Design flows to use. When detention time is mentioned, the flow associated with it is normally the average flow. As noted in Chapter 1, however, sewage flows are variable. In water treatment, this variation is, of course, not a problem, but it would be for sewage sedimentation basins. This situation causes a dilemma. A detention time, although customarily attributed to the average flow, may be attributed to other flow magnitudes as well. When the detention time to be used is, for example, 2.5 h in order to limit septicity, what is the corresponding flow? In this particular case, the flow would not necessarily be the average but a flow that would effect a detention time of 2.5 h. What flow then would effect this detention time?
The settling tank has an inherent capacity to damp out fluctuation in rate of inflow. In the present example, the flow is being damped out or sustained in a period of 2.5 h. Because this flow has taken effect during the detention time, it must be the flow corresponding to this detention time and, hence, must be the one adopted for design purposes. To size sedimentation basins, the flow to be used should therefore be the sustained flow corresponding to the detention time chosen. This detention time must, in turn, be a value that limits septicity. The method of calculating sustained flows was discussed in Chapter 1.
Suppose a sedimentation basin is to be designed for an average daily flow rate of 0.15 m /s, how would the sustained flow be calculated with this information on the average flow? It will be remembered that the average daily flow rate is the mean of all 24-h flow values obtained from an exhaustive length of flow record. The unit to be designed must meet this average flow requirement, but yet, the actual flow transpiring inside the tank is not always this average flow. This situation calls for a relationship between sustained flow and the average flow.
What sustained flow is equivalent to an average flow of 0.15 m /s? There are, actually, two sustained flows that must be determined: the sustained peak flow Qpks and the sustained low flow Qms. The sustained peak flow is used to size the tank; the sustained low is used to determine the amount of recirculation desired so as to deter septicity from occurring. The occurrence of the sustained low flow will produce a much, much longer detention time. As mentioned before, detention times greater than 2.5 h (or some other value depending upon the type of waste) can cause septicity and thus inefficiency in sedimentation basins.
Using the methods developed in Chapter 1, relationships of sustained flows and average flow can be obtained. In the present example, the sustained flows analyses should be performed concurrently with that of the average flow analysis so that the aforesaid relationship can be obtained. Considering the present example, the sustained period to be used in the analysis would be the 2.5 h.
In the absence of field data and for an order-of-magnitude design, a ratio of 3:1 (peaking factor) for the sustained one-day peak flow to average may be used. This ratio was obtained from Metcalf & Eddy, Inc. (1991), which shows a lowest sustained period available of one day. Of course, this is not the correct sustained period duration since, actually, it should be 2.5 h in the present example—not one day. Now, using the ratio, the design sustained peak flow rate is 3(0.15) = 0.45 m /s. This flow is the flow that must be used in conjunction with the 2.5-h detention time. Because this sustained flow rate bears a relationship to the average flow rate, using the sustained flow rate is equivalent to using the required average flow rate. In other words, the passage of the sustained flow rates (both high and low) through the tank along with the other larger and smaller flow rates, evens out to the average flow rate which, in the overall, must mean that the requirement of the average flow is met.
Using the same source of reference, the ratio of the sustained low flow to the average for a one-day sustained period is 0.3 (minimizing factor). Again, the sustained period should be 2.5 h and not one day. In the absence of field data, these values may, again, be used for an order-of-magnitude design to determine the amount of recirculation desired. Using this ratio, the sustained low flow is 0.3(0.15) = 0.045 m /s.
For a detention time of 2.5 h and a sustained peak flow rate of 0.45 m /s, the volume of the tank V would be 0.45(60)(60)(2.5) = 4,050 m3. Letting QR be the recirculated flow, we have QR(2.5) + 0.045(60)(60)(2.5) = 4,050, and QR = 1458
m /hr. Thus, with a recirculated flow of 1,458 m /hr = 0.405 m /s and the sustained low flow of 0.045 m /s, the detention time is brought back to 2.5 h and septicity would be avoided when the sustained low flow occurred. This illustration demonstrates that if the calculated detention time is greater than 2.5 h, a portion of the clarified liquid must be recirculated to bring the detention time to 2.5 h, thus maintaining the efficiency of the basin. The recirculation flow is the maximum and may be used to design the recirculation pump. In operation, the rate of pumping would have to be adjusted to effect the required detention time of 1.5 to 2.5 h. The equation for recirculation flow is
Design is an iterative procedure. In the above example, the detention time was chosen as 2.5 h. It could have been chosen as 1.5 h or any value within the range allowed for no septicity and similar steps followed. No matter how the calculation is started, the ultimate goal is to have no septicity in the tank and to satisfy the overflow requirement. If the final calculations show that these goals are met, then the design is correct and the sedimentation tank should operate efficiently. Remember, however, that the 2.5 is the upper limit of the range of values. In addition, a given particular waste may have a different upper limit (or a different lower limit for that matter). If this is the case, then whatever applies in reality should be used.
One other method of design is to use equalization basins ahead of the sedimentation basins. In this case, the inflow would be the equalized flow. There should be no need for recirculation, because only one flow rate is introduced into the sedimentation basin. Design of equalization basins was discussed in Chapter 3.
Example 5.9 To meet effluent limits, it has been determined that the design of a primary sedimentation basin must remove 65% of the influent suspended solids.
The average influent flow is 6,000 m /d; the peak daily flow is 13,000 m /d; and the minimum daily flow is two-thirds of the average flow. It was previously determined that 65% removal corresponds to an overflow rate of 28 m/d. Design a circular basin to meet the effluent requirement at any cost.
Solution: The data on the daily peak and daily low flow rates are not the ones that would be used in design. The peak daily flow rate is a sustained 24-h peak flow rate and the minimum daily flow rate is the sustained 24-h low flow rate. Because the detention time of the sedimentation is a mere 2.5 h, the 24-h sustained flow period is incorrect. If the maximum and minimum hourly flow values were available, however, they would be close to the sustained peak and sustained low flow values over the period of the detention time (2.5 h), respectively, and could be used. The data, as given, could not be used in design. Ans
Example 5.10 The prototype detention time and overflow rate for the design of a sewage sedimentation basin were determined to be 2.5 h and 28 m/d, respectively. The peaking factor is 3.0 and the minimizing factor is 0.3. Design a rectangular settling basin for an average daily flow rate of 20,000 m /d.
Assume solids are flocculent, hence vh should not be greater than 9.0 m/h. Use 8.0 m/h = 0.0022 m/s.
Area, vertical section = 0 0022 = 313.64 m
Considering overflow rate,
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