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vekshin1
2 years ago
13

2. Determine the surface area of a primary settling tank sized to handle a maximum hourly flow of 0.570 m3/s at an overflow rate

of 60.0 m/d. If the effective tank depth is 3.0 m, what is the effective theoretical detention time
Engineering
2 answers:
Hitman42 [59]2 years ago
7 0

Answer:

The surface area of the primary settling tank is 0.0095 m^2.

The effective theoretical detention time is 0.05 s.

Explanation:

The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.

Volumetric flow rate = 0.570 m^3/s

Overflow rate = 60 m/s

Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2

Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate

Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3

Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s

Yuki888 [10]2 years ago
7 0

Given Information:

overflow rate = v = 60 m/d

volumetric flow rate = Q = 0.570 m³/s

depth of tank = h = 3 m

Required Information:

detention time = θ = ?

Answer:

detention time = θ = 3 days or 1.2 hours or 72 minutes or 4.32x10³ seconds

Explanation:

We know that the detention time is given by

θ = V/Q

Where V is the volume of tank and Q is the volumetric flow rate of the tank.

To find out the volume of tank, first we have to find the surface area of the tank.

As we know the volumetric flow rate is given by

Q = Av

A = Q/v

Where A is the surface area of tank and v is the over flow rate and Q is the volumetric flow rate.

First convert the overflow rate from m/d into m/s

1 day has 24 hours, each hour has 60 minutes, each minute has 60 seconds.

v = 60/(24*60*60)

v = 6.94x10⁻⁴ m/s

A = Q/v

A = 0.570/6.94x10⁻⁴

A = 821.32 m²

The volume of tank is given by

V = A*h

V = 821.32*3

V = 2463.97 m³

Finally, the detention time is

θ = V/Q

θ = 2463.97/0.570

Detention time in seconds:

θ = 4.32x10³ seconds

Detention time in minutes:

θ = 4.32x10³/60 = 72 minutes

Detention time in hours:

θ = 72/60 = 1.2 hours

Detention time in days:

θ = 1.2/24 = 0.05 days

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iVinArrow [24]

Answer:

Functions to create a median and mode of a set of numbers

Explanation:

def median(list):

   if len(list) == 0:

       return 0

   list.sort()

   midIndex = len(list) / 2

   if len(list) % 2 == 1:

       return list[midIndex]

   else:

       return (list[midIndex] + list[midIndex - 1]) / 2

def mean(list):

   if len(list) == 0:

       return 0

   list.sort()

   total = 0

   for number in list:

       total += number

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def mode(list):

   numberDictionary = {}

   for digit in list:

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       if number == None:

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       else:

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   maxValue = max(numberDictionary.values())

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def main():

   print "Mean of [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]: ", mean(range(1, 11))

   print "Mode of [1, 1, 1, 1, 4, 4]:", mode([1, 1, 1, 1, 4, 4])

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3 0
2 years ago
Read 2 more answers
An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
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2 years ago
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
makkiz [27]

Answer:

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So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

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-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

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2 years ago
A plane wall of thickness 2L = 60 mm and thermal conductivity k= 5W/m.K experiences uniform volumetric heat generation at a rate
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Answer:

Explanation:

A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq  

˙

q

q

˙

​  

, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT  

∞

​  

=20  

∘

C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx  

2

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C,b=−210  

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C/m,c=−2×10  

4

C/m  

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x

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(−L) and q

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(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q

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6 0
2 years ago
A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli
monitta

The smallest allowable depth is d=16.04 \mathrm{mm} for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force,\mathbf{p}=18 \mathrm{kN}

a=30 \mathrm{mm}

=0.03 \mathrm{m}

Allowable stress,\sigma_{a l l}=135 \mathrm{MPa}

cross sectional area of bar,

A=a \times d

A=a d

e - eccentricity

e=\frac{a}{2}-\frac{d}{2}

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

M=P e

=P\left(\frac{a}{2}-\frac{d}{2}\right)

=\frac{P(a-d)}{2}

Allowable stress

\sigma=\frac{P}{A}+\frac{M c}{I}

c=\frac{d}{2}

Moment of Inertia,

I=\frac{b d^{3}}{12}

=\frac{a d^{3}}{12}

\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}

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\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0

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