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: ktCC -

0e = First, find k:

1-

0

0

0

min 0.128 0.88

1ln min 11

0.88ln

min 11 ln1 ====

CC

CC

tk

Next, calculate the grease remaining on the clothes after 5 minutes: ( ) ( )( ) g 0.264 eg 0.500 e

1-1 min .005min 128.0--0 === ktmm

The grease that is not on the clothes must be in the water, so

g/L 0.00472 L 50.0

g 0.264 - g 0.500 w ==C

Q0, C0 Qs, CsQb, C0

farm

Plateau CreekC.V.

Qf/2, Cf

1.9 Qf, C0

a. A mass balance around control volume (C.V.) at the downstream junction yields

( ) ( )( ) ( )( )

( ) mg/L 0.112 /sm 4.5

mg/L 0015.0/sm 4.0 mg/L 00.1/sm 0.5 /2 3

33

s

0bffs =

+=

+=

QCQCQC

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

( ) ( )( ) /sm 0.371 mg/L

20.04 0.0015 -

2 1.0

mg/L 0.0015 - 04.0/sm 5.0

2 -

2

- 33

s0

f

0s0f =

⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛ +

=CCC

CCQQ

1.10 Write a mass balance on a second order reaction in a batch reactor:

Accumulation = Reaction

2- ere wh kmr(m)Vr(m)dtdmV ==

∫∫ =tm

m

dtkmdmt

02 -

0

so, ktmm t

- 1 - 1

0

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

0t

1 - 11 k mmt

calculate mt=1 based on the equation’s stoichiometry that

1 mole of methanol yields 1 mole of carbon monoxide

( ) OHCH g 114.3 OHCH moleOHCH g 32

CO mole 1OHCH mole 1

CO g 28CO moleCO g 100 3

3

33 =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

so, g 85.7 g 114.3 - g 200 1 tOH,CH3

===C

and 1-1- gd 0.00667 g 200

1 - g 85.7

1d 11 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=k

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.

( ) ( ) 1-

1/2

32

p min 0.0578 min) (122ln

t2ln and ft 24,033

4ft 3400 ft 3.0 ==== kV π

For first-order decay in a steady-state PFR

( ) ( )( ) Lmg2.82 ft 0.134

galgal 10

minft 033,24min0578.0expmg/L 1.0 e 3431-

0 /=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛== −ktCC

1.15 The stomach acts like a CSTR reactor in which a first order decay reaction is occurring.

V, r(C)

Stomach Q, Ce

Gastric juices in

Q, Ci Digested food stream out

V = 1.15 L, k = 1.33 hr-1, Q = 0.012 L/min, m0 = 325 g, t = 1 hr Accumulation = In – Out + Reaction

)( - ei CVrQCQCdtdCV += where r(C) = -kC, Ci = 0, and C = Ce

so, ∫∫ ⎟⎠⎞

⎜⎝⎛ +=

t

dtkVQ

CdC

0

C

C

- 0

( )( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛==

⎟⎠⎞

⎜⎝⎛ +

hr 1hr 1.33 L 1.15

hrmin/60L/min 0.012 exp L 1.15g 325 e 1-

- tkVQ

CC

C = 40.0 g/L and then mt=1hr = (40.0 g/L)(1.15L) = 46.0 g

1.16-1.20 The solutions for these problems are the solutions for problems 1.11-1.15 in the 2nd edition Solutions Manual.

1.21 a. Calculate the volume that 1 mole of an ideal gas occupies at 1 atm and 20 °C.

( )( )( ) L 24.04

atm1K 15.293K mol atm L 082056.0mole 1

-1-1

=⋅⋅⋅

==P

nRTV then

ppmv = (mg/m3)(24.04 L/mol)(mol wt)-1 = (60 mg/m3)(24.04 L/mol)(131 mol/g)-1

= 11.0 ppmv

b. Draw a sketch of the valley as the CSTR, non-steady state control volume.

Q, Ca

V, r(C), kd, ks

Coal Valley Q, Ce

kd = radioactive decay rate constant = (ln2)/t1/2 = (ln2)/(8.1 d) = 0.0856 d-1

ks = sedimentation rate constant = 0.02 d-1

c. The mass balance for the CSTR control volume is

( )esedeaea - )r( - dd CkCkVQCQCCVQCQC

tCV +−=+=

Assuming Ca = 0 and integrating yields

( )

⎟⎠⎞

⎜⎝⎛

×

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

++×

⎟⎠⎞

⎜⎝⎛ ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛×

=⎟⎠⎞⎜

⎝⎛ ++

= ppmv101

11.0 ln

0.02d 0.0856d m 102.0

dmin 2460

minm 10 6

1 ln

1 5-

1-1-36

35

0

sdCC

kkVQ

t

t = 0.0322 d = 46.4 min

1.22 a.

( ) NaOH mg 6.53 L 00.1mmole

mg40mg 98

mmoleL

mg 8.002 NaOH =⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=m

b. The reaction (P1.3) is second-order (see the rate constant’s units) so that

2HOHH

H kC- - d

d- ++

+

== CkCt

C

After integration and noting that at t = t1/2, C = C0/2, the equation can be written

seconds 10 4.38

mg 98mmole

Lmg 00.82

sLmol 10 1.4

1 1 -8

1101/2 ×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅×

==kC

t

Therefore, neutralization occurs almost instantly.

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

So, 2

0

2

20

0

2

0

H gAl g 9

H g 2H mole

Al moleAl g 27

H moles 3Al moles 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+] [SO4

2-] = [SO42-]2

At saturation: [SO4

2-] = Ksp½ = (2 x 10-5) ½ = 0.00447 mol/L

Check if saturation is exceeded: mol/L 0.00368 g 136

mole L

g 5.0=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

Since saturation is not exceeded the sulfate concentration is 0.00368 mol/L

2.23 Assume a closed system. a. [Ca2+] = Ctot = [H2CO3] + [HCO3

-] + [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.

[ ] [ ][ ] ( )( )( ) M 10 M 10 2.24

10 4.4710 10 HCO H CO 7.65--8

7-

-5.5-8.5

1

3(aq)2 =×=

×==

−+

K

c.

[ ] ( )( ) g/L 0.20 g/L 0.19968 mol

g 100 M 10 - M 10 2 CaCO 5.5-3-(s) 3 ==⎟

⎠⎞

⎜⎝⎛×=

2.24 - 2.36 The solutions for these problems are the solutions for problems 2.20 - 2.32 in the 2nd 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.

rabbits 1,750 2rabbits 3,500

2 * ===KN

Use the midpoint of the first year population (350 rabbits) as N0

( ) ( )

( )

1-

00

0

yr 0.635

rabbits 3,500rabbits 350 - 1rabbits 350

yrrabbits 200

- 1

dd

=

⎥⎦

⎤⎢⎣

⎡=

⎟⎠⎞⎜

⎝⎛

=

KNN

tN

r

and ( )( ) rabbits/yr 556 4

rabbits 500,3yr 0.635 4

d

d -1*

===rK

tN

b.

( )( ) ( )( ) rabbits/yr 501

rabbits 3,500rabbits 1,200 - 1rabbits 200,1yr 0.635 - 1

dd 1-0

00 =⎥

⎦

⎤⎢⎣

⎡=⎟

⎠⎞

⎜⎝⎛=

KNrN

tN

3.19 a.

deer 3,500 2

deer 7,000 2

* ===KN

b. Use the mean current population of deer during the year (2,350 deer) as N0

( ) ( )

( )

1-

00

0

yr 0.192

deer 7,000deer 2,350 - 1deer 2,350

yrdeer 300

- 1

dd

=

⎥⎦

⎤⎢⎣

⎡=

⎟⎠⎞⎜

⎝⎛

=

KNN

tN

r

and ( )( ) deer/yr 336 4

deer 000,7yr 0.192 4

d

d -1*

===rK

tN

c.

( )( ) ( )

( )( )( )( ) yr 3.55

deer 3,500 - 000,7deer 2,350deer 2350 - 000,7deer 3,500ln

yr 0.1921

- - ln 1 1-

0

0 =⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

NKNNKN

rt

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.

( ) 3-10 1.05 rems 8,000deathcancer 1 yr 70

yrrem 0.12 risk Denver ×=⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

( ) 3-10 0.35 rems 8,000deathcancer 1 yr 70

yrrem 0.04 risk level Sea ×=⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

b. Denver deaths due to cosmic radiation exposure:

( ) yrdeaths 8.6

rems 000,8death 1 people 10 57.0

yr personrem 12.0 6 =⎟

⎠

⎞⎜⎝

⎛×⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

( ) yrdeaths 1,077

people 000,100yr

deaths 189 people 10 0.57 Expected 6 =

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛×=

c.

( ) ⎟⎠

⎞⎜⎝

⎛⎟⎠⎞⎜

⎝⎛==

rems 8,000deathcancer 1 yr N yr

rem 0.04 - 0.12 10 risk lIncrementa 6-

months) 2 suggests 4.3 (Tableyr 0.1 0.08

10 8,000 N-6

=×

=

d.

( ) yrdeaths 1,500

rems 000,8deathcancer 1

yrperson rem 0.04 people 10 300 6 =⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×

4.12 Radon exposure of 1.5 piC/L equivalent to 400 mrem/yr (0.4 rem/yr): a.

( ) yrdeaths 15,000 people 10 300

rems 8,000deathcancer 1

yrperson rem 4.0 6 =×⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

b.

lifetimedeathcancer 10 3.5

lifetimeyr 70

rems 8,000deathcancer 1

yrrem 0.4 Risk 3-×=⎟

⎠⎞

⎜⎝⎛⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

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. ( )( )( )( )( )( )( )durationfreqrate intakePF

yrd 365lifespanbody wtrisk

conc⎟⎠⎞⎜

⎝⎛

=

( )( )

34-32-

6-

mg/m 10 4.93

lifeyr 25

yrd 250

dm 20

mgdkg 10 2.9

yrd 365

lifeyr 70 kg 70 10

×=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅×

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=

b. ( )( )( )( )( )( )( )durationconcrate intakePF

yrd 365lifespanbody wtrisk

freq⎟⎠⎞⎜

⎝⎛

=

( )( )

d/yr 52

lifeyr 25

mmg 10 4.93

dm 20

mgdkg 10 2.9

yrd 365

lifeyr 70 kg 70 10

3

4-32-

7-

=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ×⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅×

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=

4.16 a. Use Table 4.11 oral RfDs and Table 4.10 for other factors HItotal = HIarsenic + HImethylene chloride

3.5 mg 0.060dkg

Lmg 0.560

mg 0.0003dkg

Lmg 0.070

kg 70d

L 1 HI =⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅⎟⎠⎞

⎜⎝⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

b. No, this is not a safe level, since the HI > 1. c. Use Table 4.9 for PFs and Table 4.10 for other factors risk = riskarsenic + riskmethylene chloride

( )⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅+⎟

⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=L

mg 0.560 mg

dkg 0075.0 L

mg 0.0700 mg

dkg 75.1

yrd 365

lifeyr 70kg 70

lifeyr 10

yrd 250

dL 1

risk

-410 1.8 risk ×= Since this risk is » 10-6, it would probably not be considered acceptable

4.17 a.

( )

5-

3

3

10 7

yrd 365

lifeyr 70 kg 70

lifeyr 25

yrd 250

dm 20

mmg 0.00002

mgdkg 50

risk ×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

=

b. ( ) ( ) workers15 10 7 workers220,000 -5 =×

c. The 2006 U.S. cancer rate of 24% taken from text in Section 4.2 is the same as the 1992 rate given in Table 4.1

( )( ) workers52,815 workers15 24.0 workers000,220 =+

d. From Table 4.11 RfDs for both arsenic and mercury are 0.0003 mg/kg·d

( )

0.11 m

mg 0.0001 m

mg 0.00002 kg70

dkgmg 0.0003dm 20

HI 33

3

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

Since the HI is less than 1 the non-cancer hazard is acceptable, despite there being a cancer risk greater than 10-6.

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 yr70 kg x 365 d/yr x 70 yr

x 1.1x10-2

mg/kg - d= 6.6x10−7

Δ cancer = 260x106 people x 6.6x10-7cancer / person − life

70 yr/lifetime= 2.5 cancer/yr

b. benzene at 0.005 mg/L:

Risk = 2L/d x 0.005 mg/L x 360 d/yr x 30 yr

70 kg x 365 d/yr x 70 yrx 2.9x10-2

mg/kg - d= 1.75x10−6

Δ cancer = 260x106 people x 1.75x10-6cancer / person − life

70 yr/lifetime= 6.5 cancer/yr

c. arsenic at 0.01 mg/L:

cancer/yr 785

eyr/lifetim 70lifeperson/cancer2.11x10 x people 260x10 =cancer

10x11.2d-mg/kg

1.75x yr 70d/yr x 365 x kg 70

yr 30d/yr x 360 x mg/L 0.01 x 2L/dRisk

4-6

4

=−

Δ

== −

d. carbon tetrachloride at 0.005 mg/L:

Risk = 2L/d x 0.005 mg/L x 360 d/yr x 30 yr70 kg x 365 d/yr x 70 yr

x 0.13mg/kg - d

= 7.9x10−6

Δ cancer = 260x106 people x 7.9x10 -6cancer / person − life

70 yr/lifetime= 29 cancer/yr

e. vinyl chloride at 0.002 mg/L:

Risk = 2L/d x 0.002 mg/L x 360 d/yr x 30 yr70 kg x 365 d/yr x 70 yr

x 2.3mg/kg - d

= 5.6x10−5

Δ cancer = 260x106 people x 5.6x10 -5cancer / person − life

70 yr/lifetime= 206 cancer/yr

f. PCBs at 0.0005 mg/L:

Risk = 2L/d x 0.0005 mg/L x 360 d/yr x 30 yr70 kg x 365 d/yr x 70 yr

x 7.7mg/kg - d

= 4.6x10−5

Δ cancer = 260x106 people x 4.6x10 -5 cancer / person − life

70 yr/lifetime=172 cancer/yr

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:

14

13.3

03.012.040.0

PNC

==

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

( ) ( )( )( )

( ) dm 61.2

0.0005d

m 09.00.34

dd

dd ====

Lhv'

LhvK η

b. tcontaminan

R v

v'= so d

m 0.015 6

dm 0.09

tcontaminan ===Rv'v

and the distance traveled is (0.015 m/d) (365 d) = 5.48 m c. Calculate the maximum contaminant mass dissolved based on the aqueous

solubility and density of PCE (use Table 5.14)

( ) ( ) ( ) ( ) g 2,230 m 5.48 m 4.0 m 2.0 0.34 mg 150 3max =⎟

⎠⎞⎜

⎝⎛=m

Calculate the mass leaked

( ) ( ) g 59,500 d 365 g 1.63 dmL 100 leaked =⎟

⎠⎞⎜

⎝⎛= mLm

Since the mass leaked > soluble mass, the solubility limit is exceeded and the guideline for the aqueous concentration in the presence of NAPLs (see page 258) is used: CPCE = 0.10 Aq SolPCE = 15 mg/L

d. ( ) ( )( ) ( ) d

m 0.490 dm 0.09 0.34 m 4 2 m 2 2

3capture === BvwQ

e.

( )( ) ( ) yr 22.2 d 8,100

mgg 10m

L 10Lmg 15d

m 0.490

g 59,500 3

333

PCE

PCE ==⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛

==−QC

mt

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:

( ) ( )( )

hr 4.17 d 0.174 gal 7.4805

ftgal/d 10 2.5

ft 0.10ft 43.0 36

2

==⎟⎠⎞⎜

⎝⎛×

==πθ

QV

b. The critical velocity is:

( )

( ) hrftft 2.40

ft 43.0hr 24

dgal 7.4804

ftd

gal 10 2.5 2

3

2

36

bo ⋅

=⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛ ×

==πA

Qv

c. The weir loading rate is:

( )( ) hrft

ft 51.5 ft 43.02

hr 24d

gal 7.4804ft

dgal 10 2.5

rate loadingweir 3

36

w ⋅=

⋅⋅

⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛ ×

==πL

Q

6.4 From Appendix C: At 20°C, ρ = 998.2 kg·m-3 and μ = 0.00100 kg·m-1·s-1

a. The particle settling velocity is:

( ) ( )( )( )( )s

m 10 2.18

smkg 00100.018

m 10 0.1kg/m 2.9984.0m/s 9.807 18

- 5-

2-5322pp

s ×=⎟⎠⎞

⎜⎝⎛

⋅

×==

μρρ dg

v

b.

( )

( )( ) sm 10 2.89

m 30m 10s 606024

dd/m 7,500 4-

3

bo ×=

⎟⎠⎞

⎜⎝⎛

⋅⋅==AQv

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,

( ) ( ) sμm 36.5

sm 10 3.65

mkg 2.9984.0

sm 9.807

sm10 89.2

smkg0.0010018

-

18 5-

21

32

4-2

1

p

sp =×=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ ×⎟⎠⎞

⎜⎝⎛

⋅=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

ρρμ

gvd

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:

( ) ( )m 42.7

m 10mkg 998.2 - 200,1

sm 9.807 5

smkg 00100.0

sm 0.10018

d - 5

18

21

2532

3

21

2p

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−ρρμ

gQw

6.6 From Appendix C: At 15°C, μ = 0.00114 kg·m-1·s-1

a. b

VPGμ

=

( )( ) ( )

1-

21

21 s 50.0

m 75.3m 17.4sm

kg 0.00114

W186 =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛

⋅

=G

Similarly, G2 = 20.1 s-1 and G3 = 10.0 s-1. b.

( ) ( )( )

4-3

6-153-6

03p 10 7.54

6m

mL 10mL 10 8.1m 10 0.20

6 ×=

××==Ωππ Nd

QGV

NN

πα b0 4 1 Ω

+=

( ) ( )( )( ) ( ) 17.9

s 606024d

dm16,000

m 75.3m 17.4s 0.5010 54.7 4 1.0 1 3

2-1-40 =

⎟⎠⎞

⎜⎝⎛

⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛×

+=π

NN

Therefore, there will be on average 18 singlets per aggregate.

6.7 From Appendix C: At 15°C, μ = 0.00114 kg·m-1·s-1

a.

Accumulation = Reaction: )r( NVdtdNV =

and π

α GNN Ω=

4- )r(

so, ∫∫Ω

=tN

N

dtGN

dN

0

4- 0

πα

which yields: ⎟⎠⎞

⎜⎝⎛ Ω

= πα Gt

NN4 -

0e b.

( )

1-

21

3

3

25-

s 3.78 m 004.0

smkg 0.00114

smkg 10 5.6

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛

⋅

⋅×

==VPGμ

c.

( )

4-3

735-

03p 10 1.64

6m 0040.0

10 0.1m 10 0.5

6 ×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××

==Ωπ

π Nd

( ) ( )( )( )( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ××==

⎟⎠⎞

⎜⎝⎛ Ω

ππ

α min 30mins 60s 78.310 64.141.0

- exp L 4.010 1.0 e

1-4-74 -

0

Gt

NN

L

particles 10 6.07 5×=N

d.

aggregatesinglets 4.12

L 10 07.6L 0.410 1.0

1-5

7

0 =×

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×

=NN

e. ( ) ( )( )( )( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ××==

⎟⎠⎞

⎜⎝⎛ Ω

ππ

α min 30mins 60s 78.310 64.140.2

- exp L 4.010 1.0 e

1-4-74 -

0

Gt

NN

L

particles 10 1.88 6×=N

6.8 a. ( )hrm 640 m 80

hrm 8.0

32

fa =⎟⎠⎞

⎜⎝⎛== AvQ

b. From the definition of filter efficiency: Vf – (Vb + Vr) = ηfVf (6.8.1) and the definition of the effective filtration rate gives: Vf – (Vb + Vr) = refAftc (6.8.2) Combining (6.8.1) and (6.8.2) and rearranging yields:

( )( )( )

cycle)(per m 33,370 0.96

hr 52m 80hrm 7.7

32

f

cfeff ===

ηtArv

From (6.8.1): ( ) ( )( ) cycle)(per m 1,335 0.96 - 1m 370,33 - 1 33

ffrb ===+ ηVVV Thus, the volume of water lost per cycle (Vb + Vr) is 1,340 m3.

6.9 Note that this problem incorporates a practically realistic, yet possibly confusing, nuance in that the total backwash time per cycle (tb) is double the total time that backwash water is flowing through the filter (tb’). Thus, calculation of the water volume used in backwash is based on 8 minutes of flow per cycle, whereas the time the filter is off-line for backwashing is 16 minutes.

a.

( )( )( ) L 307,200 mins 60min 0.8m 64

smL 10.0 22

'bf

f

bb =⎟

⎠⎞

⎜⎝⎛

⋅== tA

AQV

( )( )( ) L 316,800 mins 60min 15m 64

smL 5.50 22rfar =⎟

⎠⎞

⎜⎝⎛

⋅== tAvV

( ) ( ) ( ) min 1,409 min 15 - min 16 min 6024 rbcf =−⋅=−−= tttt

( )( )( ) L 10 2.98 mins 60min ,4091m 64

smL 5.50 722f

f

ff '

f×=⎟

⎠⎞

⎜⎝⎛

⋅== tA

AQV

% 98 0.979 L 10 2.98

L 316,800 - 307,200 - 10 2.98 7

7

f

rbff ==

××

=−−

=V

VVVη

b. The fraction of a filtration cycle that is not backwashing is:

( )

0.989 hr 24

min 60hrmin 16 -hr 24

c

brf =⎟⎠⎞

⎜⎝⎛

=−+

tttt

so the number of filters required is:

( )( )( ) ( )

6.89 989.0

L 1000mm 64

smL 5.50

sm 2.40

989.0

3

22

3

fa

plant =

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅

==AvQ

n

Therefore, at least 7 filters are required.

6.10 a. For a steady state PFR, the Giardia concentration in the contactor effluent is: θ*-

0e kNN =

so, ( )( )

( ) 31-

3

0*b m 156 000,1ln

min0.53min

s 60sm 0.20

ln ===NN

kQV

b. For a steady state CSTR, the mass balance is:

QN0 = QN + k*NVb

so, ( )( )

( ) ( ) 31-

3

0*b m 22,600 1 - 000,1

min 0.53min

s 60sm 0.20

1 - ==⎟⎠⎞

⎜⎝⎛=

NN

kQV

c. For a steady state series of CSTRs, the mass balance is:

( )m*0

m

11 kN

Nθ+

=

so, ( )( )

( ) ( )( ) m 67.5 1 - 000,1min 0.53

mins 60s

m 0.20 1 - 35

1

1-

31

m

0*b,1 ==

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

m

NN

kQV

and the total contactor volume is: mVb,1 = 5(67.50 m3) = 337 m3

d.

minmg

L 0.265

Lmg 2.0

min 0.53 0.1

-1*

⋅=

⎟⎠⎞⎜

⎝⎛

== nCkk

6.11 For a steady state CSTR, the mass balance is: QN0 = QN + k*NVb

so, ( )( )( ) s

L0.0156 1 - 1,000

s 80.7L 2.00 1 -

-1

0

*b ==

⎟⎠⎞⎜

⎝⎛

=

NN

kVQ

and the volume produced during ten hours of operation would be:

( ) L 562 hr

s 6060hr 10sL 0.0156 =⎟

⎠⎞

⎜⎝⎛ ⋅

⎟⎠⎞

⎜⎝⎛=Qt

6.12 Hardness in meq/L:

meq/L 7.50 mg/meq 20.0mg/L 150 Ca 2 ==+

meq/L 4.92 mg/meq 12.2mg/L 60 Mg2 ==+

Total hardness = 7.50 + 4.92 = 12.4 meq/L

Hardness as CaCO3:

33 CaCO as mg/L 621

meqCaCO as mg 50.0

Lmeq 12.4 hardness Total =×=

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)

Equiv. Weight (mg/meq)

Concentration (meq/L)

CO2 14.5 22 0.6591 Ca2+ 110.0 20 5.500 Mg2+ 50.7 12.2 4.156 Na+ 75 23.0 3.261

HCO3- 350 61.0 5.738

SO42- 85.5 48.0 1.781

Cl- 16.2 35.5 0.4563 pH 8.2 - -

9.65

6

5.73

8

0 5.50

0

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

M g-NCH

9.65

6

5.73

8

0 5.50

0

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

M g-NCH

a. Mg-CH = CH – Ca-CH = 5.738 –5.500 = 0.238 meq/L

b. Mg-NCH = TH – CH = 9.656 – 5.738 = 3.918 meq/L

c. 1 meq/L Ca(OH)2 neutralizes 1 meq/L CO2(aq)

so, Ca(OH)2 required = 0.6591 meq/L

6.18 Component Concentration

Equiv. Weight

(mg/meq) Concentration

(meq/L) CO2 6.0 mg/L 22 0.2727

Ca2+ 50.0 mg/L 20 2.500

Mg2+ 20.0 mg/L 12.2 1.639

HCO3- 3.1 mmole/L 61.0 3.100

pH 7.6 - -

4.13

9

3.10

0

0 2.50

0

0

C a2+

HCO 3-

CH

M g2+

Ca-CH

NCH

M g-NCH

4.13

9

3.10

0

0 2.50

0

0

C a2+

HCO 3-

CH

M g2+

Ca-CH

NCH

M g-NCH

Table P6.18 Components, lime and soda ash dosage, and solids generated Component (eq’n.) Concentration

(meq/L) Lime

(meq/L) Soda ash (meq/L)

CaCO3(s)(meq/L)

Mg(OH)2(s)(meq/L)

CO2(aq) (6.37) 0.2727 0.2727 0 0.2727 0 Ca-CH (6.38) 2.500 2.500 0 5.000 0 Mg-CH (6.39) 0.600 1.200 0 1.200 0.600 Ca-NCH (6.40) 0 0 0 0 0 Mg-NCH (6.41) 1.039 1.039 1.039 1.039 1.039

Excess 0.400 Totals 5.412 1.039 7.512 1.639

a. Lime required = (5.412 meq/L)(37.07 mg/meq) = 200.5 mg/L Soda ash required = (1.039 meq/L)(53.0 mg/meq) = 55.1 mg/L b. Sludge generated:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ×⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

mg 10kg

galL 785.3

dgal 10 15

meqmg 9.22

Lmeq .6391

meqmg 0.50

Lmeq 512.7 6

6

kg/d 24,040 =

6.19 Component Concentration

Equiv. Weight

(mg/meq) Concentration

(meq/L) Concentration (mg/L CaCO3)

Ca2+ 90 mg/L 20 4.500 225.0

Mg2+ 30 mg/L 12.2 2.4596 123.0

HCO3- 165 mg/L 61.0 2.705 135.3

pH 7.5 - - -

348.

0

135.

3

0 225.

0

0

C a2+

H C O 3-

CH

M g2+

C a-CH

N C H

M g-N CHC a-N CH

348.

0

135.

3

0 225.

0

0

C a2+

H C O 3-

CH

M g2+

C a-CH

N C H

M g-N CHC a-N CH

( )[ ] [ ][ ]3

4-7-

5.7

1

-3

aq2 CaCO mg/L 19.14 M 10 1.914 M 10 4.47

Lmol 10

g 61.0mol

Lg 0.165

HHCO CO =×=×

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

==

−+

K

Table P6.19 Components, lime and soda ash dosage, and solids generated Component (eq’n.) Concentration

(mg/L CaCO3)

Lime (mg/L

CaCO3)

Soda ash (mg/L

CaCO3)

CaCO3(s)(mg/L

CaCO3)

Mg(OH)2(s)(mg/L

CaCO3) CO2(aq) (6.37) 19.14 19.14 0 19.14 0 Ca-CH (6.38) 135.3 135.3 0 270.6 0 Mg-CH (6.39) 0 0 0 0 0 Ca-NCH (6.40) 90.00 0 90.00 90.00 Mg-NCH (6.41) 123.0 123.0 123.0 123.0 123.0

Excess 20 Totals 297.4 213.0

a. Lmg 220.1

CaCO mg 100Ca(OH) mg 74

LCaCO mg 297.4 Lime

3

23required =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

b. Lmg 225.8

CaCO mg 100Ca(OH) mg 061

LCaCO mg 213.0 Ash Soda

3

23required =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

6.20 The solution for this problem is the solution for problem 6.6 in the 2nd edition Solutions Manual. 6.21

Component Concentration

Equiv. Weight (mg/meq)

Concentration (meq/L)

Concentration (mg/L CaCO3)

Ca2+ 95 mg/L 20 4.75 238

Mg2+ 26 mg/L 12.2 2.13 107

HCO3- 160 mg/L 61.0 2.62 131

pH 7.0 - - -

6.88

2.62

0 4.75

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

M g-NCHCa-NCH

6.88

2.62

0 4.75

0

Ca2+

HCO 3-

CH

M g2+

Ca-CH

NCH

M g-NCHCa-NCH

( )[ ] [ ][ ] meq/L 1.17 M 10 5.868

M 10 4.47L

mol 10g 61.0

molLg 0.160

HHCO CO 4-7-

7

1

-3

aq2 =×=×

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

==

−+

K Table P6.21 Components, lime and soda ash dosage, and solids generated

Component (eq’n.) Concentration (meq/L)

Lime (meq/L)

Soda ash (meq/L)

CaCO3(s)(meq/L)

Mg(OH)2(s)(meq/L)

CO2(aq) (6.37) 1.17 1.17 0 1.17 0 Ca-CH (6.38) 2.62 2.62 0 5.24 0 Mg-CH (6.39) 0 0 0 0 0 Ca-NCH (6.40) 2.13 0 2.13 2.13 0 Mg-NCH (6.41) 2.13 2.13 2.13 2.13 2.13

Excess 0.400 Totals 6.32 4.26 10.67 2.13

a. Lmg 234

meqmg 7.073

Lmeq 6.32 Limerequired =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

b. Lmg 226

meqmg 35

Lmeq 4.26 Ash Soda required =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

6.22 The solution for this problem is the solution for problem 6.7 in the 2nd edition Solutions Manual. 6.23

Component Concentration

Equiv. Weight (mg/meq)

Concentration (meq/L)

Concentration (mg/L CaCO3)

Ca2+ 40.0 mg/L 20 2.00 100

Mg2+ 10.0 mg/L 12.2 0.820 41.0

HCO3- 110 mg/L 61.0 1.80 90.2

pH 7.0 (assumed) - - -

2.82

1.80

0 2.00

0

Ca2+

HCO3-

CH

Mg2+

Ca-CH

NCH

Mg-NCHCa-NCH

2.82

1.80

0 2.00

0

Ca2+

HCO3-

CH

Mg2+

Ca-CH

NCH

Mg-NCHCa-NCH

( )[ ] [ ][ ] meq/L 0.807 M 10 4.034

M 10 4.47L

mol 10g 61.0

molLg 0.110

HHCO CO 4-7-

7

1

-3

aq2 =×=×

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

==

−+

K Table P6.23 Components, lime and soda ash dosage, and solids generated

Component (eq’n.) Concentration (meq/L)

Lime (meq/L)

Soda ash (meq/L)

CaCO3(s)(meq/L)

Mg(OH)2(s)(meq/L)

CO2(aq) (6.37) 0.807 0.807 0 0.807 0 Ca-CH (6.38) 1.80 1.80 0 3.60 0 Mg-CH (6.39) 0 0 0 0 0 Ca-NCH (6.40) 0.200 0 0.200 0.200 0 Mg-NCH (6.41) 0.820 0.820 0.820 0.820 0.820

Excess 0.400 Totals 3.83 1.02 5.43 0.82

a. Lmg 142

meqmg 7.073

Lmeq 3.83 Limerequired =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

b. Lmg 54.1

meqmg 35

Lmeq 1.02 Ash Soda required =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

c. Sludge generated:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ×⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

mg 10kg

galL 785.3

dgal 10 73

meqmg 9.22

Lmeq .820

meqmg 0.50

Lmeq .435 6

6

kg/d 41,400 =

6.24 Qf = 5 ×106 L/d Qp = 3 ×106 L/d Cf = 1,500 mg/L Cp = 75 mg/L

Qc, Cc

QfCf = QcCc + QpCp and Qf = Qc + Qp

So,

( )( )

( )( ) ( )( )[ ]( ) mg/L 3,640

L/d 10 3 - 10 5mg/L 57L/d 10 3 - mg/L 500,1L/d 10 5

- -

66

66

pf

ppffc =

××××

==QQ

CQCQC

6.25 Qf = 5 ×106 L/d Qp = 3 ×106 L/d Cf = 1,500 mg/L Cp = 75 mg/L

Cc = 3,640 mg/L Qc = 2 ×106 L/d

a. Water recovery:

60%or 0.60 d

L 10 5d

L 10 3

6

6

F

P =×

×==

QQr

b. Salt rejection:

95%or 0.95

Lmg 1,500

Lmg 75

- 1 - 1 F

P ===CCR

c. Log salt rejection: ( ) ( ) 1.30 0.95 - 1log- - 1log-

log=== RR

6.26 a. The main constituent of concern from the perspective of configuring the POTW treatment train is biodegradable organic matter. It is, however, arguable from the perspective of human health that the main constituents of concern are human pathogens.

b. Unit operation function based on an overall aim of removing BOD:

i. The grit chamber removes the very largest and most settleable particles, which may contain a modest fraction of organic matter.

ii. The primary sedimentation basin removes most of the gravitationally settleable organic matter (as well as inorganic matter). Typically about 35% of the BOD can be removed by primary sedimentation.

iii. The bioreactor converts dissolved and fine particulate biodegradable organic matter into microbial cell mass and energy for microbial metabolism.

iv. The secondary clarifier physically removes the cell mass generated in the bioreactor by gravitational settling.

v. Digestors further degrade the organic particles in the primary clarifier sludge and/or the microbial cell mass separated into the secondary clarifier sludge by exposing it to more prolonged biodegradation.

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

well-mixed aeration pond

Q, S0, X0 Q, S, X

V, S, X, rg’, rsu

Q = 30 m3/d S0 = 350 mg/L BOD5S = 20 mg/L BOD5Ks = 100 mg/L BOD5kd = 0.10 d-1

μm = 1.6 d-1

Y = 0.60 mg VSS/mg BOD5

a. Assuming X0 = 0, the microbe mass balance yields equation 6.57:

( )( )dm

ds

- 1 - 1

kkKSθθμθ+

=

and solving for the hydraulic detention time, θ,

( )[ ]( )( ) ( )[ ] d 6.0

d 1.0BOD mg/L 100 - d 0.1 - 6.1BOD mg/L 20BOD mg/L 20 100

- - 1-

51-

5

5

sdm

s =+

=+

=dkKkS

SKθμ

b. From equation 6.60 and 6.51,

( )( )

( )( ) LVSS mg 7.07

LBOD mg 350d 6.1d 6.0

LBOD mg 20 - 350

LBOD mg 20 100

BOD mgVSS mg 0.60

- 51-

55

50s =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ +⎟⎟⎠

⎞⎜⎜⎝

⎛

=+

=KS

SSSKYXθ

and at steady state with X0 = 0, the effluent flux of microbial mass must equal the rate of microbial mass production in the pond: '

gr VQX =

and, dVSS kg 0.212

mg 01kg

LVSS mg 07.7

mL 000,1

dm 30 63

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=QX

6.31

well-mixed aeration lagoon

Q, S0, X0 Q, S, X

V, S, X, rg’, rsu

l = 60 m w = 5 m h = 2 m Q = 30 m3/d S0 = 350 mg/L BOD5S = 20 mg/L BOD5Ks = 100 mg/L BOD5kd = 0.10 d-1

μm = 1.6 d-1 a. The lagoon’s hydraulic detention time is:

( )( )( ) d 1.50 d

m 400

m 2m 5m 60 3 ===QVθ

and the steady state substrate concentration in the lagoon is:

( )( )

( )( )( )( )( ) ( )( )[ ] L

BOD mg 161 d 08.0d 1.5 - 1 - d 10.1d 1.5

d 08.0d 1.5 1LBOD mg 76.0

- 1 -

1 51-1-

1-5

dm

ds =+⎟

⎠⎞

⎜⎝⎛

=+

=k

kKSθθμθ

Therefore, the lagoon’s BOD5 removal efficiency is:

( ) ( ) 52.2% 100

LBOD mg 336

LBOD mg 161 - 336

100 - 5

5

0

0 =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

==S

SSEff

c. The lagoon’s daily oxygen demand is:

( )dO kg 70.2

mg 10kg

mL 000,1

LBOD mg 161 - 336

dm 400 - 2

635

3

0 =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=SSQ

6.32

Q, S0 Qs Qe, Xe, S0

WAS Qw, Sw, Xw

RAS Qr, Sr, Xr

V, X, S P.C.

Act. Sludge Basin

S.C. Q = 0.300 m3/s X = 2,100 mg VSS/L Xr = 10,000 mg VSS/L θc = 9.0 d S0 = 220 mg BOD5/L

dVSS mg

BOD mg 0.52 5

⋅=M

F

a. Based on the definition of the food to microbe ratio,

( )3

5

53

0 m 5,220

LVSS mg 100,2

dVSS mgBOD mg 0.52

ds 606024

LBOD mg 220

sm 0.300

=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛ ⋅⋅⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

=XM

FQSV

b. Noting that Xr = Xw, the cell retention time can be used to calculate the WAS flow rate.

( )

( ) dm 122

LVSS mg 000,10d 9.0

LVSS mg 100,2m 5220

3

3

wcw =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

==X

VXQθ

c. From a hydraulic flow balance around the activated sludge basin, recycle line and secondary clarifier,

s

m 0.299 s 606024

dd

m 122 - s

m 0.300 - 333

we =⎟⎠⎞

⎜⎝⎛

⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛== QQQ

d. First, write a microbial mass balance around the secondary clarifier (S.C.). QsX = QeXe + (Qr + Qw)Xw Noting that Xe ≈ 0 and Qr = Qs – Qe - Qw, simplify and solve for Qs.

( ) sm 0.378

LVSS mg 2,100 - 10,000

LVSS mg 000,10

sm 0.299

-

3

3

w

wes =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

==XX

XQQ

and the hydraulic detention time is then:

( ) d 0.160

ds 606024

sm 0.378

m 5220 3

3

s

=

⎟⎠⎞

⎜⎝⎛ ⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛==

QVθ

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:

( ) ( )( ) ( ) ( ) ⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=+=

LTS mg 8000.82

LTS mg 8000.82 - 800

MGD 0.070ΜGD 20 - ee0

ss CCC

QQC

LTS mg 41,800 s =C

b. The mass of solids removed annually is:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==

yrs 365606024

mg 10kg

Lmg 800,41

mL 1,000

MGDs

m 0.04381MGD 0.070 63

3

ss tCQm

yearper tonne4,040 yrTS kg 10 4.04 6 =×=m

Pg. 7.1

SOLUTIONS FOR CHAPTER 7

7.1 From (1.9), mg / m3 =ppm x mol wt

24.465 (at 1 atm and 25 o C)

a. CO2mg / m3 =5000ppm x 12 + 2x16( )

24.465 = 8992mg/m 3 ≈ 9000mg / m3

b. HCHO ppm =24.465 x 3.6 mg/m3

2x1 + 12 + 16( )= 2.94ppm

c. NO mg / m3 =25ppm x 14 +16( )

24.465 = 30.7mg/m 3

7.2 70% efficient scrubber, find S emission rate:

600 MWeη = 0.38

600/0.38=1579 MWt

9000 Btu/lb coal 1% S

Input = 600, 000 kWe0.38

x 3412 BtukWhr

x lb coal9000Btu

x 0.01 lb Slb coal

= 5986 lb S/hr

70% efficient, says release 0.3 x 5986 lbS/hr = 1796 lb S/hr ≈1800 lbS/hr

7.3 If all S converted to SO2 and now using a 90% efficient scrubber:

SO2 = 0.1 x 5986 lbShr

x (32 + 2x16) lb SO2

32 lb S= 1197 lb SO2 / hr ≈ 1200 lb SO2 / hr

7.4 70% scrubber, 0.6 lb SO2/106 Btu in, find % S allowable:

a. X lbs Slbs coal

x 0.3 lbs S out1 lb S in

x 2 lbs SO2

lb Sx lb coal

15, 000Btu=

0.6 lb SO2

106 Btu

X =15,000x0.60.3x2x106 = 0.015 = 1.5% S fuel

b. X lbs Slbs coal

x 0.3 lbs S out1 lb S in

x 2 lbs SO2

lb Sx lb coal

9, 000Btu=

0.6 lb SO2

106 Btu

X =9, 000x0.60.3x2x106 = 0.009 = 0.9% S fuel

Pg. 7.2

7.5 Compliance coal:

1.2 lbs SO2

106 Btu=

lb coal12, 000 Btu

x X lb Slb coal

x 2 lb SO2

lb S

X =1.2x12, 0002x106

= 0.0072 = 0.7%S

7.6 Air Quality Index:_________________________________________________________ Pollutant Day 1 Day 2 Day 3_________________________________________________________O3, 1-hr (ppm) 0.15 0.22 0.12CO, 8-hr (ppm) 12 15 8PM2.5, 24-hr (µg/m3) 130 150 10PM1 0, 24-hr (µg/m3) 180 300 100SO2, 24-hr (ppm) 0.12 0.20 0.05NO2, 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 NO2c. 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

e−0.402t =55.465.4

= 0.871 so t = - 10.402

ln 0.871( ) = 0.41 hr

7.9 RH to produce HCHO:RO • +O2 → HO2 • +R' CHO (7.19)for R' CHO to be HCHO, R' must be H so thatRO • + O2 →HO2 • +HCHOfor the reaction to balance , R = CH3

which says RH in (7.16) must be CH4 (methane)

Pg. 7.3

7.10 RH = propene = CH2=CH-CH3 = C3H6 so, R = C3H5

so the sequence of reactions (7.16) to (7.19) is:

C3H6 +OH•→ C3H5 • +H2OC3H5 • +O2 → C3H5O2 •

C3H5O2 • +NO→C3H5O • +NO2

C3H5O • +O2 → HO2 • +C2H3CHO

The end product is acrolein, CH2CHCHO.

7.11 A 20-µm particle blown to 8000 m. From (7.24) its settling velocity is

v = d2ρg18η

=(20x10−6 m)2 x 1.5x106g / m3 x 9.80m/s2

18 x 0.0172g/m - s= 0.019m / s

Time to reach the ground = 8000m0.019m / s x 3600s/hr x 24hr/d

= 4.87days

Horizontal distance = 4.87 days x 10 m/s x 3600s/hr x 24hr/d x 10-3km/m = 4200 km

7.12 Residence time for 10-µm particle, unit density, at 1000m:

Settling velocity = v = d2ρg18η

=(10x10−6 m)2 x 106g / m3 x 9.80m/s2

18 x 0.0172g/m - s= 0.00317m / s

Residence time = τ = hv=

1000m0.00317m / s x 3600s/hr

= 87.6hrs

7.13 Settling velocity and Reynolds numbers:

a. 1 µm: v = d2ρg18η

=(1x10−6 m)2 x 106g / m3 x 9.80m/s2

18 x 0.0172g/m - s= 3.2x10−5 m / s

Re = ρair dvη

=1.29x103 g/ m3 x 1x10-6 m x 3.17x10-5m/s

0.0172 m/s= 2.4x10−6

b. 10 µm: v = d2ρg18η

=(10x10−6 m)2 x 106g / m3 x 9.80m/s2

18 x 0.0172g/m - s= 3.2x10−3 m / s

Re = ρair dvη

=1.29x103 g/ m3 x 10x10-6 m x 3.17x10-3m/s

0.0172 m/s= 2.4x10−3

c. 20 µm: v = d2ρg18η

=(20x10−6 m)2 x 106g / m3 x 9.80m/s2

18 x 0.0172g/m - s= 0.0127m / s

Pg. 7.4

Re = ρair dvη

=1.29x103 g/ m3 x 20x10-6 m x 0.0127m/s

0.0172 m/s= 0.02

So for all of these particles, the Reynolds number is much less than 1, which means (7.24) isa reasonable approximation to the settling velocity.

7.14 Finding the percentage by weight of oxygen and the fraction (by weight) of oxygenateneeded to provide 2% oxygen to the resulting blend of gasoline.

a. Ethanol, CH3CH2OH

€

OxygenEthanol

=16

2x12 + 6x1+1x16= 0.347 = 34.7%

€

0.02 =Xg oxygenate x %O in oxygenate

Yg fuel

€

EthanolFuel blend

=XY

=2%%O

=2%

34.7%= 0.058 = 5.8% by weight

b. Methyl tertiary butyl ether (MTBE), CH3OC(CH3)3

€

OxygenMTBE

=16

5x12 +12x1+1x16= 0.182 =18.2%

€

MTBEFuel

=2%%O

=2%18.2%

= 0.11=11%

c. Ethyl tertiary butyl ether (ETBE), CH3CH2OC(CH3)3

€

OxygenETBE

=16

6x12 +14x1+1x16= 0.157 =15.7%

€

ETBEFuel

=2%%O

=2%15.7%

= 0.127 =12.7%

d. Tertiary amyl methyl ether (TAME), CH3CH2C(CH3)2OCH3

€

OxygenTAME

=16

6x12 +14x1+1x16= 0.157 =15.7%

€

TAMEFuel

=2%%O

=2%15.7%

= 0.127 =12.7%

7.15 Ethanol fraction CH3CH2OH (sg = 0.791) in gasoline (sg = 0.739) to give 2% O2:

€

OxygenEthanol

=16

2x12 + 6x1+1x16= 0.347 = 34.7%

€

0.02 =X (mL eth) x 0.791 g eth /mL x 0.347 gO/g eth

X (mL eth) x 0.791 g eth/mL + Y (mL gas) x 0.739 g gas/mL

€

0.02 =0.2745X

0.791X + 0.739Y=

0.27450.791+ 0.739Y /X

Pg. 7.5

€

YX

=0.2745 − 0.02x0.791

0.02x0.739=17.50

€

ethanolfuel blend

=X(mL eth)

X(mL eth) + Y(mL gas)=

11+Y /X

=1

1+17.50= 5.4% by volume

7.16 The CAFE fuel efficiency for flex-fuel cars that get:

a. 18 mpg on gasoline and 12 mpg on ethanol. b. 22 mpg on gasoline and 15 mpg on ethanolc. 27 mpg on gasoline and 18 mpg on ethanol

€

a. CAFE mpg = 1 mile0.5 mile

18 mile/gal gas+

0.5 mile12 mile/gal alcohol

x 0.15 gal gas1 gal alcohol

= 29.4 mpg

€

b. CAFE mpg = 1 mile0.5 mile

22 mile/gal gas+

0.5 mile15 mile/gal alcohol

x 0.15 gal gas1 gal alcohol

= 36.1 mpg

€

c. CAFE mpg = 1 mile0.5 mile

27 mile/gal gas+

0.5 mile18 mile/gal alcohol

x 0.15 gal gas1 gal alcohol

= 44.1 mpg

7.17 At 25 miles/100 ft3 of natural gas, 0.823 gallons of gasoline equivalents per 100 ft3,and each equivalent gallon counting as 0.15 gallons of gasoline gives a CAFÉ rating of

€

100 ft3 ngas25 mile

x 0.823 equiv. gal gasoline 100 ft 3 ngas

x 0.15 gal gasoline1 equiv. gal gasoline

= 0.004938 gal gasoline/mile

€

CAFE mpg = 10.004938 gal/mile

= 202.5 miles/gallon ≈ 203 mpg

7.18 The “break-even” price of E85 with gasoline at $3.50/gallon:

From Table 7.5, the energy ratio E85/gasoline = 81,630/115,400 = 0.70736So 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 cheaperc. E85 at $2.75/gal or gasoline at $4/gal

E85 < $4.00/gallon x 0.71 = $2.84 E85 is cheaper

Pg. 7.6

7.20 With a 15-gallon fuel tank and 25 mpg on gasoline.

a. E10 = 0.10 x 75,670 Btu/gal + 0.90 x 115,400 Btu/gal = 111,427 Btu/gal

€

111,427115,400

x 25 mpg = 24.14 mpg

Range = 15 gal x 24.14 mi/gal = 362 miles

b. E85 = 0.85 x 75,670 + 0.15 x 115,400 = 81,630 Btu/gal

€

81,630115,400

x 25 mpg =17.7 mpg

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

€

98,515115,400

x 25 mpg = 21.3 mpg

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 perweek, 2 days @ 25 mi /d .

a. At $3.50 per gallon and $0.08/kWh:.

€

Electric = (30 mi/d x 5 d/wk +25 mi/d x 2 d/wk) x 0.25 kWh/mi x $0.08/kWh = $4/wk

€

Gasoline = (20 mi/d x 5 d/wk)45 mi/gal

x $3.50/gal = $7.78/wk

€

PHEV cost/mile = $7.78 +$4.00(50x5 +25x2) mi

= $0.0393 = 3.93¢/mile

Original 50 mpg

€

HEV cost/mile = $3.50/gallon50 mi/gal

= 7¢/mile

b. On an annual basis:PHEV annual cost = $0.0393/mi x 52 wk/yr x 300 mi/wk = $612/yr

€

HEV annual cost = $3.50/gallon50 mi/gal

x52wk/yr x 300mi/wk = $1092 / yr

Annual savings = $1092 - $612 = $480/yr

c. At $3000 for batteries:

€

Simple payback = Extra 1st costAnnual savings

=$3000

$480/yr= 6.25 yr

Batteries would need to last 6.25 yr x 300 mi/wk x 52 wk/yr = 97,500 miles

Pg. 7.7

7.22 From Figure 7.28 the well-to-wheels CO2/mile are:

a. 25 mpg car:

€

11.2 kg CO2/gal x 1000 g/kg25 mi/gal

= 448 gCO2/mile

b. 50 mpg car:

€

11.2 kg CO2/gal x 1000 g/kg50 mi/gal

= 224 gCO2/mile

c. PHEV:

€

gasoline : 11.2 kg CO2/gal x 1000 g/kg45 mi/gal

= 248 gCO2/mile

€

electricity : 0.25 kWh/mi x 640 gCO2/mi =160 gCO2/mile

€

Half gas, half electricity : 0.5 x 248 + 0.5 x 160 = 204 gCO2/mile

d. EV:

€

electricity : 0.25 kWh/mi x 640 gCO2/mi =160 gCO2/mile

e. FCV:

€

14.4 gC/MJ0.36 mi/MJ

x (12 +2x16)gCO2

12gCx MJ CH4

0.61 MJ H2

= 240 gCO2/mile

7.23 At 0.25 kWh/mi from a 60%-efficient NGCC plant with a 96%-efficient grid, 14.4gC/MJ of n. gas and 1 kWh = 3.6 MJ:

a. The EV carbon emissions would be

€

14.4 gC/MJ in 1 MJ CH4

x 44 gCO2

12gCx 1 MJ in

0.6 MJ outx 3.6 MJ out

kWhx 0.96 grid = 304 gCO2/kWh

0.25 kWh/mi x 304 gCO2/kWh = 76 gCO2/mi

b. For the PHEV, half miles on gasoline and half on electricity:

€

gasoline : 11.2 kg CO2/gal x 1000 g/kg45 mi/gal

= 248 gCO2/mile

electricity : 0.25 kWh/mi x 304 gCO2/kWh = 76 gCO2/mi

€

Half gas, half electricity : 0.5 x 248 + 0.5 x 76 = 162 gCO2/mile

7.24 With 5.5 hr/day of sun, 17%-efficient PVs, 75% dc-ac, 0.25 kWh/mile, 30 mi/day:

Electricity needed = 0.25 kWh/mile x 30 miles/day = 7.5 kWh/day

€

Area =7.5 kWh/d

5.5 h/d x 1 kW/m2 x 0.17 x 0.75=10.7 m2 =116 ft 2

Pg. 7.8

7.25 A 50%-efficient SOFC, 50% into electricity, 20% into useful heat, compared to 30%-efficient grid electricity and an 80% efficient boiler:

Assume 100 units of input energy to the SOFC, delivering 50 units of electricity and 20units of heat.

For the grid to provide 50 units of electricity: Input energy = 50/0.30 = 167 unitsFor the boiler to provide 20 units of heat: Input energy = 20/0.80 = 25 unitsTotal for the separated system = 167 + 25 = 192 unitsEnergy savings = (192-100)/192 = 0.48 = 48%

7.26 NG CHP versus separated systems; Natural gas 14.4 gC/MJ, grid 175 gC/kWh. Thejoule 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-firedpower plant for electricity and an 80% efficient gas-fired boiler for heat.

CHP: Assume 100 MJ input to the CHP:

Carbon emissions would be 14.4 gC/MJ x 100 MJ = 1,440 gC

Separate: Coal 50 MJ/(3.6 MJ/kWh)=13.89 kWh x 280 gC/kWh = 3889 gC

Boiler = 20 MJ/0.80 = 25 MJ x 14.4 gC/MJ = 360 gC

Total carbon = 3889 + 360 = 4249 gC

Savings: (4249 – 1440)/4249 = 0.66 = 66%

Pg. 7.9

7.27 Power plants emitting 0.39 x 101 2 g particulates from 685 M tons coal with a heat contentof 10,000 Btu/kWh while generating 1400 billion kWh/yr.

heat input = 685x106 tons x 2000 lbton

x 10, 000 Btulb

= 1.37x101 6Btu

efficiency =outputinput

=1400x109 kWh x 3412Btu/kWh

1.37x101 6Btu= 0.349 ≈ 35%

At NSPS of 0.03 lb particulates per 106 Btu input, emissions would have been:

emissions = 0.03 lb 106 Btu heat input

x 1.37x101 6Btu in x 1000g2.2 lb

=1.87x101 1g

For comparison, emissions at NSPSactual emissions

= 1.87x101 1g0.39x101 2 = 0.48 = 48%

7.28 Derivation for the dry adiabatic lapse rate:

dQ = dU + dW where dU = Cvdt and dW = PdV dQ = Cvdt + PdV (1)ideal gas law says PV = nRTso, d(PV) = PdV + VdP = nRTor, PdV = nRT - VdP

plugged into (1) gives:

dQ = CvdT + nRdT −VdP

dQdT

= Cv + nR − V dPdT

(2)

at constant pressure :

dQdT

= Cv + nR = Cp

putting that into (2) gives,

€

dQdT

= Cp −VdPdT

or, dQ = CpdT −VdP which is (7.37)

Pg. 7.10

7.29 Plotting the data, extending from ground level to crossing with ambient profile at theadiabatic lapse rate, and extending from the stack height gives:

21201918171615140

200

400

600

800

Temperature (C)

Alt

itud

e (m

)

plume risemixing depth

a) mixing depth (projecting from 20oC at 0-m at slope -1o/100m) = 400 m

b) plume rise (projecting from 21oC at 100m) = 500m

7.30 From Problem 7.29, projection from the ground at 22oC crosses ambient at 500m.

Need the windspeed at 250 m (halfway up) using (7.46) and Table 7.6 for Class C,

uH

ua

=Hza

p

so, u250

4m / s=

250m10m

0.20

= 1.90

u250 =1.90x4 = 7.6m / s

Ventilation coefficient = 500m x 7.6m/s = 3.8x103 m2/s

7.31 Below the knee, the plume is fanning which suggests a stable atmosphere, which could beprofile (a), (b) or (d).

Above the knee, the plume is looping, which suggests superadiabatic, which is d.

7.32 H=50m, overcast so Class D, A at 1.2km, B at 1.4km.

a. From Fig 7.52, Class D, H=50m, max concentration occurs at 1km. Beyond 1 km, concentration decreases so the "A" will be more polluted than “B.”

b. Clear sky, wind < 5m/s: Class is now A, B or C. At 50m, Class A, B, or C, Fig 7.52 shows us that the maximum point moves closer to the stack.

c. It will still be house at site "A” that gets the higher concentration of pollution.

Pg. 7.11

7.33 Bonfire emits 20g/s CO, wind 2 m/s, H=6m, distance = 400m. Table 7.7, clear night,stability classification = F

C x, 0( ) =Q

π uσ yσ z

exp −H2

2σ z2

(7.49)

a. From Table 7.9 at 400m, σy = 15m, σz = 7m

C =20x106 µg / s

π 2m/s x15m x 7mexp −

62

2x72

= 21x103 µg / m3 = 21mg / m3

b. At the maximum point, Fig. 7.52 we can get a rough estimate of the key parameter

CuHQ

max

≈ 3.8x10−3 / m2

Cmax =QuH

CuHQ

max

=20x103mg / s2m / s

x 3.8x10−3

m2 = 38 ≈ 40mg / m3

7.34 H=100m, Q=1.2g/s per MW, uH=4m/s, u

Anemometer=3+m/s, C<365µg/m3.

The more unstable the atmosphere, the higher the peak downwind concentration (see Fig.7.51). From Table 7.7, with wind > 3m/s, B is the most unstable.

From Fig. 7.52, Xmax = 0.7 km; CuHQ

max

≈ 1.5x10−5 /m2

Cmax =QuH

CuHQ

max

= 365x10−6g / s =Q

4m / sx1.5x10

−5

m2

Q ≈4x365x10−6

1.5x10−5= 97g / s

Maximum power plant size = 97 g/s x MW1.2 g/s

= 80MW

7.35 Atmospheric conditions, stack height, and groundlevel concentration restrictions same asProb. 7.34 so that:

Emissions Q ≈ 97 g/s

97g / s = 0.6 lb SO2

106 Btu inx 1 Btu in

0.35 Btu outx 3412 Btu out

kWhx 1hr

3600sx 103g

2.2 lbxPk W

PkW =97x106x0.35x3600x2.2

0.6x3412x1000= 131,000KW =130MW

Pg. 7.12

7.36 H = 100m, ua = 4 m/s, Q = 80g/s, clear summer day so Class B:

First, find the windspeed at the effective stack height using (7.46) and Table 7.7:

uH = uaHza

p

= 4m / s 10010

0.15

= 5.65m / s

a. At 2 km, Table 7.9: σy = 290 m, σz = 234 m

C x, 0( ) =Q

π uσ yσ z

exp −H2

2σ z2

C =80x106 µg/ s

π 5.65m/s x290m x 234mexp −

1002

2x2342

= 61µg / m3

b. At the maximum point, 0.7 km (Fig. 7.52),

CuHQ

max

≈ 1.5x10−5 /m2

Cmax =QuH

CuHQ

max

=80x106 µg/ s5.65m / s

1.5x10−5

m2

= 212µg/ m3

c. At x = 2km, y = 0.1 km:

C x, y( ) =Q

π uσ yσ z

exp −H2

2σ z2

exp −y 2

2σ y2

= 61µg / m3 x exp -1002

2x2902

= 57µg / m3

7.37 For class C, notice from (7.47) and (7.48) with Table 7.9 and f =0

σy

σz

=ax0.894

cxd + f=

104x0.894

61x0.911 = 1.7x−0.017 ≈ fairly constant ≈ k (about 1.7)

So, assume σy = k σz , then from (7.49)

a. C x, 0( ) =Q

π uσ yσ z

exp −H2

2σ z2

=

Qπ ukσ z

2 exp −H2

2σ z2

To find the maximum concentration, differentiate and set equal to zero:

Pg. 7.13

dCdσ z

=Qπ uk

1σz

2−H2

2

−2σz

3

e

−H 2

2σ z2

+ e−

H2

2σz2 −2σ z

3

= 0

Multiply through by σz5 and cancel lots of terms to get,

2H2

2

− 2σ z

2 = 0 or σz =H2= 0.707H

b. Substituting the newly found value for σz,

Cmax =Q

π uσ yσz

exp −H2

2 H2

2( )

=

Qπ uσ yσ z

e−1 =0.117Q uσyσz

c. Using σy = k σz

Cmax =0.117Q ukσz

2 =0.117Q

uk 0.707H( )2 =f(Q, u)

H2 (varies as inverse of H2)

7.38 Find the effective stack height of the Sudbury stack:

130 C 20 m/s

15.2m 380m

10 Co o8 m/s

Using (7.52) for the buoyancy flux parameter

F = gr2vs 1−TaTs

= 9.8

ms2x 15.2m

2

2

x20 msx 1− 10 + 273

130 + 273

= 3370m4 / s3

The distance downwind to final plume rise xf is given on page 464 (with F>55),

xf = 120F0.4 = 120x 3370( )0.4 = 3092m

For stability classification C, use (7.54) for plume rise,

Δh = 1.6F1 3xf2 3

uh

=1.6x 3370( )1 3 3092( )2 3

8= 635m plume rise

H = effective stack height = h + Δh = 380 + 635 = 1015 m

Pg. 7.14

7.39 Repeat Problem 7.38 with a stable, isothermal atmosphere:

F = 3370 m4/s3 from Prob. 7.38. For isothermal atmosphere we can use (7.51) along astability parameter to estimate plume rise. The stability parameter (7.53) is

S = gTa

ΔTaΔz

+ 0.01K /m

=

9.8m / s2

10 + 273( )K0 + 0.01K/ m( ) = 3.46x10−4 / s2

Putting that into (7.51) for plume rise under these conditions gives

Δh = 2.6 FuhS

1 3

= 2.6 3370m4 / s3

8m / s x 3.46x10-4 / s2

1 3

= 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 C 10 m/s

2m100m

6.0 Co o5 m/s

Using (7.52) for bouyancy flux parameter

F = gr2vs 1−TaTs

= 9.8

ms2x 2m

2

2

x10 msx 1− 6 + 273

120 + 273

= 28.4m4 / s3

and distance downwind to final plume rise xf given on page 464 (with F<55),

xf = 50F5 8 = 50x 28.4( )5 8 = 406

For stability classification C, use (7.54) for plume rise,

Δh = 1.6F1 3xf2 3

uh

=1.6x 28.4( )1 3 406( )2 3

5= 54m plume rise

H = effective stack height = h + Δh = 100 + 54 = 154 m

Pg. 7.15

7.41 Power plant, find groundlevel pollution 16 km away. Need first find H.

200 MWe

100mr=2.5m

13.5 m/s145 Co 15 Co

5m/sClass E

lapse rate=5 C/kmoQ=300g/s SO2

16km

First, find bouyancy flux parameter (7.52),

F = gr2vs 1−TaTs

= 9.8

ms2x 2.5m( )2 x13.5m

sx 1 − 15 + 273

145 + 273

= 257m4 / s3

plume rise for stable (Class E) atmosphere needs S from (7.53),

S = gTa

ΔTaΔz

+ 0.01K /m

=

9.8m / s2

15 + 273( )K5o

1000m+ 0.01K/ m

= 5.1x10−4 / s2

plume rise is given by (7.51),

Δh = 2.6 FuhS

1 3

= 2.6 257m4 / s3

5m / s x 5.1x10-4 / s2

1 3

= 121m

So, the effective height is H = 100m + 121m = 221 m

Concentration downwind at 16km: (Table 7.9) σy = 602m, σz = 95m

C x, 0( ) =Q

π uσ yσ z

exp −H2

2σ z2

(7.49)

C =300x106 µg / s

π 5 m/s x602m x 95mexp −

2212

2x952

= 22µg / m3

7.42 A 20 g/s source, 5 m/s, H = 50 m want peak concentrations Class A, C, F..

Cmax =QuH

CuHQ

=

20x106µg / s5m / s

CuHQ

= 4x106 CuH

Q

µg / s

50mA,C,F5 m/s

20 g/s

Using Fig. 7.52 with 50 m and varying stability classifications gives:

Pg. 7.16

"A" CuH

Q

≈ 6x10−5 at 0.25km, Cmax = 4x106 x 6x10-5 = 240µg / m3

"C" CuH

Q

≈ 5.8x10−5 at 0.55km, Cmax = 4x106 x 5.8x10-5 = 230µg / m3

"F" CuH

Q

≈ 2.4x10−5at 3.7km, Cmax = 4x106 x 2.4x10-5 = 96µg/ m3

x (km)

C ( g/m )µ 3

240230

96

0.25 0.55 3.7

7.43 H = 50m, 100m, 200m; Class C, 20 g/s, 5 m/s wind:

Cmax =QuH

CuHQ

=

20x106µg / s5m / s

CuHQ

= 4x106 CuH

Q

µg / s

Using Fig. 7.52,

@50m: Cmax ≈ 4x106 x 5.7x10-5 = 228 µg/m3

@100m: Cmax ≈ 4x106 x 1.5x10-5 = 60 µg/m3

@200m: Cmax ≈ 4x106 x 3.4x10-6 = 14 µg/m3

Do they drop as (1/H)2? That is,

€

expectation is C(2H)C(H)

=14

and C(4H)C(H)

=1

16

Test them: C(100m)C(50m)

=6

22.8= 0.26 C(200m)

C(100m)=

1.46= 0.23 not bad!

Expect C(200m)C(50m)

=116

= 0.0625 C(200m)C(50m)

=1.422.8

= 0.061 again, not bad.

Pg. 7.17

7.44 Paper mill emitting H2S, 1km away want 0.1 x odor threshold:

1 km

40 g/s4-10m/sClass B 0.01 mg/m3

Class B, at 1 km, (Table 7.9) σy = 156m, σz = 110m

C x, 0( ) =Q

π uσ yσ z

exp −H2

2σ z2

0.01x10−3 g / m3 =40g / s

π u m/s x156m x 110mexp −

H2

2x1102

Rearranging: eH2

24,200 =40

π u 156 x 110 x 0.01x10-3 =74.2

u

or, H = 24, 200 ln 74.2u

0.5

so, at each end of the wind speed range we can find the height needed:

H u= 4 = 24, 200 ln 74.24

0.5

= 265m

H u=10 = 24, 200 ln 74.210

0.5

= 220m says to be conservative use H=265m

From Fig 7.52 at H=265, Class B, Xmax ≈1.8km.

Therefore, with the peak occurring beyond the 1 km house, the concentration will rise forbuildings located > 1km away. YES there could be a problem.

7.45 Stack under an inversion:

45m

150 g/s

5 m/sClass C

X L

L=100m

At x = XL σz = 0.47 (L-H) = 0.47 (100 - 45) = 26 m

Pg. 7.18

For class C, σz = 26m at x = 0.4 km (Table 7.9), therefore XL = 0.4 km, and alsofrom Table 7.9, σy = 46m.

At x = XL : C X L,0( ) =Q

π uσ yσz

exp −H2

2σ z2

(7.49)

= 150x103 mg / sπ 5 m/s x 46m x 26m

exp −452

2x262

=1.8mg / m3

At x = 2XL : σy = 85m (Table 7.9), so using (7.55) gives

C 2XL, 0( ) = Q2π uσ yL

=150x103mg / s

2π x 5m/s x 85m x 100m= 1.4mg / m3

7.46 Stack under an inversion layer:

50m

80 g/s 5 m/s

X L

L=250m

4 m/s

x=4km

C=?

We need the stability classification: Clear summer day, 4m/s, Table 7.7 says Class B.

at x = XL , (7.56) gives us σz = 0.47 (L-H) = 0.47 (250 - 50) = 94 m

a. From Table 7.9, at σz = 94m Class B, XL ≈ 0.9km. Since our point of interest is at 4 km, we are well past the point at which reflections first occur so we can use (7.55). We need σy at 4km, which is given in Table 7.9 as 539m:

C 4km, 0( ) =Q

2π uσ yL=

80x106µg / s2π x 5m/s x 539m x 250m

= 47µg / m3

b. Without the inversion layer, at 4km σz = 498m, σy = 539m so,

C 4km, 0( ) =Q

π uσ yσ z

exp −H 2

2σ z2

=

80x106 µg / sπ x 5m/s x 539m x 498m

exp −502

2x4982

= 19µg / m3

Pg. 7.19

7.47 Agricultural burn. Clear fall afternoon, winds 3 m/s, so stability class "C" (Table 7.7),and σz = 26m (Table 7.9). Using (7.57),

C 0.4km( ) =2q

2π uσ z

=2x300mg / m − s2π x 3m/s x 26m

= 3.0mg / m3

0.3g/s-mu=3m/s

400m

7.48 A freeway modelled as a line source:

10,000 vehicles/hr u=2m/s

200m1.5 g/mi

Clear summer, 2 m/s, Table 7.7 suggests Class A or A-B.

At 0.2 km, σz = 29m for Class A; σz = 20m for Class B. What should we use? Since itis somewhere between Class A and Class B, but closer to A, let's use σz ≈ 26m:

To find the linear emission rate:

q =10, 000 vehicleshr

x 1 hr3600s

x 1.5gmi − vehicle

x mi5280ft

x ft0.3048m

= 2.58x10−3g / m − s = 2.58mg /m − s

Then, using (7.57),

C 0.2km( ) =2q

2π uσ z

=2x2.58mg / m − s2π x 2m/s x 26m

= 0.04mg / m3

7.49 Box model, 250,000 vehicles between 4 and 6pm, driving 40km ea, emitting 4g/kmCO.

a. qs = 250, 000veh. x 40kmveh

x 4gCOkm

x 12hrs

x hr3600s

x 115x80x106 m2 = 4.6x10−6 gCO/ m2s

b. Using (7.61) with t= 2hrs x 3600s/hr =7200s,

C t( ) = qsLuH

1 − e− ut / L( )

Pg. 7.20

= 4.6x10−6 g / m2s x 15, 000m0.5m / s x 15m

1− e−0.5m / s⋅7200s / 15000m( ) = 0.002g / m3 = 2mg / m3

c. With no wind, go back to (7.58) and solve the differential equation:

LWH dC

dt= qs LW

dC =qsLWLWH

dt so, C = qs

Ht

C = qs

Ht = 4.6x10−6 gCO/ m2 ⋅ s

15mx 2hrs x 3600s

hr= 0.0022gCO / m3 = 2.2mg / m3

7.50 Box model, 105 m on a side, H=1200m, u=4m/s, SO2=20kg/s, steady state:4m/sCin=0

10 m5

510 m

1200m

20 k g/s

input rate = output rate

20 kgs

x 109µgkg

= 4 ms

x 105 m x 1200m x C µgm3( )

C = 20x109

4x105x1200= 41.7µg / m3

7.51 Assume steady-state conditions were achieved by 5pm Friday so that from Problem7.50, C(0) = 41.7 µg/m3.

With qs = 0, and Cin = 0, (7.60) gives us C(t) = C 0( )e− ut / L .

a. At midnight, t=7hrs x 3600s/hr = 2.52x104 s

C(t) = C 0( )e− ut / L = 41.7µg/ m3 ⋅ e-4m/s ⋅ 2.52x104s 1 05 m =15.2µg / m3

b. Starting up again at 8am on Monday, by 5pm (9hrs later):

first check to see concentration left from Friday at 5pm (63 hrs earlier):

C(t) = C 0( )e− ut / L = 41.7µg/ m3 ⋅ e-4m/s ⋅ 63hrx3600s / hr 1 05m = .005µg / m3 ≈ 0

Pg. 7.21

so we can ignore that and let C(0) = 0 at 8am Monday. First find the emissions per unit area,

qs =emission rate

area=

20kg / s x 109µg / kg105 m x 105 m

= 2.0µg/ m2 ⋅ s

Then use (7.57) with Cin = 0:

C t( ) = qsLuH

1 − e− ut / L( )

= 2.0µg/ m2 ⋅ s x 105 m4m / s x 1200m

1 − e−( 4m / s x 9hr x 3600s/hr/105 m)( ) = 30.2µg / m3

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( ) = qsLuH

1 − e− ut / L( ) + C(0)e−ut / L

C 2hr( ) =2.0µg / m2 ⋅s x 105 m

2m / s x 1200m1− e−( 2m / s x 2hr x 3600s/hr/105m)( ) + 41.7e− 2m / s x 2hr x3600s/hr /105m( )

= 47.3 µg/m3

7.53 Modified Prob. 7.50, now incoming air has 5 µg/m3 and there are 10 µg/m3 alreadythere at 8am. Find the concentration at noon:

4m/s

Cin=5 g/m3

10 m5

510 m

1200m

2 g/m2-s

µ

µ

C t( ) =qsLuH

+ Cin

1 − e− ut / L( ) + C(0)e−ut / L (7.60)

C 4hr( ) =2.0µg / m2 ⋅s x 105 m

4m / s x 1200m+ 5µg / m3

1 − e−( 4 m/ s x 4hr x 3600s/hr/105 m)( )

+ 10e− 4m / s x 4hr x3600s/hr /105m( )

C(4hr = noon) = 26.1 µg/m3.

Pg. 7.22

7.54 Now using conditions of Prob 7.50, but for a nonconservative pollutant withK=0.23/hr:

Rate into box = Rate out of box + Rate of decay

S = u W H C + K C V

20 kgs

x 109µgkg

= 4 ms

x 105 m x 1200m x C µgm3( )

+0.23hr

x C µgm3 x 1hr

3600sx105m x 105 m x 1200m

20x109 = 4.8x108 C + 7.6x108 C

C = 16 µg/m3

7.55 Starting with (7.64) and using the special conditions of this tracer-gas study; that is, aconservative tracer (K=0), no tracer in the air leaking into the room (Ca=0), and thetracer source turned off at t=0 (S=0) gives the exponential decay of tracer as:

€

C t( ) = C0e−nt and then taking the log:

€

ln C t( )[ ] = ln C0( ) − ntwhich is of the form y = mx + b, where y = lnC, m = n(ach), and b = lnCo

time (hr) C (ppm) ln C0 10.0 2.3030.5 8.0 2.0791.0 6.0 1.7921.5 5.0 1.6092.0 3.3 1.194

32101.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

time (hr)

ln

C

From the graph, the slope is about: slope ≈ 2.1−1.3( )2.0 − 0.5

= 0.53

Thus, the infiltration rate is about 0.53 air changes per hour.

Pg. 7.23

7.56 Infiltration 0.5ach, 500m3 volume, 200 m2 floor space, radon 0.6pCi/m2s:

0.5 ach

V=500m3

0.6pCi/m s2

K=7.6x10 /hr- 3

Using (7.63) with K = 7.6x10-3/hr (Table 7.14),

S =SV( )

I + K=

0.6pCi / m2s x 200m2

500m3

0.5 / hr + 7.6x10−3 / hr( )x 1hr3600s

= 1700pCi / m3 =1.7pCi / L

7.57 Same as Problem 7.56 but have half as much ground-floor area to let radon in, so:

S =SV( )

I + K=

0.6pCi / m2s x 100m2

500m3

0.5 / hr + 7.6x10−3 / hr( )x 1hr3600s

= 850pCi / m3 = 0.85pCi / L

7.58 From Problem 7.56, the radon concentration is 1.7 pCi/L. Using a residentialexposure factor of 350 days/yr from Table 4.10

€

Exposure =1.7pCi/L x 350 day

yrx30yr

365day/yr x 70yr= 0.70 pCi/L average over 70 yrs

From Problem 4.10 we are given a cancer death rate of 1 per 8000 rems exposure. FromProblem 4.12 a 1.5 pCi/L radon concentration yields an exposure of about 400 mrem/yr.

€

Risk = 0.70 pCi/L x 400mrem/yr1.5pCi/L

x1 cancer death8000 rem

x rem103mrem

x70yr = 0.0016 ≈ 0.16%

Pg. 7.24

7.59 A 300m3 house, 0.2ach, oven+2burners 6pm to 7pm, find CO at 7pm and and 10pm.For these circumstances, (7.65) is appropriate:

€

C t( ) =S

IV1− e−n t( ) (7.65)

From Table 7.13, the source strength S is

6 – 7 pm: Oven + 2 burners = 1900 mg/hr + 2 x 1840 mg/hr = 5580 mg/hr CO

solving for C after 1 hr:

C 1hr, 7pm( ) = 5580mg/hr

0.2 airchangehr

x300 m3

ac

1− e−0.2 / hr x 1hr( ) = 16.8mg / m3

Now turn off the burners and watch CO coast down until 10pm, 3hrs later:

€

C(10pm) = C(7pm) x e−n t =16.8 e−0.2/hr x 3hrs = 9.3mg/m3

7.60 n = 0.39 ach, V = 27m3, after 1-hr NO = 4.7ppm. Find source strength, S:

First convert NO in ppm to mg/m3 using (1.9) and assuming T=25oC,

mg / m3 =ppm x mol wt

24.465=

4.7 x 14 +16( )24.465

= 5.76mg / m3

a. To find the NO source strength, rearrange (7.65)

€

S =n V C

1− e−n t( )=

0.39 achr

x27 m3

acx5.76 mg

m3

1− e−0.39/hr x 1hr( )=188 mgNO/hr

b. 1-hr after turning off the heater,

€

C = C0 e−n t = 4.7ppm x e-0.39/hr x 1hr = 3.2 ppmNO

c. In a house with 0.2 ach, 300m3,

€

C ∞( ) =S

n V=

188 mg/hr

0.2 achr

x 300 m3

ac

= 3.1 mg/m3 NO

Using (1.9) again gives

€

ppm = 24.465ppm x mol wt

=24.465

3.1 x (14 +16)= 2.6 ppm NO

Pg. 7.25

7.61 Find the settling velocity of 2.5-micron particles having density 1.5x106 g/m3. In aroom with 2.5-meter-high ceilings, use a well-mixed box model to estimate theresidence time of these particles.

From (7.24) the settling velocity is

€

v = d2ρg18η

=2.5x10−6m( )

2⋅ 1.5x106g/m3( ) ⋅ 9.8m/s2( )

18x0.0172g/m ⋅ s= 2.97x10−4m/s

From Example 7.4, the residence time is

€

τ = hv

=2.5 m

2.97 x 10−4 m/s x 3600 s/hr= 2.3 hours

7.62 100,000 kW coal plant, 33.3% efficient, CF = 0.70,

a. Electricity generated per year,

€

100,000 kW x 24 hr/day x 365 day/yr x 0.70 = 613x106kWh/yr

b. heat input = 613x106 kWh / yr out x 3 kWht in1 kWhe out

x 3412BtukWh

= 6.28x101 2Btu / yr

c. Shut it down and sell the allowances,

SO2 saved by shutting down = 6.28x101 2 Btuyr

x 0.6 lb SO2

106 Btux ton

2000 lb= 1883 tons/yr

€

1883 tonsyr

x 1 allowanceton

x $400allowance

= $753,200/yr

SOLUTIONS FOR CHAPTER 8

8.1 From (8.1),

δ18O ooo( )=

18O16O( )

sample18O

16O( )standard

−1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ x103 =

0.00201500.0020052

−1⎡ ⎣ ⎢

⎤ ⎦ ⎥ x103 = 4.9

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 /oo( )=2H/1H( ) sample

2H/1H( ) standard−1

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ x103 =

0.000081000.00015575

−1⎛ ⎝ ⎜

⎞ ⎠ ⎟ x103 = −479.8

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 = −32.1oC

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: T oC( )= 0.1661 δD o

oo( )+ 72.45 8.6 The flat earth!

1370W/m 2 R

Eabsorbed = Eradiated

Sπ R2 = σ T4A = σ T4 2π R2( )

T =S

2σ⎛ ⎝ ⎜ ⎞

⎠ ⎟

14

=1370W / m2

2x5.67x10−8 W / m2K4

⎛ ⎝ ⎜

⎞ ⎠ ⎟

14

= 331.5K - 273.1 = 58.4o C

8.7 The basic relationship is S =kd2 . Using d and S for Earth from Table 8.2 lets us find k:

k = S d2 = 1370W / m2 x 150x106 km x 103 m / km( )2

= 3.083x1025W

a. Mercury: S =kd2 =

3.083x1025W58x106 km x 103 m / km( )2 = 9163W / m2

Pg. 8.2

b. The effective temperature (8.7) of Mercury would be:

Te =S 1− α( )

4σ⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

=9163W / m2 1− 0.06( )4x5.67x10−8 W / m2K4

⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

= 441K (168o C)

c. Peak wavelength:

λmax =2898T K( )

=2898441

= 6.6μm

8.8 Solar flux variation of ± 3.3%, gives a range of S Smax = 1370 (1+0.033) = 1415.2 W/m2 Smin = 1370 (1 -0.033) = 1324.8 W/m2

Te,max =S 1− α( )

4σ⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

=1415.2W / m2 1 − 0.31( )4x5.67x10−8 W / m2K4

⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

= 256.2K (-17o C)

Te,min =S 1− α( )

4σ⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

=1324.8W / m2 1 − 0.31( )4x5.67x10−8 W / m2K4

⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

= 252K (-21o 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 Ts

4 = 240W / m2

Pg. 8.3

Ts =240W / m2

5.67x10−8 W / m2 K4

⎡ ⎣ ⎢

⎤ ⎦ ⎥

14

= 255K (-18o C)

b. X, Atmosphere to space: Balance Incoming from space = Outgoing to space 342 = 69 + X X = 273 W/m2 c. Y, Absorbed by earth: Incoming solar has to go somewhere, 342 = 69 + 257 + Y Y = 16 W/m2 d. Z, Radiation from atmosphere to surface: Balance earth's surface radiation, Y + Z = 240 = 16 + Z Z = 224 W/m2 8.10 A 2-layer atmosphere:

342

107

X

Y

168

24 78

40

W

ZW

350

Z

T1

T2

390

a. At the surface: 168 + Z = 24 + 78 + 390 Z = 324 W/m2 b. Extraterrestrial: 342 = 107 + 40 + W W = 195 W/m2 c. Lower atmosphere: Y + 24 + 78 + 350 + 195 = 2 x 324 Y = 1 W/m2 d. Incoming: 342 = 107 + X + 1 + 168 X = 66 W/m2 e. Temperatures T1 and T2 (assuming blackbody radiation) can be found from

σ T14 = W = 195 T1 =

195W / m2

5.67x10−8 W / m2K 4

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1 4

= 242K (-31o C)

Pg. 8.4

σ T24 = Z = 324 T2 =

324W / m2

5.67x10−8 W / m2 K4

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1 4

= 275K (2o C)

8.11 Hydrologic cycle:

evaporation =78W / m2 x5.1x1014m2 x 1J/s

Wx3600 s

hrx24 hr

dx365 d

yr2465kJ / kg x 103 kg/ m3 x 103 J / kJ

= 5.1x1014 m3

Averaged over the globe, with area 5.1x1014 m2, annual precipitation is very close to 1 m 8.12 Greenhouse enhanced earth:

100342

67 24 78

30 Z

Y

X

291K

W

a. Incoming energy: 342 = 100 + 67 + W W = 175 W/m2 b. Find Z from radiation to space: 342 = 100 + 30 + Z Z = 212 W/m2 c. To find X, need the energy radiated by a 291 K surface: surface radiated = σ T 4 = 5.67x10−8 W/ m2 ⋅ K4 x 291K( )4 = 406.6W/ m2 so that, 406.6 = X + 30 X = 376.6 W/m2 d. Can find Y several ways; at the surface, or in the atmosphere, W + Y = 406.6 + 24 + 78 = 175 + Y Y = 333.6 W/m2 or, 67 + 24 + 78 + X = Y + Z 67 + 24 + 78 + 376.6 = Y + 212 (Y = 333.6) 8.13 CO2 from 10 GtC/yr to 16 GtC/yr over 50 years, with initial 380 ppm and A.F. = 40%. Since it is linear, the total emissions would be those at constant level

plus the area of a triangle rising by 6 Gt/yr:

Pg. 8.5

50 yrs x10 Gt/yr + 1/2 x 50 yrs x 6 GtC/yr = 575 GtC.

Using the 2.12 GtC/ppm ratio and the 0.40 A.F. gives

CO2( )= 380 +575 GtC x 0.402.12 GtC/ppm

= 380 +108 = 488 ppm

8.14 CO2 growing at 2 ppm/yr, fossil fuel and cement emissions at 9 GtC/yr, and A.F. of 38%. The remaining emissions due to land use changes are:

Cemissions =2 ppm/yr x 2.12GtC/ppm

0.38=11.15 GtC/yr

Land use emissions = 11.15 – 9 = 2.15 GtC/yr

8.15 With 40% oil, 23% coal, 23% gas and 14% carbon free: a. Using LHV values from Table 8.3:

Coal 23% @ 25.8 gC/MJ

Oil 40% @ 20.0 gC/MJ

Gas 23% @ 15.3 gC/MJ

Other 14% @ 0 Avg C intensity = 0.23x25.8 + 0.40x 20.0 + 0.23x15.3 + 0.14x0 = 17.45 gC/MJ b. Coal replaced by non-carbon emitting sources: Avg C intensity = 0.23x0 + 0.40x20.0 + 0.23x15.3 + 0.14x0 = 11.52 gC/MJ c. Modeled as an exponential growth function over 100 years:

C = C0e

rt

r =1t

ln CC0

⎛

⎝ ⎜

⎞

⎠ ⎟ =

1100

ln 11.5217.45

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = −0.0042 = −0.42% / yr

8.16 With resources from Table 8.4 and LHV carbon intensities from Table 8.3, A.F. = 50%:

a. All the N. Gas: 15.3 gC/MJ x 36,100 x 1012 MJ = 552,330 x 1012 gC = 552 GtC

ΔCO2 =552 GtC x 0.5

2.12 GtC/ppmCO2

=130 ppm CO2

b. All the Oil: 20.0 gC/MJ x 24,600 x 1012 MJ = 492 GtC

Pg. 8.6

ΔCO2 =492 GtC x 0.5

2.12 GtC/ppmCO2

=116 ppm CO2

c. All the Coal: 25.8 gC/MJ x 125,500/2 x 1012 MJ = 1619 GtC

ΔCO2 =1619 GtC x 0.5

2.12 GtC/ppmCO2

= 382 ppm CO2

d. All three: 130 + 116 + 382 = 628 ppm CO2. From (8.29) with ΔT2X = 2.8oC:

ΔTe =ΔT2X

ln 2 ln

CO2( )CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

2.8ln 2

ln 628 + 380380

⎡ ⎣ ⎢

⎤ ⎦ ⎥ = 3.9oC

8.17 Out of oil and gas, demand = 2 x 330 EJ/yr, 28%coal, 60% syn gas/oil@44gC/MJ, a. Carbon emission rate: Avg carbon intensity = 0.28 x 25.8 + 0.60 x 44 + 0.12 x0 = 33.6 gC/MJ

Emissions =2 x 330x1018J

yrx

MJ106 J

x33.6gC

MJx

GtC1015gC

= 22.2GtC / yr

b. Growth from 6.0 GtC/yr to 22.2 GtC/yr in 100 yrs,

r =1

100ln

22.26.0

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 0.013 = 1.3%/ yr

c. Amount remaining with 50% airborne fraction, use (8.27):

Total emitted = Ctot =C0

rer T −1( )=

6.0 GtC/yr0.01308

e0.01308/yr x 100 yr −1( )=1239 GtC

Amount remaining in atmosphere = 0.50 x 1239 = 619 GtC d. Amount in atmosphere in 100 yrs = 750 + 619 = 1369 GtC

(CO2 ) =1369GtC

2.12GtC/ ppmCO2

= 646ppm

e. Equilibrium temperature increase, with ΔT2x=3oC from (8.29):

ΔT =ΔT2x

ln 2⋅ ln

CO2

CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

3.0ln 2

⋅ ln645356

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 2.57o C

Pg. 8.7

8.18 Repeat of Prob. 8.17, but now conservation scenario: a. Carbon emission rate: Avg carbon intensity = 0.20 x 25.8 + 0.30 x 15.3 + 0.10 x20 = 11.75 gC/MJ

Emissions = 330x1018 J

yrx

MJ106 J

x11.75gC

MJx

GtC1015 gC

= 3.88GtC / yr

b. Growth from 6.0 GtC/yr to 3.88 GtC/yr in 100 yrs,

r =1

100ln

3.886.0

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = −0.0044 = −0.44% / yr

c. Amount remaining with 50% airborne fraction, use (8.27):

Total emitted = Ctot =C0

rer T −1( )=

6.0 GtC/yr−0.0044

e−0.0044/yr x 100 yr −1( )= 483 GtC

Amount remaining in atmosphere = 0.50 x 483 = 242 GtC

d. Amount in atmosphere in 100 yrs = 750 + 242= 992 GtC

(CO2 ) =992GtC

2.12GtC/ ppmCO2

= 468ppm

e. Equilibrium temperature increase, with ΔT2x=3oC,

ΔT =ΔT2x

ln 2⋅ ln

CO2

CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

3.0ln 2

⋅ ln468356

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 1.18o C

8.19. Finding LHV efficiency of a condensing furnace with 95% HHV efficiency.

From Example 8.4, HHV = 890 kJ/mol and LHV = 802 kJ/mol. The output of a HHV 95% efficient furnace burning 1 mole of methane is 0.95 x 890 kJ = 845.5 kJ. On an LHV basis, you still get the same output, but the efficiency is now

LHV efficiency=845.5 kJ delivered802 kJ LHVinput

=1.054 =105.4%

Pg. 8.8

This over 100% efficiency is one reason LHV values are sometimes avoided in the U.S.

8.20 Finding HHV carbon intensities:

a. Ethane, C2H6 : 2 x 12 gC/mol1542 kJ/mol

x103kJMJ

=15.56 gC/MJ

b. Propane, C3H8 : 3 x 12 gC/mol2220 kJ/mol

x103kJMJ

=16.36 gC/MJ

c. n - Butane, C4H10 : 4 x 12 gC/mol2878 kJ/mol

x103kJMJ

=16.68 gC/MJ

8.21 Using HHV carbon intensities from Table 8.3, the four options are:

COP=3

η =0.95100MJ1380gC

95MJ delivered 1380gC

95MJ=14.5gC/MJ

η =0.70100MJ1380gC

70MJ delivered 1380gC

70MJ=19.7gC/MJ

1) pulse

2) conv gas

3) heat pump η =0.35100MJ2420gC

35MJ

2420gC35MJ

=69.1gC/MJ

power plant

heat pump

70 from enviro.

105MJ del 2420gC105MJ

=23.0gC/MJ

4)resistance η =0.35100MJ2420gC

35MJ

power plant Notice the tremendous range: 14.5 to 69.1 gC/MJ, almost 5:1 !

8.22 Propane-fired water heater with 2200 kJ/mol vs Example 8.6:

a. Carbon intensity C3H8 : 3 x 12 gC/mol2220 kJ/mol

x103kJMJ

=16.36 gC/MJ

b. Delivering heat at 85% efficiency to hot water

Pg. 8.9

c. Savings versus 32.5 gC/MJ with a n. gas electric water heater in Example 8.6:

propaneelectric

=19.25 gC/MJ32.5 gC/MJ

= 0.59 so there is a 41% savings vs electricity

8.23 Initial CO2 = 356 ppm, 6 GtC/yr and 750 GtC; want 70 year scenario. Do it by scenario: (A) Using r = 1.0 + 0.3 - 2.0 - 0.7 = -1.4%/yr in (8.27)

Ctot =C0

rer T −1( )=

6.0 GtC/yr−0.014

e−0.014/yr x 70 yr −1( )= 268 GtC

CO2( )=750GtC + Ctot x AF2.12 GtC/ppmCO2

=750 + 268 x 0.4 GtC2.12 GtC/ppmCO2

= 404 ppm

ΔT =ΔT2x

ln 2⋅ ln

CO2

CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

3ln 2

⋅ ln404356

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 0.55o C

To find the doubling time, rearrange (8.27):

Ctot to double current 750 GtC =750 GtCAF = 0.4

=C0

rer Td −1( )

Td =1r

ln750/0.4( )r

6.0+1

⎡

⎣ ⎢

⎤

⎦ ⎥ =

1−0.014

ln750/0.4( ) −0.014( )

6.0+1

⎡

⎣ ⎢

⎤

⎦ ⎥ = never!

(B) r = 1.5 + 1.5 - 0.2 + 0.4 = 3.2%/yr

Ctot =C0

rer T −1( )=

6.0GtC/yr0.032

e0.032/yr x 70 yr −1( )=1574 GtC

CO2( )=750 GtC + Ctot x AF2.12 GtC/ppmCO2

=750 +1574 x 0.5GtC2.12 GtC/ppmCO2

= 725 ppm

Pg. 8.10

ΔT =ΔT2x

ln 2⋅ ln

CO2

CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

2ln 2

⋅ ln725356

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 2.05o C

Td =1r

ln750

AF( ) r6.0

+1⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

=1

0.032ln

7500.5( ) 0.032

6.0+1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

= 69 yrs

(C) r = 1.4 + 1.0 - 1.0 - 0.2 = 1.2%/yr

Ctot =C0

rer T −1( )=

6.0 GtC/yr0.012

e0.012/yr x 70 yr −1( )= 658 GtC

CO2( )=750 GtC + CtotxAF2.12 GtC/ppmCO2

=750 + 658 x 0.5 GtC2.12 GtC/ppmCO2

= 509 ppm

ΔT =ΔT2x

ln2⋅ ln CO2

CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

3ln2

⋅ ln 509356

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =1.55oC

Td =1r

ln750

AF( ) r6.0

+1⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

=1

0.012ln

7500.5( ) 0.012

6.0+1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

=116yrs

8.24 With 1990 6.0 GtC/yr + land use 2.5 GtC/yr and the following growth rates to 2100

Population growth rate dP/dt = 0.8% Per capita GDP growth rate d(GDP/P)/dt = 1.3% Final Energy per GDP growth rate =d(FE/GDP)/dt = - 0.7% Primary Energy to Final Energy growth rate d(PE/FE)/dt = 0.1% Carbon per unit of Primary Energy growth rate d(TC/PE)/dt = -0.2% Carbon Sequestration growth rate d(C/TC)/dt = 0.0%

Total growth rate = 0.8 + 1.3 – 0.7 + 0.1 – 0.2 + 0.0 = 1.3%/yr

a. The carbon emission rate in 2100

From energy C = C0ert = 6.0e0.013x110 = 25.1 GtC/yr

Including land use: Total emission rate = 25.1 + 2.5 = 27.6 GtC/yr

b. Total carbon emissions:

Ctot (energy) =C0

rerT −1( )=

6.00.013

e0.013x110 −1( )= 1467 GtC

Ctot (land use and industry) = 110 yrs x 2.5 GtC/yr = 275 GtC

Total emissions = 1467 + 275 = 1742 GtC

Pg. 8.11

c. The increase in CO2 concentration with A.F. = 0.5:

ΔCO2 =1742 GtC x 0.5

2.12 GtC/ppmCO2

= 410 ppm CO2

d. Estimated 2100 CO2 concentration = 360 + 410 = 770 ppm

e. With ΔT2X = 2.8oC, the global equilibrium temperature increase 2100

ΔTe =ΔT2X

ln 2 ln

CO2( )CO2( )0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

2.8ln 2

ln 770360

⎡ ⎣ ⎢

⎤ ⎦ ⎥ = 3.1o C

8.25 Identification of the halocarbons:

a. C3HF7 is an HFC (no Cl), 3 1 7 - 90 = 227, HFC-227 b. C2H3FCl2 is an HCFC, 2 3 1 - 90 = 141, HCFC-141 c. C2F4Cl2 is a CFC, 2 0 4 - 90 = 114, CFC-114 d. CF3Br is a Halon, H-1301

8.26 a. HCFC-225, 225 + 90 = 315 (3C, 1H, 5F), 8 sites - (1+5) = 2 Cl, ∴ C3HF5Cl2 b. HFC-32, 32 + 90 = 122 (1C, 2H, 2F), 4 sites, 0 Cl, CH2F2 ∴ c. H-1301, (1C, 3F, 0Cl, 1Br) CF3Br ∴ d. CFC-114, 114 + 90 = 204 (2C, 0H, 4F), 6 sites - 4 = 2 Cl, C2F4Cl2 ∴ 8.27. Finding climate sensitivity λ and varying feedback factor g. a. From (8.35) and (8.40) using ΔT2X = 2.5 oC.

λ =ΔT2X 4.2

=2.54.2

= 0.595oC

W m2 g =1- λB

λ=1−

0.27λ

=1−0.270.595

= 0.546

If g = 0.1 + 0.546 = 0.646, then

λ =λB

1− g=

0.271− 0.646

= 0.763 oC W m2( ), ΔT2X = 4.2λ = 4.2x0.763 = 3.2oC

Pg. 8.12

b. For ΔT2X = 3.5 oC

λ =ΔT2X 4.2

=3.54.2

= 0.833oC

W m2 g =1- λB

λ=1−

0.27λ

=1−0.270.833

= 0.676

If g increases to 0.776, then

λ =λB

1− g=

0.271− 0.776

=1.205 oC W m2( ), ΔT2X = 4.2λ = 4.2x1.205 = 5.1o C

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).

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 =

ΔFC − C0( )=

0.14311 − 275( )= 0.133

If N2O reaches 417 ppb, added forcing would be: ΔF = k2 C − C0( )= 0.133 417 − 311( )= 0.37W / m2 8.30 a. Combined radiative forcings from 1850 to 1992

ΔFCO2 = 6.3 lnCO2( )[ ]

CO2( )0[ ]= 6.3 ln356278

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 1.558 W/m 2

ΔFCH4= 0.031 CH4 − CH 4( )

0( )= 0.031 1714 − 700( )= 0.463 W/m2

ΔFN2O = 0.133 N 2O − N2O( )

0( )= 0.133 311 − 275( )= 0.140W / m2

ΔFCFC−11 = 0.22 CFC − 11( )− CFC −11( )0[ ]= 0.22 0.268 − 0( ) = 0.059 W / m2 ΔFCFC−12 = 0.28 CFC −12( )− CFC −12( )0[ ]= 0.28 0.503 − 0( )= 0.141 W / m2

Pg. 8.14

Combined forcing = 1.558 + 0.463 + 0.140 + 0.059 + 0.141 = 2.36 W/m2 b. From 1992 to 2100:

ΔFCO2 = 6.3 lnCO2( )[ ]

CO2( )0[ ]= 6.3 ln710356

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 4.35 W/m2

ΔFCH4

= 0.031 CH4 − CH 4( )0( )= 0.031 3616 − 1714( )= 0.581W / m2

ΔFN2O = 0.133 N 2O − N2O( )

0( )= 0.133 417 − 311( )= 0.370W / m2

ΔFCFC−11 = 0.22 CFC − 11( )− CFC −11( )0[ ]= 0.22 0.040 − 0.268( ) = −0.050 W / m2 ΔFCFC−12 = 0.28 CFC −12( )− CFC −12( )0[ ]= 0.28 0.207 − 0.503( ) = −0.083 W / m2 Combined forcing = 4.35 + 0.581 + 0.370 - 0.050 - 0.083 = 5.17 W/m2

c. From 1850 to 2100

ΔF = 6.3 ln710278

⎛ ⎝ ⎜ ⎞

⎠ ⎟ + 0.031 3616 − 700( )+ 0.133 417 − 275( )

+ 0.22x0.040 + 0.28x0.207 = 7.53 W/m2 (alternatively: ΔF = 2.36 + 5.17 = 7.53 W/m2)

8.31 From Prob. 8.30 for 1850 to 2100:

ΔF = 6.3 ln710278

⎛ ⎝ ⎜ ⎞

⎠ ⎟ + 0.031 3616 − 700( )+ 0.133 417 − 275( )

+ 0.22x0.040 + 0.28x0.207 = 7.53 W/m2 ΔTs = λ ΔF = 0.57 oC/(W/m2) x 7.53 W/m2 = 4.3 oC

8.32 Using and forcing

ratios of HFC-134a to CO

RCO2t( )dt

0

20

∫ ≈13.2yrs; RCO2t( )dt

0

100

∫ ≈ 43.1yrs; RCO2t( )dt

0

500

∫ ≈138yrs

2 of (Fg/FCO2) = 4129 and τ = 14 yrs. First simplify GWP to

Pg. 8.15

GWPg =Fg

FCO2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⋅

e− t /τ dt0

T

∫

RCO2t( )dt

0

T

∫=

Fg

FCO2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⋅

τ 1− e−T /τ( )RCO2

t( )dt0

T

∫

a. GWP(20) = 4129 ⋅14 1− e−20 /14( )

13.2= 3330 (vs 3300 in Table 8.7)

b. GWP(100) = 4129 ⋅14 1− e−100 /14( )

43=1340 (vs 1300 in table)

c. GWP(500) = 4129 ⋅14 1− e−500 /14( )

138= 420 (vs 400 in table)

8.33 For a greenhouse gas with τ = 42 years and a relative forcing of 1630 times that of CO2.

From Problem 8.32, GWP =ΔFg

ΔFCO2

⋅τ 1 − e− t

τ( )RCO2

t( )dt∫

a. The 20-year GWP would be

GWP20 = 1630 ⋅42 1− e− 20

42( )13.2

= 1965

b. The 100-year GWP would be

GWP100 = 1630 ⋅42 1− e−100

42( )43.1

=1440

c. The 500 year GWP would be

GWP500 =1630 ⋅42 1 − e−500

42( )138

= 495

8.34 Applying GWPs from Table 8.7 to the emission rates given:

Pg. 8.16

8.35 Using 100-year GWPs from Table 8.7 with emission rates of 6,000 million metric tons (Mt) of CO2, 26.6 MtCH4, and 1.2 Mt N2O. gives

6000 x 1(CO2) + 26.6 x 23(CH4) + 1.2 x 296(N2O) = 6967 MtCO2 = 6.967 GtCO2-eq

Adjusting for the ratio of C to CO2 gives

6.967 GtCO2 x (12gC/44gCO2) = 1.9 GtC-eq/yr

8.36 The actual ΔTrealized is estimated to be 0.6oC, which is 75% of the equilibrium ΔT ΔTrealized = 0.6oC = 0.75 ΔTequilibrium so, ΔTequilibrium = 0.6/0.75 = 0.8oC but, ΔTequilibrium = λ ΔFactual = 0.57 x ΔFactual = 0.8

that is, ΔFactual = =0.80.57

=1.40W / m2

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 x16 gCH4

12 gCx 23 gCO2

1 gCH4

= 703 gCO2 − eq

The actual CO2 emissions remain the same at 5525 gCO2

So, with 83.73 MJ of heat to the water, total CO2-eq emissions per MJ gives

703 gCO2 - eq + 5525 gCO 2

83.73 MJ heat to water= 74.4 gCO2−eq /MJ

8.38 Using Table 8.3 for the LHV carbon intensity of coal (25.8 gC/MJ), (3.18) to find

σ, and (3.20) to find tm, then plotting (3.17) gives for (a):

Q∞ = 200,000 EJ x 25.8 gCMJ

x 1 GtC1015gC

x1012MJEJ

= 5160 GtC

σ = Q∞

Pm 2π=

5160 GtC22 GtC/yr 2π

= 93.57 yr

Pg. 8.17

tm = σ 2ln Pm

P0

= 93.57 yr 2 ln 22 GtC/yr6.0 GtC/yr

=150.8 yr

then put these into P = Pm exp −12

t − tm

σ⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎡

⎣ ⎢

⎤

⎦ ⎥

Putting this into a spreadsheet so it can be plotted yields…

8.39 With a carbon tax of $20/mt of C (as CO2):

a. Assuming a capacity factor of 100% (plant operates all of the time):

Cemissions =50 MW

0.35x1 MJ/s

MWx 3600 s

hrx 8760 hr

yrx 24 gC

MJx 1 mtC

106gC=1.08x105mtonC/yr

Pg. 8.18

Carbon tax =1.08 x105mtC/yr x $20mton

= $2.16 million/yr

b. With carbon sequestering:

Area = 1.08x105mtonC/yr5000 kgC/yr ⋅ acre

x103 kgmton

= 21,600 acres

c. Biomass instead of paying the tax:

Forestry could cost = $2.16 million/yr21,600 acres

= $100 /yr per acre

8.40 Landfill leaking 10 tonnes (1 tonne = 1 mt = 1000 kg) CH4 per year

a. 20-year GWP for methane = 62 (Table 8.7) 10 tonnes CH4 /yr x 62 = 620 tonnes/yr b. Burning the methane CH4 + 2 O2 CO2 + 2 H2O

1molCO2

1molCH4

x12 + 2x12( )gCO2/mol12 + 4x1( )gCH4/mol

x10tonneCH4

yr= 27.5tonneCO2/yr

c. Equivalent CO2 savings = 620 – 27.5 = 592.5 tonne CO2/yr.

as C : 592.5 tonne CO2/yr x 12tonneC44tonneCO2

=161.5 tonneC/yr saved

d. Carbon tax saved = 161.5 tonne C/yr x $20/tonneC = $3232/yr saved

e. Same thing, 592.5 tonne CO2/yr x $5.45/tonneCO2 = $3229/yr saved 8.41 Gasoline C7H15 and 6.15 lbs/gal, fully combusted,

a. Gasoline =6.15 lbgas

galx

7x12 = 84( ) lbsC7x12 +15x1= 99( ) lb gas

= 5.22 lbs C/gal

C = 40,000 miles12 miles/gal

x 5.22 lbsCgal

=17,394 lbsC that will be released

b. 4000 lb car, 10,000 miles/yr

Pg. 8.19

C = 17,394 lbs C40,000 miles

x10,000 milesyr

= 4348 lbsC/yr

Carbon/yrVehicle weight

=4348 lbsC/yr

4000 lbs=1.09

the car emits slightly more carbon per year than it weighs!

c. Carbon tax =5.22 lbsC

galx $15

2000 lbsC= $0.039/gal = 3.9¢/gal

d. New car at 40 mpg, for 40,000 miles:

Carbon reduction =17,394 lbsC- 40,000 mi40 mi/gal

x 5.22 lbsCgal

=12,174 lbs C saved

e. Trading in the clunker for the 40 mpg car would save in carbon taxes

Tax savings = 12,174 lbsC x $152000 lbs C

= $91/car

That is, those C offsets would save the utility $91, which they could spend to get the clunker off the road.

8.42 Electric versus gasoline-powered cars:

a. Gas car emissions=5.22 lbsC/gal40 miles/gal

x1000g2.2lbs

= 59.3 gC/mi

b. With the very efficient natural-gas fired power plant:

N - gas plant emissions =8000kJkWh

x13.8gCMJ

x MJ103kJ

x kWh5mi

= 22.1 gC/mi

c. With the typical coal plant:

Coal plant heat rate =l kW in

0.30 kWe outx 1 kJ/s

kW heat inx 3600 s

hr= 12,000 kJ/kWhe

Coal plant emissions =12,000kJ

kWhx 24 gC

MJx MJ

103kJx kWh

5mi= 57.6 gC/mi

So, more than half of the carbon can be saved with electric cars when efficient natural gas power plants are assumed. There is even a slight advantage with an old, inefficient coal plant.

8.43 NO2 + hv NO + O

From (8.48): E J/photon( )=306,000

6.02x1023 = 5.08x10−19 J/photon

Pg. 8.20

and from (8.46): λmax =hcE

=6.626 x10−34 Js x 2.998x108m/s

5.08x10−19 J= 390x10−9 = 390 nm

8.44 O2 + hv O + O (oops… same as Example 8.13):

E J/photon( )=495,000

6.02x1023 = 8.22x10−19 J/photon

λmax =hcE

=6.626 x10−34 Js x 2.998x108m/s

8.22x10−19 J= 241.6x10−9 = 241.6 nm

Meant to do photodissociation of ozone, requiring 104.6 kJ/mol:

O3 + hv O2 + O

E J/photon( )=104,600

6.02x1023 =1.737x10−19 J/photon

λmax =hcE

=6.626 x10−34 Js x 2.998x108m/s

1.737x10−19 J=1.14x10−6m =1.14 μm

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 CO2

2000 lb/tonx 2200 lb

metric tonx 44 ton CO2

12 ton C= $40.33/MTCE

Savings = $167,531/yr (from spreadsheet) c. A $400,000/yr recycling program saves

Net benefit = $1,040,000 + $167,531 - $400,000 = $807,531/yr

9.4 Comparing a 10,000mi/yr, 20 mpg SUV burning 5.22 lbs C//gal at 125,000 Btu/gal to cardboard recovery savings.

c. The carbon savings from cardboard recycling is equivalent to carbon emissions

from how many SUVs? d. How many “average SUVs” of energy are saved by cardboard recycling?

a. How many tons of CO2 will be emitted per SUV per year?

CO2 =10,000 miles x x 5.22 lbsC/gal

20 miles/gal x 2200 lbs/tonx 44 tonCO2

12 tonC= 4.35 ton CO2/yr

b. Btus for those SUVs

Energy =10,000 miles x 125,000 Btu/gal

20 miles/gal = 62.5 million Btu/yr

c. From Table 9.10, 42 million tons of cardboard are kept out of landfills. And from Table 9.8, each ton saves 0.96 metric tons of carbon

9.2

Car equivalents =42x106ton/yr x 0.96 mtonC/ton x 1.1 ton/mton

4.32 tonCO 2/yr −SUVx 44 tonCO2

12 tonC

= 37.6 million SUVs taken off the road in carbon savings

d. From Table 9.9, cardboard recycling saves 15.65 million Btu/ton. So,

Car equivalents = 42x106ton/yr x 15.65 x10 6 Btu/ton 62.5 x 106 Btu/yr/SUV

=10.5 million SUVs

9.5 A 0.355-L (12 oz) 16 g aluminum can with 70% recycling. Need to adjust Table

9.13 which was based on 50% recycling. Each can now has 0.7x16 = 11.2 g of recycled aluminum and 0.3 x 16 = 4.8 g of new aluminum from bauxite.

New aluminum from bauxite = 4.8g x1765 kJ8g

=1059 kJ

Recycled cans to make 11.2 g of Al =11.2g x 40 kJ8g

= 56 kJ

The remaining energy for can production is the same as Table 9.13:

Total for 16g (0.355L) can = 3188 – (1765 + 40) + (1059 + 56) = 2498 kJ/can

Per liter of can = 2498 kJ/0.355L = 7037 kJ/L 9.6 Heavier cans from yesteryear, 0.0205 kg/can and 25% recycling rate.

New aluminum per can was 0.75 x 0.0205 kg = 0.015375 kg

Recycled aluminum per can was 0.25 x 0.0205 kg = 0.005125 kg From Table 9.12:

New aluminum = 0.015275 kg x 220,600 kJ/kg = 3370 kJ

Recycled aluminum = 0.005125 kg x 5060 kJ/kg = 26 kJ

Total energy = 3370 + 26 = 3396 kJ/can

Compared to today’s 1805 kJ/can (Example 9.4)

Today : 1805 kJ/canEarlier : 3396 kJ/can

= 0.53 Savings is 47%

9.7 Using 1.8 million tons/yr of aluminum cans, 63 percent recycled, and Table 9.12:

a. The total primary energy used to make the aluminum for those cans.

New aluminum = 0.37 x 1.6x106tons x 220,600 kJ/kg x kg2.2 lb

x 2000 lbton

=118.7x1012kJ

9.3

Old aluminum = 0.67 x 1.6x106tons x 5060 kJ/kg x kg2.2 lb

x 2000 lbton

= 4.9x1012kJ

Total = (118.7 + 4.9) x 1012 = 123.6 x 1012 kJ

b. With no recycling:

All new aluminum =1.6x106tons x 220,600 kJ/kg x kg2.2 lb

x 2000 lbton

= 320.9x1012kJ

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: 3.97x106metric tonsC x 44 tons CO2

12 tons C=14.6x106 metric tons/yr

or 14.6x106tonne/yr x 2200 lbstonne

x ton2000 lbs

=16 million U.S. tons/yr

9.8 With pickups from both sides of the 1-way street:

With pickups on one-side only on the 1-way street, need to make 2 passes

9.9 A 30 yd3 packer truck, 750 lb/yd3, 100 ft stops, 5 mph, 1 min to load 200 lbs:

time/stop =100 ft

stopx mi

5280 ftx hr

5 mix 60 min

hr+1 min =1.227 min/stop

9.4

9.10 Route timing:

a. Not collecting = 20min + 3x20min + 2x15min + 15min + 40min = 165 min/day

To fill a truck takes:

25yd3truck x 4 yd3curbyd3truck

x customer0.2yd3 x stop

4 customerx1.5 min

stop=187.5 min/load

Two loads per day takes:

2 loads/d x 187.5min/load +165 min(travel,breaks)[ ]x hr60min

= 9.0 hrs/day

b. Customers served:

25yd3truck x 4 yd3curbyd3truck

x customer0.2yd3 = 500 customers/load

#customers = 500 customersload

x 2 loadsday

x 5 daysweek

= 5000 customers/truck

c. Labor = $40hr

x 8 hrsday

+$60hr

x 1hrday

⎛

⎝ ⎜

⎞

⎠ ⎟ x

5 dayweek

x 52 wksyr

= $98,800/yr

Truck cost = $10,000yr

+$3500/yr

yd3 x 25yd3 = $97,500/yr

Customer cost = ($98,800 +$97,500)/yr5000 customers

= $39.26/yr

9.11 To avoid overtime pay, working 8 hrs/day and needing 165 min to make runs back and forth to the disposal site, breaks, etc (Problem 9.2):

Collection time = 8 hr/d x 60 min/hr – 165 min/d = 315 min/day

Customers = 315 minday

x stop1.5 min

x 4 customersstop

x 5 daywk

= 4200 customers

Annual cost of service per customer is now:

($40/hr x 8 hr/wk x 5 d/wk x 52 wk/yr + $97,500/yr)/4200 = $43/yr

Cheaper to pay them overtime.

9.5

9.12 So, with 8-hr days a smaller truck can be used. As in Problem 9.11:

Collection time = 8 hr/d x 60 min/hr – 165 min/d = 315 min/day

With 2 truckloads per day,

Customers = 315 minday

x stop1.5 min

x 4 customersstop

= 840 customers/day

or 420 customers per truckload. At 2 loads per day and 5 days per week, that would give 4200 customers once a week service. Truck size needed is therefore,

Truck size = 420 customerstruckload

x 0.2 yd3 at curbcustomer

x yd3 in truck4yd3at curb

= 21yd3

costing:

Truck$ =$10,000

yr+

$3500/yd3

yrx 21yd3 = $83,500/yr

Labor$ = $40hr

x 8hrday

⎛

⎝ ⎜

⎞

⎠ ⎟ x

5 dayweek

x 52 weeksyr

= $83,200/yr

Resulting in: Customer$ =$83,200 +$83,500( )/yr

4200 customers= $39.69/yr

(compared with $39.26 per customer in Problem 9.10 and $43 in Problem 9.11. 9.13 Comparing two truck sizes:

a. Customers for each truck:

(A) 27 m3 truck: 27m3 truck x 4m3curb1m3 in truck

x customer0.25m3 curb

= 432 customers/load

#customers = 432 customerstruckload

x 2 loadsday

x 5 daysweek

= 4320 customers (27m3)

(B) 15 m3 truck: 15m3 truck x 4m3curb1m3 in truck

x customer0.25m3 curb

= 240 customers/load

#customers = 240 customerstruckload

x 3 loadsday

x 5 daysweek

= 3600 customers (15m3)

b. Hours per day for the crew:

(A) 27 m3 truck: 432 customerstruckload

x 0.4 mincustomer

x 2 loadsday

= 346 min

Crew : 346 min +160 min(misc.)60 min/hr

= 8.43 hr/day

9.6

(B) 15 m3 truck: 240 customerstruckload

x 0.4 mincustomer

x 3 loadsday

= 288 min

Crew : 288 min +215 min(misc.)60 min/hr

= 8.38 hr/day

c. Cost per customer:

(A) 27 m3 truck:

Cost : $40hr

x 8.43hrday

x 5 dayweek

x 52 weekyr

+ $120,000/yr = $207,672/yr (27m3)

Customer$ = $207,672/yr4320 customers

= $48.07/yr (27m3)

(B) 15 m3 truck:

Cost : $40hr

x 8.38hrday

x 5 dayweek

x 52 weekyr

+ $70,000/yr = $157,152/yr (15 m3)

Customer$ = $157,152/yr3600 customers

= $43.65/yr (15 m3)

Cheaper to run 3 trips a day in the smaller 15 m3 truck.

9.14 A $150,000 truck, 2 gal/mi, $2.50/gal, 10,000 mi/yr, $20k maintenance:

a. Amortized at 12%, 8-yr: CRF 8yr,12%( )=i 1+ i( )n

1+ i( )n −1=

0.12 1+ 0.12( )8

1+ 0.12( )8 −1= 0.201/ yr

Amortization = $150,000 x 0.201/yr = $30,195/yr

Fuel = 10,000 mi/yr x 2 gal/mi x $2.50/gal = $50,000/yr

Total truck cost = $30,195 + $50,000 + $20,000 (maint) = $100,195/yr

b. Labor$ = $25/hr-person x 2 people/truck x 40 hr/wk x 52 wk/yr = $104,000/yr

c. Total Cost =$100,195 + $104,000

10 ton/day x 260day/yr= $78.54/ton

9.15 Reworking Examples 9.5 – 9.7 to confirm the costs in Table 9.17:

a. One-run per day, t = 150 ft/stop x 3600 s/hr5 mi/hr x 5280 ft/mi

+ 4 can/stop x 20s/can =100.5 s/stop

Time to collect = 8 – 0.4 – (2x1-1)x0.4 – 0.25 – 1- 1x0.2 = 5.75 hr/day

which allows N stops per day (l-load per day)

9.7

N = 5.75 hr/d x 3600 s/hr100.5 s/stop x 1 load/day

= 206 stops/load

Truck volume need is,

V = 4 can/stop x 4 ft 3/can x 206 stop/load27ft 3/yd3 x 3.5 yd3 curb/yd3in truck

= 34.9 yd3

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

2000 lb/ton= 3214 tons/yr

Cost/ton = (164,600 + 99,840) $/yr3214 ton/yr

= $82.25/ton (OK)

b. Three-runs per day: 100.5 s/stop

Time to collect = 8 – 0.4 – (2x3-1)x0.4 – 0.25 – 1- 3x0.2 = 3.75 hr/day

which allows N stops per day (3-load per day)

N = 3.75 hr/d x 3600 s/hr100.5 s/stop x 3 load/day

= 44.8 stops/load

Truck volume need is,

V = 4 can/stop x 4 ft 3/can x 44.8 stop/load27ft 3/yd3 x 3.5 yd3 curb/yd3in truck

= 7.6 yd3

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

2000 lb/ton= 2096 tons/yr

Cost/ton = (99,840 + 55,400) $/yr2096 ton/yr

= $74/ton (OK)

9.16 Transfer station: 200 tons/day, 5d/wk, $3 million, $100,000/yr, trucks $120,000, 20 ton/trip, $80k/yr, 4 trip/day, 5d/wk, 10%, 10-yr amortization.

Station costs: CRF 10yr,10%( )=i 1+ i( )n

1+ i( )n −1=

0.10 1+ 0.10( )10

1+ 0.10( )10 −1= 0.16275/ yr

9.8

$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

52,000 tons/yr= $11.31/ton

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 =$99530/yr

20,800 ton/yr= $4.79/ton

for a total of $4.79/ton (truck) + $11.31/ton (station) = $16.10/ton

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 kJkg H2O

x18 kgH2O2 kgH

x 0.14 kgHkg PET

x 54g PET103g/kg

=166 kJ/bottle

LHV = 43,500 kJkgPET

x 54 gPET/bottle103g/kg

−166 kJ/bottle = 2183 kJ/bottle

9.21 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:

VMSW =129x106 ton

yrx 2000 lb

tonx yd3

27 ft 3 = 8.71x109 ft 3/yr

@ 80% per cell,

Vlandfill =8.71 x 109 ft 3/yr

0.80=10.9x109 ft3/yr

Alift =10.9 x 109 ft3/yr

10 ft/liftx acre

43,560 ft2 = 24,987 acres/yr

No population is specified, but at roughly 300 million people, the area per 1000 people would be about

Alift per 1000 =24,987 acres/yr

300,000 thousand people≈ 0.08 acre/yr per 1000 people

9.24 50,000 people, 40,000 tons/yr, 22% recovery, 1000 lb/yd3, 10-ft lift, 80% MSW:

VLandfill =40,000(1- 0.22)ton

yrx 2000 lb

tonx yd3

1000 lbx 27ft 3

yd3 x ft 3 landfill0.80ft 3MSW

= 2.11x106 ft 3/yr

a. Area of lift needed each year

Alift =2.11 x 106 ft3/yr

10 ft/liftx acre

43,560 ft2 = 4.83 acre/yr

b. To complete the landfill will take:

time remaining = 40 acre/lift x 2 lifts remaining4.83 acre/yr

=16.5 yrs

9.25 By increasing the recovery rate from 22% to 40%,

VLandfill =40,000(1- 0.40)ton

yrx 2000 lb

tonx yd3

1000 lbx 27ft 3

yd3 x ft 3 landfill0.80ft 3MSW

=1.62x106 ft 3/yr

Alift =1.62 x 106 ft3/yr

10 ft/liftx acre

43,560 ft2 = 3.72 acre/yr

9.12

time remaining = 40 acre/lift x 2 lifts remaining3.72 acre/yr

= 21.5 yrs

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 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:

VCH4=

0.0224 m3CH4

mol CH4

x 6.5225 molCH4

kg"asreceived"= 0.146m3CH4/kg lawn trimmings

b. Energy content: CH4 energy =6.5225 molCH4

kg "as received"x 890 kJ

mol= 5,805 kJ/kg

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 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 5.9825 CH4 + 4.5175 CO2 + 0.66

NH3

c. Fraction of the resulting gas that is methane:

CH4 =5.9825 mol CH4

(5.9825 + 4.5175 +0.66) moles gas= 0.536 = 53.6%

d. Volume of methane per kg of food waste:

VCH4=

0.0224 m3CH4

mol CH4

x 5.9825 molCH4

kg"as received"= 0.134m3 CH4/kg food wastes

e. HHV value of methane produced:

CH4 energy =5.9825 mol CH4

kg "as received"x 890 kJ

mol= 5,324 kJ/kg food waste

9.14