Question

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

Answer 1

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