In this essay we will discuss about:- 1. Functions of Distribution Reservoirs 2. Capacity of the Reservoirs 3. Determination of Storage Capacity 4. Accessories of Reservoirs.
Essay # 1. Functions of Distribution Reservoirs:
These are also known as service reservoirs, and are mainly provided for storing the treated water, for supplying water to the town or city. These reservoirs are also provided for meeting the water demand during fires, breakdown of pumps, repair etc. The reservoirs absorb the hourly fluctuations in the water demand.
Following are the main functions of the storage and distribution reservoirs:
(i) To store the treated water till it is distributed to the city.
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(ii) To absorb the hourly variation in the water demand, and thus allowing the treatment units and pumps to work at the average constant rate. This will reduce the running maintenance operation costs of the treatment units as well as also improve their efficiency.
(iii) To maintain the constant pressure in the distribution main, because when the pressure in pipe lines decreases due to increase in demand at peak hours, the extra demand of water is fed by these reservoir, and the pumps continue their work at constant speed.
(iv) Distribution reservoirs lead to an overall economy by reducing the sizes of pumps, pipe lines and treatment units,
(v) By providing distribution reservoirs, the pumping of water in shifts is possible, because treated water will continuously flow in these reservoirs.
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Generally distribution reservoirs are located near the central portion of the distribution area. It is always better to construct them on high ground of city or at such place where it can be constructed economically.
Distribution reservoirs may be classified according to their position as ‘surface’ or Elevated reservoirs. These are also classified according to their materials of construction e.g. Steel, R.C.C. or masonry. The capacity of distribution reservoir depends on the maximum pumping capacity, minimum safe yield of source of supply, requirements for the fire extinguishing and maximum rate of consumption.
Essay # 2. Capacity of the Reservoirs:
The total capacity of the service reservoir is determined by adding the quantity of water required for various purposes.
The total capacity is made up to of the following:
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(i) Balancing Reserve:
It is the quantity of water required for balancing the variations in the demand against the constant supply from the treatment plant. This is mostly calculated by means of Mass Curve or Hydrograph of the inflow and outflow, or by analytical methods using standard tables.
(ii) Breakdown Reserve:
Sometimes there is breakdown in the pumps or power driving them. Therefore some quantity of water is required to be kept as reserve or breakdown time, so that water can be distribute even during the repair work of pumps or power.
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This quantity of water depends on the time taken during the repair work, and as such it is variable, therefore, it is very difficult to estimate. Any way the repair time may be assumed as 1½ to 2 hours of average daily supply. Sometimes human provision of 25% of the total capacity of the reservoir is done by breakdown reserve.
(iii) Fire Reserve:
The quantity of water required to be kept as reserve for fire-fighting can be obtained by deducting the reserve fire pumping capacity ‘C’ from the fire demand ‘F’ and then multiplying it by the probable duration of fire time ‘T’.
... 20 Fire reserve = (F – C) T
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The value of F mainly depends up to the properties, which are to be protected from fire and the risk of fire involved. In India 5 litres/capita is taken as fire reserve.
The total quantity of water required for fire can be calculated by the formulae 5.1, 5.2, 5.3 and 5.4.
Essay # 3. Determination of Storage Capacity:
Determination of the storage capacity of the storage reservoir can be determined by various methods, which will be clear from the following solved examples:
Example 1:
A city with a population of 1.2 million has a continuous water supply. The average demand of the town is 250 litres/capita/day. The water is supplied to the city by direct pumping.
The total supply of 250 litres is supplied as under:
Water is supplied from the treatment plant at a uniform rate of 12.5 million litres per hour, for all the 24 hours. Determine the capacity of the reservoir required for distribution of water.
Assume no loss or draw from the water main.
Solution:
Daily demand of the town
= 250 × 1.2 × 106 litres
= 300 × 106 litres
The supplied water is converted into cumulative demand as shown in Table 18.1:
The mass curve of the demand is now plotted from the data or columns (1) and (4) of Table 18.1 as shown in Fig. 18.8.
The supply of the water works is 12.5 × 106 litres/hour, the supply line is also plotted on the curve as shown in Fig. 18.8. The two maximum ordinates A and B enclosed between the demand and supply lines are read out from Fig. 18.8 as
A = 0 million litres
B = 47.5 × 106 litres
... Total storage capacity required
20 = A + B
20 = 0 + 47.5 × 106
20 = 47.5 × 106 litres
Example 2:
The population of a town is 2.5 lakh and its per capita demand is 270 litres/day/capita. The probable hourly variations in the demand rate are given below in Table 18.2.
Determine the capacity of the storage reservoir to be provided for balancing the variable demand against a constant rate of pumping:
(a) If the pumping is to be done only from 5 a.m. to 11 a.m. and 2 p.m. to 8 p.m.
(b) If the pumping is to be done for all the 24 hours. Also determine the rate of pumping in each case.
Solve this problem using:
20 (i) mass curve method,
20 (ii) analytical method.
Solution:
Using mass curve method.
Average daily requirement = 270 × 0.25 × 106 litres = 67.5 × 106 litres
Average hourly demand = 67.5/24 × 106 litres/hour = 2.8125 × 106 litres/hour
The cumulative hourly demand is calculated as shown in Table 18.2 which is self- explanatory.
(a) When pumping is done for the limited period:
The demand curve is drawn in Fig. 18.9 with the help of the columns (1) and (4) of Table 18.2. The total quantity of water required is pumped between 5 am. To 11 am. And from 2 p.m. to 8.00 p.m. (i.e., for 112 hours)
... Pumping rate = 6.75 × 106/12 liters/hour = 5.625 liter/hour
The supply is also drawn on the mass curve of the demand as shown in Fig. 18.9.
From Fig. 18.9, the maximum ordinates enclosed at A and B between supply line and demand curve are noted as follows:
A = 7.8 × 106 litres
B = 3.18 > 109 litres
Total storage required = 10.98 x 106litres
(b) When pumping is done for all the 24 hours of the day:
The mass curve of the demand is plotted from the data of table 18.2. Fig. 18.10 shows the mass curve for demand and supply lines.
The rate of pumping in this case:
= 67.5 × 106/24 = liters/hour
= 2.8125 × 106 liters/hour
From the mass curve, the two maximum ordinates A and B are measure on the curve as shown in Fig, 18.10 between the supply line and demand curve and are noted as below:
Supply line and demand curve are noted as below:
A = 14 million litres
B = 4.55 million litres
Total storage required = 18.55 million litres.
Using Analytical Method:
Case (a) When pumping is done for fixed hours. The analytical method can be easily followed from the Table 18.3 which is self-explanatory.
From the above Table 18.3, it is clear that the maximum excess demand is 3.178 and the maximum excess supply is 7.791 million litres.
... Total storage required = (3.178 + 7.791) × 106litres
= 10.969 × 106 litres
which is nearly equal to that obtained by the mass curve method
= 10.98 × 106 litres
Case (b). When the pumping is done for 24 hours.
The analytical solution is shown in Table 18.4.
From Table 18.4 it is clear that maximum excess of demand is 4.5835 million litres and maximum excess supply is 13.9785.
... Total storage required = (4.5835 + 13.9785) million litres
= 18.562 × 106 litres
which is nearly the same as obtained from the mass curve method
= 18.55 × 106.
Essay # 4. Accessories of Reservoirs:
Following are the various accessories which are commonly provided in the reservoirs:
(i) Inlet pipe for the entry of water.
(ii) Outlet pipe for the withdrawal of the water.
(iii) Overflow pipe, to prevent the overflow of reservoir. This overflow pipe is generally connected to the drain.
(iv) Float switch to stop the pump when the tank is full.
(v) Float gauge to show the depth of water in the tank.
(vi) Wash out pipe for washing out the suspended impurities in the tank.
(vii) Main-holes for providing entry in the tank.
(viii) Ladder to reach the top and bottom of the tanks.
(ix) Ventilation for fresh air circulation in tank. Steel mesh is provided which allows the entry of fresh air, as well as prevents the entry of birds etc.
(x) Chlorinator in case when the water is directly pumped in the overhead reservoir from the tube-wells.