6.4 integral approximate solution

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

Chapter 6: Melting and Solidification

6.4 Integral Approximate Solution The melting and solidification problems that can be

solved by exact solution are very limited, so it is necessary to introduce some approximate solution techniques.

The integral approximate method proposed by Goodman (1958) is one of the most attractive techniques because it is very simple and its physical concept is very clear.

After the integral approximate solution technique is introduced by solving heat conduction in a semi-infinite body, its application to various melting/solidification problems will be discussed.

6.4 Integral Approximate Solution



6.4.1 Heat Conduction in a Semi-Infinite Body The integral approximate solution will be employed to solve the heat

conduction problem in a semi-infinite body with a boundary condition of the first kind.

The physical model of the problem is illustrated in Fig. 6.7, and the governing equation of the heat conduction problem and the corresponding initial and boundary conditions are:

(6.86)

(6.87)

(6.88)


2

2( ) 1 ( ) 0 0T x t T x t x tx tα

∂ , ∂ ,= > , >∂ ∂

0( ) 0 0T x t T x t, = = , >

( ) 0 0iT x t T x t, = > , =


Chapter 6: Melting and Solidification6.4 Integral Approximate Solution

Figure 6.8 Heat conduction in a semi-infinite body with boundary condition of the first kind.



(6.89)

(6.90)

Integrating eq. (6.86) in the interval (0, δ) one obtains

(6.91)

The right-hand side of eq. (6.91) can be rewritten using Leibnitz’s rule, i.e.,

(6.92)


( ) 0 ( )T x t x tx

δ∂ , = =∂

( ) ( )iT x t T x tδ, = =

( )

0( ) 0

1 ( )t

x t x

T T T x t dxx x t

δ

δ α= =

∂ ∂ ∂ ,− =∂ ∂ ∂∫

( )0( ) 0

1x

x t x

T T d dTdx Tx x dt dt

δ

δδ

δα =

= =

∂ ∂ − = − ∂ ∂ ∫



(6.92)

(6.93)

Equation (6.93) is the integral energy equation of the conduction problem, and this equation pertains for the entire thermal penetration depth.

It follows that a temperature distribution which satisfies eq. (6.93) does not necessarily satisfy differential eq. (6.86), which describes the energy balance at any and all points in the domain of the problem.


0

( )ix

T d Tx dt

α δ=

∂− = Θ −∂

( )

0( ) ( )

tt T x t dx

δΘ = ,∫



(6.95)

where A0, A1, A2 and A4 are four constants to be determined using the boundary conditions.

Since there are only three boundary conditions available – eqs. (6.87), (6.89) and (6.90) –one more condition must be identified so that all four constants in eq. (6.95) can be determined.

The surface temperature of the semi-infinite body, T0 is not a function of time t so

(6.96)


2 30 1 2 3( )T x t A A x A x A x, = + + +

( ) 0 0T x t xt

∂ , = =∂



(6.97)

Substituting eq. (6.95) into eqs. (6.87), (6.89), (6.90) and (6.97) yields four equations for the constants in eq. (6.95).

Solving for the four constants and substituting the results into eq. (6.95), the temperature distribution in the thermal penetration depth becomes

(6.98)


3

0

( ) 3 112 2

i

i

T x t T x xT T δ δ

, − = − + −

2

2

( ) 0 0T x t xx

∂ , = =∂



equation for δ is obtained:

(6.99)

Since the thermal penetration depth equals zero at the beginning of the heat conduction, eq. (6.99) is subject to the following initial condition:

(6.100)

(6.101)


0 0tδ = =

8 tδ α=

4 0d tdt

δα δ= >



From the above analysis, we can summarize the procedure of the integral approximate solution as follows:

1. Obtain the integral equation of the problem by integrating the partial differential equation over the thermal penetration depth

2. Assume an appropriate temperature distribution – usually a polynomial function – in the thermal penetration depth, and determine the unknown constants in the polynomial by using the boundary conditions at x=0 and x= Additional boundary conditions, if needed, can be obtained by further analysis of the boundary conditions and the conduction equation.

3. Obtain an ordinary differential equation of thermal penetration depth by substituting the temperature distribution into the integral equation. The thermal penetration depth thickness can be obtained by solving this ordinary differential equation.

4. The temperature distribution in the thermal penetration depth can be obtained by combining the thermal penetration depth thickness from step 3 into the temperature distribution from step 2.


δ



6.4.2 One-Region Problem In order to solve the problem by integral

approximate solution, the thermal penetration depth must be specified.

For a one-region melting problem, only the temperature distribution in the liquid phase needs to be solved because the temperature in the solid phase remains uniformly equal to the melting point of the PCM.

Therefore, the thickness of the liquid phase is identical to the thermal penetration depth.




Integrating eq. (6.42) with respect to X in the interval of (0, S), one obtains

(6.102)

(6.103)

Substituting eq. (6.45) into eq. (6.102) yields the integral equation of the one-region problem:

(6.104)


( ) 0

( ) ( ) ( )

X S X

X X dX X dτ

θ τ θ τ ττ= =

∂ , ∂ , Θ− =∂ ∂

( )

0( ) ( )

SX dX

ττ θ τΘ = ,∫

0

1 ( ) ( )

X

dS X dSte d X d

θ τ ττ τ=

∂ , Θ− − =∂



Assume now that the temperature distribution in the liquid phase is the following second-degree polynomial:

(6.105)

where A0, A1, and A2 are three unknown constants to be determined.

Equations (6.43) and (6.44) can be used to determine the constants in eq. (6.105).

However, eq. (6.45) is not suitable for determining the constant in eq. (6.105) because in eq. (6.45) is unknown.

Another appropriate boundary condition at the solid-liquid interface is needed.


dS dτ/

2

0 1 2( ) X S X SX A A AS S

θ τ − − , = + +



Differentiating eq. (6.44), one obtains

(6.106)

i.e.,

(6.107)

(6.108)

Equation (6.108) is an additional boundary condition at the solid-liquid interface; it can be used to determine the coefficients in eq. (6.105).

Thus, the constants in eq. (6.105) can be determined by eqs. (6.43), (6.44) and (6.108).


0 ( )d dX d X SXθ θθ τ τ

τ∂ ∂= + = =∂ ∂

0 ( )dS X SX dθ θ τ

τ τ∂ ∂+ = =∂ ∂

2 2

2SteX Xθ θ∂ ∂ = ∂ ∂



After the constants in eq. (6.105) are determined, the temperature distribution in the liquid phase becomes

(6.109)

Substituting eq. (6.109) into the integral equation (6.104) leads to an ordinary differential equation for :

(6.110)

The initial condition of eq. (6.110) is

(6.111)


( )S τ1 1 2 265 1 2 2

dS Ste SteSd Ste Steτ

− + +=+ + +

( ) 0 0S τ τ= =

21 1 2 1 1 2( ) 1Ste X S Ste X SXSte S Ste S

θ τ − + − − + − , = + +



(6.112)

(6.113)

Figure 6.9 shows the comparison of λ obtained by the exact solution and the integral approximate solution.

The integral approximate solution agrees very well for a small Stefan number, but the difference increases along with increasing Stefan number.

For a latent heat thermal energy storage system, the Stefan number is usually less than 0.2, so the integral approximate solution can provide sufficiently accurate results for that case.


122S λ τ=

121 1 2 23

5 1 2 2Ste SteSte Ste

λ − + += + + +



Figure 6.9 Comparison between integral and exact solutions of the one-region conduction problem




Example 6.3 A solid PCM with a uniform initial temperature at its melting

point, Tm is in a half-space, x>0. At time t=0 a variable heat flux is suddenly applied to the surface of the semi-infinite body. Assume that the densities of the PCM for both phases are the same, and that natural convection in the liquid phase is negligible. Find the transient location of the solid-liquid interface.



Solution: This problem is the same as Example 6.1 except for the

heat flux, which is (6.114)

where is a reference heat flux and is a given function. The dimensionless governing equation and the corresponding initial and boundary conditions of the problem are

(6.115)

(6.116)

0( ) ( )q t f t q′′ ′′=

0q′′ ( )f t

2

2 0 ( ) 0X SX

θ θ τ ττ

∂ ∂= < < , >∂ ∂

( ) 0 0f XXθ τ τ∂ = − = , >

∂



(6.117)

(6.118)

where the nondimensional variables in the above eqs. (6.115) – (6.118) are defined by eq. (6.57). Since eq. (6.116) is a nonhomogeneous boundary condition, the exact solution in Example 6.1 cannot be directly applied here. The integral approximate solution, on the other hand, has the capability of dealing with nonlinear and non-homogeneous boundary conditions.

( ) 0 ( ) 0X X Sθ τ τ τ, = = , >

1 ( ) 0dS X SX Ste dθ τ τ

τ∂− = = , >∂



Integrating eq. (6.115) with respect to X in the interval (0, S), and considering the boundary conditions at the surface, the integral equation of the problem is obtained:

where(6.120)

Integrating eq. (6.119) with respect to in the interval (0, ) yields

(6.121)

1 ( ) ( )( )dS dfSte d d

τ τττ τ

Θ− + =

( )

0( ) ( )

SX dX

ττ θ τΘ = ,∫

τ

0

( ) ( ) ( )S f dSte

ττ τ τ τ+ Θ = ∫



Assuming that the temperature distribution in the liquid phase is

(6.122)where the constants, A0, A1, and A2 are three unspecified

constants that can be determined from eqs. (6.116), (6.117) and (6.108). After all unknown constants are determined, the temperature distribution in the liquid phase becomes

(6.123)where

(6.124)

2

0 1 2( ) X S X SX A A AS S

θ τ − − , = + +

21 2 1 2 21 (1 4 ) 1 (1 4 )1 1( )2 8

X S X SXSte S Ste S

µ µθ τ

/ /

− + − +− − , = +

( ) ( )f S Steµ τ τ=



Substituting eq. (6.123) into eq. (6.121), one obtains(6.125)

For the case of constant heat flux, i.e., ,eq. (6.125) can be simplified to

(6.126)

2 1 2

0( ) ( ) 5 (1 4 )

6Ste f f d

τ µτ τ τ µ µ /

= + + +∫( ) 1f τ =

1 25 (1 4 ) 6S SteS SteS Steτ/

+ + + =



6.4.3 Two-Region Problem under a Boundary Condition of the Second Kind The physical model of a melting problem is shown in Fig. 6.9 – a

solid PCM with a uniform initial temperature Ti, which is below its melting point Tm is in a half-space, x>0.

At time t=0, a constant heat flux, , is suddenly applied to the surface of the semi-infinite body.

Because the initial temperature of the PCM is below its melting point, melting does not begin, until after the wall temperature reaches the melting point.

Therefore, the problem can be divided into two sub-problems: (1) heat conduction over the duration of preheating, (2) the actual melting process.

An integral approximate method will be employed to solve both sub-problems.


0q′′



Figure 6.10 Melting in a subcooled semi-infinite body under constant heat flux heating.



6.4.3.1 Duration of Preheating(6.127)

(6.128)

(6.129)

(6.130)

where tm is the duration of preheating. This problem is solved by integral approximate method.


22 2

22

1 0 0 mT T x t tx tα

∂ ∂= < < ∞ , < <∂ ∂

20

2

1 0 0 mT q x t tx k

∂ ′′= − = , < <∂

2( ) 0i mT x t T x t t, → → ∞ , < <

2( ) 0 0iT x t T x t, = , < < ∞ , =



Assuming the temperature profile is a second-degree polynomial, one obtains the temperature profile:

(6.131)

where δ is the thermal penetration depth. It can be obtained by substituting eq. (6.131) into the integral

equation, which can in turn be obtained by integrating eq. (6.127) in the interval of (0, δ).

The result is

(6.132)

(6.133)


20

22

( ) 12iq xT x t Tkδ

δ′′ , = + −

26 tδ α=

0

2

( )2s iqT t Tkδ′′

= +



Melting occurs when the surface temperature reaches the melting point, Tm, and the corresponding thermal penetration depth is

(6.134)

Then, the duration of pre-heating can be calculated from

(6.135)

The temperature distribution at time tm is

(6.136)


2

0

2 ( )m im

k T Tq

δ −=′′

2 22

22 0

2 ( )3

m im

k T Ttqα−=′′

2

2( ) ( ) 1m i m im

xT x t T T Tδ

, = + − −



6.4.3.2 Governing Equations for the Melting Stage After melting starts, the governing equations in

the different phases must be specified separately. The temperature in the liquid phase satisfies

(6.137)

(6.138)


21 1

21

1 0 ( ) mT T x s t t tx tα

∂ ∂= < < , >∂ ∂

10

1

( ) 1 0 mT x t q x t tx k

∂ , ′′= − = , >∂



(6.139)

(6.140)

(6.141)

At the solid-liquid interface, the following boundary conditions are necessary to link solutions in the liquid and solid phases:

(6.142)

(6.143)


2

2 22

2

1 ( ) mT T s t x t tx tα

∂ ∂= < < ∞ , >∂ ∂

(6.132)

2 ( ) i mT x t T x t t, → → ∞ , > (6.133)

2

2 ( ) ( ) 1 0i m i mm

xT x t T T T x t tδ

, = + − − > , =

(6.134)

1 2( ) ( ) ( )m mT x t T x t T x s t t t, = , = = , > (6.135)

2 12 1 ( )s mT T dsk k h x s t t tx x dt

ρ∂ ∂− = = , >∂ ∂ l (6.136)



(6.144)

1 1 2 2 21 2

1 1 0 1 1 0 1 1 0

1 22

1 1 0 1 1 0 1

( ) ( ) ( )

( )

p m p m p m i

s s s

s s s

mm

s s

c T T c T T c T TSc

h h hx sX Sh q h q h q

t Nh q h q α

θ θ

δα ρ α ρ α ρ

δ α ατα ρ α ρ α

− − −= = =

= = ∆ =′′ ′′ ′′/ / /

∆ = = =′′ ′′/ /

l l l

l l l

l l



21 12 0 ( ) mX S

Xθ θ τ τ τ

τ∂ ∂= < < , >∂ ∂

1( ) 1 0 mX XX

θ τ τ τ∂ , = − = , >∂

22 22

1 ( ) mS XX Nα

θ θ τ τ ττ

∂ ∂= < < ∞ , >∂ ∂

2 ( ) mX Sc Xθ τ τ τ, → − → ∞ , >

2

2 ( ) 1 1 0 mm

XX Sc Xθ τ τ τ , = − − > , = ∆

(6.145)

(6.146)

(6.147)

(6.148)

(6.149)



(6.150)

(6.151)

where ∆m and τm can be obtained by substituting eq. (6.144) into eqs. (6.129) and (6.135), i.e.,

(6.152)

(6.153)


1 2( ) ( ) 0 ( ) mX X X Sθ τ θ τ τ τ τ, = , = = , >

1 21 ( ) mdS X S

X N X dα

θ θ τ τ ττ

∂ ∂− + = = , >∂ ∂

2m N Scα∆ =

223m N Scατ =



6.4.3.3 Integral Approximate Solution Figures 6.10 shows the physical model represented by

the above dimensionless governing equations. Integrating eq. (6.147) over the interval then applying

the definition of the thermal penetration depth and eq. (6.150), yields the integral equation of the solid phase:

(6.154)

(6.155)


( )22

1X S

d ScX N dα

θτ=

∂− = Θ + ∆∂

2 2SdXθ

∆Θ = ∫



The temperature distribution in the solid phase is assumed to be a second-degree polynomial, i.e.,

(6.156)

The constants in eq. (6.156) can be obtained from the boundary condition, eq. (6.150), and the definition of the thermal penetration depth ( and ).

(6.157)

Substituting eq. (6.157) into eq. (6.154) yields a relationship between the location of the solid-liquid interface and the thermal penetration depth:

(6.158)


2 XScθ

= ∆= −

22 0 1 2( ) ( )B B X S B X Sθ = + − + −

2

2( ) 1XX ScS

θ τ ∆ − , = − ∆ −

6 2N dS dS d d

α

τ τ∆= +

∆ −



Integrating the differential equation of liquid phase eq. (6.145) over the interval of (0, S) and applying the boundary conditions eqs. (6.146) and (6.150) yields the integral equation of the liquid phase:

(6.159)

(6.160)

Assuming that the temperature profile in the liquid phase is a third-degree polynomial function,

(6.161)


1 11X S

dX dθ

τ=

∂ Θ+ =∂

1 10

SdXθΘ = ∫

2

1 0 1 2X S X SA A AS S

θ − − = + +



(6.162)

(6.163)

(6.164)

(6.165)


1 11 0 ( )d dX d X S

Xθ θθ τ τ

τ∂ ∂= + = =∂ ∂

1 1 0 ( )dS X SX dθ θ τ

τ τ∂ ∂+ = =∂ ∂

2 21 1 2 1

2

1X N X X Xα

θ θ θ θ∂ ∂ ∂ ∂ − = ∂ ∂ ∂ ∂ 2 2 2

1 2

1( )2 2S X S X SX p

S Sθ τ − − , = −



(6.166)

(6.167)

(6.169)

(6.170)


21 12 2

N ScS N ScSp SS S

α α = − + − + ∆ − ∆ −

( ) 12 1dS d dSc

d d dτ τ τΘ+ Θ + ∆ + =

21 ( 3 2 ) ( ) 3( )2 m mS p Sc S Sc τ τ+ + + + ∆ − ∆ = −

5 1 4 6S S S τ

+ + + =



6.4.4 Ablation under Constant Heat Flux Heating Ablation is an effective means of protecting the surfaces

of missiles and space shuttles from high-rate aerodynamic heating during atmospheric reentry.

It is a sacrificial cooling method because the protective layer is partially destroyed.

The advantage of the ablative cooling process is its self regulation: the rate of ablation is automatically adjusted in response to the heating rate.

The most commonly-used materials for ablative cooling are PCMs with higher melting points (such as glass, carbon, or polymer fiber) in combination with an organic binder.




The ablation material moves with a velocity –Ua in the moving coordinate system. The energy equation in a moving coordinate is

(6.171)

Since the ablating materials usually have very low thermal conductivity, it is reasonable to assume that the ablation occurs in a semi-infinite body. Therefore, the initial and boundary conditions of eq. (6.171) are

(6.172)

(6.173)

(6.174)


a p pT T Tk U c c

x x x tρ ρ∂ ∂ ∂ ∂ = − + ∂ ∂ ∂ ∂

, 0T T t∞= =

, 0mT T x= =

, 0,TT T xx∞

∂= = → ∞∂



Figure 6.12 Physical model for ablation.




(6.175)

Shortly after ablation begins, the process enters steady-state and the ablation velocity becomes a constant. Equation (6.171) can be simplified as

(6.176)

where the thermal properties are assumed to be independent of temperature.

Equation (6.176) can be treated as a first-order differential equation of dT/dx, and its solution is

(6.177)


0

, 0a sx

Tq U h k xx

ρ=

∂′′ − = − =∂l

2

2aUd T dT

dx dxα= −

1

aU xdT C edx

α−

=



(6.178)

(6.179)

(6.180)

(6.181)


12

aU x

a

CT e CU

αα −= − +

( )aU x

mT T T T e α−

∞ ∞= + −

( )a s p a mq U h c U T Tρ ρ ∞′′ − = −l

(1 )as

qUh Scρ

′′=

+l



(6.182)

(6.183)

(6.184)


( )p m

sl

c T TSc

h∞−

=

a a sq U hq q

ρ′′=

′′ ′′l

11

aqq Sc

′′=

′′ +



6.4.5 Solidification/Melting in Cylindrical Coordinate Systems Application of the integral approximate solution to 1-D solid-liquid

phase change problems – including ablation – in Cartesian coordinate system has been discussed in the preceding sections.

Since tubes are widely used in shell-and-tube thermal energy storage devices, it is necessary to study melting and solidification in cylindrical coordinate systems as well.

The polynomial temperature distribution is a very good approximation of the 1-D problem in the Cartesian coordinate system, but it can result in very significant error if it is used to solve for the phase change heat transfer in a cylindrical coordinate system.

This is because heat transfer area for a cylindrical coordinate system varies with the coordinate instead of remaining constant.

The temperature distribution in the coordinate has to be modified by taking into account the variation of the heat transfer area.




Figure 6.13 Solidification of an infinite

liquid PCM around an ID-cooled cylinder.




Conduction controls the solidification process because the temperature in the liquid phase is uniformly equal to the melting point of the PCM.

The governing equation and the initial and boundary conditions of this problem are

(6.185)

(6.186)

(6.187)


1 1 ( ) 0iT Tr r r s t t

r r r tα∂ ∂ ∂ = < < > ∂ ∂ ∂

0( ) 0iT r t T r r t, = = >

( ) 0mT r t T r s t, = = >



(6.188)

where the subscript “1” for solid phase has been dropped for ease of notation.

(6.189)

(6.190)

(6.191)


0sT dsk h r s tr dt

ρ∂ = = >∂ l

1 1 ( ) 0R R SR R R

θ θ τ ττ

∂ ∂ ∂ = < < > ∂ ∂ ∂

( ) 1 1 0R Rθ τ τ, = = >

0

m

m

T TT T

θ −=− i

rRr

=i

sSr

= 2i

trατ = ( )0p m

sl

c T TSte

h−

=



(6.192)

(6.193)

The above eqs. (6.190) – (6.193) are also valid for melting around a hollow cylinder when used with the appropriate dimensionless variables.


( ) 0 ( ) 0R R Sθ τ τ τ, = = >

1 0dS R SR Ste dθ τ

τ∂− = = >∂



It is very useful here to recall that the logarithmic function appears in the exact solution of the steady state heat conduction problem in a cylindrical wall.

Therefore, we can assume that the temperature distribution has a second-order logarithmic function of the form:

(6.194)

where φ is an unknown variable. Equation (6.194) automatically satisfies eqs. (6.191) and

(6.192), and can be obtained by differentiating eq. (6.192):

(6.195)


2ln ln1 (1 )ln lnR RS S

θ ϕ ϕ = + − +

0dSR dθ θ

τ τ∂ ∂+ =∂ ∂



Substituting eqs. (6.190) and (6.192) into eq. (6.195) yields

(6.196)

Substituting eq. (6.194) into eq. (6.196) gives the following expression for :

(6.197)

Substituting eq. (6.194) and (6.197) into eq. (6.193)(6.198)

Subjected to the initial condition(6.199)


1 2 12 SteSte

ϕ + −+ =

1 2 1ln

dS Sted S Sτ

+ −=

( ) 1 0S τ τ= , =

2 1 0Ste RR R R Rθ θ∂ ∂ ∂ − + = ∂ ∂ ∂

R S= 0τ >



Integrating equation (6.198) over the time interval results in the following equation for the location of the solid-liquid interface:

(6.200)

which is valid for solidification around a cylinder with a constant inner temperature.

However, in practical applications, cooling inside the tube is usually achieved by the flow of cooling fluid through the tube.

Therefore, the boundary condition at the inner surface of the tube should be a convection boundary condition instead of an isothermal boundary condition.


( )2 21 1ln ( 1) 1 2 12 4S S S Ste τ− − = + −



Example 6.4 An infinite liquid PCM has a uniform initial temperature

equal to the melting point of the PCM, Tm. At time t = 0, a cooling fluid with temperature Ti flows inside the tube. The heat transfer coefficient between the cooling fluid and the inner surface of the tube is hi. Find the transient location of the solid-liquid interface.


Chapter 6: Melting and Solidification6.4 Integral Approximate Solution Solution: The governing equations of the problem can also be

represented by eq. (6.185) – (6.188) except that eq. (6.186) needs to be replaced by the following expression:

(6.201)The dimensionless governing equations of the problem are

the same as eqs. (6.190) – (6.193), but eq. (6.191) needs to be replaced by the dimensionless form of eq. (6.201), i.e.,

(6.202)

( ) 0i i iTk h T T r r tr

∂ = − = >∂

( 1), 1 0Bi RRθ θ τ∂ = − = =

∂



Bi in eq. (6.202) is the Biot number defined as

(6.203)and θ is defined as

(6.204) It is also assumed that the temperature distribution has a

second-order logarithmic function of the form

(6.205)

ihrBik

=

m

m i

T TT T

θ −=

−

2ln ln( )ln lnR RA B A BS S

θ = + − +



The two unknown variables and in eq. (6.205) can be determined by substituting eq. (6.205) into eqs. (6.202) and (6.196), with the result that

(6.206)

(6.207) Substituting eq. (6.206) into eq. (6.207), an equation for

A is obtained as follows: (6.208)

Substituting eq. (6.205) into eq. (6.193), the ordinary differential equation for the location of the solid-liquid interface is obtained:

(6.209)

(1 ) lnB Bi A S= − −1 2 12 ASteA B

Ste+ −+ =

1 2 12 (1 ) ln ASteA Bi A SSte

+ −− − =

1 2 1ln

dS ASted S Sτ

+ −=



Eq. (6.209) is subject to the initial condition specified by eq. (6.199).

The temperature of the inner surface of the tube is (6.210)

The solid-liquid interface location can be obtained by numerical solution of eq. (6.209) and (6.210). It should be noted that A becomes 1 if the Biot number becomes infinite. In that case the temperature of the inner surface becomes 1 and eq. (6.209) reduces to eq. (6.198).

(1 ) Aθ τ, =

6.4 integral approximate solution

Documents