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Paraphin [41]
3 years ago
8

Sludge wasting rate (Qw) from the solids residence time (Thetac = mcrt) calculation. Given the following information from the pr

evious problem. The total design flow is 15,000 m3/day. Theoretical hydraulic detention time (Theta) = 8 hours. The NPDES limit is 25 mg/L BOD/30 mg/L TSS.
Assume that the waste strength is 170 mg/L BOD after primary clarification.

XA=MLSS = 2200 mg/L,
Xw = Xu = XR = 6,600 mg/L,
qc = 8 days.

Make sure you account for the solids in the discharge.

What volume of sludge (Qw=m3/day) is wasted each day from the secondary clarifiers?

Engineering
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day

Explanation:

Check the file attached for a complete solution.

The volume of the aeration tank was first calculated, V = 5000 m^3 / day.

The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.

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Area = \frac{\pi d^{2} }{4}

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<h3>Solve for inertia</h3>

\frac{\pi *0.75^4}{4}

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Solve for the tensile force from here

\frac{F}{A} +\frac{Mc}{I}

30x10³ = \frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

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