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