test prep rep sanitary solutions

7/29/2019 Test Prep Rep Sanitary Solutions

1/31

PE Exam Review - Sanitary

Demonstration Problem Solutions

I. Demo Problem Solutions...................................................................... 2

1. Module 1 - Water Demand.............................................................. 2

2. Module 2 - Water Softening ............................................................ 5

3. Module 3 - Coagulation and Flocculation........................................ 9

4. Module 4 - Groundwater............................................................... 13

5. Module 5 - Trickling Filters ............................................................ 17

6. Module 6 - Stream Purification...................................................... 21

7. Module 7 - Sludge Digestion......................................................... 26

8. Module 8 - Landfills....................................................................... 29


2/31

Demonstration Problem Solutions Sanitary and Environmental Engineering 2

2003 SmartPros Ltd.

Demo Problem Solutions

Module 1 - Water Demand

SituationThe water district is under a mandate to provide water for a town whose current population is19,800.

Requirements1A) The total water use on a particular day was 4,503,000 gallons. What was the water use

expressed in gpcd?

1. Gallons per capita day:population

gpdgpcd=

2. 800,19

000,503,4

=gpcd

3. 42.227=gpcd

1B) The average daily flow (in million gallons per day) at the treatment plant is given below foreach month in the year. Using this information, calculate the annual average daily flow(AADF).

January 5.36 July 5.48

February 4.95 August 6.12

march 5.16 September 5.94April 5.44 October 5.77

May 5.62 November 5.68

June 5.79 December 5.49

1. Average Annual Daily Flow:12

montheachforflowdaily=AADF

2.12

80.66AADF=

3. gallonsmillion57.5=AADF


3/31


2003 SmartPros Ltd.

1C) The average daily flow (in million gallons per day) at the treatment plant varies accordingto the time of day and the time of the year. Using the information below, calculate howmuch more the average water demand might be in the summer versus the winter. (Answerin millions of gallons.)

consumption time/period multiplier

winter 0.80summer 1.30

maximum daily 1.50 - 1.80

maximum hourly 2.00 - 3.00

early morning 0.25 - 0.40

noon 0.50 - 2.0

1. Change in average daily flow: erwsummer ADFADFADF int=

summersummer MultiplierADFADF =

24.73.157.5 ==summerADF

erwerw MultiplierADFADF intint =

46.48.057.5int ==erwADF

2. galmil78.246.424.7 ==ADF

1D) For what demand should the water district design its water treatment facility? (Answer inmillion gallons per day.)

(use the consumption/multiplier table from the previous requirement)

1. DailyMaxMultiplierAADFDemand =

2. 8.157.5 =Demand

3. MGD03.10=Demand

1E) This town is expected to increase in population by 50% over the next decade. For whatfuture demand should the water district design its water treatment facility? (Answer inmillions of gallons.)

1. GrowthDemandandFuture Dem =

2. 5.103.10 =andFuture Dem

3. MGD05.15=andFuture Dem


4/31


2003 SmartPros Ltd.

1F) How much water should the treatment facility supply for the future population (that is to say50% population increase) if fire-fighting requirements are taken into account? (Answer ingpcd.)

1. Calculations for meeting fire-fighting water requirements apply to water that isstored, not to output of the water treatment plant.

2. MGD05.15=andFuture Dem


5/31


2003 SmartPros Ltd.

Module 2 - Water Softening

SituationWater with excess hardness is unacceptable as a drinking water and will require softening.Several methods are available to soften water. Raw water with an analysis as shown below isto be treated using the excess lime-soda ash softening treatment. The practical limits of thelime-soda precipitation softening process are 30 mg/l of calcium expressed as CaCO3 and 10mg/l of Mg(OH)2 expressed as CaCO3.

mg/lEquivalent

Weight meq/lmg/l asCaCO3

CalciumMagnesiumSodiumChlorideSulfateBicarbonateCarbon dioxide

(free)

803019186433615

20.012.223.035.548.061.022.0

4.02.50.80.51.35.50.7

20012540

2565

27535

Raw Water Analysis Bar Chart

.7 0 4.0 6.5 7.3

CO2 Ca++ Mg++ Na+

HCO3- SO4

- - Cl-

5.5 6.8 7.3

Requirements2A) Some source waters contain carbon dioxide. What happens to carbon dioxide when lime is

added to the water?

1. Carbon dioxide in the water reacts with the lime to form calcium carbonate, whichwill precipitate out. If there is a lot of carbon dioxide in the water, the water will beexpensive to treat. In this case you are using the lime not to soften the water, whichis your objective, but to get rid of the carbon dioxide. To treat the water in a cost-effective manner, you might want to use an aeration process before the limetreatment.

2. CO2 + Ca(OH)2 CaCO3 + H2O


6/31


2003 SmartPros Ltd.

2B) Complete this common lime softening equation: ? + Ca(OH)2 2 CaCO3 + 2 H2O

1. Calcium in the form of calcium bicarbonate, one of the hard water substances youare trying to remove, will react with the lime to form the insoluble calcium carbonate.The calcium carbonate precipitates out of the water.

2. Ca(HCO3)2 + Ca(OH)2 2 CaCO3 + 2 H2O

2C) What is formed from this combination? Mg(HCO3)2 + Ca(OH)2 ?

1. Magnesium, in the form of magnesium bicarbonate reacts with lime to formmagnesium carbonate. Keep in mind that magnesium carbonate is soluble. Youneed to add more lime to convert the magnesium to magnesium hydroxide, which isinsoluble and will settle out.

2. Mg(HCO3)2 + Ca(OH)2 MgCO3 + 2 H2O

2D) What would you add to remove the hardness causing a CaSO4 product?

1. Adding soda ash to calcium sulfate will produce sodium sulfate, which does notcontribute to hardness, and calcium carbonate, which precipitates out. So, you addsoda ash to remove calcium sulfate.

2E) When magnesium sulfate reacts with lime what resulting substance does not precipitateout?

1. Magnesium sulfate reacts with lime to form calcium sulfate and magnesiumhydroxide. The calcium sulfate does not precipitate out.

MgSO4 + Ca(OH)2 CaSO4 + Mg(OH)22. To remove the calcium sulfate you need to add soda ash, and that will precipitate

out as calcium carbonate.

CaSO4 + Na2CO3 Na2SO4 + CaCO3


7/31


2003 SmartPros Ltd.

2F) Based upon the analysis of the water, how much quick lime (CaO) should you add forsoftening? Assume that you are adding 35 mg/l of excess lime to accelerate the process.(answer in lbs/million gallons)

1. Determine quantity of reactants in the water, using the milliequivalents (meq) of eachsubstance.

CO2 0.7 meq

Ca(HCO3) 4.0 meq

Mg(HCO3) 1.5 meq

Mg(SO4) 1.0 meq

Total = 7.2 meq CaO required for the reaction

2. Use equivalent weight of CaO (28) to find mg/l of lime required

28

.7CaOmg/l

meq/l2 =

202CaOmg/l

3. Add excess lime required by problem

23735202CaOmg/l =+=

1,9778.34237watergalCaO/millb =

(to convert from mg/l to lb/mil gal)

2G) Based upon the analysis of the water, how much soda ash should you add for softening?(answer in lbs/mil gal).

1. You need 1.0 meq/l of soda ash (Na2CO3) to react with the 1.0 meq/l of MgSO4 inthe water.

2. The equivalent weight of soda ash is 53

mg/l soda ash = 1.0 53 = 53

53 8.34 442 lb/mil gal water


8/31


2003 SmartPros Ltd.

2H) Calculate the amount of carbon dioxide required to neutralize excess lime and convert theMg(OH2) to Mg(CO3) according to the equation Ca(OH2) + CO2 CaCO3 + H2O (answerin mg/l).

1. -OHexcessmeq/llimeexcessmeq/lCOweightequivalent

COmg/l

2

2 +=

equivalent weight CO2 = 22

25.1==28

CaOmg/l35limeexcessmeq

2.0==50

Mg(OH)mg/l10-OHmeq 2

2. 1.450.21.2522

COmg/l 2 =+=

3. mg/l CO2 = 31.9

2I) Calculate the total amount of carbon dioxide required to stabilize the water by convertingresidual carbonate to bicarbonate. Add carbon dioxide from the previous answer for thetotal amount of carbon dioxide needed. (answer in mg/l).

1. Use this equation: CaCO3 + CO2 +H2O = Ca(HCO3)2

2. To find the amount of CO2 required: 32

2 CaCOmeq/lCOweightequivalent

COmg/l=

6.0==50

30CaCOmeq/l 3

2.136.022 ==2COmg/l

3. Add the CO2 requirement you found in 2H

1.459.312.13 =+=2COmg/l


9/31


2003 SmartPros Ltd.

Module 3 - Coagulation/Flocculation

Situation

A town of 150,000 is going to build a water treatment plant to treat surface water. Thecoagulant alum (Al2(SO4)3 14 H2O) is to be added to a rapid mix unit to create a concentrationof 20 mg/l. The coagulated water goes to a flocculator unit then to a clarifier.

The flocculator is 68 feet long, 31 feet wide, and 15 feet deep. It is equipped with 12-inchpaddles supported parallel to and moved by three horizontal shafts which rotate at a speed of2.5 rpm. The radius of the paddles is 6.0 ft from the shaft, which is mid-depth of the tank. Twopaddles are mounted on each shaft, one opposite the other. The paddles are 30 feet long, sothere is a 6-in clearance on each side of the tank. The mean velocity of the water isapproximately 1/4 of the velocity of the paddles and their drag coefficient is 1.8. Assume awater temperature of 50 degrees F, and dynamic viscosity = 2.74 x 10

-5(lb. force)(sec)/ft

2.

The coagulated, flocculated water enters a sedimentation basin 50 feet wide and 200 feet long

at a rate of 15 MGD. There are 1200 feet of overflow weir and the basin has a 10 foot depth.

Requirements

3A) For a complete mix basin designed to treat 15 MGD, what volume (ft3) is needed to

completely mix the alum with the raw water. (Common detention time is 30-60 seconds.Use 60 seconds.)

1. timedetentionrateflowVtank =

/dayft348,005,2gal7.48

ft1

day

galmil15rateflow 3

3

===

33

tank ft1393s86,400

day1s60

day

ft2,005,348V ==

3B) At a flow of 15 MGD, how many pounds per day of alum must be applied to achieve therequired 20 mg/l of alum?

1. 20 mg/l 8.34 = 166.8 lb/mil gal

2. lb/day2502day

galmil15

galmil

lb166.8 =


10/31


2003 SmartPros Ltd.

3C) Determine the velocity differential between the paddles and the water.

1. Paddle velocity:60

2 rnVp

= paddle rotation speed

r = 6 ft

n = 2.5 rpm

ft/s57.160

5.262=

=

pV

2. Velocity differential: Vd = Vp (1.0 0.25)

fractional difference betweenpaddle and water velocities

Vd = 1.57 0.75 = 1.18 ft/s

3D) Determine the power requirements for turning the three flocculating paddles. Express youranswer in horsepower units.

1. Power:g2

ADVCP

3

dD=

CD (drag coefficient) = 1.8

D (density) = 62.4

g = 32.2

A (area) = LW

W = 12 in = 1 ft

L = tank width 6-in clearance on each side

= 31 1 = 30 ft

Total paddle area = 30 ft2 3 shafts 2 paddles per shaft = 180 ft2

2. 5164.64

1.1862.41801.8P

3

=

=

3. Power in horsepower =550

516= 0.94


11/31


2003 SmartPros Ltd.

3E) Determine the hydraulic detention time, td, for the flocculation unit for a flow of 15 MGD.(Answer in minutes.)

1.Q

Vtd = tank volume

flow

LWDV =

31,620153168V ==

Q = 15 mil gal/day

2. min7.22day1

min1440

ft1

gal48.7

gal15,000,000

day1ft620,31t

3

3

d ==

3F) Determine the value of the mean velocity gradient, G.

1.tankV

PG =

P (power) = 516

(dynamic viscosity) = 2.74 X 10-5

V (volume of tank) = 31,620 ft3

2.620,31

516G

=

5-102.74

3. G = 24.4

3G) Determine the mixing opportunity parameter, Gtd. Is this value within the typical range fordesign of a flocculator?

1. dt G tG d =

2. s233,33min1

s6022.724.4G

dt==

3. 4t 103.32G d =

4. Typical range: 104 105, so this result is within the typical range.


12/31


2003 SmartPros Ltd.

3H) Determine the hydraulic detention time of the sedimentation basin for a flow of 15 MGD.(Answer in hours.)

1.Q

Vtd =

LWDV = 100,0001050200V ==

Q = 15 mil gal/day

2. hr2.1day1

hr42

ft1

gal48.7

gal15,000,000

day1ft000,100t

3

3

d ==

3I) Determine the surface overflow rate for the sedimentation basin at a flow of 15 MGD.

1. Surface overflow rate =areasurface

Q

surface area = LW = 200 50 = 10,000 ft2

2. Surface overflow rate = 2gal/day/ft150010,000

15,000,000=

3J) Determine the weir overflow rate at the clarifier at a flow of 15 MGD.

1. Weir overflow rate =lengthoverflowweir

Q

2. Weir overflow rate = GPD/ft500,12ft1200

MGD15=


13/31


2003 SmartPros Ltd.

Module 4 - Groundwater

SituationA 6-inch diameter well provides drinking water to a town. The well draws water from a 200 feetthick aquifer at the rate of 100 gpm. The resulting cone of depression is a 2000-foot radiusfrom the well. The vadose zone is 50 feet thick and the drawdown at the well is 40 feet.

Requirements4A) If the water table is the upper boundary of the aquifer and is free to move up and down,

the aquifer is called:

1. When water is free to move up and down the aquifer, the aquifer is called anunconfined aquifer. The phreatic zone, or water table, is the fluctuating upperboundary of an unconfined aquifer. By contrast, water in a confined or artesianaquifer is trapped by rock and dense soil, and does not move freely up and down.

4B) The water that can be removed from an aquifer is referred to as:

1. Not all water can be removed from an aquifer. Sometimes water is trapped by rockor dense soil, so it can't be pumped to the surface. Water trapped in an aquifer iscalled specific retention. The opposite of specific retention is specific yield, which iswater that can be removed from an aquifer. Porosity is a measure of the void spacein soils or fractured rock. Hydraulic conductivity indicates the resistance to flow ofwater through soil.


14/31


2003 SmartPros Ltd.

4C) Determine the coefficient of permeability, K, for the aquifer. Assume that it is anunconfined aquifer.

1. Use the Dupuit Equation:

=

w

oe

wo

r

r

hhKQ

log

)( 22

2. Solve for K:

=

w

oe

wo r

r

hh

QK log

)( 22

Q = 100 gpm = 0.223 cfs

ho (aquifer depth) = 200 ft

hw (ho drawdown) = 200 40 = 160 ft

ro(well radius of influence) = 2000 ft

rw(pipe radius of influence) = 0.25 ft

3.

=25.0

2000log

)160200(

223.022 e

K

4. 51043.4 =K


15/31


2003 SmartPros Ltd.

4D) Due to growth of the town, the need for water will increase to 200 gpm. Determine theresulting drawdown.

1. Drawdown = ho - hw

Use the Dupuit Equation and solve for hw:

=

w

oe

wo

rr

hhKQ

log

)( 22

Q = 200 gpm = 0.446 cfs


K = 4.43 X 10-5



=

25.02000log

)200(1043.414.3446.0

225

e

wh

241039.156.50.4 wh=

106=wh

2. Drawdown = 200 106 = 94 ft


16/31


2003 SmartPros Ltd.

4E) What would be the impact on the drawdown if the well was enlarged from 6 diameter to12 diameter for a 200 gpm demand?

1. Drawdown = ho - hw

Use the Dupuit Equation and solve for hw:

=

w

oe

wo

rr

hhKQ

log

)( 22

Q = 200 gpm = 0.446 cfs


K = 4.43 X 10-5



=

5.02000log

)200(1043.414.3446.0

225

e

wh

241039.156.5699.3 wh=

116=wh

2. Drawdown = 200 116 = 84 ft


17/31


2003 SmartPros Ltd.

5. Trickling Filters

SituationA municipal wastewater with a BOD5(20 C) of 250 mg/l and suspended solids at 300 mg/l istreated with a primary clarifier which removes 35% BOD5(20C) and 60% suspended solids. Theaverage flow rate is 4 MGD with a peak flow rate of 8 MGD. Assume an effluent BOD5(20C) of30 mg/l.

Requirements5A) Determine the depth of two parallel primary clarifiers based on an average overflow rate of

1000 gpd/ft2 for each clarifier and a peak overflow rate of 2500 gpd/ft2 with a hydraulicdetention time of 2.0 hours.

1.clarifiers2

d

areasurface

tflowdailydepth

=

Find surface area for average and peak flows, than use the larger number

clarifiersbothforft40001000

4,000,000

overflowaverage

flowdailySA 2ave ===


8,000,000

overflowpeak

flowpeakSA 2peak ===

2.day

hrgal2000

4000

24,000,000depth =

=

3. ft1.11gal7.48

ft1

hr24

day1

day

hrgal2000depth

3

==


18/31


2003 SmartPros Ltd.

5B) The wastewater from the primary clarifiers is to be treated using one of two identicalparallel trickling filters. Use the National Research Council equation to determine therequired diameters for the two low rate trickling filters. Assume a depth of 6 feet for thetrickling filters.

1. Use the surface-area equation to find the diameter:4

2dSA

=

D

VSA =

Use the National Research Council equation to find Volume:

VF

WE

0085.01

1

+

=

2

2

11

0085.0

=

E

F

WV

W (BOD loading) = pounds of BOD5 coming into the filters afterpassing through the clarifiers (which remove 35% of BOD5 fromraw water)

filtersintogoinglbm/day5421)35.01(34.8250

4 ==l

mgMGDW

To find the filter efficiency, we need to know how much BOD5goes into the filters, and how much comes out. 30 mg/l come

out of the filter. 250 mg/l is the BOD5 content of raw water, andthe clarifiers remove 35%.

250 - 35% = 162 mg/l BOD5 going into the filters

Filter efficiency: 815.0162

30162=

=E

Wastewater does not recirculate, so F = 1

32

2

2

ft056,331acre

ft43,560ft-acre60.7

1815.0

11

0085.05421==

=V

176,556

056,331===

D

VSA ft2 for both filters, or 27,588 ft2 for one filter

2.4

588,272d

=

3. ft187=d


19/31


2003 SmartPros Ltd.

5C) Determine the required depth of the two secondary clarifiers operating in parallel if thedesign average overflow rate is 600 gpd/ft2 and the peak overflow rate is 1000 gpd/ft2. Thehydraulic detention time is 4 hours.

1.clarifiers2

d

areasurfacetflowdailydepth =

Find surface area for average and peak flows, than use the larger number


4,000,000

overflowaverage

flowdailySA 2ave ===


8,000,000

overflowpeak

flowpeakSA 2peak ===

2.

day

hrgal2000

8000

44,000,000depth =

=

3. ft1.11gal7.48

ft1

hr24

day1

day

hrgal2000depth

3

==


20/31


2003 SmartPros Ltd.

5D) Determine the required diameters of the two low rate trickling filters, K=0.175, using theVelz equation instead of the NRC equation. Assume the removable portion of the loaded

BODu is 90% with no recirculation, and a BODu loading rate of 20 lb per 1000 ft3 per day.

1.4

2dSA

=

BODu received/day = BODu loading rate SA D

Use the Velz equation to find depth: pKDd

L

L =10

L (removable portion of incoming BODu) = 162 0.90 = 146 mg/l

Ld (portion of BODu that remains after trickling) = 30 - (0.10 162) = 14

K = 0.175

pD175.010

146

14 =

pD= 175.0096.0log

ft8.5=pD

5,421 = 20 lb/1000 ft3/day SA 5.8

SA = 46,733 ft2 for both clarifiers, or 23,367 ft2 for one.

2.4

367,232d

=

3. d = 173 ft

5E) The BOD5 removal efficiency of two parallel low-rate trickling filters is 80%. Assuming a10% BOD5 removal by the primary clarifier what is the overall BOD5 removal by thecombined primary clarifiers and trickling filters?

1. Primary clarifiers remove 10% of BOD5. Trickling filters remove 80% of theremaining 90%.

2. 0.10 + 0.80 0.90 = 82% total BOD5 removal.


21/31


2003 SmartPros Ltd.

Module 6 - Steam Purification

SituationA city of 25,000 treats its domestic wastewater at an activated sludge treatment plant (ASTP)and discharges treated effluent into a receiving stream. The water coming into the ASTP has aBOD5 of 300. The effluent from the treatment plant is discharged at a velocity of 1.0 feet persecond through a 24-inch diameter pipe that flows full. The dissolved oxygen of the effluent is3.0 mg/l. The receiving stream has a flow of 15 cubic feet per second, a summer temperatureabove the ASTP outfall of 15 degrees C and a dissolved oxygen concentration of 9.0 mg/l. TheBOD5 of the stream above the ASTP is 2 mg/l and the ASTP effluent BOD5(20C) is 30 mg/l.

Requirements6A) Determine the flow rate in million gallons per day from the ASTP.

1. Flow: VAQ =

V = 1.0 ft/s

22

ft14.3`4

==d

A

2. MGD2.03gpd294,029,2day1

s400,86

ft1

gal48.7

s1

ft14.33

3

===Q

6B) Determine the temperature of the stream-effluent mixture immediately below the ASTPoutfall if the effluent temperature is 68 degrees F.

1. The temperature of the mix is the weighted average of the temperatures of thestream and effluent.

es

eessmix

QQ

QTQTT

++

=

)32(9

5= FTe

CTeo20)3268(

9

5==

CTmix

o

9.1514.315

14.3201515

=+

+

=


22/31


2003 SmartPros Ltd.

6C) Determine the BOD5 of the stream-effluent mixture immediately below the ASTP outfall.

1. The BOD5 content of the mix is the weighted average of the BOD5 contents of thestream and effluent.

es

eess

mixQQ

QBODQBODBOD

+

+= 555

mg/l8.614.315

14.3301525 =+

+=

mixBOD

6D) Determine the Ultimate BOD (BODu) of the stream-effluent mixture below the ASTP outfallif the stream deaeration above the ASTP outfall, kd, during the summer period (T = 15degrees C) is 0.15 per day and the laboratory kd is 0.10 for both the stream and theeffluent.

1. The BODu content of the mix is the weighted average of the BODu contents of thestream and effluent.

es

eeussu

mixuQQ

QBODQBODBOD

+

+=

)5(10.0

5u

101

BODBOD

= (0.10 is deoxygenation rate and 5 is reaction time (days))

mg/l9.2101

2BOD

)5(10.0us=

=

mg/l4410130

BOD )5(10.0ue ==

mg/l1014.315

14.344159.2BOD

mixu=

++

=


23/31


2003 SmartPros Ltd.

6E) If the stream velocity is 0.25 feet per second, and the reaeration coefficient, kr20, is 0.25per day, determine the dissolved oxygen in the stream 25 miles below the outfall. Assume

a oxygen saturation level of 10 mg/l for water that is 15.9 C.

1. tsatt DDODO =

DOsat = 10 mg/l

Use Streeter-Phelps equation to find Dt:

( ) ( )tka

tktk

dr

ud

trrdmix D

kk

BODkD +

= 101010

mg/l10=mixu

BOD

16.015.0047.1 159.15 == dk

23.025.0016.1 209.15 == rk

days1.6s86,400

day1

mile1

ft5280

ft25.0

smile25

velocity

distance

===t

Da = DOsat - DOmix

DO for the mix is the weighted average of DO for the streamand effluent

mg/l814.315

14.33159=

++

=++

=es

eessmix

QQ

QDOQDODO

Da = 10 - 8 = 2 mg/l

( ) ( ))1.6(23.01.623.01.616.0 102101016.023.0 1016.0 + =tD

Dt = 1.59 mg/l

2. mg/l41.859.110 ==tDO

6F) What treatment level was achieved at the ASTP to achieve the dissolved oxygen level atmile 25?

1. Treatment level is the difference between BOD5 content of the incoming water andthe effluent, expressed as a proportion.

90.0300

30-300= or 90%


24/31


25/31


2003 SmartPros Ltd.

6H) At what point below the outfall will the minimum dissolved oxygen be observed? Answerin miles.

1. Mile point for minimum DO is velocity times the critical time:

milesft5,280

mile

day1

sdays

s

ft61.7

1400,8686.1

25.0=


26/31


2003 SmartPros Ltd.

Module 7 - Sludge Digestion

SituationA 30 MGD wastewater treatment plant provides primary and secondary treatment to domesticwastewater for a town of 300,000. The raw wastewater suspended solids are 200 mg/l and theBOD5 (20C) is 250 mg/l. The primary clarifier removes 60% of the suspended solids and 50 mg/lof BOD5 and the secondary clarifier removes 50% of the suspended solids. The effluent fromthe wastewater treatment plant contains 20 mg/l BOD5. The sludge pumped daily to the

digesters contains 4% solids, 65% of which are volatile. Digestion at 90 F reduces the volatilesolids by 60% producing methane gas having a heat value of 600 BTU per cubic foot ofdigested sludge at a rate of 15 cubic feet per pound of volatile material destroyed.All themethane gas generated by digestion is used as fuel to heat the digesters and exchanges heatfrom the engine jacket at 75% efficiency.

Requirements7A) Determine the amount of solids removed by the primary clarifier each day.

1. You need the volume of water that is treated each day, the concentration ofsuspended solids in the water, and the removal rate of the clarifier to find how muchof solids are removed each day.

34.860.0mg200

MGD30solidsremoved =l

30,024 lb of solids/day are removed


27/31


2003 SmartPros Ltd.

7B) Determine the total amount of solids produced each day by the treatment plant. Assume ayield coefficient of 95 lb suspended solids per lbm of BOD5 utilized.

1. Total solids produced = solids from primary clarifier + solids from secondary clarifier

Solids from 1st clarifier 30,000 lb/day

Solids from 2nd

clarifier = (solids going in + solids produced inside) removal rate of 2nd clarifier

Solids going in = raw water solids (1 removal rate of 1st)

Solids going in = 200 0.40 = 80 mg/l

Solids produced inside = 0.50 BOD5 removed by 2nd clarifier

BOD5 removed = raw BOD5 BOD5 removed by 1st

effluent BOD5

BOD5 removed = 250 50 20 = 180 mg/l

Solids produced inside = 0.50 180 = 90 mg/l

Solids from 2nd

clarifier = (80 + 90) 0.50 = 85 mg/l 85 30 MGD 8.34 = 21,267 lb/day 21,000 lb/day

2. Total solids produced = 30,000 + 21,000 = 51,000 lb/day

7C) Determine the amount of volatile solids sent to the digester each day.

1. Amount sent to digester = total solids produced percent volatile solids

2. Amount sent to digester = 51,000 0.65 = 33,150 33,200 lb/day

7D) Determine the amount of volatile solids digested each day.

1. Volatile solids digested = solids sent to digester percent removed by digestion

2. Volatile solids digested = 33,200 0.60 = 19, 920 19,900 lb/day

7E) How much heat is produced each day by digestion of the volatile solids if heat is capturedfrom the engines at a 75% efficiency? Answer in BTU/day.

1. Heat produced = solids digested rate of methane produced BTU equivalent

2. Heat produced = 19,900 15 600 = 179,100,000 BTU/day


28/31


2003 SmartPros Ltd.

7F) If the specific heat of sludge is 1.0 BTU per lbm per F, how much heat would be requiredto raise the temperature of the sludge from 45 F to 90 F? Answer in BTU/hour.

1. retemperatucontentsolids

producedsolidstotalRequiredHeat =

2. BTU/hr625,390,2Flb

BTU1

hr24

day1F45

0.04

1

day

lb51,000RequiredHeat == o

o

7G) If the digester loses 10 million BTU per hour of heat through the side wall and roof, how

much supplementary heat will be needed to maintain the digestion process at 90 F?Answer in BTU/hour.

1.inputheatextraheatgeneratedbacteria

temptoheatheatlostE

+

+=

2.inputheatextraBTUM7.47

BTUM2.39BTU/hrM100.75

++=

3. extra heat input = 9.05 million BTU/hr to maintain the digestion process at 90 F


29/31


2003 SmartPros Ltd.

Module 8 - Landfills

SituationA town of 10,000 is planning to adopt a method of disposing of the solid waste (refuse)generated by the town. If the town chooses to use a landfill as the chosen method, a square 50acre site is available at the edge of town. The per capita solid waste generation rate is 1200lbm per capita year. It is estimated that the final in-place density of the well-compacted refusewill be 1000 lbm per yd3. The refuse would be applied and compacted in 24 inch lifts. The

landfill height would be limited to 25 feet. The side walls would have a 2:1 horizontal to verticalslope.

Requirements8A) Which of the refuse disposal methods available to the town would not be an appropriate

choice?

1. Three of the methods, landfilling, incineration, and composting would be appropriatechoices. Wet oxidation however, would not be appropriate because it's designed fortreating sludge.

8B) What minimum depth of daily cover is common for a landfill operation?

1. Each day you should cover the day's landfill refuse with dirt. The dirt protects therefuse from rodents, birds, and wind. If the dirt's too shallow, it won't be an effectivebarrier. If the dirt's too deep, it'll take up space that you want to use for refuse. Sixinches of dirt is the normal daily cover.

8C) The final cap for a sanitary landfill should have a minimum cover of how many inches?

1. When a section of a landfill is full, you must cover it with a thick layer of dirt topermanently seal it before you use the site for other purposes. If this final layer ofdirt's too shallow, it won't be an effective seal. If the dirt's too deep, you'll have aminimal return for the cost and effort it takes to lay the extra dirt. 24 inches of dirt isthe normal final cap for a landfill.


30/31


2003 SmartPros Ltd.

8D) If the town decides to adopt the landfill method to dispose of refuse, what volume in acre-feet would be available on the 50-acre site with a required 100 feet buffer at the edge ofthe site and an allowable height of 25 feet.

1. Find the length and width at the base of the site in ft

50 acres 43,560 ft2/acre = 2,178,000 ft2

length of 1 side = ft1476000,178,2 =

dimensions at base = 1476 ft 100 ft buffer on each side = 1276 X 1276 ft

2. Find the length and width at the top of the site in ft

50 25

1276(elevation view)

Landfill is 25-ft high; slope is 2:1, so base of each side triangle is 50 ft

Dimensions at top = 1276 50 ft on each side = 1176 X 1176

3. Find volume of center square section

V = LWH = 1176 1176 25 = 34,574,400 ft3

4. Find volume of 4 triangular sides

V = bhL

V = 50 25 1176 = 735,000 ft3 for one side

735,000 4 sides = 2,940,000 ft3

5. Find volume of 4 pyramidal corners

V = 1/3 Abh

V = 1/3 50 50 25 = 20,833 ft3 for 1 corner

20,833 4 corners = 83,333 ft3

6. Sum all the volumes for the total landfill volume

V = 34,574,400 + 2,940,000 + 83,333 = 37,597,333 ft3

37,597,333 ft3 / 43,560 ft2 = 863 acre-ft


31/31


8E) What is the estimated useful life (in years) of the landfill if the population of the town doesnot change? Assume a daily cover of 6 inches.

1.landfillannual

landfill

V

Vyrs =

2430VV placeinlandfillannual = (to account for soil cover)

densityrefuse

generationrefuseV placein =

capita/yrperrefusepopulationgenerationrefuse =

refuse generation = 10,000 12,000 = 12,000,000

3placein yd000,121000

12,000,000V ==

3landfillannual yd000,15243012,000V ==

2. yearsft

yd

yd15,000

ft37,597,333yrs

3

3

3

3

9327

1==

8F) If the town adopts a recycle program after using the landfill for 10 years, and the programreduces the solid waste generation by 5%, what is the remaining life expectancy of thelandfill?

1. Find volume of refuse after 10 years

V10 yrs = 15,000 10 = 150,000 yd3

2. Find reduced annual volume of refuse

Vreduced = 15,000 0.95 = 14,250 yd3

3. Total landfill volume equals the volume of refuse for the first ten years plus thereduced volume of refuse for n remaining years.

1,392,474 yd3 = 150,000 + 14,250 n

n = 87 years remaining landfill life

test prep rep sanitary solutions

Documents