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kaheart [24]
2 years ago
5

Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti

c wastewater from an influent BOD concentration of 250 mg/L to an effluent concentration of 10 mg/L. X (MLVSS) = 3000 mg/L, and the kinetics are first order and not variable order. The first order equation you must use to calculate the specific substrate utilization rate is q = K S where S is the effluent BOD concentration and K is the first order BOD degradation rate constant. The value of K is 0.04 L/(day*mg). What is the required reactor volume in MG (millions of gallons)? All the choices below are in units of MG.
0.4
1.0
0.2
4.8
Engineering
1 answer:
12345 [234]2 years ago
4 0

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

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Answer:

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             r is the growth rate

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                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

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So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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