Water Storage Capacity of Distribution Reservoirs

STORAGE CAPACITY OF DISTRIBUTION RESERVOIRS

The total storage capacity of a distribution reservoir is the total of

1. Balancing storage (or equalising or operating storage)

2. Breakdown storage.

3. Fire storage

 

1. Balancing or Equalising Storage:

• The primary function of a distribution reservoir is to meet the fluctuating demand.

• The quantity of water required to be stored in the reservoir for balancing the variable demand is called balancing storage of the distribution reservoir.

• This can be found by following methods :

(a) Hydrograph Method.

(b) Mass curve Method.

(c) Analytical Methods.

 

(a) Hydrograph Method :

• The water demand is not constant and varies hourly.

• The demand is more during peak hours in the morning and evening. A

• A hydrograph of hourly demand for maximum day is shown in figure 3.5.

• The pumping rate i.e. mean of hourly demand is shown by line PQ.

• The storage is obtained by determining the shaded area between curve BE and line PQ.

 

(b) Mass Curve Method:

A mass curve is the cumulative demand curve which is obtained by continuously adding the houly demands of the maximum day and plotting against time.

Figure 3.6 shows the mass demand curve CAB plotting it against hours of maximum day.

• CAB curve continuously rises.

• Steepness of curve indicates high rate demand.

• Flatness of curve indicates low rate demand.

• Line CD is the cumulative pumping at uniform rate (join ends C&D)

• Draw tangents through the lowest point A and highest point B parallel to line

CD.

• The highest vertical distance BE between the twe

• At 6 AM there is excess supply = AA' which should be stored.

• At 8.30 PM there is deficit (BB') which must be drawn from storage

Therefore, Storage = Morning excess AA' + Evening deficiency BB'

S = Ep+ Ed

Where S = Storage capacity required

Ep = maximum excess supply through pumping

Ed = maximum excess demand (maximum deficiency)

 

(c) Analytical Method

The cumulative hourly demand and cumulative hourly supplies are tabulated for 24 hrs. The summation of maximum excess demand and the maximum excess supply gives the balancing storage.

 

2. Breakdown storage (Emergency storage)

• It is the storage for emergencies due to pump failure, power failure and during repair works.

• It is difficult to determine this storage as it depends upon the frequency and extent of failures.

• Generally, 25% of total storage capacity of reservoir or about 11⁄2 to 2 times average hourly supply may be considered as breakdown storage.

 

3. Fire Storage:

• This storage is required for fire fighting, which depends on the chances of fire and duration of fire.

• The National Board of Fire Under Writers (America) recommends that the reserve should supply water for 10 hours for fire fighting in communities of 6000 people and for 8, 6 and 4 hours in places with 4000, 2000 and 1000 people respectively.

• For 10 hours of fire fighting per day, the volume of water required to be stored should be 2 million litres.

Fire reserve is determined from the formula :

R = [F-P] T

P = reserve fire pumping capacity, litres/min

T= duration of fire, minutes.

McDonand has suggested the following expression

Where,

R = total storage capacity (million litres)

D= average domestic demand for maximum month (MLD)

F = fire demand (MLD)

P = Pump capacity (MLD)

a,b = Coefficients; 0.2 and 0.1 respectively.

 

Problem 5.1:

A town with a population of one lakh is to be supplied with water daily at 200 litres per head. The variation in demand is as follows

Determine the capacity of service reservoir assuming pumping at uniform rate and the period of pumping to be from 6 am to 6 pm.

Neglect fire demand.

Solution:

Total daily requirement = 1,00,000 x 200 litres = 2 x 10⁷ litres = 20 ML

(a) Analytical Solution

Max. excess demand = 3 ML

Max. excess supply = 5 ML

Total storage = 3+5 = 8ML

 

(b) Mass curve method

Graph is plotted between time and cumulative demand.

Pumping rate = 20 ML /12 hr  = 1.667 ML | hr

The supply curve is drawn with slope as 1.667 ML/hr. Maximum ordinates are found between supply and demand lines.

Storage = 3 ML + 5 ML = 8 ML