Interval 2 00501 kg m3

There are two processes located in interval 2. As follows from Table 13.3, the influent concentration of both processes 2 and 3 should be c2in = c3 in = 0.05 kg m-3. However, the concentration coming from interval 1 is 0 kg m-3 and the flow rate is 70 m3 h-1. In order to get 0.05 kg m-3, the same flow rate of 70 m3 h-1 and an effluent concentration of process 1 and 2 of 0.1 kg m-3 must be recycled. After mixing 70/0 with 70/0.1, 140/0.05 is obtained, which is then divided into 100/0.05 for process 2 and 40/0.05 for process 3. Fundamentally, at each of the mixing points 1, 2, 3 and 4, the balances for water (the equation of continuity) and for the impurity (mass balance) can be formulated and solved to calculate the unknown flow rate or concentration. The relatively simple example presented in Fig. 13.10 makes a simpler solution possible.

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