Theoretical Physics

Cooling a Bottle of Champagne:Mathematical Models for the Cooling Process

Oskar Birkne (891020-0156)birkne@kth.se

Jana Hanke (880806-1686)janah@kth.se

SA104X Degree Project in Engineering Physics, First LevelDepartment of Theoretical Physics

Royal Institute of Technology (KTH)Supervisor: Tommy Ohlsson

May 13, 2011

Abstract

In this report, a number of mathematical models for the cooling of a bottle of champagnehave been developed. Both simple analytical calculations and more advanced numericalsimulations have been made. The models considered varied in the inclusion of differentthermodynamical phenomena, such as conduction and radiation, as well as geometricsimplifications. The impact of these as well as the influence of the different ingredientsin champagne and the cork on the bottle have been evaluated. The cooling of thechampagne has been simulated when the bottle has been placed in different environments;a refrigerator, a freezer, water and ice. The cooling times have then been comparedbetween different models and between different cooling surroundings to check the validityof the different models.

Sammanfattning

I denna rapport har ett antal matematiska modeller för kylningen av en champagne-flaska utvecklats. Både enkla analytiska beräkningar och mer avancerade numeriskasimuleringar har gjorts. De övervägda modellerna har inkluderat olika termodynamiskafenomen, såsom konduktion och strålning, såväl som olika geometriska förenklingar. De-ras inverkan såväl som effekterna av de olika ingredienserna i champagne och korken påflaskan har utvärderats. Kylningen av champagnen har simulerats då flaskan har placer-ats i olika omgivningar; ett kylskåp, en frys, vatten samt is. Kylningstiderna har jämförtsmellan olika modeller och mellan olika kylande omgivningar för att kontrollera giltighetenhos de olika modellerna.

Contents

1 Introduction and Background Material 1

2 Investigation 32.1 Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2.1 The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.2 The Transmission and Boundary Conditions . . . . . . . . . . . . 52.2.3 The Subdomains . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2.4 The Inner and Outer Boundaries . . . . . . . . . . . . . . . . . . 72.2.5 Simplifications of the General Model . . . . . . . . . . . . . . . . 7

2.3 Analytical Calculations: Model 4 . . . . . . . . . . . . . . . . . . . . . . 102.3.1 Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3.2 Wine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 Numerical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4.1 Setting up the Bottle Geometry . . . . . . . . . . . . . . . . . . . 132.4.2 Setting up the Physics . . . . . . . . . . . . . . . . . . . . . . . . 162.4.3 Performing the Simulations . . . . . . . . . . . . . . . . . . . . . 17

2.5 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5.1 Model 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5.2 Table and Diagrams of all Results . . . . . . . . . . . . . . . . . . 18

2.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.6.1 General Simplifications and Their Influence . . . . . . . . . . . . 192.6.2 Model 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6.3 Model 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.6.4 Model 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.6.5 Model 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.6.6 Comparison of our Models . . . . . . . . . . . . . . . . . . . . . . 22

3 Summary and Conclusions 23

Notations 25

Bibliography 26

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Chapter 1

Introduction and Background Material

One of the main concerns in food industry is storing its products at right temperature inorder to guarantee quality and taste. Therefore, efficient heating and cooling proceduresare needed. For heating, cooking is the most common way but some groceries can also beheated faster in a microwave oven. For cooling, the most common methods is to put thegrocery in a refrigerator or a freezer, depending on the desired temperature and effect.

One of the groceries that have a very narrow interval of serving temperature is cham-pagne and other sparkling wines, varying from 6 ◦C to 8 ◦C. These, one often wants tocool as fast as possible, either by putting it in the refrigerator or in the freezer. Thisreport aims to calculate the cooling time for different cooling methods for a filled bottleof champagne.

In this report, we will consider the following problem: We assume a filled bottle ofchampagne consisting of two parts, the champagne and the glass bottle. This bottle willthen be considered to be in a third domain, the refrigerant, which will vary. Differentmathematical models are then used to determine the cooling times, including variousthermodynamical phenomena.

Similar problems have been studied both by experiments, calculations and simulations.A simple calculation for cooling white wine was performed in Ref. [1]. In the article, somesimple approximations were made by assuming the wine to be water and neglecting con-vection. This gave some limits for the time required to cool the bottle to optimal servingtemperature. According to the article, the cooling time can be reduced considerably byusing a commercial cooling jacket instead of air cooling.

In Ref. [2], both experiments and simulations of beer bottles were used to compareglass and aluminum bottles. The calculations were made with the help of a completelydifferent model, including both conduction, convection and radiation. The experimentsagreed very well with the numerical solutions. One of the conclusions was that thethermal conductivity in the bottle only plays a minor role for the cooling process. Theimportant part of the cooling is foremost the convection, followed by radiation.

The article in Ref. [3], concerning heat transfer in refrigerators, also concludes thatconvection represents a major part of the cooling. Another result of their work wasthat for any product in a refrigerator the primary contribution to the heat transfer isconvection outside and conduction inside the product.

Another aspect is that the heat transfer coefficient may vary with the temperaturedifference between the product and the surroundings. An article that illuminates this

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fact is Ref. [4], which certainly concentrates on solid foods but the theory behind is stillsimilar. The difference between solid and fluid food is the convection occurring insidethe latter.

One difference between a wine bottle and a champagne bottle is that champagne issubject to a high pressure. One may wonder if this pressure influences the heat capacityand thereby the heat exchange, but according to Ref. [5] the difference in heat capacityis found to be so small that for our purposes we will regard it as negligible.

In this report, we are going to compare these different models by simple analyticalcalculations and numerical simulations. We will start with the most general model pos-sible for us to solve numerically. By stepwise adding simplifications, we are going to tryto find a simple “toy model” to be solved analytically. In the end, we will discuss thedifferent models and possible failures and weaknesses of these. Moreover, we will discussthe influence of the different simplifications on our results.

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Chapter 2

Investigation

In the following sections, we are going to start by describing the problem and try torepresent this by models of different complexity. The next step is to calculate the coolingtime, both analytically and numerically, with respect to these models. To complete thischapter, we will discuss our results, compare our models with each other and try to findways to improve them. All notations that will be used in this report are presented in thechapter “Notations” on page 25.

2.1 ProblemSuppose we have a bottle of champagne which is stored at room temperature. As thecommon serving temperature for champagne is roughly 7 ◦C, we have to cool it to thisdesired temperature. There are a lot of different ways to do this but we will focus onfinding the fastest way with the equipment you can find in your kitchen. We assume thatthe standard equipment in a kitchen includes a water tap, a refrigerator and a freezer.

2.2 ModelGenerally it can be said that we are looking at heat exchange through a wall of glass. Thisshows that we are to use the transient heat transfer equation, which will be dependenton several parameters:

• Physical properties, such as heat capacity, density and heat transfer coefficient, ofthe different substances (e.g. champagne, glass and air),

• Geometrical properties, such as the shape and the size of the bottle,

• The initial temperature of the bottle,

• The surrounding temperature of the bottle,

• The cooling time.

Taking these into account, the complete equation for all three subdomains, being thechampagne, the glass bottle and the refrigerant, will according to Ref. [6, p. 426f.] be

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(here written in the same form as in COMSOL Multiphysics, Ref. [7, p. 168ff.]):

ρcp∂T

∂t= ∇ · (k∇T )− ρcpu · ∇T +Q. (2.1)

We will approximate our system to exist in a bounded environment. Thus, we willhave to pose boundary conditions. We can either set Dirichlet conditions, assuming agiven temperature, or Neumann/Robin conditions, assuming a given heat flux.

Additionally, there will be transmission conditions on the inner boundaries. Theseare:{

Tl = Tr,

klnl · ∇Tl = −krnr · ∇Tr − εσd(T 4amb − T 4),

continuity in temperature,continuity in flux.

(2.2)

The subscripts identify the side of the boundary where l stands for left and r for right.It should be noted that nl = −nr. The quantity d denotes the thickness of the radiatingbody, ε the emissivity and σ the Stefan-Boltzmann constant.

As the bottle contains three different subdomains, a warm liquid filling, a glass con-tainer and a refrigerant, as well as two inner and one outer boundary, we will get asystem of three different heat equations and three different boundary and transmissionconditions.

2.2.1 The Heat Equation

Equation (2.1) is derived from the conservation of energy. Especially, the right-handside describes the different physical phenomena contributing to the change in the energydensity, ρcp ∂T∂t , in particular heat sources, conduction and convection.

It should be noted that the heat capacity will depend on the temperature and thepressure. Both T and p are time dependent and they affect the inner vibrations in theatoms and molecules. Thus, cp can be defined as

cp(p, T ) = cp(0, T ) + T · f(p, T )

as described in Ref. [5].Also, ρ will be dependent on the temperature and the pressure.

The first part of the heat equation, ∇ · (k∇T ), describes the heat conduction in eachsubdomain. Here, ρ is the density of the material in the respective area, cp is the specificheat capacity for constant pressure, p, also depending on the material, k is the thermalconductivity for each substance and t is the time.

Part two is the convection term which will occur in the liquid and gaseous subdomains.In equation (2.1), this is the term ρcpu · ∇T .

The vector u is the so called velocity field. It models the inner flow of the fluids due toa difference in the density or external forces (e.g. stirring). As the density is dependenton the temperature and the temperature in turn is not constant over a subdomain, thewarmer part of the fluid and gas will rise, against the gravity, while the cooler parts sink.

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This contribution to u can be determined by the incompressible Navier–Stokes equation,given as ρ

(∂u

∂t+ u · ∇u

)= −∇p+∇T+ f ,

∇ · u = 0,(2.3)

with ρ, p, t and u as given above, T being the stress tensor and f the forces on the bodyof fluid per unit volume. It is derived from the conservation of mass, the conservation ofmomentum and the energy equation. The whole derivation can be found in Ref. [8, p. 1ff.].

The third part of the heat equation is the part that describes the heat sources, Q. Theseheat sources are external sources and thus not due to any thermal or fluid dynamical ef-fects. They are independent of the materials and the system itself. A positive Q modelsa heat which is added to the system, a negative Q leads away heat from the system. Theterm can be dependent on the temperature and the time, as for example in a freezer, orfully independent of the system and its heat exchange.

2.2.2 The Transmission and Boundary Conditions

In our description of the transmission conditions, we have to consider the phenomena ofradiation.

As we have a warm body inside a cooler surrounding which can absorb and emit heat,we get so-called grey body radiation on the outside of our bottle.

Grey body radiation is a form of black body radiation for materials that do not radiateall of its heat to the environment. This radiation is described by the Stefan–Boltzmannlaw

P = εσAT 4 (2.4)

as found in Ref. [9, p. 139], where P is the radiated power, ε is the emissivity, σ is theStefan-Boltzmann constant, A is the area which is radiating and T the temperature.The emissivity is a dimensionless factor between 0 and 1, with 1 indicating a black bodyradiation and 0 indicating no radiation at all.

In the transmission conditions, this radiation is found in the term for radiation density,εσd(T 4

amb − T 4), with Tamb denoting the temperature of the surroundings.For the outer boundary, we will set Dirichlet conditions, giving the boundary a con-

stant temperature. This can be done when assuming the volume of the refrigerant to besufficiently large to compensate for the emitted heat from the bottle.

2.2.3 The Subdomains

There are three subdomains that are part of the system: the champagne inside the bottle,the bottle itself and the surroundings. Terms written in [ ] in the following sections areterms that may appear in certain materials but not in all.

The Champagne

The champagne is a fluid at room temperature mainly containing the following sub-stances:

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• Water

• Ethanol

• Sugar

• Carbonic acid

Inside this subdomain, we will have both conduction and convection. The conduction willbe present because of the general heat exchange taking place inside the whole system forreaching a global equilibrium. The convection will be part of the fluid dynamics inside thechampagne. As the champagne is cooled on the outside due to conduction, the warmerpart of the liquid will rise, leaving the cold fluid in the lower regions of the bottle. Theheat equation for this domain is thus

ρChcp,Ch∂T

∂t+∇ · (−kCh∇T ) = −ρChcp,Chu · ∇T. (2.5)

There are no external heat sources inside the bottle.

The Glass Bottle

The bottle itself is made of glass, a solid. Here, no fluid dynamics will occur. Conse-quently, there is no convection term in the heat transfer equation for the glass. Also, noheat sources exist and equation (2.1) can be reduced to a completely conductive term asfollows:

ρGcp,G∂T

∂t+∇ · (−kG∇T ) = 0. (2.6)

The Refrigerant

As mentioned earlier, the refrigerant will be changed several times in our calculations. Itwill be:

• air in a refrigerator (gas),

• air in a freezer (gas),

• cold water (liquid),

• ice (solid).

It will be necessary to look at all surroundings separately as they are all in different statesapart from the air which is used two times.

In the air, fluid dynamics will be important and hence we need a convection termin the description of the air. If we study the freezer and refrigerator in detail and howthey cool the products within, we see an external heat source in form of a ventilation,extracting the heat radiating from the warmer products. Conduction will also be part ofthe model.

The water, being a liquid, will also follow the laws of fluid dynamics and consequently,convection will occur. In contrast to the air cooling, there will be no external heat sources.

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The ice will have no convective term or any heat sources and be purely conductive.Equation (2.1) can generally be seen as

ρSucp,Su∂T

∂t+∇ · (−kSu∇T ) = [Q]− [ρSucp,Suu · ∇T ] (2.7)

for a general environment.

2.2.4 The Inner and Outer Boundaries

The inner boundaries of the system are: between the champagne and the bottle andbetween the bottle and the refrigerant. Additionally, we have one outer boundary wherethe surroundings of the simulation ends.

The Champagne-Bottle Boundary (Boundary 1)

This boundary is inside the system and hence needs transmission conditions to describethe heat transfer from one side to the other. Generally, one could describe the boundaryas done in equation (2.2) without a term for radiation.

The Bottle-Refrigerant Boundary (Boundary 2)

Depending on the substance of the surroundings we will have either a solid-gas, solid-liquid or solid-solid transition.

For all transitions, it can be said that we have a warm, dark body in a cooler envi-ronment. Hence, we will always have a radiation term.

This boundary is also an inner boundary and thus equation (2.2) is used.

The Boundary of the Refrigerant (Boundary 3)

As mentioned above, we choose to set Dirichlet conditions when the volume is sufficientlylarge to compensate the heat from the bottle. This gives us:

Tamb = Ts, (2.8)

where Ts is the starting temperature of the surroundings.Additionally, while including the cork, we have more inner boundaries. They are

mostly insulating and are hence seen as non-radiating. The condition of continuity stillholds.

2.2.5 Simplifications of the General Model

If we were to calculate the exact model of the heat transfer in our system, we would haveto take into account the Navier–Stokes equation (2.3), among others. This is beyond thescope of this work. Therefore, we will make some simplifications. Possible ways could beto:

• assume the physical properties of the material to be constant in time, temperatureand pressure,

• simplify the geometry of the bottle,

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• assume symmetry,

• ignore the cavity of the bottle,

• ignore the cork,

• ignore all ingredients in the champagne apart from the water and the ethanol,

• ignore the air and carbonic acid in the bottle,

• neglect the convection,

• neglect the radiation.

General Simplifications

We will make several simplifications that are used in all of our models. These are:

• constant physical properties, such as density and heat capacity,

• no cavity at the bottom of the bottle,

• approximate the champagne to be made of only water or water mixed with ethanol,

• no gaseous phase inside the bottle (e.g. air and carbonic acid).

As we have no specification for the dependence on the pressure and temperature of thedensity and heat capacity, they are taken as constant. The pressure has only a marginalimpact on the heat capacity, as described in Ref. [5]. The density is dependent onthe volume, V (T, p). A champagne bottle has only a little quantity of air inside itself.There are no very large differences in the temperature and pressure in the measuredtime intervals. Thus the volume will only change marginally and can be assumed to beconstant. Consequently, the density is constant.

Removing the cavity at the bottom of the bottle, we get a simpler geometry, and byremoving the gases, we can approximate the filling of the bottle to be entirely made ofchampagne. The total volume will rise as a result, rising the amount of warm substance,but the insulating effects of the air will also be ignored, leading to a faster cooling of theliquid.

The champagne is only defined by its main ingredients to give a good value for thetotal heat capacity and density. For every model, we will first do a simulation assumingthat the champagne is pure water and secondly a simulation where it is assumed toconsist of a mixture of water and ethanol, later called wine.

All our following models will include all simplifications of all the models previouslymentioned. Also, until mentioned otherwise, there will be a mostly insulating cork onthe bottle where essentially no heat transfer takes place.

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Without Convection (Model 1)

Our first simplification will be to neglect all forms of convection. According to article [1],this gives a fairly good approximation of the problem.

By ignoring the convection, the problem also becomes axially symmetric. We hencereduce the problem to be two dimensional with the coordinates (r, φ, z). Thereby, a newinner boundary has to be taken into account for all following models: the symmetryline at r = 0. This gives the boundary condition ∂T

∂r= 0. Here, the condition is axial

symmetry. Without this approximation, the convection had e.g. depended on whetherthe bottle was standing up or lying down.

The equation system for the subdomains will beρChcp,Ch

∂T∂t

+∇ · (−kCh∇T ) = 0,

ρGcp,G∂T∂t

+∇ · (−kG∇T ) = [Q],

ρSucp,Su∂T∂t

+∇ · (−kSu∇T ) = [Q],

inside the champagne,inside the glass,in the surroundings.

(2.9)

The transmission and boundary condition will remain as described above:Continuity without radiation,Continuity,Tamb = Ts,

boundary 1,boundary 2,boundary 3.

(2.10)

Without Radiation (Model 2)

In the second model, we will ignore the radiation, in addition to the simplifications inModel 1.

Our system of heat equations remain as in (2.9), but the boundary conditions changeto

Continuity without radiation, boundary 1,Continuity without radiation, boundary 2,Tamb = Ts, boundary 3.

(2.11)

Without an Insulating Cork (Model 3)

By taking the insulating cork away, we get a geometry made purely of glass. While theamount of heat exchange through the bottle will change, the equations for the system aswell as its boundary conditions remain as given in (2.9) and (2.11).

With a Simple Geometry (Model 4)

In our last model, we assume the bottle to be a closed, cylindrical tube made of glass.Furthermore, for all calculations, we assume infinite fast diffusion to obtain a constanttemperature in the corresponding domains. This is called a well-stirred model.

The azimuthal component φ does not need to be considered, so we only have totake the radial component, which will be denoted by r, and the height component, z,into account. We can then put the following boundary and initial conditions, whereT = T (r, z, t):

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T (r, z, 0) =

{T0, if r ≤ Rb, 0 ≤ z ≤ H,Ts, else, (2.12)

α∇T = α

(∂T

∂r+

∂T

∂z

)= γ(T − Ts), if r = Rb and z = 0 or z = H, (2.13)

where γ is a constant, T0 is the initial temperature of the bottle, Ts is the temperatureof the surroundings and Rb and H are the radius and height of the bottle, respectively.Using the purely conductive heat equation and integrating over the whole bottle we get∫

bottle

∂T

∂tdV =

∫bottle

∇ · (α∇T )dV, (2.14)

where α is a material constant and calculated as α = kρcp

. Using Gauss’ theorem and∂V∂t

= 0, we can rewrite this as

∂

∂t

∫bottle

TdV =

∫boundary

α(∇T · n)dA−∫

bottle

α∇T · (∇1)dV. (2.15)

Here, the subscript boundary means the boundary of the bottle.Using equation (2.13) and ∇1 = 0, we get

∂

∂t

∫bottle

TdV =

∫boundary

γ(T − Ts)dA, (2.16)

where γ is a constant. The temperature T is assumed to be constant over the wholebottle, which yields

V∂T

∂t= γ(T − Ts)A. (2.17)

This can be rewritten as

∂T

∂t=

γA

V︸︷︷︸− 1

τ

(T − Ts), (2.18)

which is Newton’s law of cooling. This method is used in the beginning of article [1].Having a simple geometry with a cylindrical symmetry as well as a very simplified

heat equation, this model can be solved analytically.

2.3 Analytical Calculations: Model 4The only model that can be solved analytically is Model 4, described in Section 2.2.5.

The simplification in the geometry gives us a cylindrical closed tube of glass as ourbottle. This is a quite good approximation for a standard wine bottle, but a little worseapproximation for a champagne bottle as the champagne bottle has a more defined bottle

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Geometrical property SizeCircumference 23.5 cmHeight to bottle neck 20 cmHeight to wine surface 25 cmCircumference of bottle neck 9.5 cmVolume 75 cl

Table 2.1: Measurements of a wine bottle, Bardolino Classico.

neck. We use a real (red) wine bottle to get some good measures, listed in Table 2.1.For the cylinder approximation, we take the outer height to be 20 cm and the outerradius 3.7 cm. As in article [1], we assume the thickness of the glass to be 3 mm. Thevolume is not really 75 cl as we do not take the small cavity at the bottom into account,but the relative deviation is only two percent.

For the cooling, we use Newton’s law of cooling, equation (2.18). By solving thisequation, we get

T (t) = Ts + (T0 − Ts)e− t

τ , (2.19)

where T0 is the temperature of the liquid at t = 0, which we assume to be 20◦C.According to Ref. [1], the time constant τ is depending on the material according to

τ = RC, (2.20)

where C is the heat capacity of the liquid and R is the thermal resistance of the glass.Here, R is depending on the geometry of the bottle according to R = d/(kA), where dis the thickness, A is the area of the glass wall and k is the thermal conductivity. Somethermal data is found in Table 2.2. Furthermore, one can calculate the constant γ inequation (2.18) by comparing the different expressions for τ , (2.18) and (2.20). One findsthat

γ = − V

RCA. (2.21)

2.3.1 Water

In article [1], the wine is assumed to be water for the calculations. We start off by doingthe same calculations for our approximated wine bottle, see Section 2.3.

We assume that the bottle is surrounded by a cool substance, starting with the air ina fridge. For the area, we take the outer surface area as described in the aforementionedarticle, that is A = 2πRb(Rb +H) where Rb is the outer radius and H is the height. Theheat capacity will be C = cpm, where cp is the specific heat capacity and m is the mass.Insertion in τ = RC gives

τ =mcpd

k · 2πRb(Rb +H). (2.22)

1Data from Ref. [10–15].2The data for water is SMOW - Standard Mean Ocean Water.3The data for thermal conductivity, k, and density, ρ, for wine are calculated via xwine = xwvw+xeve,

where x is the property (i.e. k or ρ), v is the volume percentage of the substance and the subscriptsw and e denotes water and ethanol, respectively. The specific heat capacity is calculated according toequation (2.24).

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Material Specific heat capacity[ J

kgK

]Thermal conductivity

[W

Km

]Density

[kg

m3

]Water2 (10 ◦C) 4192 0.5800 999.7Ethanol 2480 0.176 789Wine3 4029 0.532 974Air (1 bar, 300 K) 1007 0.0262 1.161

Glass 840 0.95 2400-2800Ice (−10 ◦C) 2030 2.3 918.7Cork (granulated) 1900 0.048 86

Table 2.2: Thermal data for used materials.1

The time t is then given by the equation

t = −τ ln

(Tserv − Ts

T0 − Ts

), (2.23)

where equation (2.23) is a result of equation (2.19) with T (t) = Tserv. Observe that Tserv

must be larger than Ts. Otherwise, the equation does not hold.

2.3.2 Wine

To approximate the champagne as water is not the most brilliant approximation. Abetter approximation would be to include the alcohol of the champagne in the water.That is a good approximation for wine. Most champagnes and sparkling wines containaround 10 - 12 volume percent of alcohol (ethanol). In our calculations we will use 12 %alcohol like many champagnes.

The heat capacity will differ from that of water:

C = cpm = cpρV = (cp,wρwvw + cp,eρeve)V, (2.24)

where the indices w and e denotes water and ethanol respectively and v denotes thevolume percentage. We can still use equations (2.22) and (2.23).

2.4 Numerical AnalysisAs most of our models are too complex to calculate analytically, we need to use a numer-ical solver. We also want to do the calculations for the actual shape of a real champagnebottle. One way to do this is to take a photo of the bottle and import it to the programin some way. One program that is able to import images is COMSOL Multiphysics, butonly when it is connected to Matlab.

COMSOL Multiphysics is a program which utilizes a Finite Element Method (FEM)to solve Partial Differential Equations (PDEs), such as the heat equation, numerically.For all our numerical calculations, we use the programmes COMSOL Multiphysics 3.5a(hereby denoted CM) and Matlab 7.10.0 (R2010a).

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2.4.1 Setting up the Bottle Geometry

As we just said, to import the geometry to CM we need to connect it to Matlab. Afterdoing this we can apply the CM function flim2curve. This function returns a CM curveobject from Matlab image data.

(a) (b) (c)

Figure 2.1: (a) shows the Lutter & Wegener bottle we use to get the contours for thesimulation. Picture taken from Ref. [16], c©1999 - 2011 by guenstiger.de Verlag GmbH.To get a good contrast between the bottle and the background we paint the whole bottleblue. As Matlab rotates the image 180◦, we also rotate it 180◦ to get the correct orientationafter importing it into Matlab. This is shown in (b). As CM works with axial symmetry,it is enough to import half the bottle as in (c).

We will use the bottle in Figure 2.1, which is a Lutter & Wegener bottle. We paintit blue to get a good contrast to the white background when importing it into Matlab.Furthermore, we cut it in half and rotate it 180◦ to get the correct bottle needed whenwe have imported it into Matlab and CM. Half the bottle will be sufficient becauseof the axial symmetry and the reason why we rotate the picture 180◦ is that Matlabautomatically turns the picture upside down. Our manipulations give us a correctlyoriented picture in the end.

The commando imread in Matlab takes an image file and converts into a matrix.This matrix can be used as the input image for the flim2curve commando. The otherinput parameters are used to create contours and control the noise levels in the image.flim2curve returns the curve2-object that we will use in CM. The following is the Matlabcode we use for our bottle:

imfile = imread(’sektblue-half2.jpg’);[c,r] = flim2curve(imfile,{[],0:50:20000},’KeepFrac’,0.10);

To use this in CM we only need to import it. We then remove all the unnecessarysubdomains that were created when importing the image, see Figure 2.2.

We will have a basic geometry for all our numerical models, which consists of twobottles, a smaller bottle inside the bigger one. These build up the bottle, the space inbetween them being the glass walls.

Figure 2.2b is rotated 180◦ about its symmetry axis and placed so that the bottomleft corner is at the origin in CM. To get the smaller bottle, we clone the first bottleand remove a stripe at the left of the bottle, see Figure 2.3, corresponding to the bottleglass thickness of 8 mm, a value found in Ref. [17]. We also need to scale it down in thez-direction.

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(a) (b)

Figure 2.2: Removal of unnecessary subdomains (denoted by CO) in CM. (a) shows thebottle before removal of the unneccesary subdomains and (b) shows the bottle afterwards.

Figure 2.3: Two identical bottles before removing the stripe on the right bottle.

When we have the two bottle shapes, we put them into each other with the straightside of both bottles aligned at the symmetry axis and so that the thickness of the glasswalls is approximately the same everywhere, as in Figure 2.4. We also make sure the sizeof the bottle matches the measurements in Table 2.3.

Figure 2.4: Two subdomains, the white corresponding to the champagne and the red to theglass walls of the bottle.

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Geometrical property SizeCircumference 26.5 cmHeight to bottle neck 15 cmHeight to wine surface 25 cmCork height 4 cmCircumference of bottle neck 10 cmVolume 75 cl

Table 2.3: Measurements of a sparkling wine bottle, Rotkäppchen Sekt.

Next, we need to define the geometry of the surroundings. To avoid the programmingof external heat sources like the air cooling in a freezer, we embed the bottle into asufficiently large volume. This will cause the warming of the surroundings at infinity tobe neglected. We introduce a rectangle with a height that is three times the height ofour bottle and a width that is ten times larger than the radius of the bottle. The finalstructure, containing three subdomains, is shown in Figure 2.5. This geometry describesModel 3.

Figure 2.5: The geometry used in Model 3.

At last, a cork is included, resulting in a fourth subdomain. This gives us our finalgeometry for Model 1 and Model 2 which can be seen in Figure 2.6.

(a) (b)

Figure 2.6: The geometry used in Models 1 and 2.

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As CM uses a FEM method, the geometry needs to be meshed to create finite elementsfor CM to calculate with. The meshed geometry is shown in Figure 2.7.

Figure 2.7: The meshed problem in CM, when the cork is included.

2.4.2 Setting up the Physics

The values used in the numerical simulations are given in Table 2.2 and Table 2.4.

Property Value CommentTserv 280.15 K Serving temperature for champagneTs,fridge 279.15 K Temperature of the refrigeratorTs,freezer 255.15 K Temperature of the freezerTs,water 277.15 K Temperature of the waterTs,ice 263.15 K Temperature of the iceT0 293.15 K Initial temperature of the champagne bottleεglass 0.95 Emissivity of glass

Table 2.4: Used constants for temperature and emissivity in CM.

The Subdomains

The three subdomains are constructed in CM as described before. Their physical prop-erties remain the same for all simulations and are defined by their corresponding valuesof cp, k and ρ from the predefined constants. In Models 1 and 2, we get an additionalsubdomain for the cork.

There are two different ways to solve Model 1. The first is to do as suggested inSection 2.2.2 and put the radiation in the boundaries. The other way is to assume thatthe glass bottle wall is thin compared to its surroundings and put the radiation termthere. This is done by putting

Q = −εσ(T 4 − T 4s )

d

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in the glass in equation (2.9). Here, d is the thickness of the bottle wall. The minus signimplies that the heat is led away from the bottle.

As mentioned earlier, we do not have any external heat sources in our simulationsexcept for the just mentioned second way to solve Model 1. Thus, in all other modelswe only take the conductive terms into account and set all convective and heat terms tobe equal to zero. The time-scaling coefficient used by CM is not relevant in our problemand is set to one.

Additionally, the initial temperature for all parts of the bottle is taken to be T0 andfor the surroundings the temperature is taken to be Ts. Here, Ts is changed in the tableof predefined constants for every simulation to account for the different cooling materials.

The Boundaries

The boundaries will be defined similarly in all models. The boundaries on the two sidesof the bottle are taken to be interior boundaries and for all models apart from Model1, the transmission condition are assumed to be continuity. The additional boundaryoccurring on the cut through the bottle is defined by axial symmetry. The boundaries(except the one at the symmetry axis) of the cork have the property of continuity in allmodels.

The outer boundary is important to describe the type of cooling with a view to thecompensation of the heat taken out of the bottle. To model this, we assume that thevolume of the surroundings is sufficiently large to compensate for all heat of the bottleand the outer boundary thus has a constant temperature.

Before simulating Model 1 we change the property of the boundary between the bottleand the surroundings. To describe the radiation, we change to a discontinuity in the heatflux and include a term for radiation according to Stefan–Boltzmann’s law, equation (2.4).The value for the emissivity is found in Table 2.4, taken from Ref. [18], and is taken tobe 0.95. All other coefficients are set equal to zero.

2.4.3 Performing the Simulations

When the geometry and the physics of the problem is correctly set up in CM the onlything left to do is performing the actual simulations. We will consider the time taken forcooling the whole bottle to the desired temperature. That means that we are looking atthe warmest point in the bottle in our simulations.

As we have a transient problem we need to tell CM how long time it should simulate.We choose to make a ”trial and error” approach to the simulations and try with anestimated time value and then try with a better estimation and shorter time step (forbetter accuracy) until we reach the desired temperature for the warmest point in thebottle.

2.5 Results

2.5.1 Model 4

In the following, our results for the analytical calculations obtained with Model 4 areshown.

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Water

With data from Table 2.2 and our earlier assumptions of the bottle size done in Sec-tion 2.3, equations (2.22) and (2.23) gives us:

τ = 130 s (2.25)t = 343 s = 5 min 43 s (2.26)

where Tserv = 7◦C. Here we assume Ts = 6◦C (the temperature of a refrigerator).If we make the same calculations for a freezer (Ts = −18◦C) instead of a refrigerator,

we get:t = 54 s (2.27)

The results in equations (2.26) and (2.27) are apparently much too low. In reality, it willtake much longer time to cool the bottle, but we save the discussions to Section 2.6.5.

Wine

With data from Table 2.2, equation (2.24) gives C = 2937.5 J/K. If we now put thebottle in a refrigerator like in the case with only water, we get the following relaxationtime and cooling time:

τ = 122 s (2.28)t = 322 s = 5 min 22 s (2.29)

In a freezer instead of a refrigerator the cooling time is:

t = 51 s (2.30)

These results are even more unrealistic than the corresponding results of a water bottle.

2.5.2 Table and Diagrams of all Results

All our results are compiled in Table 2.5. To facilitate the qualitative analysis, thediagrams in Figure 2.8 displays the results grouped by refrigerant. A picture of one ofthe results in CM is found in Figure 2.9. All our numerical calculations were made on acomputer running Intel Core2 Quad CPU Q9550 (2.83 GHz) with a 4 GB memory. Onesimulation took between one to 20 minutes depending on the mesh and the number oftime steps.

2.6 DiscussionIn the following section we will discuss our results and draw conclusions of the validityof the different models.

A reference value for the time needed to cool a champagne bottle in air is found inRef. [1]. The thermal relaxation time was found to be around 3 hours, which correspondsto cooling times for our refrigerator and freezer of 8 hours and 1 hour and 15 minutesrespectively. The other refrigerants are expected to cool faster as their thermal resistancesare lower. We will use these estimations as reference values for our further discussions ofthe validity of the models.

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Surrounding Filling Cooling timematerial Model 1S Model 1B Model 2 Model 3 Model 4

Air (6◦C) Water 9 h 38 min 10 h 48 min 83 h 36 min 83 h 6 min 5 min 43 sWine 9 h 19 min 10 h 24 min 79 h 30 min 79 h 12 min 5 min 22 s

Air (−18◦C) Water 2 h 1 min 2 h 14 min 13 h 24 min 13 h 24 min 54 sWine 1 h 58 min 2 h 10 min 12 h 48 min 12 h 48 min 51 s

Water (4◦C) Water 2 h 15 min 2 h 15 min 2 h 34 min 2 h 32 min N/AWine 2 h 10 min 2 h 11 min 2 h 28 min 2 h 26 min N/A

Ice (−10◦C) Water 38 min 12 s 38 min 6 s 38 min 36 s 38 min 6 s N/AWine 38 min 37 min 48 s 38 min 24 s 38 min 30 s N/A

Table 2.5: The table shows a compilation of all our results for the different models. Model1S denotes Model 1 when the subdomain is assumed to be radiating and Model 1B whenthe boundary radiates instead.

2.6.1 General Simplifications and Their Influence

In Section 2.2.5 we made some general simplifications. One of them was to assume thatthe champagne consisted of only water or a mixture of water and ethanol. This does nottake into account the other ingredients in the champagne, such as sugar and carbonicacid, as well as the air and pressure inside the bottle.

From our calculations we can conclude that the difference in cooling time betweenwater and what we call wine is quite small. Water and ethanol are the main ingredients

(a) (b)

(c) (d)

Figure 2.8: Overview of the results divided with respect to the different refrigerants. Model4 is only used for the refrigerator and freezer and the resulting cooling times are too low tobe viewable.

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Figure 2.9: A CM plot of the results obtained from Model 2 when the champagne is assumedto be wine and the bottle is placed in a refrigerator.

in champagne, so the results would probably not change very much if the other ingredi-ents also were included. To approximate the champagne with water is therefore a goodapproximation when there is no need for very high accuracy in the calculations.

As we always chose to measure the maximum temperature of the champagne in thenumerical calculations, all of the numerical results should be seen as upper limits for theactual cooling times.

2.6.2 Model 1

We calculated Model 1 in two different ways; once with a radiating subdomain and oncewith a radiating boundary. The approximation with a radiating subdomain holds as longas the corresponding dimension in length is small compared to the whole system.

This approximation seems to hold for water and ice as the cooling substance but notfor air cooling when compared to a radiating boundary. A reason for this deviation couldbe that the radiation plays a larger role when cooling with a gas. Hence, when includingradiation, the size of the radiating area influences the results for air cooling much more

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than in water or ice.It can also be noted that when cooling with ice, it takes longer time with a radiating

subdomain than a radiating boundary. This seems very unrealistic. On the other hand,the difference between the results is very small, less than 15 seconds from a total of 38minutes. Therefore, one reason could be computational errors in the program. To checkthis theory, we tried to refine the mesh and solve the simulation with a smaller time step.This simulation exceeded the memory of the computer making it impossible to solve it.

When comparing the results with each other they seem to be realistic as a wholealthough one should take the result from the radiating boundary when cooling with airbecause of the above mentioned reasons.

2.6.3 Model 2

In this model, we ignored the radiation. This has a great influence on the air cooledsystems. The calculated time for the cooling becomes unrealistically high, in a refrigeratormore than three days. Hence, the radiation plays a major role in these systems.

For the water and ice cooled systems, the results vary less than for the other sys-tems. For water, the radiation influences the results noticeably but the results are stillreasonable.

The radiation seems to have little influence on the ice cooled system, differing lessthan two percent from the results in the previous model.

2.6.4 Model 3

In Model 3 we removed the insulating cork and assumed a bottle purely made of glass.The deviations for the different systems when compared to the previous model are rel-atively small. For the air and water cooled systems, the deviations are less than twopercent. Hence, the insulating cork does not seem to have a large influence for thisnon-radiating model.

For the ice cooled system, we can notice two things. One is that the disregard of thecork seems to compensate for the lacking radiation when assuming a filling of water. Onthe other hand, when assuming a filling of a wine mixture, the cooling time is gettinglonger. This is unrealistic and probably is a computational error in the calculation.

2.6.5 Model 4

As we saw in Section 2.5.1, the calculated cooling times, around 5-6 minutes in therefrigerator and about 50 seconds in the freezer, were much too low compared to ourintuition and practical experience. This leads to the conclusion that our model hasseveral shortcomings.

One of the most significant shortcomings is that we do not take the material sur-rounding the bottle into account. We only assume that the surrounding has a specifictemperature but ignore its thermal properties. The error is rather large for air as it is aquite good insulator.

Essentially, we model our problem by taking a thermal barrier, here glass, and a“thermal layer” at each side of the barrier with constant temperature and describe the heattransfer between the two layers through the barrier. This leads to several complications.

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The temperature is not constant, neither inside the bottle nor outside of the bottle.On the outside, we have a thin layer of varying temperature which will be warmer thanthe rest of the air. The temperature in the system will be continous at all points withoutany abrupt changes in temperature. Also, the temperature is seen as constant but itactually varies with the radial coordinate. The closer we are to the symmetry axis, thewarmer the liquid will be. On the outside, the temperature will be higher the closer weget to the bottle.

As the temperature is varying in the separate subdomains, there also occurs convec-tion as both the gas and the liquid will try to reach thermal equillibrium. This is ignoredin our analytical calculations. The same can be said about radiation, which is also animportant factor. These two disregards will however contribute to an even shorter coolingtime which would yield even more unrealistical results.

We approximate the wine bottle with a cylinder which is causing deviations. We alsoassume that this cylinder only consists of glass, thereby disregarding the insulation nearthe cork.

2.6.6 Comparison of our Models

It can be seen when studying Table 2.5 that the cooling times for ice are by far the short-est. Thus, for cooling the champagne as fast as possible, a surroundings of ice shouldbe chosen. It should be noted that the actual time will be longer than calculated as weassumed a large ice cube in the simulation. In reality, this is difficult to accomplish andone would take small ice cubes instead. Thus, there will be a layer of air in between thecubes which will slow the process down. A way to get around this problem is to havethe ice cubes in water which apparently is a better cooling substance than air, but thetime will still be longer than if one would be able to freeze the whole bottle into a bigice block. On the other hand, as said before, we look at the maximum temperature inthe champagne which makes our results upper limits for the actual cooling time.

It can be said that not all models give reasonable results.For ice cooled systems, the model plays a minor role. The results differ with less than

a minute between all models. This means that the largest contribution to the cooling isby far conduction.

For the water-cooled system, all models give reasonable results although there is anoticeable difference between Model 1 and Models 2 and 3. The choice of way to computethe radiation in Model 1 is not important for water as both ways give the same results.The conclusion is that conduction is the biggest contributor to the cooling in water butfor an exact result radiation should also be taken into account.

For air cooled systems, only Model 1 gives realistic results. Models 2 and 3 resultin too long times and Model 4 in too short times. Also, it should be noted that theapproximation with a radiating subdomain does not seem to hold for this system. Ra-diation is far more important to consider when air (and probably most other gases) isused as refrigerant. This has to do with the large thermal resistance of air which makethe conduction more inessential in air than in water or ice.

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Chapter 3

Summary and Conclusions

In this report, our goal has been to study different ways of cooling groceries. Here, weconcentrated on the cooling of a filled champagne bottle to the optimal serving temper-ature of 7◦C. To estimate the cooling speed, we used several mathematical models ofdifferent complexity.

To determine the cooling time needed when the bottle was placed in a refrigera-tor, a freezer, a water-filled sink and a large ice cube, the heat equation was studied.Several models with different simplifications were derived for the calculations and sim-ulations. These models were later simulated in a program specialized on the solving ofpartial differential equations. All results were compiled and compared, both between thedifferent substances in each model and between different models with the same substance.

Even if the models predict very different cooling times, the overall picture is identi-cal. It was noted that the fastest way of cooling a champagne bottle was to put it ina large ice cube. For all models, the time essentially did not differ at all and gave acooling time of about 40 minutes. Practically, it would be difficult to produce this icecube. Instead, one would probably use many small ice cubes, thus introducing small airchannels in the system, resulting in a longer cooling time.

The next to fastest ways were, in the order from fast to slow: water-filled sink (twoto three hours), freezer (between two and 13 hours, model depending) and refrigerator(between ten hours and over three days, model depending). For the analytical model,using Newton’s law of cooling, the results for the refrigerator and freezer were five andone minutes respectively which were deemed unrealistic compared to the reference valuesdiscussed in the beginning of Section 2.6.

When studying the different models, it became clear that radiation plays a majorrole in air cooled systems. When ignoring radiation in these systems, the cooling timerequired could rise to being eight times slower than when including radiation. In watercooled systems, radiation seemed to play a minor role. In ice cooled systems, no realinfluence was noted at all.

For all simulated models, it could be seen that the champagne had temperature vary-ing over space.

To obtain better results in future analysis, one would have to take into account fluiddynamical effects such as convection. As mentioned above, the effects on the champagnewere very noticeable. For smaller volumes of air, the lacking convection would probably

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have been seen as well.To simulate the actual champagne better, all ingredients should be considered. Thus,

more reliable thermodynamical properties could have been obtained.When studying the ice cooled system, it should be taken into account that the actual

cooling will be done by many small ice cubes. In contrast to the results mentionedabove, the cooling time would get longer because of small air channels on the outside ofthe bottle. The cooling time could be accelerated by having water in between the icecubes.

Our results occasionally varied very much between the different models. To avoidthis, one could use a better mesh and smaller time steps. It was not possible for us todo this as the memory of our computer was not sufficient.

To check our models better, experiments should be made to get some good referencevalues for the ”real” cooling time.

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Notations

Denotation DescriptionA AreaC Heat capacitycp Specific heat capacity at constant pressured Thickness of bottlef Force per unit volumeH Heightk Thermal conductivitym Massn Normal unit vectorP Powerp PressureQ Heat sourceR Thermal resistanceRb Radius of the bottleT Stress tensorT TemperatureTamb Ambient temperatureTs Surrounding temperature at t = 0Tserv Serving temperatureT0 Initial temperaturet Timeu Velocity fieldV Volumeα Thermal diffusivityγ Material constantε Emissivityρ Mass densityσ Stefan-Boltzmann constantτ Thermal relaxation time�Ch Champagne�G Glass�Su Surroundings

The denotations used in this report. � denotes an arbitrary physical property.

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