isothermal gravity current

7/27/2019 Isothermal Gravity Current

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Isothermal Spreading Gravity Currents

Linearised finite difference method to solve nonlinear diffusion equation

Ahmos Sansom

June 13, 2010

Abstract. The linearisation method is applied to the nonlinear diffusion equation governing the isothermal flow

of viscous gravity currents to derive a finite difference numerical scheme. The numerical approach can be validated

using similarity solutions.

1. Introduction

This document discusses the nonlinear diffusion equation noted below where a numerical algo-

rithm using a finite difference approach is described.

h

t=

1

3

x

h3

h

x

. (1)

The scheme can be validated by similarity solution as discussed in this document.

The diffusion equation can be derived from the Navier-Stokes equations assuming laminar flow,small aspect ratio and dominant viscous forces, see discussion in the Very viscous flow chapter inAcheson [1] and derivation in [5] which discusses gravity currents with temperature dependentviscosity.

The diffusion equation has been applied to lava dome growth by Huppert [3] where the resultingsimilarity solutions was first discussed by Barenblatt [2].

2. Numerical Solution: Linearisation method

The isothermal equation governing the free surface can be linearised as discussed for the generalnonlinear diffusion equation in Richtmyer and Morton[4] and is the inspiration for the numericalsolution discussed here. Rewritting the diffusion equation, such that

h

t

=1

12

2h4

x2

(2)

and discretising the above equation by denoting h((i1)x, nt) by hni using a Crank-Nicolsonimplicit scheme; this gives,

hn+1i hni

t=

(h4)n+1i1 2(h4)n+1i + (h

4)n+1i+1 + (h4)ni1 2(h

4)ni + (h4)ni+1

24x2. (3)

The above results in a nonlinear system of simultaneous equations, however a linear scheme isrequired. The solution is to approximate (h4)n+1i by a linear scheme. This is obtained by time

marching using Taylors expansion of (h4)n+1i , such that

(h4

)n+1

i = (h4

)n

i + th4

tn

i + O(t2

). (4)

c 2010 . .

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2 Ahmos Sansom

Note that

h4

t= 4h3

h

t,

hence to O(t)

(h4)n+1i = (h4)ni + 4(h

3)ni

hn+1i hni

. (5)

Let wi = hn+1i h

ni and substitute (5) in (3) giving

wit

=

(h4)ni1 + 4(h3)ni1wi1 2(h

4)ni 8(h3)ni wi + (h

4)ni+1

+4(h3)ni+1wi+1 + (h4)ni1 2(h

4)ni + (h4)ni+1

/(24x2).

Rearranging gives,

2rx(h3)ni1wi1 + (1 + 4rx(h

3)ni )wi 2rx(h3)ni+1wi+1 = d

ni , (6)

where

dni = rx

(h4)ni1 2(h4)ni + (h

4)ni+1

,

and rx =t

12x2. Equation (6) can be solved for w using the Thomas algorithm taking the dni

term as known. The relationship hn+1i = hni + wi completes the time step. Taking account of

symmetry in the problem at i = 1 gives hi+1 = hi1 (hx = 0) such that,

(1 + 4rx(h3)ni )wi 4rx(h

3)ni+1wi+1 = dni , (7)

where

dni = 2rx

(h

4

)ni + (h

4

)ni+1

.

A similar boundary condition can be applied at the end of the discretised spatial domain.

To complete the numerical description, the initial condition for the height profile, at t = 0,is h = (1 x2)+ + 10

6, where a prewetting film of thickness 106 has been included to avoidthe difficulty of tracking sharp flow front.

The above nonlinear partial differential equation can also be solved by the method of linesapproach which reduces the problem to solving a system of ordinary differential equations inNAG routine D03PCF.

3. Analytical Solution: Similarity method

Similarity methods, techniques based on invariance under continuous Lie group transformations,can be used to simplify ordinary and partial differential equations. Similarity transformations re-duce either the order of the ordinary differential equation, or the number of independent variablesin a partial differential equation. A particular similarity reduction is sought by substituting

h = taf(), (8)

where = x/tb into equation:

ta1

af + b

df

d

=

1

3 t4a2b d

d

f3 df

d

. (9)



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Isothermal Gravity Currents 3

To elimate t in equation (9) we require

3a + 2b = 1. (10)

A second relation is required to fix the exponents a and b; for the two-dimensional constantmass case we have

hdx = 2Qc (constant, say) (11)

or

tba

f d = 2Qc. (12)

Hence a = b and, using equation (10), gives a = b = 15

. Hence a similarity solution is

h = t

1

5 f(), (13)

where = xt1

5 and equation (9) simplifies to

1

5

f +

df

d

=

1

3

d

d

f3

df

d

.

Integrating the above equation and using dfd = 0 at = 0 (from symmetry at x = 0) gives

3

5f = f3

df

d;

integrating again to give

f =

c2

9

102 1

3

, (14)

where c is a positive constant. The resulting height profile is given by

h =

9

10

13 t1/5

d2 x2t2/5

13 , |x| < dt1/5

0, |x| dt1/5(15)

where d is a positive constant which can be expressed in terms of Qc via (12).

Note that the axisymmtric case and point and line sources are derived in [5].



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4 Ahmos Sansom

References

1. Acheson, D. J.: 1990, Elementary Fluid Dynamics. Oxford University Press.

2. Barenblatt, G. I.: 1952, On some unsteady motions of a liquid or gas in a porous medium. Prikl. Mat.

Mech. 16, 6778.3. Huppert, H. E.: 1982, The propagation of two-dimensional and axisymmetric viscous gravity currents over

a rigid horizontal surface. J. Fluid Mech. 121, 4348.

4. Richtmyer, R. D. and K. W. Morton: 1967, Difference methods for initial-value problems. John Wiley and

Sons, 2 edition.

5. Sansom, A.: 2000, Gravity currents with temperature-dependent viscosity. PhD Thesis, University of

Nottingham.


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