xfem using 1D stefan problem

Simulation of 1-d Stefan Problem using XFEM

V. S. S. Srinivas

Outline Stefan melting problem

Physical description Analytical solution

Conventional numerical modeling

Associated issues

Review of Xfem

Implementation

Results

Unresolved issues

Challenges in Xfem-2d

Discussion and future directions

Stefan problem

T=T0 0°C

Initial Temperature = 0°C

h(t)

( )r

ﾶ ﾶ- = < < <

ﾶ ﾶ

2

20,0 ,0

T k Tx h t t

t c x( ) ( )00, , , 0T t T T l t= =

,@h T

L k x ht x

r ﾶ ﾶ= - =

ﾶ ﾶ

( )0, 0 0initialT h= =

Governing equations Initial and boundary conditions

Analytical solution

( ) 2k

h t tc

mr

=

( ) 2 2

0

0

2

, z z

xkt

c

T x t T e dz e dzm m

r

- -=

2 20

0 2z T c

e e dzL

mmm - =

Solve the transcendental equation, get m

Transcendental equation:

Temperature profile:

Interface position:

Conventional method

Always the interface has to coincide with a node

Necessitates re-meshing

T=T0

T=T0

T=T0

t=0

t=t1

t=t2

0<t1<t2

T=0

XFEM

Enrichment to elements whose support is intersected

When interface moves across elements

new nodes get enriched while the old ones shed

enrichment

No re-meshing required

Quite useful when re-meshing is costly and the interface has to be tracked explicitly

T=T0

T=T0

T=T0

t=0

t=t1

t=t2

0<t1<t2

ￄ ￄ

ￄ ￄ

ￄ ￄ

Enriched shape function( )1 1

e

xN x

l= - ( ) ( ) ( ) ( )( )3 1 0N x N x xf f= -

( ) bx x xf = -Level set function in 1d:

Interface position: xb,

Element length: le=1

Xb=0.5

Xb=0.3

Contd..

( )2e

xN x

l= ( ) ( ) ( ) ( )( )4 2 eN x N x x lf f= -

Xb=0.5

Xb=0.3

Implementation

Considered 1D as opposed to a 2D scenario

Reduced the complexities associated with level sets

Performed analytical integration in computing the element matrices using Mathematica

Initial values of the enriched degrees of freedom are computed by applying the interface conditions

The element nodes are enriched when its support is intersected by the interface and vice versa

Initial values determination-A case study

At t=0, the interface is located in the first element

Its nodes are enriched

Initial values?

T=T0

t=t1ￄ ￄ

T=0

Interface conditions:

1. T=Tm

2. Discontinuity in the gradient of temperature

T5 T6

T1

T2

T4T3

511 12 1

621 22 2

Ta a b

Ta a b

￦￦ ￦=

￨ ￨￨

Unique solution

No solution

Many solutions

Consider 2D case…

Contd..

4 unknowns 6 conditions This is unlike 1D case where there are 2 unknowns and 2

conditions

ￄ

ￄ

ￄ ￄ

Important features of XFEM

Enriched dofs are determined automatically

Don’t need explicit attention as in determining initial values

Matrices that have to be inverted remain symmetric

ￄ

ￄ

ￄ

ￄ ￄ

ￄ

ￄ

ￄ

Positions of enriched dofs changing

Number of enriched dofs is also changing

ￄ

A peep into the matrices

( ) ( )( ) ( )1 1 11

q

n n n nT cT cT d T k T d Tq dtd r d d+ + +

W W G

- W + W = - GD

1 11 1n n nqt t

+ +￦ + = +D D￨

*M K T M T f

( )1 1 1Tn n nc dr+ + +

W

= WM N N ( )* 1 Tn n nc dr+

W

= WM N N

( )1 1 1Tn n nk d+ + +

W

= WK N N ( )1 1 1

q

Tn n nqf q d+ + +

G

= - GN

Governing equation after applying weighted residuals

Discretized equation:

UnsymmetricSymmetric

Validation: Numerical against Analytical

Validation of XFEM

0

0.005

0.01

0.015

0.02

0.025

0 5000 10000 15000 20000

Time [s]

Inte

rfac

e p

osi

tio

n [

m]

Analytical

Numerical-XFEM

Convergence check

Convergence check

0

0.005

0.01

0.015

0.02

0.025

0 2000 4000 6000 8000 10000 12000 14000 16000 18000

Time [s]

Inte

rface p

os

itio

n [

m]

Nelem:11

Nelem: 9

Nelem: 6

Nelem: 4

Nelem: 3

Nelem: 2

T vs x @ t= 0-16000s

-2

0

2

4

6

8

10

12

0 0.005 0.01 0.015 0.02 0.025 0.03

x [m]

T [

C]

T versus x at different times

Issues-1D

Problems in determining the initial values of enriched dof for certain positions of the interface

Nodes are a set of examples May be endemic to 1D

During the interface propagation, if it falls on a node; in one instance it caused the matrix to become singular

Challenges XFEM-2D

Level set schemes for embedding the interface in an implicit function

Requires stabilization schemes

May cause rank deficient matrices

Pros and Cons

Don’t need re-meshing

Interfaces can be tracked explicitly

May cause rank deficient matrices

Discussion

• What is our objective?

• Will this idea serve to be useful?

• What are the various possibilities in modeling our problem.

• Will this method score over others?

Future directions

Check if the singularity issues related to determining initial values persist in 2D

If yes, ways to eliminate them Extend this model to 2D

Explore the possibility of using Abaqus with XFEM capability

Thank you