The ultimate BOD of a river at the confluence of the river and a sewage outfall is 50 mg/L and the DO is at the saturation value of 9.0 mg/L after mixing. The deoxygenation (or decay) rate coefficient (Kd) is 0.3 d-1 and the reaeration rate coefficient (Kr) is 0.9 d-1. The river is flowing at a velocity of 48 miles/day and the only source of BOD is from the sewage discharge.

Required:
a. Find the critical distance downstream (in miles) at which DO is a minimum.
b. Find the minimum DO (in mg/L)
c. If a wastewater treatment plant is to be built, what fraction of the BOD would have to be removed from the sewage to assure a minimum of 5.0 mg/L everywhere downstream?

Answer:

the estimated time of submersion is 17.7 years

Explanation:

Given the data in the question;

estimate the time of submersion in years.

we write down the relation between time of submersion and corrosion penetration as follows;

CPR(mpy) = K × W(mg) / [ A(in²) × p(g/cm³) × t(hr) ]

we solve for t

t = (K × W) / ( AP × CPR )

given that;  

Area A = 5 in²

W = 2.3 kg = 2.3 × 10⁶ mg

density of steel p = 7.9 g/cm³

CPR = 200

we know that K is 534

so we substitute

t = (534 × 2.3 × 10⁶ mg) / ( 5 in² × 7.9 g/cm³ × 200 mpy )

t = 1,228,200,000 / 7900

t = 155468.3544 hr

t = 155468.3544 hr × ( 1 yrs / ( 365 × 24 hrs )

t = 17.7 years

Therefore, the  estimated time of submersion is 17.7 years

Answer:

I am thick but I dont know the anwser

The time that it will take for the outer surface of the wall to reach a temperature of 750 degree C will be 33800 seconds.

Using the approximation methods, Fo will be;

= In(0.215/1.262)/(1.4289)²

= 0.867

Then, the time taken will be:

= 0.867(0.15)²/(1.5/2600 × 1000)

= 33800 seconds.

In conclusion, the time taken is 33800 seconds.

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Answer:

b. 336,000.

Explanation:

Step 1: Calculate the molecular weight of the monomer

Polyethilene is a polymer with the formula (C₂H₄)ₙ, where C₂H₄ is the monomer and n is the number of monomers in the polymer. We can calculate the molecular weight of the monomer by addition of the weights of the atoms that form it.

MC₂H₄ = 2 × MC + 4 × MH

MC₂H₄ = 2 × 12 + 4 × 1 = 28

Step 2: Calculate the average molecular weight of polyethylene

The average degree of polymerization (DP) of polyethylene is 12,000. We can calculate the average molecular weight of polyethylene using the following expression.

DP = M(C₂H₄)ₙ/MC₂H₄

M(C₂H₄)ₙ = DP × MC₂H₄

M(C₂H₄)ₙ = 12,000 × 28 = 336,000

a bc if the bulbs are in a bad conditio. than u know that u dont have to remove it but only repair it.

Answer:

A) 105.7 rpm

B) 11.32 kw

C) 20.85 NPa

Explanation:

Number of solid rods to be manufactured = 3600

a) Determine the RPM of the screw

we will apply the relation below

discharge rate ( Qd ) = 0.5 π^2 * D^2 * N di * sinA * cos A  ------- ( 1 )

where : D = 50 mm , di = 5 mm , N = ?

Tan A = p / πD = 50 / π*50 ∴ A = 17.65°

Insert values into equation ( 1 )

Qd = 17.83 * 10^-6 * N

required discharge rate ( Q ) =  [tex]\frac{\frac{\pi D^2}{4}*L*N }{Time}[/tex]   ------ ( 2 )

where : D = 0.01 , L = 25 * 10^-2 , N = 3600 , time = 10 * 3600

input value into 2

Q = 31.415 * 10^-6 m^3/s

Hence the RPM of the screw ( N )

= Q / Qd =  31.415 * 10^-6 / 17.83 * 10^-6  = 1.76 rev/s = 105.7 rpm

b) Determine the power requirements of the extruder

max power requirement  = Pm * A * πDN / 60

                                         = ( 20.85 * π * ( 50 )^2 / 4 ) * π * 150 *1.76

max power requirement = 11.32 kw

c) What is the pressure buildup within the extruder

Pressure buildup within the extruder  = ( 6π*D*N*L* η * cot A ) / di^2  

= ( 6π * 0.05 * 1.76 * 1 * 100 * cot17.65 ) / ( 5 * 10^-3 )^2

therefore ; Pm = 20.85 NPa

Answer:

Solvents can damage certain plastics and rubbers found in the engine

Explanation:

the answer is 2

kids simple just multiply 3 by 2 which gets the answer as 6 then divide 6 by 2 then it comes to again 2 and that is a simple answer...

The output of the given code, i.e., print( (3**2)//2)​ will be 4.

Coding, also known as computer programming, is the method by which we communicate with computers.

Code tells a computer what to do, and writing code is similar to writing a set of instructions. You can tell computers what to do or how to behave much more quickly if you learn to write code.

When you choose an instructional programming language, coding is simple to learn. It can be difficult to learn to code if you begin with a more complex coding language.

The ** operator is used to multiply a number by a power.

The // operator calculates how many times the right number can fit into the left.

Here it is given that

[tex]print(3^2)/2[/tex]

So, the round division will be 4.

Thus, the output will be 4.

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Answer:

Crankshaft position sensor - F     I can't quite make out the letter but it's the thing at the bottom almost touching the notched wheel.

Coil Module - B  

Knock Sensor - D

Coil Pack -E

Fuse Block - A

Powertrain Control Module - C

Answer:

Critical Flaw Length=17.08 mm

The Critical flaw Length > 4mm, It means it is detectable.

Explanation:

Given Data:

Fracture Toughness=[tex]K_{tc}[/tex]=88MPa

Yield Strength=σ=710 MPa

Y=1.07

Solution:

Formula:

[tex]Critical\ Length=\frac{1}{\pi } *(\frac{K_{tc}}{Y*\sigma} )^2[/tex]

Since yield Strength is half, Critical Length will be:

[tex]Critical\ Length=\frac{1}{\pi } *(\frac{K_{tc}}{\frac{\sigma}{2} *Y} )^2\\Critical\ Length=\frac{1}{\pi } *(\frac{88MPa}{\frac{710MPa}{2} *1.07} )^2\\\\Critical\ Length=0.01708\ m[/tex]

Critical Flaw Length=17.08 mm

The Critical flaw Length > 4mm, It means it is detectable.

Answer:

Explanation:

1)  Pump does not extract energy from a flowing fluid instead is converts mechanical energy to fluid energy  ( i.e. initial statement is  FALSE )

2) Flow within a boundary layer cannot be considered an inviscid flow because flow within a boundary layer is considered a viscid flow  (  i.e. initial statement is FALSE )

3) Laminar boundary layer velocity profile does not have more mixing than the velocity profile in a turbulent boundary layer because in a turbulent boundary layer the mixing of velocity profile is higher/more  ( i.e. initial statement is FALSE )

4 ) Flow within the log-law region is considered Turbulent and not Laminar( i.e. initial statement is FALSE )

5) In a physical modeling of air flow over an aircraft the Reynold number is more important than the Froude number (  i.e. initial statement is FALSE )

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:    

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ =  0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

p = 8 bar = 800 kPa

Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ +  17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

Answer:

In the 1940s and 50s, a handful of scientists from a variety of fields (mathematics, psychology, engineering, economics and political science) began to discuss the possibility of creating an artificial brain. The field of artificial intelligence research was founded as an academic discipline in 1956

Explanation:

Answer:

fracture will occur since ( 31.8 Mpa√m ) is greater than the [tex]K_{Ic[/tex] of the material ( 30 Mpa√m )

Explanation:

Given the data in the question;

To determine whether the aircraft component will fracture, given a fracture toughness of 30 Mpa√m, stress level of 355 and maximum internal crack length of 1.39 mm.

On a similar component, it has been said that fracture results at a stress of 237 MPa when the maximum (or critical) internal crack length is 2.78 mm.

so we first of all solve for the parameter Y in the condition where fracture occurred.

[tex]K_{Ic[/tex] = 30 Mpa√m

σ = 237 MPa

2α = 2.78 mm = 2.78 × 10⁻³ m  

so

Y = [tex]K_{Ic[/tex] / σ√πα

we substitute

Y = (30 Mpa√m) / (237 MPa)√(π(2.78 × 10⁻³ m / 2 ) )

Y =  (30 Mpa) / (237)( 0.06608187 )

Y = 30 / 15.6614

Y = 1.9155

Next we solve for Yσ√πα for the second case;

σ = 355 Mpa, 2α = 1.39 mm = 1.39 × 10⁻³ m

so

Yσ√πα = 1.9155 × 355 Mpa × √( π × (1.39 × 10⁻³ m / 2) )

= 1.9155 × 355 × 0.0467269

= 31.8 Mpa√m

so

( 31.8 Mpa√m ) > [tex]K_{Ic[/tex] ( 30 Mpa√m )

Therefore, fracture will occur since ( 31.8 Mpa√m ) is greater than the [tex]K_{Ic[/tex] of the material ( 30 Mpa√m )

Answer:

yeah

Explanation:

The rate at which steam is produced is equal to 701 kg/hour.

A Boiler may be characterized as a type of device or instrument that significantly transforms water into steam. There are two types of boiler are found. They are water tube boilers and fire tube boilers.

According to the question,

The power generated by combustion, W = 813kW.

The efficiency of the boiler, η = 65% = 0.65.

Temperature, To = 650°C.

Water enters the boiler tubes as a liquid, T1 = 20°C.

Water leaves the tubes as saturated steam, P2 = 20 bar.

The enthalpy of water at 20°C, [tex]h_1[/tex] = 83.9kJ/kg.

The enthalpy of water at 20 bar pressure, [tex]h_2[/tex] = 2797.29kJ/kg.

Enthalpy change can be calculated by ΔH = [tex]h_2-h_1[/tex]

= 2797.29kJ/kg - 83.9kJ/kg = 2713.3 kJ/kg.

The total energy that can be developed can be calculated by the formula:

The mass of the flow rate of the rate at which steam is produced is calculated by the following formula:

             = 528.45 kW/2713.3 kJ/kg.

             = [tex]\frac{528.45kW }{2713.3kJ/kg} *\frac{3600kJ/h}{1kW}[/tex] = 701 kg/hour.

Therefore, the rate at which steam is produced is equal to 701 kg/hour.

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Answer:

Hello your question is incomplete attached below is the complete question

answer :

a) 95.80°C

b) 8.23 MW

Explanation:

Convection heat transfer coefficient = 860 W/m^2 . k

a) Calculate for the temp of sheet metal when it leaves the oil bath

first step : find the Biot number

Bi = hLc / K  ------- ( 1 )

where : h = 860 W/m^2 , Lc = 0.0025 m ,  K = 60.5 W/m°C

Input values into equation 1 above

Bi = 0.036 which is < 1  ( hence lumped parameter analysis can be applied )

next : find the time constant

t ( time constant ) = h / P*Cp *Lc  --------- ( 2 )

where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C

Input values into equation 2 above

t ( time constant ) = 0.10092 s^-1

Determine the elapsed time

T = L / V = 9/20 = 0.45 min

∴   temp of sheet metal when it leaves the oil bath

= (T(t) - 45 ) / (820 - 45)  = e^-(0.10092 * 27 )

T∞ =  45°C

Ti = 820°C

hence : T(t) = 95.80°C

b) Calculate the required rate of heat removal form the oil

Q = mCp ( Ti - T(t) ) ------------ ( 3 )

m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s

Cp = 434 J/kg°C

Ti =  820°C

T(t) = 95.80°C

Input values into equation 3 above

Q = 8.23 MW

Answer:

Explanation:

When used as nouns,  schematic diagram means a plan, drawing, sketch or outline to show how something works, or show the relationships between the parts of a whole, whereas schematic means a simplified line-drawing generally used by engineers and technicians to describe and understand how a system works at an abstract level.

Hope this helped!!!