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Pavel [41]
3 years ago
11

A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl

ow to the pond has a flow rate of Q=4,000 m3 /day and a BOD concentration of Cin = 25 mg/L. The volume of the pond is 20,000 m3 . The purpose of the pond is to allow time for the decay of BOD to occur before discharge into the environment. BOD decays in the pond with a first-order rate constant equal to 0.25/day. What is the BOD concentration at the outflow, Cout of the pond, in units of mg/L?
Engineering
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

<em>Cout = 11mg/L</em>

Explanation:

Assume Steady State and CSTR, this means Qin = Qout

Given: Qin = 4 000m3/day

Cin = 25mg/L

V = 20 000m3

k = 0.25/day

Find: Cout = ?

outStarting from the mass balance equation for steady state 0 = QCin - QCout - kCoutV

manipulate the equation to get:

\(Cout=Cin(Q/Q+kV)\)  

<em>Then simple plug in your givens: </em>

<em>Cout = (25mg/L)((4000m3/day)/(4000m3/day + 0.25/day(20000m3)) </em>

<em>Cout = 11mg/L</em>

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A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
Cat is stuck in 10 foot hole how do i get it unstuck
Semenov [28]

Answer:

you fill the hole so there is no more cat

7 0
2 years ago
Read 2 more answers
Use the graph to determine which statement is true about the end behavior of f(x).
Airida [17]

Answer:

As the x-values go to negative infinity, the function’s values go to positive infinity.

Explanation:

if the ans choices are:

As the x-values go to negative infinity, the function’s values go to negative infinity.

As the x-values go to negative infinity, the function’s values go to positive infinity.

As the x-values go to positive infinity, the function’s values go to negative infinity.

As the x-values go to positive infinity, the function’s values go to zero.

the ans is the 2nd choice

4 0
3 years ago
Read 2 more answers
An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

3 0
3 years ago
A building wall has dimensions of 3 m tall and 10 m wide. It is constructed of 2 cm. wallboard (k = 0.5 W/m-C) on the inside, 3
Art [367]

Answer: heat loss through wall is 16.58034kW

Temperature of inside wall surface is 47°c

Temperature of outside wall surface is -2.7°c

Explanation:detailed calculation and explanation is shown in the image below.

4 0
3 years ago
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