(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Wrote a code using Fortran 95 to solve for an advection-dispersion equation but at the spatial steps specified at dx = 20 m over a total length, L of 20000 m, I keep getting an arithmetic overflow error.

I have run this same program at smaller spatial intervals (dx = 200 m and dx = 2000 m) and they worked perfectly. There must be something about the total number of iterations that Fortran is trying to perform that makes it incapable of computing it (I'm not actually sure).

I'm a novice at using Fortran but curious if this has anything to do with the size of the concentration values that are being computed being too small (i.e. smaller than 10^-38). Does this have anything to do with double precision?

2. Relevant equations

The scheme used to solve the numerical solution to the ADE is an finite differenece upwind scheme.

The first equation in the code under the two "do" statements defines this upwind scheme within the internal bounds of the system.

The second equation in the code outside of the two "do" statements defines the boundary condition at X = L.

3. The attempt at a solution

module global ! Module is used to define arrays and use in main program blocks

implicit none

real len, dx, k, Dl

real, allocatable, dimension (:, :) :: conc ! I used allocatable arrays so that the size of

! the array can be changed, not necessary for this

! simple of a problem

real, allocatable, dimension (:) :: tim

integer ndx, i, n

end module global

program main

use global

real dt , tottime, u

This is where you write out what you're going to be doing. You can just link to where a

! source file is located, but make sure you do it.

dt = 30. ! unit is seconds

len = 20020. ! unit is meters

tottime = 44000. ! unit is seconds (43200 seconds in half a day)

ndx = 1001 ! number of grid points - we want 10 cells so we start at 1.

u = 0.25 ! unit is m/s

Dl = 50. ! unit is m^2/sec

k = 0.1 / 86400. ! unit is sec^-1

nts = (tottime) / dt ! number of time steps

dx = len / real(ndx) ! grid spacing

! Allocate the arrays, conc(entration), and tim(e).

allocate (conc ( ndx , nts ), tim ( nts ))

conc = 0.0 ! set initial concentration elements to 0 mg/L

tim = 0.0 ! initial time is zero

conc (1, :) = 100. ! set BC on LHS to 100 mg/l for all time

! Run the internal finite difference equation and populate the array conc ()

do n = 1, nts - 1

tim (n) = dt * (n - 1)

do i = 2, ndx - 1

conc (i, n + 1) = conc (i, n) - u * (dt / dx) * (conc (i, n) - conc (i - 1, n)) &

+ Dl * (dt / (dx**2)) * (conc (i + 1, n) - 2 * conc (i, n) + conc (i - 1, n)) - k * dt * conc (i, n)

end do

conc (ndx, n + 1) = conc (ndx, n) - u * (dt / dx) * (conc (ndx, n) - conc (ndx - 1, n)) &

+ Dl * (dt / dx**2) * (2 * conc (ndx - 1, n) - 2 * conc (ndx, n)) - k * dt * conc (ndx, n)

end do

! Set BC on RHS where c (ndx + 1, nts) = c (ndx - 1, nts)

! Write the array to a file

open (10, file = 'C:\g95\fortran\CEE 573\Problem Set 5\upwind20.csv', status = 'unknown')

! Write the column headings

write (10, 100) (tim (n) / 86400., n = 1, nts, 10)

100 format ('X, ',2x, 10000 ('t=', f8.4, ','))

! Write the columns

do i = 1, ndx

x = dx * (i - 1)

write (10, 110) x, (conc (i, n), n = 1, nts, 10)

110 format (f10.1, ',', 10000 (f8.4, ','))

end do

end program main

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# Arithmetic overflow in Fortran 95

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