## Reservoir design (Storage)

The capacity of reservoir of reservoirs depends upon the rates of inflow, losses, and demand. For determining the capacity of the storage reservoir to be constructed, these are to be calculated first.

The inflow and demand values are to be determined for various months of the year. The deficits and surpluses of water are calculated and the storage capacity is made equal to the total deficits.

## Methods of estimating reservoir capacity

- Mass curve method
- Hydrographic method
- Peak demand method

#### a) Mass curve method for reservoir capacity

The mass curve method is used to determine the storage capacity of the reservoir. This is determined by the principle of the mass diagram. A mass curve showing the rate of demand drawn with time as abscissa and the total flow of water during the period as ordinate.

This is represented by the line p as shown in figure 2.1. If the ends of the line p are joined by a straight-line marked Q, then line Q represents the mass diagram of pumping into the tank and its slope represents the rate of pumping.

From the graph, it may be observed that the tank is full at the point â€˜aâ€™ and empty at the point â€˜bâ€™, so that its tangent is drawn at points a and b parallel to the line Q. The vertical intercept bc on the scale would represent the required capacity of the reservoir.

If now it is desired to determine the capacity of the reservoir when it is not required to pump water for all the 24 hours as in the above case but for a certain period, say 6 hours to 18 hours (6 AM to 6 PM), the method would be as follows:

Draw a straight-line marked R from the point of 6 hours on the abscissa with the total flow zero to the point of 18 hours on the line with the total flow of the day. The slope of this line represents the rate of pumping during the aforesaid period of pumping parallel to line R.

Draw another line S, intersecting line P at 6 hours and the ordinate at 18 hours at point d. Then the vertical distance de (e being the point of intersection of a vertical line from d to the line P) represents on the scale the required capacity of the reservoir.

#### b) Hydrographic method

The hydrographic method is also used to determine of storage capacity of the reservoir. This method is rarely used. In this method daily or monthly stream flow data are determined depending upon the size of the reservoir.

The quantity of consumption is determined by corresponding to daily or monthly time. These stream flow and consumption are plotted on ordinate while time is plotted on the abscissa. The curve so obtained is called hydrograph. The quantity of the reservoir is computed in the following steps:

- If the yield value curve is above the consumption curve at a certain time, it indicates the inflow of water is more than the demand.
- If the yield value curve is below the consumption curve at a certain time, it indicates the inflow of water is less than the demand, and hence storing of water is required.
- If the yield value curve just meets the consumption curve at a certain time, it indicates the inflow of water is just sufficient to meet the demand. In this case, the reservoir may be required for the emergency period.

The storage capacity of the reservoir is determined by computing the maximum cumulative total deficit between certain time intervals. In some cases, allowances for seepage and evaporation are also considered.

#### c) Peak demand method

The demand for water required by the consumers does not remain constant. It varies from hour to hour, day to day, week to week, and year to year. If the supply of water is enough to fulfill the demand of consumers, then storage of water is not required otherwise required.

The storage of the reservoir is obtained by the following formula,

**Storage capacity of reservoir = average demand of water x peak factor -minimum supply of water at that duration.**

**= Peak demand of water- minimum supply at that duration.**

**Q. Determine the capacity of the distribution reservoir, if the pump installed can supply the water in the reservoir at a uniform rate of 1.45 cumec/sec.**

Sol: -Rate of pumping =1.45 cumec/sec.

=1.45 Ã— 60 Ã— 60 Ã— 1000 lit/sec

=5.22 million liter/sec

#### Calculation table

Hour | Demand/ consumption in million liters | Pumping of water in million-liter hr. | Water required from storage reservoir in million lit/hr |

1 | 7.55 | 5.22 | 2.33 |

2 | 6.35 | 5.22 | 1.13 |

3 | 5.95 | 5.22 | 0.73 |

4 | 5.75 | 5.22 | 0.53 |

5 | 3.50 | 5.22 | â€“ |

Total | 4.72 |

Hence, the required capacity of the reservoir **= 4.72 million lit/hr.**

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