CE 260 2013 - Lecture 21- Wastewater Treatment Problems

7/30/2019 CE 260 2013 - Lecture 21- Wastewater Treatment Problems

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WasteWater Treatment Processes: Example

Problems

CE 260 Lecture 19

05/31/2013
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Primary Clarifier

30000 people, 0.5m3 / persone / day. Design a conventional

primary clarifier to have a 2hrs detention time and a 40m/doverflow rate and a high-rate clarifier with a 1500m/d overflow.What are the dimension of the conventional vs high rate?
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Primary Clarifier

A = Q/v = 375m2

diameter=

(4A

pi

) = 22m

depth=QA

= 15000237524

= 3.3m

high rate: A=Qv

= 15000/1500 = 10m2
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Secondary treatment: Pond problem
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Secondary treatment: Pond problem

Mass balance:dX

dt =QX0 QXf

+Vr

Insert the Monod expression for the rate r, with also a die off ofthe bacteria, and consider the steady state:

dX

dt = QX0

QX + V

(

kmaxXS

Ks + S

kdX) = 0

Considering X0 = 0, simplify by Q:

X +V

Q (

kmaxXS

Ks +

S kdX) = 0

Factor X out and consider = VQ

:

X((kmaxS

Ks + S kd)) = 0
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This is true if:

X((kmaxS

Ks + S kd)) = 0

for any X. Solving for S gives:

S =Ks(1 + kd)

kmax 1 kd

Note that S, the amount of BOD (or equivalently of organicmatter) in the system does not depend on the input concentration

but only on microbiological information (and residence time)!
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The same approach leads to an expression for X, the microbeconcentration:

dS

dt= QS0 QS V (

kXS

Ks + S) = 0

Note that k is the substrate utilisation rate here (hence the minussign), not the microbial population growth rate

(kmaxintheslidesabove).Again, eliminating Q and solving for X gives:

X =(Ks + S)(S0 S)

kS

Using design values (table 6E2 in Nazaroff) for Ks = 60,kd = 0.06 and kmax = 3 yields S=6.9mg/L. With k=5 and S thatwe just calculated, we obtain X=43mg/L.
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Secondary treatment: Pond performance

Matlab code to plot S and X as a function of.

1 S=@(Ks,kd,tau,km,Y) ...

Ks.*(1+kd.*tau)./(km*Y.*tau(1+kd.*tau));2

3

X=@(Sin,S,tau,Y,k) ...((SinS)./tau).*((Y.*tau)./(1+k.*tau));

4

5 Ks=60;kd=0.05;km=3;k=5;Y=0.6;Sin=200;

6

7 tau=0.1:0.1:10;

8

9 Sresult=S(Ks,kd,tau,km,Y);

10 Xresult=X(Sin,Sresult,tau,Y,kd);
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Matlab code to plot S and X as a function of(continued).

1

2 figure

3 plot(tau,Sresult,'linewidth',2);

4 hold on

5 plot(tau,Xresult,'r','linewidth',2)

6

7 set(gca,'ylim',[0 100],'xlim',[1 ...

10],'box','off','linewidth',2)

8 set(gcf,'color','w')

9

10 xlabel('Residence Time (days)')

11 ylabel('Concentration (mg/L)')

12 legend({'Substrate';'Bacteria'})
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Note that the bacterial concentration decreases as the residencetime increases in this case because of the die off parameter. Thatswhy for an activated sludge reactor, we recirculate some of themass to maintain an optimal concentration. Similar mass balanceapproach can be used to derive expressions for a complete

activatedd sludge reactor (see Nazaroff pp 349 to 352 for this).
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CE 260 2013 - Lecture 21- Wastewater Treatment Problems

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