approximate solution methods for one-dimensional

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Applied Mathematics and Computation xxx (2008) xxx–xxx

www.elsevier.com/locate/amc

Approximate solution methods forone-dimensional solidification from an incoming fluid

S.L. Mitchell a,*, T.G. Myers b,1

a MACSI, Department of Mathematics and Statistics, University of Limerick, Limerick, Irelandb Department of Mathematics and Applied Mathematics, University of Cape Town, Rondebosch 7701, South Africa

Abstract

This paper concerns a one-dimensional model for solidification due to incoming supercooled liquid impacting on a sub-strate that is maintained at a fixed temperature. Using a boundary immobilisation method, and assuming that both thesolid and liquid layers remain thin throughout the process, a second-order accurate perturbation expansion is determined.An alternative approximate solution, found using the heat balance integral method, is also described to analyse the prob-lem, and the liquid height and temperatures in the solid and liquid are subsequently found for both approximate solutions.These are then compared with a numerical scheme which solves the full Stefan problem. The perturbation solution isshown to be more accurate, but the HBI method is simpler to implement and avoids complications that arise in the order-ing of terms in the perturbation expansion as the difference between the substrate and melting temperature changes.� 2008 Elsevier Inc. All rights reserved.

Keywords: Solidification; Heat balance integral; Boundary immobilisation; Thin film; Stefan problem

1. Introduction

The solidification of a molten material sprayed onto a substrate that is maintained at a temperature belowthe melting temperature has numerous natural and industrial applications. Perhaps the most common examplebeing when atmospheric water freezes on a structure. This has been studied in the context of icing on powertransmission and generating equipment, aircraft and seacraft, for example, see [6,12,13,16,17,20]. In an indus-trial setting solidification from a flowing liquid or a droplet spray is of interest in the casting of metals and sprayforming, lava flows and hydrate build-up in oil pipelines [4,5,9]. In this paper, we focus on a particular situation,relevant to atmospheric icing, where the liquid spray is supercooled. Consequently, when it impacts on a surfacethat is below the solidification temperature the liquid immediately solidifies. If there is sufficient energy in thesystem then, once an insulating layer has built up, a liquid layer can form on the top of the solid layer. For iceaccretion the initial layer is known as rime ice, the ice formed from a liquid layer is known as glaze ice.

0096-3003/$ - see front matter � 2008 Elsevier Inc. All rights reserved.

doi:10.1016/j.amc.2008.02.031

* Corresponding author.E-mail address: [email protected] (S.L. Mitchell).

1 Present address: Department of Mathematical Sciences, KAIST 373-1 Guseong-dong, Yuseong-gu, Daejeon 305-701, South Korea.

Please cite this article in press as: S.L. Mitchell, T.G. Myers, Approximate solution methods for ..., Appl. Math. Com-put. (2008), doi:10.1016/j.amc.2008.02.031

mailto:[email protected]

2 S.L. Mitchell, T.G. Myers / Applied Mathematics and Computation xxx (2008) xxx–xxx

ARTICLE IN PRESS

One of the most challenging aspects of modelling ice accretion on structures is when there is a significantwater layer [6,21]. This occurs when temperatures are relatively high and can lead to ice forming away fromthe main impact region. In a series of papers Myers and co-workers have developed a hierarchy of modelsstarting with one-dimensional ice growth from an incoming supercooled water spray [14,18] to three-dimen-sional accretion and flow on a flat surface, which has then been extended to an arbitrary shaped surface [15–17]. In the current work we return to the basic problem, namely one-dimensional accretion, since an accuratemodel at this stage can be implemented at all subsequent stages of the hierarchy.

In the case of ice accretion on aircraft, the ice layer is never permitted to become large. Based on theassumption that both the ice and water layers remain thin throughout the process the analysis in [14–18] usesa leading order perturbation to determine the temperatures in the two layers. In the following analysis, weemploy a boundary immobilisation method (BIM) which has been applied numerically to problems with amoving boundary by Crank [3] and more recently by Kutluay et al. [11]. This technique has also been usedto obtain a perturbation solution to Stefan problems, see [2,10] and references therein. By changing the timevariable to the ice thickness we can determine the perturbation solution up to second-order, and so improve onthe accuracy of previous work. We also analyse the same problem using a heat balance integral (HBI) method,see [7,8]. Finally a numerical scheme is described, adapted from that of Brakel et al. [1], which is used to com-pare the accuracy of the perturbation and HBI solutions.

Although we are interested in general solidification problems we will confine our discussion to ice and waterunder aircraft icing conditions. Data for this situation are readily available, see [1,18]. We will also restrict ourresults to relatively high temperatures (typically down to around �6 �C).

2. Governing equations

Consider a situation where a supercooled spray impacts on a surface. If the surface is below the solidifica-tion temperature then the initial droplets will freeze almost instantaneously. In mild temperatures, or when thesolidified layer is sufficiently thick, a fluid layer may subsequently appear. In the following work we will focuson an ice and water system, although the model is equally applicable to other materials. The problem config-uration is depicted in Fig. 1.

The temperatures in the ice and water are denoted by bT ðz; tÞ and hðz; tÞ, respectively, and the thickness ofeach layer is bðtÞ and hðtÞ. The problem is described by two heat equations:

Pleaput.

obTot¼ ki

qici

o2bToz2

; 0 6 z 6 b; ð1Þ

oh

ot¼ kw

qwcw

o2hoz2

; b 6 z 6 bþ h; ð2Þ

Fig. 1. Schematic of the 1D model problem.

se cite this article in press as: S.L. Mitchell, T.G. Myers, Approximate solution methods for ..., Appl. Math. Com-(2008), doi:10.1016/j.amc.2008.02.031

S.L. Mitchell, T.G. Myers / Applied Mathematics and Computation xxx (2008) xxx–xxx 3

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where k, q and c are the conductivity, density and specific heat capacity, respectively; a Stefan condition

Pleaput.

qiLfdbdt¼ ki

obToz� kw

ohoz; at z ¼ bðtÞ; ð3Þ

where Lf is the latent heat of freezing; and a mass balance,

qi

dbdtþ qw

dhdt¼ _m; ð4Þ

where _m is the rate at which mass enters from the spray.We impose a fixed temperature boundary condition at the substrate z ¼ 0
bT ð0; tÞ ¼ bT s; ð5Þ
where bT s is the substrate temperature. This is a reasonable approximation to the true state when the substrateis a good conductor with a relatively high thermal mass. A more general cooling condition is dealt with in[1,18], but we concentrate on this case here since it reduces the number of ice growth regimes and thereforeallows us to focus on the solution methods. The extension to a cooling condition is straightforward. Theremaining boundary conditions for (1)–(4) depend on the type of ice growth, either rime or glaze. To simplifythe algebra we will restrict attention to the case, where bT s is the same as the air temperature, bT a. This corre-sponds to the substrate being a good conductor with a region exposed to the air (as occurs with aircraft icing).

We now consider the rime and glaze ice growth separately.

� Rime ice growth: In this case there is no water layer present and so h � 0. We therefore only need to solve(1) over the domain ½0; b�, where b ¼ _mt=qi is determined from the integrated form of (4) with bð0Þ ¼ 0. Anenergy balance gives the boundary condition at the top of the ice layer z ¼ b (at the ice–air interface). Weconsider two different conditions, either a fixed or variable energy condition:

ðiÞ ki

obToz¼ Qi or ðiiÞ ki

obToz¼ Qi þ qlðbT s � bT Þ; at z ¼ b; ð6Þ

where the terms in Qi may represent droplet kinetic energy, aerodynamic heating and latent heat, respectively,and ql is composed of the droplet thermal energy, evaporation and convective heat transfer. The energy termsare discussed in detail in [1].The rime phase ends at time t ¼ tw, when the ice surface reaches the melting tem-perature bT ðbw; twÞ ¼ bT f where bw ¼ bðtwÞ.� Glaze ice growth: In this phase h > 0 and we must solve (1)–(5) with the following boundary conditions. At

the ice/water interface z ¼ b, the temperatures are constant and equal to the melting temperature of the icebT f , i.e.
bT ðb; tÞ ¼ bT f ¼ hðb; tÞ: ð7Þ At the top of the water layer z ¼ bþ h, similar conditions hold to those in (6), namely
ðiÞ kwohoz¼ Qw or ðiiÞ kw

ohoz¼ Qw þ qmðbT s � hÞ: ð8Þ

Provided _m is constant, Eq. (4) integrates to

qiðb� bwÞ þ qwh ¼ _mðt � twÞ; ð9Þ
using the initial condition hðtwÞ ¼ 0, and this allows us to eliminate h in subsequent calculations.
3. Non-dimensional analysis

The system is now non-dimensionalised in order to identify the dominant terms. We set

z ¼ zH; t ¼ t

s; b ¼ b

H; h ¼ h

H; T ¼

bT � bT f

DT; h ¼ h� bT f

DT; ð10Þ



ARTICLE IN PRESS

where DT ¼ bT f � bT s. Applying this scaling to the mass balance relation (4) gives

Pleaput.

dbdtþ q

dhdt¼ 1; q ¼ qw

qi

; ð11Þ

provided we choose the timescale s ¼ qiH= _m. The Stefan condition (3) is now

dbdt¼ oT

oz� k

ohoz

� �z¼b

; k ¼ kw

ki

; ð12Þ

where the height-scale is

H ¼ kiDTLf _m

: ð13Þ

The choice of height and timescales is motivated by the fact that the mass balance and Stefan conditions obvi-ously define the growth rates. The heat equations (1) and (2) therefore become

�i

oTot¼ o2T

oz2; �i ¼

ciH _mki

; 0 6 z 6 b; ð14Þ

�wohot¼ o

2hoz2

; �w ¼ qcwH _m

kw

; b 6 z 6 bþ h: ð15Þ

The non-dimensional boundary condition (5) at the substrate–ice interface is now

T ð0; tÞ ¼ �1: ð16Þ

The rime boundary conditions (6) at the ice–air interface, z = b, are

ðiÞ oToz¼ P i; or ðiiÞ oT

oz¼ P i � pið1þ T Þ: ð17Þ

The glaze boundary conditions (7) and (8) become

T ðb; tÞ ¼ 0 ¼ hðb; tÞ; ð18Þ

and at z = b + h

ðiÞ ohoz¼ P w or ðiiÞ oh

oz¼ P w � pwð1þ hÞ; ð19Þ

where

P i ¼QiH

kiDT; pi ¼

qlH

ki

; P w ¼QwH

kwDT; pw ¼

qmH

kw

: ð20Þ

The non-dimensional problem is now set up, so we move on to describe two approximate solutions, onefound using a perturbation method and one found using a heat balance integral (HBI) method, and comparethem to a numerical solution adapted from that given in [1].

4. Rime problem 0 < t < tw

In the rime period there is no water layer and so b(t) = t. Then from the previous section we have to solve

o2Toz2¼ �i

oTot; 0 < z < bðtÞ; ð21Þ

T ð0; tÞ ¼ �1; and ðiÞ oToz¼ P i

��z¼b

or ðiiÞ oToz¼ P i � pið1þ T Þ

��z¼b

: ð22Þ



ARTICLE IN PRESS

4.1. The perturbation solution

The glaze problem, which is dealt with in the following section, is simplified by using b as the time variable(since b(t) is monotonic it is valid to set t = t(b)). For the rime problem, b = t and so this transformation isirrelevant. However, for consistency with the following section we will work with t(b). We also use a co-ordi-nate system moving with the freezing boundary y = z � b(t), otherwise known as the BIM [2,3,10,11], anddenote T(z, t) = S(y, b).

Under this co-ordinate change Eqs. (21) and (22) become

Pleaput.

o2Soy2¼ �i

oSob� oS

oy

� �dbdt¼ �i

oSob� oS

oy

� �; �b < y < 0 ð23Þ

Sð�b; bÞ ¼ �1; and ðiÞ oSoy¼ P i

��y¼0

or ðiiÞ oSoy¼ P i � pið1þ SÞ

��y¼0

: ð24Þ

Provided �i is small, S may be expanded in powers of �i as

S ¼ S0 þ �iS1 þ �2i S2 þ � � � ð25Þ

and substituted into (23) and (24) to give

o2S0

oy2¼ 0;

o2Sk

oy2¼ oSk�1

ob� oSk�1

oy; ð26Þ

for k = 1,2, . . . , with S0(�b, b) = �1, Sk(�b, b) = 0 and

ðiÞ oS0

oy¼ P i

��y¼0

;oSk

oy¼ 0

��y¼0

;

ðiiÞ oS0

oy¼ P i � pið1þ S0Þ

��y¼0

;oSk

oy¼ �piSk

��y¼0

:

Let us first consider the fixed energy condition (i) at the ice–air interface. The solution is given by

S0 ¼ P iðy þ bÞ � 1; S1 ¼ S2 ¼ � � � ¼ 0; ð27Þ
and so S(y, b) = Pi(y + b) � 1. The rime period ends when T(bw, tw) = 0 or, equivalently, S(0, bw) = 0. Hencethe ice thickness and time at which the rime period ends occurs at bw = tw = 1/Pi. In terms of the physicalparameters this is tw ¼ Lf _m=Qi. It should be noted that back in the original co-ordinate system, the perturba-tion solution is T(z, t) = Piz � 1 which is the exact solution in this case.
For the cooling condition (ii) at the ice–air interface, the leading order solution S0 is

S0ðy; bÞ ¼ A0ðbÞðy þ bÞ � 1; A0ðbÞ ¼P i

1þ pib: ð28Þ

In this case Sk 6¼ 0 for k > 0. To determine S1 we must solve

o2S1

oy2¼ oS0

ob� oS0

oy¼ A00ðbÞðy þ bÞ; S1ð�b; bÞ ¼ 0;

oS1

oy¼ �piS1

��y¼0

;

which leads to

S1ðy; bÞ ¼ A1ðbÞðy þ bÞ þ 1

6A00ðbÞðy þ bÞ½y2 þ 2bðy � bÞ�; A1ðbÞ ¼

piA00ðbÞb3

3ð1þ pibÞ: ð29Þ

Also, the second-order solution S2 is

S2ðy; bÞ ¼ A2ðbÞðy þ bÞ þ 1

6ðA01ðbÞ � A00ðbÞbÞðy þ bÞ½y2 þ 2bðy � bÞ� þ 1

120A000ðbÞðy þ bÞ½y2 þ 2bðy � 2bÞ�2;

ð30Þ



ARTICLE IN PRESS

where

Pleaput.

A2ðbÞ ¼ �2piA

000ðbÞb5

15ð1þ pibÞþ piðA01ðbÞ � A00ðbÞbÞb3

3ð1þ pibÞ: ð31Þ

The time tw when the rime period ends is then the solution of

S0ð0; twÞ þ �iS1ð0; twÞ þ �2i S2ð0; twÞ ¼ 0;

which becomes

tw ¼ A0ðtwÞ þ �i A1ðtwÞ �1

3A00ðtwÞt2

w

� �þ �2

i A2ðtwÞ �1

3ðA01ðtwÞ � A00ðtwÞtwÞt2

w þ2

15A000ðtwÞt4

w

� �� 1

¼ P itw

1þ pitw

þ �ipiP it3w

3ð1þ pitwÞ3� �

2i piP it4

wð5þ pitw þ p2i t2

wÞ15ð1þ pitwÞ5

( )�1

: ð32Þ

We can approximate tw to the same level of accuracy as the temperature by taking

tw ¼ tw0þ �itw1

þ �2i tw2þ � � � ¼ tw0

��ipit

3w0

3ð1þ pitw0Þ þ

�2i pit

4w0ð15þ 18pitw0

þ 8p2i t2

w0Þ

45ð1þ pitw0Þ3

; ð33Þ

where

tw0¼ 1

P i � pi

: ð34Þ

The leading order term is singular when Pi = pi. This reflects the possibility that a water layer may never ap-pear and occurs when the incoming energy balances the heat loss. Using the expressions for Pi and pi in (20)leads to the condition DT < Qi/ql (this may be seen as the requirement to have a positive temperature gradientin (6)). Since the fixed energy condition (24i) has no cooling, pi = 0, and so provided Pi > 0 water will alwaysappear in that case.

4.2. The quadratic HBI solution

We now consider the quadratic HBI solution of the systems (21) and (22). In [19] a cubic HBI solution, withno quadratic term, is used to approximate the temperature in a finite ice block. That choice was motivated bycomparison with an exact solution, since the small z expansion gave this cubic form. Furthermore, a pertur-bation solution in the thin water layer also indicated the cubic profile was appropriate. In the current situationour perturbation solution includes a quadratic term at first-order, and so we stick with the standard quadraticHBI approximation. The solution method is now described for the boundary condition (ii) at the ice–air inter-face and then the solution for condition (i) follows directly by setting pi = 0. In general, the quadratic heatbalance integral method [7,8] involves assuming a solution of the form,

T ðz; tÞ ¼ a0ðtÞ þ a1ðtÞ 1� zb

þ a2ðtÞ 1� z

b

2

; ð35Þ

where the coefficients ai(t) are unknown. We can immediately eliminate two of these by applying the boundaryconditions in (22). Thus (35) reduces to

T ðz; tÞ ¼ �1þ P ib� ð2þ pibÞa2

1þ pib

� �zbþ a2

z2

b2: ð36Þ

The remaining unknown coefficient a2(t) must now be determined by integrating the heat equations (2) fromz = 0 to z = b,

oToz

��z¼b

� oToz

��z¼0

¼ �i

d

dt

Z b

0

T ðz; tÞdz� dbdt

T ðbðtÞ; tÞ� �

: ð37Þ



ARTICLE IN PRESS

Then substituting T from (36) gives an ODE to solve for a2(t). At this stage Goodman and Shea [8] introducethe function wðtÞ ¼

R b0

T ðz; tÞdz, which has initial condition w(0) = 0, as there is not always an initial conditionfor the remaining unknown coefficient. In the present case we can determine that a2(0) = 0 and so could solvefor a2(t) directly. However, for the glaze problem it is more convenient to work with w, and the correspondingintegral of h, and so for consistency we also use this form here. Therefore (37) becomes

Pleaput.

dwdt¼ 2

�ib� 1

1þ pib

� �a2 þ

P ib1þ pib

� 1: ð38Þ

Before solving (38) we must eliminate a2(t) by re-writing it in terms of w as

a2ðtÞ ¼�6ð1þ pibÞðbþ wðtÞÞ þ 3P ib

2

bð4þ pibÞ: ð39Þ

Since b = t Eq. (38) is now a first-order differential equation for the single unknown w(t). Once w is known wemay determine a2(t) and hence T(z, t) via Eq. (36).

Finally, the rime period ends at t = b = tw, where T(tw, tw) = 0. Using (36) this leads to

1 ¼ P itw � ð2þ pitwÞa2ðtwÞ1þ pitw

� �þ a2ðtwÞ: ð40Þ

The solution for boundary condition (i) is simply found by setting pi = 0 in the above equations. In fact, forthe fixed boundary condition, it can be shown that a2(t) = 0 for all t. The simplest way to deduce this it tosolve the ODE for a2(t) found from (37). The general solution, when b = t and pi = 0, is given bya2ðtÞ ¼ C

ffiffitp

e3=ð�itÞ and we must set C = 0 to ensure a finite solution as t ? 0. Then the expression for T in(36) reduces to T = �1 + Piz which is identical to the perturbation solution (and the exact solution).

4.3. The numerical solution

In this section we describe a semi-implicit finite difference scheme which is used to solve the systems (21) and(22), as developed for a related problem in [1].

The problem involves solving the heat equation in (21) with a moving boundary z = b(t) and so the tem-perature profile T is calculated on a moving grid with a constant number of equally spaced mesh points. Sincethe height of the ice layer varies with time, the size of the mesh cell must be recalculated at each time step. Thiseffect is dealt with by incorporating a convection term into the heat equation. Then the total derivativereplaces the time derivative in the following way:

DTDt¼ oT

otþ oT

ozozot¼ 1

�i

o2Toz2þ oT

ozozot; ð41Þ

where ozot is the speed of the moving grid and oT

ot has been eliminated using (14). The discretised form of (41) is

T nþ1j � T n

j

Dt¼ 1

�i

T nþ1jþ1 � 2T nþ1

j þ T nþ1j�1

Dznþ1i

2

!þ 1

2

T nþ1jþ1 � T nþ1

j

Dznþ1i

ozi

ot

��jþ1

2

þT nþ1

j � T nþ1j�1

Dznþ1i

ozi

ot

��j�1

2

" #; ð42Þ

which holds for j = 1, . . . , Ji � 1 and n = 0, . . . ,N � 1. The subscript ‘‘i”, denoting ice, is introduced here forconvenience as it is necessary to distinguish between the meshes for ice and water in the numerical solution ofthe glaze problem, described in Section 5.3.

We write the discretisation in (42) as

ri þmi

2

ozi

ot

��jþ1

2

" #T nþ1

jþ1 � 1þ 2ri þmi

2

ozi

ot

� ��jþ1

2

� ozi

ot

��j�1

2

!" #T nþ1

j þ ri �mi

2

ozi

ot

��j�1

2

" #T nþ1

j�1 ¼ �T nj ; ð43Þ

where

ri ¼Dt

�iDznþ1i

2; mi ¼

DtDznþ1

i

: ð44Þ



ARTICLE IN PRESS

Since dbdt ¼ 1 in the rime period we have bn+1 = bn + Dt and so the parameters ozi

ot and Dznþ1i are evaluated as

Pleaput.

ozi

ot

��j

¼ jJ i

dbdt¼ j

J i

; Dznþ1i ¼ bnþ1

Dt: ð45Þ

We must now approximate the boundary conditions in (22). At the substrate–ice interface the conditionT(0, t) = �1 is simply T nþ1

0 ¼ �1. To discretise condition (ii) at the ice–air interface we set j = Ji in (42)and eliminate the first occurrence of T nþ1

J iþ1 using the discretised form,

T nþ1J iþ1 � T nþ1

J i�1

2Dznþ1i

¼ P i � pið1þ T nþ1J iÞ:

The second occurrence of T nþ1J iþ1 is approximated slightly differently using

T nþ1J iþ1 � T nþ1

J i

Dznþ1i

ozi

ot

��J iþ1

2

¼ ½P i � pið1þ T nþ1J iÞ�ozi

ot

��J i

¼ P i � pið1þ T nþ1J iÞ;

since ozi

ot

��J i¼ 1. Then the boundary condition (ii) becomes

� 1þ 2ri þ 2piDznþ1i ri þ piDznþ1

i

mi

2� mi

2

ozi

ot

��J i�1

2

" #T nþ1

J iþ 2ri �

mi

2

ozi

ot

��J i�1

2

" #T nþ1

J i�1

¼ �T nJ i� 2ðP i � piÞDznþ1

i ri � ðP i � piÞDznþ1i

mi

2ð46Þ

with (i) recovered by setting pi = 0 in (46). This solution is valid whilst T nþ1J i

< 0, and the time at which thiscondition is violated gives the numerical value of tw.

5. Glaze ice growth t > tw

During this phase the growth is governed by Eqs. (11), (12), (14), (15) subject to conditions (16), (18), (19).The initial condition t = tw, b = bw, h = 0 is determined from the solution in the rime stage.

5.1. The perturbation solution

With the change of variables y = z � b(t) and t = t(b), we write T(z, t) = S(y, b), h(z, t) = /(y, b). The ther-mal problem for the variable energy boundary condition (ii) may then be written as

o2S

oy2¼ �i

oSob� oS

oy

� �dbdt; Sð0; bÞ ¼ 0; Sð�b; bÞ ¼ 0; ð47Þ

o2/oy2¼ �w

o/ob� o/

oy

� �dbdt; /ð0; bÞ ¼ 0;

o/oy¼ P w � pwð1þ /Þ

��y¼h

ð48Þ

with the problem for the simpler fixed energy boundary condition (i) the same but with pw = 0. The Stefancondition (12) is now

dbdt¼ oS

oy� k

o/oy

� �y¼0

ð49Þ

and the mass balance (11) still holds which relates h to b. Note that we obviously require dhdt > 0 and the solu-

tion below satisfies this at leading order for all DT P 0.Before finding a perturbation solution we need to think about the size of the small parameters, �i and �w,

which are plotted against DT in Fig. 2. Clearly different terms dominate over different temperature regions,indicating that a variety of expansions are needed. For DT 2 [0, 1] both �i and �w are very small, with �w � 10�i.In this range, a first-order expansion in �w should provide accurate results (with errors of around 0.4%). ForDT 2 [1, 5.5] we note �2

w � �i and an expansion to Oð�2wÞ leads to errors typically less than 0.7%. In this case we

may set �i ¼ b�2w. In the range DT 2 [5.5, 8.5] we may set �i ¼ b1�

3w and so on.


0 2 4 6 8 10 12 14 16 180

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

temperature difference

εw

εw2

εi

εiε

w

εi2

Fig. 2. Variation of �i, �w and second-order terms, plotted against Dt.


ARTICLE IN PRESS

As discussed in Section 1 our primary interest is in relatively high temperatures, where a significant waterlayer appears. For this reason we will analyse DT 2 [1, 5.5]. The solution for DT < 1 can be taken from theleading order terms of this solution. Expanding the temperatures as

Pleaput.

S ¼ S0 þ �2wS2 þ � � � ; / ¼ /0 þ �w/1 þ �2

w/2 þ � � �
and substituting into (47) and (48) leads to
o2S0

oy2¼ 0; S0ð0; bÞ ¼ 0; S0ð�b; bÞ ¼ �1

o2S2

oy2¼ b

oS0

ob� oS0

oy

� �oS0

oy� k

o/0

oy

� �y¼0

; S2ð0; bÞ ¼ 0; S2ð�b; bÞ ¼ 0

and

o2/0

oy2¼ 0; /0ð0; bÞ ¼ 0;

o/0

oy¼ P w � pwð1þ /0Þ

��y¼h

;

o2/1

oy2¼ o/0

ob� o/0

oy

� �oS0

oy� k

o/0

oy

� �y¼0

; /1ð0; bÞ ¼ 0;o/1

oy¼ �pw/1

��y¼h

;

o2/2

oy2¼ o/0

ob� o/0

oy

� ��k

o/1

oy

� �y¼0

þ o/1

ob� o/1

oy

� �oS0

oy� k

o/0

oy

� �y¼0

; /2ð0; bÞ ¼ 0;o/2

oy¼ �pw/2

��y¼h

:

For brevity, in this section we only give the results for boundary condition (i). The expansions for boundarycondition (ii) are quoted in Appendix.

The leading order solutions with pw = 0 are therefore

S0 ¼ A0ðbÞy; A0ðbÞ ¼1

b; and /0 ¼ B0y; B0 ¼ P w; ð50Þ

and the Oð�wÞ solution is given by



ARTICLE IN PRESS

Pleaput.

/1 ¼ B1ðhÞy �k2

B0y2; B1ðb; hÞ ¼ B0kh; ð51Þ

where

k � kðbÞ ¼ oS0

oy� k

o/0

oy

� �y¼0

¼ A0ðbÞ � kB0 ¼1

b� kP w: ð52Þ

Then the Oð�2wÞ solutions are

S2 ¼ A2ðbÞy þkb6½A00ðbÞy3 � 3A0ðbÞy2�; ð53Þ

/2 ¼ B2ðb; hÞy þ1

2B1ðb; hÞðkB0 � kÞy2 þ k

6ðB1b þ B0kÞy3 � k

24B0kby4 ð54Þ

with

A2ðbÞ ¼kb6½A00ðbÞb2 þ 3A0ðbÞb�;

B2ðb; hÞ ¼ �B1ðb; hÞðkB0 � kÞh� k2ðB1b þ B0kÞh2 þ k

6B0kbh3;

where we have used B1b to denote oB1

ob . Substitution of expansions S and / into the Stefan condition (49) gives

dbdt¼ oS0

oyþ �2

w

oS2

oy� k

o/0

oyþ �w

o/1

oyþ �2

w

o/2

oy

� �� y¼0

¼ A0ðbÞ þ �2wA2ðbÞ � kfB0 þ �wB1ðhÞ þ �2

wB2ðhÞg; ð55Þ

subject to the initial condition b = bw at t = tw = bw. The substitution h = (t � b)/q reduces Eq. (55) to a first-order ordinary differential equation for b. Note that the expression for /2 contains the term hb = (tb � 1)/q,where tb may be replaced using tb = 1/k. This is a leading order approximation which is sufficient here sincehb only appears at second-order. However, for the boundary condition (ii) this quantity appears at first-orderand so we must expand tb as tb,0 + �wtb,1. This is discussed in more detail in Appendix.

Once Eq. (55) is solved, we can calculate h(t) and consequently the temperature profiles in the two layers.

5.2. The quadratic HBI solution

The method for determining the quadratic HBI solution for glaze growth is similar to that discussed in Sec-tion 4.2. We briefly describe the solution method for the boundary condition (ii) at the water–air interface andthen the solution for condition (i) follows directly by setting pw = 0. The solution for T is again of the form(35) and similarly for h we assume

hðz; tÞ ¼ c0ðtÞ þ c1ðtÞbþ h� z

h

� �þ c2ðtÞ

bþ h� zh

� �2

; ð56Þ

where the coefficients ai(t) and ci(t) are unknown. Note that we have divided by h in (56) as this is the width ofthe water region, and it simplifies the algebra when dealing with the boundary condition at z = b + h.

We can immediately eliminate four of the coefficients by applying the boundary conditions in (16), (18),(19ii). Thus T in (35) is simply

T ðz; tÞ ¼ �1þ ð1� a2Þzbþ a2

z2

b2; ð57Þ

and h in (56) becomes

hðz; tÞ ¼ ðP w � pwÞh� c2

1þ pwh� ðP w � pwÞhþ c2pwh

1þ pwhbþ h� z

h

� �þ c2

bþ h� zh

� �2

: ð58Þ



ARTICLE IN PRESS

The remaining unknown coefficients a2(t) and c2(t) must now be determined by integrating the heat equation(14) from z = 0 to z = b and (15) from z = b to z = b + h. Hence

Pleaput.

dwdt¼ 2a2

�ib;

dvdt¼ 2c2

�whþ db

dtþ dh

dt

� �ðP w � pwÞh� c2

1þ pwh

� �; ð59Þ

where

wðtÞ ¼Z b

0

T dz; vðtÞ ¼Z bþh

bhðz; tÞdz: ð60Þ

As discussed for the HBI solution in the rime region in Section 4.2, we find it convenient to use these integralsas it is not clear how to choose initial conditions for a2 and c2 and the resulting ODEs are more complicated.Using these expressions we can eliminate a2 and c2 from (59) by re-writing them in terms of w and v, i.e.

a2 ¼ �3� 6wb; c2 ¼

3ðP w � pwÞh4þ pwh

� 6ð1þ pwhÞvhð4þ pwhÞ : ð61Þ

The pair of first-order differential equations in (59) must be coupled with (11) and (12) to give a system of fourequations for the four unknowns w, v, b and h. However, h can immediately be eliminated by writing it interms of b as h = (t � b)/q. The Stefan condition (12) becomes

dbdt¼ 1þ a2

b� k

ðP w � pwÞh� ð2þ pwhÞc2

hð1þ pwhÞ

� �ð62Þ

and the system (59) and (62) can be solved subject to initial conditions w(tw) = ww (found from the rime solu-tion), v(tw) = 0 and b(tw) = tw.

Again note that condition (i) at the water–air interface is recovered by setting pw = 0 in the above analysis.

5.3. The numerical solution

We now explain how to extend the semi-implicit finite difference scheme described in Section 4.3 to includethe water layer which develops in the glaze period. The rime model is run until T exceeds zero. The final valueprofile in the rime region which satisfies T nþ1

J i< 0 is the initial condition for the glaze solution. At the top of the

ice layer the boundary condition (17) is replaced by (18) and so hnþ10 ¼ 0.

The ice and water growth thicknesses b and h are determined by

bnþ1 ¼ bn þ Dtbnt ; hnþ1 ¼ hn þ Dthn

t ; ð63Þ
for n = 0, . . . ,N � 1, where bn
t and hnt are found by discretising (11) and (12) as follows: the former is simply

hnt ¼ ð1� bn

t Þ=q and then

bnt ¼

3T nJ i� 4T n

J i�1 þ T nJ i�2

2Dzni

� k�3hn

0 þ 4hn1 � hn

2

2Dznw

� �: ð64Þ

Note that to determine b1 in (63) we need to use the final profile in the rime region which satisfies T nþ1J i

< 0.This gives T 0

J i, T 0

J i�1 and T 0J i�2 in b0

t , where the second term is zero since h = 0 at z = b, and so b1 is known sinceb0 is simply bw = t.

The numerical scheme (43) remains valid but now

ozi

ot

��j

¼ jJ i

bnt ; ð65Þ

where bnt is given in (64). Inside the water layer the numerical scheme is derived identically to that for the ice

layer (43), namely

rw þmw

2

ozw

ot

��jþ1

2

" #hnþ1

jþ1 � 1þ 2rw þmw

2

ozw

ot

� ��jþ1

2

� ozw

ot

��j�1

2

!" #hnþ1

j þ rw �mw

2

ozw

ot

��j�1

2

" #hnþ1

j�1 ¼ �hnj ; ð66Þ



ARTICLE IN PRESS

which holds for j = 1, . . . ,Jw � 1 and where

TablePhysic

ci

ki

qi

Lf

Qi

qlbT f

TableNume

Mesh

m = 1m = 2m = 4m = 8m = 16

The nu

Pleaput.

rw ¼Dt

�wDznþ1w

2; mw ¼

DtDznþ1

w

; Dznþ1w ¼ hnþ1

J w

;ozw

ot

��j

¼ jJ w

hnt : ð67Þ

Finally, the boundary condition (ii) at the water/air interface is similar to (46), and so

� 1þ 2rw þ 2pwDznþ1w rw þ pwDznþ1

w hnt

mw

2� mw

2

ozw

ot

��Jw�1

2

" #hnþ1

Jwþ 2rw �

mw

2

ozw

ot

��Jw�1

2

" #hnþ1

Jw�1

¼ �hnJw� 2ðP w � pwÞDznþ1

w rw � ðP w � pwÞDznþ1w hn

t

mw

2; ð68Þ

with (i) again found from setting pw = 0.

6. Results

6.1. Rime results

Table 1 shows standard physical parameter values for ice and water, applicable to aircraft icing conditions,taken from [1]. We begin with the case of the fixed energy boundary condition (22i). As discussed above, boththe perturbation and HBI solutions in fact equal the exact solution in this case, and we therefore use this caseto test the validity of the numerical scheme. The end of the rime phase is tw = 1/Pi = 0.893533 which is inde-pendent of DT. Table 2 gives the numerical predictions of tw as we reduce the mesh size (i.e. we double m

where Dt = 1 10�4/m and Ji = 40m) for two values of DT. Observe that both columns converge to theexact tw. As discussed in Section 6.1 we only consider DT 2 [1, 5.5] since this is the range of validity for theperturbation solution in the glaze phase. However, in the rime phase the restriction is DT < Qi/ql � 18.89,which is needed to ensure that there is a positive gradient in the variable energy condition (ii), as discussedat the end of Section 5.1. Although not shown in Table 2, choosing larger values of DT in the numerical solu-tion also shows the same predictions. When m = 16 we find agreement with the exact solution to 4 d.p. and sowe use m = 16 in numerical calculations from now on.

The results for the variable energy boundary condition (22ii) are of more interest since tw now changes asDT varies. In Table 3 we present values of �i, H , Pi and pi for DT = 5,2. We refer to the perturbation solutionsT = T0, T = T0 + �iT1 and T ¼ T 0 þ �iT 1 þ �2

i T 2 as ‘‘Pert0”, ‘‘Pert1” and ‘‘Pert2”, respectively (after convert-

1al parameter values for ice and water for aircraft icing conditions [1]

2050 J/kg K cw 4218 J/kg K2.18 W/m K kw 0.57 W/m K917 kg/m3 qw 1000 kg/m3

3.34 105 J/kg _m 0.05 kg/m2 s1.869 104 J/m2 s Qw 1990 J/m2 s989.3 W/m2 K qm 956.9 W/m2 K273 K

2rical prediction of tw for DT = 5 and DT = 2

size DT = 5 DT = 2

tw = 0.8932000 tw = 0.8934000tw = 0.8933500 tw = 0.8934500tw = 0.8934500 tw = 0.8935000tw = 0.8934875 tw = 0.8935125tw = 0.8935125 tw = 0.8935250

merical time-step is Dt = 1 10�4/m and mesh spacing is Ji = 40 m.


Table 3Scaling parameters in the rime period for various DT

�i H Pi pi

DT = 5 0.03069 0.0006527 1.1191528 0.29620DT = 2 0.01228 0.0002611 1.1191528 0.11848


ARTICLE IN PRESS

ing back from S(y, b) to T(z, t)). For DT = 5 the numerical solution gives tw = 1.211219 whilst the predictionsfor the approximate solutions (and the differences from the numerical solution) are: tw = 1.215137 (0.32%),tw = 1.211171 (0.0040%), tw = 1.211258 (0.0032%) and tw = 1.21221 (0.082%) for the Pert0, Pert1, Pert2and HBI solutions, respectively. We can also look at the errors between the temperatures T for the variousapproximate solutions. If we define E� = jTnum � T�j and calculate the L2 norm L2;� ¼ kE�k2 ¼

PE2�

� 1=2

then L2,pert0 = 1.566 10�2, L2,pert1 = 4.78 10�4, L2,pert2 = 1.838 10�4 and L2,hbim = 2.192 10�3. Usingsmaller values of DT gives slightly smaller times to melting tw, i.e. tw � 0.9989 for DT = 2. However, the result-ing errors all show the same trend as for DT = 5, and in particular that Pert2 is the most accurate solution.

6.2. Glaze results

Typical solutions for the temperatures and ice heights b are shown in Figs. 3 and 4, respectively, for bothcases pi, pw = 0 and pi, pw 6¼ 0, with DT = 5. Table 4 shows the L2 errors between each of the approximatesolutions, Pert0, Pert1, Pert2 and HBI, and the numerical solution. In both cases, the L2 errors for tempera-ture in the ice, T, show that Pert2 is significantly more accurate than the other approximations. The HBI solu-tion is better than both Pert0 and Pert1 which is also true for the errors in the ice height b. However, for theerrors in the water layer h, the HBI solution is very inaccurate for boundary condition (i) but much better forboundary condition (ii). The errors for T for Pert0 and Pert1 are identical which is due to the fact that both ofthese expansions use T = T0 (i.e. S = S0) because the next order corrections are Oð�2

wÞ. The high errors in b aredue to the fact that the glaze calculation starts at a different tw for each solution.

The errors in Table 4 and the profiles in Figs. 3 and 4 seem to indicate that it is acceptable to use the HBIsolution in the ice layer but the perturbation solution (up to second-order accuracy) must be used in the waterlayer. This will avoid some of the lengthy calculations required for the perturbation expansion. In [19] a sim-ilar approximate solution was determined for the one-dimensional melting of a finite thickness layer. A per-turbation expansion was used in the thin liquid layer and an HBI solution used in the solid thicker layer.

Eq. (55) only involves first-order derivatives of b. If we use t as the time variable, the second-order term inthe corresponding version of (55) involves btt and b2

t so we end up with a second-order equation, with a small

1.8 2 2.2 2.4 2.6 2.8

–0.1

–0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

ice

heig

ht b

(i)

ice

water

pi, p

w = 0

position z 2 2.2 2.4 2.6 2.8

–0.22

–0.2

–0.18

–0.16

–0.14

–0.12

–0.1

–0.08

–0.06

–0.04

–0.02

ice

water

pi, p

w≠ 0

(ii)

position z

Fig. 3. Plots of the temperatures T and h for glaze growth for the two boundary conditions (19) with DT = 5. Shown are the numericalsolution (solid line), Pert1 (dotted line), Pert2 (dashed line) and HBI (dot-dashed line).

Please cite this article in press as: S.L. Mitchell, T.G. Myers, Approximate solution methods for ..., Appl. Math. Com-put. (2008), doi:10.1016/j.amc.2008.02.031

2 2.2 2.4 2.6 2.8 31.6

1.65

1.7

1.75

1.8

1.85

1.9

1.95

2

2.05

2.1

t

ice

heig

ht b

(i)

pi, p

w = 0

2 2.2 2.4 2.6 2.8 31.8

1.9

2

2.1

2.2

2.3

2.4

2.5

t

pi, p

w≠ 0

(ii)

Fig. 4. Plots of ice height b for glaze growth for the two boundary conditions (19) with DT = 5. Shown are the numerical solution (solidline), Pert1 (dotted line), Pert2 (dashed line) and HBI (dot-dashed line).

Table 4Table showing the L2 errors for T, h and b (between each approximate solution and the numerical solution) from Figs. 3 and 4 in the glazeperiod

Error pi = 0, pw = 0 pi 6¼ 0, pw 6¼ 0

Pert0 Pert1 Pert2 HBI Pert0 Pert1 Pert2 HBI

T 1.90e�2 1.90e�2 9.02e�5 2.42e�3 3.17e�2 3.17e�2 1.70e�4 3.91e�3

h 4.32e�2 4.45e�2 3.69e�2 2.33e�1 5.06e�2 4.30e�2 2.05e�2 3.59e�2

b 0.350 0.283 0.097 0.128 0.578 0.592 0.292 0.422


ARTICLE IN PRESS

parameter in front of the leading order term. We also only have a single initial condition on b. Carrying on tohigher order terms then introduces higher order derivatives. Using t = t(b) results in a first-order equation, andthis is the reason for making this transformation.

Lower temperature ranges generally involve one more calculation. For example if we look at DT 2 [5.5, 8.5]and set �i ¼ b�3

i then

Pleaput.

S ¼ S0 þ �3wS3 þ � � � ;

/ ¼ /0 þ �w/1 þ �2w/2 þ �3

w/3 þ � � � :

We find S3 = S2 from the solution above, the / terms remain the same and we are left to calculate /3.

7. Conclusion

In this paper we have investigated two approximate solution methods to describe one-dimensional solidi-fication due to incoming supercooled liquid impacting on a fixed temperature substrate, which is kept belowthe melting temperature. Firstly, we applied a perturbation expansion to the non-dimensional system. Usingthe boundary immobilisation method allowed us to carry out the expansion to second-order, making it moreaccurate than in previous investigations. The use of the solid thickness as the time variable ensures that thegoverning ODEs are first-order and linear. Secondly, we used an alternative approximate solution, knownas the heat balance integral method which is a popular technique for solving one-dimensional heat conductionproblems involving a change of phase. We have also described a numerical scheme which solves the full Stefansystem, and this was used to test the accuracy of the approximate perturbation and HBI solutions.

We showed that the second-order perturbation solution gave the most accurate predictions for both thetemperature and height of the solid (and consequently liquid) layer. However, the perturbation solution is ana-



ARTICLE IN PRESS

lytically more involved since the second-order correction must be included to ensure the best accuracy, andthis requires the use of the boundary immobilisation method and a change in time variable. In addition, extracomplications arise with �i and �w changing orders depending on the size of DT. The HBI method is mucheasier to implement, though from examining the results in Section 7 it is clear that it is not accurate in thewater layer. Therefore, it seems that the best compromise between simplicity and accuracy is to use a pertur-bation solution in the liquid layer and the HBI solution in the solid layer.

Acknowledgements

T.M. acknowledges the support of the National Research Foundation of South Africa, under Grant Num-ber 2053289 as well as the Department of Mathematical Sciences at the Korean Advanced Institute of Scienceand Technology (KAIST) where a large part of this work was carried out. SM acknowledges the University ofCape Town Post-Doctoral Fellowship and KAIST.

Appendix. Glaze solution for the variable energy boundary condition

In Section 6.1 we determined the perturbation solutions for both the temperatures in the water and ice lay-ers, given by T and h, respectively, (or S and / after the transformation y = z � b(t) and t = t(b)). We wrotedown results for the simpler fixed energy boundary condition (i) and here we give the results for condition (ii).These are far more complicated, especially the second-order correction /2, but this term is shown to be nec-essary to give a more accurate approximate solution than the HBI solution, as discussed in the results sectionin Section 7.

The leading order solutions corresponding to (50) are

Pleaput.

S0 ¼ A0ðbÞy; A0ðbÞ ¼1

b; and /0 ¼ B0ðhÞy; B0ðhÞ ¼

P w � pw

1þ pwh; ð69Þ

with the Oð�wÞ solution given by

/1 ¼ B1ðb; hÞy �k2

B0ðhÞy2 þ k6

B00ðhÞhby3; ð70Þ

where

B1ðb; hÞ ¼k2

B0ðhÞh2þ pwh1þ pwh

� k6

B00ðhÞhbh2 3þ pwh1þ pwh

; ð71Þ

and

k � kðb; hÞ ¼ oS0

oy� k

o/0

oy

� �y¼0

¼ A0ðbÞ � kB0ðhÞ ¼1

b� kðP w � pwÞ

1þ pwh: ð72Þ

Note that the expression in (70) involves hb and so we will have hbb terms in /2. The second-order solution S2 isidentical to that in (53) but with k replaced by (72). However, /2 is now

/2 ¼ B2ðb; hÞy þ1

2B1ðb; hÞðkB0ðhÞ � kÞy2 þ 1

6½kðB1b þ B0kÞ � kB1ðb; hÞB00ðhÞhb�y3 � k

24½kbB0ðhÞ

þ 2kB00ðhÞhb�y4 þ k120½kbB00ðhÞhb þ kðB000ðhÞh2

b þ B00ðhÞhbbÞ�y5; ð73Þ

where

B2ðhÞ ¼ �1

2B1ðb; hÞðkB0ðhÞ � kÞh 2þ pwh

1þ pwh� 1

6½kðB1b þ B0kÞ � kB1ðb; hÞB00ðhÞhb�h2 3þ pwh

1þ pwhþ k

24

½kbB0ðhÞ þ 2kB00ðhÞhb�h3 4þ pwh1þ pwh

� k120½kbB00ðhÞhb þ kðB000ðhÞh2

b þ B00ðhÞhbbÞ�h4 5þ pwh1þ pwh

: ð74Þ



ARTICLE IN PRESS

The Stefan condition is again given by (55) and so b can be determined using the initial condition b(tw) = tw.Note that the expressions for /1 and /2 both involve hb and the expression for /2 also involves hbb. Hence wemust determine hb to first-order and hbb to leading order. As discussed in Section 7, hb = (tb � 1)/q. Sincetb ¼ db

dt

� �1we use (55) to give

Pleaput.

tb ¼ fA0ðbÞ � kB0ðhÞ � �wkB1ðb; hÞg�1: ð75Þ

However, B1(b, h) also involves hb and so we write tb = tb,0 + �wtb,1. After substitution of expressions for A0(b),B0(h) and B1(b, h) into (75) we find that

tb;0 ¼1

k; tb;1 ¼

1

kk2

B0ðhÞh2þ pwh1þ pwh

� k6q

B00ðhÞðtb;0 � 1Þh2 3þ pwh1þ pwh

� �:

Also, to leading order in �w it can easily be shown that

tbb ¼ �1

ðA0ðbÞ � KB0ðhÞÞ2A00ðbÞ �

kq

B00ðhÞ1

ðA0ðbÞ � KB0ðhÞÞ2� 1

!" #:

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2939.[13] L. Makkonen, M. Auttti, Wind Energy-technology and Implementation, The Effects of Icing on Wind Turbines, Elsevier Science

Publishing, 1991.[14] T.G. Myers, An extension to the Messinger model for aircraft icing, AIAA J. 39 (2) (2001).[15] T.G. Myers, J.P.F. Charpin, A mathematical model for atmospheric ice accretion and water flow on a cold surface, Int. J. Heat Mass

Transfer 47 (2007) 5483–5500.[16] T.G. Myers, J.P.F. Charpin, S.J. Chapman, The flow and solidification of a thin fluid film on an arbitrary three-dimensional surface,

Phys. Fluids 14 (8) (2002) 2788–2803.[17] T.G. Myers, J.P.F. Charpin, C.P. Thompson, Slowly accreting ice due to supercooled water impacting on a cold surface, Phys. Fluids

14 (1) (2002) 240–256.[18] T.G. Myers, D.W. Hammond, Ice and water film growth from incoming supercooled droplets, Int. J. Heat Mass Transfer 42 (1999)

2233–2242.[19] T.G. Myers, S.L. Mitchell, G. Muchatibaya, M.Y. Myers, A cubic heat balance integral method for one-dimensional melting of a

finite thickness layer, Int. J. Heat Mass Transfer 50 (2007) 5305–5317.[20] G.I. Poots, Ice and Snow Accretion on Structures, Research Studies Press, 1996.[21] S.K. Thomas, R.P. Cassoni, C.D. MacArthur, Aircraft anti-icing and de-icing techniques and modelling, J. Aircraft 33 (5) (1996) 841–

854.


approximate solution methods for one-dimensional

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