Example 8.4
Design a septic tank suitable for a household with up to eight occupants in a lowdensity housing area in which the houses have full plumbing, all household wastes go to the septic tank and the nominal water supply is 200 l per person per day. Water is used for anal cleaning and the ambient temperature is not less than 25°C for most of the year.
Stage 1
Volume of liquid entering the tank each day
A = P × q
where 
A = volume of liquid to be stored in the septic tank 

P = number of people using the tank 

q = sewage flow = 90% of the daily water consumption per person (Q). 
q 
= 0.9 × Q = 0.9 × 200 

= 180 litres per person per day. 
Therefore A = 8 × 180 = 1440 litres
Stage 2
The volume of sludge and scum is given by
B = P × N × F × S
where 
B = volume of sludge and scum 

P = number of people using the tank 

N = period between desludgings 

F = sizing factor (see Table 6.2) 

S = sludge and scum accumulation rate (see Chapter 6) 
Assume N is 3 years; from Table 6.2, F = 1.0; as all wastes go to septic tank S = 40 l per person per year.
Therefore:
B 
= 8 × 3 × 1.0 × 40 

= 960 litres 
Stage 3
Total tank volume 
= A + B 

= 1440 + 960 

= 2400 litres (2.4 m^{3}) 
Stage 4
Assume liquid depth = 1.5 m
Assume tank width is W m
Fig. 8.1. Internal dimensions of the septic tank designed in example 8.4
WHO 91496
Assume two compartments,
length of first = 2W
length of second = W
This tank is illustrated in Fig. 8.1.
Volume of tank (V) 
= 1.5 × (2W + W) × W 

= 4.5 W^{2} 
Thus 4.5 W^{2} 
= 2.4 m^{3} 
W 
= 0.73 
Therefore:
width of tank 
= 0.73 m 
length of first compartment 
= 1.46 m 
length of second compartment 
= 0.73 m 
Depth of tank from floor to soffit of cover slab
= liquid depth + freeboard
= 1.5 + 0.3
= 1.8 m
Flotation
Since septic tanks have sealed walls and floor, the design must be checked to make sure that the tank does not float out of the ground. Flotation will occur if the total mass of the empty septic tank is less than the mass of the water it displaces. This will only happen if the groundwater level is higher than the bottom of the tank.
Calculate the mass of the walls, floor, roof and any baffle walls (concrete: 2400 kg/m^{3}; brickwork: 1500 kg/m^{3}). Measure the volume of the tank (outside dimensions) between the highest groundwater level and the bottom of the tank. Multiply the volume by the density of water (1000 kg/m^{3}). This gives the mass of water displaced.
If the mass of water displaced is greater than the total mass of the empty septic tank then the tank may float. This can be prevented by increasing the mass of the structure (e.g., by increasing the thickness of the floor or walls) or reducing the amount of the tank that is below the water table.
Soil pressure
For large tanks, such as for a school or a number of houses, it is necessary to check that the side walls of the tank are not likely to collapse owing to the outside soil and water pressure. This is most likely when the tank is empty. Such a calculation is beyond the scope of this book, and reference should be made to a manual on reinforced concrete or masonry design.
Example 8.5
Design a septic tank for a household having five occupants in a mediumdensity housing area in which the houses have full plumbing. Only WC wastes go to the septic tank, and paper is used for anal cleaning. The ambient temperature is more than 10°C throughout the year.
Stage 1
Daily volume of liquid
A = P × q
If the WC has a 10litre cistern and each person flushes it four times a day, the sewage flow q = 4 × 10 = 40 litres per person per day and A = 5 × 40 = 200 litres.
Stage 2
Volume for sludge and scum
B = P × N × F × S
Assume N is 3 years; from Table 6.2, F = 1.0; as only WC wastes go to septic tank S = 25 litres per person per year.
So
B 
= 5 × 3 × 1.0 × 25 

= 375 litres 
Stage 3
Total tank volume V 
= A + B 

= 200 + 375 

= 575 l (0.575 m^{3}) 
As this is less than the minimum recommended volume of 1.0 m^{3}, the dimensions for the minimum volume should be calculated.
Stage 4
Assume liquid depth = 1.5 m.
Assume tank width is W m.
Assume two compartments:
length of first 
= 2W 
length of second 
= W 
Volume of tank 
= 1.5 × (2W + W) × W 

= 4.5 W^{2} 
If 4.5 W^{2}= 1.0 m^{3},
then W = 0.47 m
As this is less than the recommended minimum width of 0.6 m, assume W = 0.6 m.
Length of first compartment (2W) = 1.2 m
Length of second compartment (W) = 0.6 m
Depth of tank from floor to soffit of cover slab
= 1.5 m (liquid depth) + 0.3 m (freeboard)
= 1.8 m
The tank volume (excluding freeboard) is:
(1.2 + 0.6) × 0.6 × 1.5 = 1.62 m^{3}
which is larger than the required volume calculated in stage 3. This is no disadvantage; in practice the minimum retention time will be greater than 24 hours or the tank will provide longer service than three years before requiring desludging.