Example 1: Q7-15, page 646

Assume the NH4-N is oxidized to NO2-N and the cell yield is 0.12 g VSS/g NH4-N oxidized.

Other kinetic coefficients related to substrate utilization and growth are:

Maximum specific bacterial growth rate=0.50 g VSS(new cells)/gVSS(cells)*d

Half-velocity constant Ks=0.50 mg/l

Endogenous decay coefficient kd=0.08 gVSS/gVSS*d

What is the substrate and biomass concentration at 0.80 days? Plot the substrate and biomass concentration versus time up to 1.0 day.

Solution:

  1. The substrate utilization rate (g/m3*d)

The net biomass production rate (g VSS/m3*d)

  1. S and X change with time
  2. and also change with time. But we can ignore their changes at very small time increments.
  1. Set =0.2 day, then t0, t1, t2, t3 and t4 are respectively 0, 0.2, 0.4, 0.6, 0.8 days

First, calculate the change of substrate and biomass from t0 to t1

Fro substrate:

So,

For biomass

So,

  1. Similarly, we can get following results:

Time / / / /
t1=0.2 / -41.15 / 4.84 / 31.77 / 10.97
t2=0.4 / -44.99 / 5.29 / 22.777 / 12.03
t3=0.6 / -49.03 / 5.76 / 12.97 / 13.18
t4=0.8 / -52.87 / 6.21 / 2.397 / 14.42

6. To plot the substrate and biomass concentrations versus time

Example 2: Q7-32, page 651

An anoxic suspended growth reactor is operated at a SRT of 5.0 d and acetate is added as the electron donor. Given the following kinetic coefficients for acetate under nitrate reduction conditions, determine how much acetate is needed, in kg/d, to remove 20 mg/l of nitrate in a treatment flow rate of 4000m3/d.

Y=0.3 g VSS/g COD removed

kd=0.08 gVSS/gVSS*d

Solution:

  1. bsCODr=bsCODsyn+bsCOD0 where bsCODr =bsCOD removed

bsCODsyn=bsCOD incorporated into cell synthesis

bsCOD0=bsCOD oxidized, gbsCOD/d

  1. bsCOD0 is the COD oxidized and is equal to the oxygen equivalent of the NO3-N used for

bsCOD oxidation, hence

bsCOD0=2.86 NOx

(Where 2.86 =O2 euqvalent of NO3-N, gO2/gNO3-N)

bsCODsyn can be calculated from the net biomass yield and the ratio of 1.42 gO2/gVSS

bsCODsyn= 1.42 Yn bsCODr

Hence,

where, Yn is the net biomass yield, gVSS/gCODr

Since

  1. To determine how much acetate is needed, need to know the COD of acetate

60 32

The Acetate needed=bsCODr/COD of Acetate=(18.42gCOD/m3)/(1.06gO2/gAcetate)

=19.53 gAcetate/m3

Since Q=4000m3/d,

Total Acetate needed= 19.53gAcetate/m3*4000m3/d=78120 g Acetate/d= 78.12kg/day

Example 3

A wastewater treatment process is designed to achieve nitrification and
denitrification using the organic material in the wastewater as the electron
donor in the latter process. The wastewater contains 30 mg/L of NH4+-N.
Determine the decrease in COD concentration that will occur in the
denitrification process if 100% of the NH4+ is removed and 100% of the NO3-
is reduced. Assume typical values for parameters where necessary.

Assume

SRT of 5.0 d

Y=0.3 g VSS/g COD removed

kd=0.08 gVSS/gVSS*d

Solution:

1. The reaction of nitrification and denitrification

NH4+ + 2O2 → NO3 - + 2H+ + H2O (1)

  1. To determine the decrease in COD concentration that will occur in the
    denitrification process