3.10 CONDUCTION-CONTROLLED FREEZING AND MELTING Heat conduction with freezing melting occurs in a number of applications, such as ice formation, permafrost melting, metal casting, food p
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and the steady periodic temperature distribution is given by
(Bi2+ 2Bi + 2)1/2 e−√πXcos
where
θ =T − T ∞,m
X = ω 2πα
1/2
τ = ωt
Bi=h
k
2 α
1/2
Note that ash −→ ∞, Bi −→ ∞, and β −→ 0, eqs (3.333) reduce to eq (3.328)
withT ∞,m = T i The presence ofthe factor Bi/(Bi+2Bi+2)1/2in eq (3.333) shows
that convection enhances the damping effect and that it also increases the phase angle
by an amountβ = arctan 1/(1 + Bi).
3.9.5 Finite Plane Wall with Periodic Surface Temperature
Consider a plane wall ofthicknessL with the face at x = 0 insulated and the face at
x = L subjected to a periodic temperature change ofthe form
whereT i is the initial temperature ofthe wall and the insulated boundary condition
atx = 0 gives
∂T (0, t)
The steady periodic solution is
T (x, t) = T i + aφ1
x
L ,
ω 2α
1/2
L
cos
ωt + φ2
x
L ,
ω 2α
1/2
L
(3.337)
where the numerical values ofφ1as a function ofx/L and (w/2α)1/2 L are supplied
in Table 3.13, andφ2, which is also a function ofx/L and (w/2α)1/2 L, is given by
φ2= arctanφaφd− φbφc
φaφb− φcφd (3.338)
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TABLE 3.13 Values of the Amplitude Decay Function,φ as a Function of x/L
andσ = (ω/2α)1/2 L
σ x/L = 0 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000
∞ 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
where
φa = cos ω
2α
1/2
L
cosh ω 2α
1/2
L
(3.339a)
φb = cos ω
2α
1/2
x
cosh ω
2α
1/2
x
(3.339b)
φc= sin ω
2α
1/2
L
sinh ω 2α
1/2
L
(3.339c)
φd = sin ω
2α
1/2
x
sinh ω 2α
1/2
x
(3.339d)
3.9.6 Infinitely Long Semi-infinite Hollow Cylinder with Periodic Surface Temperature
Consider an infinitely long cylinder ofinside radius,r i, extending to infinity in the radial direction The inner surface is subjected to a periodic temperature variation of the form
T (r i , t) = T i + a cos ωt (3.340) whereT i is the initial temperature ofthe cylinder The equation governing the
tem-perature distribution is
∂2T
∂r2 +1
r
∂T
∂r =
1 α
∂T
The other boundary condition is
T (∞, t) = T i , ∂T (∞, t) ∂r = 0 (3.342)
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and the initial condition is
An application ofthe method ofcomplex combination gives the steady-state periodic solution as
θ =T − T i
where
Λ1= ker
√
ω/α r i ker√
ω/α r+ kei√
ω/α r i kei√
ω/α r
ker2√
ω/α r i+ kei2√
Λ2= ker
√
ω/α r i kei√
ω/α r− kei√
ω/α r i ker√
ω/α r
ker2√
ω/α r i+ kei2√
and ker and kei are the Thomson functions discussed in Section 3.3.5
As indicated in Section 3.9.1, the method ofcomplex combination is described by Arpaci (1966), Myers (1998), Poulikakos (1994), and Aziz and Lunardini (1994) The method may be extended to numerous other periodic heat problems ofengineering interest
3.10 CONDUCTION-CONTROLLED FREEZING AND MELTING
Heat conduction with freezing (melting) occurs in a number of applications, such as ice formation, permafrost melting, metal casting, food preservation, storage of latent energy, and organ preservation and cryosurgery Books and review articles on the subject include those ofLunardini (1991), Cheng and Seki (1991), Rubinsky and Eto (1990), Aziz and Lunardini (1993), Viskanta (1983, 1988), and Alexiades and Solomon (1993) Because ofthe vastness ofthe literature, only selected results that are judged to be offundamental importance are discussed in this section
3.10.1 One-Region Neumann Problem
The one-region Neumann problem deals with a semi-infinite region ofliquid initially
at its freezing temperature,T f At timet > 0, the face at x = 0 is suddenly reduced
and kept atT0such thatT0 < T f, as shown in Fig 3.39 This initiates the extraction ofheat by conduction from the saturated liquid to the surface and the liquid begins to freeze As the cooling continues, the interface (assumed sharp) between the solid and liquid phases penetrates deeper into the liquid region The prediction ofthe location ofthe interface calls for determination ofthe one-dimensional transient temperature distribution in the solid assuming that the liquid continues to remain atT fat all times.
The governing partial differential equation is
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Figure 3.39 One-region Neumann problem (freezing)
∂2T
with the boundary conditions
wherex fdenotes the location ofthe interface, which is not known a priori and must
be determined as part ofthe solution An energy balance at the interface gives
k ∂T
∂x
x=x f
= ρL ∂x f
wherek and ρ are the thermal conductivity and density ofthe solid phase, respectively,
andL is the latent heat.
The temperature distribution in the solid is given as
T f − T
T f − T0
= 1 − erf
x/2√αt
erf
wherex f /2√αt, denoted by λ, is a root ofthe transcendental equation
√
π λ erfλeλ 2
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TABLE 3.14 Interface Location Parameter λ
and
St=c(T f − T0)
is the Stefan number, the ratio ofthe sensible heat to the latent heat For water, St
is about 0.10, for paraffin wax about 0.90, for copper about 2.64, and for silicon dioxide about 436 Table 3.14 gives selected values ofλ and St that satisfy eq (3.350)
Viskanta (1983) reports that the Neumann model accurately predicts the solidification
of n-octadecane on a horizontal plate.
The solution presented here applies to the one-region melting problem ifT f is replaced by the melting temperatureT m WithT0> T m, eq (3.349) gives the temper-ature in the liquid region
3.10.2 Two-Region Neumann Problem
The two-region Neumann problem allows for heat conduction in both the solid and liquid phases For the configuration in Fig 3.40, the mathematical description ofthe problem is
∂2T s
∂x2 = α1
s
∂T s
∂t (0 < x < x f andt > 0) (3.352)
∂2T l
∂x2 = α1
l
∂T l
∂t (x f < x < ∞ and t > 0) (3.353)
with initial and boundary conditions
T s (x f , t) = T l (x f , t) = T f (3.354c)
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k s ∂T s
∂x − k l
∂T l
∂x
x=x f
= ρL dx f
where the subscriptss and l refer to the solid and liquid phases, respectively.
The solutions forT sandT lare
T s − T0
T f − T0
= erf(x/2
√
αs t)
erf(x f /2√αs t) (3.355)
T i − T l
T i − T f =
erfc(x/2√αl t)
erfc(x f /2√αs t) (3.356)
With x f /2√αs t denoted by λ, the interface energy balance given by eq (3.354d)
leads to the transcendental equation forλ:
e−λ 2
λ erf(λ) −
T i − T f
T f − T0
(kρc) l
(kρc) s
1/2
e−λ 2(α s /α l )
λ erfc λ(α s /α l )1/2
√
π L
c(T f − T0) (3.357)
Churchill and Evans (1971) noted thatλ is a function of three parameters:
θ∗ =T T i − T f
f − T0
(kρc)
l
(kρc) s
1/2
α∗ =ααs
l St=c(T f − T0)
L
and solved eq (3.357) for a range ofvalues ofthese parameters Table 3.15 summa-rizes these results forλ
T f
T l
x f
T0
T s
k s s,␣
k l, ␣l
,L
Figure 3.40 Two-region Neumann problem (freezing)
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TABLE 3.15 Values of λ
θ∗
3.10.3 Other Exact Solutions for Planar Freezing
Besides the one- and two-region Neumann solutions, several other exact solutions for planar freezing problems are available These include:
1 The two-region problem with different solid- and liquid-phase densities (Lu-nardini, 1991)
2 The two-region problem with phase change occurring over a temperature range (Cho and Sunderland, 1969)
3 The one-region problem with a mushy zone separating the pure solid and pure liquid phases (Solomon et al., 1982)
4 The two-region problem with temperature-dependent thermal conductivitiesk s
andk l(Cho and Sunderland, 1974)
5 The two-region problem with arbitrary surface temperature and initial condi-tions (Tao, 1978)
3.10.4 Exact Solutions in Cylindrical Freezing
Carslaw and Jaeger (1959) give an exact solution for the freezing of a subcooled liquid while the solid phase remains at the freezing temperature The latent heat released is used to bring the subcooled liquid to its freezing temperature The process is described
by the differential equation
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1
r
∂
∂r
r ∂T
∂r
α
∂T
and the initial boundary conditions
lim
wherer f represents the radial growth ofthe solid phase andT0< T fis the subcooled liquid temperature
The solution ofeq (3.358) satisfying the conditions ofeqs (3.359) is
T = T0+T f − T0
Ei− r2/4αt
Ei
− r2
whereλ = r2
f /4αt is given by
λ2· Ei− λ2
eλ 2
In eqs (3.360) and (3.361), Ei is the exponential integral function discussed in Section 3.3.4, and in eq (3.361), St is the Stefan number Table 3.16 provides the solution of
eq (3.361)
Another situation for which an exact solution is available is shown in Fig 3.41
A line heat sink ofstrengthQ s(W/m) located atr = 0 and activated at time t = 0
causes the infinite extent ofliquid at a uniform temperatureT i (T i > T f ) to freeze.
The interface grows radially outward The mathematical formulation for the solid and liquid phases leads to
1
r
∂
∂r
r ∂T s
∂r
αs
∂T s
∂t (0 < r < r f ) (3.362)
1
r
∂
∂r
r ∂T ∂r l
=α1
l
∂T l
∂t (r f < r < ∞) (3.363)
with initial and boundary conditions
T s (r f , t) = T l (r f , t) = T f (3.364c)
k s ∂T s
∂r − k l
∂T l
∂r
r=r f
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TABLE 3.16 Stefan Number and Interface Location Parameter
Ozisik (1993) gives the solution as
T s = T f+4πkQ s
s
Ei
−4αr2
s t
− Ei
2
f
4αs t
(0 < r < r f ) (3.365)
T l = T i− T i − T f
Ei
− r2
f /4α l tEi
− r2
4αl t
(r f < r < ∞) (3.366)
whereλ = r f /2√αs t is obtained from the transcendental equation
Q s
4πe−λ
2
+ k l (T i − T f )
Ei(−λ2αs /α l ) e−λ
2αs /αl = λ2αs ρL (3.367)
The solution presented by eqs (3.365) through (3.367) has been extended by Ozisik and Uzzell (1979) for a liquid with an extended freezing temperature
Figure 3.41 Cylindrical freezing due to a line strength of fixed strength
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3.10.5 Approximate Analytical Solutions
Because ofthe mathematical complexity and the restrictive nature ofexact analyti-cal solutions, several approaches have been employed to generate approximate an-alytical solutions that provide rapid results in a number ofpractical situations The methods used are the quasi-steady solution, the heat balance integral approach of Goodman (1958) and Lunardini (1991), and the perturbation method ofAziz and
Na (1984), Aziz and Lunardini (1993), and others A collection ofsuch solutions is provided next
One-Region Neumann Problem The quasi-steady-state solution where St= 0 for this problem is
T = T0+ (T f − T0) x
x2
f = 2k(T0− T f )
One-Region Neumann Problem with Surface Convection With convective cooling, the boundary condition ofeq (3.347a) is replaced by
k ∂T
∂x
where T∞ is now the coolant temperature The quasi-steady-state approach with
St= 0 yields
T = T f+h(T f − T∞)(x − x f )
t = ρLx f h(T f − T∞)
1+hx f
2k
(3.372)
as the results
Outward Cylindrical Freezing Consider a saturated liquid at the freezing tem-peratureT f, surrounding a cylinder ofradiusr0whose outer surface is kept at a sub-freezing temperature,T0 < T f Ifa quasi-steady-state assumption (St= 0) is used, the solution is
T = T0+ T f − T0
ln(r f /r0)ln
r
r0
(3.373)
t = k(T ρL
f − T0)
1
2r2
flnr f
r0 −1 4
r2
f − r2 0
(3.374)