Shown below are initial concentrations (mg/liter) at 4 nodes of a 1 dimensional

Question

Shown below are initial concentrations (mg/liter) at 4 nodes of a 1 dimensional transport system. We wish to use flux boundary conditions to solve this system. In this problem, you need to find the values of the imaginary nodes to finish the line of initial conditions (time, t=0). At the right side, there is no transport across the boundaries. On the left side, the transport into the system is constant at 0.8 mg/s. Fick's law is Q = A*a*dC/dx, diffusivity is 0.2 cm^2/s, dx = 2 cm and A - 1 cm^2.

Explanation / Answer

Flux equation is given by

Q= A.a.dc/dx

Now on the left side

Q= 0.8 mg/s

dc= Cleftimag -4*10-3 mg/cm3 .....................(since direction of transport from left imaginary to system)

dx = 2 cm

A= 1 cm2

Putting these values in flux equation we get

Q= A.a.dc/dx = [1 cm2 * 0.2 cm2/s * (Cleftimag - 4*10-3)mg/cm3] / 2 cm=0.8 mg/s

Solve for Cleftimag

(Cleftimag -4*10-3)mg/cm3 = (0.8 * 2 )/(1 * 0.2) = 8

Cleftimag -4*10-3 = 8

Cleftimag = 8 + 4*10-3 = 8.004 mg/ml = 8004 mg/L ..........(1 cm3=1 ml= 10-3 L )

Hence at left imaginary node, at t=0, cocentration is 8004 mg/L

Now at right side, there no transport across the boundry.

Hence Q= 0 mg/s

Then the flux equation become

Q= [1*0.2*(11*10-3 - Crightimag)]/2 = 0

Simplification will give

1*0.2*(11*10-3 - Crightimag) = 0

11*10-3 - Crightimag= 0

Crightimag = 11*10-3 mg/ml = 11 mg/L

Hence at right imaginary node, at t=0, cocentration is 11 mg/L

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.