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1
Two-Phase Flow Equations
n
n
n
c a
z
k
kckk
Note: Walls are drawn such that xA = constant but this doesn’t have to hold true
zn = Unit vector in the flow direction
kn = Unit outward normal vector to the phase surface
ckn = Unit outward normal to the interface contained within the cross section (⊥ to zn )
2
Transverse View
ka = Cross sectional area of phase k
ic = Contour between phases
kc = Contour between phase k and the wall Limiting forms of Leibnitz Rule
1) ( , )k k
i kk kc
a a c
F dCF r t dS ds Fv nt t n n∂ ∂
= + ⋅∂ ∂ ⋅∫ ∫ ∫
and Gauss’s Theorem
2) k k i k
c
z kk kc
a a c c
dCBdS B n dS B nz n n
+
∂∇ ⋅ = ⋅ + ⋅
∂ ⋅∫ ∫ ∫
Similar to that shown for single phase flow
3) k z kz
c k kc
a n nB n dCz n n
∂ ⋅= = = −
∂ ⋅∫
Begin with the general balance equation written for phase k
4) ( )k k k sk gkvtψ ψ ψ ψ∂
+∇⋅ = −∇ ⋅ +∂
Integrate the general balance equation over the area ka
5) ( )k k k k
k k k sk gk
a a a a
dS v dS dS dStψ ψ ψ ψ∂
+ ∇⋅ = −∇⋅ +∂∫ ∫ ∫ ∫
and apply Gauss’s Theorem and Leibnitz Rule
3
6)
( )k k i k
k i k c
k k k z k k i kk kc
a a c c
sk z sk k gkk kc
a c c a
dCdS v n dS v v nt z n n
dCn ds n dSz n n
ψ ψ ψ
ψ ψ ψ
+
+
∂ ∂+ ⋅ + − ⋅ =
∂ ∂ ⋅
∂− ⋅ − ⋅ +∂ ⋅
∫ ∫ ∫
∫ ∫ ∫
Note: ( ) 0k
k k i kk kc
c
dCv v nn n
ψ − ⋅ =⋅∫ such that
7)
( )k k i
k i k c
k k k z k k i kk kc
a a c
sk z sk k gkk kc
a c c a
dCdS v n dS v v nt z n n
dCn ds n dSz n n
ψ ψ ψ
ψ ψ ψ+
∂ ∂+ ⋅ + − ⋅ =
∂ ∂ ⋅
∂− ⋅ − ⋅ +∂ ⋅
∫ ∫ ∫
∫ ∫ ∫
Mass Conservation
k kψ ρ= 0skψ = 0gkψ =
1) ( ) 0k k i
k k k z k k i kk kc
a a c
dCdS v n dS v v nt z n n
ρ ρ ρ∂ ∂+ ⋅ + − ⋅ =
∂ ∂ ⋅∫ ∫ ∫
Also
a
a
k1
k2
ak2
ak2
Under conditions such as sub-cooled nucleate boiling, we could have a situation such that at some interfaces we would have evaporation and at others condensation. To model such a situation, we would need mass, energy and momentum equations for each “species” of a phase, as well as terms to describe how one species transitioned to the other, i.e. 1 2k ka a→
4
Define ( )k k k i kv v nρΓ ≡ − ⋅ which represents mass exchange at the interface due to evaporation and condensation
2) 0k k i
k k k z kk kc
a a c
dcdS v n dSt z n n
ρ ρ∂ ∂+ ⋅ + Γ =
∂ ∂ ⋅∫ ∫ ∫
In terms of our averaging notation
3) 2 2 0i
k k k z k kk kc
c
dca v at z n n
ρ ρ∂ ∂< > + < > + Γ =
∂ ∂ ⋅∫
Let 2k k xa Aα=< >
i
k kk kc
c
dcn n
δ ′ ≡ Γ⋅∫
4) 2 2 2 2 0x k k k k k x kA v At z
ρ α ρ α δ∂ ∂ ′< > < > + < > < > + =∂ ∂
Momentum Conservation
k kvψ ρ=
( )sk k k kT P Iψ σ= = −
gk k gψ ρ=
5)
( )i
k k
k i k k i k k
kk k k k k z k k k i
c k kca a
k kk z k k z k k
k kc k kca c c a c c a
nv dS v v n dS v v v dCt z n n
n nP I n dS P I dC n dS dC gdSz n n z n n
ρ ρ ρ
σ σ ρ+ +
∂ ∂+ ⋅ + − ⋅ =
∂ ∂ ⋅
∂ ∂− ⋅ − ⋅ + ⋅ + ⋅ +∂ ⋅ ∂ ⋅
∫ ∫ ∫
∫ ∫ ∫ ∫ ∫
Project along the flow direction by dotting against the unit vector zn
6)
( )i
k k
k i k k i k k
kk k k k k k k k i
c k kca a
k z k kk k zzk k k z
k kc k kca c c a c c a
nv dS v v dS v v v dCt z n n
n n n nP dS P dC dS dC g dSz n n z n n
ρ ρ ρ
σ σ ρ+ +
∂ ∂+ + − ⋅ =
∂ ∂ ⋅
⋅ ⋅∂ ∂− − + + ⋅ +∂ ⋅ ∂ ⋅
∫ ∫ ∫
∫ ∫ ∫ ∫ ∫
5
7)
2 2 2 2
2 2 2 2
2 2
i
k i
x k k k k k k k x k kk kc
c
z kk k x k zzk k x
k kcc
k z k zk k k x k z
k kc k kcc c
dcA v v v A vt z n n
n nP A P dC Az n n z
n n n ndC dC A gn n n n
ρ α ρ α
α σ α
σ σ ρ α
∂ ∂< > < > + < > < > + Γ =
∂ ∂ ⋅
⋅∂ ∂− < > < > − + < > < >∂ ⋅ ∂
⋅ ⋅+ ⋅ + + < > < >
⋅ ⋅
∫
∫∫ ∫
Let: ˆi
k k k kk kc
c
dcv vn n
δ ′Γ =⋅∫
1 1 1 2z k z k k
k k k k k xk kc k kc
c c
n n n n aP dC P dC P P An n n n z z
α⋅ ⋅ ∂ ∂=< > = − < > = − < > < >
⋅ ⋅ ∂ ∂∫ ∫
1
k
k zk wk wk
k kcc
n n dC Pn n
σ τ⋅⋅ = − < >
⋅∫
1
i
k zk ik i
k kcc
n n dC Pn n
σ τ⋅=< >
⋅∫
where iP is the interfacial perimeter. Substituting into the area averaged momentum equation gives
8) { }
2 2 2 2
2 2 2 1 2 2 2
1 1 2 2
ˆx k k k k k k k x k k
k x k k k k x zzk k x
wk wk ik i k x k z
A v v v A vt z
A P P P A Az z z
P P A g
ρ α ρ α δ
α α σ α
τ τ ρ α
∂ ∂ ′< > < > + < > < > + =∂ ∂
∂ ∂ ∂− < > < > − < > − < > < > + < > < >
∂ ∂ ∂− < > + < > + < > < >
Internal Energy
9) ( ) )k k k k k k k ku u v P v qtρ ρ∂ ′′+∇ ⋅( = − ∇ ⋅ −∇ ⋅
∂
Again, the internal energy equation does not satisfy the general balance equation, but we can still employ our area averaging techniques. Integrate over the area ka , applying Leibnitz Rule and Gauss’s Theorem
6
10) ( ) ( )( )k k i
kk k k k k z k k k i
k kca a c
nu dS u v n dS u v v dCt z n n
ρ ρ ρ∂ ∂+ ⋅ + − ⋅
∂ ∂ ⋅∫ ∫ ∫
k
k k k z
a
P v ds q n dSz∂ ′′= − ∇ ⋅ − ⋅∂∫
k k i
k kk k
k kc k kca c c
n nq dC q dCn n n n
′′ ′′− ⋅ − ⋅⋅ ⋅∫ ∫ ∫
11)
2 2 2 2
i
k
x k k k k k k k x k kk kc
c
k k wk wk ik i
a
dCA u u v A vt z n n
P v dS q P q P
ρ α ρ α∂ ∂< > < > + < > < > + Γ =
∂ ∂ ⋅
′′ ′′− ∇ ⋅ + +
∫
∫
Let
2
k k
k k k k
a a
P v ds P v ds∇⋅ =< > ∇⋅∫ ∫
Note, the weighting is different thus the pressure is different than the other area averaged pressure.
2 2
k k k
kk k k k k z k
k kca a a c
nP v dS P v dS P v n dS v dCz n n
⎡ ⎤∂⎢ ⎥∇ ⋅ =< > ∇ ⋅ =< > ⋅ + ⋅⎢ ⎥∂ ⋅⎣ ⎦
∫ ∫ ∫ ∫
2 ( )k
k kk z k i k i
k k kc k kca c c
ndCP v n dS v v n v dCz n n n n
ρρ
⎡ ⎤∂⎢ ⎥=< > ⋅ + − ⋅ +⎢ ⎥∂ ⋅ ⋅⎣ ⎦∫ ∫ ∫
22 2 2 2
i
k kk k x k k i
k k kc k kcc c
P nd dCP v A P v dCdz n n n n
αρ
< >⎡ ⎤=< > < > < > + Γ + < > ⋅⎢ ⎥ ⋅ ⋅⎣ ⎦ ∫ ∫
Recall Leibnitz Rule
( , ) ( , ) ( , )k k
i k
k kca a c
v nF r t dS F r t dS F r t dCt t n n
⋅∂ ∂= +
∂ ∂ ⋅∫ ∫ ∫
2 2 2
k k
i kk k k
k kcc a a
v nP dC P dS P dSn n t t⋅ ∂ ∂
< > = < > − < >⋅ ∂ ∂∫ ∫ ∫
2 2k k k kP a a Pt t∂ ∂
= < > − < >∂ ∂ 2
kk
aPt
∂=< >
∂
7
Such that the internal energy equation is
12) 2 2 2
2 2 2 2 2
kx k k k k k k k x k k
k k kcc
k k k x k k x wk wk ik i
P dCA u u v A ut z n n
P v A P A q P q Pz t
ρ α ρ αρ
α α
⎡ ⎤< >∂ ∂< >< > + < > < > + Γ + =⎢ ⎥∂ ∂ ⋅⎣ ⎦
∂ ∂ ′′ ′′− < > < > < > − < > < > + +∂ ∂
∫
Jump Conditions Jump conditions provide the coupling of the phasic equations across the interface. The general form of the jump condition is 13) [ ] [ ]( ) ( )a a i sa a b b i sb bv v n v v nψ ψ ψ ψ− + ⋅ = − − + ⋅
Continuity (Mass)
0k k
sk
ψ ρψ
==
( ) ( )a a i a b b i bv v n v v nρ ρ− ⋅ = − − ⋅
a bΓ = −Γ
Note: a ac b bcn n n n⋅ = ⋅
a ba ac b bc
c c
dC dCn n n n
Γ = − Γ⋅ ⋅∫ ∫
a bδ δ′ ′= −
8
Momentum
k k kvψ ρ=
sk k kP Iψ σ= −
( ) ( )a a a i a a a b b b i b b bv v v P I n v v v P I nρ σ ρ σ⎡ ⎤ ⎡ ⎤− + − ⋅ = − − + − ⋅⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
Project along flow direction by dotting against zn
( ) ( )a a a i a z a z a b b b i b z b z bv v v P I n n n v v v P I n n nρ σ ρ σ⎡ ⎤ ⎡ ⎤− + ⋅ − ⋅ ⋅ = − − + ⋅ − ⋅ ⋅⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
a a a z a a z a b b b z a b z bv P n n n n v P n n n nσ σΓ + ⋅ − ⋅ ⋅ = −Γ − ⋅ + ⋅ ⋅
a a a z a a z a a a b z b b z ba ac b bc
dC dCv P n n n n v P n n n nn n n n
σ σ⎡ ⎤ ⎡ ⎤Γ + ⋅ − ⋅ ⋅ ⋅ = − Γ + ⋅ − ⋅ ⋅ ⋅⎣ ⎦ ⎣ ⎦⋅ ⋅
ˆ ˆ i i i i i
a a a z a a z a b b b z b b z ba ac c ac b bc b bc
c c c c c
dC dc dC dCv P n n n n v P n n n nn n n n n n n n
δ σ δ σ′ ′→ + ⋅ − ⋅ = − − ⋅ + ⋅⋅ ⋅ ⋅ ⋅∫ ∫ ∫ ∫ ∫
Assuming that pressure is continuous on the interface
1ˆ z aa a ia i a
a ac
n nv P Pn n
δ τ ⋅′ − < > +⋅ 1ˆ
i
z bb b ib i b
a acc
n ndC v P Pn n
δ τ ⋅′= − + < > −⋅∫
ic
dC∫
1 1ˆ ˆa a ia i b b ib iv P v Pδ τ δ τ′ ′− < > = − + < >
Total Energy
( ) ( )( ) ( )a a a i a a a a a b b b i b b b b be v v q P I v n e v v q P I v nρ σ ρ σ⎡ ⎤ ⎡ ⎤′′ ′′− + + − ⋅ ⋅ = − − + + − ⋅ ⋅⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
Note
2k k
k kv ve u ⋅
= +
9
( ) ( ) ( )2
aa a a i a a a i a a a i a i a a au v v v v v v q P v v P v v nρρ σ⎡ ⎤′′− + ⋅ − + + − + − ⋅ ⋅⎢ ⎥⎣ ⎦
( ) ( ) ( )2
bb b b i b b b i b b b i b i b b bu v v v v v v q P v v P v v nρρ σ⎡ ⎤′′= − − + ⋅ − + + − + − ⋅ ⋅⎢ ⎥⎣ ⎦
( ) ( )2
aa a a i a a a i a a i a a ah v v v v v v q P v v nρρ σ⎡ ⎤′′− + ⋅ − + + − ⋅ ⋅⎢ ⎥⎣ ⎦
( ) ( )2
bb b b i b b b i b b i b b bh v v v v v v q P v v nρρ σ⎡ ⎤′′= − − + ⋅ − + + − ⋅ ⋅⎢ ⎥⎣ ⎦
Assume the kinetic energy and viscous terms balance
( ) ( )2 2
a ba a a i a a b b b i b bv v v v v v v v v vρ ρσ σ⋅ − − ⋅ = ⋅ − − ⋅
such that [ ]( )a a a i a a i ah v v q P v nρ ′′− + + ⋅ [ ]( )b b b i b b i bh v v q P v nρ ′′= − − + + ⋅
Multiply both sides by k kc
dCn n⋅
and integrate over the interfacial contour
i i i
i i i
a a a a a i aa ac a ac a ac
c c c
b b b b b i bb bc b bc b bc
c c c
dC dC dCh q n P v nn n n n n n
dC dC dCh q n P v nn n n n n n
′′Γ + ⋅ + ⋅ =⋅ ⋅ ⋅
′′− Γ − ⋅ − ⋅⋅ ⋅ ⋅
∫ ∫ ∫
∫ ∫ ∫
Assume pressure is continuous on the interface, such that
ˆ
i
a a ia i a i aa ac
c
dCh q P P v nn n
δ ′ ′′− + ⋅⋅∫ ˆ
i
b b ib i b i bb bc
c
dCh q P P v nn n
δ ′ ′′= − + − ⋅⋅∫
ˆ ˆ
a a ia i b b ib ih q P h q Pδ δ′ ′′ ′ ′′− = − +
10
Two Phase Equation Summary Apply the area averaged two-phase equations to a liquid vapor system. We will again assume all average pressures are the same, the average of products is the product of averages and we can drop the average notation. Liquid Mass
x xA v At zα ρ α ρ δ∂ ∂ ′+ = −
∂ ∂
Vapor Mass
x g g g g g x gA v At zα ρ α ρ δ∂ ∂ ′+ = −
∂ ∂
Liquid Internal Energy
ˆx x x x w w i iA u u v A h P v A P A q P q P
t z z tα ρ α ρ δ α α∂ ∂ ∂ ∂′ ′′ ′′+ + = − − + +
∂ ∂ ∂ ∂
Vapor Internal Energy
ˆx g g g g g g g x g g g g g x g g x wg w ig iA u u v A h P v A P A q P q P
t z z tα ρ α ρ δ α α∂ ∂ ∂ ∂′ ′′ ′′+ + = − − + +
∂ ∂ ∂ ∂
Liquid Momentum
ˆ( ) ( )x x x w w i i x zPA v v v A v A P P A g
t z zα ρ α ρ δ α τ τ α ρ∂∂ ∂ ′+ = − − − + +
∂ ∂ ∂
Vapor Momentum
ˆ( ) ( ) gx g g g g g g g x g g g x wg wg ig i g x g z
PA v v v A v A P P A g
t z zα ρ α ρ δ α τ τ α ρ
∂∂ ∂ ′+ = − − − + +∂ ∂ ∂
Jump Conditions
gδ δ′ ′= −
ˆ ˆi i g g ig ih q P h q Pδ δ′ ′′ ′ ′′− = − +
ˆ ˆi i g g ig iv P v Pδ τ δ τ′ ′− = − +
11
The area averaged two-phase equations constitute the “Six Equation” model for modeling two phase systems as it contains six conservation equations, one for each phase. Assuming the interfacial and wall interaction terms can be expressed in terms of the following “fundamental” variables, we have as unknowns Fundamental Variables
, , , , ,k k k k k ku v P Tρ α → 12 Equations: Conservation Equations → 6 State Equations: ( , )k k k ku Pρ ρ= → 2
( , )k k k kT T u P= → 2 Volume Constraint 1k
kα =∑ → 1
An additional equation is require that relates the phase pressures
( )gP P P= → 1 which closes the system. A common assumption is that the phase pressures are equal, i.e.
gP P P= = which leads to the six equation, single pressure model that is the basis for most design codes. Liquid Mass
x xA v At zα ρ α ρ δ∂ ∂ ′+ = −
∂ ∂
Vapor Mass
x g g g g g x gA v At zα ρ α ρ δ∂ ∂ ′+ = −
∂ ∂
Liquid Internal Energy
ˆx x x x w w i iA u u v A h P v A P A q P q P
t z z tα ρ α ρ δ α α∂ ∂ ∂ ∂′ ′′ ′′+ + = − − + +
∂ ∂ ∂ ∂
12
Vapor Internal Energy
ˆx g g g g g g g x g g g g x g x wg w ig iA u u v A h P v A P A q P q P
t z z tα ρ α ρ δ α α∂ ∂ ∂ ∂′ ′′ ′′+ + = − − + +
∂ ∂ ∂ ∂
Liquid Momentum
ˆ( ) ( )x x x w w i i x zPA v v v A v A P P A g
t z zα ρ α ρ δ α τ τ α ρ∂ ∂ ∂′+ = − − − + +
∂ ∂ ∂
Vapor Momentum
ˆ( ) ( )x g g g g g g g x g g g x wg wg ig i g x g zPA v v v A v A P P A g
t z zα ρ α ρ δ α τ τ α ρ∂ ∂ ∂′+ = − − − + +
∂ ∂ ∂
Jump Conditions
gδ δ′ ′= −
ˆ ˆi i g g ig ih q P h q Pδ δ′ ′′ ′ ′′− = − +
ˆ ˆi i g g ig iv P v Pδ τ δ τ′ ′− = − +
13
Mixture Equations The Mixture Equations are a convenient form of the two-phase equations which are obtained by adding the phasic equations. The Mixture Equations provide global balances on mass, energy and momentum as opposed to balances on each phase. Mixture Mass
0
( ) ( ) 0x g g g g g x gA v v At zα ρ α ρ α ρ α ρ δ δ∂ ∂ ′ ′+ + + + + =
∂ ∂
Mixture Momentum
0
( ) ( ) ( )
ˆ ˆ
x g g g g g g g x x wg wg wl wl g g x z
i i g g ig i
PA v v v v v v A A P P A gt z z
v P v P
α ρ α ρ α ρ α ρ τ τ α ρ α ρ
δ τ δ τ
∂ ∂ ∂+ + + = − − − + +
∂ ∂ ∂′− + − +
Mixture Internal Energy
0
( ) ( ) ( )
ˆ ˆ
x g g g g g g g x g g x wg wg wL wL
i i g g ig i
A u u u v u v A P v v A q P q Pt z z
h q P h q P
α ρ α ρ α ρ α ρ α α
δ δ
∂ ∂ ∂ ′′ ′′+ + + = − + + +∂ ∂ ∂
′ ′′ ′ ′′− + − +
Mixture Variable Definitions
g gρ α ρ α ρ≡ +
g g gv v v Gρ α ρ α ρ≡ + =
g g gu u uρ α ρ α ρ≡ +
vv ρρ
≡
w w wg wg wl wlP P Pτ τ τ≡ +
w w wg wg w wq P q P q P′′ ′′ ′′= +
The mixture equations then simplify to
14
Mixture Mass
0x xA vAt zρ ρ∂ ∂+ =
∂ ∂
Mixture Momentum
( )x g g g g x x w w x zPA v v v v v A A P A g
t z zρ α ρ α ρ τ ρ∂ ∂ ∂
+ + = − − +∂ ∂ ∂
Mixture Internal Energy
( ) ( )x g g g g x g g x w wA u u v u v A P v v A q Pt z zρ α ρ α ρ α α∂ ∂ ∂ ′′+ + = − + +
∂ ∂ ∂
The convective terms in the momentum and internal energy equations are all of the form
g g g gv vα ρ φ α ρ φ+ . We wish to express these convective terms as mixture quantities plus correction terms, i.e.
+ Correction Termsg g g gv v vα ρ φ α ρ φ ρφ+ = ⋅ Define
g g gρφ α ρ φ α ρ φ≡ + Note: This is already consistent with our definitions of uρ and vρ above.
g g g
g g g
v v
vv
ρφ α ρ φ α ρ φ
ρρφ α ρ φ α ρ φρ
⎡ ⎤⋅ = +⎣ ⎦
⎡ ⎤⋅ = +⎣ ⎦
[ ]g g gg g g
v vv
α ρ α ρρφ α ρ φ α ρ φ
ρ+
⋅ = +
[ ] [ ]g gg g g g g g g gv v v v v
α ρα ρρφ α ρ φ α ρ φ α ρ φ α ρ φρ ρ
⋅ = + + +
[ ( )] [ ( )]g gg g g g g g g g g g g g gv v v v v v v v v
α ρα ρρφ α ρ φ α ρ φ α ρ φ α ρ φ α ρ φ α ρ φρ ρ
⋅ = + − − + + + −
15
( )g gvα ρ α ρ
ρφρ+
⋅ = [ ] ( ) ( )g g g gg g g g g g gv v v v v v
α ρ α ρ α ρ α ρα ρ φ α ρ φ φ φ
ρ ρ+ − − + −
( )( )g gg g g g g gv v v v v
α ρ α ρρφ α ρ φ α ρ φ φ φ
ρ⋅ = + − − −
Internal Energy Equation a) Convective Energy Term g g g g g g g gv v v u v uα ρ φ α ρ φ α ρ α ρ+ = + uφ = : g guφ = uφ = b) Pressure Work Term g g g g g gv v v vα ρ φ α ρ φ α α+ = + φ υ= : g gφ υ= φ υ=
( ) ( ) ( ) ( )( )
( )( )
g gx x x w w g g x
g gg g x
dA u uvA P vA q P u u v v At z dz z
P v v Az
α ρ α ρρ ρ
ρ
α ρ α ρυ υ
ρ
⎧ ⎫∂ ∂ ∂′′+ = − + − − −⎨ ⎬∂ ∂ ∂ ⎩ ⎭⎧ ⎫∂
− − −⎨ ⎬∂ ⎩ ⎭
Momentum Equation Momentum Flux Term g g g g g g g gv v v v v vα ρ φ α ρ φ α ρ α ρ+ = + vφ = : g gvφ = vφ =
2( ) ( ) ( )g gx x x w w x z g x
PA v vvA A P A g v v At z z z
α ρ α ρρ ρ τ ρ
ρ⎧ ⎫∂ ∂ ∂ ∂
+ = − − + − −⎨ ⎬∂ ∂ ∂ ∂ ⎩ ⎭
The Mixture Equations are not linearly independent from the phasic equations, and as such can be used to replace a phasic equation, but can not be used in addition to the phasic equations. Since the mixture equations reduce to the single phase equations for 0kα = , they are convenient for handling phase appearance and disappearance. Combinations of Mixture Equations and phasic equations can be produce various simplifications to the Six Equation Models.
16
5 Equation Models Five Equation Models are composed of 3 mixture and 2 phasic equations. Two common Five Equation models are: a) Phasic Equations = Mass + Momentum
Since we only have the mixture energy equation, information regarding energy distribution among the phases has been lost. Assume least massive phase to be at saturation (could be liquid or vapor
( , ) ( )k ksu P T u P= and ( , ) ( )k ksP u Pρ ρ= b) Phasic Equations = Mass + Energy
Since we have only the mixture momentum equation, information regarding relative phase velocity has been lost. The general approach it to correlate the relative velocity in terms of the other system variables, i.e.
( , , , )r g r k k kv v v v u Pα ρ= − = → Correlation
Drift Flux Models are a common example of this approach.
Four Equation Models
Four Equation Models are composed of three mixture equations and one phasic equation. The most common of the four equation models consist of the three mixture equations and the liquid phase mass equation. The liquid phase mass equation is
x xA v At zα ρ α ρ δ∂ ∂ ′+ = −
∂ ∂
Since liquid phase velocity alone does not appear in the mixture equations, we can eliminate liquid phase velocity in favor of mixture velocity and relative velocity by noting
g g gv vvvα ρ α ρρ
ρ ρ+
= =
g g g g g g gv v v v
vα ρ α ρ α ρ α ρ
ρ+ − +
=
( )g gvα ρ α ρ
ρ+
= ( )g ggv v v
α ρρ
+ −
17
( )g ggv v v v
α ρρ
= − −
Substituting into the liquid phase mass equation gives
( )g gx x g xA vA v v A
t z zα ρ α ρ
α ρ α ρ δρ
⎧ ⎫∂ ∂ ∂′+ = − + −⎨ ⎬∂ ∂ ∂ ⎩ ⎭
which is of the same general form as the mixture equations. As we have only the mixture momentum equation, a correlation is required for relative velocity. In addition, since we have only the mixture energy equation, the vapor phase is usually taken to be at saturation. The four equation model does allow for the liquid phase to be subcooled.
Three Equation Models
Three equation models are based solely on the mixture equations. A correlation is required for relative velocity and when two phases are present, they are both assumed to be at saturation. In either the 3, 4 or 5 equation models homogeneous flow is obtained by simply setting 0rv = .
18
Drift Flux Models The Drift Flux is defined in terms of the relative velocity as
( )g g g g rj v v vα α α α= − =
We can also define a Drift Velocity such that
( )gj g rV v v vα α= − =
For gravity dominated flows in the absence of wall shear, an equation which has been found to correlate a wide variety of data gives
ng gj vα α ∞=
where v∞ is the terminal rise velocity of a single bubble in an infinite fluid. The relative velocity, Drift Velocity and Drift Flux are then all proportional to the terminal rise velocity. One model for terminal rise velocity can be obtained from the following force balances. Buoyancy Force: ( )b gV gρ ρ− Surface Tension: bPσ
Drag Force: 2
2d bvC Aρ ∞
Balancing the Buoyancy and Surface Tension forces gives
( )b g bV g Pρ ρ σ− =
2ˆ( )
bb
b g
V dP g
σρ ρ
= ∝−
ˆ( ) b
g
dg
σρ ρ
∝−
Balancing the buoyancy and drag forces gives
2
( )2b g d bvV g C Aρ ρ ρ ∞− =
19
and solving for v∞
1
2
ˆ
( )2
b
gb
d b
C d
gVvC A
ρ ρρ∞
−=
2
2
2 2
( )2( )
( )2
g
d g
g
d
gv C
C g
gC
C
ρ ρσρ ρ ρ
σ ρ ρρ
∞
−=
−
−=
1/4
2
( )g gv C
σ ρ ρρ∞
−⎛ ⎞= ⎜ ⎟
⎝ ⎠
20
Numerical Solution of the Three Equation Model Numerical solution of the two phase equations is based on the assumption of Global Compressibility, i.e. the spatial pressure distribution is unimportant and system parameters can be evaluated at the global system pressure. This eliminates the need for a spatially discretized momentum equation.
Examine the open channel illustrated above. Fluid enters subcooled and can leave subcooled, a two phase mixture or superheated. We assume the 3 equation model is valid where the relative velocity r gv v v= − is available by an appropriate correlation. Consistent with the three equation model, both phases are at equilibrium when two phases are present. The boundary conditions are inlet velocity, pressure, density and internal energy. We also assume a know exit pressure. Equations: Mixture Mass
( ) 0x xA uvAt zρ ρ∂ ∂+ =
∂ ∂
Mixture Internal Energy
( ) ( ) ( ) ( )( )
( )( )
g gx x x w w g g x
g gg g x
dA u uvA P vA q P u u v v At z dz z
P v v Az
α ρ α ρρ ρ
ρ
α ρ α ρυ υ
ρ
⎧ ⎫∂ ∂ ∂′′+ = − + − − −⎨ ⎬∂ ∂ ∂ ⎩ ⎭⎧ ⎫∂
− − −⎨ ⎬∂ ⎩ ⎭
21
Integrate the mass and energy equations over a node centered at j and bounded by 1/ 2j ± Mass
1/2 1/2( ) ( ) 0j j x j x jdV vA vAdtρ ρ ρ+ −+ − =
Internal Energy
1/2 1/2 1/2 1/2
1/2 1/2
1/2 1/2
( ) ( ) ( ) [( ) ( ) ]
( )( ) ( )( )
j
j j x j x j x j x j j
j jg g g g
g g x g g xj j
S
dV u uvA uvA P vA vA qdt
u u v v A P v v A
ρ ρ ρ
α ρ α ρ α ρ α ρυ υ
ρ ρ
+ − + −
+ +
− −
+ − = − − +
⎧ ⎫ ⎧ ⎫− − − − − −⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭
The term jS contains the two phase correction terms and is zero for single phase flow.
Semi-Implicit Time Discretization W assume a Semi-Implicit time discretization, where velocities are evaluated at new time, and convected properties are evaluated at old time Mass
1/2 1/2 1/2 1/2 1/2 1/2 0t t tj j t t t t t t
j j j j j j jV v A v At
ρ ρρ ρ
+Δ+Δ +Δ
+ + + − − −
⎧ ⎫−⎪ ⎪+ − =⎨ ⎬Δ⎪ ⎪⎩ ⎭
Internal Energy
1/2 1/2 1/2 1/2 1/2 1/2
( ) ( )[( ) ] [( ) ]
t t tj j t t t t t t t t t t t t t
j x j x j j xj j xj j j
u uV u v A u v A P v A v A q S
tρ ρ
ρ ρ+Δ
+Δ +Δ +Δ +Δ+ − + + − −
⎧ ⎫−⎪ ⎪ ⎡ ⎤+ − = − − + +⎨ ⎬ ⎣ ⎦Δ⎪ ⎪⎩ ⎭ Consistent with our treatment of convected properties in the single phase equations, we assume any boundary valued property Ψ can be represented in terms of the cell centered properties by
{ } { }1/21/2 1 1
1/2
| |1 12 2
t tjt t t t t
j j j j jt tj
vv
+Δ+
+ + ++Δ+
Ψ = Ψ +Ψ + Ψ −Ψ
22
State Equations a) Single Phase
( , )j ju Pρ ρ=
( ) ( , )j j ju u P uρ ρ= b) Two Phase
( ) [ ( ) ( )]j g g
f gj g fP P P
ρ α ρ α ρ
ρ α ρ ρ
= +
= + −
( )
( ) ( ) ( ) ( ) ( ) ( )j g g g
f f gj f g f f
u u u
P u P P u P P u P
ρ α ρ α ρ
ρ α ρ ρ
= +
⎡ ⎤= + −⎣ ⎦
These state equations can be expressed in general as
( , )j j Pρ ρ ψ= ( ) ( , )j ju u Pρ ρ ψ= For Single Phase uψ = and Two Phase gψ α= The Mass and Internal Energy Equations are linear in the new time values. The equations are nonlinear as a result of the state equations. Linearize the equations via the Newton-Raphson technique Mass
11 1
1/2 1/2 1/2 1/2 1/2 1/2 0k tj j t k t k
j j j j j j jV v A v At
ρ ρρ ρ
++ +
+ + + + + +
⎧ ⎫−⎪ ⎪+ − =⎨ ⎬Δ⎪ ⎪⎩ ⎭
Internal Energy
11 1 1 1
1/2 1/2 1/2 1/2 1/2 1/2
( ) ( )[( ) ] [( ) ]
k tj j t k t k t k k t t
j x j x j j xj j xj j j
u uV u v A u v A P v A v A q S
tρ ρ
ρ ρ+
+ + + ++ − + + − −
⎧ ⎫−⎪ ⎪ ⎡ ⎤+ − = − − + +⎨ ⎬ ⎣ ⎦Δ⎪ ⎪⎩ ⎭
23
State
1 1 1( , ) ( ) ( )kj
k kk k k k k k kj j j j
jj
P P PP
ρ
ρ ρρ ρ ψ ψ ψψ
+ + +∂ ∂= + − + −
∂ ∂
1 1 1( , ) ( ) ( )kj
k kk k k k k k kj j j j
jju
u uu u P P PP
ρ
ρ ρρ ρ ψ ψ ψψ
+ + +∂ ∂= + − + −
∂ ∂
where the state equation derivatives are a) Single Phase
( , )u Pρ ρ=
uψ = Pu
ρ ρψ∂ ∂
=∂ ∂
uP P
ρ ρ∂ ∂=
∂ ∂
( , )u u P uρ ρ=
( , )P
u u P uu
ρ ρρψ
∂ ∂= +
∂ ∂
u
u uP Pρ ρ∂ ∂
=∂ ∂
b) Two Phase
( , ) ( ) [ ( ) ( )]g f g g fP P P Pρ α ρ α ρ ρ= + −
gψ α= g fρ ρ ρψ∂
= −∂
f ggP P P
ρ ρρ α α∂ ∂∂
= +∂ ∂ ∂
24
( , ) ( ) ( ) [ ( ) ( ) ( ) ( )]g f f g g g f fu P P u P P u P P u Pρ α ρ α ρ ρ= + −
( ) ( ) ( ) ( )g g f fu P u P P u Pρ ρ ρψ
∂= −
∂
f f g g
f f g g g
u uu u uP P P P P
ρ ρρ α ρ α ρ∂ ∂ ∂ ∂⎧ ⎫ ⎧ ⎫∂
= + + +⎨ ⎬ ⎨ ⎬∂ ∂ ∂ ∂ ∂⎩ ⎭ ⎩ ⎭
The linearized equations can be written in the form Mass
1 1 11 1/2 2 1/2 1
k t k t k tj j j j j ja v a v Sρ + + +
+ −+ + = Internal Energy ( ) 1 1 1
1 1/2 2 1/2 2k t k t k t
j j j j jju b v b v Sρ + + +
+ −+ + =
State
i) 1k k
k kj j j
jj
PP
ρ ρρ ρ δψ δψ
+ ∂ ∂= + +
∂ ∂
ii) 1( )k k
kkjj j
jj
u uu u PP
ρ ρρ ρ δψ δψ
+ ∂ ∂= + +
∂ ∂
Substitute i) into the Mass Equation, and ii) into the Internal Energy Equation
1 11 1/2 2 1/2 1
k kt k t k t k
j j j j j j jjj
pP a v a v SP
ρδψ δ ρψ
+ ++ −
∂ ∂+ + + = −
∂ ∂
1 11 1/2 2 1/2 2
k kkt k t k tjj j j j j j
jj
u uP b v b v S uP
ρ ρδψ δ ρψ
+ ++ −
∂ ∂+ + + = −
∂ ∂
Divide the Mass Equation by ρψ∂∂
, divide the Internal Energy Equation by uρψ
∂∂
and subtract to
eliminateδψ
25
iii) 1 1
1/2 1/2k k k k k kj j j j j ja v b v c Pδ ξ+ +
+ −+ + = For J nodes, the number of unknowns are J velocities for inlet velocity known and P (or Pδ ). We have the J equations implied by iii) above. One more equation is required. We assume a simplified Momentum Boundary Condition of the form
21/2 1/2( )
2
t t tJ J
exitvP P K ρ +Δ
+ += +
which can be linearized to give
( )1 11/21/2 1/2 1/22
2
tk k k kJ
exit J J JP P K v v vρ+ +++ + += + −
1 1
1/2k k k
J JP P P dv Cδ + ++= − = +
The linear system of equations may be written in matrix form, where the matrix structure is illustrated below where 1/2n nv v +=
1 1/21 1 1
22 2 2 2
33 3 3 3
1 J
va c vb a c v
b a c v
Cd P
ξξξ
δ
−⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
= ↓↓ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ↓↓⎜ ⎟⎜ ⎟⎜ ⎟
↓↓ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠
While not tridiagonal, solution algorithms can be developed for this structure that are similar to the Thomas Algorithm for tridiagonal systems.
26
Recirculating Systems The flow paths within U-Tube steam generators and Boiling Water Reactors may be approximated as the recirculating system illustrated below.
For the one dimensional segments contained between the inlet manifold (node 1) and the exit manifold (node J) the mass and internal energy equations are of the same form as the previous open channel problem
1/2 1/2j j j j ja v b v Pδ ξ+ −+ + = [2, 1]j J∈ − For the exit manifold, the mass and internal energy equations reduce to
1J exit J j J Ja v b v c v Pδ ξ+ + + = and for the inlet manifold (node 1)
2 3 2 2 2 1j ja v b v c v Pδ ξ+ + + = In addition, we have the momentum boundary condition at the steam outlet
ex exit Jd v P Cδ− + = Assuming the velocity 2jv is known, the unknowns are:
3jv , jv ( [2, 1]j J∈ − ), exitv , v , Pδ 2J→ +
27
The number of equation is J+1, the last equation is an integrated momentum equation around the entire loop. The mixture momentum equation (valid for both single and two phase conditions) is
2 220
2
1 [ ] )2 2
1sin ( )
x j jjx e
g gg x
x
P f v vv vvA k z zt A z z D
g v v AA z
ρ ρρ ρ φ δ ψ
α ρ α ρρ θ
ρ
∂ ∂ ∂+ = − − − ( −
∂ ∂ ∂
⎧ ⎫∂− − −⎨ ⎬∂ ⎩ ⎭
∑
or in non conservative form
2 220
2
)2 2
1sin ( )
j jje
g gg x
x
v v P f v vv k z zt z z D
g v v AA z
ρ ρρ ρ φ δ ψ
α ρ α ρρ θ
ρ
∂ ∂ ∂+ = − − − ( −
∂ ∂ ∂
⎧ ⎫∂− − −⎨ ⎬∂ ⎩ ⎭
∑
1
j
j
z
j z −
→∑∫ ∫ (integrate from center of one node to center of the next)
Results in an equation for the new iterate values of velocity of the form
1/2j j jj
B v e+ =∑
which closes the system of equations.