Comparing the largest and the smallest triangle in Fig. 13.14 and using cout = 0.01 • cin, it follows that:
17.8 + a a a and it can be calculated that: a = 0.18 kg h-1 COD.
From the upper smaller and the largest triangles: c -0 1 c ri v« J. v,-. ri
cin = 0.300 kg m 3 COD is obtained as the allowed influent concentration for the flow rate of:
Only 59.33 « 60 m3 h-1, respectively ~ 66.6% must be treated.
An obvious solution would be to treat the effluent from process 1. But the flow rate is too low and the concentration is too high. A better strategy is to use a part of the effluent from process 2 for mixing with the effluent of process 1 and a second part for mixing with the effluent from process 3 (Fig. 13.15).
Starting with cin = 0.3 kg m-3 DOC and a = 0.99, cout is obtained from:
a = _in—out, cout = cin (1 - a), cout = 0.003 kg m-3
To obtain Q12, the flow rate from process 2 is mixed with that of process 1; and a further mass balance at mixing point M1 is:
Problem 13.2 353
QM1,in cin Ql C1
With two further mass balances, we are able to test whether the limit of ce —
0.020 kg m-3 DOC is upheld:
1. The concentration after mixing effluent from processes 2 and 3 at mixing point M2 is:
20 • 0.03 + 10 • 0.1 — 30 • cin2 20-0.03 + 10-0.1
2. The concentration after mixing at mixing point M3 is: 30 • 0.053 + 60 • 0.003 — 90 • ce 30-0.053 + 60-0.003
The proposed network for the treatment of three wastewater streams is suitable to undercut the required limit of:
Thus, 30 m3 of the wastewater must not be treated.
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