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olasank [31]
2 years ago
9

A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and th

e return activated sludge concentration is 8000 mg/L. The biomass concentration is 3500 mg/L. The combined design flow rate of the primary and secondary clarifiers is 2.5 MGD. After primary treatment, the wastewater has an influent BOD concentration of 200 mg/L and an influent suspended solids concentration of 200 mg/L. Two secondary clarifiers, each 28 ft in diameter, are then used. After secondary treatment, the effluent BOD concentration is 15 mg/L, and the effluent suspended solids concentration is 20 mg/L. The volume of sludge produced is 0.5 MGD. What is most nearly the solids residence time
Engineering
1 answer:
Inessa05 [86]2 years ago
3 0

Answer:

11 hours approximately

Explanation:

We are to calculate mean cell residence time mcrt

= Mass of solid in reactor/mass of solid wasted in a day

Q = Qe + We

Q = 2.5

Qw = 0.5

Qe = 2.5 - 0.5

= 2 MGD

10⁶/svi

= 10⁶/125

= 8000

X = 3500

Xe = 20mg/

1MGD = 0.1337million

Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337

= 262500000/[4000000000+40000000} x 0.1337

= 262500000/574800000

= 0.45668 days

= 0.45668 x 24 hours

= 10.9603 hours

Approximately 11 hours

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Which situation is an enabler for the rise of Artificial Intelligence (AI) in recent years?
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The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc
Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

\theta = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑f_{y} = 0, ∑M = 0

If ∑M_{A} = 0

⇒F_{BC}×0.9 - P × 0.6 = 0

⇒F_{BC}×3 - P × 2 = 0

⇒F_{BC} = \frac{2P}{3}

If ∑M_{B} = 0

⇒F_{AD}×0.9 = P × 0.3

⇒F_{AD} ×3 = P

⇒F_{AD} = \frac{P}{3}

Now,

Area, A = \frac{\pi }{4} X (0.012)^{2} = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , \delta = \frac{P l}{A E}

Now,

\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 9.1626 × 10⁻⁹ P

\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 1.83253 × 10⁻⁸ P

Given that,

\theta = 0.015°

⇒\theta = 2.618 × 10⁻⁴ rad

So,

\theta =  \frac{\delta_{BC} - \delta_{AD}}{0.9}

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0
2 years ago
The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards
lesya [120]

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

a_r = \frac{V^2}{R}

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

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2 years ago
The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely w
vodka [1.7K]

Answer:

The density of the unknown liquid is 1.025 kg/m³ (considering the density of the water as 1.000 kg/m³)

Explanation:

The hydrometer works by the Archimedes principle. The cylinder floats in the liquid because the hydrostatic thrust is equal to the weight force. This means:

Tr-W=0N\\Tr=W\\\delta_{fl} \cdot Vol \cdot g =W_{hydr}

If we measure 2 fluids, the weight of the hydrometer is the same, so:

\delta_{fl1} \cdot Vol_1 \cdot g =W_{hydr}=\delta_{fl2} \cdot Vol_2 \cdot g\\\delta_{fl1} \cdot Vol_1 =\delta_{fl2} \cdot Vol_2\\\delta_{fl1} H_1 \pi R^2 =\delta_{fl2} H_2 \pi R^2\\\delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}

If the original watermark height is 12.3cm (H₁) and the mark for water has risen 0.3 cm above the unknown liquid–air interface, the height of the unknown liquid mark is 12cm (H₂). Therefore:

delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}\\1.000\frac{kg}{m^3} \frac{12.3cm}{12cm}=\delta_{fl2}=1.025\frac{kg}{m^3}

8 0
2 years ago
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