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4-Activated Sludge F11

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Phosphorous Removal in

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CHAPTER 6 SOLUTIONS

Sheet Music

6-1

Highway rest area septic tank and tile field Given: traffic data; soil percolation rate = 5 min/cm; and GWT at 4.2 m below grade Solution: a. Compute volume of wastewater at 10% turn in Avg. day = (6,000 v/d)(0.10)(20.0 L/turn in) = 12,000 L/d Max. day = 2.5(12,000) = 30,000 L/d b. Septic tank design Assume volume = 24 h detention of max day flow. -3 3 3 V = (30,000 L/d)(1 d)(10 m /L) = 30.0 m c. Tile field design Area of trench Application rate is found from Table 6-7 at 5 min/cm 3 2 (= 0.5 min/mm) to be 0.03 m /m A=

30.0m 3 3

0.03 m m

2

= 1000m 2

Master your With semester a 1 m wide trench, withneed Scribd 1,000 m length. Use 10 trenches of 100 m length.  Read Free Foron 30this Days Sign up to vote title  & The New York usefuland tile field  Nottanks NOTE: Times it is obvious from the tank and tile field  sizeUseful why septic 3

Special offer for students: Only $4.99/month. limited to flows of 40 m /d.

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Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.

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b. Check area of tile field Sheet Music 3

2

From Table 6-7 with a sandy loam, the application rate is 0.02 m /m A=

4.0m 3 3

0.02 m m

2

= 200m 2

Therefore, the tile field is too small.

6-3

Terminal settling velocity 3

o

Given: particle radius = 0.0170 cm; density = 1.95 g/cm ; water temperature = 4 C Solution: a. Convert to SI units -2

-4

diameter = 2r = 2(0.017 cm)(10 m/cm) = 3.40 x 10 m 3

-3

6

3

3

density = (1.95 g/cm )(10 kg/g)(10 cm /m ) = 1950 kg/m

3

0

b. From Appendix A, Table A-1 at 4 C find

µ = 1.567 x 10-3 Pa-s -3

(NOTE: factor of 10 to convert from mPa-s to Pa-s) c. Solve Eqn. 4-98

Master your semester with Scribd 9.80  1950 − 1000  v =  (3.40 × 10 ) & The New York Times 18  1.567 × 10 

−4 2

s

−3

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-2

-2

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Read Free Foron 30this Days Sign up to vote title 5 = ( )(6.063 )(1.156 × 10−7 ) 0.5444 10useful Useful  ×Not Cancel anytime.

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-4

velocity = (0.0950 cm/s)(10 m/cm) = 9.50 x 10 m/s Sheet Music 3

-3

6

3

3

density = (2.05 g/cm )(10 kg/g)(10 cm /m ) = 2050 kg/m

3

0

b. From Appendix A, Table A-1 at 15 C find

µ = 1.139 x 10-3 Pa-s -3

(NOTE: factor of 10 to convert from mPa-s to Pa-s) c. Solve Eqn. 4-98 for d

 µ   18   d = (v s )  ρ − ρ 9 . 80    s  

12

−3   − 4  18  1.139 × 10   d = (9.50 × 10 )   9.80  2050 − 1000  

6-5

12

= 4.35 × 10 −5 m

Horizontal flow, gravity grit chamber in winter and summer 3

Given: diameter = 0.020 cm; particle density = 1.83 g/cm ; water temperature = 1 grit chamber depth = 1.0 m; detention time in grit chamber = 60 s; assume de 3 of water = 1000 kg/m Solution: For winter conditions

Master youra.semester with Scribd Convert diameter to m & The New York Times 0.020cm

Special offer for students: Only $4.99/month. = 0.00020 m

100 cm m

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vs =

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g(ρ s − ρ)d 2 18µ 2

vs =

9.80(1830 − 1000 )(0.00020 ) 18(1.1235 × 10

−3

= 0.01464 m s

)

e. In 60 s (the detention time) the particle will fall: h = (0.01464m/s)(60s) = 0.88 m This is less than 1.0 m depth of the grit chamber. The particle will captured in winter. For summer conditions a. Data for (a) and (b) in winter still apply o

b. From Appendix A at 25 C

µ = 0.890 × 10 −3 Pa ⋅ s c. Solving 2

vs =

9.80(1830 − 1000 )(0.00020 ) 18(0.890 × 10 −3 )

In 60 s Master yourd.semester with Scribd = (0.02031m/s)(60s) = 1.2 m & The New York hTimes Special offer for students: Only $4.99/month.

= 0.02031 m s


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Useful

This is greater than 1.0 m depth. The particle will be captured.

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beginning of the cycle. Sheet Music 3

3

The "volume-out" = (0.400 m /s)(3,600 s/h)(1 s/h)(1 h) = 1440 1440 m . Because of space limitations volume out is not shown in the table below. Time 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 0000 0100 0200 0300 0400 0500 0600 0700 0800

Flow Volume-in 3 Accumulated dS/dt (m ) 3 3 3 (m /s) (m ) dS/dt (m ) 0.446 1605.6 165.6 165.6 0.474 1706.4 266.4 432.0 0.482 1735.2 295.2 727.2 0.508 1828.8 388.8 1116.0 0.526 1893.6 453.6 1569.6 0.530 1908.0 468.0 2037.6 0.552 1987.2 547.2 2584.8 0.570 2052.0 612.0 3196.8 0.596 2145.6 705.6 3902.4 0.604 2174.4 734.4 4636.8 0.570 2052.0 612.0 5248.8 0.552 1987.2 547.2 5796.0 0.474 1706.4 266.4 6062.4 0.412 1483.2 43.2 6105.6 0.372 1339.2 -100.8 6004.8 0.340 1224.0 -216.0 5788.8 0.254 914.4 -525.6 5263.2 0.160 576.0 -864.0 4399.2 0.132 475.2 -964.8 3434.4 0.132 475.2 -964.8 2469.6 0.140 504.0 -936.0 1533.6 0.160 576.0 -864.0 669.6 0.254 914.0 -525.0 144.0 0.360 1296.0 -144.0 0.0

Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  c. The maximum volume and the volume of theequalization is 6105.6 plus 2 & The New York Times Useful  basin Not useful excess = 7630 m Special offer for students: Only $4.99/month.

3

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entry. This occurs just before the "volume-in" begins to exceed the "volume-out". is the beginning of the cycle. The "volume-out" is equal to the average volume-in = 3 352.215 m . Because of space limitations the volume out is not shown in the table bel

Time 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 0000 0100 0200 0300 0400 0500 0600 0700

Flow Volume-in Accumulated 3 dS/dt (m ) 3 3 3 (m /s) (m ) dS/dt (m ) 0.1130 406.80 54.585 54.585 0.1310 471.60 119.385 173.970 0.1350 486.00 133.785 307.755 0.1370 493.20 140.985 448.740 0.1350 486.00 133.785 582.525 0.1290 464.40 112.185 694.710 0.1230 442.80 90.585 785.295 0.1110 399.60 47.385 832.680 0.1030 370.80 18.585 851.265 0.1040 374.40 22.185 873.450 0.1050 378.00 25.785 899.235 0.1160 417.60 65.385 964.620 0.1270 457.20 104.985 1069.605 0.1280 460.80 108.585 1178.190 0.1210 435.60 83.385 1261.575 0.1100 396.00 43.785 1305.360 0.0875 315.00 -37.215 1268.145 0.0700 252.00 -100.215 1167.930 0.0525 189.00 -163.215 1004.715 0.0414 149.04 -203.175 801.540 0.0334 120.24 -231.975 596.565 0.0318 114.48 -237.735 331.830 0.0382 137.52 -214.695 117.135 0.0653 235.08 -117.135 0.000 Read Free Foron 30this Days Sign up to vote title

Master your semester with Scribd   c. The maximum basin 1305.36 plus 25 & The New York Timesvolume and the volume of the equalization Notisuseful  Useful  3

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Time 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 0000 0100 0200 0300 0400 0500 0600 0700


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Flow Volume-in Volume-out 3 Accumulated dS/dt (m ) 3 3 3 3 (m /s) (m ) (m ) dS/dt (m ) 0.0033 11.88 10.875 1.005 1.005 0.0039 14.04 10.875 3.165 4.17 0.0047 16.92 10.875 6.045 10.215 0.0044 15.84 10.875 4.965 15.18 0.0041 14.76 10.875 3.885 19.065 0.0041 14.76 10.875 3.885 22.95 0.0042 15.12 10.875 4.245 27.195 0.0038 13.68 10.875 2.805 30 0.0033 11.88 10.875 1.005 31.005 0.0039 14.04 10.875 3.165 34.17 0.0046 16.56 10.875 5.685 39.855 0.0046 16.56 10.875 5.685 45.54 0.0044 15.84 10.875 4.965 50.505 0.0034 12.24 10.875 1.365 51.87 0.0031 11.16 10.875 0.285 52.155 0.002 7.2 10.875 -3.675 48.48 0.0012 4.32 10.875 -6.555 41.925 0.0011 3.96 10.875 -6.915 35.01 0.0009 3.24 10.875 -7.635 27.375 0.0009 3.24 10.875 -7.635 19.74 0.0009 3.24 10.875 -7.635 12.105 0.0013 4.68 10.875 -6.195 5.91 0.0018 6.48 10.875 -4.395 1.515 0.0026 9.36 10.875 -1.515 7.99361E-15

Average = Maximum volume = Design volume =

3

10.875 m 3 52.155 m 3 65.19375 m



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MBOD-in = (Q)(So)(t) 3

3

-3

= (0.446 m /s)(170 g/m g/m )(1 h)(3,600 s/h)(10 kg/g) = 272.952 kg S=

S=

Vi So + VSS Vi + VS

(1605.6)(170) + (0) 1605.6 + 0

= 170 g m 3

MBOD-out = (Q)(S)(t) 3

3

-3

= (0.400 m /s)(170 g/m g/m )(1 h)(3,600 s/h)(10 kg/g) = 244.80 kg

At 1000 S=

(1706.4)(220) + (165.6)(170) 1706.4 + 165.6

= 215.577 g m 3

c. Tabulation of BOD mass flow



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Time 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 0000 0100 0200 0300 0400 0500 0600 0700 0800 Average

of 66

MBOD in (kg) 272.95 375.41 433.80 490.12 534.00 534.24 532.57 513.00 439.85 365.30 287.28 258.34 249.13 234.35 206.24 150.55 107.90 54.72 38.02 40.39 47.88 57.60 107.90 176.26

 

S 170.00 215.58 243.14 260.93 274.19 277.38 272.75 262.68 239.51 213.93 191.25 174.43 167.96 166.00 163.84 156.92 151.62 146.03 39.59 132.96 126.53 119.28 118.54 34.25

271.16

July 2011 Issue


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MBOD out (kg) 244.80 310.43 350.12 375.73 394.83 399.42 392.76 378.26 344.90 308.05 275.39 251.17 241.86 239.04 235.94 225.98 218.33 210.29 201.02 191.46 182.20 171.77 170.70 193.33 271.16

Master yourd.semester with Scribd Average BOD concentration flowing out of equalization basin Read Free Foron 30this Days Sign up to vote title & The New York Times (271.16kg )(1000 g kg)  Useful  Not useful Special offer for students: Only $4.99/month. BOD AVG =

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(0.400 m s )(1h )(3600 s h ) 3

= 188.303 mg L

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Given: Data from Problem 6-7 Solution: a. The solution was computed using a spreadsheet program. Time 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 0000 0100 0200 0300 0400 0500 0600 0700

MBOD in (kg) 50.85 66.02 72.90 76.45 77.76 69.66 61.99 53.95 48.20 44.93 47.25 62.64 91.44 99.07 74.05 51.48 34.65 20.41 10.02 5.22 3.85 4.81 9.08 21.63

S 125.00 138.44 146.95 151.91 156.12 153.40 148.19 143.74 139.51 133.55 130.97 137.00 157.26 174.64 173.39 163.02 152.72 140.83 128.59 116.50 105.48 94.86 86.40 90.14

MBOD out (kg) 44.03 48.76 51.76 53.50 54.99 54.03 52.19 50.63 49.14 47.04 46.13 48.25 55.39 61.51 61.07 57.42 53.79 49.60 45.29 41.03 37.15 33.41 30.43 31.75

Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Average 48.26

48.26


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Compare equalized and unequalized BOD loading Given: Excel data from Problem 6-8 Solution: See following spreadsheet

Time 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 0000 0100 0200 0300 0400 0500 0600 0700

Flow 3 (m /s) 0.0033 0.0039 0.0047 0.0044 0.0041 0.0041 0.0042 0.0038 0.0033 0.0039 0.0046 0.0046 0.0044 0.0034 0.0031 0.002 0.0012 0.0011 0.0009 0.0009 0.0009 0.0013 0.0018 0.0026

Volume-In 3 (m ) 11.88 14.04 16.92 15.84 14.76 14.76 15.12 13.68 11.88 14.04 16.56 16.56 15.84 12.24 11.16 7.2 4.32 3.96 3.24 3.24 3.24 4.68 6.48 9.36

BOD In (mg/L) 150 195 235 265 290 290 275 225 170 180 190 190 190 160 125 80 50 34 30 30 33 55 73 110

MBOD (kg) 1.78 2.74 3.98 4.20 4.28 4.28 4.16 3.08 2.02 2.53 3.15 3.15 3.01 1.96 1.40 0.58 0.22 0.13 0.10 0.10 0.11 0.26 0.47 1.03

Volume-Out 3 (m ) 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875 10.875

Master your semester with Scribd & The New York Times Average 10.875 151 2.03 Special offer for students: Only $4.99/month. 3 Maximum volume 52.155 m

dS/dt Accumulated S 3 3 3 (m ) dS/dt (m ) (g/m 1.005 1.01 150.00 3.165 4.17 191.99 6.045 10.22 226.50 4.965 15.18 249.90 3.885 19.07 269.67 3.885 22.95 278.54 4.245 27.20 277.14 2.805 30.00 259.69 1.005 31.01 234.25 3.165 34.17 217.34 5.685 39.86 208.41 5.685 45.54 203.01 4.965 50.51 199.65 1.365 51.87 191.92 0.285 52.16 180.07 -3.675 48.48 167.93 -6.555 41.93 158.28 -6.915 35.01 147.56 -7.635 27.38 137.60 -7.635 19.74 126.21 -7.635 12.11 113.07 -6.195 5.91 96.88 -4.395 1.52 84.39 -1.515 0.00 106.43 Read Free Foron 30this Days Sign up to vote title

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Primary settling tank design

Sheet Music

Given: Graph of batch settling data Solution: Note: This solution follows example 4-23

Prob. 6-12 Solution Batch Settling Curves 0 0.5 1 m1.5 , h t 2 p e 2.5 D 3 3.5 4

% Removal

100% Line

0

50

100

55% 60% 65% 70% 75% 80% 90% 150

Time, min

Master your semester with Scribd Figures S-6-12a: Batch settling curves & The New York Times a. Only Required removal is Special offer for students: $4.99/month.


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v0 =

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4.0m 50 min




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(1440 min d ) = 87.3 m d

2. Detention time at 60% removal t0 = 50 min 3. % removal (from vertical line at 50 min)

R T 50 = 60 +

3.4 4.0

(65 − 60) +

2.5 4.0

(70 − 65) +

1.95 4.0

(75 − 70) +

1.6 4.0

(80 − 75) +

1.1 4.0

(90 − 80 ) +

0.4 4.0

(100 − 90 )

= 75.55%

This is too high. Plot another vertical at 55%. The overflow rate and dete time are: v0 =

4.0m 40 min

(1440 min d ) = 144.0 m d

t0 = 40 min The % removal (from vertical line at 40 min)

R T 50 = 55 +

3.2 4.0

( 60 − 55) +

2.0 4.0

( 65 − 60) +

1.5 4.0

( 70 − 65) +

1.15 4.0

( 75 − 70) +

0.85 4.0

( 80 − 75) +

0.55 4.0

( 90 − 80) +

0.2 4.0

= 67.74% Plot % removal vs v and % removal vs t Master yourc.semester with Scribd & The New York Times o


0


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% Removal vs. Vo 76 75 74 73 l a v 72 o m e R 71 %

70 69 68 67 0

20

40

60

80

100

120

140

160

overflow rate, Vo (m/d)

Figure S-6-12b: % Removal vs. v o Read vo = 125 m/d

% Removal vs. to

76 75 74

Master your semester with Scribd & The New York Times 73

l a v 72 o m Special offer for students: eOnly $4.99/month. R 71 %


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Read to = 43 min Sheet Music

d. Apply scale up factors (See example 4-23) vo = (125 m/d)(0.65) = 81.25 or 80 m/d to = (43 min)(1.75) = 75.25 or 75 min 6-13

Primary settling tank design Given: Settling column data Solution:

This problem is virtually the same as Problem 4-98. The only difference is that the influent suspended solids concentration is a factor of ten higher. The solution is the s as that for Problem 4-98. 6-14

Primary settling tank design Given: Settling column data Solution:

This problem is virtually the same as Problem 4-99. The only difference is that the influent suspended solids concentration is a factor of ten higher. The solution is the s as that for Problem 4-99. 6-15

Size primary tanks for Cynusoidal City

Master yourGiven: semester with Scribd  Average flow in Problem 6-9 (which is average flow in Read Free ForProblem 30this Days Sign up to vote on title6-6 = 0.400 /s) is design design flow, overflow overflow rate is 26.0 m/d, detention detention time is 2.0 h.  Ass & The New York msedimentation Times  Useful  Not useful tanks with length to width ratio of 4.7. 3


Solution:

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c. Compute volume of one tank

 1  1  3   = 192.0m  24 h d  15 tan ks 

V = (Q )(t 0 ) = (34560 m 3 d )(2h ) d. Compute depth of tank h=

V As

=

192.0m

3

88.62m 2

= 2.17m

e. Compute width of tank using assumption that length to width ratio is 4.7 (w)(4.7 w) = As = 88.62 m 2 w =

88.62 4.7

2

= 18.8553 1/2

w = (18.8553) = 4.34 m f. Compute length of tank l = (4.7)(4.34 m) = 20.398 or 20.4 m g. Tank dimensions = 2.17 m x 4.34 m x 20.4 m h. Maximum overflow rate

Master your semester Q with (0.604Scribd m s )(86400 s d ) Foron 30this Days to vote title = = 39Read v = .Sign 3 m up dFree ( ) A ( ) 15 tan ks 88 . 62 m & The New York Times  Useful  Not useful 3

max

0

2

s


6-16

Surface area of tank and detention time

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As =

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Q v0

=

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49248 m 3 d 60 m d

July 2011 Issue


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= 821m 2

c. Detention time t0 = 6-17

V As

(821m )(3m ) (24 h d ) = 1.20h = 2

49248 m 3 d

New overflow rate and detention time for Problem 6-16 3

2

Given: Flow rate = 0.400 m /s, surface surface area = 821 821 m , depth = 3.0 m Solution: 3

a. Convert flow rate to m /d 3

3

Q = (0.400 m /s)(86,400 s/d) = 34,560 m /d b. Compute overflow rate v0 =

Q As

=

34560 m 3 d 821m 2

= 42.1 m d

c. Detention time t0 =

V As

(821m )(3m ) (24 h d ) = 1.71h = 2

3

34560 m d

Master your semester with Scribd 6-18 BOD removal in primary tank & The New York Times 5


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3

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Given: Influent BOD 5 = 345 mg/L; average flow rate = 0.050 m /s; removal efficien Special offer for students: Only $4.99/month. 30%

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M removed = (0.30)(1490.4 kg d ) = 447.12 kg d

Sheet Music

6-19

Suspended solids removal in primary tank 3

Given: Influent suspended solids = 435 mg/L; average flow rate = 0.050 m /s; remov efficiency = 60% Solution: a. Mass of suspended solids entering primary tank 3 (NOTE: 1.0 mg/L = 1.0 g/m ) M ss = (435 g m 3 )(0.050 m 3 s )(86400 s d )(10 −3 kg g ) = 1879.2 kg d

b. Mass of suspended solids removed M removed = (0.60)(1879.2 kg d ) = 1127.5 kg d 6-20

Number of generations 5

Given: P0 = 3.0 x 10 ; P at 36 hours = 9.0 x 10

8

Solution: a. Solve Eqn. 6-2 for n n = 3.3 log

P P0

Master your semester with Scribd b. Substitute values & The New York Times Special offer for students: Only $4.99/month. 9.0 × 10 8

n = 3.3 log

3 0 × 10 5


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= 11.47 or 11 generations

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Problem 6-21 Log Growth Curve 1.00E+07 t 1.00E+06 n u o C 1.00E+05 l a i r e 1.00E+04 t c a B 1.00E+03

1.00E+02 0

10

20

30

40

50

Time, h Figure S-6-21: Log growth curve log growth starts at 10 h log growth stops at 35 h b. Number of generations by solving Eqn. 6-2 as in Prob. 6-20 n = 3.3 log 6-22

E. Coli growth

1.05 × 10 6 1.5 × 10 3

= 9.39 or 9 generations

curve

Master your semester with Scribd Given: Time, bacterial count, pH data & The New York Times Special offer for students: Only $4.99/month. Solution:


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Problem Problem 6-22 Log Growth Curve 1.00E+09 Death

t 1.00E+08 n u 1.00E+07

Stationary

o C 1.00E+06 l a i r 1.00E+05 e t c 1.00E+04 a B

Log growt g rowth h

1.00E+03 1.00E+02 0

100

200

30

Time, h Figure S-6-22: E. Coli log growth curve

6-23

Volume of aeration tank

Given: Example 6-5 assumptions, BOD from Problem 6-6, Q from Problem 6-6, 32% removal of BOD in primary tank  Read Free Foron 30this Days Sign up to vote title  Solution:  Useful  Not useful


a. From Problem 6-6

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S = 30.0 -(0.63)(30.0) = 11.1 mg/L d. Calculate mean cell residence time from Eqn 6-21 and assumptions for growth constants from Example 6-5 11.1 =

100(1 + 0.05θ C )

θ C (2.5 − 0.05) − 1

11.1[θc (2.45) - 1] = 100 + 5.0 (θc) 21.195 θc - 11.1 = 100 + 5.0 (θc) 22.195 θc = 111.1

θc = 5.0 d e. Calculate hydraulic residence time using Eqn. 6-23 and assumptions for growth constants from Example 6-5 2000 =

θ=

(5.0)(0.5)(127.8 − 11.1) θ(1 + (0.05)(5.0 ))

291.75

(2000 )(1.25)

= 0.1167 d or 2.80 h

f. Calculate volume using Eqn 6-18

Master your semester with Scribd V =θQ = (2.80 h)(0.400 m /s)(3,600 s/h) s/h) = 4,032 or or 4,000 m Read Free Foron 30this Days Sign up to vote title & The New  Useful  Not useful 6-24 York VolumeTimes of aeration tank 3


3

 

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Given: Example 6-5 assumptions, BOD from Problem 6-7, Q from Problem 6-7, 32%

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b. BOD from primary tank Sheet Music

So = (1 - 0.32)(137) = 93.16 mg/L c. From Example 6-5 S = 30.0 -(0.63)(30.0) = 11.1 mg/L d. Calculate mean cell residence time from Eqn 6-21 and assumptions from Example 11.1 =

100(1 + 0.05θ C )

θ C (2.5 − 0.05) − 1

11.1[θc (2.45) - 1] = 100 + 5.0 (θc) 21.195 θc - 11.1 = 100 + 5.0 (θc) 22.195 θc = 111.1

θc = 5.0 d e. Calculate hydraulic residence time using Eqn. 6-23 and assumptions in Table 6-11

2000 =

θ=

(5.0)(0.5)(93.16 − 11.1) θ(1 + (0.05)(5.0 ))

205.15

= 0.0821d or 1.97 h

(2000 )with (1.25) Scribd Master your semester

& The New York Times f. Calculate volume using Eqn 6-18





3


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V = θQ = (1.97 h)(0.0978 m /s)(3,600 s/h) s/h) = 693.6 or 700 m

3

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3

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Sheet Music

Qavg = 0.0030 m /s BODavg after equalization = 187 mg/L b. BOD from primary tank So = (1 – 0.32)(187) = 127.16 c. From Example 6-5 S = 30.0 – (0.63)(30.0) = 11.1 mg/L d. Calculate mean cell residence time from Equation 6-21 11.1 =

100(1 + 0.05θ C )

θ C (2.5 − 0.05) − 1

11.1[θc (2.45) - 1] = 100 + 5.0 (θc) 21.195 θc - 11.1 = 100 + 5.0 (θc) 22.195 θc = 111.1

θc = 5.0 d

e. Calculate hydraulic residence time using Equation 6-23 and assumptions in Table 6 2000 =

(5.0)(0.5)(127.16 − 11.1) θ(1 + (0.05)(5.0 ))

Master your semester with Scribd 290.15 θ= = 0.116d or 2.785 h ( ) ( ) 2000 1 . 25 & The New York Times Special offer for students: Only $4.99/month.

f. Calculate volume using Equation 6-18


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Solution: a. Calculate allowable S S = (25.0 mg/L BOD 5) – (0.70)(30 mg/L suspended solids) = 4 mg/L b. Calculate S o So = (240 mg/L) – (240 mg/L)(0% removed) = 240 mg/L c. Calculate θc

θc = θc =

Ks + S

(µ m )(S) − (S)(k d ) − (K s )(k d ) 100 mg L + 4 mg L

(10d − )(4 mg L) − (4 mg L)(0.025d − ) − (100 mg L )(0.025d − ) 1

1

1

= 2.78d

d. Calculate θ

θ=

θ c Y(So − S) MLVSS(1 + (k d )(θ c ))

θ=

(2.78d )(0.8)(240 − 4) = 0.164d 3000(1 + (0.025)(2.78))

e. Calculate the volume

Master your semester with Scribd V = (0.029 m /s)(0.164 d)(86400 d)(86400 s/d) s/d) = 410 m upFree Read Foron 30this Days Sign to vote title & The New York Times  Useful  Not useful 3

6-27

Rework Example 6-5 using spreadsheet


3

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1000

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MLVSS concentration 2500

1500

T otal BOD allowed Sus. Solids allowed % of SS = BOD

30 mg/L 30 mg/L 63 %

30 mg/L 30 mg/L 63 %

30 mg/L 30 mg/L 63 %

Flow rate Soluble BOD raw % Removed in P.S.

3 0.15 m /s 84 mg/L 0

3 0.15 m /s 84 mg/L 0

3 0.15 m /s 84 mg/L 0

100 mg/L BOD 0.05 d

100 mg/L BOD 0.05 d

100 mg/L BOD 0.05 d

Growth constants Ks kd

-1

2.5 d 0.5 mg mg VSS/mg BOD

µm Y



-1


300 3 3

0.15 8

10 0.05

-1


2.5 0

Design MLVSS

1000 mg/L

1500 mg/L

2500 mg/L

300

Allowable S

11.1 mg/L

11.1 mg/L

11.1 mg/L

11

84 mg/L

84 mg/L

84 mg/L

So

8

θc

5.0056 d

5.006 d

5.0056 d

5.005

θ

0.1459 d

0.097 d

0.0584 d

0.048

Volume of A.T.

3 1891.3 m

3 756.51 m

630.42

6-28

3 1261 m

Effect of MLVSS on effluent soluble BOD Given: Example 6-5 and MLVSS values from 6-27

Master yourSolution: semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times  Useful  Not useful a. The solution was computed using a spreadsheet program Special offer for students: Only $4.99/month.

b. Results from spreadsheet

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

Volume of A.T . Flow rate So

97 0 m 3 0.15 m3/s 84 mg/L

97 0 m 3 0.15 m3/s 84 m g / L

9 70 m 3 0.15 m3/s 84 mg/L

97 0 m 0.15 m 84 m

Growth constants Ks

100 mg/L BOD

100 mg/L BOD

100 mg/L BOD

10 0 m

kd

0.05 d(-1)

0.05 d(-1)

0.05 d(-1)

0.05 d

µm Y

2.5 d(-1) 0.5 mg mg VSS/mg BOD

Design MLVSS

2.5 d(-1) 0.5 mg mg VSS/m g BOD

2.5 d(-1) 0.5 mg mg VSS/mg BOD

2 .5 d 0.5 m

1000 mg/L

1 5 0 0 m g /L

2500 mg/L

5d

5 d

5 d

5d

θ

0 . 07484 6 d

0 . 0748 46 d

0.074846 d

0 . 07484 6 d

S

46 . 577 1 6

27 . 86 574

θc

6-29

-9.557099

30 0 0 m

-28.26852

Volume of aeration tank using F/M 3

-1

Given: Q = 0.4380 m /s, F/M = 0.200 d , BOD = 150 mg/L, MLVSS = 2200 mg/L Solution: 3

a. Using Eqn. 6-26 (Note: mg/L = g/m )

(0.4380 m s)(86400 s d )(150 g = (2200 g m )(V) 3

0.200d

V=

−1

m

3

)

3

5676480

(0.200)(2200 )

= 1.29 × 10 4 m 3

6-30 Settled volume with Scribd Master your semester Given: Sludge in Example 6-8; X = 5700 mg/L & The New York Times r

Solution: Special offer for students: Only $4.99/month.


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Reduce return sludge flow 3

3

Given: Example 6-8; Q r = 0.150 m /s,Qr desired = 0.0375 m /s Solution:

a. Derive an expression for Q w in terms of SVI using Eqn 6-19 and the ratio of MLSS MLVSS from Example 6-8 and a conversion factor to make units consistent QW =

QW =

VX

θc X r

=

6.4213 X ′r

(970)( )(2000 ) 

 1    X ′r   86400 s d  5   1.43 

=

6.4213(SVI ) 10 6

b. Using Eqn 6-33 in Eqn 6-29 with the derived expression above

 10 6   10 6   + Q W   (Q + Q r )(X ′) = Q r  SVI SVI    

(0.0375)(10 6 ) (6.4213)(SVI )(10 6 ) (0.150 + 0.0375)(2860 ) = + SVI (10 6 )(SVI) (0.0375)(10 6 )

+ 6.4213 536.25 = with Scribd Master your semester SVI & The New York Times

(

(0.0375) 10 Special offer for students: Only $4.99/month. SVI = 529.827

6

) = 70.78 or 70 mL/g


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θ=

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(7.0m )(30.0m )(4.3m )(2 tan ks ) 3

0.0796 m s



= 22688s or 6.3 h 3

b. F/M ratio (Eqn. 6-26 and note that mg/L = g/m )

(0.0796 m s )(86400 s d )(130 g = M (1806m )()(1500 g m ) 3

F

3

3

m

3

) = 0.33 mg

mg ⋅ d

c. SVI (Eqn. 6-27) SVI =

230.0 mL L

(1.40)(1500 mg L )

(1000 mg g ) = 109.52 or 110 mL/g

d. Solids concentration in return sludge (Eqn. 6-20) Xr = 6-33

10 6 109.5

= 9132.4 or 9130 mg/L

Evaluation of Lotta Hart hospital activated sludge plant Given: size and operating characteristics Solution: a. Aeration period (Eqn. 6-18)

(10.0m )(10.0m )(4.5m ) = 0.750d or 18 h 500beds )(1.200 m 3 bed ⋅ d ) ( Master your semester with Scribd θ=

Read Free Foron 30this Days Sign up to vote title 3 b. F/M ratio (Eqn. 6-26 and note that mg/L = g/m ) Useful  Not useful

& The New York Times

Cancel anytime.


F

(750 m

3

d )(500 g m 3 )

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Xr =

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6-34

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10 6 66.67

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

= 14999.25 or 15000 mg/L

Evaluation of Jambalaya activated sludge plant Given: size and operating characteristics Solution: a. Aeration period (Eqn. 6-18)

θ=

(8.0m )(8.0m )(5.0m ) 0.012 m 3 s

= 26666.67s or 7.4 h or 0.31 d

b. F/M ration (Eqn. 6-26) Note: mg/L = g/m

(0.012 m s )(86400 s d )(966 g = M (320m )(2000 g m ) 3

F

3

3

3

m3 )

= 1.565 or 1.6 mg/mg-d

c. SVI (Eqn. 6-27) SVI =

225.0mL

(1.25)(2000 mg L )(1.0L )

(1000 mg g ) = 90.0 mL g

d. Solids concentration in return sludge (X r) (Eqn. 6-33) Xr =

10

6

= 11111 or 11000 mg/L

Master your semester with Scribd 6-35 York Sludge age and wastage at Turkey Run & The New Times 90






Useful

Given: Data from Problem 6-32 and operating assumptions

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b. Sludge wasting (Eqn. 6-19) Sheet Music

Recognize that wasting from aeration tank means that X r = X so Eqn 6-19 reduces to

θc =

V QW

and QW =

V

θc

=

(7.0m )(30.0m )(4.3m )(2 tan ks ) 11.5d 3

3

Qw = 157.04 m /d or 0.00182 0.00182 m /s c. Return sludge flow rate (Eqn. 6-30) From Problem 6-32 X’ = MLSS = (1.40)(1,500) = 2100 mg/L Xr’ = 9132.4 mg/L 3 Q = 0.0796 m /s Then Qr =

(0.0796 )(2100) − (0.00182 )(9132.4) 9132.4 − 2100 3

Qr = 0.02141 or 0.0214 m /s

Master your semester with Scribd 6-36 Solids retention and wastage at Lotta Hart Read Free Foron 30this Days Sign up to vote title & The New York Times  Useful  Not useful Given: Data from Problem 6-33 and operating assumptions Special offer for students: Only $4.99/month.

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2500 =

1875 =

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θ c (0.60)(500 − 16.675) (0.75)(1 + (0.06)θ c ) 289.995θ c 1 + 0.06θ c

177.495θ c = 1875

θ c = 10.56d

c. Sludge wasting (Eqn. 6-20)

Note that the assumption in deriving Eqn. 6-11 , namely that X e = 0 is not true becau the effluent suspended solids are 30.0 mg/L. Thus, Eqn. 6-11 must include this loss.

θc =

VX Q w X r + (Q − Q w )X e

The effluent suspended solids is Xe = (1.000 - 0.6667)(25.0 mg/L) = 8.3325 mg/L

The return sludge MLVSS is computed using the ratio of MLVSS to MLSS from Pro 33 Xr = 15000/1.2 = 12500 mg/L

Master your semester with Scribd Free Foron 30this Days (450m )(2500 g m ) Read Sign up to vote title 10.56 = & The New York Times  Useful Q (12500 g m ) + (750 − Q )(8.3325 g m ) Not useful 3

3

3


3

w

w

3

3

Q = 8.03 m /d or 0.000093 0.000093 m /s

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Qr =

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(0.00868)(3000 ) − (0.000093)(15000 ) 15000 − 3000



= 0.0023 m 3 s

Solids retention and wastage at Jambalaya Given: Data from 6-34 and operating assumptions Solution: a. Find S from effluent suspended solids and inert fraction s in Example 6-5 S = 25.0 − [(1.0 − 0.30)(30.0)] = 4.0 b. Sludge age 2000 =

620 =

θ c (0.50)(966 − 4.0) (0.31)(1 + 0.075 ⋅ θ c )

(481.0)θ c 1 + 0.075 ⋅ θ c

θ c = 1.427 or 1.43 d c. Sludge wasting (Eqn. 6-20) The effluent suspended solids is

X = (1.00 − 0.30)(30.0 mg L ) = 21.0 mg L Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  The return sludge flow is computed using the ratioofUseful MLVSS MLSS & The New York Times usefulfrom Problem toNot e

34


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3



QW = 48.6 m /d Sheet Music

d. Return sludge flow rate X’ = MLSS = (1.25)(2000) = 2500 mg/L Xr’ = 11000 mg/L 3

Q = 0.012 m /s QW =

Qr =

6-38

48.61 m 3 d 86400 s d

= 0.0056 m 3 s

(0.012 )(2500 ) − (0.0056)(11000 ) 11000 − 2500

= 0.0028 m 3 s

Evaluation of secondary settling tanks at Turkey Run Given: Problem 6-32; tanks are 16.0 m diameter and 4.0 m deep at side wall. Solution: a. Overflow rate The design standard is 33 m/d. The overflow Q = Q of wastewater into the plant = 3 0.0796 m /s from Prob. Prob. 6-32. The radius radius of the tank is 16.0/2 = 8.0 m. m.

(0.0796 m s )(86400 s d ) = 17.1 m d = 3

vo

8.0m ) (2Scribd tan ks ) Master your semesterπ(with & The New York Times This is less than 33 m/d, therefore okay. Special offer for students: Only $4.99/month.

b. Side water depth

2


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3

From Prob. 6-35, Q r = 0.0214 m /s and then r=

Qr Q

=

0.0214 0.0796

= 0.2688

So

(1 + 0.2668)(0.0796 )(1.40)(1500 )(86400 ) 1.833 × 10 7 SL = = = 45.57 kg m 2 5 (402.12)(1000 ) 4.021 × 10 2

Checking with Figure 6-28, we find this is much less than the 253 kg/m -d allowed. d. Weir loading

(0.0796 m s)(86400 s d ) = 68.4 m WL = 3

3

π(16.0m )(2 tan ks )

d⋅m 3

3

This is less than the design standard of 125 m /d-m to 250 250 m /d-m. 6-39

Evaluation of secondary settling tank at Lotta Hart Given: Problem 6-33, 10.0 m diameter, 3.4 m deep tank Solution: a. Overflow rate

Master your semester with Scribd  Foron 30this Days Sign to vote title The design standard is 33 m/d. The overflow Q =Read Q ofupFree wastewater into the plant  from & The New York Times Prob. 6-33. The radius of the tank is 10.0/2 = 5.0 m.Useful  Not useful Special offer for students: Only $4.99/month.

(500beds )(1 2 m 3 d bed )

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SL =

July 2011 Issue

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(1 + r )(Q)(X )(86400 s d ) (A s )(1000 g kg )

where r = recycle ratio; X = MLSS = (1.20)(MLVSS) 3

From Prob. 6-36, Q r = 0.0023 m /s and then r=

Qr Q

=

0.00157 0.00694

= 0.3314

So

(1 + 0.3314 )(0.00694 )(1.20)(2000 )(86400 ) 1.916 × 10 6 = = 24.395 kg SL = (78.54)(1000 ) 7.854 × 10 4 2

Checking with Figure 6-28, we find this is much less than the 300 kg/m -d allowed. d. Weir loading

(0.00694 m s )(86400 s d ) = 599.62 = 19.1 m WL = 3

π(10.0m )

31.42

3

d⋅m

3

3

This is less than the design standard of 125 m /d-m to 250 250 m /d-m. 6-40

Evaluation of secondary settling time at Jambalaya

Problem 6-34; tank diameter = 5.0 m; 2.5 m depth at side wall Master yourGiven: semester with Scribd Read Free Foron 30this Days Sign up to vote title Solution: & The New York Times  Useful  Not useful Special offer for students: Only $4.99/month.

a. Overflow rate

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b. Sidewater depth Sheet Music

Recommended depth = 3.4 m from Table 6-12 The actual depth (2.5 m) is too small. c. Solids loading SL =

(1 + r )(Q)(X )(86400 s d ) (A s )(1000 g kg )

From Problem 6-37 r=

Qr Q

3

=

0.0028 m s 3

0.012 m s

= 0.2333

So SL =

(1 + 0.2333)(0.012)(1.25)(2000 )(86400 ) = 162.81 kg m 2 ⋅ d (19.63)(1000 )

Check with Figure 6-28 with SVI = 90 mL/g from Problem 6-34. The loading 2 162.81 is much less than 300 kg/m -d allowed. d. Weir Loading

(0.012 m s )(86400 s d ) = 66.0 m WL = 3

π(5.0)

3

d⋅m

Master your semester with Scribd This is less than the design standard of 125Read to 250 /d-m. Free Foron 30this Days Sign up tomvote title & The New York Times  Useful  Not useful 6-41 Determine treatability treatabilit y factor 3


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 − k (0.786)(2.00) = 0.645 = exp   Li (41.1)0.5  

Le

Sheet Music

0.645 = exp[-k(0.245)] ln(0.645) = ln(exp[-k(0.245)]) -0.4385 = -k(0.245) k = 1.7898 or 1.79 d 6-42

-1

Depth of Envirotech filter to achieve 82.7% removal -1

o

o

Given: k for Problem 6-41 = 1.79 d at 20 C; wastewater temperature is 20 C Solution: a. Note that for 82.7% removal the amount remaining is 1 – 0.827 = 0.173 which is equal to L e /Lo b. Solve Envirotech equation for D

 (− 1.79)(1)D   0.5 ( ) 4 . 1  

0.173 = exp 

D Master your semester with  − 1.79Scribd = exp[− 0.279D] 0.173 = exp   6.41  & The New York Times Special offer for students: Only $4.99/month.

ln 0.173 = ln(exp[− 0.279D ])


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a. Compute temperature correction factor Sheet Music 16 − 20

θ = (1.035)

= 0.871

b. Compute efficiency

 (− 1.79)(0.871)(1.8)  (5)0.5 Le   = Li   (− 1.79)(0.871)(1.8)  (1 + 2.00) − 2.00exp  (5)0.5     exp 

Le Li

=

0.285 3.0 − 2(0.285)

= 0.117

 L  Efficiency = 1 − e (100% ) = (1 − 0.117 )(100) = 88.3%  L i  6-44

Trickling filter effluent BOD 5

Given: NRC equations apply, two stage trickling filter, wastewater temperature = 17 3 Q = 0.0509 m /s, Cin = 260 mg/L, diameter of each filter = 24.0 m, depth of e 3 filter = 1.83 m, Q r = 0.0594 m /s Solution: Computer the volume of each filter Master youra.semester with Scribd  π(24.0m )  & The New York Times (1.83m ) = 827.87m V=


2

 Special offer for students: Only $4.99/month. 4 

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3

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d. The efficiency of the first filter is E1 =

E1 =

1

 (0.0509 )(260)   1 + 4.12 ( ) ( ) 827 . 87 1 . 7377   1 0.5

1 + 4.12(0.0092 )

=

0.5

1 1 + 0.3952

= 0.7168

e. Correct the efficiency for temperature using Eqn 6-42 E17 = (0.7168)(1.035)

17 - 20

= (0.7168)(0.9019) = 0.6465

f. The effluent concentration from the first stage is Ce = (1 - 0.6465)(260) = 91.91 mg/L g. The efficiency of the second stage is E2 =

E2 =

1

 (0.0509 )(91.91)    1+ (827.87 )(1.7377 )  1 − 0.6465  4.12

1 0.5

1 + 11.6549(0.0033)

=

0.5

1 1 + 11.6549(0.0570 )

= 0.6007

Master yourh.semester withforScribd Correct the efficiency temperature using EqnRead 6-42Free For 30 Days Sign up to vote on this title & The New York ETimes  Useful  Not useful = (0.6007)(1.035) = (0.6007)(0.9019) = 0.5418 17


i. The final effluent BOD

17 - 20

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Influent BOD5 =

260 260 24 1.83 17

Filter diameter = Filter depth = W W Temp.

3

Volume of ea. Fltr (m ) 3

Recirc. Flow rate (m /s) 3

July 2011 Issue




 Search document

mg/L mg/L m m o C

827.8744 0.0594

Flow rates (m /s)

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Recirc. ratio

2.97

1.98

1.49

1.19

0.99

0.85

0.74

0.66

Recirc. Factor (F)

2.36

2.08

1.88

1.75

1.65

1.57

1.51

1.46

1st Filter Efficiency

0.82

0.78

0.75

0.72

0.69

0.67

0.65

0.64

Corr. For Temp.

0.74

0.71

0.67

0.65

0.63

0.61

0.59

0.57

66.60

76.44

84.53

91.35

97.22

102.34

106.87

110.93

2nd Filter Efficiency

0.70

0.66

0.63

0.60

0.58

0.56

0.55

0.53

Corr. For Temp.

0.64

0.60

0.57

0.54

0.52

0.51

0.49

0.48

Final effluent BOD5 (mg/L) Flow rates

24.3 0.02

30.8 0.03

36.6 0.04

41.7 0.05

46.3 0.06

50.4 0.07

54.2 0.08

57.6 0.09

Eff. From 1st filter (mg.L)

Effluen Effluen t BOD5 vs. Flow R ate 70.0 L 60.0 / g m50.0 , 5 D40.0 O Special offer for students: Only $4.99/month. 30.0 B t

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6-46

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Diameter of single stage rock filter

Sheet Music 3

Given: Applied BOD 5 = 125 mg/L, effluent BOD 5 = 25 mg/L, Q = 0.14 m /s, R = 12 and depth = 1.83 m. Assume NRC equations apply and the wastewater temperature is °C. NOTE: In the first printing of the 3rd edition, a hydraulic loading rate rather than flow rate was given and no temperature was given. Solution: a. Calculate E 1 E1 =

125 − 25 125

= 0.80

b. Calculate recirculation factor F=

1 + 12.0

[1 + 0.1(12.0)]2

= 2.686

c. Solve Eqn 6-39 for volume 0.80 =

0.80 =

1

 (0.14)(125)   1 + 4.12 ( ) V 2 . 686  

0.5

1

 1  1 + 4.12(2.5525 )   V 

0.5

Master your semester with Scribd & The New York Times   1   0.5

0.801 + 10.5163 Special offer for students: Only $4.99/month.



  =1  V  

=

1

 1  1 + 10.5163   V 

0.5


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V = 1769.48 m

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

3

d. The area of the filter is then A=

1769.48m 1.83m

3

= 966.93m 2

e. The diameter of the filter is

πD 2

= 966.93m 2

4

 (966.93m 2 )(4)  D=  π  

6-47

0.5

= 35m

Effect of loading rate on trickling filter performance Given: Data from Problem 6-4, filter diameter of 35.0 m, loading rates Solution: Notes: NRC eqns., single stage filter, constant recirculation ratio = 12

Master yourInfluent semester with125 Scribd BOD 12 5 mg mg//L Filter Filter diamete diameterr 35 m & The New York Times Filter Filter dep depth th 1.83 1.8 3m 5

ecirc$4.99/month. ulation ratio Special offer for students:ROnly W W T em p .

12 o 20 C


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Effluent BOD5 vs Loading Rate 35.00

L / g 30.00 m , 25.00 5 D 20.00 O B 15.00 t n e 10.00 u l f f 5.00 E

0.00

0

5

10

15

20

25

Loading Rate, m/d Figures S-6-47: Effluent BOD 5 vs. loading rate 6-48

Evaluation of oxidation pond 2

3

Given: As = 90,000 m , Q = 500 m /d, 180 kg/d kg/d of BOD, BOD, operating depth depth ranges ranges from m to 1.6 m Solution: a. Loading rate

Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title 180 kg d = = ⋅ LR 20 . 0 kg ha d & The New York Times  Useful  Not useful (90000m )(1× 10 ha m ) 2


−4

2

Cancel anytime.

This is less than the allowable of 22 kg/ha-d, so okay.

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3

Given: Q = 3,800 m /d, BOD5 = 100 mg/L Solution: 3

a. Loading rate (Note: 1.0 mg/L = 1.0 g/m ) 3

3

-3

L = (3,800 m /d)(100 g/m g/m )(1 x 10 kg/g) = 380.0 kg/d b. Surface area (at 22.0 kg/ha -d) SA =

380.0 kg d 22.0 kg ha ⋅ d

= 17.2727 ha 4

2

5

2

SA = (17.2727 ha)(1 x 10 m /ha) = 172,727 172,727 or 1.73 x 10 m 6-50

Alum required to remove P Given: Example 6-15, alum as Al 2(SO4)3-18 H2O Solution: a. The gram molecular weights are Alum = 666.4094 or 666.41 P = 30.97376 or 30.97 b. Compute theoretical amount of alum

Eqn. 6-47 each mole Scribd of alum removes one mole of P, thus, the theoretical amou Master yourFrom semester with  Read Free Foron 30this Days Sign up to vote title of alum is  & The New York Times  Useful  Not useful Cancel anytime.

666.41  (4.00 mg L ⋅ of ⋅ P )   = 86.07 or 86.1 mg/L 30 97  


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Ca(OH)2 = 74.0946 or 74.09 P = 30.97376 or 30.97

Sheet Music

b. Compute theoretical amount of alum

From Eqn. 6-48 the ratio is 5 moles of Ca(OH) 2 to remove three moles of P, and one mole of CaO yields one mole of Ca(OH) 2 from Chapter 3. Thus, the theoretical amo of lime is

(5)(56.08)  (4.00 mg L ⋅ of ⋅ P )   = 3.02 mg L  30.97 

6-52

Spray irrigation storage (Wheatville, (Wheatvill e, Iowa)

Given: Pop. = 1000, 280.0 Lpcd, maximum application rate is 27.74 mm/mo, percola rate = 150 mm/mo Solution:

a. Assume "spray season" is when when temperature is above 0 °C. In fact spraying can continue to about - 4 °C but once spraying has stopped it may not recommence until temperatures exceed + 4 °C. With this assumption, the "season" excludes JAN. Sinc runoff is contained and reapplied, R = 0 in Eqn. 3-3. The wastewater available for application (in mm) in any month (WW m) is computed from the following based on t available area of 40 ha:

(population )(Lpcd )(days ⋅ in ⋅ mo)(1 × 10 6 mm 3 L ) WWm = (40ha )(1 × 1010 mm 2 ha )

Master your semester with Scribd For example for JAN & The New York Times


 Useful  Not useful 6 3 Cancel anytime. ( ) ( ) ( ) ( ) × 1000 280 31 1 10 mm L Special offer for students: Only $4.99/month. WWm = = 21.70 mm 3 mm 2 ⋅ mo 10 2 (40ha )(1 × 10 mm ha )

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Population WW Area No. of mo. can spray

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July 2011 Issue

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

1000 280 Lpcd 40 ha 10 3

Max. Appl. Rate

2

25.55 mm /mm - mo

Water WW WW Evapotranspiration Precip. Month losses Available Applied (mm) (mm) (mm) (mm) (mm) JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC


N/A N/A 43 79 112 155 203 198 152 114 64 25

N/A N/A 193 229 262 305 353 348 302 264 214 175

N/A N/A 63 90 112 116 81 96 83 73 46 39

N/A N/A 63 58.45 54.6 50.05 46.2 46.2 42.35 38.5 34.65 30.8 26.95

N/A N/A 25.55 25.55 25.55 25.55 25.55 25.55 25.55 25.55 25.55 25.55

Total Water Avail. (mm) N/A N/A 88.55 115.55 88.55 115.55 137.55 141.55 106.55 121.55 108.55 98.55

Monthly Water Balance (mm) N/A N/A -104.45 -113.45 -173.45 -189.45 -215.45 -206.45 -195.45 -142.45 -105.45 -76.45

ds/dt 21.7 19.6 -3.85 -4.55 -3.85 -4.55 -3.85 -3.85 -4.55 -3.85 -4.55 -3.85

Note: The last value in the Σds/dt column is a positive value of 1.4 mm, therefore an additionally 1.4 mm must be sprayed sometime during the spray season. This addition amount will not result in exceeding the nitrogen limit of 27.74 mm.

Max.sum from table

41.3 Look up in Σds/dt column. Not calculated

Storage volume m Scribd Master your semester16,520 with c. Explanation of columns & The New York Times 3


(2) Evapotranspiration = from data in table


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V=

6-53

of 66

July 2011 Issue

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(41.3mm)(40.0ha )(1 × 1010 mm 2 ha )

(1× 10

6

mm L )(1000 L m 3


3

)



= 16520 or 16500 m 3

Spray irrigation storage

Given: Pop. = 8800, 485.0 Lpcd, percolation rate = 150 mm/mo, available area = 150 ha, no nitrogen limit Solution:

a. Assume "spray season" is when when temperature is above 0 °C. In fact spraying can continue to about - 4 °Cbut once spraying has stopped it may not recommence until temperatures exceed + 4 °C. With this assumption, the "season" excludes DEC throu MAR. Since runoff is contained and reapplied, R = 0 in Eqn. 2-2. The average mont wastewater application rate (in mm) to dispose of the annual generation in 8 months (WWm) is computed from the following based on the available area of 125 ha:

(population )(Lpcd )(days ⋅ in ⋅ y)(1× 10 6 mm 3 L ) WWm = (125.0ha )(1 × 1010 mm 2 ha )(8months ) (8800 )(485)(365)(1 × 10 6 mm 3 L) WWm = = 155.8 mm 3 mm 2 ⋅ mo 10 2 (125.0ha )(1 × 10 mm ha )(8mo)

The maximum WW application rate is not limited by nitrogen balance. The differenc between the maximum allowed and the wastewater available for application allows fo application of the WW stored during the non-spray season.

Master yourb.semester with To determine the storageScribd required a water balance table is constructed as follows.  Read Free Foron 30this Days Sign up to vote title  Note that the table begins in APR with an accumulated storage of 413.15 mm from D & The New York Times Useful Not useful   through MAR because no spraying can occur when the temperature is below 0 °C and Cancel anytime.

Special offer for students: Only $4.99/month. cannot begin until the temperature is over + 4 °C. Explanation of table is in following followin

paragraph.

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Water Evapotranspiration Month losses (mm) (mm) DEC JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV

N/A N/A N/A N/A 58 89 117 142 130 104 76 41

N/A N/A N/A N/A 258 289 317 342 330 304 276 241

 

July 2011 Issue

 Search document

WW WW Precip. Available Applied (mm) (mm) (mm) N/A N/A N/A N/A 59 102 106 100 73 67 54 63


N/A N/A N/A N/A 515.57 465.64 412.29 362.35 312.42 262.48 212.55 162.61

N/A N/A N/A N/A 155.8 155.8 155.8 155.8 155.8 155.8 155.8 155.8

Total Water Avail. (mm) N/A N/A N/A N/A 214.8 257.8 261.8 255.8 228.8 222.8 209.8 218.8

Monthly Water Balance (mm) N/A N/A N/A N/A -43.2 -31.2 -55.2 -86.2 -101.2 -81.2 -66.2 -22.2



ds/dt 105.85 105.85 95.60 105.85 -53.35 -49.94 -53.35 -49.94 -49.94 -53.35 -49.94 -53.35

Note: The last value in the Σds/dt column is a positive value of 6.8 mm, therefore an additionally 6.8 mm must be sprayed sometime during the spray season. There is no nitrogen limit for this problem. Max.sum frm table Storage volume

413.1 Look up in Σds/dt column. Not calculated 516,375 m

3

c. Explanation of columns (2) Evapotranspiration = from data in table (3) Water losses = sum of (1) and 150 mm/mo percolation (4) Precip. = from data in table (5) WW available = ( Σds/dt) + WW avail for month (WW m) ReadupFree Foron 30this Days vote title (6) WW Appl. = wastewater applied (up to max. ofSign 27.74tomm/mo)  Useful  Not useful (7) Total Water Avail. = (4) + (6) Cancel anytime. Special offer for students: $4.99/month. (8)Only Balance = (7) - (3) (If balance is + then must reduce WW applied until balance is 0.0)

Master your semester with Scribd & The New York Times

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Given: population = 10080; 385 Lpcd; area = 200.0 ha, percolation rate = 150 mm/mo Sheet Music

Solution: o

a. Assume “spray season” is when temperature is above 0 C and once spraying is sto o it may not recommence until the temperature exceeds 4 C. With this assumption, assumption, the “season” excludes JAN, FEB, FEB, and MAR. The wastewater available for application (i mm) in any month (WW m) is computed from the following based on the available are 200.0 ha.

(population )(Lpcd )(days ⋅ in ⋅ mo)(1 × 10 6 mm 3 L ) WWm = (200ha )(1 × 1010 mm 2 ha ) Note: The 200.0 ha is the land available For example in JAN (10080 )(385.0 )(31)(1 × 10 6 mm 3 L ) = 64.45 mm 3 mm 2 ⋅ mo WWm = 10 2 (200.0ha )(1 × 10 mm ha )

b. To determine the storage required a water balance table is constructed as follows. Note that the table begins in APR with an accumulated storage of 165.17 mm from JA through MAR because no spraying can occur when the temperature is below 0 °C and cannot begin until the temperature is over + 4 °C. Explanation of table is in following followin paragraph. Population WW Area No. of mo. can spray

10,800 385 Lpcd 200 ha 9

Master yourMax. semester with Scribd Appl. Rate 84.315 mm /mm - mo & The New York Times 3

Special offer for students: Only $4.99/month. Water Evapotranspiration Month losses

2


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Useful

WW WW Precip. Available Applied

Total Water

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Monthly Water

ds/d

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OCT NOV DEC

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8.4 4.3 3.1

July 2011 Issue


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158.4 154.3 153.1

16 12 12

130.28 110.42 90.55

84.3 84.3 84.3

100.3 96.3 96.3



-58.1 -58.0 -56.8

-19 -21 -19

Note: The last value in the Σds/dt column is a positive value of 6.2 mm, therefore an additional 6.2 mm must be sprayed sometime sometime during the spray season. There is no nitr limit for this problem. Max.sum frm table Storage volume

6-55

187.11 Look up in Σds/dt column. Not calculate 374,220 m

3

Daily and annual sludge production Given: Operating data from WWTP Solution: a. Specific gravity of solids (Eqn. 6-54)

As in example problem, recognize that mass fraction may be used instead of actual m of volatile and fixed solids. Ss =

  (2.50)(1000 )(0.970)(1000 )   1000  (0.970)(0.3)(1000 ) + (2.50)(0.7 )(1000 ) 1

All the 1000's cancel out and then Ss =

2.425

= 1.188

Master your semester with Scribd b. Specific gravity of sludge (Eqn. 6-59) & The New York Times 0.291 + 1.75


S

1.188

1 007


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Vsl =

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 

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354.89 kg d

(1000 kg

m

3

)(1.007)(0.0450 )

July 2011 Issue


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

= 7.83 m 3 d

d. Annual sludge production 3

3

Vsl = (7.83 m /d)(365 d/y) d/y) = 2,857.95 or or 2,860 m /y 6-56

Daily and annual sludge production Given: Operating data from WWTP Solution: a. Specific gravity of solids (Eqn. 6-54)

As in example problem, recognize that mass fraction may be used instead of actual m of volatile and fixed solids. Ss =

(2.50)(1000 )(0.999)(1000 )     1000  (0.999)(0.32)(1000 ) + (2.50)(0.68)(1000 ) 1

All the 1000's cancel out and then Ss =

2.4975 0.3197 + 1.70

= 1.237

b. Specific gravity of sludge (Eqn. 6-59)

Master your semester with Scribd 1.237 = 1.0098 S = 0.0520 + (1.237)(0.948) & The New York Times sl





c. Mass of sludge per day at 47% removal efficiency


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d. Annual sludge production Sheet Music 3

5

3

Vsl = (277 m /d)(365 d/y) d/y) = 101,105 or or 1.01 x 10 m /y 6-57

Sludge production as a function of efficiency Given: Data for Problem 6-56 and removal efficiency of 40, 45, 50, 55, 60, and 65% Solution: 2 m /s 179 mg/L 2.5 0.999 0.32 0.68 5.2 %

Flow Influent suspended solids Sp. grav. of fixed solids Sp. grav. of volatile solids Fixed solids fraction Volatile solids fraction Sludge concentration Sp. grav. of solids Sp. grav. of sludge

1.236582 1 .0 1 0 0 4 9

Removal efficiency Mass of sludge (kg/d) 3 Daily sludge volume (m /d) 3 Annual sludge volume (m /y) Removal efficiency

0.4 1 2 ,3 7 2 23 6 8 5 ,9 8 1 40


0.45 13,919 26 5 96,729 45

0 .5 15,466 294 107,477 50

0.55 17,012 324 118,224 55

0 .6 18,559 353 128,972 60


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Sludge Production vs Efficieny 160,000

y / 140,000 3 m , 120,000 e m100,000 u l o 80,000 V e 60,000 g 40,000 d u l 20,000 S

0 0

20

40

60

80

Removal efficiency, efficiency, % 6-58

Figure S-6-57: Sludge production vs. Efficiency Sludge mass balance Given: Figure 6-36, Table 6-14 and efficiencies Solution:

Effluent from primary sedimentation tank Master youra.semester with Scribd 185.686 & The New York Times E=

 1 Special offer for students: Only $4.99/month. 




Not useful = 190 Mg d Cancel anytime.

Useful

  − 0.0 − 0.150[(1 − 0.250 − 0.0 )(1 + 0.190 )] 0 . 900  

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d. Solids destroyed Sheet Music

J = (0.250)(190.011) = 47.503 or 47.5 Mg/d e. Solids to dewatering K = (190.011)(1 - 0.250 - 0.0) = 142.508 or 143 Mg/d f. Solids to ultimate disposal L = 142.508(1 + 0.190)(1 - 0.150) = 144.147 or 144 Mg/d 6-59

Rework 6-58 with dewatering removed Given: Prob. 6-58, K = L Solution: a. With no dewatering n P = 0, so E=

A 1 nE

=

185.686 1 0.900

= 167.117

b. Solids from digestion K = E(1 - nJ - 0.0) K = 167.117(1 - 0.250 -0.0) = 125.34 or 125 Mg/d 6-60 Sludge mass balance Master your semester with Scribd Given: Problem & The New York Times6-58, n = 0.5 E


Solution:


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M=

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99.5 0.9

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July 2011 Issue


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

= 110.56

c. Effluent from primary sedimentation B = (1 - 0.50)(185.686 + 110.56) = 148.122 Mg/d d. Solids destroyed J = (0.250)(99.5) = 24.9 Mg/d e. Solids dewatering K = (99.5)(1 – 0.250 – 0.0) = 74.6 Mg/d f. Solids to ultimate disposal L = 74.625(1 + 0.190)(1-0.150) = 75.5 Mg/d

6-61

Doubtful WWTP mass flow to ultimate disposal

Given: Flow chart in Figure P-6-61, efficiencies from Figure 6-37, A = 7.250, X = 1.2 N = 0.0

Master yourSolution: semester with Scribd a. Revised flow chart & The New York Times Special offer for students: Only $4.99/month.

A


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b. Table 6-15 with with appropriate assumptions may be used to solve mass balances. Sin we need K, note that the following functional dependencies exist: K = f(E,H), E = f(A H = f(F), F = f(E,X). Since A and X are given given the solution sequence is: E, F, H and K. Note that since N = 0.0 then nN = 0.0 and that no filtration and no dewatering mea that nR = 0.0 and n P = 0.0. c. Primary sludge

α = 0.0 + 0.0 = 0.0 β=

(1 − 0.650)(1 − 0.08) 0.650

= 0.4954

γ = 0.0 + 0.0(1 − 0.0) = 0.0

 1.288  (0.0 − 0.0 ) − 1 0 . 0   = 4.7125 E=  1    − 0.0 − (0.4954 )(0.0 )  0.650  7.250 − 

d. Secondary sludge Since there is no thickening H = F. H = F = (0.4954 )(4.7125) −

1.288

= 1.0465 Master your semester with Scribd 1 − 0.0 Read Free Foron 30this Days Sign up to vote title & The New York Times  Useful  Not useful e. Sludge to ultimate disposal Special offer for students: Only $4.99/month.

Cancel anytime.

K = (1 - 0.350 - 0.0)(4.7125 + 1.0466) = 3.743 Mg/d

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nE =

of 66

E A+N

=

 

July 2011 Issue


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8.910 7.280 + 9.428



= 0.533

b. Aeration tank solids destruction nD =

D B

=

0.390 7.798

= 0.050

c. Anaerobic digester supernatant nN =

N

9.428

=

E+F

= 0.595

8.910 + 6.940

d. Anaerobic digester solids destruction nJ =

J E+F

=

4.755 8.190 + 6.940

= 0.595

e. Secondary clarifier nX =

6-63

X B−D

=

0.468 7.798 + 0.390

= 0.063

Doubtful thickening and dewatering Given: Figure 6-37, values for n and A = 7.250 Mg/d, X = 1.288 Mg/d

Master yourSolution: semester with Scribd  Read Free Foron 30this Days Sign up to vote title  a. TableTimes 6-15 with appropriate assumptions may be to solve theuseful mass balances. & The New York Useful used  Not

anytime. Note the following functionally dependencies exist: Cancel L = f(K), K = f(E,H), H = f(F Special offer for students: Onlythat $4.99/month.

= f(E,X), E = f(A,X). A and X are given. The solution sequence is then E, F, H, K

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γ = 0.150 + 0.1214(1 − 0.150) = 0.2532 Sheet Music

 1.288  (0.2532 − 0.0 )  1 − 0.0 

7.250 −  E=

 1    − 0.1214 − (0.4954 )(0.2532 ) 0 . 650  

= 5.3605

c. Underflow from secondary settling 1.288 = 1.3675 F = (0.4954 )(5.3605 ) − 1 − 0.0 d. Thickened sludge to digester H = (1 − 0.150)(1.3675) = 1.1624 e. Digester effluent to dewatering K = (1 − 0.350 − 0.05)(5.3605 + 1.1624 ) = 3.9138 f. To ultimate disposal L = 3.9138(1 + 0.190)(1 − 0.100 ) = 4.19 Mg d 6-64

Gravity thickening for WAS 3

Given: WAS flow = 3255 m /d, thicken thicken from 10600 10600 mg/L to 2.50%,

Master yourSolution: semester with Scribd a. Compute points for batch flux curve & The New York Times Special offer for students: Only $4.99/month. 3 2 SS (kg/m ) v (m/d) Fs (kg/d-m )


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c. Note that sludge is to be thickened from 10600 mg/L to 2.50%. Plot tangent from 2 to find: Fs = 14.5 kg/d-m

2

3

d. Solids to thickener (Note: mg/L = g/m ) 3

3

-3

Ms = (10600 g/m )(3255 m /d)(10 kg/g) = 34,503 kg/d e. Surface area required As =

34503 kg d 14.5 kg d ⋅ m

2

= 2379.5m 2

f. Diameter of tank

 (2379.5)(4 )  D=  π  

12

= 55.04

This exceeds the 30 m maximum. g. Try 4 tanks 2379.5 4

= 594.87 m 2

 (594.87 )(4 )  D=  π  

12

= 27.5m



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Problem 6-64 Batch Flux Curve 30 25 2

m d 20 / g k , x 15 u l f s d 10 i l o S

5 0 0

1

2

3

4

Suspended solids conncentration, % Figure S-6-64: Batch flux curve

Master your semester with Scribd Read Free Foron 30this Days 6-65 Gravity thickening for mixed WAS and primary sludge Sign up to vote title & The New York Times  Useful  Not useful 3

3 Cancel anytime.

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Given: WAS flow = 3255 m /d, primary flow = 710 m /d, thicken from 2.00% to 5.0 Special offer for students: Only $4.99/month.

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b. Plot as shown below Sheet Music

Problem 6-65 Batch Flux Curve 50 45 2

40

m35 d / g k 30 , x 25 u l f s 20 d i l 15 o S

10 5 0 0

2

4

6

Suspended solids conncentration, % Figure S-6-65: Batch flux curve

Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  c. Note that sludge is to be thickened from 2.00% to 5.00%. Plot tangent from 5.00% & The New York Times Useful  Not useful  find: Cancel anytime.


F = 24 kg/d-m

2

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f. Diameter of tank Sheet Music

 (635.8)(4 )  D=  π   6-66

12

= 28.5m

Surface area for gravity thickener for Little Falls 3

Given: Q = 7.33 m /d, final sludge sludge concentration concentration of 3.6%, 3.6%, settling test data Solution: a. Compute the flux (spreadsheet) SS Conc. 3

(kg/m ) 4 6 8 14 29 41

Init. Sett. SS conc. (%) F(s) (kg/d-m2) Vel. (m/d) 58.5 0 .4 234 36.6 0 .6 219.6 24.1 0 .8 192.8 8.1 1 .4 113.4 2.2 2 .9 63.8 0.73 4.1 29.93

Batch Flux Curve 250 2

m200 d Master your / semester with g k 150 Times & The New ,York x Only $4.99/month. Special offer for students: u l f 100

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b. From the tangent line read a solids flux of 180 kg/d-m



2

c. Compute the mass loading 3

3

-3

Mse = (36000 g/m )(733 m /d)(10 kg/g) = 26388 kg/d d. Surface area of gravity thickener As = 6-67

26388 kg d 180 kg d ⋅ m

2

= 146.6m 2

Pomdeterra sludge dewatering volume savings

Given: Suspended solids = 3.8%, filter press yields 24% solids, current sludge volum 3 33 m /d Solution: a. Sludge volume after filter press 33 m 3 d V2

=

0.24 0.038

Solve for V2 V2 =

(33)(0.038) 0.24

= 5.225 m 3 d

Master yourb.semester with Scribd Annual volume savings & The New York Times 3

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V = (33 m /d - 5.225 m /d)(365 d/y) d/y) = 10,137.8 10,137.8 or 10,000 m Special offer for students: Only $4.99/month.

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Solve for V2

(13)(0.078)

V2 =

0.35

= 2.897 m 3 d

b. Annual volume 3

3

V = (2.897 m /d)(365 d/y) d/y) = 1,057 or 1,000 m /y 6-69

Concentration to reduce sludge volume 3

Given: 30 m /mo of sludge, sludge, suspended suspended solids concentration concentration of 2.5% Solution: a. Solids concentration 3

30 m mo 3

3.0 m mo

=

X 0.025

X = 10(0.025) = 0.25 In percent X = (0.25)(100%) = 25% solids



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DISCUSSION QUESTIONS Sheet Music

6-1

Electron acceptor Given: Biological reactors with and without odor Solution:

Reactor A operating at 35 °C, with a strong odor is probably anaerobic. The potential electron acceptors are sulfate, carbon dioxide, and organic compounds that can be reduced.

Reactor B, also operating at 35 °C, is either aerobic or anoxic. There are not enough d given to differentiate between the two reactor types. If the reactor is aerobic, the elect acceptor is oxygen. If it is anoxic a potential electron acceptor is nitrate. 6-2

Processes preceding tertiary Given: Regulatory agency requires tertiary treatment Solution: Probable treatment processes: bar rack, grit chamber, activated sludge including secondary settling tank or trickling filter with secondary settling tank.

6-3

Recirculation versus return sludge Given: Differentiate between two processes

Master yourSolution: semester with Scribd  Read Free Foron 30this Days Sign up to vote title  The purpose of recirculation is to reduce the organic loading on the trickling filter. Re & The New York Times  Useful  Not useful sludge is to return biomass to the activated sludge process. They differ in that Special offer for students: Only $4.99/month.

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recirculation is pumped from the supernatant of the secondary clarifier while return sludge is pumped from the bottom of the secondary clarifier.

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Removal of NH 4

Sheet Music

Given: Industrial wastewater containing only NH 4 at pH = 7.00, being stripped with oxygen Solution: It cannot be denitrified. The NH 4 cannot be stripped without raising the pH.



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