The considerations here are similar to those made in the case of the facultative ponds (Section 2.8). The accumulation rate is in the order of 0.03 to 0.10 m3/ inhab.year (Mendon¸ca, 1990; Gon¸calves, 2000), and the lower range is more usual in warm-climate areas. Other data available for accumulation rates are 2 to 8 cm/year (Silva, 1993; CETESB, 1989; Gon¸calves, 2000). These values of yearly increases in the thickness of the sludge layer correspond to accumulation rates lower than 0.03 m3/inhab.year.
The aspects of sludge management in anaerobic ponds are different from fac- ultative ponds. In the latter, the system can operate for several years, eventually during all of the design period, without needing to remove sludge (provided there is a good grit removal in the preliminary treatment). However, because of the smaller volume of the anaerobic ponds, the sludge accumulation manifests itself more rapidly, bringing about the need of an appropriate planning related to the sludge management (see Chapter 11). The anaerobic ponds should be cleaned according
to one of the following strategies:
• when the sludge layer reaches approximately 1/3 of the liquid depth • annual removal of a certain volume, in a pre-determined month, to include
the cleaning stage in a systematic way in the operational strategy of the pond
If the removal is not by emptying and drying inside the pond, the whole sludge mass should not be removed, since this would lead to a total loss of the biomass, requiring the anaerobic pond to start up again.
Example 3.1
Design an anaerobic – facultative pond system using the same data from Example 2.3:
• Population= 20,000 inhabitants
• Influent flow= 3,000 m3/d
• Influent BOD: So= 350 mg/L
• Temperature: T= 23◦C (mean liquid temperature in the coldest month) Solution:
a) Influent BOD load
From Example 2.3:
L= 1,050 kgBOD5/d
Design of the anaerobic pond
b) Adoption of a value for the volumetric loading rate Lv Lv = 0.15 kgBOD/m3.d
This is a conservative value (see Section 3.3.a). However, higher values would lead to a smaller pond volume and, as a result, to low detention times (see section d below).
c) Calculation of the required volume
volume= load volumetric load −→ V = L Lv = 1,050 kgBOD/d 0.15 kgBOD/m3.d = 7,000 m 3
d) Verification of the detention time
t=V
Q =
7,000 m3
3,000 m3/d= 2.3 d OK!
Ponds with such a low detention time should have the inlet at the bottom, in contact with the settled sludge.
Example 3.1 (Continued)
e) Determination of the required area and dimensions
Depth H= 4.5 m (adopted) area= volume depth −→ A = V H= 7,000 m3 4.5 m = 1,556 m 2 Adopt 2 ponds
Area of each pond: 1,556 m2/2 = 778 m2
Possible dimensions of each pond: 34 m× 23 m
f) Concentration of effluent BOD
BOD removal efficiency: E= 60% (see Section 3.4)
BODeffl= (1 − E/100).So= (1 − 60/100) × 350 = 0.4 × 350 = 140 mg/L
The effluent from the anaerobic pond is the influent to the facultative pond.
g) Sludge accumulation in the anaerobic pond
Adopting an accumulation rate of 0.04 m3/inhab.year (see Section 3.6): Annual accumulation= 0.04 m3/inhab.year× 20,000 inhab = 800 m3/year Thickness of the sludge layer in 1 year:
thickness= Annual accumulation× time
pond area =
800 m3/year × 1 year 1,556 m2 = 0.51 m/year = 51cm/year
This annual accumulation rate, expressed in cm/year, is greater than the values mentioned in Section 3.6, probably because the pond in the present example is deep and has a small detention time (smaller surface area for the sludge to spread itself).
Time to reach 1/3 of the pond depth:
time= H/3
yearly thickness =
4.5 m / 3
0.51 m / year= 2.9 year
The sludge volume accumulated during this period corresponds to 1/3 of the net pond volume, that is, 7,000 m3/3 = 2,333 m3of sludge.
The sludge should be removed approximately every 3 years (volume of 2,333 m3) or, annually (removal of 800 m3).
Design of the facultative pond
h) Influent load to the facultative pond
The effluent load from the anaerobic pond is the influent load to the fac- ultative pond. With the removal efficiency of 60% in the anaerobic pond, the
Example 3.1 (Continued)
influent load to the facultative pond is: L= (100− E).Lo
100 =
(100− 60) × 1,050
100 = 420 kg BOD/d
i) Adoption of the surface loading rate
Ls= 220 kgBOD/ha.d (equal to the value adopted in Example 2.3)
j) Required area A = L Ls = 420 kgBOD/d 220 kgBOD/ha.d = 1.9 ha (19,000 m 2 ) Adopt two ponds
Area of each pond: 19,000 m2/2 = 9,500 m2
Possible dimensions of each pond: L= 155 m and B = 62 m (L/B ratio = 2.5) k) Adoption of a value for the depth
H= 1.80 m (adopted)
l) Calculation of the resulting volume
V= A.H = 19,000 m2× 1.80 m = 34,200 m3
m) Calculation of the resulting detention time
t=V
Q =
34,200 m3
3,000 m3/d = 11.4 d
n) Adoption of a value for the BOD removal coefficient (K)
• Complete-mix regime, at 20◦C: K= 0.27 d−1(adopted – see Section 3.5) • Correction for the temperature of 23◦C:
KT= K20.θ(T−20)= 0.25 × 1.05(23−20)= 0.31 d−1
o) Estimation of the effluent soluble BOD
Using the complete-mix model, since the pond is not predominantly longitudi- nal (length/breadth ratio of 2.5):
S= So
1+ K.t =
140
1+ 0.31 × 11.4 = 31 mg/L
p) Estimation of the effluent particulate BOD
Assuming an effluent SS concentration equal to 80 mg/L, and considering that each 1 mgSS/L leads to a BOD5of around 0.35 mg/L (see Section 2.6.2):
Example 3.1 (Continued)
It should be remembered that the particulate BOD is detected in the BOD test, but it may not be exerted in the receiving body, depending on the survival conditions of the algae.
q) Total effluent BOD
total effluent BOD= soluble BOD + particulate BOD
Total effluent BOD= 31 + 28 = 59 mg/L
r) Calculation of the total BOD removal efficiency of the anaerobic–facultative pond system
E=(So− BODeffl)
So .100 =
350− 59
350 × 100 = 83%
s) Total net area (anaerobic+ facultative pond)
Total net area= 0.16 ha + 1.9 ha = 2.1 ha t) Total area required
The total area is in the order of 25% to 33% greater than the required net area. Thus, the total area occupied by the system of ponds and auxiliary structures is approximately:
Total area= 1.3 × 2.1 = 2.7 ha
With primary facultative ponds (Example 2.3), the total area required is 6.2 ha. Therefore, there is a substantial economy of area (56%). The total detention time in the present example is 2.7 d (= 2.3 + 11.4), much lower than that for a primary facultative pond (28.8 m).
It should be remembered that these land requirements are applicable to the current example, which is associated to a relatively high temperature of the liquid, which allows high loading rates and removal efficiencies. In applications in colder places, the required area will be naturally larger.