Fowler A. Mathematical Geoscience

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840 E Averaged Equations in Two Phase Flow

To interpret the interfacial advective derivative of X

, we have, for test functions

φ(x,t) which vanish both at x →∞ and t →±∞,





∂X

∂t

.∇X



dV dt

=−





∂φ

∂t

.∇φ



dV dt

=−



∞

−∞



(t)



∂φ

∂t

.∇φ



dV dt

=−



∞

−∞



(t)

φdV dt =−





(t)

φdV



∞

−∞

=0. (E.14)

The averaged form of (E.12) is now derived in terms of the averaged volume

fraction α

, density ρ

, velocity u

, species ψ

, ﬂux J

, and source f

, deﬁned as

,α

ρ, α

ρu,

ρψ, α

J,α

ρf ,

(E.15)

and the conservation law (E.12) then takes the form

∂

∂t

(α

) +∇.







=−∇.(α

) +α



ρψ(u −u

) +J



.∇X

, (E.16)

where the proﬁle velocity U

is deﬁned by

ρψu −α

. (E.17)

In one-dimensional ﬂows, U

=(D

−1)u

, and D

is called a proﬁle coefﬁcient.

Apart from this, the last term in (E.16), representing interfacial transfer of ψ ,must

be constituted.

E.3 Mass and Momentum Equations

Conservation of mass is determined from (E.11) by putting

ψ =1, J =0,f=0. (E.18)

The corresponding equations for each phase are, from (E.16), with ψ

=1, J

and U

=0,

∂

∂t

(α

) +∇.[α

]=Γ

, (E.19)

E.3 Mass and Momentum Equations 841

where

=ρ(u −u

).∇X

, (E.20)

and Γ represents a mass source due to phase change (without which u =u

at the

interface).

Next, consider momentum conservation. With appropriate interpretation of ten-

sor notation, we put

ψ =u, J =−T ≡pI −τ ,f=g, (E.21)

where p is the pressure, τ is the deviatoric stress tensor, and g is gravity. In addition,

we write

ρuu = α

−α



; (E.22)

the second term can be interpreted as (minus) the Reynolds stress (cf. (B.7)). The

momentum equation can thus be written as

∂

∂t

(α

) +∇.[α

]=∇.





)



+α

g +M

, (E.23)

where

(−pI +τ ), M

=(pI −τ ).∇X

ρu(u −u

).∇X

ρ(u −u

).∇X

(E.24)

We deﬁne the average pressure and deviatoric stress in phase k to be

, τ

. (E.25)

It is conventional to separate the local interfacial stresses from those due to large

scale variations in α

by writing the interfacial momentum source as

∇α



, (E.26)

where



=(p −p

)∇X

−τ .∇X

, (E.27)

is the average interfacial pressure in phase k, and we use the fact that ∇X

∇α

. Thus the momentum equation can be written as

∂

∂t

(α

) +∇.(α

) =−α

∇p

+(p

−p

)∇α

+∇.[α

]

+∇.[α



]+α

g +M



. (E.28)

842 E Averaged Equations in Two Phase Flow

Often we may ignore the Reynolds stresses as well as the macroscopic viscous

stresses, and if we ignore surface energy effects, we may take p

= p

.Theterm



is the interfacial force, and is generally much larger than the other stress terms.

In this case, the momentum equation becomes

∂

∂t

(α

) +∇.(α

) =−α

∇p

+α

g +M



. (E.29)

The interfacial force M



includes the important interfacial drag, as well as other

forces, in particular the virtual mass force. Interfacial drag is due to friction at the

interface, while virtual mass terms are associated with relative acceleration. There

are various other forces which are sometimes included, also (see Drew and Wood

1985). The momentum source from phase change u

is often ignored. In condi-

tions of slow ﬂow, constitution of the interfacial drag as a term proportional to the

velocity difference between the phases leads to Darcy’s law.

E.4 Energy Equation

The point form of the energy equation is given in (E.6)or(D.12); we use the form

of (D.14), speciﬁcally

∂

∂t

(ρe) +∇.(ρeu) =−∇.q +T:∇u. (E.30)

To derive the averaged version, we put ψ = e, J = q, ρf = T:∇u in (E.12). By

analogy with (E.22), we deﬁne the turbulent heat transport q



via

ρeu =α

+α



; (E.31)

we then obtain (cf. (E.16)) the averaged energy equation

∂

∂t

(α

) +∇.{α

}=−∇.





)



+α

, (E.32)

where

T:∇u

ρe(u −u

).∇X

ρ(u −u

).∇X

=q.∇X

, (E.33)

and are respectively the average viscous dissipation, the interfacial internal energy

transfer, and the interfacial heat transfer. The ﬁrst two of these are generally negligi-

ble, while the third is usually large, at least if the two phases have different average

temperatures. It is because of this that typically temperature does not vary locally,

so that it sufﬁces to consider total energy conservation. To see why this should be,

we need to consider the averaged jump conditions between the phases.

E.5 Jump Conditions 843

E.5 Jump Conditions

The jump conditions for the point forms of the conservation laws in (E.7) have their

counterpart in the averaged equations. For the general conservation law (E.11), the

corresponding jump condition at an interface is



−



ρψ(u −u

) +J





−

, (E.34)

where n =n

−

points from − to +, and m

represents a surface production term,

which is normally zero. From (E.13), we can identify



−



ρψ(u −u

) +J







∂V



ρψ(u −u

) +J



.∇X

, (E.35)

where n

points out of phase k, and the angle brackets denote a speciﬁc surface

average (i.e., a surface integral over the interface divided by the volume); thus (with

no surface source term) the jump conditions for the averaged equations take the

form





ρψ(u −u

) +J



.∇X

=0, (E.36)

bearing in mind that n

=−n

Mass and momentum jump conditions are quite straightforward. Consulting

(E.18) and (E.20), we have



=0; (E.37)

consulting (E.21) and (E.24), we have







=0. (E.38)

Energy is slightly more opaque, since we have to go back to the conservation form

of the equation in (E.6) to derive the appropriate jump condition. This takes the form











=0, (E.39)

where the extra terms not deﬁned in (E.33)aregivenby





ρu

(u −u

).∇X

ρ(u −u

).∇X

=−T.u.∇X

, (E.40)

representing the interfacial kinetic energy transport and the interfacial work.

844 E Averaged Equations in Two Phase Flow

E.5.1 Practical Approximations

Generally speaking, the interfacial momentum ﬂux u

can be neglected, so that

(E.38) reduces to the force balance



≈0. (E.41)

The interfacial kinetic energy and interfacial work terms in (E.40) are generally

small, and additionally we suppose e

≈e

, so that (E.39) becomes the Stefan con-

dition



) =0. (E.42)

We can normally also neglect the dissipation term in (E.32). If we suppose that

the interfacial transport terms E

, typically proportional to the difference in tem-

perature between the phases, are large, then the conclusion is that the temperatures

must be equal, and a single equation for the temperature then follows from adding

the energy equations for the two phases. Adopting the jump condition (E.42), this

leads to

∂

∂t







+∇.







=−∇.







)



. (E.43)

Generally, we are only concerned with energy conservation when there is phase

change, i.e., Γ

= 0. In this case, the assumption of local thermodynamic equi-

librium prescribes the local temperature as the freezing or boiling temperature as

appropriate. Thus the energy equation does not in fact determine the temperature,

but serves to determine the mass source due to phase change, Γ

. To see how this

happens, we need to relate the internal energies e

to temperature T .

E.5.2 Thermodynamics

Quite generally, (D.3), (D.9) and (D.10) imply that the enthalpy and internal energy

satisfy

∂h

∂T



∂e

∂T



, (E.44)

and one usually takes

h =c

T, e=c

T. (E.45)

Commonly one rewrites the energy equation in terms of the enthalpy, since in phase

change problems the latent heat is deﬁned (at a ﬁxed pressure and temperature) by

L =T

s =e +pv =h. (E.46)

E.6 Nye’s Energy Equation in a Subglacial Channel 845

Using (E.19) and (E.37), we can write the energy equation in the form





+α



−



=∇.[

K∇T ], (E.47)

where

K is the phase-averaged thermal conductivity, including both molecular and

turbulent conductivities,

K =



, (E.48)

and

∂

∂t

.∇. (E.49)

For example, consider a vapour–liquid ﬂow, with h

−h

=L. Denoting speciﬁc

heats as c

, we can write (E.47) in the form

L +





−



∂

∂t

(α

) +∇.(α

)



=∇.[K∇T

], (E.50)

and this determines the mass source term Γ

(which is positive for boiling, and

negative for condensation).

E.6 Nye’s Energy Equation in a Subglacial Channel

A particular variant of the procedures outlined above is Nye’s derivation of the

energy equation governing water ﬂow in a sub-glacial channel. In particular, the

variables and thus also the equations are cross-sectionally averaged. Nye (1976)

provided his Eq. (11.4) with the minimum of fuss. Let us now try and derive this

equation using the principles enunciated above. The equation is



∂θ

∂t

∂θ

∂x





g sin α −

∂p

∂x



−m



L +c

(θ

−θ

)



, (E.51)

in which θ

is the water temperature, S is the channel cross-sectional area, Q =Su

is the volume ﬂux, p is the channel pressure, and θ

is the surrounding ice tempera-

ture, taken as constant. See Chap. 11 for further details.

To derive this in detail, we need to derive also the appropriate forms of the con-

servation of mass and momentum equations in the channel. To begin with, we note

the general relation



V(t)

LdV =



V(t)

∂L

∂t

dV +



∂V

dS, (E.52)

where V

is the normal velocity of the moving boundary ∂V of the time dependent

volume V . This applies whether or not the volume V is a material volume. If it is,

846 E Averaged Equations in Two Phase Flow

then V

, the ﬂuid normal velocity. For a subglacial channel, this is not the case.

We can then relate the rate of change of the integral of L over V(t)to that over the

material volume which is instantaneously coincident with V :



V(t)

LdV =



(t)

LdV −



∂V

L(u

−V

)dS, (E.53)

where V

is the corresponding material volume.

The equation of conservation of mass follows from putting L = ρ (= ρ

). The

mass



ρdV is conserved, and mass conservation takes the form

∂

∂t



V(t)

ρdV +



∂V



ρu

dS =−



∂V

⊥

ρ(u

−V

)dS, (E.54)

where we take the volume V to be the cross section of the channel times a small

(ﬁxed) increment δx in the downstream direction, ∂V



denotes the end faces of the

volume (on which V

= 0), and ∂V

⊥

denotes the ice-water interface. Dividing by

δx and letting δx →0, we obtain conservation of mass in the form

∂

∂t

(ρS) +

∂

∂x

(ρSu) =m, (E.55)

where u is the average velocity and ρ is the average density over the cross section,

and

m =



∂S

−

Γds=−



∂S

−



ρ(u

−V

)



−

ds; (E.56)

∂S

−

is the perimeter of the cross section S, taken on the inside.

This same procedure allows us to form averaged momentum and energy equa-

tions. The basic momentum equation in integral form is



(t)

ρu

dV =



∂V

.ndS +



ρf

dV, (E.57)

and performing the same reduction as above leads to

∂

∂t



ρu

dV +



∂V



ρu

=−



∂V

⊥

ρu

−V

)dS −



∂V

dS +



∂V

.ndS +



ρf

dV, (E.58)

wherewewriteσ =−pδ + τ , δ being the unit tensor and τ being the deviatoric

stress tensor. We now use the divergence theorem on the pressure term to write this

∂

∂t



ρu

dV +



∂V



ρu

=−



∂V

⊥

ρu

−V

)dS −



∂

∂x

(p +ρχ)dV +



∂V

.ndS, (E.59)

E.6 Nye’s Energy Equation in a Subglacial Channel 847

in which we suppose that ρ is constant; χ is the gravitational potential energy. Tak-

ing i =1(thex direction) and averaging, we obtain the momentum equation in the

form

∂

∂t

(ρSu) +

∂

∂x



ρSu



=−S

∂

∂x

(p +ρχ) −τ

l, (E.60)

where l is the wetted perimeter, τ

is the wall stress. Importantly, no slip at the

wall implies u

= 0on∂V

⊥

(if we assume downstream ice velocity is negligible).

We have neglected deviatoric longitudinal stress on the ends of ∂V. Note that it is

important to convert the surface integral in pressure in (E.58) to the volume integral

in (E.59) before deriving (E.60) (otherwise we would be tempted to put the S coef-

ﬁcient of the pressure term inside the derivative). Speciﬁcally, (E.59) takes the form

∂

∂t

(ρuS δx) +···=−

∂

∂x

(p +ρχ)S δx +···=−Sδ(p+ρχ) +···, (E.61)

and on dividing by δx, we obtain (E.60). (This is analogous to the absorption of the

term p

∇α

into the interfacial term in (E.26).)

Note also that taking i =3(thez direction) gives us the hydrostatic condition (if

we neglect deviatoric normal stress)

∂

∂z

(p +ρχ) =0. (E.62)

The energy equation is derived in a similar way. We take the integral form of the

third equation in (E.6), and apply the same procedure as above. This leads us to

∂

∂t





e +

+χ



dV +



∂V





e +

+χ



=−



∂V

⊥



e +

+χ



−V

)dS +



∂V

(σ

−q

)dS. (E.63)

We split the stress tensor up as before and conﬂate the pressure and potential energy

term. Averaging the consequent result, putting u

= 0on∂V

⊥

, assuming ρ

= 0,

and neglecting deviatoric longitudinal stress, then leads (after a good deal of manip-

ulation) to the averaged energy equation

ρS



∂

∂t



e +



∂

∂x



e +



=−



∂S

ds −Su

∂

∂x

(p +ρχ) −

(E.64)

This is essentially Nye’s equation (E.51), if we neglect the kinetic energy terms

proportional to

, and put e = c

and χ =g(zcos α −x sin α). The details of

the algebraic manipulation form the substance of Question E.1.

To complete the derivation of (E.51), we need to constitute the heat ﬂux term



∂S

ds. In Nye’s equation, this is given by



∂S

ds =m



L +c

(θ

−θ

)



. (E.65)

848 E Averaged Equations in Two Phase Flow

To derive this, we go back to the jump conditions (E.9) and (E.10). First we note

that the right hand side of (E.65) is the jump in enthalpy −[h]

−

−h

, where

the enthalpy is h = e +

. Next we assume that the ice is at the melting point, so

that θ

is constant and there is no heat ﬂux from the ice to the interface. Therefore

the heat ﬂux in (E.65)is−



∂S

]

−

ds, and thus (E.65) will follow from the result

that

Γ [h]

−

=[q]

−

, (E.66)

where Γ is given by (E.8). The jump in enthalpy is related to the jump in internal

energy by the relation

[h]

−

=[e]

−

−pv, (E.67)

where v =1/ρ is the speciﬁc volume, and we deﬁne the change of volume v on

melting as

v =−[v]

−

. (E.68)

From (E.8), we derive

Γv=[u

]

−

, (E.69)

and thus from (E.9), we have

=−p −Γ(u

−u

−

) =−p −Γ

v. (E.70)

Hence we obtain

[σ

]

−

=−pΓ v −Γ

vu

, (E.71)

and thus (E.10) implies, using (E.67),

Γ [h]

−

=[q

]

−

+Γ

vu

, (E.72)

where we take [

]

−

= 0 assuming no slip at the interface; (E.66) and then also

(E.65) follow on neglecting the term Γ

vu

, which is comparable to the kinetic

energy of the ice.

E.7 Exercises

E.1 Consider the energy equation in the form of (E.63):

∂

∂t





e +

+χ



dV +



∂V





e +

+χ



=−



∂V

⊥



e +

+χ



−V

)dS +



∂V

(σ

−q

)dS,

E.7 Exercises 849

where the volume V is a short cylindrical segment of length δx, with the

ice/water interface being denoted as ∂V

⊥

with normal in the (e

, e

) plane,

and the ends being denoted as ∂V



, with normal in the e

direction. By using

the relationship that



V(t)

LdV =



V(t)

∂L

∂t

dV +



∂V

dS,

where V

is the normal velocity of ∂V , show that the energy equation can be

written in the form

∂

∂t





e +



dV +



∂V





e +



dS +



χdV

=−



∂V

⊥

ρ(u

−V

)



e +



dS −



∂V

⊥

−



∂V

⊥

(p +ρχ)u

dS −



∂V



(p +ρχ)u

dS,

where we take ∂V



to be ﬁxed in space, write σ

=−pδ

+τ

, and assume

that deviatoric longitudinal stress and longitudinal heat ﬂux are negligible, and

that u

=0on∂V

⊥

. What does the term



χdV represent physically?

Assuming now that ρ is constant, show that the averaged energy equation

can be written as

∂

∂t



ρSe +

ρSu



∂

∂x



ρSeu +

ρSu



=me −

∂

∂x



(p +ρχ)Su



−



∂S

(p +ρχ)u

ds −



∂S

ds,

where S is the cross-sectional area, and

m =−



∂S

ρ[u

−V

]ds.

Using this last equation, show that

=−



∂S

ds +S

and by assuming that p +ρχ is constant (why?) on ∂S, show that

−



∂S

(p +ρχ)u

ds =(p +ρχ )(Su)

where you should assume conservation of mass in the form

(ρS)

+(ρSu)

=m.