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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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Pepsi [2]

Answer:

ok

Explanation:

thx for points

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4 years ago
A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______
lbvjy [14]

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

  • Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
  • The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.
5 0
3 years ago
Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton
SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude q_{1},q_{2} the magnitude of force between them is given by

F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}

where

k_{e} is constant

r is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}

Applying value of the constant we get

F_{1}=\frac{62.208}{0.7^{2}}

Thus F_{2}=126.955Newtons

5 0
3 years ago
Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Di
Tresset [83]

Answer:

0.556 Watts

Explanation:

w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=\frac{2\times (s-ut)}{t^2}\\\Rightarrow a=\frac{2\times (5-0)}{29^2}=\frac{10}{481}

Mass of the body

m=\frac{w}{g}=\frac{762}{9.81}

Force required to move the body

F=ma\\\Righarrow F=\frac{762}{9.81}\times \frac{10}{481}

Velocity of object

v=u+at\\\Rightarrow v=0+\frac{10}{481}\times 29\\\Rightarrow v=\frac{10}{29}

Power

P=Fv\\\Rightarrow P=\frac{762}{9.81}\times \frac{10}{481}\times \frac{10}{29}=0.556\ W

∴ Amount of power required to move the object is 0.556 Watts

7 0
3 years ago
How many ( 1/8") in one inch?
Anna35 [415]

Answer:

1 inch

Explanation:

There for there are 8 1/8 inch that make up 1 inch so the answer is 1 inch

4 0
3 years ago
Read 2 more answers
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