A food processing plant deposits 40 cfs (cubic feet per second) of processing wa

Question

A food processing plant deposits 40 cfs (cubic feet per second) of processing water having an ultimate BOD (L0) of 25 mg/L and a DO 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 (kd) is 0.61/day and the reaeration coefficient (kr) is 0.76/day. Assume complete and instantaneous mixing of the processing water and the river.

a) What is the initial oxygen deficit and uBOD just downstream of the outfall?

b) Find the time and distance to reach the minimum DO in the river.

c) Find the DO that could be expected 10 miles downstream.

Explanation / Answer

(a). Initial conditions:

DO = (40 cfs * 1.8 mg/L) + (260 cfs * 7.6 mg/L) / (40 + 260 cfs)

= 6.826 mg/L

Initial oxygen defict, Do = 8.5 mg/L - 6.826 mg/L

= 1.674 mg/L

Initial BOD, Lo = (40 cfs * 25 mg/L) + (260 cfs * 3.6 mg/L) / (40 + 260 cfs)

= 6.453 mg/L

(b). Time to reach minimum DO:

tc = 1 / (kr - kd) ln {(kr / kd) [1 - Do(kr - kd) / kd*Lo]}

= 1 / (0.76 - 0.61) ln {(0.76 / 0.61) [1 - 1.674(0.76 - 0.61) / 0.61*6.453]}

= (1/ 0.15) * ln 1.158

= 0.973 days

Distance to reach minimum DO:

Xc = { (1.1 ft/s) * 3600 s/hr * 24 hr/day / 5280 miles/ft} * 0.973 days

= 17.5 miles

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