Determine the equalization basin volume required for the following cyclic flow pattern. Provide a 25% excess capacity for equipment, unexpected flow variations, and solids accumulation. Evaluate the impact of equalization on the mass loading of BOD5 .

Question

Determine the equalization basin volume required for the following cyclic flow pattern. Provide a 25% excess capacity for equipment, unexpected flow variations, and solids accumulation. Evaluate the impact of equalization on the mass loading of BOD[latex]_{5}[/latex] .

Accepted Answer

Design of the equalization basin volume.

a. Because of the repetitive and tabular nature of the calculations, a spreadsheet is ideal for this problem. The spreadsheet solution is easy to verify if the calculations are set up with judicious selection of the initial value. If the initial value of the first flow rate is greater than the average after the sequence of nighttime low flows, then the last row of the computation should result in a storage value of zero for a perfect sinusoidal flow pattern.

b. The first step is to calculate the average flow. In this case it is 0.05657 m³/s. Next, the flows are arranged in order beginning with the time and flow that first exceeds the average. In this case it is at 0900 h with a flow of 0.0631 m³/s.
The tabular arrangement is shown in Table 20-13. An explanation of the calculations for each column is given in the following steps.

c. In the third column, the flows are converted to volumes using the time interval between flow measurements:
[latex]\sout{V} [/latex] = (0.0631 m³/s)(1 h)(3, 600 s/h) = 227.16 m³

d. The average volume that leaves the equalization basin is calculated in the fourth column. It is the average flow rate computed on an hourly basis.
[latex]\sout{V} [/latex] = (0. 05657 m³/s)(1 h)(3, 600 s/h) = 203.655 m³

e. The fifth column is the difference between the inflow volume and the outflow volume.
[latex]dS =\sout{V} _{in} - \sout{V}_{out} =[/latex] 227.16 m³ − 203.655 = 23.505 m³

f. The required storage is computed in the sixth column. It is the cumulative sum of the difference between the inflow and outflow. For the second time interval, it is
Storage = Σ dS = 37.55 m³ + 23.51 m³ = 61.06 m³

Note that the last value for the cumulative storage is 0.12 m³ . It is not zero because of round-off truncation in the computations. At this point the equalization basin is empty and ready to begin the next day's cycle.

g. The required volume for the equalization basin is the maximum cumulative storage. It is the shaded value. With the requirement for 25% excess, the volume would then be
Storage = (863.74 m³)(1.25) = 1, 079.68 or 1, 080 m³

Evaluation of the impact on BOD[latex]_{5}[/latex] loading.

h. The mass of BOD[latex]_{5}[/latex] entering the equalization basin is the product of the inflow (Q), the concentration of BOD[latex]_{5}[/latex] (S[latex]_{0}[/latex] ), and the integration time (Δt):
[latex]M_{BOD-in} = (Q)(S_{0})(Δ t)[/latex]

The mass of BOD[latex]_{5}[/latex] leaving the equalization basin is the product of the average outflow (Q[latex]_{avg}[/latex] ), the average concentration (S[latex]_{avg}[/latex]) in the basin, and the integration time (Δt):
[latex]M_{BOD-out} = (Q_{avg})(S_{avg})(Δ t)[/latex]

The average concentration is determined as:
[latex] S_{avg} = \frac{(\sout{V}_{i})(S_{0}) + (\sout{V}_{S})(S_{prev})}{\sout{V}_{i} + \sout{V}_{s} }[/latex]

where [latex] \sout{V}_{i}[/latex]= volume of inflow during time interval Δt, m³
[latex] S_{0}[/latex] = average BOD[latex]_{5}[/latex] concentration during time interval Δt, g/m³
[latex]\sout{V}_{S}[/latex] = volume of wastewater in the basin at the end of the previous time interval Δt, m³
[latex] S_{prev}[/latex] = concentration of BOD[latex]_{5}[/latex] in the basin at the end of the previous time interval Δt, g/m³

i. Noting that 1 mg/L = 1 g/m³ , find that the first row (0900 h time) computations in columns 8, 9, and 10 are

[latex] M_{BOD-in} [/latex] = (0.0631 m³/s)(140 g/m³)(1 h)(3, 600 s/h)([latex]10^{−3} [/latex] kg/g) = 31.8 kg
[latex] S_{avg} = \frac{(227.16 m^{3})(140 g/m^{3}) + 0}{227.16 m^{3} + 0} = [/latex] 140 g/m³
[latex] M_{BOD-out} [/latex] = (0.05657 m³/s)(140 g/m³)(1 h)(3, 600 s/h)( [latex]10^{−3} [/latex] kg/g) = 28.5 kg

Note that the zero values in the computation of [latex]S_{avg}[/latex] are valid only at start-up of an empty basin. Also note that in this case MBOD-in and MBOD-out differ only because of the difference in flow rates.

j. For the second row (1000 h), the computations are
[latex] M_{BOD-in} [/latex]= (0.0670 m³/s)(150 g/m³)(1 h)(3, 600 s/h)([latex]10^{−3} [/latex] kg/g) = 36.2 kg
[latex] S_{avg} = \frac{(241.2 m^{3})(150 g/m^{3}) + (23.51 m^{3})(140 g/m^{3})}{241.2 m^{3} + 23.51 m^{3}} =[/latex] 149.11 mg/L
[latex] M_{BOD-out} [/latex] = (0.05657 m³/s)(149.11 g/m³)(1 h)(3, 600 s/h)([latex]10^{−3} [/latex] kg/g) = 30.37 kg

Note that
[latex]\sout{V}_{s} [/latex] is the volume of wastewater in the basin at the end of the previous time interval. Therefore, it is equal to the accumulated dS. The concentration of BOD[latex]_{5} (S_{prev}[/latex]) is the average concentration at the end of previous interval (S[latex]_{avg}[/latex]) and not the influent concentration for the previous interval (S[latex]_{0}[/latex]).

k. For the third row (1100 h), the concentration of BOD[latex]_{5}[/latex] is
[latex] S_{avg} = \frac{(245.52 m^{3})(155 g/m^{3}) + (61.06 m^{3})(149.11 g/m^{3})}{245.52 m^{3} + 61.06 m^{3}} [/latex] = 153.83 mg/L

l. The ratio of the maximum BOD mass to the minimum BOD mass drops from
56.23 kg : 2.15 kg or 26.15 : 1

to

33.97 kg : 18.55 kg or 1.83 : 1

Comments:

1. The BOD calculations assume a completely mixed basin.

2. The volume accumulation at the end of the diurnal cycle is unlikely to be exactly zero because the diurnal flow pattern is not a perfectly symmetrical sinusoidal curve. Therefore, the pumping rate out of the equalization basin cannot be constant.

Time, h	Flow, m³/s	BOD $_{5}$ , mg/L	Time, h	Flow, m³/s	BOD $_{5}$ , mg/L
0000	0.0481	110	1200	0.0718	160
0100	0.0359	81	1300	0.0744	150
0200	0.0226	53	1400	0.0750	140
0300	0.0187	35	1500	0.0781	135
0400	0.0187	32	1600	0.0806	130
0500	0.0198	40	1700	0.0843	120
0600	0.0226	66	1800	0.0854	125
0700	0.0359	92	1900	0.0806	150
0800	0.0509	125	2000	0.0781	200
0900	0.0631	140	2100	0.0670	215
1000	0.0670	150	2200	0.0583	170
1100	0.0682	155	2300	0.0526	130

Question 20.6: Determine the equalization basin volume required for the fol...

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