Question

A food processing plant discharges 40 cfs (cubic feet per second) of process water containing an ultimate BOD (L0) of 25 mg/L and a DO concentration of 1.8 mg/L into a river that has a flow rate of 260 cfs and a velocity of 1.1 ft/s. Just upstream of the release point, the river has an ultimate BOD of 3.6 mg/L and a DO of 7.6 mg/L. The saturation value of DO in the river is 8.5 mg/L. The deoxygenation coefficient in the river (kd) is 0.61/day and the reaeration coefficient for the river (kr) is 0.76/day. Assume complete and instantaneous mixing of the process water and the river water.

Required:
a. What are the initial oxygen deficit and ultimate BOD concentration just downstream of the outfall?
b. What are the time and distance downstream to the minimum DO concentration?
c. What DO concentration could be expected 10 miles downstream?

Answer 1

Answer:

The right solution according to the question is provided below.

Explanation:

According to the question,

(a)

The initial conditions will be:

DO = $\frac{(40\times 1.8)+(260\times 7.6)}{40+260}$

      = $\frac{2048}{300}$

      = $6.826 \ mg/L$

The initial oxygen defict will be:

Do = $8.5-6.826$

     = $1.674 \ mg/L$

The initial BOD will be:

Lo = $\frac{(40\times 25)+(260\times 3.6)}{40+260}$

    = $\frac{1936}{300}$

    = $6.453 \ mg/L$

(b)

The time reach minimum DO:

tc = $\frac{1}{(kr-kd)} ln{(\frac{0.76}{0.61} )[1-\frac{1.674(0.76-0.61)}{0.61\times 6.453} ]}$

   = $\frac{1}{0.15}\times ln \ 1.158$

By putting the values of log, we get

   = $0.973 \ days$

The distance to reach minimum DO will be:

Xc = $\frac{1.1\times 3600\times 24}{5280}\times 0.973 \ days$

    = $18\times 0.973$

    = $17.5 \ miles$