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3241004551 [841]
3 years ago
8

You are provided the following information about a municipal wastewater treatment plant. This plant uses the traditional activat

ed-sludge process.
Assume the microorganisms are 55 percent efficient at converting food to biomass, the organisms have a first-order death rate constant of 0.05/day, and the microbes reach half of their maximum growth rate when the BOD5 concentration is 10 mg/L. There are 150,000 people in the community (their wastewater production is 225 L/day-capita, 0.1 kg BOD5/capita-day). The effluent standard is BOD5 = 20 mg/L and TSS = 20 mg/L.

Suspended solids were measured as 4,300 mg/L in a wastewater sample obtained from the biological reactor, 15,000 mg/L in the secondary sludge, 200 mg/L in the plant influent, and 100 mg/L in the primary clarifier effluent. SRT is equal to 4 days.

(a) What is the design volume of the aeration basin (m3)?

(b) What is the plant

Engineering
1 answer:
3241004551 [841]3 years ago
5 0

Answer:

Explanation:

Attached is the solution

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Answer:

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Explanation:

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For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio
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This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

x^2 / Dt = constant

where D is constant

then

x^2 / t = constant

x^2_1 / t₁ = x^2_2 / t₂

x^2_1 t₂ = t₁x^2_2

t₂ = t₁x^2_2 / x^2_1

t₂ = (x^2_2 / x^2_1)t₁

t₂ = ( x_2 / x_1 )^2 × t₁

so we substitute

t₂ = ( 0.0049  / 0.0018  )^2 × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

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