Solution Manual
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:
C0
1 C
1
1
1
k = ln
=
=
= 0.128 min -1
ln
ln
t C 0 1 min 0.88C 0 1 min 0.88
Next, calculate the grease remaining on the clothes after 5 minutes:
-1
1
m = m0 e - kt = (0.500 g )e - (0.128 min )(5.00 min ) = 0.264 g
The grease that is not on the clothes must be in the water, so
Cw =
0.500 g - 0.264 g
= 0.00472 g/L
50.0 L
1.9
Plateau Creek
Q0, C0
Qf, C0
C.V.
Qb, C0
Qs, Cs
Qf/2, Cf
farm
a. A mass balance around control volume (C.V.) at the downstream junction
yields
Cs
(Q /2) Cf + Qb C0
= f
Qs
(0.5 m /s)(1.00 mg/L) + (4.0 m /s)(0.0015 mg/L) = 0.112 mg/L
=
(4.5 m /s)
3
3
3
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
Qf =
Q0 (Cs - C 0 )
C
⎛ Cf
- C0 + s
⎜
2
⎝ 2
⎞
⎟
⎠
=
(5.0 m /s)(0.04 - 0.0015 mg/L) = 0.371 m /s
3
3
0.04
⎛ 1.0
⎞
- 0.0015 +
mg/L ⎟
⎜
2
⎝ 2
⎠
1.10 Write a mass balance on a second order reaction in a batch reactor:
Accumulation = Reaction
V
mt
dm
= Vr(m)
dt
where r(m) = - km 2
t
dm
∫m m 2 = - k ∫0 dt
0
so,
1⎛ 1 1 ⎞
⎟
k = ⎜⎜
t ⎝ mt m0 ⎟⎠
calculate mt=1 based on the equation’s stoichiometry that
1 1
= - kt
m0 m t
1 mole of methanol yields 1 mole of carbon monoxide
⎛
⎞
(100 g CO )⎛⎜⎜ mole CO ⎞⎟⎟⎛⎜ 1 mole CH 3OH ⎞⎟⎜⎜ 32 g CH 3OH ⎟⎟ = 114.3 g CH 3OH
⎝ 28 g CO ⎠⎝
so,
and k =
1 mole CO
⎠⎝ mole CH 3 OH ⎠
CCH3OH, t =1 = 200 g - 114.3 g = 85.7 g
1 ⎛ 1
1 ⎞
⎟⎟ = 0.00667 d -1g -1
⎜⎜
1 d ⎝ 85.7 g 200 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.
Vp =
π (3.0 ft )2 (3400 ft )
4
24,033 ft 3
and k =
ln 2
ln 2
=
= 0.0578 min -1
t 1/2 (12 min)
For first-order decay in a steady-state PFR
⎡
C0 = Ce -kt = (1.0 mg/L)exp⎢ 0.0578 min −1 24,033 ft 3
⎣
(
)(
gal
)⎛⎜⎜ 10mingal ⎞⎟⎟⎛⎜ 0.134
ft
⎝
4
⎠⎝
3
⎞⎤
⎟⎥ = 2.82 mg/L
⎠⎦
1.15 The stomach acts like a CSTR reactor in which a first order decay reaction is
occurring.
Stomach
Gastric
Q, Ci
Digested food
Q, Ce
juices in
V, r(C)
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
V
dC
= QCi - QCe + Vr (C )
dt
dC
⎛Q
⎞
∫C C = - ⎜⎝ V + k ⎟⎠∫0 dt
0
t
C
so,
where r(C) = -kC, Ci = 0, and C = Ce
C = Ce
⎛Q
⎞
-⎜ + k ⎟ t
⎝V
⎠
⎞
⎛ ⎛ (0.012 L/min )(60 min/ hr )
⎛ 325 g ⎞
⎞
+ 1.33 hr -1 ⎟(1 hr )⎟⎟
=⎜
⎟ exp ⎜⎜ − ⎜
1.15 L
⎝ 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.
V=
(
)
nRT (1 mole) 0.082056 L ⋅ atm ⋅ mol -1 ⋅ K -1 (293.15 K )
=
= 24.04 L then
1 atm
P
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, Ce
Q, Ca
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
V
dC
= QCa - QCe + Vr(C ) = QCa - QCe − V (k d Ce + k s Ce )
dt
Assuming Ca = 0 and integrating yields
t=
C
1
1
ln 0 =
3
Q
C
⎛⎛
⎛⎜
⎞
⋅
+ k d + k s ⎞⎟
⎜ ⎜ 6 × 10 5 m ⎟ ⎛⎜ 60 24 min ⎞⎟
⎝ V
⎠
⎟
⎜
min
d
⎜⎝
⎠
⎠⎝
+ 0.0856d
⎜
6
3
×
2.0
10
m
⎜
⎜
⎝
(
t = 0.0322 d = 46.4 min
)
-1
⎛ 11.0
⎞
ln ⎜
ppmv ⎟
-5
⎞
⎝ 1 × 10
⎠
⎟
⎟
+ 0.02d -1 ⎟
⎟
⎟
⎠
1.22
a.
⎛ 8.00 mg ⎞⎛ mmole ⎞⎛ 40mg ⎞
⎟⎟⎜
mNaOH = 2⎜
⎟(1.00 L ) = 6.53 mg NaOH
⎟⎜⎜
⎝ L ⎠⎝ 98 mg ⎠⎝ mmole ⎠
b. The reaction (P1.3) is second-order (see the rate constant’s units) so that
dC H +
dt
= - kC H + C OH- = - kC H +
2
After integration and noting that at t = t1/2, C = C0/2, the equation can be written
t1/2 =
1
1
= 4.38 × 10 -8 seconds
=
kC0 ⎛
⎛ ⎛ 8.00 mg ⎞⎛ mmole ⎞ ⎞
11 mol ⎞
⎟⎟
⎜1.4 × 10
⎟⎜⎜ 2⎜
⎟⎜
L ⋅ s ⎠⎝ ⎝ L ⎠⎜⎝ 98 mg ⎟⎠ ⎟⎠
⎝
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 edition Solutions Manual.
2.1 – 2.14 The solutions for these problems are the solutions for problems 2.1 - 2.14 in
the 2nd edition Solutions Manual.
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)
⎛ 2 moles Al0 ⎞ ⎛ 27 g Al0 ⎞ ⎛ mole H 2 ⎞ 9 g Al0
⎟⎟ ⎜⎜
⎟⎟ =
⎟⎜
So, ⎜⎜
0 ⎟⎜
⎝ 3 moles H 2 ⎠ ⎝ mole Al ⎠ ⎝ 2 g H 2 ⎠ g H 2
2.17 - 2.21 The solutions for these problems are the solutions for problems 2.15 - 2.19 in
the 2nd edition Solutions Manual.
2.22
From Table 2.3, Ksp,CaSO4 is 2 x 10-5 and Ksp = [Ca2+] [SO42-] = [SO42-]2
At saturation: [SO42-] = Ksp½ = (2 x 10-5) ½ = 0.00447 mol/L
⎛ 0.5 g ⎞ ⎛ mole ⎞
⎟⎟ = 0.00368 mol/L
Check if saturation is exceeded: ⎜
⎟ ⎜⎜
⎝ L ⎠ ⎝ 136 g ⎠
Since saturation is not exceeded the sulfate concentration is 0.00368 mol/L
2.23
Assume a closed system.
a. [Ca2+] = Ctot = [H2CO3] + [HCO3-] + [CO32-]
eqn. 1
Charge balance: [H+] + 2[Ca2+] = [HCO3-] + 2[CO32-] + [OH-]
eqn. 2
At pH 8.5: Ctot ≈ [HCO3-] » [CO32-] (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.
[CO
] = (10 )(10 ) = 2.24 × 10
] = [H ][HCO
K
(4.47 × 10 )
−
+
-8.5
-5.5
3
2 (aq)
-7
-8
M = 10 -7.65 M
1
c.
[CaCO ] = ((2 × 10
3 (s)
-3
)
)
⎛ 100 g ⎞
M - 10 -5.5 M ⎜
⎟ = 0.19968 g/L = 0.20 g/L
⎝ mol ⎠
2.24 - 2.36 The solutions for these problems are the solutions for problems 2.20 - 2.32 in
the 2nd edition Solutions Manual.
3.1 – 3.17 The solutions for these problems are the solutions for problems 3.1 - 3.17 in
the 2nd edition Solutions Manual.
3.18 a.
N* =
K 3,500 rabbits
=
= 1,750 rabbits
2
2
Use the midpoint of the first year population (350 rabbits) as N0
dN 0
r=
and
b.
N 0 ⎛⎜1 ⎝
dt
N0
⎞
K ⎟⎠
=
200 rabbits
yr
= 0.635 yr -1
(350 rabbits)⎡⎢1 - (350 rabbits) ⎤⎥
⎣ (3,500 rabbits) ⎦
(
)
dN * rK 0.635 yr -1 (3,500 rabbits )
=
=
= 556 rabbits/yr
dt
4
4
dN 0
⎡ (1,200 rabbits) ⎤
⎛ N ⎞
= rN 0 ⎜1 - 0 ⎟ = 0.635 yr -1 (1,200 rabbits)⎢1 ⎥ = 501 rabbits/yr
dt
⎝ K ⎠
⎣ (3,500 rabbits)⎦
(
)
3.19 a.
N* =
K 7,000 deer
=
= 3,500 deer
2
2
b. Use the mean current population of deer during the year (2,350 deer) as N0
dN 0
r=
and
N 0 ⎛⎜1 ⎝
dt
N0
⎞
K ⎟⎠
(
=
300 deer
yr
= 0.192 yr -1
(2,350 deer )⎡⎢1 - (2,350 deer )⎤⎥
⎣ (7,000 deer ) ⎦
)
dN * rK 0.192 yr -1 (7,000 deer )
=
=
= 336 deer/yr
dt
4
4
c.
⎡ (3,500 deer )(7,000 - 2350 deer ) ⎤
1
1 ⎡ N (K - N 0 ) ⎤
ln ⎢
t = ln ⎢
⎥=
⎥ = 3.55 yr
-1
r ⎣ N 0 (K - N ) ⎦ 0.192 yr
⎣ (2,350 deer )(7,000 - 3,500 deer ) ⎦
(
)
3.20 – 3.28 The solutions for these problems are the solutions for problems 3.18 - 3.26 in
the 2nd edition Solutions Manual.
4.1 – 4.10 The solutions for these problems are the solutions for problems 4.1 - 4.10 in
the 2nd edition Solutions Manual.
4.11
a.
⎛ 0.12 rem ⎞
⎛ 1 cancer death ⎞
⎟⎟ (70 yr ) ⎜
Denver risk = ⎜⎜
⎟ = 1.05 × 10 -3
yr
8,000
rems
⎠
⎝
⎠
⎝
⎛ 0.04 rem ⎞
⎛ 1 cancer death ⎞
⎟⎟ (70 yr ) ⎜
Sea level risk = ⎜⎜
⎟ = 0.35 × 10 -3
yr ⎠
⎝ 8,000 rems ⎠
⎝
b. Denver deaths due to cosmic radiation exposure:
⎛ 0.12 rem ⎞
⎛ 1 death ⎞
⎜⎜
⎟⎟ 0.57 × 10 6 people ⎜
⎟ = 8.6 deaths yr
⋅
person
yr
8
,
000
rems
⎝
⎠
⎝
⎠
(
)
⎛ 189 deaths
⎞
⎜
yr ⎟
deaths
Expected = 0.57 × 10 people ⎜
⎟ = 1,077
yr
100
,
000
people
⎜
⎟
⎝
⎠
(
6
)
c.
⎛ 1 cancer death ⎞
Incremental risk = 10 -6 = ⎛⎜ 0.12 - 0.04 rem ⎞⎟ (N yr ) ⎜
⎟
yr
⎝
⎠
⎝ 8,000 rems ⎠
N=
8,000 × 10 -6
= 0.1 yr (Table 4.3 suggests 2 months)
0.08
d.
(300 × 10
6
⎛ 0.04 rem ⎞ ⎛ 1 cancer death ⎞
⎟⎟ ⎜
people) ⎜⎜
⎟ = 1,500 deaths yr
⎝ person ⋅ yr ⎠ ⎝ 8,000 rems ⎠
4.12
Radon exposure of 1.5 piC/L equivalent to 400 mrem/yr (0.4 rem/yr):
a.
⎛ 0.4 rem ⎞ ⎛ 1 cancer death ⎞
⎜⎜
⎟⎟ ⎜
⎟ (300 × 10 6 people) = 15,000 deaths yr
⎝ person ⋅ yr ⎠ ⎝ 8,000 rems ⎠
b.
⎛ 0.4 rem ⎞ ⎛ 1 cancer death ⎞ ⎛ 70 yr ⎞
-3
⎟⎟ ⎜
Risk = ⎜⎜
⎟⎜
⎟ = 3.5 × 10 cancer death lifetime
⎝ yr ⎠ ⎝ 8,000 rems ⎠ ⎝ lifetime ⎠
4.13-4.14
The solutions for these problems are the solutions for problems 4.13 - 4.14
in the 2nd edition Solutions Manual.
4.15
Data is taken from Tables 4.9 and 4.10
a. conc =
(risk )(body wt )(lifespan )⎛⎜ 365 d yr ⎞⎟
⎝
(PF)(intake rate)(freq )(duration )
⎠
(10 ) (70 kg ) ⎛⎜ 70lifeyr ⎞⎟ ⎛⎜⎜ 365yr d ⎞⎟⎟
-6
=
b. freq =
⎝
⎠⎝
⎠
= 4.93 × 10 -4 mg/m 3
3
⎛ 2.9 × 10 kg ⋅ d ⎞ ⎛ 20 m ⎞ ⎛ 250 d ⎞ ⎛ 25 yr ⎞
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ ⎜
⎟
mg
⎝
⎠ ⎝ d ⎠ ⎝ yr ⎠ ⎝ life ⎠
-2
(risk )(body wt )(lifespan )⎛⎜ 365 d yr ⎞⎟
⎝
⎠
(PF)(intake rate)(conc)(duration )
d⎞
⎟
(10 ) (70 kg ) ⎛⎜ 70lifeyr ⎞⎟ ⎛⎜⎜ 365
yr ⎟
-7
=
⎝
⎠⎝
⎠
= 25 d/yr
-4
⎛ 2.9 × 10 kg ⋅ d ⎞ ⎛ 20 m ⎞ ⎛ 4.93 × 10 mg ⎞ ⎛ 25 yr ⎞
⎜⎜
⎟⎟ ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ ⎜
⎟
mg
m3
⎝
⎠⎝ d ⎠⎝
⎠ ⎝ life ⎠
-2
3
4.16
a.
Use Table 4.11 oral RfDs and Table 4.10 for other factors
HItotal = HIarsenic + HImethylene chloride
⎛ 1 L ⎞ ⎡ 0.070 mg ⎛ kg ⋅ d ⎞ 0.560 mg ⎛ kg ⋅ d ⎞⎤
⎞
⎞
⎛
d ⎟ ⎛⎜
⎟⎟ + ⎜
⎟⎟⎥ = 3.5
HI = ⎜⎜
⎟ ⎜⎜
⎟ ⎜⎜
⎢
⎟
70 kg ⎣⎝
L
L
⎠ ⎝ 0.0003 mg ⎠ ⎝
⎠ ⎝ 0.060 mg ⎠⎦
⎝
⎠
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
⎛ 1 L ⎞⎛ 250 d ⎞⎛ 10 yr ⎞
⎟⎟⎜
⎜ ⎟⎜⎜
⎟
⎝ d ⎠⎝ yr ⎠⎝ life ⎠ ⎡⎛ 1.75 kg ⋅ d ⎞ ⎛ 0.0700 mg ⎞ ⎛ 0.0075 kg ⋅ d ⎞ ⎛ 0.560 mg ⎞⎤
⎟⎟ ⎜
⎟⎟ ⎜
risk =
⎟ + ⎜⎜
⎟⎥
⎢⎜⎜
L
mg
L
⎠ ⎝
⎝
⎠⎦
⎛ 70 yr ⎞⎛ 365 d ⎞ ⎣⎝ mg ⎠ ⎝
⎠
⎟⎟
(70 kg )⎜
⎟⎜⎜
⎝ life ⎠⎝ yr ⎠
risk = 1.8 × 10 -4
Since this risk is » 10-6, it would probably not be considered
acceptable
4.17
a.
⎛ 50 kg ⋅ d ⎞ ⎛ 0.00002 mg ⎞ ⎛ 20 m 3 ⎞ ⎛ 250 d ⎞ ⎛ 25 yr ⎞
⎟⎟ ⎜⎜
⎜⎜
⎟⎟ ⎜
⎟⎟ ⎜
⎟ ⎜⎜
⎟
3
d
yr
mg
m
⎝
⎠
⎝ life ⎠
⎝
⎠
⎝
⎠
⎝
⎠
risk =
= 7 × 10 -5
⎛
⎞
(70 kg ) ⎛⎜ 70 yr ⎞⎟ ⎜⎜ 365 d ⎟⎟
⎝ life ⎠ ⎝ yr ⎠
b.
(220,000 workers ) (7 × 10 -5 ) = 15 workers
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
(220,000 workers)(0.24) + 15 workers = 52,815 workers
d.
From Table 4.11 RfDs for both arsenic and mercury are 0.0003 mg/kg·d
⎡
⎤
⎛ 20 m 3 ⎞
⎜⎜
⎟⎟
⎢
⎥
d ⎠
⎝
⎢
⎥ ⎡⎛⎜ 0.00002 mg ⎞⎟ + ⎛⎜ 0.0001 mg ⎞⎟⎤ = 0.11
HI =
⎥
⎢ ⎛ 0.0003 mg ⎞
⎥ ⎢⎣⎝
m3
m3
⎠ ⎝
⎠⎦
⎟⎟(70kg ) ⎥
⎢ ⎜⎜
⎣⎢ ⎝ kg ⋅ d ⎠
⎦⎥
Since the HI is less than 1 the non-cancer hazard is acceptable, despite
there being a cancer risk greater than 10-6.
4.18 – 4.26 The solutions for these problems are the solutions for problems 4.15 - 4.23 in
the 2nd edition Solutions Manual.
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 1.1x10-2
x
= 6.6x10 −7
70 kg x 365 d/yr x 70 yr
mg/kg - d
Δ cancer =
260x106 people x 6.6x10 -7 cancer / person − life
= 2.5 cancer/yr
70 yr/lifetime
b. benzene at 0.005 mg/L:
Risk =
2L/d x 0.005 mg/L x 360 d/yr x 30 yr 2.9x10 -2
x
= 1.75x10 −6
70 kg x 365 d/yr x 70 yr
mg/kg - d
260x106 people x 1.75x10 -6 cancer / person − life
Δ cancer =
= 6.5 cancer/yr
70 yr/lifetime
c. arsenic at 0.01 mg/L:
2L/d x 0.01 mg/L x 360 d/yr x 30 yr
1.75
Risk =
x
= 2.11x10 − 4
70 kg x 365 d/yr x 70 yr
mg/kg - d
260x10 6 people x 2.11x10 -4 cancer / person − life
Δ cancer =
= 785 cancer/yr
70 yr/lifetime
d. carbon tetrachloride at 0.005 mg/L:
Risk =
2L/d x 0.005 mg/L x 360 d/yr x 30 yr
0.13
x
= 7.9x10 −6
70 kg x 365 d/yr x 70 yr
mg/kg - d
Δ cancer =
260x106 people x 7.9x10 -6 cancer / person − life
= 29 cancer/yr
70 yr/lifetime
e. vinyl chloride at 0.002 mg/L:
Risk =
2L/d x 0.002 mg/L x 360 d/yr x 30 yr
2.3
x
= 5.6x10 −5
70 kg x 365 d/yr x 70 yr
mg/kg - d
Δ cancer =
260x106 people x 5.6x10 -5 cancer / person − life
= 206 cancer/yr
70 yr/lifetime
f. PCBs at 0.0005 mg/L:
Risk =
2L/d x 0.0005 mg/L x 360 d/yr x 30 yr
7.7
x
= 4.6x10 −5
70 kg x 365 d/yr x 70 yr
mg/kg - d
Δ cancer =
260x106 people x 4.6x10 -5 cancer / person − life
= 172 cancer/yr
70 yr/lifetime
4.28 – 4.33 The solutions for these problems are the solutions for problems 4.25 - 4.30 in
the 2nd edition Solutions Manual.
5.1 – 5.33 The solutions for these problems are the solutions for problems 5.1 - 5.33 in
the 2nd edition Solutions Manual.
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:
C 0.40 13.3
=
=
N 0.12
4
P 0.03
1
Therefore, carbon is limiting.
5.35 – 5.53 The solutions for these problems are the solutions for problems 5.34 - 5.52 in
the 2nd edition Solutions Manual.
5.54
a. From Table 5.11, η = 0.34 for a sand aquifer
K=
b. R =
(
)
(0.34) 0.09 m d
v
ηv'
=
=
= 61.2 m
d
dh
dh
(0.0005)
dL
dL
(
) (
v'
)
so vcontaminant =
vcontaminant
m
v' 0.09 d
=
= 0.015 m
d
6
R
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)
mmax = ⎛⎜150 g 3 ⎞⎟ (0.34)(2.0 m )(4.0 m )(5.48 m ) = 2,230 g
m ⎠
⎝
Calculate the mass leaked
(
mleaked = 100 mL
)
⎛1.63 g
⎞ (365 d ) = 59,500 g
mL ⎟⎠
d ⎜⎝
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.
(
Q = wcapture 2 Bv = (2 m ) 2 (4 m )(0.34 ) 0.09 m
e.
t=
mPCE
=
QC PCE 0.490 m 3
(
)
(59,500 g )
(
)= 0.490 m
d
)
3
⎛15 mg ⎞⎟ 10 3 L
⎛⎜10 −3 g
⎞
d ⎜⎝
L⎠
mg ⎟⎠
m3 ⎝
d
= 8,100 d = 22.2 yr
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:
V
π (43.0 ft ) (10.0 ft )
θ= =
= 0.174 d = 4.17 hr
Q 2.5 × 10 6 gal/d ⎛ ft 3
⎞
⎜ 7.4805 gal ⎟
⎝
⎠
2
(
)
b. The critical velocity is:
(
)
⎛⎜ 2.5 × 10 6 gal ⎞⎟⎛⎜ ft 3
⎞d
d ⎠⎝ 7.4804 gal ⎟⎠ 24 hr
ft 3
Q ⎝
=
=
2.40
vo =
Ab
ft 2 ⋅ hr
π (43.0 ft )2
c. The weir loading rate is:
(
)
⎛⎜ 2.5 × 10 6 gal ⎞⎟⎛⎜ ft 3
⎞d
d ⎠⎝ 7.4804 gal ⎟⎠ 24 hr
Q ⎝
ft 3
weir loading rate =
=
= 51.5
Lw
π ⋅ 2 ⋅ (43.0 ft )
ft ⋅ hr
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:
vs =
g (ρ p - ρ )d p2
18μ
(9.807 m/s )(0.4)(998.2 kg/m )(1.0 × 10
=
2
3
-5
m
)
kg ⎞
⎛
18⎜ 0.00100
⎟
m⋅s ⎠
⎝
2
= 2.18 × 10 -5 m
s
b.
(7,500 m / d )⎛⎜ 24 ⋅ 60d ⋅ 60 s ⎞⎟
3
vo =
Q
=
Ab
⎝
(10 m )(30 m )
⎠ = 2.89 × 10 -4 m
s
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,
⎡ 18μvs ⎤
dp = ⎢
⎥
⎢⎣ g (ρ p - ρ )⎥⎦
1
2
⎡ ⎛
kg ⎞⎛
-4 m ⎞ ⎤
⎢18⎜ 0.00100 m ⋅ s ⎟⎜ 2.89 × 10 s ⎟ ⎥
⎠⎝
⎠⎥
=⎢ ⎝
kg
m
⎞
⎛
⎞ ⎥
⎢ ⎛
⎢ ⎜⎝ 9.807 s 2 ⎟⎠(0.4)⎜⎝ 998.2 m 3 ⎟⎠ ⎥
⎣
⎦
1
2
= 3.65 × 10 -5
m
μm
= 36.5
s
s
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:
⎡ 18Qμ
⎤
w=⎢
2 ⎥
⎣⎢ 5 g (ρ p - ρ )d ⎦⎥
1
2
⎡
⎤
⎛
kg ⎞
m 3 ⎞⎛
⎜
⎟⎟⎜ 0.00100
18
0.100
⎟
⎢
⎥
⎜
m ⋅s ⎠
s ⎠⎝
⎝
⎢
⎥
=
⎢ ⎛
kg ⎞ −5 2 ⎥
m ⎞⎛
⎢ 5 ⎜ 9.807 2 ⎟⎜1,200 - 998.2 3 ⎟ 10 m ⎥
m ⎠
s ⎠⎝
⎥⎦
⎣⎢ ⎝
(
)
1
2
= 42.7 m
6.6
From Appendix C: At 15°C, μ = 0.00114 kg·m-1·s-1
G=
a.
P
μVb
⎡
⎤
⎢
⎥
(186 W )
⎥
G1 = ⎢
kg ⎞
2
⎢⎛
⎥
⎢ ⎜⎝ 0.00114 m ⋅ s ⎟⎠(4.17 m ) (3.75 m ) ⎥
⎣
⎦
1
2
= 50.0 s -1
Similarly, G2 = 20.1 s-1 and G3 = 10.0 s-1.
b.
Ω=
πd p3 N 0
6
(
π (20.0 × 10 -6 m ) (1.8 × 105 mL-1 )10 6 mL
3
=
6
)
m 3 = 7.54 × 10 -4
α 4ΩGVb
N0
=1+
N
πQ
(
)(
)
N0
(
1.0) 4 7.54 × 10 -4 50.0 s -1 (4.17 m ) (3.75 m )
=1+
= 17.9
N
⎛
m 3 ⎞⎛
d
⎞
π ⎜⎜16,000 ⎟⎟⎜
⎟
d ⎠⎝ 24 ⋅ 60 ⋅ 60 s ⎠
⎝
2
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: V
and r( N ) = -
α 4ΩGN
π
dN
α 4ΩG
∫N N = - π ∫0 dt
0
N
so,
dN
= Vr( N )
dt
t
N = N 0e
which yields:
⎛ α 4 ΩGt ⎞
-⎜
⎟
⎠
⎝ π
b.
⎡
kg ⋅ m 2
6.5 × 10 -5
⎢
P
s3
G=
=⎢
kg ⎞
μV ⎢ ⎛
3
⎢ ⎜⎝ 0.00114 m ⋅ s ⎟⎠ 0.004 m
⎣
(
⎤
⎥
⎥
⎥
⎥
⎦
)
1
2
= 3.78 s -1
c.
3
Ω=
N = N 0e
⎛ α 4 ΩGt ⎞
-⎜
⎟
⎝ π
⎠
πd p3 N 0
6
=
⎛ 1.0 × 10 7 ⎞
⎟
3 ⎟
⎝ 0.0040 m ⎠
π (5.0 × 10 -5 m ) ⎜⎜
6
(
)(
= 1.64 ×10 -4
)(
)
⎛ (1.0)4 1.64 × 10 -4 3.78 s -1 60 s
(30 min ) ⎞⎟
1.0 × 10 7
min
=
exp ⎜⎜ ⎟
π
4.0 L
⎝
⎠
N = 6.07 × 10 5
particles
L
d.
⎛ 1.0 × 10 7 ⎞
⎜⎜
⎟
4.0 L ⎟⎠
N0
⎝
=
= 4.12 singlets
5 -1
aggregate
N 6.07 × 10 L
e.
N = N 0e
⎛ α 4 ΩGt ⎞
-⎜
⎟
⎝ π
⎠
(
)(
)(
)
⎛ (0.2)4 1.64 × 10 -4 3.78 s -1 60 s
(30 min ) ⎞⎟
1.0 × 10 7
⎜
min
=
exp ⎜ ⎟
π
4.0 L
⎝
⎠
N = 1.88 × 10 6
particles
L
6.8
(
)
m⎞
m3
⎛
a. Q = va Af = ⎜ 8.0 ⎟ 80 m 2 = 640
hr ⎠
hr
⎝
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:
2
m
ref Af t c 7.7 hr 80 m (52 hr )
=
= 33,370 m 3 (per cycle)
vf =
ηf
0.96
(
)(
(
)
)
From (6.8.1): Vb + Vr = Vf (1 - η f ) = 33,370 m 3 (1 - 0.96) = 1,335 m 3 (per cycle)
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.
Vb =
(
(
)
)
Qb
L ⎞
⎛
= 307,200 L
Af t b' = ⎜10.0 2 ⎟ 64 m 2 (8.0 min ) 60 s
min
Af
m ⋅s ⎠
⎝
(
(
)
)
L ⎞
⎛
= 316,800 L
Vr = va Af t r = ⎜ 5.50 2 ⎟ 64 m 2 (15 min ) 60 s
min
m ⋅s ⎠
⎝
t f = t c − t b − t r = (24 ⋅ 60 min ) − (16 min ) - (15 min ) = 1,409 min
Vf =
ηf =
(
(
)
)
Qf
L ⎞
⎛
Af t ' = ⎜ 5.50 2 ⎟ 64 m 2 (1,409 min ) 60 s
= 2.98 × 10 7 L
min
f
Af
m ⋅s ⎠
⎝
Vf − Vb − Vr 2.98 × 10 7 - 307,200 - 316,800 L
=
= 0.979 = 98 %
Vf
2.98 × 10 7 L
b. The fraction of a filtration cycle that is not backwashing is:
⎛ hr ⎞
24 hr - (16 min )⎜
⎟
tf + tr − t b
60 min ⎠
⎝
=
= 0.989
tc
24 hr
so the number of filters required is:
n=
Qplant
va Af (0.989 )
(2.40 m s )
3
=
(
L ⎞
⎛
2
⎜ 5.50 2 ⎟ 64 m
m
s
⋅
⎝
⎠
Therefore, at least 7 filters are required.
⎛ m3 ⎞
⎟⎟(0.989 )
⎜⎜
⎝ 1000 L ⎠
)
= 6.89
6.10
a. For a steady state PFR, the Giardia concentration in the contactor effluent is:
*
N = N 0 e -k θ
)(
(
)
m 3 60 s
Q N 0 0.20
s
min ln (1,000 ) = 156 m 3
so, Vb = * ln
=
N
k
0.53 min -1
b. For a steady state CSTR, the mass balance is:
QN0 = QN + k*NVb
(
)(
)
m 3 60 s
Q ⎛ N 0 ⎞ 0.20
s
min (1,000 - 1) = 22,600 m 3
- 1⎟ =
so, Vb = * ⎜
k ⎝ N ⎠
0.53 min -1
(
)
c. For a steady state series of CSTRs, the mass balance is:
Nm
1
=
N 0 1 + θk *
(
Q ⎛⎛ N ⎞
so, Vb,1 = * ⎜⎜ ⎜⎜ 0 ⎟⎟
k ⎜⎝ Nm ⎠
⎝
1
)
m
m
(
)(
)(
⎞ 0.20 m 3 60 s
s
min (1,000) 15 - 1 = 67.5 m 3
- 1⎟⎟ =
-1
0.53 min
⎟
⎠
(
)
)
and the total contactor volume is: mVb,1 = 5(67.50 m3) = 337 m3
d.
k=
k*
0.53 min -1
L
=
= 0.265
n
1.0
mg ⋅ min
C
⎛⎜ 2.0 mg ⎞⎟
L⎠
⎝
6.11
For a steady state CSTR, the mass balance is:
QN0 = QN + k*NVb
Vb k *
(
2.00 L )(7.80 s -1 )
so, Q =
=
= 0.0156 L
s
(
1,000 - 1)
⎛ N 0 - 1⎞
⎜ N ⎟
⎝
⎠
and the volume produced during ten hours of operation would be:
L⎞
⎛
⎛ 60 ⋅ 60 s ⎞
Qt = ⎜ 0.0156 ⎟(10 hr )⎜
⎟ = 562 L
s⎠
⎝
⎝ hr ⎠
6.12 Hardness in meq/L:
Ca 2+ =
150 mg/L
= 7.50 meq/L
20.0 mg/meq
Mg 2+ =
60 mg/L
= 4.92 meq/L
12.2 mg/meq
Total hardness = 7.50 + 4.92 = 12.4 meq/L
Hardness as CaCO3:
Total hardness = 12.4
mg as CaCO 3
meq
× 50.0
= 621 mg/L as CaCO 3
L
meq
Table 6.4 would classify this as very hard water.
6.13 – 6.16 The solutions for these problems are the solutions for problems 6.2 - 6.5 in
the 2nd edition Solutions Manual.
6.17
Component
Concentration
(mg/L)
14.5
110.0
50.7
75
350
85.5
16.2
8.2
CO2
Ca2+
Mg2+
Na+
HCO3SO42ClpH
Equiv. Weight
(mg/meq)
22
20
12.2
23.0
61.0
48.0
35.5
-
CH
Concentration
(meq/L)
0.6591
5.500
4.156
3.261
5.738
1.781
0.4563
-
NCH
M g-N C H
9.656
0
5.500
C a-C H
C a 2+
M g 2+
5.738
0
H C O 3-
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
6.0 mg/L
Equiv. Weight
(mg/meq)
22
CO2
2+
50.0 mg/L
20
2.500
2+
Ca
Mg
Concentration
(meq/L)
0.2727
20.0 mg/L
12.2
1.639
HCO3-
3.1 mmole/L
61.0
3.100
pH
7.6
-
-
CH
NCH
M g-N C H
4.139
0
2.500
C a-C H
C a 2+
M g 2+
0
3.100
H C O 3-
Table P6.18 Components, lime and soda ash dosage, and solids generated
Component (eq’n.)
CO2(aq) (6.37)
Ca-CH (6.38)
Mg-CH (6.39)
Ca-NCH (6.40)
Mg-NCH (6.41)
Excess
Totals
Concentration
(meq/L)
0.2727
2.500
0.600
0
1.039
Lime
(meq/L)
0.2727
2.500
1.200
0
1.039
0.400
5.412
Soda ash
(meq/L)
0
0
0
0
1.039
CaCO3(s)
(meq/L)
0.2727
5.000
1.200
0
1.039
Mg(OH)2(s)
(meq/L)
0
0
0.600
0
1.039
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:
⎡⎛
meq ⎞⎛
mg ⎞ ⎛
meq ⎞⎛
mg ⎞⎤⎛ 15 × 10 6 gal ⎞⎛ 3.785 L ⎞⎛ kg ⎞
⎟⎟⎜⎜
⎟⎟
⎜
⎟
⎜
⎟⎟⎥⎜⎜
⎟⎟⎜⎜ 6
7
.
512
50
.
0
1
.639
2
9.2
+
⎟⎜
⎟⎜
⎢⎜
⎟ ⎜
L
meq
L
meq
d
gal
10
mg
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
= 24,040 kg/d
6.19
Component
Concentration
Ca2+
2+
Concentration
(meq/L)
4.500
Concentration
(mg/L CaCO3)
225.0
30 mg/L
12.2
2.4596
123.0
HCO3-
165 mg/L
61.0
2.705
135.3
pH
7.5
-
-
-
NCH
CH
C a-N C H
M g-N C H
0
225.0
C a-C H
C a 2+
348.0
Mg
90 mg/L
Equiv. Weight
(mg/meq)
20
M g 2+
0
135.3
H C O 3-
[CO ( ) ] = [HCOK ][H ]
3
2 aq
1
+
g ⎞⎛ mol ⎞⎛ −7.5 mol ⎞
⎛
⎟⎜10
⎟
⎜ 0.165 ⎟⎜⎜
L ⎠⎝ 61.0 g ⎟⎠⎝
L ⎠
⎝
= 1.914 × 10 -4 M = 19.14 mg/L CaCO 3
=
-7
4.47 × 10 M
Table P6.19 Components, lime and soda ash dosage, and solids generated
Component (eq’n.)
CO2(aq) (6.37)
Ca-CH (6.38)
Mg-CH (6.39)
Ca-NCH (6.40)
Mg-NCH (6.41)
Excess
Totals
Concentration
(mg/L
CaCO3)
19.14
135.3
0
90.00
123.0
Lime
(mg/L
CaCO3)
19.14
135.3
0
0
123.0
20
297.4
Soda ash
(mg/L
CaCO3)
0
0
0
90.00
123.0
CaCO3(s)
(mg/L
CaCO3)
19.14
270.6
0
90.00
123.0
Mg(OH)2(s)
(mg/L
CaCO3)
0
0
0
123.0
213.0
⎛ 297.4 mg CaCO 3 ⎞⎛ 74 mg Ca(OH) 2 ⎞
⎟⎟ = 220.1 mg
a. Limerequired = ⎜
⎟⎜⎜
L
L
⎝
⎠⎝ 100 mg CaCO 3 ⎠
⎛ 213.0 mg CaCO 3 ⎞⎛ 106 mg Ca(OH) 2 ⎞
⎟⎟ = 225.8 mg
b. Soda Ash required = ⎜
⎟⎜⎜
L
L
100
mg
CaCO
⎝
⎠⎝
3 ⎠
6.20 The solution for this problem is the solution for problem 6.6 in the 2nd edition
Solutions Manual.
6.21
Component
Concentration
Ca2+
2+
Concentration
(meq/L)
4.75
Concentration
(mg/L CaCO3)
238
26 mg/L
12.2
2.13
107
HCO3-
160 mg/L
61.0
2.62
131
pH
7.0
-
-
-
CH
NCH
C a-N C H
M g-N C H
0
4.75
C a-C H
C a 2+
6.88
Mg
95 mg/L
Equiv. Weight
(mg/meq)
20
M g 2+
2.62
0
H C O 3-
[CO ( ) ] = [HCOK ][H ]
3
2 aq
1
+
g ⎞⎛ mol ⎞⎛ −7 mol ⎞
⎛
⎟⎜10
⎟
⎜ 0.160 ⎟⎜⎜
L ⎠⎝ 61.0 g ⎟⎠⎝
L ⎠
⎝
= 5.868 × 10 -4 M = 1.17 meq/L
=
4.47 × 10 -7 M
Table P6.21 Components, lime and soda ash dosage, and solids generated
Component (eq’n.)
CO2(aq) (6.37)
Ca-CH (6.38)
Mg-CH (6.39)
Ca-NCH (6.40)
Mg-NCH (6.41)
Excess
Totals
Concentration
(meq/L)
1.17
2.62
0
2.13
2.13
Lime
(meq/L)
1.17
2.62
0
0
2.13
0.400
6.32
Soda ash
(meq/L)
0
0
0
2.13
2.13
CaCO3(s)
(meq/L)
1.17
5.24
0
2.13
2.13
Mg(OH)2(s)
(meq/L)
0
0
0
0
2.13
4.26
10.67
2.13
⎛ 6.32 meq ⎞⎛ 37.07 mg ⎞
⎟⎟ = 234 mg
a. Limerequired = ⎜
⎟⎜⎜
L
L
⎝
⎠⎝ meq ⎠
⎛ 4.26 meq ⎞⎛ 53 mg ⎞
⎟⎟ = 226 mg
b. Soda Ash required = ⎜
⎟⎜⎜
L
L
meq
⎝
⎠⎝
⎠
6.22 The solution for this problem is the solution for problem 6.7 in the 2nd edition
Solutions Manual.
6.23
Component
Concentration
Ca2+
2+
Mg
40.0 mg/L
Equiv. Weight
(mg/meq)
20
Concentration
(meq/L)
2.00
Concentration
(mg/L CaCO3)
100
10.0 mg/L
12.2
0.820
41.0
HCO3-
110 mg/L
61.0
1.80
90.2
pH
7.0 (assumed)
-
-
-
NCH
CH
Ca-NCH
M g-NCH
Ca 2+
2.82
0
2.00
Ca-CH
M g 2+
0
1.80
HCO 3 -
[CO ( ) ] = [HCOK ][H ]
3
2 aq
1
+
g ⎞⎛ mol ⎞⎛ −7 mol ⎞
⎛
⎟⎜10
⎟
⎜ 0.110 ⎟⎜⎜
L ⎠⎝ 61.0 g ⎟⎠⎝
L ⎠
⎝
= 4.034 × 10 -4 M = 0.807 meq/L
=
4.47 × 10 -7 M
Table P6.23 Components, lime and soda ash dosage, and solids generated
Component (eq’n.)
CO2(aq) (6.37)
Ca-CH (6.38)
Mg-CH (6.39)
Ca-NCH (6.40)
Mg-NCH (6.41)
Excess
Totals
Concentration
(meq/L)
0.807
1.80
0
0.200
0.820
Lime
(meq/L)
0.807
1.80
0
0
0.820
0.400
3.83
Soda ash
(meq/L)
0
0
0
0.200
0.820
CaCO3(s)
(meq/L)
0.807
3.60
0
0.200
0.820
Mg(OH)2(s)
(meq/L)
0
0
0
0
0.820
1.02
5.43
0.82
⎛ 3.83 meq ⎞⎛ 37.07 mg ⎞
⎟⎟ = 142 mg
a. Limerequired = ⎜
⎟⎜⎜
L
L
⎝
⎠⎝ meq ⎠
⎛ 1.02 meq ⎞⎛ 53 mg ⎞
⎟⎟ = 54.1 mg
b. Soda Ash required = ⎜
⎟⎜⎜
L
L
meq
⎝
⎠⎝
⎠
c. Sludge generated:
⎡⎛
meq ⎞⎛
mg ⎞ ⎛
meq ⎞⎛
mg ⎞⎤⎛ 37 × 10 6 gal ⎞⎛ 3.785 L ⎞⎛ kg ⎞
⎟⎟⎜⎜
⎟⎟
⎜
⎟
⎜
⎟⎥⎜
⎟⎟⎜⎜ 6
5
.43
50
.
0
0
.82
2
9.2
+
⎜
⎟
⎜
⎟
⎢
L ⎠⎜⎝
meq ⎟⎠ ⎝
L ⎠⎜⎝
meq ⎟⎠⎦⎜⎝
d
⎠⎝ gal ⎠⎝ 10 mg ⎠
⎣⎝
= 41,400 kg/d
6.24
Qf = 5 ×106 L/d
Cf = 1,500 mg/L
Qp = 3 ×106 L/d
Cp = 75 mg/L
Qc, Cc
QfCf = QcCc + QpCp
So,
Cc =
(Q C
(Q
f
f
f
- Qp C p )
- Qp )
=
[(5 × 10
6
)
and Qf = Qc + Qp
(
)
]
L/d (1,500 mg/L) - 3 × 10 6 L/d (75 mg/L)
= 3,640 mg/L
5 × 10 6 - 3 × 10 6 L/d
(
)
6.25
Qf = 5 ×106 L/d
Cf = 1,500 mg/L
Qp = 3 ×106 L/d
Cp = 75 mg/L
Qc = 2 ×106 L/d
Cc = 3,640 mg/L
a. Water recovery:
6 L
QP 3 × 10 d
r=
=
= 0.60 or 60%
Q F 5 × 10 6 L
d
b. Salt rejection:
75 mg
CP
L = 0.95 or 95%
=1R =1mg
CF
1,500
L
c. Log salt rejection:
R log = - log(1 - R ) = - log(1 - 0.95) = 1.30
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).
6.28 – 6.29 The solutions for these problems are the solutions for problems 6.9 - 6.10 in
the 2nd edition Solutions Manual.
6.30
Q, S0, X0
well-mixed
aeration pond
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:
S=
K s (1 + θk d )
(θμ m - 1 - θk d )
and solving for the hydraulic detention time, θ,
θ=
(100 + 20 mg/L BOD 5 )
Ks + S
=
= 6.0 d
[S (μ m - k d ) - K s k d ] 20 mg/L BOD 5 (1.6 - 0.1 d -1 ) - 100 mg/L BOD 5 (0.1 d -1 )
[
]
b. From equation 6.60 and 6.51,
⎛
mg BOD 5 ⎞⎛
mg BOD 5 ⎞
mg VSS ⎞⎛
⎜⎜ 0.60
⎟⎟⎜100 + 20
⎟⎜ 350 - 20
⎟
mg BOD 5 ⎠⎝
L
L
Y (K s + S )(S 0 - S ) ⎝
mg VSS
⎠⎝
⎠
X=
=
= 7.07
θKS
L
(6.0 d ) 1.6 d -1 ⎛⎜ 350 mg BOD 5 ⎞⎟
L
⎝
⎠
(
)
and at steady state with X0 = 0, the effluent flux of microbial mass must equal the
rate of microbial mass production in the pond:
QX = Vrg'
⎛ m 3 ⎞⎛
L ⎞⎛
mg VSS ⎞⎛ kg ⎞
kg VSS
⎟⎟⎜1,000 3 ⎟⎜ 7.07
⎟⎟ = 0.212
and, QX = ⎜⎜ 30
⎟⎜⎜ 6
d ⎠⎝
L ⎠⎝ 10 mg ⎠
d
m ⎠⎝
⎝
6.31
Q, S0, X0
well-mixed
aeration lagoon
l = 60 m
w=5m
h=2m
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
Q, S, X
V, S, X, rg’, rsu
a. The lagoon’s hydraulic detention time is:
θ=
V (60 m )(5 m )(2 m )
=
= 1.50 d
3
Q
400 m
d
and the steady state substrate concentration in the lagoon is:
(
(
))
)]
mg BOD5 ⎞
⎛
-1
⎟ 1 + (1.5 d ) 0.08 d
⎜ 76.0
mg BOD 5
K s (1 + θk d ) ⎝
L
⎠
= 161
=
S=
-1
-1
(θμ m - 1 - θk d ) (1.5 d ) 1.10 d - 1 - (1.5 d ) 0.08 d
L
[
(
)
(
Therefore, the lagoon’s BOD5 removal efficiency is:
mg BOD 5 ⎞
⎛
⎟
⎜ 336 - 161
S0 - S
L
⎠ (100) = 52.2%
⎝
(100) =
Eff =
mg BOD 5 ⎞
S0
⎛
⎟
⎜ 336
L
⎠
⎝
c. The lagoon’s daily oxygen demand is:
⎛
mg BOD 5 ⎞⎛ 1,000 L ⎞⎛ kg ⎞
kg O 2
m 3 ⎞⎛
⎟⎟⎜ 336 - 161
⎟⎟ = 70.2
Q(S 0 - S ) = ⎜⎜ 400
⎟⎜
⎟⎜⎜ 6
3
d ⎠⎝
L
d
⎠⎝ m ⎠⎝ 10 mg ⎠
⎝
6.32
P.C.
Q, S0
V, X, S
Qs
S.C.
Qe, Xe, S0
Act. Sludge Basin
RAS
Qr, Sr, Xr
Q = 0.300 m3/s
X = 2,100 mg VSS/L
θc = 9.0 d
S0 = 220 mg BOD5/L
F = 0.52 mg BOD 5
M
mg VSS ⋅ d
WAS
Qw, Sw, Xw
Xr = 10,000 mg VSS/L
a. Based on the definition of the food to microbe ratio,
⎛
mg BOD 5 ⎞⎛ 24 ⋅ 60 ⋅ 60 s ⎞
m 3 ⎞⎛
⎜⎜ 0.300
⎟⎟⎜ 220
⎟⎜
⎟
s ⎠⎝
L
d
QS 0 ⎝
⎠
⎠⎝
V=
= 5,220 m 3
F
⎛
mg BOD5 ⎞⎛
mg VSS ⎞
X
M
⎜⎜ 0.52
⎟⎟⎜ 2,100
⎟
mg VSS ⋅ d ⎠⎝
L ⎠
⎝
( )
b. Noting that Xr = Xw, the cell retention time can be used to calculate the WAS
flow rate.
(5220 m )⎛⎜ 2,100 mg LVSS ⎞⎟
3
VX
Qw =
=
θc X w
3
⎠ = 122 m
d
(9.0 d )⎛⎜10,000 mg VSS ⎞⎟
L ⎠
⎝
⎝
c. From a hydraulic flow balance around the activated sludge basin, recycle line
and secondary clarifier,
⎛
m3 ⎞ ⎛
m 3 ⎞⎛
d
m3
⎞
⎟ - ⎜122
⎟⎜
Qe = Q - Qw = ⎜⎜ 0.300
⎟ = 0.299
s ⎟⎠ ⎜⎝
d ⎟⎠⎝ 24 ⋅ 60 ⋅ 60 s ⎠
s
⎝
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.
⎛
m 3 ⎞⎛
mg VSS ⎞
⎜⎜ 0.299
⎟⎟⎜10,000
⎟
s ⎠⎝
L
Qe X w
m3
⎠
⎝
= 0.378
Qs =
=
( X w - X ) ⎛10,000 - 2,100 mg VSS ⎞
s
⎜
⎟
L
⎝
⎠
and the hydraulic detention time is then:
(
)
V
5220 m 3
=
θ=
= 0.160 d
Qs ⎛
m 3 ⎞⎛ 24 ⋅ 60 ⋅ 60 s ⎞
⎜⎜ 0.378
⎟⎜
⎟
s ⎟⎠⎝
d
⎠
⎝
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:
Cs =
Q
(C0 - Ce ) + Ce = (20 ΜGD ) ⎛⎜ 800 - (0.82)800 mg TS ⎞⎟ + (0.82)⎛⎜ 800 mg TS ⎞⎟
(0.070 MGD ) ⎝
L ⎠
L ⎠
Qs
⎝
Cs = 41,800
mg TS
L
b. The mass of solids removed annually is:
⎛ 0.04381 m 3
⎜
s
m = Qs Cs t = (0.070 MGD)⎜
MGD
⎜
⎝
m = 4.04 × 10 6
⎞
mg ⎞⎛ kg ⎞⎛ 24 ⋅ 60 ⋅ 60 ⋅ 365 s ⎞
⎟⎛ 1,000 L ⎞⎛
⎟⎟
⎟⎜ m 3 ⎟⎜ 41,800 L ⎟⎜⎜ 10 6 mg ⎟⎟⎜⎜
yr
⎠⎝
⎠⎝
⎟⎝
⎠
⎠⎝
⎠
kg TS
= 4,040 tonne per year
yr
SOLUTIONS FOR CHAPTER 7
7.1
From (1.9), mg / m 3 =
a. CO2 mg / m 3 =
ppm x mol wt
(at 1 atm and 25 o C)
24.465
5000ppm x (12 + 2x16)
= 8992mg/m 3 ≈ 9000mg / m 3
24.465
24.465 x 3.6 mg/m 3
b. HCHO ppm =
= 2.94ppm
(2x1 + 12 + 16)
c. NO mg / m 3 =
25ppm x (14 +16 )
= 30.7mg/m 3
24.465
7.2 70% efficient scrubber, find S emission rate:
600/0.38=1579 MWt
η = 0.38
600 MWe
9000 Btu/lb coal 1% S
Input =
600, 000 kWe 3412 Btu lb coal 0.01 lb S
x
x
x
= 5986 lb S/hr
0.38
kWhr
9000Btu lb coal
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 (32 + 2x16) lb SO2
x
= 1197 lb SO2 / hr ≈ 1200 lb SO2 / hr
hr
32 lb S
7.4 70% scrubber, 0.6 lb SO2/106 Btu in, find % S allowable:
a.
X lbs S 0.3 lbs S out 2 lbs SO2
lb coal
0.6 lb SO2
x
x
x
=
lbs coal
1 lb S in
lb S
15, 000Btu
106 Btu
X=
b.
15,000x0.6
= 0.015 = 1.5% S fuel
0.3x2x106
X lbs S 0.3 lbs S out 2 lbs SO2
lb coal
0.6 lb SO2
x
x
x
=
lbs coal
1 lb S in
lb S
9, 000Btu
10 6 Btu
X=
9, 000x0.6
= 0.009 = 0.9% S fuel
0.3x2x106
Pg. 7.1
7.5 Compliance coal:
1.2 lbs SO2
lb coal
X lb S 2 lb SO2
=
x
x
6
10 Btu
12, 000 Btu lb coal
lb S
X=
1.2x12, 000
= 0.0072 = 0.7%S
2x10 6
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
3
PM2.5, 24-hr (µg/m )
130
150
10
PM1 0, 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, PM1 0 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
55.4
1
e −0.4 0 2t =
= 0.871
so t = ln( 0.871) = 0.41 hr
65.4
0.402
7.9 RH to produce HCHO:
RO • +O 2 → 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. 7.2
7.10
RH = propene = CH2=CH-CH3 = C3H6
so, R = C3H5
so the sequence of reactions (7.16) to (7.19) is:
C3 H6 + OH• → C3 H5 • +H 2 O
C3 H5 • +O 2 → C3 H5 O2 •
C3 H5 O2 • +NO → C 3 H5O • +NO2
C3 H5 O • +O 2 → HO2 • +C2 H 3CHO
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 (20x10−6 m)2 x 1.5x10 6 g / m3 x 9.80m/s 2
=
= 0.019m / s
18η
18 x 0.0172g/m - s
Time to reach the ground =
8000m
= 4.87days
0.019m / s x 3600s/hr x 24hr/d
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 =
Residence time = τ =
7.13
d2 ρg (10x10−6 m)2 x 10 6 g / m 3 x 9.80m/s2
=
= 0.00317m / s
18η
18 x 0.0172g/m - s
h
1000m
=
= 87.6hrs
v 0.00317m / s x 3600s/hr
Settling velocity and Reynolds numbers:
a. 1 µm:
v=
d2 ρg (1x10−6 m)2 x 10 6 g / m 3 x 9.80m/s2
=
= 3.2x10 −5 m / s
18η
18 x 0.0172g/m - s
Re =
b. 10 µm: v =
ρair dv 1.29x103 g/ m 3 x 1x10-6 m x 3.17x10-5 m/s
=
= 2.4x10 −6
η
0.0172 m/s
d2 ρg (10x10−6 m)2 x 10 6 g / m 3 x 9.80m/s2
=
= 3.2x10 −3 m / s
18η
18 x 0.0172g/m - s
ρair dv 1.29x103 g/ m 3 x 10x10-6 m x 3.17x10-3 m/s
=
= 2.4x10 −3
η
0.0172 m/s
2
−6
2
6
d ρg (20x10 m) x 10 g / m3 x 9.80m/s 2
c. 20 µm: v =
=
= 0.0127m / s
18η
18 x 0.0172g/m - s
Re =
Pg. 7.3
Re =
ρair dv 1.29x103 g/ m 3 x 20x10-6 m x 0.0127m/s
=
= 0.02
η
0.0172 m/s
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, CH3 CH2 OH
Oxygen
16
=
= 0.347 = 34.7%
Ethanol 2x12 + 6x1+ 1x16
Xg oxygenate x %O in oxygenate
Yg fuel
Ethanol
X 2%
2%
= =
=
= 0.058 = 5.8% by weight
Fuel blend Y %O 34.7%
0.02 =
€
b.€Methyl tertiary butyl ether (MTBE), CH3 OC(CH3 )3
Oxygen
16
=
= 0.182 = 18.2%
€
MTBE 5x12 + 12x1+ 1x16
MTBE 2%
2%
=
=
= 0.11 = 11%
Fuel
%O 18.2%
€
c. Ethyl tertiary butyl ether (ETBE), CH3 CH2 OC(CH3 )3
Oxygen
16
=
= 0.157 = 15.7%
€
ETBE 6x12 + 14 x1+ 1x16
ETBE 2%
2%
=
=
= 0.127 = 12.7%
Fuel %O 15.7%
d.€Tertiary amyl methyl ether (TAME), CH3 CH2 C(CH3 )2 OCH3
€
Oxygen
16
=
= 0.157 = 15.7%
TAME 6x12 + 14 x1+ 1x16
TAME 2%
2%
=
=
= 0.127 = 12.7%
Fuel
%O 15.7%
€
7.15 Ethanol fraction CH3 CH2 OH (sg = 0.791) in gasoline (sg = 0.739) to give 2% O2 :
€
Oxygen
16
=
= 0.347 = 34.7%
Ethanol 2x12 + 6x1+ 1x16
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
0.2745
=
0.791X + 0.739Y 0.791+ 0.739Y / X
€
€
Pg. 7.4
€
Y 0.2745 − 0.02x0.791
=
= 17.50
X
0.02x0.739
€
ethanol
X(mL eth)
1
1
=
=
=
= 5.4% by volume
fuel blend X(mL eth) + Y(mL gas) 1+ Y / X 1+ 17.50
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
1 mile
= 29.4 mpg
 0.5 mile

0.5 mile
0.15 gal gas 
+
x


18 mile/gal gas  12 mile/gal alcohol 1 gal alcohol 
1 mile
b. CAFE mpg =
= 36.1 mpg
 0.5 mile

0.5 mile
0.15 gal gas 
+
x


 22 mile/gal gas 15 mile/gal alcohol 1 gal alcohol 
1 mile
c. CAFE mpg =
= 44.1 mpg
 0.5 mile

0.5 mile
0.15 gal gas 
+
x


27 mile/gal gas  18 mile/gal alcohol 1 gal alcohol 
a. CAFE 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 ft 3 ngas 0.823 equiv. gal gasoline
0.15 gal gasoline
x
x
= 0.004938 gal gasoline/mile
3
25 mile
100 ft ngas
1 equiv. gal gasoline
1
CAFE mpg =
= 202.5 miles/gallon ≈ 203 mpg
0.004938 gal/mile
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. 7.5
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
x 25 mpg = 24.14 mpg
115,400
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
x 25 mpg = 17.7 mpg
115,400
Range = 15 gal x 17.7 mi/gal = 266 miles
c. E85/gasoline = 0.5 x 81,630 + 0.5 x 115,400 = 98,515 Btu/gal
98,515
x 25 mpg = 21.3 mpg
115,400
Range = 15 gal x 21.3 mi/gal = 320 miles
€
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
(20 mi/d x 5 d/wk)
Gasoline =
x $3.50/gal = $7.78/wk
45 mi/gal
$7.78 + $4.00
PHEV cost/mile =
= $0.0393 = 3.93¢/mile
(50x5 + 25x2) mi
$3.50/gallon
€
Original 50 mpg HEV cost/mile =
= 7¢/mile
50 mi/gal
€
b. On an annual basis:
PHEV annual cost = $0.0393/mi x 52 wk/yr x 300 mi/wk = $612/yr
€
$3.50/gallon
HEV annual cost =
x52wk/yr x 300mi/wk = $1092 / yr
50 mi/gal
Annual savings = $1092 - $612 = $480/yr
€
c. At $3000 for batteries:
Simple payback =
Extra 1st cost
$3000
=
= 6.25 yr
Annual savings $480/yr
Batteries would need to last 6.25 yr x 300 mi/wk x 52 wk/yr = 97,500 miles
€
Pg. 7.6
7.22 From Figure 7.28 the well-to-wheels CO2 /mile are:
a. 25 mpg car:
11.2 kg CO 2 /gal x 1000 g/kg
= 448 gCO 2 /mile
25 mi/gal
b. 50 mpg car:
11.2 kg CO 2 /gal x 1000 g/kg
= 224 gCO 2 /mile
50 mi/gal
€
11.2 kg CO 2 /gal x 1000 g/kg
= 248 gCO 2 /mile
45 mi/gal
electricity : 0.25 kWh/mi x 640 gCO 2 /mi = 160 gCO 2 /mile
c. PHEV: gasoline :
€
€
d.€ EV:
Half gas, half electricity : 0.5 x 248 + 0.5 x 160 = 204 gCO 2 /mile
electricity : 0.25 kWh/mi x 640 gCO 2 /mi = 160 gCO 2 /mile
€
14.4 gC/MJ (12 + 2x16)gCO 2
MJ CH 4
e. FCV:
x
x
= 240 gCO 2 /mile
0.36 mi/MJ
12gC
0.61 MJ H 2
€
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 44 gCO 2
1 MJ in
3.6 MJ out
x
x
x
x 0.96 grid = 304 gCO 2 /kWh
1 MJ CH 4
12gC
0.6 MJ out
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 :
11.2 kg CO 2 /gal x 1000 g/kg
= 248 gCO 2 /mile
45 mi/gal
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 gCO 2 /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
= 10.7 m2 = 116 ft 2
5.5 h/d x 1 kW/m2 x 0.17 x 0.75
€
Pg. 7.7
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. 7.8
7.27
Power plants emitting 0.39 x 101 2 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
efficiency =
lb
Btu
x 10, 000
= 1.37x101 6 Btu
ton
lb
output 1400x10 9 kWh x 3412Btu/kWh
=
= 0.349 ≈ 35%
input
1.37x101 6 Btu
At NSPS of 0.03 lb particulates per 106 Btu input, emissions would have been:
emissions =
0.03 lb
1000g
x 1.37x101 6 Btu in x
= 1.87x101 1g
10 Btu heat input
2.2 lb
6
emissions at NSPS 1.87x101 1g
For comparison,
=
= 0.48 = 48%
actual emissions
0.39x101 2
7.28
Derivation for the dry adiabatic lapse rate:
dQ = dU + dW
where dU = Cv dt and dW = PdV
dQ = Cv dt + PdV
(1)
ideal gas law says PV = nRT
so,
or,
d(PV) = PdV + VdP = nRT
PdV = nRT - VdP
plugged into (1) gives:
dQ = Cv dT + nRdT − VdP
dQ
dP
= Cv + nR − V
(2)
dT
dT
at constant pressure :
dQ
= Cv + nR = C p
dT
putting that into (2) gives,
dQ
dP
= Cp − V
dT
dT
or,
dQ = C p dT − VdP
which is (7.37)
€
Pg. 7.9
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:
Altitude
(m)
800
600
mixing depth
plume rise
400
200
0
14
15
16
17
18
19
Temperature (C)
20
21
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  H 
= 
u a  za 
p
so,
u2 5 0  250m  0.20
 = 1.90
=
4m / s  10m 
u 2 5 0 = 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. 7.10
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 ) =
 H2 
Q
exp  −
2
π uσ yσ z
 2σ z 
(7.49)
a. From Table 7.9 at 400m, σy = 15m, σz = 7m
C=
 62 
20x106 µg / s
3
3
3
exp  −
 = 21x10 µg / m = 21mg / m
π 2m/s x15m x 7m
 2x72 
b. At the maximum point, Fig. 7.52 we can get a rough estimate of the key parameter
 CuH 

 ≈ 3.8x10−3 / m 2
 Q  max
C max =
7.34
Q  CuH 
20x10 3 mg / s 3.8x10 −3


=
x
= 38 ≈ 40mg / m3
uH  Q  max
2m / s
m2
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.
 Cu 
From Fig. 7.52, Xmax = 0.7 km;  H  ≈ 1.5x10 −5 / m 2
 Q  max
C max =
Q  CuH 
Q
1.5x10 −5
= 365x10−6 g / s =
x


uH  Q  max
4m / s
m2
4x365x10 −6
Q≈
= 97g / s
1.5x10 −5
Maximum power plant size = 97 g/s x
MW
= 80MW
1.2 g/s
7.35 Atmospheric conditions, stack height, and groundlevel concentration restrictions same as
Prob. 7.34 so that:
Emissions Q ≈ 97 g/s
97g / s =
Pk W
0.6 lb SO2
1 Btu in
3412 Btu out 1hr
10 3 g
x
x
x
x
xP
106 Btu in 0.35 Btu out
kWh
3600s 2.2 lb k W
97x106 x0.35x3600x2.2
=
= 131,000KW = 130MW
0.6x3412x1000
Pg. 7.11
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:
p
0.15
 H
 100 
u H = u a   = 4m / s
= 5.65m / s
 10 
 za 
a. At 2 km, Table 7.9: σy = 290 m, σz = 234 m
C(x, 0 ) =
C=
 H2 
Q
exp  −

π uσ yσ z
 2σ z 2 
 1002 
80x10 6 µg/ s
3
exp  −
 = 61µg / m
π 5.65m/s x290m x 234m
 2x2342 
b. At the maximum point, 0.7 km (Fig. 7.52),
 CuH 

 ≈ 1.5x10 −5 / m 2
 Q  max
C max =
Q  CuH 
80x106 µg/ s  1.5x10 −5 
3


=

 = 212 µg/ m
uH  Q  max
5.65m / s  m 2 
c. At x = 2km, y = 0.1 km:
C(x, y ) =
 −y 2 
 H2 
Q
exp  −
exp


2
2
π uσ yσ z
 2σ z 
 2σ y 
 -1002 
3
= 61µg / m 3 x exp
 = 57µg / m
 2x2902 
7.37
For class C, notice from (7.47) and (7.48) with Table 7.9 and f =0
σy
ax 0.894 104x0.894
−0.017
= d
=
≈ fairly constant ≈ k (about 1.7)
0.911 = 1.7x
σ z cx + f
61x
So, assume σy = k σz , then from (7.49)
a. C(x, 0 ) =
 H2 
 H2 
Q
Q
exp  −
2 =
2 exp −
2
π uσ yσ z
 2σ z  π ukσ z
 2σ z 
To find the maximum concentration, differentiate and set equal to zero:
Pg. 7.12
H2
H2
−
 −2  
dC
Q  1  −H 2   −2  − 2σ z2
2σz 2
=

+e


e
 3  = 0
dσ z π uk σ z 2  2   σ z 3 
 σ z  
Multiply through by σz5 and cancel lots of terms to get,
 2H2 
2

 − 2σ z = 0
 2 
or σ z =
H
= 0.707H
2
b. Substituting the newly found value for σz,
C max

Q
H2

=
exp −
π uσ y σ z
H2
2
2

( )

Q
0.117Q
=
−1
 π uσ yσ z e = uσ y σ z

c. Using σy = k σz
C max =
7.38
0.117Q
0.117Q
f(Q, u)
(varies as inverse of H2)
2 =
2 =
ukσ z
uk( 0.707H)
H2
Find the effective stack height of the Sudbury stack:
130 o C
20 m/s
15.2m
10 oC
8 m/s
380m
Using (7.52) for the buoyancy flux parameter

T
m  15.2m  2
m
10 + 273 
 x20 x 1−
 = 3370m 4 / s3
F = gr 2 vs  1− a  = 9.8 2 x




T
s
2
s
130
+
273

s
The distance downwind to final plume rise xf is given on page 464 (with F>55),
x f = 120F0.4 = 120x(3370 )
0. 4
= 3092 m
For stability classification C, use (7.54) for plume rise,
Δh =
1.6F1 3 xf 2 3 1.6x(3370)1 3 (3092 )2 3
=
= 635m plume rise
uh
8
H = effective stack height = h + Δh = 380 + 635 = 1015 m
Pg. 7.13
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  ΔTa
9.8m / s2


(0 + 0.01K/ m) = 3.46x10 −4 / s2
+ 0.01K / m  =
 (10 + 273)K
Ta  Δz
Putting that into (7.51) for plume rise under these conditions gives
 F 
Δh = 2.6

 uh S
13


3370m 4 / s3
= 2.6
-4
2
 8m / s x 3.46x10 / s 
13
= 278m
H = effective stack height = h + Δh = 380 + 278 = 657 m
(Notice the atmospheric stability lowered effective stack height vs Prob. 7.38)
7.40
Cloudy summer day, stability classification C (Table 7.7),
120 o C
6.0o C
5 m/s
10 m/s
2m
100m
Using (7.52) for bouyancy flux parameter

T
m  2m  2
m
6 + 273 
 x10 x 1−
 = 28.4m 4 / s3
F = gr 2 vs  1− a  = 9.8 2 x



Ts 
s
2
s
120 + 273 

and distance downwind to final plume rise xf given on page 464 (with F<55),
x f = 50F5 8 = 50x(28.4 )
58
= 406
For stability classification C, use (7.54) for plume rise,
1.6F1 3 xf 2 3 1.6x(28.4)1 3 (406)2 3
Δh =
=
= 54m plume rise
uh
5
H = effective stack height = h + Δh = 100 + 54 = 154 m
Pg. 7.14
7.41
Power plant, find groundlevel pollution 16 km away. Need first find H.
Q=300g/s SO2
13.5 m/s
145o C
Class E
5m/s
o
lapse rate=5 C/km
15 oC
100m
r=2.5m
200 MWe
16km
First, find bouyancy flux parameter (7.52),

T
m
m 
15 + 273 
2
 = 257m 4 / s3
F = gr 2 vs  1− a  = 9.8 2 x(2.5m ) x13.5 x 1 −

Ts 
s
s
145 + 273 

plume rise for stable (Class E) atmosphere needs S from (7.53),
S=
g
Ta

9.8m / s2  5o
 ΔTa

+ 0.01K / m  =
+ 0.01K/ m = 5.1x10 −4 / s2

 Δz
 (15 + 273)K  1000m

plume rise is given by (7.51),
 F 
Δh = 2.6

 uh S
13


257m 4 / s3
= 2.6
-4
2
 5m / s x 5.1x10 / s 
13
= 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 ) =
C=
 H2 
Q
exp  −
2
π uσ yσ z
 2σ z 
(7.49)
 2212 
300x10 6 µg / s
3
exp  −
2  = 22µg / m
π 5 m/s x602m x 95m
 2x95 
7.42 A 20 g/s source, 5 m/s, H = 50 m want peak concentrations Class A, C, F..
20 g/s
5 m/s
A,C,F
C max
 Cu 
Q  CuH  20x10 6 µg / s  Cu H 
=

=

 = 4x10 6  H  µg / s
uH  Q 
5m / s
 Q 
 Q 
Using Fig. 7.52 with 50 m and varying stability classifications gives:
Pg. 7.15
50m
 Cu 
"A"  H  ≈ 6x10 −5 at 0.25km, C max = 4x106 x 6x10 -5 = 240 µg / m 3
 Q 
 Cu 
"C"  H  ≈ 5.8x10−5 at 0.55km, Cmax = 4x10 6 x 5.8x10-5 = 230µg / m 3
 Q 
 Cu 
"F"  H  ≈ 2.4x10 −5at 3.7km, C max = 4x106 x 2.4x10 -5 = 96 µg/ m 3
 Q 
240
230
C ( µ g/m3)
96
0.25 0.55
7.43
3.7
x (km)
H = 50m, 100m, 200m; Class C, 20 g/s, 5 m/s wind:
C max =
 Cu 
Q  CuH  20x10 6 µg / s  Cu H 

=

 = 4x10 6  H  µg / s
uH  Q 
5m / s
 Q 
 Q 
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
Test them:
€
Expect
C(2H) 1
=
C(H) 4
and
C(100m)
6
=
= 0.26
C(50m) 22.8
C(200m) 1
=
= 0.0625
C(50m) 16
C(4H) 1
=
C(H) 16
C(200m) 1.4
=
= 0.23 not bad!
C(100m)
6
C(200m) 1.4
=
= 0.061 again, not bad.
C(50m) 22.8
Pg. 7.16
7.44
Paper mill emitting H2S, 1km away want 0.1 x odor threshold:
40 g/s
4-10m/s
Class B
0.01 mg/m3
1 km
Class B, at 1 km, (Table 7.9) σy = 156m, σz = 110m
C(x, 0 ) =
 H2 
Q
exp  −
2
π uσ yσ z
 2σ z 

40g / s
H2 
0.01x10 g / m =
exp −

π u m/s x156m x 110m
 2x1102 
3
−3
Rearranging: e
H2
24,200
=
40
74.2
-3 =
π u 156 x 110 x 0.01x10
u

 74.2  
or, H =  24, 200 ln 
 u  

0. 5
so, at each end of the wind speed range we can find the height needed:

74.2  
H u= 4 =  24, 200 ln 
 4  

0 .5

 74.2  
H u=1 0 =  24, 200 ln 
 10  

= 265m
says to be conservative use H=265m
0. 5
= 220m
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:
150 g/s
45m
L=100m
5 m/s
Class C
XL
At x = XL σz = 0.47 (L-H) = 0.47 (100 - 45) = 26 m
Pg. 7.17
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(X L ,0 ) =
 H2 
Q
exp −
2
π uσ yσ z
 2σ z 
(7.49)
 452 
150x103 mg / s
3
=
exp −
2  = 1.8mg / m
π 5 m/s x 46m x 26m
 2x26 
At x = 2XL : σy = 85m (Table 7.9), so using (7.55) gives
Q
150x10 3 mg / s
=
= 1.4mg / m3
2π uσ y L
2π x 5m/s x 85m x 100m
C(2XL , 0) =
7.46
Stack under an inversion layer:
80 g/s
5 m/s
L=250m
50m
C=?
4 m/s
XL
x=4km
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
80x10 6 µg / s
=
= 47µg / m 3
2π uσ y L
2π x 5m/s x 539m x 250m
b. Without the inversion layer, at 4km σz = 498m, σy = 539m so,
C(4km, 0 ) =
 −H 2 
 −502 
Q
80x10 6 µg / s
3
exp 
=
exp


2
2  = 19µg / m
π uσ yσ z
2
σ
π
x
5m/s
x
539m
x
498m

2x498

 z 
Pg. 7.18
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
2x300mg / m − s
=
= 3.0mg / m 3
2π uσ z
2π x 3m/s x 26m
0.3g/s-m
u=3m/s
400m
7.48
A freeway modelled as a line source:
10,000
vehicles/hr
u=2m/s
1.5 g/mi
200m
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:
vehicles 1 hr
1.5g
mi
ft
x
x
x
x
hr
3600s mi − vehicle 5280ft 0.3048m
−3
= 2.58x10 g / m − s = 2.58mg / m − s
q = 10, 000
Then, using (7.57),
C(0.2km ) =
7.49
2q
2x2.58mg / m − s
=
= 0.04mg / m 3
2π uσ z
2π x 2m/s x 26m
Box model, 250,000 vehicles between 4 and 6pm, driving 40km ea, emitting 4g/km
CO.
a. q s = 250, 000veh. x
40km 4gCO
1
hr
1
−6
2
x
x
x
x
6
2 = 4.6x10 gCO/ m s
veh
km 2hrs 3600s 15x80x10 m
b. Using (7.61) with t= 2hrs x 3600s/hr =7200s,
C(t ) =
qs L
1 − e − ut / L )
(
uH
Pg. 7.19
=
4.6x10−6 g / m 2 s x 15, 000m
1− e −0.5m / s⋅7200s / 15000m ) = 0.002g / m 3 = 2mg / m 3
(
0.5m / s x 15m
c. With no wind, go back to (7.58) and solve the differential equation:
dC
= qs LW
dt
q LW
dC = s
dt
LWH
LWH
C=
7.50
so, C =
qs
t
H
qs
4.6x10−6 gCO/ m 2 ⋅ s
3600s
t=
x 2hrs x
= 0.0022gCO / m 3 = 2.2mg / m 3
H
15m
hr
Box model, 105 m on a side, H=1200m, u=4m/s, SO2=20kg/s, steady state:
4m/s
Cin=0
1200m
10 5 m
105m
20 k g/s
input rate = output rate
20
kg 109 µg
m
x
=4
x 105 m x 1200m x C µg m 3
s
kg
s
C=
7.51
(
)
20x10 9
= 41.7µg / m3
4x105 x1200
Assume steady-state conditions were achieved by 5pm Friday so that from Problem
7.50, C(0) = 41.7 µg/m3.
− ut / L
With qs = 0, and Cin = 0, (7.60) gives us C(t) = C(0)e
.
a. At midnight, t=7hrs x 3600s/hr = 2.52x104 s
4
5
C(t) = C(0)e − ut / L = 41.7µg/ m 3 ⋅ e -4m/s ⋅ 2.52x10s 1 0 m = 15.2 µg / 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):
5
C(t) = C(0)e − ut / L = 41.7µg/ m 3 ⋅ e -4m/s ⋅ 63hrx3600s/ hr 1 0 m = .005µg / m 3 ≈ 0
Pg. 7.20
so we can ignore that and let C(0) = 0 at 8am Monday. First find the emissions per
unit area,
qs =
emission rate 20kg / s x 10 9 µg / kg
=
= 2.0 µg/ m 2 ⋅ s
5
5
area
10 m x 10 m
Then use (7.57) with Cin = 0:
qs L
1 − e − ut / L )
(
uH
5
2.0 µg/ m 2 ⋅ s x 10 5 m
=
1 − e −( 4m / s x 9hr x 3600s/hr/10m) = 30.2µg / m3
4m / s x 1200m
C(t ) =
(
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 ) =
C(2hr) =
qs L
1 − e − ut / L ) + C(0)e −ut / L
(
uH
5
5
2.0µg / m 2 ⋅s x 105 m
−( 2m / s x 2hr x3600s/hr /10
m)
1− e −( 2m / s x 2hr x 3600s/hr/10m) + 41.7e
2m / s x 1200m
(
)
= 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 µ g/m3
1200m
10 5 m
105m
2 µ g/m2-s
q L

C(t ) =  s + C in (1 − e − ut / L ) + C(0)e −ut / L
 uH

(7.60)
 2.0 µg / m 2 ⋅s x 105 m

5
C(4hr) = 
+ 5µg / m 3  1 − e −( 4 m/ s x 4hr x 3600s/hr/10m)
 4m / s x 1200m

5
−( 4m / s x 4hr x3600s/hr /10
m)
+ 10e
(
C(4hr = noon) = 26.1 µg/m3.
Pg. 7.21
)
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 = uWHC + KCV
kg 109 µg
m
x
=4
x 105 m x 1200m x C µg m 3
s
kg
s
µg 1hr
 0.23

+
xC 3x
x10 5 m x 105 m x 1200m 
 hr

m 3600s
9
8
8
20x10 = 4.8x10 C + 7.6x10 C
(
20
)
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 ) = C 0e−nt and then taking the log: ln[C(t )] = ln(C 0 ) − nt
which is of the form y = mx + b, where y = lnC, m = n(ach), and b = lnCo
time (hr)
0
0.5
1.0
1.5
2.0
C (ppm)
10.0
8.0
6.0
5.0
3.3
ln C
€
2.303
2.079
1.792
1.609
1.194
2.4
2.2
2.0
ln C
€
1.8
1.6
1.4
1.2
1.0
0
1
2
time
3
(hr)
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. 7.22
7.56
Infiltration 0.5ach, 500m3 volume, 200 m2 floor space, radon 0.6pCi/m2s:
0.5 ach
V=500m3
0.6pCi/m 2 s
K=7.6x10 - 3/hr
Using (7.63) with K = 7.6x10-3/hr (Table 7.14),
 0.6pCi / m 2 s x 200m 2

3

500m

S= V =
= 1700pCi / m 3 = 1.7pCi / L
1hr
−3
I + K (0.5 / hr + 7.6x10 / hr )x
3600s
( )
S
7.57
Same as Problem 7.56 but have half as much ground-floor area to let radon in, so:
 0.6pCi / m 2 s x 100m 2

3

500m 
S= V =
= 850pCi / m 3 = 0.85pCi / L
1hr
I + K (0.5 / hr + 7.6x10 −3 / hr )x
3600s
( )
S
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
day
x30yr
yr
= 0.70 pCi/L average over 70 yrs
365day/yr x 70yr
1.7pCi/L x 350
Exposure =
€
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 cancer death
rem
x
x 3
x70yr = 0.0016 ≈ 0.16%
1.5pCi/L
8000 rem
10 mrem
€
Pg. 7.23
A 300m3 house, 0.2ach, oven+2burners 6pm to 7pm, find CO at 7pm and and 10pm.
For these circumstances, (7.65) is appropriate:
7.59
C(t ) =
S
1− e−n t )
(
IV
(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 / hr x 1hr
= 16.8mg / m 3
)
3 (1− e
airchange
m
0.2
x300
hr
ac
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
n = 0.39 ach, V = 27m3, after 1-hr NO = 4.7ppm. Find source strength, S:
7.60
€
First convert NO in ppm to mg/m3 using (1.9) and assuming T=25oC,
mg / m 3 =
ppm x mol wt 4.7 x (14 +16 )
=
= 5.76mg / m 3
24.465
24.465
a. To find the NO source strength, rearrange (7.65)
S=
nVC
=
(1− e−n t )
0.39
ac
m3
mg
x27
x5.76 3
hr
ac
m = 188 mgNO/hr
−0.39/hr x 1hr
(1− e
)
b. 1-hr after turning off the heater,
€
C = C 0 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
=
nV
188 mg/hr
3
3 = 3.1 mg/m NO
ac
m
0.2
x 300
hr
ac
Using (1.9) again gives
€
ppm =
24.465
24.465
=
= 2.6 ppm NO
ppm x mol wt 3.1 x (14 +16)
Pg. 7.24
€
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
2
−6
6
3
2
d 2 ρg (2.5x10 m) ⋅ (1.5x10 g/m ) ⋅ (9.8m/s )
v=
=
= 2.97x10−4 m/s
18η
18x0.0172g/m⋅ s
From Example 7.4, the residence time is
€
7.62
τ=
h
2.5 m
=
= 2.3 hours
−4
v 2.97 x 10 m/s x 3600 s/hr
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 = 613x10 6 kWh/yr
b. heat input = 613x106 kWh / yr out x
€
3 kWht in
3412Btu
x
= 6.28x101 2 Btu / yr
1 kWhe out
kWh
c. Shut it down and sell the allowances,
SO2 saved by shutting down = 6.28x101 2
1883
Btu 0.6 lb SO2
ton
x
x
= 1883 tons/yr
6
yr
10 Btu
2000 lb
tons 1 allowance
$400
x
x
= $753,200/yr
yr
ton
allowance
€
Pg. 7.25
SOLUTIONS FOR CHAPTER 8
8.1 From (8.1),
(
( )
(
)
)
⎡ 18 O
⎤
16
⎢
⎥
⎡ 0.0020150 ⎤ 3
O
sample
δ18O o oo = ⎢ 18
−1⎥ x10 3 = ⎢
−1 x10 = 4.9
⎣ 0.0020052 ⎥⎦
O
⎢
16
O standard ⎥⎦
⎣
Since the sample has more 18O in it, there would be more glaciation since ice
selectively accumulates 16O, increasing the concentration of 18O left behind in
seawater.
8.2 Plotting delta D versus temperature gives
As shown, δD changes by 6.19 per mil per oC.
.
8.3 An ice core with (2H/1H) = 8.100 x 10-5.
a. Using VSMOW 0.00015575 for deuterium in (8.1) gives
δD (
o
⎡ (2 H/1 H) sample
⎤
⎛ 0.00008100 ⎞ 3
/oo )= ⎢ 2 1
−1⎥ x10 3 = ⎜
−1⎟ x10 = −479.8
⎝ 0.00015575 ⎠
⎢⎣ ( H/ H) standard ⎥⎦
b. With δD(0/00) = –435 per mil, and 6 per mil change in δD(0/00) per oC, the rise in
δD from –479.8 to – 435 (44.8 per mil) gives a temperature rise of 44.8/6 = 7.5oC.
8.4 From the equation given for the ice core,
( )
T (o C) = 1.5 δ 18O 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 δ 18O o oo means colder temperatures.
( )
8.5 Plotting the ice core data for T(oC) and δD
So:
( oo)+ 72.45
T(o C)= 0.1661 δD o
8.6 The flat earth!
R
1370W/m 2
E absorbed = E radiated
Sπ R = σ T A = σ T (2π R
2
⎛ S⎞
T =⎜ ⎟
⎝ 2σ ⎠
4
1
4
4
2
)
⎛
⎞
1370W / m 2
=⎜
−8
2 4⎟
⎝ 2x5.67x10 W / m K ⎠
8.7 The basic relationship is S =
1
4
o
= 331.5K - 273.1 = 58.4 C
k
2 . Using d and S for Earth from Table 8.2 lets us find k:
d
k = S d 2 = 1370W / m 2 x (150x10 6 km x 103 m / km ) = 3.083x10 25 W
2
a. Mercury: S =
k
3.083x10 25 W
2
=
2 = 9163W / m
2
6
3
d
(58x10 km x 10 m / km )
Pg. 8.2
b. The effective temperature (8.7) of Mercury would be:
⎡ S(1− α ) ⎤
Te = ⎢
⎣ 4σ ⎥⎦
1
4
⎡ 9163W / m 2 (1− 0.06) ⎤
=⎢
⎣ 4x5.67x10−8 W / m 2 K4 ⎥⎦
1
4
o
= 441K
(168 C)
= 256.2K
(-17o C)
c. Peak wavelength:
λmax =
8.8
2898 2898
=
= 6.6 μm
T (K) 441
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
⎡ S(1− α )⎤
Te ,max = ⎢
⎣ 4σ ⎥⎦
⎡ S(1− α )⎤
Te ,min = ⎢
⎣ 4σ ⎥⎦
1
1
4
4
⎡1415.2W / m 2 (1 − 0.31)⎤
=⎢
⎣ 4x5.67x10−8 W / m 2 K4 ⎥⎦
⎡1324.8W / m 2 (1 − 0.31)⎤
=⎢
⎣ 4x5.67x10 −8 W / m 2 K4 ⎥⎦
1
1
4
4
= 252K
o
(-21 C)
The variation from –17oC to –21oC is a difference of about 4oC, or about ± 2oC.
8.9 After a nuclear war:
2
a. Surface temperature,
σ Ts4 = 240W / m 2
Pg. 8.3
⎡
⎤
240W / m 2
Ts = ⎢
−8
2 4⎥
⎣ 5.67x10 W / m K ⎦
1
4
o
= 255K
(-18 C)
b. X, Atmosphere to space: Balance Incoming from space = Outgoing to space
X = 273 W/m2
342 = 69 + X
c. Y, Absorbed by earth: Incoming solar has to go somewhere,
Y = 16 W/m2
342 = 69 + 257 + Y
d. Z, Radiation from atmosphere to surface: Balance earth's surface radiation,
Z = 224 W/m2
Y + Z = 240 = 16 + Z
8.10 A 2-layer atmosphere:
342
40
107
W
T1
X
W
Y
24 78
168
350
390
Z
T2
Z
a. At the surface: 168 + Z = 24 + 78 + 390
Z = 324 W/m2
b. Extraterrestrial:
W = 195 W/m2
342 = 107 + 40 + W
c. Lower atmosphere:
Y + 24 + 78 + 350 + 195 = 2 x 324
d. Incoming:
342 = 107 + X + 1 + 168
Y = 1 W/m2
X = 66 W/m2
e. Temperatures T1 and T2 (assuming blackbody radiation) can be found from
σ T = W = 195
4
1
⎛
⎞
195W / m 2
T1 = ⎜
−8
2 4⎟
⎝ 5.67x10 W / m K ⎠
Pg. 8.4
14
o
= 242K (-31 C)
⎛
⎞
324W / m 2
T2 = ⎜
−8
2 4⎟
⎝ 5.67x10 W / m K ⎠
σ T2 = Z = 324
4
14
o
= 275K (2 C)
8.11 Hydrologic cycle:
1J/s
s
hr
d
x3600 x24 x365
W
hr
d
yr
14
3
= 5.1x10 m
3
3
3
2465kJ / kg x 10 kg/ m x 10 J / kJ
78W / m 2 x5.1x1014 m 2 x
evaporation =
Averaged over the globe, with area 5.1x1014 m2, annual precipitation is very close to 1 m
8.12 Greenhouse enhanced earth:
342
30
100
67
24
78
Z
X
Y
W
291K
W = 175 W/m2
a. Incoming energy: 342 = 100 + 67 + W
b. Find Z from radiation to space:
Z = 212 W/m2
342 = 100 + 30 + Z
c. To find X, need the energy radiated by a 291 K surface:
surface radiated = σ T = 5.67x10 W / m ⋅ K x (291K) = 406.6W / m
−8
4
2
4
4
2
X = 376.6 W/m2
so that, 406.6 = X + 30
d. Can find Y several ways; at the surface, or in the atmosphere,
W + Y = 406.6 + 24 + 78 = 175 + Y
or,
67 + 24 + 78 + X = Y + Z
67 + 24 + 78 + 376.6 = Y + 212
8.13
Y = 333.6 W/m2
(Y = 333.6)
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
GtC x 0.40
= 380 +108 = 488 ppm
(CO2 )= 380 + 575
2.12 GtC/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
= 11.15 GtC/yr
0.38
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 = C0e rt
1 ⎛C⎞
1 ⎛11.52 ⎞
r = ln⎜ ⎟ =
ln⎜
⎟ = −0.0042 = −0.42% / yr
t ⎝ C0 ⎠ 100 ⎝17.45 ⎠
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
ΔCO 2 =
552 GtC x 0.5
= 130 ppm CO 2
2.12 GtC/ppmCO 2
b. All the Oil: 20.0 gC/MJ x 24,600 x 1012 MJ = 492 GtC
Pg. 8.6
ΔCO 2 =
492 GtC x 0.5
= 116 ppm CO 2
2.12 GtC/ppmCO 2
c. All the Coal: 25.8 gC/MJ x 125,500/2 x 1012 MJ = 1619 GtC
ΔCO 2 =
1619 GtC x 0.5
= 382 ppm CO 2
2.12 GtC/ppmCO 2
d. All three: 130 + 116 + 382 = 628 ppm CO2. From (8.29) with ΔT2X = 2.8oC:
ΔTe =
8.17
⎡ (CO ) ⎤ 2.8
⎡628 + 380 ⎤
ΔT2X
2
ln ⎢
ln ⎢
= 3.9 oC
⎥=
⎥
⎣ 380 ⎦
ln 2
⎣ (CO 2 )0 ⎦ ln 2
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
2 x 330x1018 J MJ 33.6gC GtC
Emissions =
x 6 x
x 15
= 22.2GtC / yr
10 J
MJ
10 gC
yr
b. Growth from 6.0 GtC/yr to 22.2 GtC/yr in 100 yrs,
r=
1 ⎛ 22.2 ⎞
⎟ = 0.013 = 1.3%/ yr
ln ⎜
100 ⎝ 6.0 ⎠
c. Amount remaining with 50% airborne fraction, use (8.27):
Total emitted = C tot =
C0 r T
6.0 GtC/yr 0.01308/yr x 100 yr
e −1)=
−1)= 1239 GtC
(
(e
0.01308
r
Amount remaining in atmosphere = 0.50 x 1239 = 619 GtC
d. Amount in atmosphere in 100 yrs = 750 + 619 = 1369 GtC
(CO2 ) =
1369GtC
= 646ppm
2.12GtC/ ppmCO2
e. Equilibrium temperature increase, with ΔT2x=3oC from (8.29):
ΔT =
⎡ CO ⎤ 3.0
ΔT2x
⎛ 645 ⎞
2
⎟ = 2.57 o C
⋅ln ⎢
⋅ ln ⎜
⎥=
⎝
⎠
ln
2
356
ln 2
CO
⎣ ( 2 )0 ⎦
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 MJ 11.75gC GtC
x 6 x
x 15
= 3.88GtC / yr
10 J
MJ
10 gC
yr
b. Growth from 6.0 GtC/yr to 3.88 GtC/yr in 100 yrs,
r=
1 ⎛ 3.88 ⎞
⎟ = −0.0044 = −0.44% / yr
ln ⎜
100 ⎝ 6.0 ⎠
c. Amount remaining with 50% airborne fraction, use (8.27):
Total emitted = C tot =
C0 r T
6.0 GtC/yr −0.0044/yr x 100 yr
e −1)=
−1)= 483 GtC
(
(e
r
−0.0044
Amount remaining in atmosphere = 0.50 x 483 = 242 GtC
d. Amount in atmosphere in 100 yrs = 750 + 242= 992 GtC
(CO2 ) =
992GtC
= 468ppm
2.12GtC/ ppmCO2
e. Equilibrium temperature increase, with ΔT2x=3oC,
ΔT =
⎡ CO ⎤ 3.0
ΔT2x
⎛ 468⎞
2
⎟ = 1.18o C
⋅ln ⎢
⋅ ln ⎜
⎥=
ln 2
⎣ (CO2 )0 ⎦ ln 2 ⎝ 356 ⎠
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
= 1.054 = 105.4%
802 kJ LHVinput
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, C 2H 6 :
2 x 12 gC/mol 10 3 kJ
x
= 15.56 gC/MJ
MJ
1542 kJ/mol
b. Propane, C 3H 8 :
3 x 12 gC/mol 10 3 kJ
x
= 16.36 gC/MJ
MJ
2220 kJ/mol
c. n - Butane, C 4 H10 :
8.21
4 x 12 gC/mol 10 3 kJ
x
= 16.68 gC/MJ
MJ
2878 kJ/mol
Using HHV carbon intensities from Table 8.3, the four options are:
1) pulse
2) conv gas
100MJ
η
95MJ
delivered
=0.95
η
=14.5gC/MJ
95MJ
1380gC
100MJ
1380gC
70MJ
delivered
=0.70
1380gC
=19.7gC/MJ
70MJ
1380gC
heat pump
3) heat pump
100MJ
2420gC
4)resistance
100MJ
2420gC
η
35MJ
COP=3
=0.35
power plant
η
105MJ del
105MJ
70 from enviro.
35MJ
=0.35
2420gC
35MJ
=69.1gC/MJ
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:
3 x 12 gC/mol 10 3 kJ
x
= 16.36 gC/MJ
a. Carbon intensity C 3H 8 :
2220 kJ/mol
MJ
b. Delivering heat at 85% efficiency to hot water
Pg. 8.9
2420gC
=23.0gC/MJ
c. Savings versus 32.5 gC/MJ with a n. gas electric water heater in Example 8.6:
propane 19.25 gC/MJ
=
= 0.59 so there is a 41% savings vs electricity
electric 32.5 gC/MJ
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)
C tot =
C0 r T
6.0 GtC/yr −0.014/yr x 70 yr
e −1)=
−1)= 268 GtC
(
(e
r
−0.014
(CO 2 ) =
ΔT =
750GtC + C tot x AF 750 + 268 x 0.4 GtC
=
= 404 ppm
2.12 GtC/ppmCO 2
2.12 GtC/ppmCO 2
⎡ CO ⎤
3
ΔT2x
⎛ 404 ⎞
2
⎟ = 0.55o C
⋅ln ⎢
⋅ ln ⎜
⎥=
ln 2
⎣ (CO2 )0 ⎦ ln 2 ⎝ 356 ⎠
To find the doubling time, rearrange (8.27):
C tot to double current 750 GtC =
750 GtC C 0 r Td
=
(e −1)
AF = 0.4
r
⎡(750 /0.4)(−0.014) ⎤
1
1 ⎡(750 /0.4)r ⎤
Td = ln⎢
+ 1⎥ =
ln⎢
+ 1⎥ = never!
r ⎣
6.0
6.0
⎦ −0.014 ⎣
⎦
(B)
r = 1.5 + 1.5 - 0.2 + 0.4 = 3.2%/yr
C tot =
C0 r T
6.0GtC/yr 0.032/yr x 70 yr
e −1)=
−1)= 1574 GtC
(
(e
0.032
r
(CO 2 ) =
750 GtC + C tot x AF 750 + 1574 x 0.5GtC
=
= 725 ppm
2.12 GtC/ppmCO 2
2.12 GtC/ppmCO 2
Pg. 8.10
ΔT =
⎡ CO ⎤
2
ΔT2x
⎛ 725⎞
2
⎟ = 2.05o C
⋅ln ⎢
⋅ ln ⎜
⎥=
⎝
⎠
ln
2
356
ln 2
CO
⎣ ( 2 )0 ⎦
(
)
(
)
⎡ 750
⎡ 750
r ⎤
0.032 ⎤
1 ⎢
1
AF
0.5
⎥
⎢
Td = ln
+1 =
ln
+ 1⎥ = 69 yrs
⎥ 0.032 ⎢
⎥
r ⎢
6.0
6.0
⎣
⎦
⎣
⎦
(C)
r = 1.4 + 1.0 - 1.0 - 0.2 = 1.2%/yr
C0 r T
6.0 GtC/yr 0.012/yr x 70 yr
e −1)=
−1)= 658 GtC
(
(e
0.012
r
750 GtC + C tot xAF 750 + 658 x 0.5 GtC
=
= 509 ppm
(CO 2 ) =
2.12 GtC/ppmCO 2
2.12 GtC/ppmCO 2
⎡ CO ⎤ 3
⎛ 509 ⎞
ΔT
o
2
⋅ ln⎜
ΔT = 2x ⋅ ln⎢
⎥=
⎟ = 1.55 C
ln2
⎣(CO 2 )0 ⎦ ln2 ⎝ 356 ⎠
C tot =
(
)
(
)
⎡ 750
⎡ 750
r ⎤
0.012 ⎤
1 ⎢
1
AF
0.5
⎥
⎢
Td = ln
+1 =
ln
+ 1⎥ = 116yrs
⎥ 0.012 ⎢
⎥
r ⎢
6.0
6.0
⎣
⎦
⎣
⎦
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 = C0e rt = 6.0e 0.013x110 = 25.1 GtC/yr
Including land use: Total emission rate = 25.1 + 2.5 = 27.6 GtC/yr
b. Total carbon emissions:
Ctot (energy ) =
6.0 0.013x110
C0 rT
e −1)=
−1)= 1467 GtC
(
(e
0.013
r
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:
ΔCO 2 =
1742 GtC x 0.5
= 410 ppm CO 2
2.12 GtC/ppmCO 2
d. Estimated 2100 CO2 concentration = 360 + 410 = 770 ppm
e. With ΔT2X = 2.8oC, the global equilibrium temperature increase 2100
ΔTe =
8.25
⎡ (CO ) ⎤ 2.8
⎡ 770 ⎤
ΔT2X
o
2
ln ⎢
ln ⎢
⎥=
⎥⎦ = 3.1 C
⎣
ln
2
360
ln 2
CO
(
)
⎣
2 0⎦
Identification of the halocarbons:
a. C3HF7 is an HFC (no Cl), 3 1 7 - 90 = 227,
b. C2H3FCl2 is an HCFC,
2 3 1 - 90 = 141,
c. C2F4Cl2 is a CFC, 2 0 4 - 90 = 114,
HFC-227
HCFC-141
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 oC.
λ=
o
2.5
ΔT2X
=
= 0.595 C
W m2
4.2
4.2
g = 1-
0.27
0.27
λB
= 1−
= 1−
= 0.546
0.595
λ
λ
If g = 0.1 + 0.546 = 0.646, then
λ=
λB
1− g
=
0.27
= 0.763 o C (W m2 ),
1− 0.646
Pg. 8.12
ΔT2 X = 4.2λ = 4.2x0.763 = 3.2 oC
b. For ΔT2X = 3.5 oC
λ=
o
3.5
ΔT2X
=
= 0.833 C
W m2
4.2
4.2
g =1-
0.27
0.27
λB
= 1−
= 1−
= 0.676
0.833
λ
λ
If g increases to 0.776, then
λB
0.27
= 1.205 o C (W m2 ),
ΔT2 X = 4.2λ = 4.2x1.205 = 5.1o C
1− g 1− 0.776
Notice ΔT2X becomes more sensitive as the feedback factor increases (0.7oC increase when g
changes from 0.546 to 0.646 versus 1.9oC increase when g changes from 0.676 to 0.776).
λ=
8.28
=
Using Figure 8.39:
a. The AS probability that ΔT2X is less than 2.5oC. Answer: 20%
b. The WR probability that ΔT2X is greater than 3oC. Answer: 40%
c. The AS probability that ΔT2X is between 3oC and 4oC. Answer: 50%
d. The WR probability that ΔT2X is between 3oC and 4oC. 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 / m 2
8.30
a. Combined radiative forcings from 1850 to 1992
ΔFCO2 = 6.3 ln
[(CO )] = 6.3 ln ⎛⎜ 356 ⎞⎟
2
[(CO ) ]
2 0
( CH
ΔFCH4 = 0.031
(
4
−
ΔFN2 O = 0.133 N 2 O −
⎝ 278 ⎠
= 1.558 W/m
(CH 4 ) 0 )= 0.031( 1714 −
(N2 O)0 )= 0.133(
2
700 ) = 0.463 W/m2
)
311 − 275 = 0.140W / m 2
ΔFCFC−11 = 0.22[(CFC − 11) − (CFC − 11)0 ] = 0.22(0.268 − 0) = 0.059 W / m 2
ΔFCFC−12 = 0.28[(CFC − 12) − (CFC − 12 )0 ]= 0.28(0.503 − 0 ) = 0.141 W / m 2
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
[(CO )] = 6.3 ln ⎛⎜ 710 ⎞⎟
[
2
(CO2 )0
( CH
]
⎝ 356 ⎠
= 4.35 W/m
2
−
(CH 4 ) 0 )= 0.031(
3616 − 1714 )= 0.581W / m
ΔFN2 O = 0.133 N 2 O −
(N2 O)0 )= 0.133(
417 − 311) = 0.370W / m 2
ΔFCH4 = 0.031
4
(
2
ΔFCFC−11 = 0.22[(CFC − 11) − (CFC − 11)0 ] = 0.22(0.040 − 0.268) = −0.050 W / m 2
ΔFCFC−12 = 0.28[(CFC − 12) − (CFC − 12 )0 ]= 0.28(0.207 − 0.503) = −0.083 W / m 2
Combined forcing = 4.35 + 0.581 + 0.370 - 0.050 - 0.083 = 5.17 W/m2
c.
From 1850 to 2100
(
)
(
)
⎛ 710 ⎟⎞
ΔF = 6.3 ln ⎜
+ 0.031( 3616 − 700 )+ 0.133 417 − 275
⎝ 278⎠
+ 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:
⎛ 710 ⎟⎞
ΔF = 6.3 ln ⎜
+ 0.031( 3616 − 700 )+ 0.133 417 − 275
⎝ 278⎠
+ 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
20
8.32
Using
∫
0
RCO 2 (t )dt ≈ 13.2 yrs;
100
∫
RCO 2 (t )dt ≈ 43.1yrs;
0
500
∫ R (t )dt ≈ 138 yrs and forcing
CO 2
0
ratios of HFC-134a to CO2 of (Fg/FCO2) = 4129 and τ = 14 yrs. First simplify GWP to
Pg. 8.15
T
⎛ F ⎞
GWPg = ⎜⎜ g ⎟⎟ ⋅
⎝ FCO2 ⎠
∫e
−t / τ
⎛ F ⎞ τ (1− e−T / τ )
= ⎜⎜ g ⎟⎟ ⋅ T
T
F
∫ RCO2 (t)dt ⎝ CO2 ⎠ ∫ RCO2 (t )dt
0
a. GWP(20) = 4129 ⋅
14 (1− e
b. GWP (100) = 4129 ⋅
c. GWP (500) = 4129 ⋅
8.33
dt
0
0
−20 /14
) = 3330
13.2
14 (1− e−100 /14 )
43
14 (1− e−500 /14 )
138
(vs 3300 in Table 8.7)
= 1340 (vs 1300 in table)
= 420 (vs 400 in table)
For a greenhouse gas with τ = 42 years and a relative forcing of 1630 times that of CO2.
(
−t
)
τ
ΔFg τ 1 − e
From Problem 8.32, GWP =
⋅
ΔFCO2 ∫ R CO (t )dt
2
a. The 20-year GWP would be
− 20
42 1− e 42
GWP20 = 1630 ⋅
= 1965
13.2
(
)
b. The 100-year GWP would be
−100
42 1− e 42
GWP100 = 1630 ⋅
= 1440
43.1
(
)
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
2
= 1.40W / m
0.57
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 gCH 4 23 gCO 2
x
= 703 gCO 2 − eq
12 gC
1 gCH 4
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 gCO 2 - eq + 5525 gCO 2
= 74.4 gCO 2 −eq /MJ
83.73 MJ heat to water
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
σ=
25.8 gC 1 GtC 1012 MJ
x 15 x
= 5160 GtC
MJ
10 gC
EJ
Q∞
5160 GtC
=
= 93.57 yr
Pm 2π 22 GtC/yr 2π
Pg. 8.17
t m = σ 2ln
Pm
22 GtC/yr
= 93.57 yr 2 ln
= 150.8 yr
6.0 GtC/yr
P0
⎡ 1 ⎛ t − t ⎞2 ⎤
m
then put these into P = Pm exp⎢− ⎜
⎟⎥
⎣ 2⎝ σ ⎠ ⎦
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):
C emissions =
50 MW 1 MJ/s 3600 s 8760 hr 24 gC 1 mtC
x
x
x
x
x 6
= 1.08x105 mtonC/yr
MW
hr
MJ 10 gC
0.35
yr
Pg. 8.18
Carbon tax = 1.08 x10 5 mtC/yr x
$20
= $2.16 million/yr
mton
b. With carbon sequestering:
Area =
1.08x10 5 mtonC/yr 10 3 kg
x
= 21,600 acres
5000 kgC/yr ⋅ acre mton
c. Biomass instead of paying the tax:
Forestry could cost =
8.40
$2.16 million/yr
= $100 /yr per acre
21,600 acres
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 Æ CO2 + 2 H2O
1molCO 2 (12 + 2x12)gCO 2 /mol 10tonneCH 4
x
x
= 27.5tonneCO 2 /yr
1molCH 4 (12 + 4x1)gCH 4 /mol
yr
c. Equivalent CO2 savings = 620 – 27.5 = 592.5 tonne CO2/yr.
as C : 592.5 tonne CO 2 /yr x
12tonneC
= 161.5 tonneC/yr saved
44tonneCO 2
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 =
C=
6.15 lbgas
(7x12 = 84) lbsC = 5.22 lbs C/gal
x
gal
(7x12 + 15x1 = 99) lb gas
40,000 miles 5.22 lbsC
x
= 17,394 lbsC that will be released
12 miles/gal
gal
b. 4000 lb car, 10,000 miles/yr
Pg. 8.19
C=
17,394 lbs C 10,000 miles
x
= 4348 lbsC/yr
40,000 miles
yr
4348 lbsC/yr
Carbon/yr
=
= 1.09
Vehicle weight
4000 lbs
the car emits slightly more carbon per year than it weighs!
c. Carbon tax =
5.22 lbsC
$15
x
= $0.039/gal = 3.9¢/gal
gal
2000 lbsC
d. New car at 40 mpg, for 40,000 miles:
Carbon reduction =17,394 lbsC -
40,000 mi 5.22 lbsC
x
= 12,174 lbs C saved
40 mi/gal
gal
e. Trading in the clunker for the 40 mpg car would save in carbon taxes
Tax savings = 12,174 lbsC x
$15
= $91/car
2000 lbs C
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 1000g
x
= 59.3 gC/mi
40 miles/gal 2.2lbs
b. With the very efficient natural-gas fired power plant:
N - gas plant emissions =
8000kJ 13.8gC MJ kWh
x
x 3 x
= 22.1 gC/mi
kWh
MJ
10 kJ 5mi
c. With the typical coal plant:
Coal plant heat rate =
l kW in
1 kJ/s
3600 s
x
x
= 12,000 kJ/kWhe
0.30 kWe out kW heat in
hr
Coal plant emissions =
12,000kJ 24 gC MJ kWh
x
x
x
= 57.6 gC/mi
kWh
MJ 10 3 kJ 5mi
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 Æ NO + O
From (8.48): E (J/photon) =
306,000
= 5.08x10−19 J/photon
23
6.02x10
Pg. 8.20
and from (8.46): λmax
8.44
hc 6.626 x10−34 Js x 2.998x108 m/s
=
=
= 390x10−9 = 390 nm
−19
E
5.08x10 J
O2 + hv Æ O + O (oops… same as Example 8.13):
E (J/photon) =
λmax =
495,000
= 8.22x10−19 J/photon
23
6.02x10
hc 6.626 x10−34 Js x 2.998x108 m/s
=
= 241.6x10−9 = 241.6 nm
E
8.22x10−19 J
Meant to do photodissociation of ozone, requiring 104.6 kJ/mol:
O3 + hv Æ O2 + O
E (J/photon) =
λmax =
104,600
= 1.737x10−19 J/photon
6.02x10 23
hc 6.626 x10−34 Js x 2.998x108 m/s
=
= 1.14x10−6 m = 1.14 μm
−19
E
1.737x10 J
Pg. 8.21
SOLUTIONS FOR CHAPTER 9
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 CO 2 2200 lb 44 ton CO 2
x
x
= $40.33/MTCE
2000 lb/ton metric ton 12 ton C
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?
CO 2 =
10,000 miles x x 5.22 lbsC/gal 44 tonCO 2
x
= 4.35 ton CO 2 /yr
20 miles/gal x 2200 lbs/ton
12 tonC
b. Btus for those SUVs
Energy =
10,000 miles x 125,000 Btu/gal
= 62.5 million Btu/yr
20 miles/gal
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
42x10 6 ton/yr x 0.96 mtonC/ton x 1.1 ton/mton 44 tonCO 2
Car equivalents =
x
12 tonC
4.32 tonCO 2 /yr − SUV
= 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 =
42x10 6 ton/yr x 15.65 x10 6 Btu/ton
= 10.5 million SUVs
62.5 x 10 6 Btu/yr/SUV
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.
1765 kJ
= 1059 kJ
8g
40 kJ
= 56 kJ
Recycled cans to make 11.2 g of Al =11.2g x
8g
The remaining energy for can production is the same as Table 9.13:
New aluminum from bauxite = 4.8g x
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
= 0.53
Earlier : 3396 kJ/can
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 1.6x10 6 tons x 220,600 kJ/kg x
9.3
kg
2000 lb
x
= 118.7 x1012 kJ
2.2 lb
ton
Old aluminum = 0.67 x 1.6x10 6 tons x 5060 kJ/kg x
kg
2000 lb
x
= 4.9 x1012 kJ
2.2 lb
ton
Total = (118.7 + 4.9) x 1012 = 123.6 x 1012 kJ
b. With no recycling:
All new aluminum = 1.6x10 6 tons x 220,600 kJ/kg x
kg
2000 lb
x
= 320.9x1012 kJ
2.2 lb
ton
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: 3.97x10 6 metric tonsC x
or 14.6x10 6 tonne/yr x
44 tons CO 2
= 14.6x10 6 metric tons/yr
12 tons C
2200 lbs
ton
x
= 16 million U.S. tons/yr
tonne 2000 lbs
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
mi
hr 60 min
x
x
x
+ 1 min = 1.227 min/stop
stop 5280 ft 5 mi
hr
9.4
9.10 Route timing:
a. Not collecting = 20min + 3x20min + 2x15min + 15min + 40min = 165 min/day
To fill a truck takes:
25yd 3truck x
4 yd 3curb customer
stop
1.5 min
x
x
x
= 187.5 min/load
3
3
yd truck
0.2yd
4 customer
stop
Two loads per day takes:
[2 loads/d x 187.5min/load
+165 min(travel,breaks)]x
hr
= 9.0 hrs/day
60min
b. Customers served:
25yd 3truck x
4 yd 3curb customer
x
= 500 customers/load
yd 3truck
0.2yd 3
# customers =
500 customers 2 loads 5 days
x
x
= 5000 customers/truck
load
day
week
⎛ $40 8 hrs $60 1hr ⎞ 5 day 52 wks
c. Labor = ⎜
x
+
x
x
= $98,800/yr
⎟x
hr day ⎠ week
yr
⎝ hr day
Truck cost =
$10,000 $3500/yr
+
x 25yd3 = $97,500/yr
3
yr
yd
Customer cost =
($98,800 + $97,500)/yr
= $39.26/yr
5000 customers
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
stop
4 customers 5 day
x
x
x
= 4200 customers
day
1.5 min
stop
wk
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,
stop
4 customers
315 min
x
x
= 840 customers/day
Customers =
day
1.5 min
stop
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 0.2 yd 3 at curb yd 3 in truck
x
x
= 21yd 3
customer
4yd 3at curb
truckload
costing:
$10,000 $3500/yd 3
Truck$ =
+
x 21yd 3 = $83,500/yr
yr
yr
⎛ $40 8hr ⎞ 5 day 52 weeks
Labor$ = ⎜
x
x
= $83,200/yr
⎟x
yr
⎝ hr day ⎠ week
Resulting in: Customer$ =
($83,200 +$83,500)/yr = $39.69/yr
4200 customers
(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
# customers =
4m3curb
customer
x
= 432 customers/load
3
1m in truck 0.25m3 curb
432 customers 2 loads 5 days
x
x
= 4320 customers (27m3 )
truckload
day
week
(B) 15 m3 truck: 15m3 truck x
# customers =
4m3curb
customer
x
= 240 customers/load
3
1m in truck 0.25m3 curb
240 customers 3 loads 5 days
x
x
= 3600 customers (15m3 )
truckload
day
week
b. Hours per day for the crew:
(A) 27 m3 truck:
432 customers 0.4 min 2 loads
x
x
= 346 min
customer
truckload
day
Crew :
346 min +160 min(misc.)
= 8.43 hr/day
60 min/hr
9.6
(B) 15 m3 truck:
240 customers 0.4 min 3 loads
x
x
= 288 min
customer
truckload
day
Crew :
288 min + 215 min(misc.)
= 8.38 hr/day
60 min/hr
c. Cost per customer:
(A) 27 m3 truck:
Cost :
$40 8.43hr 5 day 52 week
x
x
x
+ $120,000/yr = $207,672/yr (27m3 )
hr
day week
yr
Customer$ =
$207,672/yr
= $48.07/yr (27m3 )
4320 customers
(B) 15 m3 truck:
Cost :
$40 8.38hr 5 day 52 week
x
x
x
+ $70,000/yr = $157,152/yr (15 m3 )
hr
day week
yr
Customer$ =
$157,152/yr
= $43.65/yr (15 m 3 )
3600 customers
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
= $78.54/ton
10 ton/day x 260day/yr
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
+ 4 can/stop x 20s/can = 100.5 s/stop
5 mi/hr x 5280 ft/mi
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
5.75 hr/d x 3600 s/hr
= 206 stops/load
100.5 s/stop x 1 load/day
Truck volume need is,
4 can/stop x 4 ft 3 /can x 206 stop/load
V=
= 34.9 yd 3
3
3
3
3
27ft /yd x 3.5 yd curb/yd in truck
N=
With economics,
Labor = $99.840/yr as in Example 9.7
Truck = $25,000/yr + 4000$/yd3yrx34.9 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
= 3214 tons/yr
2000 lb/ton
Cost/ton =
(164,600 + 99,840) $/yr
= $82.25/ton (OK)
3214 ton/yr
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)
3.75 hr/d x 3600 s/hr
= 44.8 stops/load
100.5 s/stop x 3 load/day
Truck volume need is,
4 can/stop x 4 ft 3 /can x 44.8 stop/load
V=
= 7.6 yd 3
27ft 3 /yd 3 x 3.5 yd 3 curb/yd 3in truck
N=
With economics,
Labor = $99.840/yr as in Example 9.7
Truck = $25,000/yr + 4000$/yd3yr 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
= 2096 tons/yr
2000 lb/ton
Cost/ton =
(99,840 + 55,400) $/yr
= $74/ton
2096 ton/yr
(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.
n
10
i(1+ i)
0.10(1+ 0.10)
=
= 0.16275 / yr
Station costs: CRF(10yr,10%) =
n
10
(1+ i) −1 (1+ 0.10) −1
9.8
$3 million x 0.16275/yr + $100,000/yr = $588,236/yr
to handel:
5 ton/d x 5 d/wk x 52 wk/yr = 52,000 tons/yr
which is
cost =
$588,236/yr
= $11.31/ton
52,000 tons/yr
Truck costs:
Depreciation = $120,000 x CRF = $120,000 x 0.16275 = $19,530/yr
(Driver+Maint)+Depreciation = $80,000 + $19,530 = $99,530
to haul:
20 tons/trip x 4 trip/d x 5 d/wk x 52 wk/yr = 20,800 ton/truck-yr
which is
total truck cost =
for a total of
$4.79/ton (truck) + $11.31/ton (station) = $16.10/ton
$99530/yr
= $4.79/ton
20,800 ton/yr
9.17 Distance for an economic transfer station:
a. Cost of direct haul to the disposal site,
$40 + 30 t1 = 40 + 30x1.5 = $85/ton
b. Transfer station 0.3 hr from a collection route,
$40 + 30 t1 = 40 + 30x0.3 = $49/ton to get to the transfer station
for the transfer station
$10 + 10 t2 = 10 + 10(1.5 – 0.3 hr) = $22/ton
total cost with transfer station = $49 + $22 = $71/ton
c. Minimum distance from the transfer station to the disposal site,
direct haul $ = $ to transfer station + $ for transfer station
$40 + $30/hr x 1.5 hr = ($40 + 30 t1) + $10 + 10(1.5 - t1)
85 = 40 + 10 + 15 + 30 t1 – 10 t1 = 65 + 20 t1
t1 = 1 hr
t2 = 1.5 – 1 = 0.5 hr
That is, the transfer station must be no more than 1 hour away from the
collection route (or, the transfer station should be more than 0.5 hr from the
disposal site).
9.9
9.18 Newsprint: 5.97% moisture, HHV = 18,540 kJ/kg, 6.1% H.
Starting with 1 kg of “as received” waste:
Energy to vaporize moisture = 0.0597 kg H2O x 2440 kJ/kg = 145.6 kJ
Dry weight = 1 – 0.0597 = 0.9403 kg
Hydrogen in the dry waste = 0.061 x 0.9403 = 0.0574 kg
As H becomes H2O = 0.0574 kgH x 9 kgH2O/kgH x 2440 kJ/kgH2O = 1259.6 kJ
Total energy lost in water vapor = 145.6 + 1259.6 = 1405 kJ per kg of newsprint
LHV = HHV – 1405 = 18,540 – 1,405 = 17,135 kJ/kg
9.19 Corrugated boxes, 5.2% moisture, HHV = 16,380 kJ/kg, 5.7% H in dried material:
Could use the procedure shown in Prob. 9.18, or use (9.7)
LHV = HHV – 2440(W+9H)
W = 0.052 kgH2O/kg waste
H = (1-0.052)x 0.057 = 0.054 kgH/kg waste
LHV = 16,380 – 2440 (0.052 + 9x0.054) = 16380 – 1313 = 15,067 kJ/kg
9.20
2 L PET bottle, 54 g, 14% H, HHV = 43,500 kJ/kg,
QL = energy lost in vaporized H2O
QL =
2440 kJ 18 kgH 2O 0.14 kgH 54g PET
x
x
x
= 166 kJ/bottle
kg H 2O
kg PET 10 3 g/kg
2 kgH
LHV =
9.21
43,500 kJ 54 gPET/bottle
x
−166 kJ/bottle = 2183 kJ/bottle
kgPET
10 3 g/kg
Energy estimates based on HHV = 339(C ) + 1440 (H) – 139 (O) + 105 (S)
a. Corrugated boxes: based on dry weight,
HHV(dry)= 339x43.73 + 1440x5.70 –139x44.93 + 105x0.21 = 16,809 kJ/kg
There are (1-0.052) = 0.948 kg dry material per kg “as received”
HHV as received = 0.948 kg(dry) x 16,809 kJ/kg(dry) = 15,935 kJ/kg
b. Junk mail:
HHV(dry)= 339x37.87 + 1440x5.41 –139x42.74 + 105x0.09 = 14,697 kJ/kg
HHV as received = (1 - 0.0456) x 14,697 = 15,935 kJ/kg
c. Mixed garbage:
HHV(dry)= 339x44.99 + 1440x6.43 –139x28.76 + 105x0.52 = 20,568 kJ/kg
HHV as received = (1 - 0.72) x 20,568 = 5,759 kJ/kg
9.10
d. Lawn grass:
HHV(dry)= 339x46.18 + 1440x5.96 –139x36.43 + 105x0.42 = 19,218 kJ/kg
HHV as received = (1 - 0.7524) x 19,218 = 4,758 kJ/kg
e. Demolition softwood:
HHV(dry)= 339x51.0 + 1440x6.2 –139x41.8 + 105x0.1 = 20,417 kJ/kg
HHV as received = (1 - 0.077) x 20,417 = 18,845 kJ/kg
f. Tires:
HHV(dry)= 339x79.1 + 1440x6.8 –139x5.9 + 105x1.5 = 35,944 kJ/kg
HHV as received = (1 - 0.0102) x 35,944 = 35,578 kJ/kg
g. Polystyrene:
HHV(dry)= 339x87.10 + 1440x8.45 –139x3.96 + 105x0.02 = 41,147 kJ/kg
HHV as received = (1 - 0.002) x 41,147 = 41,064 kJ/kg
9.22
Draw the chemical structures:
a. 1,2,3,4,7,8-hexachlorodibenzo-p-dioxin
b. 1,2,3,4,6,7,8-heptachlorodibenzo-p-dioxin
c. Octachlorodibenzo-p-dioxin
d. 2,3,4,7,8-pentachlorodibenzofuran
9.11
e. 1,2,3,6,7,8-hexachlorodibenzofuran
9.23
U.S. 129 million tons, 800 lb/yd3, cell 10-ft, 1 lift/yr, 80% is MSW, 1000 people:
129x10 6 ton 2000 lb yd 3
VMSW =
x
x
= 8.71x10 9 ft 3 /yr
3
27 ft
yr
ton
@ 80% per cell,
Vlandfill =
Alift =
8.71 x 109 ft 3 /yr
= 10.9x10 9 ft 3 /yr
0.80
10.9 x 109 ft 3 /yr
acre
x
= 24,987 acres/yr
43,560 ft 2
10 ft/lift
No population is specified, but at roughly 300 million people, the area per 1000
people would be about
Alift per 1000 =
9.24
24,987 acres/yr
≈ 0.08 acre/yr per 1000 people
300,000 thousand people
50,000 people, 40,000 tons/yr, 22% recovery, 1000 lb/yd3, 10-ft lift, 80% MSW:
40,000(1- 0.22)ton 2000 lb
yd 3
27ft 3 ft 3 landfill
VLandfill =
x
x
x
x
= 2.11x10 6 ft 3 /yr
3
3
0.80ft MSW
1000 lb yd
yr
ton
a.
Area of lift needed each year
Alift =
2.11 x 106 ft 3 /yr
acre
x
= 4.83 acre/yr
43,560 ft 2
10 ft/lift
b. To complete the landfill will take:
time remaining =
9.25
40 acre/lift x 2 lifts remaining
= 16.5 yrs
4.83 acre/yr
By increasing the recovery rate from 22% to 40%,
VLandfill =
40,000(1- 0.40)ton 2000 lb
yd 3
27ft 3 ft 3 landfill
x
x
x
x
= 1.62x10 6 ft 3 /yr
1000 lb yd 3 0.80ft 3MSW
yr
ton
Alift =
1.62 x 106 ft 3 /yr
acre
x
= 3.72 acre/yr
43,560 ft 2
10 ft/lift
9.12
time remaining =
40 acre/lift x 2 lifts remaining
= 21.5 yrs
3.72 acre/yr
So the landfill will last 5 more years because of the program (21.5 – 16.5 yrs).
9.26
Lawn trimmings: Per kg, 620g moisture and 330g decompostables represented by
C12.76H21.28O9.26N0.54.
1 mol trimmings = 12x12.76 + 1x21.28 + 16x9.26 + 14x0.54 = 330.2 g//mol
That is, 1 kg of as received trimmings has 330g of decompostibles, which turns
out to be1 mole of decompostibles. Using (9.8) gives
C12.76H21.28O9.26N0.54 + n H2O Æ m CH4 + s CO2 + d NH3
where m = (4x12.76 + 21.28 –2x9.26 –3x0.54)/8 = 6.5225
So, 6.5225 moles of CH4 are produced per mole (330g) of decompostibles. So, 1
kg of lawn trimmings, with 330g of decomposbiles, results in 6.5225 moles of
CH4.
a. Volume of methane:
VCH 4 =
0.0224 m3CH 4 6.5225 molCH 4
x
= 0.146m3CH 4 /kg lawn trimmings
mol CH 4
kg"asreceived"
b. Energy content: CH 4 energy =
6.5225 molCH 4 890 kJ
x
= 5,805 kJ/kg
kg "as received" mol
9.27 1 kg of food wastes, with 720g water and 280g of CaHbOcNd:
a.
C
45%
0.45 x 280 = 126g
H
6.4% 0.064 x 280 = 17.92g
O
28.8% 0.288 x 280 = 80.64g
N
3.3% 0.033 x 280 = 9.24g
Total = 233.8 g/mol
C:
12g/mol x a mol = 126g
so a = 126/12 = 10.5 mol
H:
1 g/mol x b mol = 17.92g
so b = 17.92/1 = 17.92 mol
O:
16g/mol x c mol = 80.64g
so c = 80.64/16 = 5.04 mol
N:
14g/mol x d mol = 9.24g
so d = 9.24/14 = 0.66 mol
The chemical formula for dry food wastes: C10.5H17.92O5.04N0.66
b. Chemical reaction:
C10.5H17.92O5.04N0.66 + n H2O Æ m CH4 + s CO2 + d NH3
9.13
where
n = (4x10.5 – 17.92 – 2x5.04 + 3x0.66)/4 = 3.995
m = (4x10.5 + 17.92 – 2x5.04 –3x0.66)/8 = 5.9825
s = (4x10.5 – 17.92 + 2x5.04 + 3x0.66)/8 = 4.5175
d = 0.66
Methane producing reaction is therefore:
C10.5H17.92O5.04N0.66 + 3.995 H2O Æ 5.9825 CH4 + 4.5175 CO2 + 0.66
NH3
c. Fraction of the resulting gas that is methane:
CH 4 =
5.9825 mol CH 4
= 0.536 = 53.6%
(5.9825 + 4.5175 +0.66) moles gas
d. Volume of methane per kg of food waste:
VCH 4 =
0.0224 m3CH 4 5.9825 molCH 4
x
= 0.134m3 CH 4 /kg food wastes
mol CH 4
kg"as received"
e. HHV value of methane produced:
CH 4 energy =
5.9825 mol CH 4 890 kJ
x
= 5,324 kJ/kg food waste
kg "as received" mol
9.14