Volume of flow to be diverted per time period, m
Flow rate diverted, m/min
Time of diversion, hr
Frequency of diversion, number/day
Conversion constant for unit, min/hr
Therefore, VD = (0.473 m3/min) (3 hr) (2/day) (60 min/hr) = 170.28 m3/day.
The control discharge rate can be established as follows:
fC = Controlled discharge rate, m3/min VD = Volume diverted, m3 T = Time period for return, hr k = Conversion factor for unit
Therefore, fC = (170.28 m3/24 hr)(1 hr/60 min) = 0.118 m3/min.
The volume of the surge basin can now be calculated. As calculated in Equation 7.13(2), 170.28 m3 of the total flow will be diverted and fed back to the stream at a constant rate. Therefore, the average flow for the remainder of the time is (360.09 - 170.28) = 189.81 m3 for the 18hr period. This amount equates to 0.1318 m3/min on a 24hr basis. Correspondingly, maintaining this flow for the 6-hr diversion period requires a surge basin equal to the volume for the diversion time frame (6 hr in this case) at an average flow rate for the remaining period. This volume can be calculated as follows:
VS = Volume of the surge basin, m3
Fa = Average flow rate without diversion flow, m3/min Td = Diversion time period, hr k = Conversion factor
Therefore, VS = (0.175 m3/min)(6 hr)(60 min/hr) = 63.22 m3.
Any excesses in design capacity determined by management as part of the design criteria are not represented in calculation.
Combining the return of the diverted flow with the mainstream can be accomplished with in-line mixing or flash mixing just before downstream processes. The total combined flow (fT) is calculated as follows:
fA = Average flow rate without diversion, m3/min fC = Controlled discharge rate, m3/min
Therefore, fT = (0.118 + 0.176) m3/min = 0.294 m3/min.
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