Summerschool

Modelling of Mass and Energy Transport

Black Box Analogy

)(teRidt

diL

i: current, e(t): input voltage.

Characteristic Volume

Material Mix

Storage Capacity - Porosity

Grainsize, Ordering, Secundary Processes

Internal Structure

Anisotropy + Secondary Pathways

Mass and Energy Transport

ExerciseGiven a flux vector approaching an oblique oriented surface element (line) of your choice. Separate it graphically into the flux components

parallel and oblique to the surface

Normal Vector and FluxesThe normal is perpendicular to the surface.

The normal vector points to the outside of the closed are.The closed area is always surrounded anticlockwise.

)1,0(),1,0(),0,1(),0,1( in

),0(),,0(),0,(),0,( aaaavi The fluxes for the left case are

iiiWithin Lnvchange

S

Within dsnvchang

or

Fluxes Derived from a Potential

0

0

0

0

gradj

gradv

change = iii Lngrad

)( iii Lngrad

)(=

change= S

dsngrad

)( = S

dsngrad

)(

.

Gauss Theorem

SV

dsnvdVv

Green‘s Formulae,,, vuorHWGVFU

HGF ,,,

G G S

dsnvvdgdgv )(are functions of (x,y,z),

In case the vector v(F,G,H) is the gradient of a scalar field

vdsndgdg

G G S

*)()()(2 .constc

G S

dsndg *)(2

.

Mass and Energy Balance Continued

V S

dsnvudVdt

d

V S

dsnvdVt

u

0))((

dVvdivt

u

V

In this case there is no gain or loss within the control volume

Gain or loss within the control volume

VV V

dVvdivQdVdVt

u)(

0))((

dVQvdivt

u

V

Key QuestionUnder which conditions will the equality in the equations

0))((

dVvdivt

u

V

and

0))((

dVQvdivt

u

V

hold, independent of the size of the control volume?

Remark

V S

dsnvdVQt

u )(

The equation

where Q may be zero, provides a weak formulation of the problem to solve transport processes. For that purpose subdivide the area under consideration into aerial subsets, by example triangles as illustrated in the sketch

Then we can evaluate the volume integral by using the Gauss theorem and we can formally evaluate the fluxes through the bounding lines or surfaces:

FV-Method

The Differential Equation

If the flux is derived from a potential

0)(

Qvdivt

u 0 Qvut

or

0)(

Qgraddivt

uor 02 Qut

Without temporal change within the black box

02

Boundary and Initial Conditions

1st kind or DIRICHLET’s condition: Find a harmonic solution for the interior of the considered area so that the solution ),,( zyx

)(sf

2nd kind or NEUMANN’s condition: Find a harmonic solution for the interior of the considered area so that the solution ),,( zyx

satisfies given values of the normal derivative )(sfn

3rd kind or mixed condition: Find a solution for the interior of the considered area so that the solution ),,( zyxsatisfies the equation

)(sfn

and

)0,,( 22 const

satisfies given values of

Exercises

1)Find (qualitatively) the solution for the 1-D problem 0

2

2

x

u

with u=a at x=0 and u=b at x=1.

2) What would be the solution over a square with values of u given at the four corner points?

3) How can the boundary condition 0n

u be interpreted (2 possibilities).

4) Consider a well located within a homogeneous aquifer with free surface. The well has diameter r and you are pumping continuously Q [m3 ] water. The free water surface within the well than will have dropped to height H. Through the surface area of the well water will flow v [m/s]which is proportional

to the slope of the free water surface or x

yKv

with y the height of the free water surface. At the boundary of the wellthe water level in the aquifer will be equal to the free water level within the well. Derive the differential equation, the required boundaryconditions and the solution. Sketch the solution.

Solution Exercise 4

Through any cylindrical surface area xy2

surrounding the well we have the flux sec]/[mv

and the total flux becomes xyvQ 2

.

By experience, the flux is dxcdyv /

We find the differential equation '2 cxyyQ

With the boundary conditions y=H for x=r

Solution: ))(/()/ln( 22 HyQcrx

'2 cxyyQ Differential equation:

For rx and Hy for rx

Fick‘s 1st law

CDjc

D is mostly given as the diffusion coefficent in pure water. Provide an estimate for porous media.

The flux is related to the temporal change by

)( CDt

C

Or if chemical reactions are involved within the porous medium

)( CDQt

C

Heat conductionTjT

The change of energy within the control volume becomes

)( Tt

Tc

Or if sources or sinks are vailable (chemical reactions, radioactive heat production)

)( TQt

Tc

Observe that )(Tcc and )(T

,

and that the specific volume

will also depend on the temperature. In detail the size of the control volume changes

[J/s]

Exercises

1) Linearize the equation )( Tt

Tc

by partially differentiating the time derivative with respect to the parameters, observing that c=c(T) and )(T

.

2) Consider the heat flow from the interior of the earth which mostly is considered stationary, i.e. it is governed by the equation 0)( T QT )(

At the surface an average thermal gradient of 30°C is observed. At about 100 km depth the boundary between the lithosphere and the astenosphere is reached, the latter behaves like a fluid due to seismological data, the temperature at this boundary is approximately 1300°C, due to experiments.

a) Compute the expected temperature at the base of the lithosphere based on the average surface gradient.

b) Compute the expected temperature gradient throughout the lithosphere, based on a surface temperature of 0°C.

c) Discuss for a layered crust qualitatively how the thermal conductivity should vary with depth.

d) Determine for a homogenous lithosphere the heat production required to fit the boundary conditions

or

Darcy‘s Law

)(

gpk

qM

[kg m/s]

)( zgpKqD

[m/s]

)(* hKqh

or

l

hKqh

*

or´ )/( zpKqD

[m/s]

))((

gpk

Qt

))(( hg

k

t

hS

Coupled equations

0)())((

CQqCCD

t

C

0)())((

Tff

bb QqTcTt

Tc

0

f

f Qqt

)(

gpk

q

),,( cQTp

D=D(T)

),,( cbb QTpcc

),,( cff QTpcc

),,( fff QTp

),,,( cfbb QQTp

)(T

).,( iQTpk

Mathematical formulation of the thermohaline flow problem in FEFLOW

0 BoussinesqS +div( )t

Q

q

0

0

( ) f f

f

q K grad

+div div ( ) C

CC C Q

t

q Dgrad

(1 ) +div( ) div ( ) Tf f s s f fc c T c T T Q

t

q λgrad

0

( , )f

f

g

C T

k

Κ

0 0 0 00

1 ( , )( ) ( , )( ) ( )f f

sat

T p T T T p p p C CC C

f

ffsat

0

0

Stability criteria

TdK

RaT

Solutal Rayleigh number

Thermal Rayleigh number

d

sats D

CdKCC

Ra

0

The solutal and thermal Rayleigh numbers are related by

Ts RaLeNRa

Buoyancy ratio (Turner) T

CCC

N sat

0

Lewis number dD

Le

24 sTC RaRaRa

CRa

• The monotonic instability (or stationary convection) boundary is a • straight line defined by

is the critical Rayleigh number.

.

The region delimited by 24 sT RaRa

is a stable regime characterized by pure conduction and no convection.

In a range between 3002404 2 sT RaRasteady state convective cells develop

For 2csT RaRaRa the convection regime is unstable

Cross section through the Büsum diapir and adjacent rim synclines with temperature isolines (left) and vitrinite reflectance isolines (right)

Type of large-scale gw flow

Type of thermohaline flow

Thermally induced brine plumes developping on a deep salt sheet

Brine lenses, gravitational convection from a shallow salt sheet

Dasehd lines: isotherms (°C)

Thermohaline flow in the NE German BasinStratigraphic units

Thermohaline flow: Zoom salt dome

Mixed convection in the NE German Basin

Bold vectors represent the topography-driven flow (i.e. regional flow)

Thermal convection in a shallow salt dome

Thermohaline flow in a shallow salt dome

Dashed lines: isotherms (°C)