1. Theoretical Framework -...

Forced air flow and distribution in landfilled wastes

Defra Research Contract WR0302 Waste Management Research GroupJuly 2008 University of Southampton

1

Annex C The LDAT gas and leachate flow algorithm

1. Theoretical Framework

For a constant volume element there is no net change in volume following changes in the constituent volumes of solids, gases and liquids as the result of transport and degradation.

Thus

0

G

G

L

L

S

S m

t

m

t

m

t

(C1.1)

im is the mass in phase i, i is the density of phase i and the suffices S, L and G refer

to the solid, liquid and gas phases respectively. t is time.

The changes in mass and density with time are separated by differentiating equation (C1.1).

t

m

t

m

t

m

t

m

t

m

t

m G

G

GL

L

LS

S

SG

G

L

L

S

S

222

111 (C1.1a)

Both the bulk moduli of the solid and liquid phases, SK and LK , take into account the

changes in their respective densities as a result of a change in pore pressure p . In the gas phase changes in density are taken into account using the gas law

)100(1000

pRTG . Thus

2G

GGm

becomes, pp

VG 100

or pSG . GV is the volume of

gas in m3, and the units of GS are m3/kN/m2.

Equation (C1.1a) then becomes,

t

pS

K

V

K

V

t

m

t

m

t

mG

L

L

S

SG

G

L

L

S

S

111(C1.2)

Where SV and LV are the volumes of the solid and liquid phases respectively. Re-

defining the terms S

S

K

V and

L

L

K

V as storage coefficients SS and LS allows equation

(C1.2) to be written as,

t

pSSS

t

m

t

m

t

mGLS

G

G

L

L

S

S

111(C1.3)



2

The change of mass in the solid phase Sm will have a degradation component DSm ,

a solubility component PSm , a consolidation component C

Sm , and a re-fill

component RSm . The change of mass in the liquid phase Lm will have a degradation

component DLm , a solubility product component P

Lm , a transport component TLm ,

and a re-charge component RLm . The change of mass in the gas phase Gm will have

a solubility component PGm , a transport component T

Gm , and a re-charge component RGm . Thus multiplying out t for the time being, equation (C1.4) becomes,

L

RL

TL

PL

DL

S

RS

CS

PS

DS mmmmmmmm

pSSSmmm

GLSG

RG

TG

PG

(C1.4)

An assumption that is implied in this development is that there is no immediate change in the mass of the gas phase as the result of degradation reactions. That is gaseous compounds, arising from solid or liquid degradation, are assumed to remain dissolved in the liquid phase pending a solubility calculation. Following the degradation calculation a solubility calculation is carried out to determine the mass changes to maintain solubility equilibrium and this changes the phase terms in equation 1.4, including P

Gm .

Following Verruijt (1969) it may be assumed that consolidation takes place by means of a vertical solids mass transfer which maintains the density at a level appropriate to

the prevailing effective stress. Therefore, in an element that is full, S

CSm

, the net

change in the volume of solids as the result of consolidation, together with any change in the volume of solids as the result of solids re-fill and solubility, will need to

compensate the volume change following solids degradationS

DSm

, and the volume

change caused by any compression of the material. Additional compensation will also be required for any increase in solids volume DV following any change in effective stress and dry density which may take place.

Thus in a full element,

DSS

DS

S

RS

S

PS

LS

CS

US

CS

S

RS

S

PS

S

CS VpS

mmmmmmmm

(C1.5)

where US

CSm

and LS

CSm

are the vertical downward volumetric mass transfers

across the upper and lower boundaries of the element.



3

For the purpose of modelling the settlement of waste, it is assumed that it is possible to obtain a reasonably accurate estimate for the waste density from the local effective stress. At present

the LDAT model is coded using the empirical relationship between dry density /S kg/m3,

and effective stress p kN/m2 proposed by Powrie and Beaven (1999), a

S pA )(/ . A and a are empirical coefficients which will be sensitive to the

condition and nature of the particular waste being modelled.

Changes in effective stress will thus cause changes in dry density and there will thus be consequential changes in the overall solids volume. This component of volume change is designated DV in equation (C1.5).

For a given value of the dry density S and element volume V , we know that the mass of

solids in the element is VS . If the density of the solid material is S the volume of the

element occupied by the solids is VS

S

. The volume of solids in the element SV is therefore

given by, VVS

SS

.

Thus following a change in pore pressure there will be a change in SV , SVcomposed of two components, one due to particle compression, or changes in S ,

pSS , and one due to changes in S , DV .

Thus since,

ppaAVV

V a

SS

SD

1

or pSpap

VV D

SD

/)( if is constant. The units of DS are m3/kN/m2.

Inserting these assumptions into equation (C1.4) gives,

L

RL

TL

PL

DL mmmm

pSSSmmm

GDLG

RG

TG

PG

(C1.6)

Equation (C1.5) may be written,

pSSmmmmm

SDS

DS

S

RS

S

PS

LS

CS

US

CS

(C1.7)



4

Thus by introducing the term pSD , equations (C1.6) and (C1.7) separate equation (C1.4) into two coupled equations, a liquid and gas transport equation, equation (C1.6), and a solids transport equation, equation (C1.7).

TGLV / may be put equal to the change in volume as the result of liquid and gas flows

across the element boundaries into the element, that is,

G

TG

L

TLT

GL

mmV

/

In addition, DLV may be put equal to the change in the volume of the liquid phase as

the result of degradation, that is,

L

DLD

L

mV

Equation (C1.6) then becomes,

pSSSVVVV GDLP

GLR

GLD

LT

GL /// (C1.8)

and equation (C1.7) becomes

pSSVVVmm

SDD

SP

SR

S

LS

CS

US

CS

(C1.9)

where YXV is the change in volume of phase X due to process Y , and Y is either a

transport, degradation, recharge or phase change process.

Equation (C1.8) will be valid in all elements. Equation (C1.9) has two constraints. Generally,

1

i

US

CS

i

LS

CS mm

(C1.10)

That is the consolidation mass transfer across the lower boundary of element i is equal to the mass transfer across the upper boundary of the element 1i below. In the lowest element however,

0

1

LS

CSm

(C1.11)

In either the uppermost part full element or the element below, it is likely that the total volumetric mass transfer into the element,



5

RS

US

CS V

m

that is required to balance the mass transfers caused by consolidation, degradation and phase changes will not be fully available.

If the volume of solids available is S

Am

then,

RS

US

CS

S

A Vmm

(C1.12)

and equation (C1.5) gives,

RS

LS

CS

S

A

S

CS V

mmm

(C1.13)

Substituting equation (C1.13) into equation (C1.4) gives for the upper part full element,

L

RL

TL

PL

DL

LS

CS

S

A

S

PS

S

DS mmmmmmmm

pSSSmmm

GLSG

RG

TG

PG

(C1.14)

or,

pSSSVVmm

VV GLSP

GLSR

GL

LS

CS

S

ADLS

TGL

///// (C1.15)

Note that in a part full element there is no change in volume as the result of density compression. Thus, 0DS .

From equations (C1.9) and (C1.10),

1

i

US

CS

i

LS

CS mm

1111

11

iiS

iD

iDPRS

i

LS

CS

i

US

CS pSSV

mm



6

21

i

US

CS

i

LS

CS mm

and so on until

1111

1

pSSVm

SDDPR

S

US

CS

Adding the above sequence of equations gives,

1

1

1

1

Nii

SiD

NiDPR

S

N

LS

CS pSSV

m

(C1.16)

N being the number of the uppermost part full element, counting upward from the bottom of the stack.

Substituting equation (C1.16) in equation (C1.15) gives,

NNG

NL

NS

Nii

SiD

NiDPR

SS

NANP

GLSNRGL

NDLS

NTGL pSSSpSSV

mVVVV

1

1

1

1/////

(C1.17)

From equations (C1.9) and (C1.12), the maximum value of S

Am

is,

NNS

ND

DS

PS

LS

CS

N

MAXS

A pSSVVmm

NNS

ND

DS

PS

Nii

SiD

NiDPR

S pSSVVpSSV

1

1

1

1

Substituting this into equation (C1.17) reduces it to equation (C1.8) for a full elementwhich verifies the algebraic process that has led to obtaining equation (C1.17) for the upper part full elements.

Summary of theoretical framework

Equation (C1.8) is generally used as the constitutive equation for liquid and gas flowin LDAT except in the upper part full elements where equation (C1.17) is used. Equation (C1.9) is used to calculate settlement of the solid phase.

pSSSVVV GDLPR

GLD

LT

GL // (C1.8)



7

NNG

NL

NS

Nii

SiD

NiDPR

SS

NANP

GLSNRGL

NDLS

NTGL pSSSpSSV

mVVVV

1

1

1

1/////

(C1.17)

pSSVmm

SDDPR

S

LS

CS

US

CS

(C1.9)



8

2. Finite difference representation

For a time period from t to tt equation (C1.8) becomes,

ti

ttiGDL

PRGL

DL

GE

GW

LE

LW

GN

GS

LN

LS ppSSSVVqqqqqqqq

/

(C2.1)

where,

LSq is the volume of leachate entering element i from below,LNq is the volume of leachate leaving element i from the top,GSq is the volume of gas entering element i from below,GNq is the volume of gas leaving element i from the top,LWq is the volume of leachate entering element i from the left,LEq is the volume of leachate leaving element i from the right,GWq is the volume of gas entering element i from left,GEq is the volume of gas leaving element i from the right,

DLV is the liquid volume change due to degradation,

PRGLV / is the liquid and gas volume change due to recharge and phase change,

qEqW

qN

qS

Element i qEqW

qN

qS

Element iqW

qN

qS

Element i



9

GDLS // are storage coefficients,

p is pore pressure in kN/m2.

The volumetric flow rate of both leachate, LQ , and gas, GQ , across a boundary between two elements of the model is calculated using Darcy’s equation in the form,

hTQ (C2.2)

T is a transmission coefficient and h is the head difference between the mid-points of the two elements. The volume q of leachate or gas transferred in time t is given by tQq

Calculation of T

The transmission coefficient is calculated by taking the reciprocal of the sum of the resistances to flow between the mid-points and the boundary of the elements. Thus if these resistances are iR and 1iR ,

1

1

ii RRT , (C2.3)

and typically,

ii

ii KA

LR (C2.4)

In equation (C2.4), iL is the distance between the midpoint of the element and the

boundary, iA is the area of flow and iK is the permeability. iK is calculated using

the Powrie and Beaven (1999) relationship, equation (C2.5), multiplied by the relative permeability – see section 5. below.

npBK (C2.5)

The value of K may then be adjusted to suit the conditions being modelled, relating it to model temperature through the viscosity and density of the modelled leachate, using the following approach (described in Das (1983), equation 2.6).

If RL and R

L are the reference density and viscosity at which the permeability calculated by equation (C2.5) has been measured, then the permeability to a liquid or gas with density and viscosity is given by,

RL

RLRKK (C2.6)



10

The area of flow, iA in equation (C2.4), is nominally the area of the boundary

interface between the elements. This is reduced in the model to allow for the presence of gas. If 0A is the nominal area, then the area of flow for leachate LA is calculated

using the proportional relationship,

3/23/2

3/2

0

G

G

L

L

L

L

L

MM

M

A

A

(C2.7)

where LM and GM are the total mass of leachate and gas contained in the element.

The volumes of leachate and gas in equation (C7) are raised to the power 2/3 to make them more representative of the areas of leachate and gas at the element boundary.

Similarly the area of flow to gas is calculated as follows.

3/23/2

3/2

0

G

G

L

L

G

G

G

MM

M

A

A

(C2.8)

Calculation of h

h in equation (C2.2) is the head difference between the two elements calculated in terms of the density of the fluid for which the flow rate is being calculated.

In the case of gas,

21000

1

10001

100010001

,

SiGSiSG

S

SiG

SiSGS

iG

SiSG

SSGS

GiS

ddgpp

g

zzgppg

gzpgzpg

h

(C2.9)

iWGW

GiW pp

gh

1000

, (C2.10)



11

21000

1

10001

100010001

,

iNGNNiG

N

iNG

NNiGN

NG

NNiG

NiGN

GNi

ddgpp

g

zzgppg

gzpgzpg

h

(C2.11)

EiGE

GEi pp

gh

1000

, (C2.12)

Similar equations can be written for liquid head differences, so that,

21000

1,,,

SiLSiCiSCSL

S

LiS

ddgpppp

gh

(C2.9a)

iCiWCWLW

LiW pppp

gh ,,,

1000

(C2,10a)

21000

1,,,

iNLNNCNiCiL

N

LNi

ddgpppp

gh

(C2.11a)

ECEiCiLE

LEi pppp

gh ,,,

1000

(C2.12a)

Where the effective density (see section 4. below) of fluid phase F in direction X iscalculated as,

iX

iF

iXF

XFX dd

dd

iCp , is the capillary pressure, which is discussed further in section 3. below.

Thus,

21000

1,,

SiGSiSG

S

GiS

GiS

ddgpp

gTtq

21000

1,,

iNGNNiG

N

GNi

GNi

ddgpp

gTtq

iWGW

GiW

GiW ppTtq

1000,,



12

EiGE

GEi

GEi pp

gTtq

1000,,

21000

1,,,,

SiLSiCiSCSL

S

LiS

LiS

ddgpppp

gTtq

21000

1,,,,

iNLNNCNiCiL

N

LNi

LNi

ddgpppp

gTtq

iCiWCWLW

LiW

LiW pppp

gTtq ,,,,

1000

ECEiCiLE

LEi

LEi pppp

gTtq ,,,,

1000

(C2.13)

where iST , , NiT , , iWT , and EiT , are the lower, upper, left and right transmission

coefficients for element i respectively, t is the value of the incremental time step, and id is the depth of element i. The values of pore-pressure in equations (C2.13) are

the average values over the time increment t in kN/m2.

By substituting equations (C2.13) into equation (C2.1) and defining the following coefficients,

GS

GiS

LS

LiS

S

TT

g

ta

,,1000

m3/kN/m2

GN

GNi

LN

LNi

N

TT

g

ta

,,1000

m3/kN/m2

GW

GiW

LW

LiW

W

TT

g

ta

,,1000

m3/kN/m2

GE

GEi

LE

LEi

E

TT

g

ta

,,1000

m3/kN/m2

SCiCSiL

SLS

LiS

SL ppg

ddTtb ,,

,,

1000

2

2,

,SiG

SGS

GiS

SG

ddTtb



13

iCNCNiL

NLN

LNi

NL ppg

ddTtb ,,

,,

1000

2

2,

,NiG

NGN

GNi

NG

ddTtb

iCWCLW

LiW

WL ppg

Ttb ,,

,,

1000

ECiCLE

LiE

EL ppg

Ttb ,,

,,

1000

WLELSGSLNGNL bbbbbbb ,,,,,,

PRGL

DL VVc / m3

GDL SSSd m3/kN/m2

we obtain,

ti

ttiiEEiWWiNNiSS ppdcbppappappappa m3

or,

ti

ttiiEWNSEEWWNNSS ppdcbpaaaapapapapa m3

(C2.12)

But since,

2

ti

tti

i

ppp

equation (C2.12) becomes,

t

itt

itt

iti

EWNS

ttE

tE

EttW

tW

WttN

tN

NttS

tS

S

ppdcbppaaaa

ppa

ppa

ppa

ppa

2

2222 m3

or,



14

cbp

aaaap

ap

ap

ap

adp

dppaaaa

pa

pa

pa

pa

ti

EWNStE

EtW

WtN

NtS

Sti

tti

tti

EWNSttE

EttW

WttN

NttS

S

22222

22222 (C2.13)

or, the linear equation,

iiiiEEiWWiNNiSSi Apapapapapa ,,,,, m3 (C2.14)

This equation is just for element i. The equations for all the elements make up the set of linear simultaneous equations which, together with known boundary conditions, may be solved for the pore pressures, ip , by inverting the matrix equation.

Apa

Note that in the case of the uppermost part full element, element N, equation (C2.13) needs to be replaced by an equation corresponding to equation (C1.20).

The following changes have to be made,

GLS SSSd

1

1

1

1

////

Nti

iS

iD

NiDPR

S

S

NA

NPGLS

NRGL

NDLS

pSS

V

m

VVVc

Together with additional coefficients,

kS

kDki SSa ,

for all pore-pressures kp from 1k to N-1.

If an element is empty c reduces to NPRGLV /



15

3. Capillary pressure

The conventional concept of pore-liquid behaviour under partially saturated conditions is that the liquid retreats into the smaller pore-spaces and that a capillary pressure is set up across the meniscus interface between the liquid and gas. This creates a difference between the pressures in the liquid and gas phases. It follows that the smaller the degree of saturation, the smaller the pores occupied by the liquid phase and the greater the capillary pressure. It is assumed that under all conditions the meniscus acts to resist gas entry into the pores or so as to encourage invasion of the liquid content into the gas space.

Thus at equilibrium LGC ppp .

If CGL ppp the material will ‘imbibe’ and saturation will increase causing a

corresponding decrease in Cp as the interface invades larger pore spaces. Similarly if

CLG ppp the gas will enter the smaller pore spaces decreasing saturation and increasing

Cp .

When the gas pressure is zero relative to atmospheric pressure CL pp and negative. Thus

conventionally Cp is regarded a the suction pressure required to reduce the degree of

saturation in a material to . It could equally well be thought of as the gas entry pressure at .

An example of the relationship Cp is given in Figure C1. This is derived from the van

Genuchten (1980) expression,

g

p

L

CWR

WRWE

1

1

1(C3.1)

E is the effective degree of saturation which falls to zero when the actual degree of

saturation, W , reaches the residual degree of saturation, WR , for the material. The residual

degree of saturation accommodates the concept that no matter what the applied suction pressure, or air entry pressure, there will always be a minimum amount of liquid present in the material.

Cp is the capillary pressure relative to atmospheric pressure, and when 0Cp equation

(C3.1) gives 1E . As the value of becomes more negative, and the suction pressure

increases, the value of E falls below 1 and approaches the residual degree of saturation as

the suction pressure becomes infinitely large. It is assumed that 1E when 0Cp .

It is generally accepted that the shape of the relationship Cp will change depending on

whether or not the material is in the process of increasing or decreasing in saturation (imbibing or draining). This hysterisis effect is very complicated and it is not presently planned to attempt to model this in LDAT.



16

In equation (C3.1)

11 so that the function has essentially two parameters and .

For the purposes of producing Figure C1 the values 2.0 m-1 and 5 have been assumed.



17

4. Effective fluid density

Equations (C2.9) to (C2.12a) require an estimation of the effective fluid density. Under partially saturated conditions in which the degree of saturation is , if the gas and liquid phases are assumed to be fully mixed, the effective density for both phases will be,

GLGL 1/ (C4.1)

However it is conceivable that the phases will not fully mix but will remain separate to a certain degree. Suppose that when the degree of saturation is , the degree of saturation of

the gas is actually G . This means that if the volume of gas present is GV , GGV of the gas is

replaced by liquid and the mass in the gas phase becomes GLGGGG VV

Thus the effective density of the fluid fraction controlled by the gas phase pressure is given by,

GGWGG 1 (C4.2)

The corresponding mass in the liquid volume LV , will become GLGGLL VV and,

1GLGLGL

L

GGLL V

V

Thus,

GLLLL 1 (C4.3)

Where,

1

1 GL (C4.4)

These equations give a range of effective densities from conditions where the phases are completely separated to fully mixed, as G varies from 0 to . The characteristic of this

relationship can be varied by choosing values for the parameters value 0a and n in the

following equation.

0G 0an

G a

a

0

0

1

10 a (C4.5)

An illustration of the shape of these functions is given in Figures (C2) and (C3) for 2.00 a

and 5n .



18

5. Relative permeability

It can be seen from equations (A2.3) and A(2.4) that the transmission coefficient T in equation (A2.2) requires the evaluation of resistances and the product of flow area and permeability iiKA . When there is two phase flow in the fluid phase and the material is

partially saturated the area of liquid flow reduces because of the presence of gas, and the permeability reduces because of capillarity effects.

The van Genuchten (1980) expression that covers both of these effects by evaluating a relative permeability value is,

25.0

1

11

EERk (C5.1)

This is plotted in Figure 4. The expression for relative permeability to gas which is plotted in Figure 5, and is almost the mirror image to this is,

2

5.0 1

11 EEgasRk (C5.2)

The actual source of this expression is unclear.

Once calculated Rk or gasRk is used to modify iiKA by simple multiplication.



19

Figure C1 Relationship between capillary pressure and effective degree of saturation Cp

Figure C2 Equation (C4.5) for 2.00 a and 5n .

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Degree of saturation overall

Gas

ph

ase

satu

rati

on

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

-200.0 -150.0 -100.0 -50.0 0.0 50.0

Capillary pressure (kPa)

Eff

ecti

ve d

egre

e o

f sa

tura

tio

n



20

Figure C3 Equations (C4.1), (C4.3) and (C4.4).

Figure C4 Equation (C5.1)

0

200

400

600

800

1000

1200

0 0.2 0.4 0.6 0.8 1

Degree of saturation

Den

sity

of

gas

an

d li

qu

id p

has

es

ro gas

ro liq

ro_mixture

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.2 0.4 0.6 0.8 1.0


Rel

ativ

e p

erm

eab

ilit

y

krw



21

Figure C5 Equation (C5.2)

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.2 0.4 0.6 0.8 1.0


Rel

ativ

e p

erm

eab

ility

(g

as)

kra

1. Theoretical Framework -...

Documents

Transcript of 1. Theoretical Framework -...