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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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What are the four basic parts of process plan
Rus_ich [418]

Answer:

1. Goal setting

2. Developing the planning premises

3. Reviewing Limitations

4. Deciding the planning period

5. Formulation of policies and strategies

6. Preparing operating plans

7. Integration of plans

Explanation:

I don't have an explanation but i thru in a few extra answers for like extra credit or something if your teacher does that

8 0
3 years ago
Wheel diameter 150 mm, and infeed 0.06 mm in a surface grinding operation. Wheel speed 1600 m/min, work speed 0.30 m/s, and cros
gogolik [260]

Answer:

a) the average length per chip is 3 mm

b) the metal removal rate MR is 90 mm³/sec

c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

Explanation:

Given that;

wheel diameter = 150 mm

infeed = 0.06 mm

wheel speed = 1600 m/min = 16000000 mm/s

work speed = 0.30 m/s = 300 mm/s

cross feed = 5 mm

active grits per area = 50 grits/cm²

a)

Average length per chip

Average chip length is given by

Lc = √fd

f is infeed and d is diameter of the wheel

so we substitute

Lc = √( 0.06 × 150

Lc = √ 9

Lc = 3 mm

Therefore the average length per chip is 3 mm

b)

metal removal rate MR is expressed as;

MR = fvc

v is work speed, c is cross feed and f is infeed

so we substitute

MR = 0.06 × 300 × 5

MR = 90 mm³/sec

Therefore the metal removal rate MR is 90 mm³/sec

c)

number of chips formed per unit time is expressed as

No = Vw × c × G

Vw is wheel speed and G is active grits per area

so we substitute

No = 1600000 × 5 × 50/10²

= 4,000,000 chips/min

Therefore number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

4 0
3 years ago
Consider a cubic crystal with the lattice constant a. Complete the parts (a)-(c) below. (a) Sketch the crystallographic planes w
Anna [14]

Answer:

(a) See attachment

(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.

(c) d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} =   0,707

Explanation:

(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:

For [110]

X = \frac{1}{1} = 1; Y = \frac{1}{1} = 1; Z = \frac{1}{0} = \inf.

For [220]

X = \frac{1}{2} = 0,5;Y = \frac{1}{2} = 0,5;Z = \frac{1}{0} = \inf.

The drawn of the planes is shown in the attachments.

(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:

For (1,1):

K_1 = \frac{1}{1} = 1

For (0.5, 0.5):

K_2 = \frac{0.5}{0.5} = 1

As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.

(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:

The Miller indices are already given in the statement. Then, the distance is:

\frac{1}{d^{2}} = \frac{h^{2} + k^{2} + l^{2}}{a^{2}}

d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} =   0,707

7 0
3 years ago
What tools do you need to change a flat tire
Lyrx [107]

Answer:

  • Spare tire (if you’re dealing with a flat and not just rotating tires or accessing the brakes)
  • Carjack
  • Lug wrench
  • Wheel wedges
  • Work gloves
  • Tire repair kit (if you’re attempting to fix your flat tire)
  • Flare/reflective triangles (if you’re changing the tire on the side of the road or in a parking lot)
6 0
4 years ago
Read 2 more answers
Malia is working with aluminum wire. while working with this type of wire, she must remember to a. always use pressure-type term
Y_Kistochka [10]

Answer:

B

Explanation:

Aluminium doesn't rust unless exposed to copper for a duration of time.

4 0
3 years ago
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