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Homework Help: LaPlace equation in 1D

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the Laplace equation in one dimension (x, i.e. (∂^2h)/(∂x^2)= 0)
    Boundary conditions are as follows:
    h= 1m @ x=0m
    h= 13m @ x=10m
    For 0≤x≤5 K1= 6ms^-1
    For 5≤x≤10 K2 = 3ms^-1

    What is the head at x = 3, x = 5, and x = 8?

    What is the Darcy velocity (specific discharge)?


    NOTE: There are multiple steps that will need to be done. Realize that system is heterogeneous. In a multiple layer system with steady-state conditions, Darcy velocity in one layer must equal the Darcy velocity in the other layers

    2. Relevant equations

    h(x) = ho - [(h0 - hD )/D]*x

    3. The attempt at a solution

    I tried to use the equation above subbing in 3, 5, and 8 for the x and using 10m as D, 1m as h0, and 13m for hD

    Then I used the specific specific discharge for the Darcy's velocity (q=K(dh/dL))

    That was apparently all wrong. Apparently this needs to be broken into 2 systems, coupled. Each individual system can be treated as homogenous. So it need two separate LaPlace equations? I really don't know what to do with this problem, please help!
     
  2. jcsd
  3. Feb 24, 2012 #2
    Since the discharge goes through both portions of the domain, you have continuity of the dependent variable h at the interface. In addition you also have continuity of discharge at the interface.

    When you integrate the Laplace equation in 1-D, you have two constants of integration for each section of the domain. The matchings of discharge and h serve to eliminate some of the integration constants.
     
  4. Feb 24, 2012 #3
    So, what I've done is set q(in) = q(out) => -K1(dh/dl)=K2(dh/dl) => 6((h-1)/5) = 3((13-h)/5) => h(@x=5) = 5m

    Then I set the used set of boundary Conditions (BC) to interpret the solution to LaPlace in 1D (h=cx+D):
    For BC @ x=0 and h=1 =>1=c(0)+D => 1=D
    For BC @ x=5 and h=5 => 5=c(5)+1 => c=4/5
    the I used h=cx+D again to solve for the head at x=3
    =>h(@x=3)=(4/5)(3)+1=3.4

    For the second system I used the point slope eq (y1-y)=m(x1-x) to get c
    => (5-13)/(5-10) = m = 8/5 = c
    Then I used h=cx+D to solve for D
    => 5=(8/5)(5) + D => D=-3
    Then to solve for h at x=8
    h=cx+D => h(@x=8) => (8/5)(8)-3 = 9.8


    To get Darcy's velocity I just used q=K(dh/dl) for each system
    System 1 = 6(4/5) = 4.8 = q
    System 2 = 4(8/5) = 4.8 = q

    Yes?
     
  5. Feb 24, 2012 #4
    I get the same answers as you but went about the problem a little differently.

    I solve the ODE over two separate domains:

    h1(x)=c1+c2*x
    h2(x)=c3+c4*x

    I used the Dirichlet boundary conditions along with continuity of head and continuity of Darcy velocity to solve for the 4 constants of integration. Equations worked out to be:

    h(x) = 1 + .8x 0<x<5

    h(x) = -3 + 1.6x 5<x<10

    The < signs should be weak inequalities. I cannot make them on my computer.
     
    Last edited: Feb 24, 2012
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