO2 savings = 620 – 27.5 = 592.5 tonne CO2/yr.
as C : 592.5 tonne CO2/yr x
12tonneC
44tonneCO2
=161.5 tonneC/yr saved
d. Carbon tax saved = 161.5 tonne C/yr x $20/tonneC = $3232/yr saved
e. Same thing, 592.5 tonne CO2/yr x $5.45/tonneCO2 = $3229/yr saved
8.41 Gasoline C7H15 and 6.15 lbs/gal, fully combusted,
a. Gasoline =
6.15 lbgas
gal
x
7x12 = 84( ) lbsC
7x12 +15x1= 99( ) lb gas
= 5.22 lbs C/gal
C =
40,000 miles
12 miles/gal
x
5.22 lbsC
gal
=17,394 lbsC that will be released
b. 4000 lb car, 10,000 miles/yr
Pg. 8.19

C =
17,394 lbs C
40,000 miles
x
10,000 miles
yr
= 4348 lbsC/yr
Carbon/yr
Vehicle weight
=
4348 lbsC/yr
4000 lbs
=1.09
the car emits slightly more carbon per year than it weighs!
c. Carbon tax=
5.22 lbsC
gal
x
$15
2000 lbsC
= $0.039/gal= 3.9¢/gal
d. New car at 40 mpg, for 40,000 miles:
Carbon reduction =17,394 lbsC-
40,000 mi
40 mi/gal
x
5.22 lbsC
gal
=12,174 lbs C saved
e. Trading in the clunker for the 40 mpg car would save in carbon taxes
Tax savings = 12,174 lbsC x
$15
2000 lbs C
= $91/car
That is, those C offsets would save the utility $91, which they could spend to get
the clunker off the road.
8.42 Electric versus gasoline-powered cars:
a. Gas car emissions=
5.22 lbsC/gal
40 miles/gal
x
1000g
2.2lbs
= 59.3 gC/mi
b. With the very efficient natural-gas fired power plant:
N - gas plant emissions =
8000kJ
kWh
x
13.8gC
MJ
x
MJ
103
kJ
x
kWh
5mi
= 22.1 gC/mi
c. With the typical coal plant:
Coal plant heat rate =
l kW in
0.30 kWe out
x
1 kJ/s
kW heat in
x
3600 s
hr
=12,000 kJ/kWhe
Coal plant emissions =
12,000kJ
kWh
x
24 gC
MJ
x
MJ
103
kJ
x
kWh
5mi
= 57.6 gC/mi
So, more than half of the carbon can be saved with electric cars when efficient
natural gas power plants are assumed. There is even a slight advantage with an
old, inefficient coal plant.
8.43 NO2 + hv NO + O
From (8.48): E J/photon( )=
306,000
6.02x1023
= 5.08x10−19
J/photon
Pg. 8.20

and from (8.46): λmax =
hc
E
=
6.626 x10−34
Js x 2.998x108
m/s
5.08x10−19
J
= 390x10−9
= 390 nm
8.44 O2 + hv O + O (oops… same as Example 8.13):
E J/photon( )=
495,000
6.02x1023
= 8.22x10−19
J/photon
λmax =
hc
E
=
6.626 x10−34
Js x 2.998x108
m/s
8.22x10−19
J
= 241.6x10−9
= 241.6 nm
Meant to do photodissociation of ozone, requiring 104.6 kJ/mol:
O3 + hv O2 + O
E J/photon( )=
104,600
6.02x1023
=1.737x10−19
J/photon
λmax =
hc
E
=
6.626 x10−34
Js x 2.998x108
m/s
1.737x10−19
J
=1.14x10−6
m =1.14 μm
Pg. 8.21

9.1 Analysis of the recycling rates using Table 9.8 data and prices from Table 9.18
a. Carbon savings is 747MTCE
b. At $50/ton, tipping fee savings is
936 tons/yr x $50/ton = $46,800/yr
c. Revenue generated is
$118,845/yr
d. With a carbon tax of $50 per metric ton of carbon-equivalents, savings due to
recycling would be
747 MTCE/yr x $50/MTCE = $37,350/yr
9.2 Analyzing the energy side of the airport recycling program described in Problem 9.1
using Table 9.9
a. Annual energy savings is: 21,494 million Btu
b. At $5 per million Btu, the dollar savings is
21,494 million Btu/yr x $5/million Btu = $107,470/yr
c. Dollar savings per ton would be: $107,470 / 936 tons = $114.81/ton
9.1

9.3 Spreadsheet analysis for the recycling program using Tables 9.8 and 9.18.
a. Avoiding $120/ton for pick up and selling these recyclables at half the Table 9.18
market price for recyclables saves
Avoided pick up charges = 5300 ton/yr x $120/ton avoided = $636,000/yr
Revenue from sale of recyclables = $404,000/yr
Total savings = $636,000 + $404,000 = $1,040,000/yr
b. With a $10/ton of CO2 tax, recycling saves
Carbon tax =
$10/ton CO2
2000 lb/ton
x
2200 lb
metric ton
x
44 ton CO2
12 ton C
= $40.33/MTCE
Savings = $167,531/yr (from spreadsheet)
c. A $400,000/yr recycling program saves
Net benefit = $1,040,000 + $167,531 - $400,000 = $807,531/yr
9.4 Comparing a 10,000mi/yr, 20 mpg SUV burning 5.22 lbs C//gal at 125,000 Btu/gal
to cardboard recovery savings.
c. The carbon savings from cardboard recycling is equivalent to carbon emissions
from how many SUVs?
d. How many “average SUVs” of energy are saved by cardboard recycling?
a. How many tons of CO2 will be emitted per SUV per year?
CO2 =
10,000 miles x x 5.22 lbsC/gal
20 miles/gal x 2200 lbs/ton
x
44 tonCO2
12 tonC
= 4.35 ton CO2/yr
b. Btus for those SUVs
Energy =
10,000 miles x 125,000 Btu/gal
20 miles/gal
= 62.5 million Btu/yr
c. From Table 9.10, 42 million tons of cardboard are kept out of landfills. And from
Table 9.8, each ton saves 0.96 metric tons of carbon
9.2

Car equivalents =
42x106
ton/yr x 0.96 mtonC/ton x 1.1 ton/mton
4.32 tonCO2/yr −SUV
x
44 tonCO2
12 tonC
= 37.6 million SUVs taken off the road in carbon savings
d. From Table 9.9, cardboard recycling saves 15.65 million Btu/ton. So,
Car equivalents =
42x106
ton/yr x 15.65 x10 6
Btu/ton
62.5 x 106
Btu/yr/SUV
=10.5 million SUVs
9.5 A 0.355-L (12 oz) 16 g aluminum can with 70% recycling. Need to adjust Table
9.13 which was based on 50% recycling. Each can now has 0.7x16 = 11.2 g of
recycled aluminum and 0.3 x 16 = 4.8 g of new aluminum from bauxite.
New aluminum from bauxite= 4.8g x
1765 kJ
8g
=1059 kJ
Recycled cans to make 11.2 g of Al=11.2g x
40 kJ
8g
= 56 kJ
The remaining energy for can production is the same as Table 9.13:
Total for 16g (0.355L) can = 3188 – (1765 + 40) + (1059 + 56) = 2498 kJ/can
Per liter of can = 2498 kJ/0.355L = 7037 kJ/L
9.6 Heavier cans from yesteryear, 0.0205 kg/can and 25% recycling rate.
New aluminum per can was 0.75 x 0.0205 kg = 0.015375 kg
Recycled aluminum per can was 0.25 x 0.0205 kg = 0.005125 kg
From Table 9.12:
New aluminum = 0.015275 kg x 220,600 kJ/kg = 3370 kJ
Recycled aluminum = 0.005125 kg x 5060 kJ/kg = 26 kJ
Total energy = 3370 + 26 = 3396 kJ/can
Compared to today’s 1805 kJ/can (Example 9.4)
Today : 1805 kJ/can
Earlier : 3396 kJ/can
= 0.53 Savings is 47%
9.7 Using 1.8 million tons/yr of aluminum cans, 63 percent recycled, and Table 9.12:
a. The total primary energy used to make the aluminum for those cans.
New aluminum = 0.37 x 1.6x106
tons x 220,600 kJ/kg x
kg
2.2 lb
x
2000 lb
ton
=118.7x1012
kJ
9.3

Old aluminum = 0.67 x 1.6x106
tons x 5060 kJ/kg x
kg
2.2 lb
x
2000 lb
ton
= 4.9x1012
kJ
Total = (118.7 + 4.9) x 1012
= 123.6 x 1012
kJ
b. With no recycling:
All new aluminum =1.6x106
tons x 220,600 kJ/kg x
kg
2.2 lb
x
2000 lb
ton
= 320.9x1012
kJ
c. Using Table 9.8 CO2 emissions that result from that recycling.
Recycling = 0.67 x 1.6x106
tons Al x 3.71 MTCE/ton = 3.97 x 106
MTCE
As CO2: 3.97x106
metric tonsC x
44 tons CO2
12 tons C
=14.6x106
metric tons/yr
or 14.6x106
tonne/yr x
2200 lbs
tonne
x
ton
2000 lbs
=16 million U.S. tons/yr
9.8 With pickups from both sides of the 1-way street:
With pickups on one-side only on the 1-way street, need to make 2 passes
9.9 A 30 yd3
packer truck, 750 lb/yd3
, 100 ft stops, 5 mph, 1 min to load 200 lbs:
time/stop =100
ft
stop
x
mi
5280 ft
x
hr
5 mi
x
60 min
hr
+1 min =1.227 min/stop
9.4

9.10 Route timing:
a. Not collecting = 20min + 3x20min + 2x15min + 15min + 40min = 165 min/day
To fill a truck takes:
25yd3
truck x
4 yd3
curb
yd3
truck
x
customer
0.2yd3
x
stop
4 customer
x
1.5 min
stop
=187.5 min/load
Two loads per day takes:
2 loads/d x 187.5min/load +165 min(travel,breaks)[ ]x
hr
60min
= 9.0 hrs/day
b. Customers served:
25yd3
truck x
4 yd3
curb
yd3
truck
x
customer
0.2yd3
= 500 customers/load
#customers =
500 customers
load
x
2 loads
day
x
5 days
week
= 5000 customers/truck
c. Labor =
$40
hr
x
8 hrs
day
+
$60
hr
x
1hr
day
⎛
⎝
⎜
⎞
⎠
⎟x
5 day
week
x
52 wks
yr
= $98,800/yr
Truck cost =
$10,000
yr
+
$3500/yr
yd3
x 25yd3
= $97,500/yr
Customer cost =
($98,800 +$97,500)/yr
5000 customers
= $39.26/yr
9.11 To avoid overtime pay, working 8 hrs/day and needing 165 min to make runs back
and forth to the disposal site, breaks, etc (Problem 9.2):
Collection time = 8 hr/d x 60 min/hr – 165 min/d = 315 min/day
Customers =
315 min
day
x
stop
1.5 min
x
4 customers
stop
x
5 day
wk
= 4200 customers
Annual cost of service per customer is now:
($40/hr x 8 hr/wk x 5 d/wk x 52 wk/yr + $97,500/yr)/4200 = $43/yr
Cheaper to pay them overtime.
9.5

9.12 So, with 8-hr days a smaller truck can be used. As in Problem 9.11:
Collection time = 8 hr/d x 60 min/hr – 165 min/d = 315 min/day
With 2 truckloads per day,
Customers =
315 min
day
x
stop
1.5 min
x
4 customers
stop
= 840 customers/day
or 420 customers per truckload. At 2 loads per day and 5 days per week, that would
give 4200 customers once a week service. Truck size needed is therefore,
Truck size =
420 customers
truckload
x
0.2 yd3
at curb
customer
x
yd3
in truck
4yd3
at curb
= 21yd3
costing:
Truck$ =
$10,000
yr
+
$3500/yd3
yr
x 21yd3
= $83,500/yr
Labor$ =
$40
hr
x
8hr
day
⎛
⎝
⎜
⎞
⎠
⎟x
5 day
week
x
52 weeks
yr
= $83,200/yr
Resulting in: Customer$ =
$83,200 +$83,500( )/yr
4200 customers
= $39.69/yr
(compared with $39.26 per customer in Problem 9.10 and $43 in Problem 9.11.
9.13 Comparing two truck sizes:
a. Customers for each truck:
(A) 27 m3
truck: 27m3
truck x
4m3
curb
1m3
in truck
x
customer
0.25m3
curb
#customers =
432 customers
truckload
x
2 loads
day
x
5 days
week
= 4320 customers (27m3
)
(B) 15 m3
truck: 15m3
truck x
4m3
curb
1m3
in truck
x
customer
0.25m3
curb
#customers =
240 customers
truckload
x
3 loads
day
x
5 days
week
= 3600 customers (15m3
)
b. Hours per day for the crew:
(A) 27 m3
truck:
432 customers
truckload
x
0.4 min
customer
x
2 loads
day
= 346 min
Crew :
346 min +160 min(misc.)
60 min/hr
= 8.43 hr/day
9.6

(B) 15 m3
truck:
240 customers
truckload
x
0.4 min
customer
x
3 loads
day
= 288 min
Crew :
288 min +215 min(misc.)
60 min/hr
= 8.38 hr/day
c. Cost per customer:
(A) 27 m3
truck:
Cost :
$40
hr
x
8.43hr
day
x
5 day
week
x
52 week
yr
+ $120,000/yr = $207,672/yr (27m3
)
Customer$ =
$207,672/yr
4320 customers
= $48.07/yr (27m3
)
(B) 15 m3
truck:
Cost :
$40
hr
x
8.38hr
day
x
5 day
week
x
52 week
yr
+ $70,000/yr = $157,152/yr (15 m3
)
Customer$ =
$157,152/yr
3600 customers
= $43.65/yr (15 m3
)
Cheaper to run 3 trips a day in the smaller 15 m3
truck.
9.14 A $150,000 truck, 2 gal/mi, $2.50/gal, 10,000 mi/yr, $20k maintenance:
a. Amortized at 12%, 8-yr: CRF 8yr,12%( )=
i 1+ i( )
n
1+ i( )
n
−1
=
0.12 1+ 0.12( )
8
1+ 0.12( )
8
−1
= 0.201/yr
Amortization = $150,000 x 0.201/yr = $30,195/yr
Fuel = 10,000 mi/yr x 2 gal/mi x $2.50/gal = $50,000/yr
Total truck cost = $30,195 + $50,000 + $20,000 (maint) = $100,195/yr
b. Labor$ = $25/hr-person x 2 people/truck x 40 hr/wk x 52 wk/yr = $104,000/yr
c. Total Cost =
$100,195 + $104,000
10 ton/day x 260day/yr
= $78.54/ton
9.15 Reworking Examples 9.5 – 9.7 to confirm the costs in Table 9.17:
a. One-run per day, t =
150 ft/stop x 3600 s/hr
5 mi/hr x 5280 ft/mi
+ 4 can/stop x 20s/can =100.5 s/stop
Time to collect = 8 – 0.4 – (2x1-1)x0.4 – 0.25 – 1- 1x0.2 = 5.75 hr/day
which allows N stops per day (l-load per day)
9.7

N =
5.75 hr/d x 3600 s/hr
100.5 s/stop x 1 load/day
= 206 stops/load
Truck volume need is,
V =
4 can/stop x 4 ft3
/can x 206 stop/load
27ft3
/yd3
x 3.5 yd3
curb/yd3
in truck
= 34.9 yd3
With economics,
Labor = $99.840/yr as in Example 9.7
Truck = $25,000/yr + 4000$/yd3
yrx34.9 yd3
=$164,600/yr
206 stops x 2 cust/stop x 1 load/d x 5 d/wk = 2060 customers
Refuse =
2060 homes x 60 lb/home x 52 weeks/yr
2000 lb/ton
= 3214 tons/yr
Cost/ton =
(164,600 + 99,840) $/yr
3214 ton/yr
= $82.25/ton (OK)
b. Three-runs per day: 100.5 s/stop
Time to collect = 8 – 0.4 – (2x3-1)x0.4 – 0.25 – 1- 3x0.2 = 3.75 hr/day
which allows N stops per day (3-load per day)
N =
3.75 hr/d x 3600 s/hr
100.5 s/stop x 3 load/day
= 44.8 stops/load
Truck volume need is,
V =
4 can/stop x 4 ft3
/can x 44.8 stop/load
27ft3
/yd3
x 3.5 yd3
curb/yd3
in truck
= 7.6 yd3
With economics,
Labor = $99.840/yr as in Example 9.7
Truck = $25,000/yr + 4000$/yd3
yr x7.6yd3
=$55,400/yr
44.8 stops x 2 cust/stop x 3 load/d x 5 d/wk = 1344 customers
Refuse =
1344 homes x 60 lb/home x 52 weeks/yr
2000 lb/ton
= 2096 tons/yr
Cost/ton =
(99,840 + 55,400) $/yr
2096 ton/yr
= $74/ton (OK)
9.16 Transfer station: 200 tons/day, 5d/wk, $3 million, $100,000/yr, trucks $120,000,
20 ton/trip, $80k/yr, 4 trip/day, 5d/wk, 10%, 10-yr amortization.
Station costs: CRF 10yr,10%( )=
i 1+ i( )
n
1+ i( )
n
−1
=
0.10 1+ 0.10( )
10
1+ 0.10( )
10
−1
= 0.16275/yr
9.8

Solution for Introduction to Environment Engineering and Science 3rd edition by Gilbert M. Masters

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