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8.3 Service Storage 275

Table 8.1 Pumping Characteristics of System in Example 8.4
Pumps in service, No. 1 2 3 1 & 2 1 & 3 2 & 3
Rate of pumping, MGD 10 21 33.5 27 36 42
Head, ft 81 83 88 85 90 93

Efficiency, % 80 88 88 71, 86 35, 87 68, 64

Rate of pumping, MGD 15 25 37 34 43.5 49.5
Heat, ft 66 78 84 80 85 89
Efficiency, % 89 89 89 82, 88 71, 89 79, 87

Rate of pumping, MGD 16.5 28.5 40.5 40 49.5 56.5
Heat, ft 62 66 73 73 79 84

Efficiency, % 88 84 86 88, 88 83, 88 79, 89
3
Conversion factors: (1 MGD 0.0438 m /s 43.8 L/s; 1 ft 0.3048 m).

8.3 SERVICE STORAGE
The three major components of service storage are as follows:
1. Equalizing, or operating, storage
2. Fire reserve
3. Emergency reserve.

8.3.1 Equalizing, or Operating, Storage
Required equalizing, or operating, storage can be read from a demand rate curve or, more
satisfactorily, from a mass diagram. As shown in Fig. 8.6 for the simple conditions of
steady inflow, during 12 and 24 h, respectively, the amount of equalizing, or operating,
storage is the sum of the maximum ordinates between the demand and supply lines. To
construct such a mass diagram, proceed as follows:
1. From past measurements of flow, determine the draft during each hour of the day
and night for typical days (maximum, average, and minimum).
2. Calculate the amounts of water drawn up to certain times, that is, the cumulative
draft.
3. Plot the cumulative draft against time.
4. For steady supply during 24 h, draw a straight line diagonally across the diagram,
as in Fig. 8.6a. Read the storage required as the sum of the two maximum ordinates
between the draft and the supply line.
5. For steady supply during 12 h (by pumping, for example) draw a straight line diag-
onally from the beginning of the pumping period to its end—for example, from
6 a.m. to 6 p.m., as in Fig. 8.6b. Again read the storage required as the sum of the
two maximum ordinates.