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photoshop1234 [79]
1 year ago
14

The town of Camp Verde has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of

25.0 mg/L BOD5 and 30 mg/L suspended solids. They have selected a completely mixed activated sludge system for the upgrade. The existing primary treatment plant has a flow rate of 2,506 m3/d. The effluent from the primary tank has a BOD5 of 240 mg/L. Using the following assumptions, estimate the required volume of the aeration tank:
BOD5 of the effluent suspended solids is 70% of the allowable suspended solids concentration.
Growth constant values are estimated to be: Ks = 100 mg/L BOD5; kd = 0.025 d–1; μm = 10 d–1; Y = 0.8 mg VSS/mg BOD5 removed.
The design MLVSS is 3,000 mg/L.
Engineering
1 answer:
Artist 52 [7]1 year ago
4 0

For efficient treatment, the dissolved oxygen concentration in the aeration tank must be kept between 1-3 mg/L. The microbial biomass will die at low DO levels, and it will take time and money to rebuild it.

The depth, which normally ranges from 3 to 4.5 meters, determines how effectively air is aerated. The breadth, which is typically maintained between 5 and 10 meters, regulates the mixing. The ideal width-to-depth ratio is 1.2 to 2.2. The distance shouldn't be shorter than 30 meters or longer than 100 meters. Bacteria used in wastewater treatment and stabilization are given oxygen by aeration. The bacteria require oxygen for biodegradation to take place.

Learn more about efficient here-

brainly.com/question/13154811

#SPJ4

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What phenomenon allows water to reach the top of a building?
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Water pressure allows water to reach the top of a building.

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List 6 different mechanisms in the Rube Goldberg cartoon and predict the purpose of each. Does the mechanism change speed, force
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7 0
3 years ago
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold
PIT_PIT [208]

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle (\eta_{r}), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\% (1)

Where:

T_{C} - Temperature of the cold reservoir, in Kelvin.

T_{H} - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: (T_{C} = 400\,K, T_{H} = 1200\,K)

\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%

\eta_{r} = 66.667\,\%

In exchange, real efficiency for a power cycle (\eta), no unit, is defined by this expression:

\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\% (2)

Where:

Q_{C} - Heat released to cold reservoir, in kilojoules.

Q_{H} - Heat gained from hot reservoir, in kilojoules.

W_{C} - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for \eta < \eta_{r}, reversibily for \eta = \eta_{r} and it is impossible for \eta > \eta_{r}.

Now we proceed to solve for each case:

a) Q_{H} = 900\,kJ, W_{C} = 450\,kJ

\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 50\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

b) Q_{H} = 900\,kJ, Q_{C} = 300\,kJ

\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 66.667\,\%

Since \eta = \eta_{r}, the power cycle operates reversibly.

c) W_{C} = 600\,kJ, Q_{C} = 400\,kJ

\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%

\eta = 60\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

d) Since \eta >  \eta_{r}, the power cycle is impossible.

8 0
2 years ago
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