Land treatment of wastewater 427Therefore,
Area = (3.65x3, 000 m^3 /day)/(3 cm/day x 365 days/year)
= 10 haFrom the required area values obtained using both design criteria, it can be
seen that the mathematical model presented based on application rate, slope
length and application period (Equation 8.14), gives an economical value of 8.2
ha compared to an area of 10 ha with a 3 cm/day hydraulic loading.
Example 8.5
A wastewater has the following characteristics.
Flow rate = 100 m^3 /day
BOD 5 = 250 mg/L
Fecal coliforms = 6 x 10^5 no./100 mL
Design precipitation = 1000 mm/year
Percolation = 400 mm/year
Evapo-transpiration = 1200 mm/year
a) If the soil type is clay loam, which type of land treatment could be
suitable?
b) Determine whether there will be run-off from this land treatment or
not, if so calculate the run-off in m^3 /day.
c) From the experimental results it was found that the wastewater
application cycle to this land should be 3 days of application and 11
days of resting. Application area of the site is 14 ha. Determine the
approximate wastewater application rate to be applied, which will
not produce run-off.Solution:
a) As the soil type is clay loam, the suitable type of land treatment will be the
OF system (according to Figure 8.1).
b) From Figure 8.1:
Choose a liquid loading rate = 4 in/week
= 4 x 25.4 x 52 mm/year
= 5283 mm/year.
General mass balance equation for land treatment of wastewater is given in
equation 8.3:
Design precipitation + wastewater applied = percolation + evapotranspiration + runoff.
1000 + 5283 = 400 + 1200 + Runoff