FIGURE 1.10 Hydrograph for deriving inflow.

The time of extension should be at least equal to the time of the hydrograph of the rainfall. Therefore, two hydrographs now exist: the actual hydrograph that contains the rainfall event and the extended hydrograph that assumes there was no rainfall event. The two hydrographs are then superimposed; the hydrograph with rainfall event will be above that of the other. This difference represents the inflow.

Figure 1.10 shows a hydrograph for two days of flow that captures the direct inflow into the sewer. Note the pattern of the hydrograph on the first day. The 0 on the abscissa represents midnight of the first day. At around 25 hours from the previous midnight, a burst of inflow is recorded continuing until approximately 34 hours. The inflow peak is recorded at around 30 hours. The flow pattern of the previous day has been extended as shown by the dotted line. From the indicated construction, the direct inflow is approximately 47.8 m /h.

1.7.7 Summary Comments for Deriving Flow Rates by the Probability Distribution Analysis

The technique of obtaining the elements in the probability distribution table is important. Each of these elements should have equal likelihood of occurring. For example, consider element 4 (or order 4) in Table 1.13, which is 44.92. The value of n for this element is 2, which means that therearetwomembersofthiselement.

Above element 4 are 45.00, 45.20, and 45.21 numbering a total of 7 members. Thus, 7 element members are above 44.92makingatotalof9indicatedunderthe column of Zn. Now, the cumulative probability is calculated as

The use of 9 (which is the total number of element members up to element 4) means that all the members are considered having equal likelihood of occurring. That is, all the element members in a probability distribution analysis have equal probability. Verify this by studying how the calculations are done in the rest of the table and, in fact, all the tables of probability distribution.

The assumption of equal probability is correct. Table 1.13 has 61 members. If you were to determine the probability of element number 16 in this table, for example, how would you do it? Unless it is known that this element is biased, its probability has to be 24 in 62 = 0.39.

The requirement of equal likelihood of occurrence means that the elements must be sampled "uniformly and at equal intervals." For example, for the elements of Table 1.12, if the sampling were done irregularly such as taking measurements in the first two days of a week, skip the next week, resume in the last three days of the fourth week and so on, the resulting element members will not have equal likelihood of occurring and the probability distribution analysis will not apply. The sampling should be done uniformly and at equal intervals. If this is followed, every element will have equal probability. Of course, the more elements there are, the more accurate will be the prediction of the probability distribution analysis.

To repeat, sampling must be done uniformly and at equal intervals for the probability distribution analysis to be applicable. Otherwise, the application of the method will be difficult, since the probability of each element will be undetermined or, if determinable, is calculated with difficulty. Of course, the probability of each element can always be assumed equal, even if the sampling is irregular. If you see this done in practice, the concept upon which the probability distribution analysis is based has been violated and you are guaranteed that whatever is the result, it is wrong.

Example 1.8 Assuming the contributing population in the field survey of Figures 1.6 and 1.7 is 2000, what will be the design flow to size a sewer for a population of 10,000? Sizing of sewers is based on peak flows.

Solution: The peak hourly flow rate was 68.2 m /h. Therefore, the design flow to size sewer, Q^ hour, is

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