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Sergeeva-Olga [200]
2 years ago
15

(2 points) A perfectly mixed aeration pond with no recycle serves as the biological reactor for a small community. The pond rece

ives 30 m3 /d of influent with a BOD of 350 mg/L that must be reduced to 20 mg/L before discharge. It has been found that the kinetic constants for this system are: Ks = 100 mg BOD/L, kd = 0.10 day-1 , μm = 1.6 day-1 , Y = 0.60 mg SS/mg BOD. Assume Xo = 0. a. What must the hydraulic retention time (θ = V/Q) be in the aeration pond? b. What mass of microbes will be produced in the pond each day?
Engineering
1 answer:
FromTheMoon [43]2 years ago
3 0

Answer:

a)  t = 165 days

b) 9.9 kg/day

Explanation:

Given data:

final lelvel of BOD is 20 mg/l

Ks = 100 mg BOD/L,

kd = 0.10 day-1 ,

μm = 1.6 day-1 ,

Y = 0.60 mg SS/mg BOD

a) we know that

c_{out} = \frac{c_{in}}{1 + kt}

20 = \frac{350}{1 + 0.1 t}

solving for t

t  =  165 days

b) mass of microbes = Q(mlD) × C(mg/l)

= 30\times 10^{-3} (350-20) = 9.9 kg/day

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Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

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                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

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Explanation:

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Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.
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Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

     = ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

           

7 0
2 years ago
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