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morpeh [17]
2 years ago
11

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of

1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is: dC
Engineering
1 answer:
konstantin123 [22]2 years ago
6 0

This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady  state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

a)

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b)

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1  { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2    ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

 C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 +  1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

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2 years ago
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Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m^
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Answer:

The answer is "\bold{ 259.2 \times 10^{11} }".

Explanation:

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D =1.0 \times 10^{8} \ \ \ \frac{m^2}{s} \\\\ A = 0.20 \ m^2\\\\t = 1\ \ h = 3600 \ \   sec \\\\

calculating the value of \Delta C:

\Delta C =C_A -C_B

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calculating the value of \Delta X:

\Delta x = x_{A} -x_{B}

     = 0 - (5\ mm) \\\\ = - 5 \ \ mm\\\\= - 5 \times 10^{-3} \ m

M = -(1.0 \times 10^{8}  \times 0.20 \times 3600 \times  (\frac{1.8}{-5 \times 10^{-3}})) \\\\

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3 0
3 years ago
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Ideally, the supply frequency will be equal to the ripple frequency.

But in a half - way rectifier, the out put frequency is twice the input sinusoidal voltage frequency.

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4 0
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Answer:

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b. Pressure Head: -1.97,

Velocity Head: 6.53

Explanation:

a.

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rate = 0.2 m3/s

Pressure =275 kPa

elevation =25 m.

We'll consider 3 points as the water flow through the pipe

1. At the entrance

2. Inside the pipe

3. At the exit

At (1), the velocity can be found using continuity equation.

V1 = ∆V/A

Where A = Area = πr² = π(0.075)² = 0.017678571428571m²

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The value of pressure at point 1, is given by Bernoulli equation between point 1 and 2:

P1/yH20 + V1²/2g + z1 = P2/yH20 + V2²/2g + z2

Substitute in the values

P1/yH20 + 20 = (275 * 10³Pa)/yH20 + 25

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P1/yH20 = 33.03

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b.

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 3:

P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3

Substitute in the values

33.03 + 20 = P3/yH20 + 55

P3/yH20 = 33.03 + 20 - 55

=> P1/yH20 = -1.97

The velocity head at point three is then given by

V2²/2g = V3²/2g = 6.53

4 0
3 years ago
Read 2 more answers
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