i) A river flowing at 1.52 m3/s with an ambient ultimate CBOD = 3.5 mg/L and NBODu = 0.45 CBODu (DO = 6.5 mg/L) receives an input of 140 L/s with an ultimate CBOD of 150 mg/L and DO = 7.0 mg/L (Point A). The river flows at an average depth of 1.524 m (5 ft) and a velocity of 0.1524 m/s. Assume Kr = Kd = 0.2/day (base 10). Suppose right at the point where the DO deficit is worst (at Point B somewhere downstream of A), another 1.42 m3/s flow is added with CBODu = 0 mg/L and saturated DO. If the velocity of the stream does not change because of this flow, estimate the concentration of the stream DO at a point 16 km downstream of the 1.42 m3/s loading. The average temperature downstream of the first outfall is 23oC. (Use DOsat = 7.6 mg/L, assume NBOD = 0.45 BOD)