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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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Find the capacitance reactance of a 0.1 micro frequency capacitor 50Hz and at 200Hz​
Paraphin [41]

Complete Question:

Find the capacitive reactance of a 0.1 microfarad capacitor with frequency 50Hz and at 200Hz​.

Answer:

I. Capacitive reactance = 31826.86 Ohms.

II. Capacitive reactance = 7956.72 Ohms.

Explanation:

<u>Given the following data;</u>

Capacitance = 0.1 uF = 0.0000001 Farad

Frequency = 50 Hz and 200 Hz

To find the capacitive reactance;

Mathematically, the capacitive reactance of an electronic circuit is given by the formula;

X_{c} = \frac {1}{2 \pi fc}

Where;

  • Xc is capacitive reactance.
  • f is the frequency.
  • c is the capacitance.

Substituting into the formula, we have;

I. At frequency, f = 50 Hz

X_{c} = \frac {1}{2 * 3.142 * 50 * 0.0000001}

X_{c} = \frac {1}{314.2 * 0.0000001}

X_{c} = \frac {1}{0.00003142}

X_{c} = 31826.86

II. At frequency, f = 200 Hz

X_{c} = \frac {1}{2 * 3.142 * 200 * 0.0000001}

X_{c} = \frac {1}{1256.8 * 0.0000001}

X_{c} = \frac {1}{0.00012568}

X_{c} = 7956.72

5 0
2 years ago
Oliver is designing a new children’s slide to increase the speed at which a child can descend. His first design involved steel b
AVprozaik [17]

Answer:

The correct option is;

A) Steel becomes too hot in the Sun and can burn the children

Explanation:

The properties of steel includes;

Low specific heat capacity, high thermal and electrical toughness, high hardness, high tensile strength, high yield strength, appreciable elongation, high fatigue strength, can easily corrode, high malleability and ability to creep

Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact

8 0
3 years ago
A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a
Xelga [282]

Answer:

Explanation:

Given the data in the question;

L = 46 in

Ga = 5 × 10³ ksi

Gs = 11 × 10³ ksi

Outside diameter da = 5 in

ds = 4 in

Tb = 3 kip.in

Now,

Ja = polar moment of Inertia of Aluminum;

Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 ) in^u

Js = polar moment of inertia of steel

Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )

Ta is torque transmitted by Aluminum  

Ts is torque transmitted by steel  

{composite member }

T = Ta + Ts ------ let this be equation m1

Now, we use the relation;

T/J = G∅/L

JG∅ = TL

∅ = TL/GJ

so, for aluminum rod ∅_{alu = TaLa/GaJa

for steel rod ∅_{steel = TsLs/GsJs

but we know that, ∅a = ∅s = ∅_B

so

[TaLa/GaJa]  =  [TsLs/GsJs]

also, we know that, La = Ls = L

∴ [Ta/GaJa]  =  [Ts/GsJs]

we solve for Ta

TaGsJs = TsGaJa  

Ta = TsGaJa / GsJs

we substitute

Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]

Ta = 0.66Ts

now, we substitute 0.66Ts for Ta and 3 for T in equation 1

T = Ta + Ts

3 = 0.66Ts + Ts

3 = 1.66Ts

Ts = 3 / 1.66

Ts = 1.8072 ≈ 1.81 kip-in

so

∅_{steel = TsLs / GsJs

we substitute

∅_{steel = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅_{steel = 83.26 / 276460.1535

∅_{steel  = 0.000301

∅_{steel = 3.01 × 10⁻⁴ rad

so

∅_{steel = ∅_B = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴  rad

5 0
2 years ago
How could the location of tests affect the performance of a catapult ?
Luba_88 [7]

Answer:

It could affect how far the projectile travels

Explanation:

Facing Uphill: Moves less far

Downhill: Moves further

3 0
3 years ago
Read 2 more answers
A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i
Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature T_{1} = 300 F  = 422.04 K

Duct temperature T_{2} = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant \sigma = 5.67 × 10^{-8} \frac{W}{m^{2} K^{4}  }

Heat transfer  per foot length is given by

Q = ∈ \sigma A ( T_{1}^{4} - T_{2} ^{4} ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × 10^{-8} × 3.14 × 0.0508 × 1 × ( 422.04^{4} - 294.26^{4} )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0
3 years ago
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