Fundamentals of convectionRichard F Katz – University of Oxford – March 23, 2012

(Based on Geodynamics by Turcotte & Schubert, and Heat Generation and Trans-port in the Earth by Jaupart and Mareschal, both from Cambridge University Press.)

We are interested in studying the fluid flows driven by temperature variationswithin the mantle. We have learned that creeping flows, because of their highReynolds number, satisfy Stokes equation

∇p = µ∇2v + ρg. (1)

Note that in the absence of flow v = 0, the pressure gradient is lithostatic, p(z) =ρgz, where z is depth into the Earth. Here ρg is the body force of gravity. Convectionis driven by lateral variations in the body force, as we shall see. Also note that theviscosity µ has been assumed constant and taken outside the divergence operatoron the stress tensor.

To study the fundamental physics of convection, we will consider a simpler prob-lem than full mantle convection, developed in detail below, of a fluid that is heatedfrom below and cooled from above. Thermal expansion leads to a vertical densitygradient, with cold, dense fluid above hot, ligher fluid. As we shall see, this situa-tion is unstable: small perturbations to the stillness of the fluid grow exponentially,leading to convection.

1 Convection in experiments

See figure Figure 1 for an example of a laboratory experiment on convection. Theliquid is a viscous syrup containing temperature-sensitive liquid crystals. The crys-tals become visible at certain temperatures, hence the green lines are isotherms.The tank is heated from below.

2 Density variations

To a good approximation, the density of a fluid is linearly proportional to its tem-perature. The constant of proportionality is typically called α. Hence we can writethe density as

ρ = ρ0[1− α(T − T0)], (2)

where ρ0 and T0 are reference values of density and temperature.Substitution of (2) into (1) gives

∇p = µ∇2v + ρ0[1− α(T − T0)]g. (3)

. We know that the background density ρ0 does not cause flow, but merely leads tothe lithostatic pressure gradient. This motivates to write the pressure as p = P+ρ0gz

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Figure 1: A tank of viscous fluid heated from below with a thermal plume formingin the boundary layer. The green lines along the bottom of the tank and aroundthe plume are isotherms. From the lab of Anne Davaille.

(where we have taken z to be depth, and hence increasing downward). Substitutingthis into (3) gives

∇P = µ∇2v− ρ0α(T − T0)g. (4)

Variations in density arising from thermal expansion should appear in the con-servation of mass equation, which reads

∂ρ

∂t+ ∇· (ρv) = 0. (5)

As an important and provably accurate approximation, we assume that the thermaldensity variations can be neglected everywhere except in the body-force terms. Thisis because these density changes are small relative to the fluid’s density. If weneglected them in the body-force terms, however, we’d lose the driving force forconvection, and hence would be unable to make further progress. This is called theBoussinesq approximation. With it, the mass conservation equation becomes

∇· v = 0. (6)

3 The energy equation

Since density variations are due to temperature, we require an equation to computethe temperature variations through space and time. This can be derived from thefirst law of thermodynamics, which states that energy is conserved, and that heatand work are both forms of energy:

dU = dQ− dW. (7)

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In this case, we are only concerned with the thermal energy. For our idealisedsystem, it can be shown that the work done by convection is negligible, and hencewe have dU = dQ. This states that changes in internal energy are the result ofaddition or subtraction of heat.

The internal energy is related to the temperature by

dU = ρcdT, (8)

where c is the specific heat capacity. Substituting (8) into (7) and dividing by aninfinitesimal increment of time dt gives

ρcdT

dt=

dQ

dt, (9)

the rate of change of temperature is proportional to the rate of addition of heat.How is heat added? To answer this, consider a two-dimensional “volume” with

sides of length δx and δz. The mass within the box (per unit distance into the plane)is δxδzρ. Heat can enter/leave the box in one of three ways: by diffusion throughthe sides, by advection through the sides, or by radioactive decay within the box.

Diffusion Consider diffusion first, and recall Fourier’s law, which states that theheat flux q is proportional to the temperature gradient:

q = −k∇T. (10)

Along the edges of the box in the x-direction, this leads to a net rate of heat loss of

− dQ

dt

∣∣∣∣diffusion, x

= δz (qx(x+ δx)− qx(x)) ,

= δz

[(qx(x) +

∂qx∂x

δx

)− qx(x)

],

= −k∂2T

∂x2δzδx, (11)

where we have used a Taylor series, truncated at first order, to approximate thechange in diffusive heat flux across the box. Also we have assumed that k is constant.The same applies in the z direction and hence the contribution from diffusion of heatis

− dQ

dt

∣∣∣∣diffusion

= −k(∂2T

∂x2− ∂2T

∂z2

)δzδx. (12)

Advection Motion of the fluid also transports heat; this is called thermal ad-vection or advective heat transfer. To quantify it, we again consider our “controlvolume,” which, in this case, is a two-dimensional box with sides of length δx andδz.

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The advective flux of heat at a point in the domain is a = ρcTv, so the net heatadvected out of the box in the x-direction is

− dQ

dt

∣∣∣∣advection, x

= δz [ax(x+ δx)− ax] ,

= δz

[(ρcTu+ δx

∂

∂x(ρcTu)

)− ρcTu

],

= ρc∂(Tu)

∂xδxδz. (13)

Here we have assumed the ρ and c are constants. The same logic applies to thez-direction, and so we have

− dQ

dt

∣∣∣∣advection

= ρc

(∂(Tu)

∂x+∂(Tw)

∂z

)δxδz. (14)

We can now combine the first law (9) with our expressions for heat transportfrom diffusion (12) and advection (14) to obtain

ρc∂T

∂tδxδz = −ρc

(∂(Tu)

∂x+∂(Tw)

∂z

)δxδz + k

(∂2T

∂x2− ∂2T

∂z2

)δzδx. (15)

Dividing through by ρc δxδz and using vector notation gives

∂T

∂t+ ∇· (vT ) = κ∇2T, (16)

where κ = k/(ρc) is the thermal conductivity. We can simplify this further by usingthe chain rule to expand ∇· (vT ) and noting that the conservation of mass equationtells us that ∇· v = 0. Thus we have

∂T

∂t+ v ·∇T = κ∇2T. (17)

This equation governs the evolution of temperature within the fluid. It says thatchanges in temperature are due to advection by the flow and diffusion down thermalgradients.

4 The Rayleigh number and the Nusselt number

Consider a layer of fluid that is heated from below and cooled from above to createa vertical temperature gradient. The equation for density variations tells us thathotter fluid below has lower density than the colder fluid above. Hence the systemhas available potential energy that could be used to drive fluid flow in which the hotfluid rises and the cold fluid sinks. This is known as thermal convection.

We have learned, however, that viscous stresses resist fluid flow. So a primaryquestion is whether the driving force for convection is sufficiently large to overcomethe resistance of viscosity. Does flow occur?

Two things could happen: the viscous resistance to flow could overcome thedensity inversion, keeping the fluid static. In this case, heat would be conducted

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down the temperature gradient from hot to cold. On the other hand, if the den-sity variation is large enough, it can overcome the viscous resistant to flow, andheat is transported by fluid motion. Even the latter case, conduction still plays animportant role, as we shall see.

We can understand and quantify this interplay of driving and resisting forces forconvection with a simple scaling analysis.

The Rayleigh number Consider a layer of fluid of height h. The boundary atthe top of the fluid is held at temperature T0 and the boundary at the bottom isheld at the higher temperature T0 + ∆T . Now consider a sphere of cold fluid withradius a that collects near the top boundary. Because it is cold, it is denser andhence it sinks at a speed given by

W ∝ g∆ρa2

µ∝ gρ0α∆Ta2

µ, (18)

where we have used Stokes solution for the speed of settling of a dense sphericalobject, and equation (2) to eliminate ∆ρ. Since this is a scaling analysis, we arenot concerned with the exact value of W , but rather how it scales with the problemparameters (such as ∆T ).

As the sphere sinks, it is heated by the surrounding fluid. The time-scale overwhich this heating occurs is given by

τ ∝ a2

κ. (19)

How far has the cold sphere travelled in time τ? It has travelled a distance d, givenby

d ∝ Wτ ∝ ρ0gα∆Ta4

κµ. (20)

Note the a4 dependence on the radius of the sphere: this means that a small spherecan only move a very short distance before it loses its relative coldness (and hencerelative denseness) to diffusion. In other words, small temperature perturbationsget effectively damped out by the system. For convection to occur, the sphere offluid must be able to travel across the entire layer, so we must have d ≈ h, andtemperature perturbations must persist across the entire layer, so a ≈ h. Hence wehave

h ∝ ρ0gα∆Th4

κµ, (21)

or, rearranging,ρ0gα∆Th3

κµ= const ≡ Ra. (22)

where we have defined the constant to be called the Rayleigh number, Ra, afterJohn William Strutt, 3rd Baron Rayleigh. As we shall see in the next section, whenthis number exceeds a critical value (dependent on the specifics of the experiment),driving forces exceed viscous resistance and the fluid convects. When the Rayleighnumber is smaller than the critical value, the fluid remains static and heat is trans-ported by conduction only.

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The Nusselt number Whether by conduction, convection, or both, there is a fluxof thermal energy from the hot boundary to the cold boundary. This flux maintainsthe temperature difference ∆T between the two sides. Let’s denote the actual heatflux qtotal. We can compare qtotal to the heat flux that would be required if the fluidwas still and heat transport was by convection only. This is given by Fourier’s law,integrated over the height of the fluid, |qconduct| = k∆T/h.

The ratio of the actual to the conductive heat fluxes is called the Nusselt number,

Nu ≡ qtotal

k∆T/h. (23)

When qtotal > k∆T/h we have Nu > 1, which means that heat is being transportedby conduction and convection. We can obtain qtotal from an experiment (or numericalsimulation) by direct measurement.

5 The onset of convection

The Rayleigh number gives us a means of assessing the size of the driving force forconvection relative to the viscous resistance: when Ra is large enough, we expectconvection to occur. But how large is large enough? Determining this is our nextgoal. We seek to find the minimum value of Ra when the system begins to convect.We can think of this as looking for the condition when the static, conductive solutionis unstable to perturbations. That is, a small disturbance in temperature or flow willgrow and lead to full-blown convection.

Amazingly, we can do this with a clever analysis of the governing equations. Thetrick is that we know the static, conductive solution analytically, and the solutionwe seek is only a very small perturbation of that known solution. Because of this, wecan linearise the governing equations around the known state, which is also calledthe base state. Once we have linearised the equations, we will be able to solve themanalytically for the convective disturbance.

The governing equations The equations that govern the pressure, temperature,and flow within the fluid layer were given above but are collected here:

0 = −∇P + µ∇2v− ρ0α(T − T0)g, (24a)

0 = ∇· v, (24b)

0 =∂T

∂t+ v ·∇T − κ∇2T. (24c)

This is the system that we seek to understand in this analysis. Solving it analyticallyin this form is not possible because it is non-linear: the term v ·∇T involves theproduct of two unknowns. We therefore take a more clever approach that workswhen the non-linear term can be approximated by a linear term.

In what follows we will use a coordinate system that goes from z = 0 at the top,cold boundary, to z = h at the bottom, top boundary.

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The base state When the Rayleigh number is sufficiently small, we know thatthe viscous resistance to flow will prevent fluid motion, and that conduction willtransport heat across the layer of fluid. In steady-state (∂T/∂t = 0), the equationsare therefore solved by v(x) = 0 with P (x) = P0. Using this in the energy equation(24)c gives

∂2T

∂z2= 0, (25)

and hence for the case of no fluid motion, we can integrate this and apply boundaryconditions to obtain

Tc(x) = T0 +∆T

hz, (26)

where the subscript c indicates that this is the conductive solution.

The perturbed state If the Rayleigh number is just slightly above critical, weexpect very weak convection to occur, and we expect the solution to the equationsto be different than, but very close to, the base-state solution. Hence we can expandthe variables in a series about the base-state solution:

v = 0 + εv1 + ε2v2 + ... (27a)

P = P0 + εP1 + ε2P2 + ... (27b)

T = Tc + εT1 + ε2T2 + ... (27c)

where ε is a constant that is arbitrarily small and represents the size of the pertur-bations to the system. Since ε� 1 we have that ε0 � ε1 � ε2..., at each order, theimportance of the solution at that order is much smaller than the importance of thepreceding order. The leading order O (ε0) is the base state: it is the most importantbut also boring (because it has no flow). The perturbation to the base state is firstorder O (ε1): it is small but interesting to us. The second order perturbation O (ε2)is extremely small pertains to questions beyond what we are trying to discover.

Hence we can truncate the series at order ε1 and substitute into equations (24)to obtain

0 = −∇P0 − ε∇P1 + εµ∇2v1 − ρ0α(Tc − T0)g− ερ0αT1g +O(ε2

), (28a)

0 = ε∇· v1 +O(ε2

), (28b)

0 = ε∂T1

∂t+ εv1 ·∇Tc − κ∇2Tc − εκ∇2T1 +O

(ε2

). (28c)

We can simplify these equations using the fact that

• the O (ε0) terms balance (i.e. they cancel each other),

• the O (ε2) terms are tiny and hence can be dropped,

• the base-state temperature gradient is non-zero only in the vertical direction:∇Tc = (∆T/h)k.

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and we obtain equations for the first-order perturbations,

0 = −∇P1 + µ∇2v1 − ρ0αT1g, (29a)

0 = ∇· v1, (29b)

0 =∂T1

∂t+ w1

∆T

h− κ∇2T1. (29c)

So now our problem is to solve equations (29) for the convective perturbation to thebase state, P1,v1 = (u1, w1), T1. Fortunately, these equations are linear!

Boundary conditions We will assume that the boundaries are at constant tem-perature and that no flow crosses them. Hence we have

w1 = T1 = 0 on z = 0, h. (30)

We need a boundary condition on the horizontal component of the perturbationvelocity, u1. One possible choice is u1(z = 0, h) = 0. However, it is mathematicallyconvenient (and physically realistic, in comparison with the mantle) to instead as-sume that there are no shear stresses on the top and bottom boundaries. This iswritten

τ (1)xz =

∂u1

∂z+∂w1

∂x= 0 on z = 0, h. (31)

Since the perturbation to the vertical velocity vanishes on the boundaries (as givenin (30)), we can simplify this to

∂u1

∂z= 0 on z = 0, h. (32)

The stream function As you have seen previously, the stream-function is usefulfor incompressible flow, because it automatically satisfies the continuity equation.Recall that it is given by

u1 = −∂ψ1

∂z, w1 =

∂ψ1

∂x. (33)

You should verify that this gives ∇· v1 = 0.Introducing the stream-function into equations (29) gives

0 = −∂P1

∂x− µ

(∂3ψ1

∂x2∂z+∂3ψ1

∂z3

), (34a)

0 = −∂P1

∂z+ µ

(∂3ψ1

∂z2∂x+∂3ψ1

∂x3

)− ρ0gαT1, (34b)

0 =∂T1

∂t+

∆T

h

∂ψ1

∂x− κ∇2T1. (34c)

Eliminating the pressure from equations (34a) and (34b) gives the system

0 = µ∇4ψ1 − ρ0gα∂T1

∂x, (35)

0 =∂T1

∂t− κ∇2T1 +

∆T

h

∂ψ1

∂x, (36)

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where

∇4 = ∇2∇2 =∂4

∂x4+ 2

∂4

∂x2∂z2+

∂4

∂z4

is the biharmonic operator (two-dimensional, in this case). The boundary conditionsbecome

∂2ψ1

∂z2= T1 = 0 on z = 0, h. (37)

The problem is now to solve for only two variables, ψ1 and T1 using two linearequations, (35) and (36).

A trial solution Here’s what we know about the solution:

• it must consist of functions for ψ1 and T1, and it must depend on x, z, and t;

• because of the boundary conditions, it must go to zero on z = 0, h;

• because the equation are linear, it must be separable: e.g. ψ1(x, z, t) = F (x)G(z)H(t);

• because the equations are linear, each dependence on the independent variablesmust be either sinusoidal or exponential.

To satisfy all of these conditions, we can propose the following ansatz1:

ψ1 = ψ∗ sin(πzh

)sin

(2πx

λ

)eσt, (38a)

T1 = T ∗1 sin

(πzh

)cos

(2πx

λ

)eσt. (38b)

In this trial solution, ψ∗ and T ∗1 are unknown but constant coefficients; they are

of no interest to us. Of principal interest are λ, the horizontal wavelength of theperturbation, and σ, the growth rate of the perturbation.

The solution specifies a periodic array of convective rolls, but doesn’t specify theaspect ratio of those rolls (which is given by A = 2πh/λ, with λ as yet unknown).

If σ < 0 then we expect that perturbation to decay away, hence convection doesnot occur and the system is termed as stable. If σ > 0 then we expect the arbitrarilysmall perturbations to grow exponentially into a convective flow. In this case thesystem is called unstable. The state where σ = 0 and the system is perched on theedge between stability and instability is called the state of neutral stability.

Solving for λ and σ To obtain a solution we substitute (38) into (35) and (36)

µ

(π2

h2+

4π2

λ2

)2

ψ∗ = −2πρ0gα

λT ∗

1 , (39a)(σ +

κπ2

h2+κ4π2

λ2

)T ∗

1 = −2π∆T

λhψ∗. (39b)

1Ansatz is german for educated guess. It is the word typically used by physicists when theypropose a trial solution.

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These two algebraic equations can be combine to eliminate ψ∗ and T ∗1 giving the

growth rate,

σ =κ

h2

[Ra

(A2

(A2 + π2)2

)−

(A2 + π2

)]. (40)

Recall that the aspect ratio of the convection rolls and the Rayleigh number aregiven by

A =2πh

λand Ra =

ρ0gα∆Th3

µκ, (41)

respectively.

0 1 2 3 4 5 6 70

500

1000

1500

2000

2500

3000

3500

4000

A = 2πh/λ

Ra

cr

Unstable

Stable

σ=

0

Figure 2: A plot of the critical Rayleigh number versus the aspect ratio of theconvection rolls

We seek the values of the Rayleigh number Ra that, for any given value of A, isexactly neutrally stable; i.e. we seek the critical Rayleigh number Racr below whichthe system is stable, and above which it is unstable. This occurs precisely when thegrowth rate vanishes, so setting σ = 0 in (40) and solving for Ra ≡ Racr gives

Racr =(π2 + A2)

3

A2. (42)

A plot of this curve is given in Figure 2.The minimum value of the Rayleigh number at which convection (at any aspect

ratio) can occur is obtained by solving dRacr/dA = 0. The solution is

A =π√2, (43)

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or λ = 2√

2h. Substitution of this result back into (42) gives the minimum value ofthe Rayleigh number for convection,

min(Racr) =27π4

4= 657.5. (44)

For a stable, quiescent system where the Rayleigh number is slowly increased overtime, it is at this value of the Rayleigh number that convection begins, and with anaspect ratio of π/

√2.

Applicability to the mantle Although a good illustration of the onset of con-vection, the above analysis may not be applicable to the mantle for several reasons.The most important of these is that the mantle has internal heating by radioactivedecay, which was not included in the analysis above.

A calculation along the same lines as the above, but with internal heating, a coldupper surface, and a thermally insulating bottom boundary leads to a definition ofthe Rayleigh number as

RaH =αρ2

0gHh5

kµκ, (45)

where H is the rate of internal heating and k is the thermal conductivity. Theminimum critical Rayleigh number for this system is 867.8, and the associated valueof the aspect ratio is 1.79.

We can estimate the value of the Rayleigh number in the mantle. Taking µ = 1021

Pa-sec, k = 4 W/m/K, κ = 1 mm2/sec, α = 3×10−5 K−1, g = 10 m/sec2, ρ0 = 4000kg/m3, H = 9× 10−12 W/kg, and h = 2880 km, we find that RaH = 2× 109. Thisvalue is obviously much higher than the minimum critical value. Convection isexpected in the mantle.

6 Convection at high Rayleigh number

When Ra � Racr, thermal convection is rapid and disordered, with plumes risingand falling from thermal boundary layers along the top and bottom of the fluidlayer. Within the thermal boundary layers, there is a sharp temperature gradientand little or no flow. Between the thermal boundary layers, within the interior ofthe fluid, convection is very vigorous, rapidly transporting heat from the bottomto the top of the layer. The interior of the domain is therefore well-mixed, with aroughly constant temperature, as shown in Figure 3.

In this regime, it is impossible to obtain analytical solutions for the details ofthe flow. For many purposes, however, the details of the flow are not important.What we care about is the heat transported by convection and how it scales withthe problem parameters. From looking at Figure 3, we can see that heat enters andleaves the fluid by conduction through the boundaries and the thermal boundarylayers—hence we know that conduction is important. We also know that in theinterior, heat is transported by convection, and the vigor of convection scales withthe Rayleigh number. A dimensionally correct guess for the scaling of the heat fluxis thus

qtotal ∝ k∆T

hRaβ, (46)

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Figure 3: Average temperature profile through a convecting layer with Ra = 105.

where β an unknown exponent.In experiments and numerical simulations of convection at high Rayleigh number,

it is observed that the heat flow qtotal does NOT depend on the height h of the fluidlayer. Since the Rayleigh number contains h3, and the conductive scaling containsh−1, this observation can be accommodated if β ≈ 1/3.

Evidence from laboratory experiments (e.g. Figure 4) indicates that Nu ∝ Ra1/3

is a good rule-of-thumb. It is useful in analysing the thermal history of the Earth.

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Figure 4: Scaling of Nu versus Ra from experiments on Helium gas by Niemela etal, Nature 2000.

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