Sedimentation tanks

leaks and any splashes from the air valve, avoiding flooding. If the pipe is surrounded by stagnant water, the spring becomes a source of contamination.

Valve boxes are dealt with in more depth in the next chapter.

11. 7 SEDIMENTATION TANKS

Surface water tends to contain many suspended solids which give it a turbid appearance. Aside from the appearance, these particles are problematic in terms of the smell and taste they leave in the water and the wear they cause on the pipes and accessories. Once inside the pipes, they sediment and accumulate in the low points, reducing the effective diameter.

If water is allowed to settle for some time, these particles settle and the end result is clearer water. Sand has a sedimentation velocity in water of 0.5 to 6 m/min. Finer particles and bacteria do not sediment. For practical purposes, when water is at rest for an hour, it has lost the majority of suspended solids.

In systems with reservoir tanks, a small sedimentation tank with a detention time of 15 minutes allows for the largest, more abrasive particles to sediment out before entering the pipe. The remaining sedimentation will take place in the reservoir tank. A system with no storage tanks require times closer to an hour.

Measure of turbidity

Turbidity is measured in a transparent tube of water. At the bottom there is a cross or circle 2mm wide. The water is emptied from the tube until the marking at the bottom

can be seen clearly. The tube is calibrated and allows a reading to be made. The unit of measurement is NTU. For treated water, turbidity should be less than 5 NTU.

The reading in the photo is approximately 190 NTU. Note that you can see the cross on the floor of the tube, and that the measurement is taken in daylight over a white background. This sample, taken from a public tap stand of a gravity flow system in Tanzania, has an unacceptably high turbidity and indicates the absence of effective filtration.

Calculations for a sedimentation tank

The goal is to have a tank large enough to allow the water to remain there sufficient time for sedimentation to take place: 60 minutes for systems with no reservoir tank, and 15 minutes for systems with. The volume of the tank in m3

t

Q

V

=3.6*

*

is determined by: Q, flow in l/s; t, retention time in hours.

Nonetheless, not every kind of tank works for sedimentation. The internal velocity should be less than 0.005 m/s to avoid currents which impede sedimentation, and the length should be at least four times the width, to absorb incoming turbulence. To calculate the velocity:

Calculation example:

Work out the dimensions of a sedimentation tank from an intake, feeding a gravity flow system with no reservoir tank, with an average future population demand that is estimated to be around 2 l/s. Work out the same for a system with a reservoir tank.

No reservoir tank:

The depth of the tank is established as 0.75m, according to the data in the section below.

The temporal demand variations will affect the input velocity at the intake. In the absence of data on temporal variations, take a value four times higher as the peak flow rate (see section 2.8):

Q= 2l/s*4 = 8 l/s

The required volume is: V = 3.6 * Q * t = 3.6* 8l/s * 1h = 28.8m3 For a velocity of 0.005 m/s or less:

v = Q / 100 * A * h  A = Q /1000 * v * h

A= 8l/s / 1.000 * 0.005 m/s * 0.75m = 2.13m or greater: 2.2m. The required tank length for 28.8m3 is:

L = V / A * h = 28.8m3 / 2.2m * 0.75m = 17.45m

Lastly, the length, 17.45m, is verified to be more than 4 times the width, 4 * 2.2m = 8.8m.

The required tank measures 17.45m long, 2.2m wide, with a depth of 0.75m. With reservoir tank:

With storage, the detention time is 0.25 hours. Also, as the tank absorbs temporal variations, the input flow used is the original 2 l/s.

The required volume is: V = 3.6 * Q * t = 3.6* 2l/s * 1h = 7.2m3 The width is A = Q /1000 * v * h = 2l/s / 1000 * 0.005 m/s * 0.75m = 0.53m

Construction details

• The turbid input water should enter at mid-height, evenly dispersed over the entire width of the tank. This can be achieved by perforating small holes in the pipe at regular intervals.

• The clean water exit pipe should be as high as possible, just below the overflow pipe.

• The optimum depth is between 0.7m and 1m.

• There should be a washout to allow for complete emptying of the tank for cleaning, together with a valve to shut off the incoming flow.

It´s fairly easy to mix the concepts and end up making a hybrid intake, which fuses a standard intake with a sedimentation tank. In the intake shown in the photo, the sedimentation tank doesn´t have the required dimensions or shape. The water comes in from above, and is taken out from below. The input flow goes directly to the intake (in the foreground) where sedimentation occurs. Finally, the construction doesn´t allow for cleaning and fails to protect the valve box from interference.

Taking the width as 0.6m, the required length for 7.2m3 is: L = V / A * h = 7.2m3 / 0.6m * 0.75m = 16m Again, it is more than four times the width.

The required tank measures 16m long, 0.6m wide, with a depth of 0.75m. Note that the second tank is smaller and cheaper. The money saved could be used to build a reservoir tank.

A length of 16m is important and not all locations will allow for this. This can be achieved by partitioning the tank, providing considerable material savings.