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AURORKA [14]
3 years ago
10

What is acid mine drainage

Engineering
1 answer:
avanturin [10]3 years ago
7 0
Acid mine drainage is the formation and movement of highly acidic water rich in heavy metals. This acidic water forms through the chemical reaction of surface water (rainwater, snowmelt, pond water) and shallow subsurface water with rocks that contain sulfur-bearing minerals, resulting in sulfuric acid.
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A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m
kondaur [170]

Answer:

a) 6.4  mg/l

b) 5.6 mg/l

Explanation:

Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3  mg/l

a) Ultimate BOD of mixture= \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}

                                         = \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l

b) utlimate BOD at 10,000 m downstream

t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times  \frac{day}{24 hr}

putting Q_r + Q_w = 1+ 10 = 11 m^3/s

t = 0.578  days

we know

L_t = L_o e^{-kt}

L_t = 6.4 \times e^{-0.22 \times 0.578}

L_t = 5.6 mg/l

7 0
2 years ago
Choose the correct word or phrase to complete the sentence to explain human intervention in a machine system.
maksim [4K]

Answer:

Fully Automated

Periodic Maintenance Activities

6 0
3 years ago
Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu
aksik [14]

Answer:

Para x=0:

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-2.4 W/m^{2}  

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

\phi=-k\frac{dT}{dx}(1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

\frac{dT(x)}{dx}=300x-30

Remplezando esta derivada en (1):

\phi=-0.04(300x-30)

Para x=0:

\phi=0.04(30)

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-0.04(300*0.3-30)

\phi=-2.4 W/m^{2}    

Espero que te haya ayudado!

4 0
3 years ago
A large building will need several different types of workmen to install and repair pipes for water, heating,
siniylev [52]

Answer:

Plumber and pipefitters

Explanation:

3 0
2 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
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