CE 260 2013 - Lecture 21- Wastewater Treatment Problems
Transcript of 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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Secondary treatment: Pond performance
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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Secondary treatment: Pond performance
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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