A primary sedimentation basin has an influent flow of 8400 m3/d, which has an in

Question

A primary sedimentation basin has an influent flow of 8400 m3/d, which has an
influent suspended solids concentration of 185 mg TSS/L. To improve this process, 5 mg/L of
coagulant (a chemical used to bind particles, allowing them to grow and improve settling) is
added to the influent. Assume all the coagulant ends up in the sludge. The chemical addition
results in a removal of 87% of the influent solids. The resulting underflow sludge stream has a
7% solids concentration and a density of 1,028 kg/m3.

Calculate
(a) the mass flux of sludge (kg sludge/d = kg of TSS & coagulant/d) produced in the
underflow from this tank,
(b) the volumetric flow rate of sludge (m3/d) produced in the
underflow from this tank, and
(c) the concentration of suspended solids in the overflow effluent
(mg TSS/L)? Hint: the water added with the coagulant is negligible compared to the influent
flow rate.

Explanation / Answer

in primary sedimentation basin has influent flow of 8400 m3/d which has 185 mg TSS/L of suspended solids.

Qin = 8400 m3/d ; Suspended Solids = 185 mg TSS/L = 0.185 kg TSS/m3

5 mg/L of coagulant added in the basin and it completely mixed and produces 87% of influent suspended solids.

coagulant concentration = 5 mg/L = 0.005 kg/m3

Suspended solids that changed into sludge = 185*0.87= 160.95 mg TSS/L = 0.16095 kg TSS/m3

sludge contains 7% of solids only and density of the solids = 1028 kg/m3

a). the mass flux of sludge produced in the underflow from this basin:

mass of the solids (sludge solids)= mass of the coagulant + mass of the suspended solids that changed into sludge.

mass of the sludge solids = 8400*0.005 + 8400*0.16095= 1393.98 kg TSS & Coagulant/d

but this solids mass is only 7% of the total sludge.therefore, total sludge = mass of the sludge solids/0.07

total sludge produced= 1393.98/0.07 = 19914 kg TSS & Coagulant/d

b). the volumetric flow rate of sludge produced in the underflow from this basin:

total sludge produced = 19914 kg TSS & Coagulant/d

density of the solids = 1028 kg/m3

volumetric flow = mass per day/density = 19914/1028 = 19.372 m3/d

c). the concentration of suspended solids in the overflow effluent:

from the mass balance equation,

Qin*Cin = Qeff*Ceff + sludge solids

8400*0.185 = 8400*Ceff + 1393.98

or Ceff = 0.01905 kg TSS/m3 = 19.05 mg TSS/L

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