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Sergeeva-Olga [200]
3 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]3 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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What’s the number of gold atoms in a nanogram? a picogram?
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The number of gold atoms in picogram is, 3.057\times 10^{9}

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As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

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As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

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The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

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