8 12 Midnight

FIGURE 3.9 Long-term extreme sewage flow and BOD pattern in a sewage treatment plant.

difficulty imposed by these extreme variations, an equalization basin should be provided. Equalization is a unit operation applied to a flow for the purpose of smoothing out extreme variations in the values of the parameters.

In order to produce an accurate analysis of equalization, a long-term extreme flow pattern for the wastewater flow over the duration of a day or over the duration of a suitable cycle should be established. By extreme flow pattern is meant diurnal flow pattern or pattern over the cycle where the values on the curve are peak values— that is, values that are not equaled or exceeded. For example, Figure 3.10 is a flow pattern over a day. If this pattern is an extreme flow pattern then 18 m /h is the largest of all the smallest flows on record, and 62 m /h is the largest of all the largest flows on record. A similar statement holds for the BOD. To repeat, if the figure is a pattern for extreme values, any value on the curve represents the largest value ever recorded for a particular category. In order to arrive at these extreme values, the probability distribution analysis discussed in Chapter 1 should be used. Remember, that in an array of descending order, extreme values have the probability zero of being equaled or exceeded. In addition to the extreme values, the daily mean of this extreme flow pattern should also be calculated. This mean may be called the extreme daily mean and is needed to size the pump that will withdraw the flow from the equalization unit. In the figure, the extreme daily means for the flow and the BOD are identified by the label designated as average.

Now, to derive the equalization required, refer to Figure 3.10. The curve represents inflow to an equalization basin. The unit on the ordinate is m /h and that on the abscissa is hours. Thus, any area of the curve is volume. The line identified as average represents the mean rate of pumping of the inflow out of the equalization basin. The area between the inflow curve and this average (or mean) labeled B is the area representing the volume not withdrawn by pumping out at the mean rate;

FIGURE 3.10 Determination of equalization basin storage.

it is an excess inflow volume over the volume pumped out at the time span indicated (9:30 a.m. to 10:30 p.m.). The two areas below the mean line labeled A and D represent the excess capacity of the pump over the incoming flow, also, at the times indicated (12:00 a.m. to 9:30 a.m. and 10:30 p.m. to 12:00 a.m.).

The excess inflow volume over pumpage volume, area B, and the excess pumpage volume over inflow volume, areas A plus D, must somehow be balanced. The principle involved in the sizing of equalization basins is that the total amount withdrawn (or pumped out) over a day or a cycle must be equal to the total inflow during the day or the cycle. The total amount withdrawn can be equal to withdrawal pumping at the mean flow, and this is represented by areas A, C, and D. Let these volumes be VA, VC, and VD, respectively. The inflow is represented by the areas B and C. Designate the corresponding volumes as VB and VC. Thus, inflow equals outflow,

From this result, the excess inflow volume over pumpage, VB, is equal to the excess pumpage over inflow volume, VA + VD. In order to avoid spillage, the excess inflow volume over pumpage must be provided storage. This is the volume of the equalization basin—volume VB. From Equation (3.23), this volume is also equal to the excess pumpage over inflow volume, VA + VD.

Let the total number of measurements of flow rate be | and Qt be the flow rate at time t. The mean flow rate, Qmean, is then

t$ = time of sampling of the last measurement. Qmean is the equalized flow rate. Considering the excess over the mean as the basis for calculation, the volume of the equalization basin, Vbasin is

where pos of ((Qi + Qi-1)/2 - Qmean) means that only positive values are to be summed. By Equation (3.23), using the area below the mean, Vbasni may also be calculated as

neg of ((Qi + Qi-1)/2 - Qmean)(ti - ti-1) means that only negative values are to be summed. The final volume of the basin to be adopted in design may be considered to be the average of the "posof" and "negof" calculations.

Examples of quality parameters are BOD, suspended solids, total nitrogen, etc. The calculation of the values of quality parameters should be done right before the tank starts filling from when it was originally empty. Let Ci-1i be the quality value of the parameter in the equalization basin during a previous interval between times ti-1 and ti and Ci i+1 during the forward interval between times ti and ti+1. Let the corresponding volumes of water remaining in the tank be Vrem_ and Vremi i+1, respectively. Also, let Ci be the quality value of the parameter from the inflow at time ti, Ci+1 the quality value from the inflow at ti+1, Qt the inflow at ti, and Qi+1 the inflow at ti+1. Then,

( Vremi-1,i) + (^jr^ )(ti+1- h) - Q mean(ti+1 - ti)

Vremi-1,i is the volume of wastewater remaining in the equalization basin at the end of the previous time interval, ti-1 to ti and, thus, the volume at the beginning of the forward time interval, ti to ti+1. Ci-1i (Vremi-1i) is the total value of the quality inside the tank at the end of the previous interval; thus, it is also the total value of the quality at the beginning of the forward interval. (C' + Ci+1)/2 is the average value of the parameter in the forward interval and Q + Qi+1)/2 is the average value of the inflow in the interval. Thus, ((C' + Ci+1)/2)((Qi + Qi+1)/2)(ti+1 - ti) is total value of the quality coming from the inflow during the forward interval. Ci,i+1 is the equalized quality value during the time interval from tt to ti+1. Ci i+1 Qmean (ti+1 - ti) is the value of the quality withdrawn from the basin during the interval to tt to ti+1.

The sizing of the equalization basin should be based on an identified cycle. Strictly speaking, this cycle can be any length of time, but, most likely, would be the length of the day, as shown in Figure 3.10. Having identified the cycle, assume, now, that the pump is withdrawing out the inflow at the average rate of Qmean. For the pump to be able to withdraw at this rate, there must already have been sufficient water in the tank. As the pumping continues, the level of water in the tank goes down, if the inflow rate is less than the average. The limit of the going down of the water level is the bottom of the tank. If the inflow rate exceeds pumping as this limit is reached, the level will start to rise. The volume of the basin during the leveling down process starting from the highest level until the water level hits bottom is the volume Vbasin.

Let tibot be this particular moment when the water level hits bottom and the inflow exceeds pumping. Then at the interval between tibot-1 and tibot, the accumulation of volume in the tank, Vremi-1,i = Vremibot_1iibot is 0. At any other interval between ti-1 and ti when the tank is now filling,

The value of Vremi-1,i will always be positive or zero. It is zero at the time interval between tibot and tibot-1 and positive at all other times until the water level hits bottom again.

The calculation for the equalized quality should be started at the precise moment when the level hits bottom or when the tank starts filling up again. Referring to Figure 3.10, at around 10:30 p.m., because the inflow has now started to be less than the pumping rate, the tank would start to empty and the level would be going down. This leveling down will continue until the next day during the span of times that the inflow is less than the pumping rate. From the figure, these times last until about 9:30 a.m. Thus, the very moment that the level starts to rise again is 9:30 a.m. and this is the precise moment that calculation of the equalized quality should be started, using Equations (3.27) and (3.28).

Example 3.8 The following table was obtained from Figure 3.10 by reading the flow rates at 2-h intervals. Compute the equalized flow.

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