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 on BOD mass - loading rate

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

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 m3 / 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 m3 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 m3) / 24 = 1,105.5 m3

Step - 3 : Calculation of the cumulative influent volumes (column 4 in table given below)
Example for the time period 01 - 02 : 990.0 m3 + 795.6 m3 = 1,785.6 m3

Step - 4 : Calculation of the cumulative effluent volumes (column 5 in table given below)
Example for the time period 01 - 02 : 1,105.5 m3 + 1,105.5 m3 = 2,211.0 m3

Step - 5 : Calculation of the cumulative differences (column 6 in table given below)
Example for the time period 24 - 01 : 990.0 m3 - 1,105.5 m3 = - 115.5 m3

(1) (2) (3) (4) (5) (6)
Time period Influent volume (m3) Effluent volume (m3) Cumulative influent volume (m3) Cumulative effluent volume (m3) Cumulative difference (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
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
VEqualization = (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 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 m3
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, Vsp : volume in the equalization tank at the end of previous time period, Vic : volume of inflow during the current time period and Voc : 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 m3 - 1,105.5 m3 = 168.9 m3
For the time period 09 - 10 ;
Vsc = 168.9 m3 + 1,479.6 m3 - 1,105.5 m3 = 543.0 m3
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 = [ ( Vic ) ( 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), Vic : volume of wastewater inflow during the current period (m3), Xic : average concentration of BOD in the inflow wastewater volume (mg / L), Vsp : volume of wastewater in the equalization tank at the end of the previous time period (m3) and Xsp : concentration of BOD in wastewater in the equalization tank at the end of the previous time period (mg / L).

(1) (2) (3) (4) (5) (6)
Time period Influent volume (m3) Volume in storage at end of time period (m3) Average BOD concentration during time period (mg / L) Equalized BOD concentration during time period (mg / L) Equalized BOD mass loading during time period (kg / h)
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 ;
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 assesing 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