Equalization Example

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Equalization Example Determination of Flowrate - Equalization Volume Requirements and Effects on BOD Mass Loading Data and Questions... For the flowrate and BOD concentration data given in the table determine ; (1) The in-line storage volume required to equalize the flowrate (2) Time period when the equalization tank is empty (3) The effect of the flow equalization equalizati on on BOD mass - loading rate

Time  period

Average Average BOD flowrate during concentration during the period (L / the period (mg / L) s)

24 - 01

275

150

01 - 02

221

115

02 - 03

164

75

03 - 04

130

50

04 - 05

105

45

05 - 06

99

60

06 - 07

119

90

07 - 08

204

130

08 - 09

354

175

09 - 10

411

200

10 - 11

425

215

11 - 12

430

220

12 - 13

425

220

13 - 14

405

210

14 - 15

385

200

15 - 16

351

190

16 - 17

326

180

17 - 18

326

170

18 - 19

328

175

19 - 20

365

210

20 - 21

399

280

21 - 22

399

305

22 - 23

379

245

23 - 24

345

180

Average

307.0833

170

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Solution... (1) The In - Line Storage Volume Required to Equalize the Flowrate...

Step - 1 : Calculation of wastewater volumes entering the equalization tank during the each time  period (column 2 in table given below) Example for the time period 24 - 01 : (275 L / s) (3,600 s / h) (10 -3 m3 / L) = 990.0 m 3 / h Step - 2 : Calculation of wastewater volumes pumping out from the equalization tank (column 3 in table given below) Total wastewater volume entering to the equalization tank is 26,532.0 m 3  during 24 h period Wastewater volume should be pumped out from the equalization tank must be equal to this amount during 24 h period Wastewater volume should be pumped out from the equalization tank during the each time  period = (26,532.0 m 3) / 24 = 1,105.5 m 3 Step - 3 : Calculation of the cumulative influent volumes (column 4 in table given below) Example for the time period 01 - 02 : 990.0 m 3 + 795.6 m 3 = 1,785.6 m 3 Step - 4 : Calculation of the cumulative effluent volumes (column 5 in table given below) Example for the time period 01 - 02 : 1,105.5 m 3 + 1,105.5 m 3 = 2,211.0 m 3 Step - 5 : Calculation of the cumulative differences (column 6 in table given below) Example for the time period 24 - 01 : 990.0 m 3 - 1,105.5 m 3 = - 115.5 m 3 (1)

(2)

(3)

(4)

(5)

(6)

Time period

Influent volume (m3)

24 - 01

990.0

1,105.5

990.0

1,105.5

- 115.5

01 - 02

795.6

1,105.5

1,785.6

2,211.0

- 425.4

02 - 03

590.4

1,105.5

2,376.0

3,316.5

- 940.5

03 - 04

468.0

1,105.5

2,844.0

4,422.0

- 1,578.0

04 - 05

378.0

1,105.5

3,222.0

5,527.5

- 2,305.5

05 - 06

356.4

1,105.5

3,578.4

6,633.0

- 3,054.6

06 - 07

428.4

1,105.5

4,006.8

7,738.5

- 3,731.7

07 - 08

734.4

1,105.5

4,741.2

8,844.0

- 4,102.8

08 - 09

1,274.4

1,105.5

6,015.6

9,949.5

- 3,933.9

09 - 10

1,479.6

1,105.5

7,495.2

11,055.0

- 3,559.8

10 - 11

1,530.0

1,105.5

9,025.2

12,160.5

- 3,135.3

11 - 12

1,548.0

1,105.5

10,573.2

13,266.0

- 2,692.8

12 - 13

1,530.0

1,105.5

12,103.2

14,371.5

- 2,268.3

13 - 14

1,458.0

1,105.5

13,561.2

15,477.0

- 1,915.8

Effluent Cumulative Cumulative Cumulative volume influent effluent difference 3 3 3 (m ) volume (m ) volume (m ) (m3)

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14 - 15

1,386.0

1,105.5

14,947.2

16,582.5

- 1,635.3

15 - 16

1,263.6

1,105.5

16,210.8

17,688.0

- 1,477.2

16 - 17

1,173.6

1,105.5

17,384.4

18,793.5

- 1,409.1

17 - 18

1,173.6

1,105.5

18,558.0

19,899.0

- 1,341.0

18 - 19

1,180.8

1,105.5

19,738.8

21,004.5

- 1,265.7

19 - 20

1,314.0

1,105.5

21,052.8

22,110.0

- 1,057.2

20 - 21

1,436.4

1,105.5

22,489.2

23,215.5

- 726.3

21 - 22

1,436.4

1,105.5

23,925.6

24,321.0

- 395.4

22 - 23

1,364.4

1,105.5

25,290.0

25,426.5

- 136.5

23 - 24

1,242.0

1,105.5

26,532.0

26,532.0

0.0

-

-

-

Total

26,532.0 26,532.0

Step - 6 : Calculation of the equalization tank volume Equalization tank volume = Absolute value of the smallest negative difference + The largest  positive difference VEqualization = abs(- 4,102.8 m3) + none = 4,102.8 m3

Step - 7 : Dimensioning of the equalization tank Equalization tank volume determined from the calculations given in the table shown above, should be increased at least 10 % as a safety factor V Equalization = (1.112) (4,102.8 m3) = 4,562.5 m3 L = 50.00 m, h = 3.65 m and B = 25.00 m (2) Time Period when the Equalization Tank is Empty...

To determine the effect of the equalization tank on the BOD mass - loading rate, there are a number of ways. Perhaps the simplest way is to perform the necessary computations, starting

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with the time period when the equalization tank is empty. As it can be seen from the figure shown above, the equalization tank is empty at 08 am.

(3) The Effect of the Flow Equalization on BOD Mass - Loading Rate... Because the equalization tank is empty at 08 am, the necessary computations will be performed starting with the 08 - 09 time period. Step 1 : The first step is to compute the liquid volume in the equalization tank at the end of each time period. This is done by subtracting the equalized hourly flowrate expressed as a volume from the inflow flowrate also expressed as a volume. The volume corresponding to the equalized flowrate for a period of 1 h is (307.0833 L / s) (3,600 s / h) (10 -3 m3 / L) = 1,105.5 m 3 Using this value, the volume in storage is computed using the following equation ; Vsc = Vsp + Vic - Voc where ; Vsc : volume in the equalization tank at the end of current time period, V sp : volume in the equalization tank at the end of previous time period, V ic : volume of inflow during the current time period and V oc  : volume of outflow during the current time period. Thus, using the values in the original data table, the volume in the equalization tank for the time  period 08 - 09 is as follows ; Vsc = 0 + 1,274.4 m 3 - 1,105.5 m 3 = 168.9 m 3 For the time period 09 - 10 ; Vsc = 168.9 m 3 + 1,479.6 m 3 - 1,105.5 m 3 = 543.0 m 3 The volume in storage at the end of each time period has been computed in a similar way. Step - 2 : The second step is to compute the average concentration leaving the storage tank. This is done by using the following equation, which is based on the assumption that the contents of the equalization tank are mixed completely ; Xoc = [ ( V ic ) ( Xic ) + ( Vsp ) ( Xsp ) ] / ( Vic + Vsp ) where ; Xoc : average concentration of BOD in the outflow from the equalization tank during the current time period (mg / L), V ic : volume of wastewater inflow during the current period (m 3), Xic : average concentration of BOD in the inflow wastewater volume (mg / L), V sp : volume of wastewater in the equalization tank at the end of the previous time period (m 3) and Xsp : concentration of BOD in wastewater in the equalization tank at the end of the previous time  period (mg / L).

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(1)

Time period

(2)

(3)

(4)

(5)

(6)

Volume Equalized in Average BOD BOD Influent storage concentration Equalized BOD concentration volume at end of during time mass loading during during time 3 (m ) time  period (mg / time period (kg / h)  period (mg /  period L) L) 3 (m )

08 - 09

1,274.4

168.9

175

175

193.5

09 - 10

1,479.6

543.0

200

197

217.8

10 - 11

1,530.0

967.5

215

211

233.3

11 - 12

1,548.0 1,410.0

220

218

241.0

12 - 13

1,530.0 1,834.5

220

220

243.2

13 - 14

1,458.0 2,187.0

210

216

238.8

14 - 15

1,386.0 2,467.5

200

206

227.7

15 - 16

1,263.6 2,625.6

190

197

217.8

16 - 17

1,173.6 2,693.7

180

187

206.7

17 - 18

1,173.6 2,761.8

170

177

195.7

18 - 19

1,180.8 2,837.1

175

171

189.0

19 - 20

1,314.0 3,045.6

210

186

205.6

20 - 21

1,436.4 3,376.5

280

232

256.5

21 - 22

1,436.4 3,707.4

305

287

317.3

22 - 23

1,364.4 3,966.3

245

289

319.5

23 - 24

1,242.0 4,102.8

180

229

253.2

24 - 01

990.0

3,987.3

150

174

192.4

01 - 02

795.6

3,677.4

115

144

159.2

02 - 03

590.4

3,162.3

75

109

120.5

03 - 04

468.0

2,524.8

50

72

79.6

04 - 05

378.0

1,797.3

45

49

54.2

05 - 06

356.4

1,048.2

60

47

52.0

06 - 07

428.4

371.1

90

69

76.3

07 - 08

734.4

0.0

130

117

129.3

Average

-

-

-

-

192.5

Using the data given in column 2, 3 and 4 of the above computation table, the effluent concentration is computed as follows ; For the time period 08 - 09 ; X08 - 09 = [ ( 1,274.4 ) ( 175 ) + ( 0.0 ) ( 0 ) ] / ( 1,274.4 + 0.0 ) = 175 mg / L For the time period 09 - 10 ;

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X09 - 10 = [ ( 1,479.6 ) ( 200 ) + ( 168.9 ) ( 175 ) ] / ( 1,479.6 + 168.9 ) = 197 mg / L Step - 3 : The third step is to compute the hourly mass - loading rate using the following equation Li = ( Ci ) ( QORT ) For example, for the time period 08 - 09, the mass - loading rate is ; L08 - 09 = ( C08 - 09 ) ( QORT ) = ( 175 mg / L ) ( 307.0833 L / s ) ( 10 -6 kg / mg ) ( 3,600 s / h ) = 193.5 kg / h (4) The effect of the flow equalization on BOD mass - loading rate... The effect of flow equalization can best be shown graphically by plotting the hourly unequalized and equalized BOD mass - loading on the plot prepared in step 2. The following flowrate ratios, derived from the data presented in the table given in the problem statement and the computation table prepared in step 2a, are also helpful in assessing the benefits derived from flow equalization. Time  period

Unequalized mass - loading (kg / h)

Equalized mass - loading (kg / h)

24 - 01

148.5

192.4

01 - 02

91.5

159.2

02 - 03

44.3

120.5

03 - 04

23.4

79.6

04 - 05

17.0

54.2

05 - 06

21.4

52.0

06 - 07

38.6

76.3

07 - 08

95.5

129.3

08 - 09

223.0

193.5

09 - 10

295.9

217.8

10 - 11

329.0

233.3

11 - 12

340.6

241.0

12 - 13

336.6

243.2

13 - 14

306.2

238.8

14 - 15

277.2

227.7

15 - 16

240.1

217.8

16 - 17

211.2

206.7

17 - 18

199.5

195.7

18 - 19

206.6

189.0

19 - 20

275.9

205.6

20 - 21

402.2

256.5

21 - 22

438.1

317.3

22 - 23

334.3

319.5

23 - 24

223.6

253.2

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Source: http://web.deu.edu.tr 

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