## Solution

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:

17.8

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

c in

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

with:

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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