Problems 2 and 3 refer to the data below. Maintenance of an adequate dissolved o

Question

Problems 2 and 3 refer to the data below. Maintenance of an adequate dissolved oxygen concentration (DO) in natural waters is im Restriction important to sustain aquatic life. er hand, may be desirable to limit the growth of aquatic plants he lake illustrated below, with the characteristics and parameters listed in the accompanying table. the cases examined below, the area across which fluxes occur may be calculated based on the volume and of the phosphorus concentration, on the oth nsider t average depth of the lake, and that the lake may be taken as a steady state system. Lake Characteristics V 2 x 10m d 10 m Volume Average depth Incoming River Q0.6 mls Flow rate Dissolved oxygen Phosphorus concentrationn DOO = 9.0 mg/L Po DO Do Transport Parameters DO removal coefficient ko 0.025 mg/Ld kg 4 x 10 cm/s Oxygen interfacial mass Conversion Factors transfer coefficient DO saturation concentration DOsot 9.5 mg/L Phosphorus settling rate 18:10' mg-10° ?8 1 m 102 cm 1 m3 103L 106 cm vs 12 m/y (35%) Phosphorus in the lake does not undergo any net chemical transformation, but, because it is in the form of fine particulates, undergoes settling within the lake. a. Set up a mass balance on phosphorus in the lake. Write out all the terms and coefficients b. Solve the phosphorus mass 2, balance to determine the phosphorus concentration in the river exiting the lake.

Explanation / Answer

Lake volume = 2*10^7 m3

Lake depth = 10 m

=> Lake area = 2*10^6 m2

For Phosphorus,

Mass balance

=> Qin*Po-Qout*P-Area*Phosphorus settling velocity*density=0

Phosphorus density=1800 kg/m3

Phosphorus In flowrate = .6*18*10^-6/(10^-3)=.0108 g/s

Phosphorus settling rate =12/(365*24*60*60)*2*10^6*1800*1000=1369 Kg/s

Because, Settling rate is very high than incoming rate of Phosphorus rate

=> Whole Phosphorus coming in will settle down and nothing go out

=> P=0

b) DO transfer rate into atomosphere = Kgf*Area=4*10^-5*2*10^6=80 cm3/s

DO consumption rate in atmosphere=Ko*River volume = .025*2*10^7*10^3/(24*60*60)=5787 mg/s

DO inlet flowrate = DOo*flowrate=9*.6*10^3=5400 mg/s

=> DO inlet<DO consumption rate

=>Whole DO consumes in lake

=> DO outlet =0

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