66868-Chptr9PrblmSolns

7/29/2019 66868-Chptr9PrblmSolns

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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 torecycling 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 9.1using Table 9.9

a. Annual energy savings is: 21,494 million Btub. 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

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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 9.18market 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 CO22000 lb/ton

x2200 lb

metric tonx

44 ton CO212 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

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

c. The carbon savings from cardboard recycling is equivalent to carbon emissionsfrom 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 tonCO212 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 fromTable 9.8, each ton saves 0.96 metric tons of carbon

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

4.32 tonCO2/yrSUVx

44 tonCO212 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 x106 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 Table9.13 which was based on 50% recycling. Each can now has 0.7x16 =11.2 g ofrecycled aluminum and 0.3 x 16 =4.8 g of new aluminum from bauxite.

New aluminum from bauxite= 4.8g x1765 kJ

8g=1059 kJ

Recycled cans to make 11.2 g of Al=11.2g x40 kJ

8g= 56 kJ

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

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

Per liter of can =2498 kJ/0.355L =7037 kJ/L

9.6 Heavier cans from yesteryear, 0.0205 kg/can and 25% recycling rate.

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

Recycled aluminum per can was 0.25 x 0.0205 kg =0.005125 kg

From Table 9.12:

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

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

Total energy =3370 +26 =3396 kJ/can

Compared to todays 1805 kJ/can (Example 9.4)

Today: 1805 kJ/can

Earlie : 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 xkg

2.2 lbx

2000 lb

ton=118.7x1012kJ

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Old aluminum=0.67 x 1.6x106tons x 5060 kJ/kg xkg

2.2 lbx

2000 lb

ton= 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 xkg

2.2 lbx

2000 lb

ton= 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 106MTCE

As CO2: 3.97x106metric tonsC x

44 tons CO212 tons C

=14.6x106 metric tons/yr

or 14.6x106tonne/yr x2200 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=100ft

stopx

mi

5280 ftx

hr

5 mix

60 min

hr+1 min=1.227 min/stop

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9.10 Route timing:

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

To fill a truck takes:

25yd3truck x4 yd3curb

yd3truckx

customer

0.2yd3x

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 h60min

= 9.0 hrs/day

b. Customers served:

25yd3truck x4 yd3curb

yd3truckx

customer

0.2yd3= 500 customers/load

#customers=500 customers

loadx

2 loads

dayx

5 days

week= 5000 customers/truck

c. Labor =$40

hrx

8 hrs

day+

$60

hrx

1hr

day

x

5 day

weekx

52 wks

yr= $98,800/yr

Truck cost =$10,000

yr+

$3500/yr

yd3x 25yd3 = $97,500/yr

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

5000 customers= $39.26/yr

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

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

Customers=315 min

dayx

stop

1.5 minx

4 customers

stopx

5 day

wk= 4200 customers

Annual cost of service per customer is now:

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

Cheaper to pay them overtime.

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

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

With 2 truckloads per day,

Customers=315 min

day

xstop

1.5 min

x4 customers

stop

= 840 customers/day

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

Truck size=420 customers

truckloadx

0.2 yd3 at curb

customerx

yd3 in truck

4yd3at curb= 21yd3

costing:

Truck$=$10,000

yr+

$3500/yd3

yrx 21yd3 = $83,500/yr

Labor$= $40hr

x8hrday

x

5 dayweek

x52 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 x4m3curb

1 3 in truckx

customer

0.25 3 curb= 432 customers/load


truckloadx

2 loads

dayx

5 days

week= 4320 customers (27m3)

(B) 15 m3 truck: 15m3 truck x4m3curb

1 3 in truckx

customer

0.25 3 curb= 240 customers/load


truckloadx

3 loads

dayx

5 days

week= 3600 customers (15m3)

b. Hours per day for the crew:

(A) 27 m3 truck:432 customers

truckloadx

0.4 min

customerx

2 loads

day= 346 min

Crew:346 min+160 min(misc.)

60 min/hr= 8.43 hr/day

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(B) 15 m3 truck:240 customers

truckloadx

0.4 mi

customerx

3 loads

day= 288 min

Crew:288 min+215 min(misc.)

60 min/h= 8.38 hr/day

c. Cost per customer:

(A) 27 m3 truck:

Cost:$40

hrx

8.43hr

dayx

5 day

weekx

52 week

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

Customer$=$207,672/yr

4320 customers= $48.07/yr (27m3)

(B) 15 m3 truck:

Cost:$40

hrx

8.38hr

dayx

5 day

weekx

52 week

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

Customer$=$157,152/y

3600 customers= $43.65/yr (15 m3)

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

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

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

1+ i( )n 1=

0.121+ 0.12( )8

1+ 0.12( )8 1= 0.201/yr

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

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

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

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

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

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

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

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

5 mi/hr x 5280 ft/mi+ 4 can/stop x 20s/can=100.5 s/stop

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

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

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N =5.75 hr/d x 3600 s/hr

100.5 s/stop x 1 load/day= 206 stops/load

Truck volume need is,

V =4 can/stop x 4 ft3/can x 206 stop/load

27ft3/yd3 x 3.5 yd3 curb/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) $/yr

3214 ton/yr= $82.25/ton (OK)

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

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

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

N =3.75 hr/d x 3600 s/hr

100.5 s/stop x 3 load/day= 44.8 stops/load

Truck volume need is,

V =4 can/stop x 4 ft3/can x 44.8 stop/load

27ft3/yd3 x 3.5 yd3 curb/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/y

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=

0.101+ 0.10( )10

1+ 0.10( )10 1= 0.16275/yr

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$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 thecollection route (or, the transfer station should be more than 0.5 hr from thedisposal site).

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9.18 Newsprint: 5.97% moisture, HHV =18,540 kJ/kg, 6.1% H.

Starting with 1 kg of as received waste:

Energy to vaporize moisture =0.0597 kg H2O x 2440 kJ/kg =145.6 kJ

Dry weight =1 0.0597 =0.9403 kg

Hydrogen in the dry waste =0.061 x 0.9403 =0.0574 kg

As H becomes H2O =0.0574 kgH x 9 kgH2O/kgH x 2440 kJ/kgH2O =1259.6 kJ

Total energy lost in water vapor =145.6 +1259.6 =1405 kJ per kg of newsprint

LHV =HHV 1405 =18,540 1,405 =17,135 kJ/kg

9.19 Corrugated boxes, 5.2% moisture, HHV =16,380 kJ /kg, 5.7% H in dried material:

Could use the procedure shown in Prob. 9.18, or use (9.7)

LHV =HHV 2440(W+9H)

W =0.052 kgH2O/kg waste

H =(1-0.052)x 0.057 =0.054 kgH/kg waste

LHV =16,380 2440 (0.052 +9x0.054) =16380 1313 =15,067 kJ/kg

9.20 2 L PET bottle, 54 g, 14% H, HHV =43,500 kJ/kg,

QL =energy lost in vaporized H2O

QL =2440 kJ

kg H2Ox18 kgH2O

2 kgHx

0.14 kgH

kg PETx

54g PET

103g/kg=166 kJ/bottle

LHV = 43,500 kJkgPET

x54 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


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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


f. Tires:

HHV(dry)=339x79.1 +1440x6.8 139x5.9 +105x1.5 =35,944 kJ /kg


g. Polystyrene:

HHV(dry)=339x87.10 +1440x8.45 139x3.96 +105x0.02 =41,147 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


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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 ft3= 8.71x109 ft3/yr

@ 80% per cell,

Vlandfill =8.71 x 109 ft3/yr

0.80=10.9x109 ft3/yr

A lift =10.9 x 109 ft3/yr

10 ft/lift xacre

43,560 ft2 = 24,987 acres/yr

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

A lift per 1000 =24,987 acres/yr

300,000 thousand people 0.08acre/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

yr x2000 lb

ton xyd3

1000 lbx27ft3

yd3 xft3 landfill

0.80ft3MSW = 2.11x106

ft3

/yr

a. Area of lift needed each year


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 remaining

4.83 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

27ft3

yd3x

ft3 landfill

0.80ft3MSW=1.62x106 ft3/yr


10 ft/liftx

acre

43,560 ft2= 3.72 acre/yr

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time remaining=40 acre/lift x 2 lifts remaining

3.72 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 turnsout to be1 mole of decompostibles. Using (9.8) gives

C12.76H21.28O9.26N0.54 +n H2Om 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 ofCH4.

a. Volume of methane:

VCH4 =0.0224 m3CH4

mol CH4x

6.5225 molCH4kg"asreceived"

= 0.146m3CH4/kg lawn trimmings

b. Energy content: CH4 energy=6.5225 molCH4kg "as received"

x890 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 H2Om CH4 +s CO2 +d NH3

9.13


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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 CH4x

5.9825 molCH4kg"as received"

= 0.134m3 CH4/kg food wastes

e. HHV value of methane produced:

CH4 energy=5.9825 mol CH4kg "as received"

x890 kJ

mol= 5,324 kJ/kg food waste

9.14

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66868-Chptr9PrblmSolns