5.5 transport effects at the interface chapter 5: solid

Transport Phenomena in Multiphase Systems by A. Faghri & Y. Zhang 1

Chapter 5: Solid-Liquid-Vapor Phenomena and Interfacial Heat and Mass Transfer

5.5 Transport Effects at the Interface5.5.1 Interfacial Mass, Momentum, and Energy Balances

In fluid mechanics and heat transfer, the conservation laws can be reduced to local partial differential equations if they are considered at a point that does not belong to a surface of discontinuity, such as an interface.

When considering a discontinuous point, appropriate jump conditions relating the values of the fundamental quantities on both sides of the interface should be considered.

Jump conditions at an interface were discussed in Section 3.2.6, but the effects of surface tension and disjoining pressure associated with a liquid-vapor interface were not taken into account.

It is the objective of this subsection to specify mass, momentum, and energy balance at a liquid-vapor interface.

5.5 Transport Effects at the Interface




At a liquid-vapor interface, the mass balance is(5.61)

The velocity components should be defined according to these directions, as follows:

(5.62)(5.63)(5.64)

( ) ( )I v v Imδ ρ ρ′′ = − ⋅ = − ⋅V V n V V nl l&

V⋅ = nV n11 V⋅ = tV t

22 V⋅ = tV t



For a liquid-vapor interface, the mass balance is(5.65)

The momentum balance at the liquid-vapor interface(5.66)

Including the Marangoni effect the momentum balance at the interface becomes

(5.67)

In this equation, the tangential direction, t, can either be t1 or t2. The stress tensor is:

(5.68)

( ) ( )I v v Imδ ρ ρ′′ = − = −V V V Vl l&

( ) ( )v vmδ′′− ⋅ = −τ τ n V Vl l&

( ) 1 1( ) ( )v d vI II

dT p T mR R dT δ δ

σσ ′′− ⋅ + + − − ∇ = −

τ τ n n n V Vl l&


( )223

p µ µ′ = − + − ∇ ⋅I D V Iτ




The deformation tensor can be written for a reference frame that is adjusted to the interface:

(5.69)

( )

1 2

1 2

1 1 1 2

1 1 2 1

2 1 2 2

2 2 1 2

1 12 2

1 1 12 2 2

1 12 2

T

V VV V Vx x x x x

V V V VVx x x x x

V V V VVx x x x x

∂ ∂∂ ∂ ∂+ + ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ = ∇ + ∇ = + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ + + ∂ ∂ ∂ ∂ ∂

t tn n n

n t n t n

t t t tn

t n t t t

t t t tn

t n t t t

D V V




The normal direction of the interface is [1 0 0], the first tangential direction is [0 1 0] and the second tangential direction is [0 0 1]. Therefore,

(5.70)

This can be reduced to the three components to obtain:

(5.71)(5.72)

(5.73)

[ ]

[ ]1 2

1 2

0 0

1 1 22 0 02 2 3

p

V VV V Vx x x x x

µ µ

′ ⋅ = − +

∂ ∂∂ ∂ ∂+ + − ∇ ⋅ ∂ ∂ ∂ ∂ ∂

t tn n n

n t n t n

n

V

τ

1 2

1 2

2 4 223 3 3

V VV Vp px x x x

µ µ µ µ ∂ ∂∂ ∂′ ⋅ ⋅ = − + − ∇ ⋅ = − + − + ∂ ∂ ∂ ∂

t tn n

n n t tn n Vτ

1

1

1

VVx x

µ ∂∂′ ⋅ ⋅ = + ∂ ∂

tn

t nn tτ

2

2

2

VVx x

µ ∂∂′ ⋅ ⋅ = + ∂ ∂

tn

t nn tτ




The momentum equation balance at the interface is then broken into its three components, as follows: Normal Direction

(5.74)

Tangential 1

(5.75)

(5.76)

( )

1 2 1 2

1 2 1 2

, ,

, ,

, , , ,

, ,

43

23

1 1

vv v

v

v vv

d vI II

V Vp p

x x

V V V Vx x x x

p m V VR R δ

µ µ

µ µ

σ

∂ ∂− + + − ∂ ∂

∂ ∂ ∂ ∂− + − + ∂ ∂ ∂ ∂

′′+ + − = −

n n

n n

t t t t

t t t t

n n

ll l

l

l ll

l&

( )1 1

1 1

1 1 1

, ,, ,, ,

vvv v

V VV V Td m V Vx x x x dT x

δδ

σµ µ ∂ ∂∂ ∂ ∂ ′′+ − + − = − ∂ ∂ ∂ ∂ ∂

t tn nt t

t n t n t

lll l&

( )2 2

2 2

2 2 2

, ,, ,, ,

vvv v

V VV V Td m V Vx x x x dT x

δδ

σµ µ ∂ ∂∂ ∂ ∂ ′′+ − + − = − ∂ ∂ ∂ ∂ ∂

t tn nt t

t n t n t

lll l&




The non-slip condition at the liquid-vapor interface requires that and The momentum balance at the tangential directions becomes

(5.77)

(5.78)

If the liquid and vapor phases are further assumed to be inviscid, the momentum equation at the interface can be reduced to

(5.79) The energy balance at the interface can be obtained from eq.

(3.180), i.e.,(5.80)

1 1, ,vV V=t tl

2 2, , .vV V=t tl

1 1

1 1 1

, ,, , vvv

V VV V Tdx x x x dT x

δσµ µ ∂ ∂∂ ∂ ∂ + = + + ∂ ∂ ∂ ∂ ∂

t tn n

t n t n t

lll

2 2

2 2 2

, ,, , vvv

V VV V Tdx x x x dT x

δσµ µ ∂ ∂∂ ∂ ∂ + = + + ∂ ∂ ∂ ∂ ∂

t tn n

t n t n t

lll

1 1( )v dI II

p p T pR R

σ

− = + −

l

( )2 2, ,

, ,( ) ( )2 2v rel rel

v rel v v rel vm e eδ

′′ ′′ ′ ′ ′′− ⋅ − ⋅ ⋅ + ⋅ ⋅ = + − +

V Vq q n nτ V n τ V ll l l l&



If the velocity of the reference frame is taken as the interfacial velocity, eq. (5.80) can be rewritten as

(5.81)

Eq. (5.81) can be rewritten in terms of enthalpy

(5.82)

The stress tensor in eq. (5.82) can be expressed as

(5.83)

( )2 2

( ) ( ) ( ) ( )

( ) ( )2 2

v v I v v I

v I Iv

k T k T

m e eδ

∇ − ∇ ⋅ − ⋅ ⋅ − + ⋅ ⋅ −

− − ′′= + − +

n nτ V V n τ V V

V V V Vl l l l

ll&

( )2 2

( ) ( ) ( ) ( )

1 12 2

v v I v v I

vv v v I I

v

k T k T

p pm hδ ρ ρ

∇ − ∇ ⋅ − ⋅ ⋅ − + ⋅ ⋅ −

′′= − − + − ⋅ − + ⋅

n nτ V V n τ V V

V V V V V V

l l l l

ll l l

l

&


( )223

p pµ µ′ = − + − ∇ ⋅ = − +I D V I Iτ τ




Since the relative velocities at the interface satisfy and , we have

(5.84) Substituting eqs. (5.83) and (5.84) into eq. (5.82), the energy

balance can be written as

(5.85)

To simplify the energy equation, the kinetic energy terms are considered negligible and no-slip conditions are assumed at the interface,

(5.86)

( ) /I mδ ρ′′− ⋅ =V V nl l& ( ) /v I vmδ ρ′′− ⋅ =V V n &

( ) ( ) vI v v I

v

p pp p mδ ρ ρ ′′− ⋅ − − ⋅ = − −

V V n V V n ll l

l

&

( ) ( ) ( ) ( ) ( )2 21 1

2 2

v v I v v I

v v v I I

k T k T

m hδ

∇ − ∇ ⋅ − ⋅ − + ⋅ −

′′= + − ⋅ − + ⋅

n n V V n V V

V V V V V V

l l l l

l l l&

τ τ

, , ,v IV V V= =t t tl




Therefore, the energy equation can be rewritten as

(5.87)

where are kinematic viscosities of liquid and vapor phases, respectively.

The energy balance at the interface can be simplified by assuming that the change in the kinetic energy across the interface is negligible, i.e.,

(5.88)

1 2

1 2

1 2

1 2

, ,,

, ,,

4 23 3

4 2 3 3

vv

v

v vvv v

T Tk kx x

V VVm h

x x x

V VVx x x

δ ν ν

ν ν

∂ ∂− ∂ ∂

∂ ∂∂′′= + − + ∂ ∂ ∂

∂ ∂∂− + + ∂ ∂ ∂

n n

t tn

n t t

t tn

n t t

ll

l lll l l&

( )v v vk T k T m hδ′′∇ − ∇ ⋅ =nl l l&




The total species mass flux at an interface is:(5.89)

The velocity of species i in phase k and phase j is and, respectively. These velocities are defined as:

(5.90)(5.91)

Using the interfacial species mass balance, and substituting the definition of the species velocity, the interfacial species mass flux is:

(5.92) Remembering the overall mass conservation at the interface,

(5.93) The interfacial species mass flux is:

(5.94)

( ) ( ), , , ,i k i k i I j i j i Im ρ ρ′′ = − ⋅ = − ⋅V V n V V n&

, , , ,k i k i k i k i k kρ ω ρ= +V J V

, , , ,j i j i j i j i j jρ ω ρ= +V J V

( ) ( ), , , ,i k i k i k k I j i j i j jk Im ω ρ ω ρ′′ = ⋅ + − = ⋅ + −J n V V J n V V&

( ) ( )k k I j j Im ρ ρ′′ = − ⋅ = − ⋅V V n V V n&

, , , ,i k i k i j i j im m mω ω′′ ′′ ′′= ⋅ + = ⋅ +J n J n& & &




In some problems, the species mass flux will be specified, and the total mass flux is simply a sum of all the species mass fluxes.

(5.95) If the species mass flux is not specified, the total mass flux at an

interface can be calculated from the interfacial species mass flux equation, eq. (5.94):

(5.96)

(5.97)(5.98)

For a binary mixture, the species diffusion flux, J, can be calculated by Fick’s law.

(5.99)(5.100)

ii

m m′′ ′′= ∑& &

( ), ,

, ,

j i k i

k i j im

ω ω

− ⋅′′ =

−

J J n&

( ), , ,k i k i j iω ω ω=

( ), , ,j i j i k iω ω ω=

,1 ,12 ,1k k k kDρ ω⋅ = − ∇ ⋅J n n,1 ,12 ,1j j j jDρ ω⋅ = − ∇ ⋅J n n




(a) Liquid to Gas (b) Solid to Gas

Figure 5.12 Species concentration and mass transfer from solid and or liquid to a gas mixture.

GasLiquid

y

GasSolid

y ,I. A gx

,II. A gx

,II. A sx

,I. A gx

,II. A gx

,I. A sx,I. Ax l

xA

0

1




The partial vapor pressure of A in the gas mixture at the interface can be approximated from Raoult’s Law:

(5.101)

(5.102)If the solid or liquid is made of pure species A, then

and eqs. (5.101) and (5.102) reduce to (5.103)

Knowing and total pressure p, one can easily calculate the mole fraction and mass fraction of species A at the interface on the gas side by the following relation

(5.104)

(5.105)

( ), ,0 0Liquid to gasA A A saty y

p x p− += == l

( ), ,0 0Solid to gasA A s A saty y

p x p− += ==

, ,0 01A A sy y

x x= =

= =l

,0A A satyp p−= =

0A yp −=

0,

A yA g

px

p−==

,,

, ,

A g AA g

A g A B g B

x Mx M x M

ω =+




Mass transfer from a solid to a gaseous state sometimes requires the specification of diffusion molar flux, rather than concentration, at the solid surface.

(5.106) Catalytic surfaces are used to promote heterogeneous reactions

(see Chapter 3), which occur at the surface; the appropriate boundary condition is

(5.107) In the liquid, Henry’s law will relate the mole fraction of species A in

the liquid to the partial vapor pressure of A in the gas mixture at the interface by following relation

(5.108)

* AA AB

xJ cDy

∂= −∂

( )*1 0

n

A A yJ k c =′= −

0, 0

A yA y

px

H−

+

==

=l




a. Gas to Liquid b. Gas to Solid

Figure 5.13 Species concentration and mass transfer from gas mixture to liquid or solid

GasLiquid

0

,I. A sx

,I. A gx

,II. Ax l

,I. A gx

,II. A gx

Gas

Solid

y

,II. A gx

,II. A sx

,I. Ax l

1

Ax




The concentration of gas in a solid at the interface is usually obtained by the use of a property known as the solubility, S, defined below

(5.109) For a liquid/solid interface during melting and solidification, the ratio

of the species concentration of the solid in liquid phases is called the partition ratio, Kp.

(5.110)

, ,0 0A s A gy yc S p+ −= =

=

,

,

A sp

A

cK

c=

l




Example 5.4 Write the continuity, momentum, energy, and species equations and the

necessary boundary conditions for the horizontal evaporating capillary tube in Fig. 5.15, which is open at one end and closed at the other. The capillary tube is on the order of 100 μm. The evaporation is driven by the concentration gradient of vapor in the air. The evaporation cools the interface while the wall heats the fluid, causing a temperature gradient along the interface. Since the surface tension is a function of temperature, Maragoni stresses are created due to the temperature gradient along the interface. The wall is at a constant temperature, which is the ambient air temperature. The meniscus of a volatile liquid in ambient conditions with no forced heating recedes into the tube because of evaporation. The fluid density and viscosity are constant. Gravitational effects are negligible because the tube lies horizontally and the pore diameter is small enough that free convection can be considered negligible. The disjoining pressure effects are put into the contact angle, because the thin film region of the capillary diameter has minimum effects on the total heat transfer and evaporation. The gas in the vapor region is assumed to be an ideal gas. Assume quasi-steady state and that the specific heats of the vapor and air are the same.




Solution: Figure 5.16 shows the control volume, which were chosen due to

symmetry, and the boundaries identified with numerical indices of the problem. The incompressible Navier-Stokes equations are solved, and the fluid is considered to be Newtonian. The continuity and momentum equations are

(5.111)(5.112)

The energy equation in the liquid region is(5.113)

If the solid wall is being modeled, only the conduction in the solid is modeled, since only the steady-state solution is of interest.

(5.114)

0∇ ⋅ =V( )D p

Dtρ µ= − ∇ + ∇ ⋅ ∇V V

( ) ( )h h k Tt

ρ ρ∂ + ⋅ ∇ = ∇ ⋅ ∇∂

V

2 0T∇ =




The species equation for vapor phase is(5.115)

In the energy equation, the enthalpy, h, is(5.116)

This condition is used for both the liquid and the gaseous regions, because the vapor has a specific heat similar to air. The density in the liquid is constant, and the density of the gas is determined from the ideal gas law.

(5.117) The diffusion coefficient, D, in the species equation is

(5.118)

( ) ( )Dt

ρ ω ρ ω ρ ω∂ + ⋅ ∇ = ∇ ⋅ ∇∂

V

( )p refh c T T= −

1ref

uv air

p

R TM M

ρω ω

= −+

3/ 2TD Bp

=




The viscosity and thermal conductivity are considered to be constant in the liquid region, while the mass weighted average of these properties is used in the gaseous region.

(5.119) At the liquid/vapor interface (boundary 1 in Fig. 5.13), the interfacial

boundary conditions are Conservation of Mass

(5.120) The subscript g refers to the mixture of vapor and air.

Conservation of Normal Momentum(5.121)

Conservation of Tangential Momentum(5.122)

where γ is the surface temperature change with temperature .

( )1v airφ ω φ ω φ= + −

( ) ( )1 I g g Im D ωρ ρ ρω

∇ ⋅′′ = = − ⋅ = − ⋅−n V V n V V nl l&

( ) ( )1 2I g gm p p K Kσ′′ − ⋅ + − = +V V nl l&

( ) ( )( )g g Tµ µ γ− ∇ ⋅ + ∇ ⋅ ⋅ = ∇ ⋅V t V t n tl l




Conservation of Energy(5.123)

Saturation Mass Fraction(5.124)

No-Slip Velocity in the Tangential Direction(5.125)

Continuity of Temperature(5.126)

The surface tension is a function of temperature and is approximated by the following equation:

(5.127)

g vk T k T h m′′− ∇ ⋅ + ∇ ⋅ =n nl l &

satω ω=

g=V Vl

gT T=l

( )refT T refT Tσ σ γ= + ⋅ −




The mass fraction at the liquid/vapor interface was found using the molar concentration of an ideal gas

(5.128)

(5.129) The partial pressure of the vapor is a very crucial parameter in the

evaporation process. The function of pressure related to temperature from Yaws (1992) is used.

(5.130) Since the free surface is always a function of the radial location, f,

the two curvatures, K1 and K2, are:

(5.131)

satg

pxp

=

( )1g g

satg g g air

x Mx M x M

ω =+ −

( )log132.953

p baT c

= −+

( ) 32

12 1

fK

f

′′= −

′ +




(5.132) The first and second derivates of f should be taken by differentiating

the face center locations of the interface. The pressure at the face closest to the wall is calculated with a central differences scheme, but the point of intersection between the wall and the liquid/vapor interface is calculated from the prescribed contact angle, θ.

(5.133)

where a is the radial distance from the wall to its adjacent face and b is the radial distance from the wall to the second face from the wall.

2 2 1

fKr f

′= −

′ +

2 2

2

2

2

1tan

1

a b

w

a aa x xb b

xab

θ

− + − =

−




If the interface is close to the tube mouth, and xw exceeds xm, then the prescribed location of intersection between the interface and the wall to calculate the pressure is xm. The conditions at the inner wall (the boundaries labeled 3, 5, and 8-b in Fig. 5.16) are

(5.134)(5.135)(5.136)

Far from the tube mouth, ambient air conditions of pressure, temperature, and mass fraction are constants, These conditions apply to a quarter circle with a radius of 6r0, as seen in Fig. 5.16 (boundaries 6 and 7):

(5.137)(5.138)(5.139)

refT T=0=V

0ω∇ ⋅ =n

refp p=If 0 else 0refT T T⋅ ≤ → = ∇ ⋅ =V n n

0ω =




At the tube liquid entrance (boundary 10) the boundary conditions are:

(5.140)(5.141)

At the axis of axi-symmetric geometry (boundaries 2, 4, and 8-a), the boundary conditions are

(5.142)(5.143)(5.144)

,

20

I j jj

m A

rρ π

′′

⋅ =∑

V n&

0T∇ ⋅ =n

( ) 0∇ ⋅ ⋅ =V t n0⋅ =V n

0T∇ ⋅ =n



5.5 Transport Effects at the Interface( X Y ) 2 7 S e p 2 0 0 4

0 10 20 30 40 50time (s)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

Dist

ance

from

tube

mout

h(m

m)

θ =π /24, numerical (present)θ =π /12, numerical (present)θ =π /6, numerical (present)Experimental (Buffone and Sefiane 2004)

( X Y ) 2 7 S e p 2 0 0 4

Figure 5.17 Distance of meniscus center distance from mouth vs. time for

acetone, for a tube with a diameter of 600 µm (Rice

and Faghri, 2005a).



5.5.2 Interfacial Resistance in Vaporization and Condensation High-heat transfer coefficients, typically associated with

evaporation and condensation processes in heat transfer devices, are restricted by interfacial resistance.

From kinetic theory, the mass flow rate (of molecules) passing in either direction (right or left) through an imagined plane is given by

(5.145)

where j is the flux of molecules. The net molecular flux through an interface is


1/ 2

1/ 22v

u

M pmR Tδ π

=

&



The net molecular flux through an interface is(5.146)

Net mass flux at the interface(5.147)

where α is the accommodation coefficient, Ru is the universal gas constant, Mv is the molecular weight of the vapor, and the function Γ is given by

(5.148)

(5.149)

2v v

v u v

q M p pmh R T T

δδ α

π ′′ Γ′′ = = −

l

l l

&

( ) ( ) ( )[ ]aaaa erf1exp 2 ++=Γ π

( ) ( ) ( )[ ]aaaa erf1exp 2 −−=−Γ π


m m mδ δ δ+ −′′ ′′ ′′= −& & &



where(5.140)

and(5.151)

Assuming pl and pv are the saturation pressures corresponding to Tl and Tv eq. (5.147) can be represented in the following form

(5.152)

2v

v v u v

q Mah R Tδ

ρ′′

=l

( ) ∫ −=a x dxea

0

22erfπ

( ) ( )sat sat

2vv

vu v

p T p TMq hR T Tδ α

π Γ

′′ = −

ll

l




For evaporation and condensation of working fluids at moderate and high temperatures eq. (5.146) can be approximated by

(5.153) Substituting eq. (5.153) into eq. (5.147)

(5.154)

Assuming and using Clausiius-Clapeyron relation

(5.155)

Heat transfer coefficient is obtained by(5.156)

πa+=Γ 1

22 2

v v

v u v

q M p pmh R T T

δδ

αα π

′′ ′′ = = − − l

l l

&

( )22 1

2 2 2v v v v

vv v u v v

h M p vq T TT v R T hδ

αα π

′′ = − − − l l

ll l

( )22 1

2 2 2v v v v

v v v u v v

q h M p vhT T T v R T h

δδ

αα π

′′ = = − − − l l

l l l


( ) 1/ < <− vv ppp ( ) 1/ < <− vv TTT

,



5.5.3 Formation of and Heat Transfer Through Thin Liquid Films


Figure 5.18 Cross-section of the characteristic element of (a) an axially-grooved condenser, and (b) an evaporator



Local heat flux through the film(5.157)

Fully-developed laminar liquid flow velocity profile(5.158)

Assuming constant vapor pressure, and liquid flow is driven by surface tension and adhesion forces

(5.159)

The continuity and energy equations for the evaporating liquid layer(5.160)

wT Tq k δ

δ−′′ = l

( )21 22

dpuds

η δ ηµ

= − −ll

l

( )

−+−+−=

ρρρσσ δ

δ

δ

112,

2

vvv

d vdsd

dsdT

dTdK

dsdp

dsdK

dsdp

0v

d qu dds h

δη

ρ′′

=∫ ll l




Substituting eqs. (5.157-5.159) into eq. (5.160) gives the relation for the thickness of the evaporating film

(5.161)

The film surface curvature K expressed in terms of the solid surface curvature Kw and film thickness

(5.162)

Assume that the vapor core pressure is related to vapor temperature by the saturation conditions

(5.163) For small heat flux, interfacial resistance is defined as

(5.164)

( ) ( )313

wd

v

k T Td d p Kds ds h

δδ σµ ρ δ

− − = l

l l l

2/32

2

2

1−

++=

dsd

dsdKK w

δδ

( )v sat vp p T=

( )sat22 2

v v

g v

ph pqR T T

δ

δ

αα π

′′ = − − − l




Relation between vapor pressure and saturation pressure is given by the extended Kelvin equation

(5.165)

Combining eq. (5.157) and eq. (5.158)

(5.166)

Under specific conditions, a non-evaporating film thickness is present theat gives the equality of the liquid-vapor interface and solid surface temperatures

(5.167) Substituting eq. (5.167) into eq. (5.166)

(5.168)

( ) ( )( ) ( )exp sat sat dsat sat

g

p p T p Kp p TR T

δ δδ δ

δ

σρ

− + −= l

( )22 2

v v satw

g v

h p pT Tk R T T

δδ

δ

δ αα π

= + − − l

l

wT Tδ =

( ) wsat v

v

Tp pTδ =




Substituting eq. (5.168) and eq. (5.165)

(5.169)

At non-evaporating film thickness, the disjoining pressure can be obtained by eq. (5.51)

(5.170)

Combining eq. (5.169) and eq. (5.170)

(5.171)

For water, logarithmic dependence of disjoining pressure on liquid film thickness

(5.172)

( ) ( )lnw v w

d v sat w g wv sat w v

T p Tp p p T R T KT p T T

ρ σ

= − + + + l

0B

dp A δ −′= −

( ) ( )

B

v

w

wsat

vwgwsat

v

wv K

TT

TppTRTp

TTp

A

/1

0 ln'

1−

−

−−= σρδ

=

b

gd aTRp3.3

ln δρ δ





Example 5.5 Derive the expression for the equilibrium film thickness of

water on a wall. If the wall temperature Tw = 101 °C and the saturated vapor temperature Tv = 100 °C, find the non-evaporating film thickness δ0. The density of the liquid water can be assumed to be .




Solution: The disjoining pressure for water at nonevaporating film thickness is

expressed by eq. (5.172). The nonevaporating film thickness can be obtained by rearranging eq. (5.172), i.e.,

Substituting eq. (5.169) into the above equation, one can obtain the non-evaporating film thickness for water as follows:

1/

013.3 exp

b

d

g

pa R T bδ

δρ

= l

1/

0( ) /13.3 exp ln

( )

b

sat w v w v v w

g w sat w v

p T p T T K p Ta R T p T T

σδ

ρ

− + = + l




Since the molecular weight of water is 18, the gas constant for the water vapor is Rg = 8314/18 = 461.9 J/kgK. The wall and vapor temperatures are respectively and Tw = 101 + 273 = 374K and Tv = 100 + 273 = 373 K. The curvature of the liquid film is assumed to be negligible, i.e., K = 0.

The vapor pressure is the saturation pressure corresponding to the vapor temperature, i.e.,

5( ) 1.013 10 Pav sat vp p T= = ×




The saturation pressure corresponding to the wall temperature can be obtained from the Table B.48 as psat (Tw) = 1.0547x105 Pa. Therefore, the nonevaporating film thickness is

It can be seen that the non-evaporating film is extremely thin.

1/

0( ) /13.3 exp ln

( )

b

sat w v w v v w

g w sat w v

p T p T T K p Ta R T p T T

σδ

ρ

− + = + l

5 5

1/0.02435

85

1 1.0547 10 1.013 10 374 / 3733.3 exp1.5336 1000 461.9 374

1.013 10 374ln 1.51 10 m 0.0151μm3731.0547 10

−

× − × ×= × × ×

× + = × = ×



5.5.4 Heat Transfer in the Thin-Film Region of an Axially-Grooved Evaporator Khrustalev and Faghri (1995) modeled an

evaporating extended meniscus in a capillary groove, shown in Fig. 5.10(b).

Most of the heat is transferred through the region where the liquid layer is extremely thin.

The significance of the temperature difference between the saturated vapor core and the interface has been stressed in the model.




Generalized capillary pressure(5.174)

Total heat flow rate per unit length in the microfilm region

(5.175)

The following first order equations should be considered with their respective boundary conditions

(5.176)

(5.177)


dpKp −≡ σcap

( ) 1 1

mic 1 0 0'

/s swT Tq s ds q ds

kδ

δ− ′′= ≡∫ ∫l

'δδ =dsd

( )

+

−+=

−

r

B

RAp

dsd 1'

'1' cap2/32

σδ

δδ




Figure 5.19 Thin evaporating film on a fragment of the rough solid surface



(5.178)

(5.179)

(5.180)

(5.181)

(5.182)

(5.183)


( )capmic3

3 'v

dpq s

ds hν

δ= − l

l

mic'/

wdq T Tds k

δ

δ−=l

00δδ =

=s

0'0

==s

δ

B

rs

AR

p −

=+

+−= 0

00cap 'δ

δσ

mic 0' 0

sq

==



Though the initial-value problem described by eqs. (5.176-5.183) is completely determined, its solution must satisfy one more condition

(5.184)

Based on the geometry in Fig. 5.18(5.185)

The heat flow rate per unit groove length in the transition region

(5.186)

The heat flow rate per unit groove length in both the microfilm and transition region

(5.187)


mencap

1 Rp

ss

σ==

( ) 2/1

men22

menmen22

0 sin2 frr xRxRRxRR θδδ +++−−−+=

'/

tr

f

x wtr x

T Tq dxk

δ

δ−= ∫l

mic 0 /trx w

trT Tq q dx

kδ

δ−′ ′+ = ∫l




Figure 5.20 Characteristics of the evaporating film along the

solid-liquid interface (ammonia, Tv=250 K): Free liquid surface temperature; (b) Thickness of film; (c) Generalized capillary

pressure




Figure 5.21 Heat flux through the evaporating film (ammonia,

Tv=250 K, α=1): (a) Along the solid-

liquid interface (microfilm region); (b) Along the fin

axis (Rr=33µm, ∆T=1K)




Example 5.6 A schematic of a pulsating heat pipe (PHP) with one end

sealed and the other end open is shown in Figs. 5.22 and 5.23 (Zhang and Faghri, 2002). The evaporator section of the heat pipe is near the closed end, and its length is Lh. The condenser section with a length of Lc is near the open end of the heat pipe, and the adiabatic section with length La is located between the evaporator and condenser sections. When the evaporator is heated, the vapor pressure increases. The liquid plug moves toward the open end because the vapor pressure, pv, is higher than the environment pressure, pe. As a result, the volume of the vapor slug is increased and part of the vapor slug is exposed to the cooled section of the heat pipe, where vapor is condensed to liquid.




When the rate of condensation exceeds the rate of evaporation, the vapor pressure will decrease. When the vapor pressure, pv, is decreased to a value below the environment pressure, pe, the liquid plug will be pushed back to the closed end. At this point, the rate of evaporation again exceeds the rate of condensation, which enables the vapor pressure to increase and pushes the liquid plug to the open end. This process is repeated, and the oscillation of the liquid plug can be maintained. As the liquid slug moves toward the open end of the pulsating heat pipe, the trailing edge of the liquid plug leaves a thin liquid film on the pipe wall. Evaporation and condensation over this thin liquid film are the driving forces of pulsation flow in a PHP with an open end. Obtain the equation that governs the liquid film thickness in 0<x<L1.




Solution: Figure 5.23 shows the physical model of the evaporator section of a

PHP with an open end. The continuity equation of the liquid film is

(5.188)where q is the rate of heat through a given cross section due to phase

change and is defined as follows(5.189)

The revised latent heat of evaporation is defined as(5.190)

to account for the contribution of the sensible heat.For small Reynolds numbers (less than unity), a fully-developed

laminar liquid velocity profile can be assumed.(5.191)

0 ,

12 v e

qu d mR h

δη

π ρ

= − ′ ∫ l l

l l

&

0

( )2s h ck T Tq R dsπ

δ−

= ∫ l

, 0.68 ( )v e v p h vh h c T T′ = + −l l l

( )21 22

dpudx

η δ ηµ

= − −ll

l




Substituting eq. (5.191) into eq. (5.188), the continuity equation in the liquid film becomes

(5.192) The pressures in the vapor and liquid phases have the following

relationship:(5.193)

where the curvature is

(5.194)and the disjoining pressure is

(5.195)where a = 1.5336 and b = 0.0243.

3,

32 v e

dp qmdx hR

µπ ρ δ

− = − ′

l ll

ll

&

v dp p K pσ− = −l

3/ 222

211 cos arctand d dK

dx R dxdxδ δ δ

δ

− = + − −

ln3.3

b

d g vp R T a δρ =

l




It is assumed that the pressure in the vapor phase is constant. By differentiating eq. (5.193), we have

(5.196) Combining eqs. (5.192) and (5.196), one obtains

(5.197) The boundary conditions of eq. (5.197) are

(5.198)

(5.199)

( )ddp d K pdx dx

σ− = −l

3,

3( )2d

v e

d qK p mdx hR

µσπ ρ δ

− = − ′

ll

ll

&

0, 0d xdx

δ = =

2

2

1 , 0

0,

p htr

p h

x Ld R xdx x L

δ δ < −= = <




(5.200)

(5.201)where is nonevaporating liquid film thickness. The liquid film

thickness can be found by numerical solution or by an approximate solution of eq (5.197). (Zhang and Faghri, 2002).

0 1, x Lδ δ= =

1tan ,d x Ldx

δ α= =

0δ

5.5 transport effects at the interface chapter 5: solid

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