Flow Equalization Volume

V = max \int (I - O) d t from the cumulative (mass-diagram) plot

A constant-outflow equalizer needs storage equal to the maximum cumulative surplus of inflow vs the average (or target) outflow. Same mass-diagram idea as earthwork.

$I$: Inflow hydrograph
$O$: Target outflow (often the average)

Step-by-step solved example

Four 6-h inflows 2, 6, 5, 3 MGD. Constant O=average. Min V in MG?

Mass diagram of I vs O; the max intercept is V_eq.

1. Average
(2+6+5+3)/4=4 MGD. Volumes per 0.25 d: 0.5, 1.5, 1.25, 0.75 MG.
2. Cumulative surplus vs 1 MG per period
Cum I: 0.5, 2.0, 3.25, 4.0; cum O: 1, 2, 3, 4. Need 0.5 MG to cover the first deficit then peak lag — max storage from mass diagram ≈ 0.5 MG.

Answer: V ≈ 0.5 MG (from the mass diagram)

10 practice questions

0/10 correct

1.If I is already constant, V_eq =

2.Online equalization sits

3.The mass diagram for earthwork is analogous because

4.Equalization helps the plant by

5.Mixing in the eq tank is to

6.Average of 3, 5, 7 is

7.Storm tanks at a WWTP are

8.If O is a pump of capacity 6 MGD and peaks are 10, V still from

9.Units: 1 MGD × 1 day =

10.Diurnal municipal Q typically peaks