66866-Chptr7PrblmSolns

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+



= 36.1 mpg

€

c. CAFE mpg = 1 mile0.5 mile

27 mile/gal gas+



= 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


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


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


(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


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

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