Continuum mechanics problems

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Problems in Continuum Mechanics

Hans Petter Langtangen

Simula Research Laboratory

University of Oslo

September 2007


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Table of Contents

1 Heat Transfer Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Steady Heat Conduction in an Insulated Rod . . . . . . . . . . . . . . . . 11.2 Transient Heat Conduction in an Insulated Rod . . . . . . . . . . . . . . 31.3 Heat Transfer in a Two-Material Domain . . . . . . . . . . . . . . . . . . . . 51.4 Transient Heat Conduction in Soil . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Transient Heat Conduction in a 2D Geometry. . . . . . . . . . . . . . . . 121.6 Heat Transfer in Pipeflow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 Transient Heat Conduction in a 2D Geometry. . . . . . . . . . . . . . . . 161.8 Cooling of an Object; PDE Models . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.9 Cooling of an Object; Averaged Model . . . . . . . . . . . . . . . . . . . . . . 231.10 Diffusion of Ink in a Water Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 Fluid Flow Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.1 Pressure in a Fluid at Rest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.2 Pressure Force on a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.3 Pressure Force on a Cylinder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.4 Pressure Force on a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.5 Stationary Channel Flow; Newtonian Fluid . . . . . . . . . . . . . . . . . . 392.6 Channel Flow Specified as a 2D/3D Problem . . . . . . . . . . . . . . . . 422.7 Flow over a Backward-Facing Step . . . . . . . . . . . . . . . . . . . . . . . . . . 442.8 Transient Channel Flow; Newtonian Fluid . . . . . . . . . . . . . . . . . . . 442.9 Sudden Movement of Lubricated Surfaces . . . . . . . . . . . . . . . . . . . 47

2.10 Transient Channel Flow; Generalized Newtonian Fluid . . . . . . . . 482.11 Pulsatile Blood Flow in a Straight Artery . . . . . . . . . . . . . . . . . . . 502.12 Oil and Water Films Between Moving Surfaces . . . . . . . . . . . . . . . 522.13 Flow of Oil and Water in a Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.14 Flow in a Pipe with a Non-Circular Cross Section . . . . . . . . . . . . 572.15 Stationary Flow in an Open Inclined Channel . . . . . . . . . . . . . . . . 602.16 Different Cross Sections in Open Channel Flow . . . . . . . . . . . . . . 632.17 Transient Flow in an Open Inclined Channel . . . . . . . . . . . . . . . . . 642.18 Flow in a Channel with Varying Width . . . . . . . . . . . . . . . . . . . . . 652.19 Spindown of a Well Bore; Newtonian Fluid . . . . . . . . . . . . . . . . . . 732.20 Spindown of a Well Bore; Power-Law Fluid . . . . . . . . . . . . . . . . . . 78

3 Solid Deformation Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 803.1 Heavy Box on an Elastic Layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.2 Deformation of an L-Shaped Beam . . . . . . . . . . . . . . . . . . . . . . . . . 813.3 Deformation of an Elastic Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.4 Plate with a Hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 853.5 Torsion of a Hollow Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87


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IV Table of Contents

3.6 Development of Torsion Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933.7 Hollow Sphere with Inner Pressure . . . . . . . . . . . . . . . . . . . . . . . . . 102

3.8 Two-Material Hollow Sphere with Inner Pressure . . . . . . . . . . . . . 103

4 Coupled Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1084.1 Heat Transfer in a Tube with Oscillating External Temperature 1084.2 Transient Thermo-Elasticity in a Two-Material Spherical Con-

tainer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104.3 Simplified Analysis of a Thermo-Elastic Container . . . . . . . . . . . . 1134.4 Heat and Flow in a Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

5 General Modeling Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 1165.1 Symmetry of a scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1165.2 Symmetry of a vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1185.3 Given a Model, What Is the Problem? . . . . . . . . . . . . . . . . . . . . . . 120

A Mathematical Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122A.1 Useful Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122A.2 Scaling and Dimensionless Variables . . . . . . . . . . . . . . . . . . . . . . . . 125


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Chapter 1

Heat Transfer Problems

1.1 Steady Heat Conduction in an Insulated Rod

We consider a homogeneous solid rod with length L and radius R, see Fig-ure 1.1. The end surfaces x = 0 and x = L have controlled temperatures, T 0and T L, respectively. The rest of the surface is insulated (no heat is allowedto escape). We are interested in the temperature distribution in the wholerod. The problem is stationary.

L

z

x

T=T T=T 0 L

insulated surface

Fig.1.1. Sketch of a rod with controlled temperatures T 0 and L at the end surfaces,while the circular surface is insulated. (Problem 1.1)

First formulate a three-dimensional mathematical model for this problem.Thereafter, argue why we expect the temperature to vary with x only, andshow mathematically that T = T (x) fulfills the three-dimensional boundary-value problem, i.e., derive the one-dimensional model. Calculate the temper-ature distribution. Then scale the one-dimensional model, find the solution,and insert physical quantities in the scaled temperature expression. Is theone-dimensional model and its solution limited to the geometry in Figure 1.1or could it be valid for other geometries as well?

Solution of Problem 1.1

Basic Equations and Boundary Conditions. This problem is about heat con-duction in a solid. Neglecting internal heat sources and effects of deformingthe material, the governing PDE is

C v∂T

∂t = k∇2T,


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2 1. Heat Transfer Problems

when the material is homogeneous with respect to heat conduction ( constantk). Since the problem is stationary, ∂T/∂t = 0. The governing PDE is to be

solved in a domain Ω , which is the cylindrical rod geometry.One boundary condition is needed at each point on the boundary; on

x = 0 and x = L, T is fixed at T 0 and T L, while the rest of the boundary(r = R) has no heat flux: q · n = −k∂T/∂n = 0, or just ∂T/∂n = 0 forsimplicity.

The complete three-dimensional boundary-value problem becomes

k∇2T = 0, in Ω , (1.1)T = T 0, x = 0, (1.2)

T = T L, x = L, (1.3)

∂T

∂n = 0, r = R . (1.4)

Simplifications. To understand how T will vary througout space, we need tosee how the input data, i.e., geometry, coefficients, and boundary conditions,vary in space. The coefficients in the PDE are constant, the geometry isa cylinder, and the boundary conditions are constant on each of the threesurfaces. There will be an x variation since T goes from T 0 to T L as we movealong the x axis. On the other hand, there are no variations in the geometryor the no-flux boundary condition that may cause variations of T with respectto y and z. We may therefore assume T = T (x). To check that the assumptionis correct, we can check if T (x) is a solution of the boundary-value problem.The PDE becomes

kT (x) = 0 .

The boundary conditions on the end surfaces are expressed as T (0) = T 0 andT (L) = T L. The condition on r = R needs slightly more calculations. Thenormal vector to the curved surface can be written as

n = y j + zk

y2 + z2.

The condition ∂T/∂n = 0 becomes

∂T

∂n = n · ∇T = 1

y2 + z2

y

∂T

∂y + z

∂T

∂z

= 0 .

Hence, T = T (x) is compatible with ∂T/∂n = 0 at r = R.The one-dimensional boundary value problem reads

T (x) = 0, x ∈ (0, L), T (0) = T 0, T (L) = T L . (1.5)The solution is obtained by integrating twice,

T (x) = Ax + B,


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1.2. Transient Heat Conduction in an Insulated Rod 3

and determining the integration constants A and B from the boundary con-ditions. This gives B = T 0 and A = (T L − T 0)/L. The total temperaturedistribution becomes

T (x) = T 0 + (T L − T 0) xL

.

Scaling. The one-dimensional model 1.5 can be scaled according to

x̄ = x

L, T̄ =

T − T 0T L − T 0 .

Inserting these expressions in 1.5 and dropping the bars as usual result in thescaled model

T (x) = 0, x ∈ (0, 1), T (0) = 0, T (1) = 1 . (1.6)The solution is now found to be T (x) = x. The transformation back to

physical quantities goes as follows (now we need the bars again since we needto mix quantities with and without dimension):

T̄ = x̄ ↔ T − T 0T L − T 0 =

x

L ↔ T = T 0 + (T L − T 0) x

L,

which coincides with the solution found by solving the unscaled problem.Examining the validity of the T = T (x) assumption carefully, we see that

only the boundary condition ∂T/∂n = 0 at r = R brings in the geometricshape of the domain. The condition ∂T/∂n = 0 will be valid for any n thatdoes not have an x component. This means that the domain Ω must haveshape of a cylinder, but the cross section geometry can be arbitrary. (Anarbitrary cross section will have n = ny j + nzk and n · ∇T = 0 since T ’sderivatives in the y and z directions vanishes.)

1.2 Transient Heat Conduction in an Insulated Rod

We consider the same physical case as in Problem 1.1. However, now therod has the temperature T 0 at time t = 0. The, the temperature at thesurface x = L is suddenly changed to T = T L. We are interested in howthe temperature develops througout time and space. The arguments leadingto a one-dimensional mathematical model are valid also in the present time-dependent case. Set up a complete one-dimensional model for the temperaturedistribution T = T (x, t). Scale the problem. Discuss qualitatively how thetemperature will develop in time and space.


Basic Equations and Boundary Conditions. As in Problem 1.1, the governingPDE reads

C v∂T

∂t = k∇2T,


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but now we cannot make the assumption that ∂T/∂t = 0. The initial condi-tion reads T = T 0, while the boundary conditions for t > 0 (when the PDE

applies) are as in Problem 1.1.

Simplifications. Inserting T = T (x, t) in the governing PDE leads to thesimplified PDE

C p∂T

∂t = k

∂ 2T

∂x2 .

The boundary condition ∂T /∂n = 0 is fulfilled as explained in Problem 1.1(only the fact that T does not vary with y and z is important for this conditionto hold). The complete one-dimensional initial-boundary value problem canbe written as

C p∂T

∂t = k

∂ 2T

∂x2 , x ∈ (0, L), t > 0, (1.7)

T (0, t) = T 0, t > 0, (1.8)

T (L, t) = T L, t > 0, (1.9)

T (x, 0) = T 0, x ∈ [0, L] . (1.10)

Scaling. A standard scaling is

x̄ = x

L, T̄ =

T − T 0T L − T 0 , t̄ =

t

tc.

Inserting these expressions in the governing PDE leads to

∂ T̄

∂ ̄t

= α∂ 2 T̄

∂ ̄x2

, α = tck

C pL2

.

Requiring that the two terms in the PDE are of the same size (or, in otherwords, that also the derivatives of T̄ are of unit magnitude), implies α = 1and

tc = C pL2/k .

The initial condition becomes T̄ = 0, and the two boundary conditions readT̄ (0, t̄) = 0 and T̄ (1, t̄) = 1. The scaled problem can now be summarized as(dropping bars as usual)

∂T

∂t =

∂ 2T

∂x2 , x ∈ (0, 1), t > 0, (1.11)

T (0, t) = 0, t > 0, (1.12)

T (1, t) = 1, t > 0, (1.13)

T (x, 0) = 0, x ∈ [0, 1] . (1.14)


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1.3. Heat Transfer in a Two-Material Domain 5

The Temperature Evolution. We base the discussion of the temperature evo-lution in space and time on the scaled model. Initially, T = 0 in the whole

domain. Then the boundary value at x = 1 is switched to T = 1. This causesa very thin layer where the temperature goes from 0 to 1. As time increases,heat conduction leads to flow of heat, inwards from the hot boundary x = 1,and the thin layer will grow in size. As it gets thicker, it approaches thestationary solution T = x found in Problem 1.1. Rapid changes take placeat early times, while the convergence towards the stationary solution is veryslow. Figure 1.2 shows T (x, t) for some different points in time. Already fort = 0.2 we see that the graph is close to the stationary solution T = x. Thedata in Figure 1.2 were produced by an explicit finite difference scheme (firstorder in time, second order in space), using a uniform grid of 200 cells andthe largest allowable time step.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x,t=1.25000e-05)u(x,t=2.62500e-04)u(x,t=2.51250e-03)u(x,t=1.00125e-02)u(x,t=2.50125e-02)u(x,t=2.00012e-01)

Fig.1.2. Snapshot of the solution of (1.11)–(1.14) at five points of time. The plotshows how the thickness of the initially very thin boundary layer close to x = 1grows in time. (Problem 1.2)

1.3 Heat Transfer in a Two-Material Domain

Figure 1.3 shows a two-material structure, where the heat conduction coef-ficient is k1 (low) in one of the materials and k2 (high) in the surroundingmaterial. The densities and heat capacity in the two media also differ. Att = 0 we assume a constant temperature T 0 in the whole domain. Then the


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temperature is suddenly increased to T 1 > T 0 at the black segment on the leftboundary. At the right boundary, the temperature is fixed at T 0. The other

boundaries are insulated (no heat flux across the boundaries). Reduce the sizeof the domain by taking symmetry into account and specify the boundaryconditions to be used in the reduced domain.

Introduce a scaling of the independent and dependent variables. Supposeyou have a program solving the unscaled problem but want to use it toinvestigate the scaled model (which has less parameters). How can you setthe parameters in the unscaled model (program) to mimic that you solve thescaled problem?

k

k 2

1T 0

y

x1T

Fig.1.3. Sketch of the heat conduction case in Problem 1.3. The domain consistsof two materials with differing heat conduction properties. The other material pa-rameters, and C p, also differ in the two materials. The temperature is kept at T 1and T 0 at certain parts of the boundaries.


Basic Equations and Boundary Conditions. This problem concerns heattransfer in a solid. There are no heat generation sources so the appropri-ate governing equation reads

C p∂T

∂t = ∇ · (κ∇T ) . (1.15)

We have used that the material is heterogeneous with a space-varying heatconduction coefficient κ(x, y), which equals k1 and k2 in the two differentdomains, as depicted in Figure 1.3. Equation (1.16) is also valid for space-varying density and heat capacity C p. In general, also these parameters willbe different for the two materials.


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1.3. Heat Transfer in a Two-Material Domain 7

The x axis divides the domain into two equal-sized parts, see Figure 1.3.We realize that y = 0 is a symmetry line along which the symmetry condition

∂T/∂n = 0 applies.Introducing ∂ Ω 1 as the part of the boundary where T = T 1, ∂Ω 0 as the

part where T = T 0, ∂Ω N as the insulated part, and ∂Ω S as the symmetryline, we can set up the whole initial-boundary value problem for T (x,y,t) asfollows.

(x, y)C p(x, y)∂T

∂t =

∂

∂x

κ(x, y)

∂T

∂x

+

∂

∂y

κ(x, y)

∂T

∂y

, (1.16)

T (x,y, 0) = T 0, (1.17)

T = T 1, on ∂ Ω 1, t > 0, (1.18)

T = T 0, on ∂ Ω 0, t > 0, (1.19)

∂T

∂n = 0, on ∂Ω N , t > 0, (1.20)

∂T

∂n = 0, on ∂Ω S , t > 0 . (1.21)

Scaling. Let L be a characteristic length of the domain. Here we may take2L as the width of the boundary where T = T 0 applies (this allows us toexpress the width of the conducting channel, the gray area in Figure 1.3, asa fraction of L). The space and time coordinates are scaled as

x̄ = x

L, ȳ =

y

L, t̄ =

t

tc.

Variable coefficients are scaled according to the general rule that we (may)subtract a reference value and divide by the typical range of variation. Forκ, , and C p we typically just divide by the maximum value of the function.Since it is unclear here whether (say) 1 or 2 is the maximum value, wedivide by 1. The scaling can be expressed as scaling of the type

q̄ = q

q 1,

where q can be either κ, , or C p.The scaling of T is naturally taken as

T̄ = T − T 0T 1 − T 0 .

Inserting the scaling in the PDE (1.16) results in

̄C̄ p∂ T̄

∂ ̄t =

tcκ11(C p)1

∂

∂ ̄x

κ̄

∂ T̄

∂ ̄x

+

∂

∂ ̄y

κ̄

∂ T̄

∂ ̄y

.

There is only one candidate for the time scale in the present problem, namelya balance between the time and space derivatives in the PDE (there is no


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time variation in input data, i.e., the geometry, the boundary conditions, orthe coefficients in the PDE). Hence, we choose

tcκ11(C p)1

= 1 .

In material 1, κ̄, C̄ p, and ̄ all equal unity, while in material 2 they equal theratio of the values in material 2 and 1.

It might be tempting to introduce the ratio

β = κ21(C p)1κ12(C p)2

and work with the PDE∂ T̄

∂ ̄t = ∇2 T̄

in material 1 and

∂ T̄ ∂ ̄t

= β ∇2 T̄ in material 2. This is correct if each of the two PDEs is restricted to onematerial. In some numerical approaches this is inconvenient compared to aformulation where we have mathematically space varying functions κ̄, C̄ p,and ̄ (in the finite element method, for example, the formulation with twoPDEs can be utilized with some extra algebra –

The text is to be completed....).The complete scaled initial-boundary value problem becomes (dropping

bars as usual)

C p∂T

∂t =

∂

∂x

κ

∂T

∂x

+

∂

∂y

κ

∂T

∂y

, (1.22)

T (x,y, 0) = 0, (1.23)T = 1, on ∂Ω 1, t > 0, (1.24)

T = 0, on ∂Ω 0, t > 0, (1.25)

∂T

∂n = 0, on ∂Ω N , t > 0, (1.26)

∂T

∂n = 0, on ∂Ω S , t > 0 . (1.27)

The variable coefficients are

C p =

1, material 1γ 1, material 2

κ =1, material 1

γ 2, material 2

where two dimensionless numbers γ 1 and γ 2 have been introduced for conve-nience:

γ 1 = 2(C p)21(C p)1

, γ 2 = κ2κ1

.


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1.4. Transient Heat Conduction in Soil 9

Solving the Scaled Problem with the Unscaled Formulation. In the scaledproblem there are only two physical parameters to vary, γ 1 and γ 2. If we want

to take advantage of this reduction of parameters in a systematic investigationof the problem, and our tool at disposal solves the unscaled problem, we canset

– κ1 = 1 = (C p)1 = 1

– κ2 = γ 1

– 2(C p)2 = γ 2

– T 0 = 0, T 1 = 1

Moreover, we work in a geometry where the right boundary has length 1. Tocheck if these choices of the parameters are correct, we insert the choices inthe unscaled problem and see if we can recover the scaled formulation.

1.4 Transient Heat Conduction in Soil

When the temperature oscillates due to day and night or seasonal variationsat the surface of the earth, how far downwards (into the soil or litosphere) aretemperature oscillations of a significant size? Discuss how a one-dimensional mathematical model for this problem can be developed, and how the modelcan be used to answer the specific question.


Basic Equations and Boundary Conditions. This is a transient heat conduc-

tion problem in a solid. The governing PDE is

C v∂T

∂t = κ∇2T

if we neglect any heat sources (e.g., no significant radioactive heat generation)and a constant heat conduction coefficient κ. A constant κ is probably areasonable assumption for day–night variations in an upper soil layer, butfor seasonal variations that influence the temperature in deeper regions, oneshould incorporate a variable κ and replace the last term in the PDE by∇ · (κ∇T ).

Using a single heat conduction PDE we implicitly neglect convective heattransport due to air flow in the porous ground.

It is reasonable to model the soil or litosphere as a thick and wide do-main with a sinusoidal temperature variation on the top and some (simple)boundary conditions at the other artificial boundaries. Let us choose the x topoint downwards, and let x = 0 denote the surface of the land. At boundaries


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y = const and z = const (of a 3D box) we set ∂T/∂n = 0 as a symmetry or“no change” condition. At the surface we prescribe an oscillating temperature

T = T 0 + A sin ωt,

with 2π/ω as the period of the oscillations (typically 24 h for day–nightvariations and months for seasonal variations). At the bottom of the domainthe appropriate boundary condition depends on how deep we go; if we arefar away from the mantle and sufficiently below the x level where the surfaceoscillations are hardly noticable, ∂T/∂ = 0 as a “no change” condition mayapply. Closer to the mantle there may be a known heat flux or temperature.Here we go for the “no change” condition.

The initial condition is required mathematically, but the particular valueof T at t = 0 is immaterial since the effect of the initial state will die outas time increases. The steady state variations of T are solely driven by the

surface temperature and the PDE.

Simplifications. Since we are interested in the variation in x direction, weassume ∂/∂y = ∂/∂z = 0 in the whole domain, which leaves us with apure 1D problem. The bottom of the domain is supposed to be located at a“large” x value, and we may in the mathematical simplification let x → ∞.Numerically, a sufficiently large value of x is chosen such that this value doesnot influence the solution. This is a reasonable approximation since we expectoscillations at x = 0 will decay with the depth x.

Assuming T = T (x, t) we get

∂T

∂t

= λ∂ 2T

∂x2

,

with

T (0, t) = T 0 + A sin ωt, limx→∞

∂T

∂x = 0

as boundary conditions. The parameter λ equals κ/(C v). Any initial condi-tion can be used, but if discontinuities at the boundaries appear (cf. Prob-lem 1.2), thin boundary layers will occur, and these may cause numericaldifficulties if the spatial resolution near the boundaries is too coarse. Wetherefore suggest T (x, 0) = T 0 as a good candidate for the initial condi-tion, since the boundary condition T (0, 0) equals T 0 and increases smoothly(A sin ωt) from this value.

For numerical solution methods we impose the second boundary condition

as ∂T

∂n = 0, x = x∞,

where x∞ is a sufficiently large value, in the sense that x∞ does not influ-ence the temperature field significantly. Here this means that T (x∞, t) ≈ T 0(only the surface condition changes the heat state, and the surface condition


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1.4. Transient Heat Conduction in Soil 11

should not be noticable at x = x∞, such that the initial temperature T 0 is pre-served during the simulation). Clearly, we could also impose T (x∞, t) = T 0

as boundary condition. The “no change” condition is “milder”, i.e., a toosmall x∞ will have less effect on T (x, t) if ∂T/∂x = 0 is used than if we fixT (x∞, t) at T 0.

Scaling. Initial-boundary value problems of this (simple) kind can benefitfrom scaling in the way that the number of free parameters can be reduceddramatically. The choice of scales are, however, not trivial. As length scalewe may use x∞, but we may also use a characteristic length L of the spatialtemperature variations. The value of L is not known on beforehand andits estimation will require more analysis of the problem. For simplicity wetherefore use x∞ as length scale. The typical scale for T is 2A or simply A,and a reference temperature is T 0. We therefore introduce

x̄ = xx∞

, T̄ = T − T 0A

, t̄ = ttc

.

There are two candidates for the time scale: (i) balance of the two terms inthe PDE, leading to tc = x

2∞

/λ, or (ii) the period of oscillations, leading totc = 2π/ω or simply tc = ω−1. Using tc = x2∞/λ we get the scaled problem(dropping bars as usual)

∂T

∂t =

∂ 2T

∂x2 , x ∈ (0, 1), t > 0, (1.28)

T (0, t) = sin γt, t > 0, (1.29)

∂T

∂xx=x∞

= 0, (1.30)

where γ is a dimensionless number:

γ = x2∞

ω

λ .

The other time scale, tc = ω−1, leads to

∂T

∂t =

1

γ

∂ 2T

∂x2 , x ∈ (0, 1), t > 0, (1.31)

T (0, t) = sin t, t > 0, (1.32)

∂T

∂xx=x∞

= 0 . (1.33)

The latter scaling is superior to the former if ω λ/x2∞

(γ 1), i.e., veryslow surface oscillations, since we then see that the ∂T /∂t term in the PDEcan be omitted (quasi-stationary problem). For very fast oscillations (γ 1)the temperature variations will only be noticable in a thin layer close to


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the surface (there is not enough time for the conduction to transport heatinto the ground), one may question the choice of length scale; x∞ is then

much larger than the characteristic spatial variations of the solution. Also,γ 1 indicates from the PDE that ∂T/∂t ≈ 0, which is correct outside thethin surface layer, but totally wrong inside this layer. We therefore shouldwork with a smaller domain (x∞), covering the part of the x axis where T undergoes changes. A smaller x∞ then gives a smaller γ and balance betweentime rate of change of T and heat conduction.

The discussion of scaling here shows that scaling is a highly non-trivialissue.

The scaling is not sound; γ = 1 gives a spatial variation much further than x = 1. Or is it ok? NO, a better length scale is needed. However, γ = 1indicates that x∞ is too small, enlarging this value, γ is increased and the solution is better.

Numerical Solution.

Analytical Solution. However, it is also possible to find an analytical solutionin the form of a damped traveling heat wave. It appears that the solution

T (x, t) = T 0 + A exp

−x

ω

2λ

cos

ωt − x

ω

2λ

is the steady state variation of T as t → ∞.This damping is given by the factor exp (−x

ω/(2λ)). To reduce the

surface amplitude A by a factor of (say) e−3 ≈ 0.05, we get −xc

ω/(2λ) =−3, i.e.,

xc = 3

2λω

.

For x > xc, the amplitude of the surface oscillations in T is reduced by 95percent.

Quick analysis to do the length scale estimation of κ/ω (p. 156):

T (x, t) ∼ T 0 + X (x)eiωt

inserted gives two coupled equations (real and imag part)...

1.5 Transient Heat Conduction in a 2D Geometry

Figure 1.4a shows the cross section of a long isolated circular tube with anouter (homogeneous) isolating material. The goal is to compute the tem-perature distribution in the isolating material, bounded by the tube and asquare-shaped outer boundary. Water is flowing through the inner tube witha sufficiently high velocity such that we can assume constant temperature


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1.5. Transient Heat Conduction in a 2D Geometry 13

T f throughout the fluid. Outside the square-shaped boundary the temper-ature is oscillating, e.g. due to day and night variations, here modeled as

T o = T m + A sin ωt. At the inner boundary we assume that the temperatureis constant, T = T f , since new water at temperature T f is constantly en-tering the system. At the outer boundary we may employ Newton’s coolinglaw. Work with as small computational domain as possible by exploiting thesymmetry in the problem (identify the symmetry lines and their associatedboundary conditions).

We are primarily interested in the steady-state oscillating behavior of thetemperature accross the isolating material to make sure that the materialis really isolating. Numerical simulations, however, must be started at someinitial time. Comment upon how to choose an appropriate initial condition.

In Figure 1.4b the pipe is partially digged into the ground such thata part of the outer boundary is in contact with soil while the rest of theouter boundary is in contact with air. How does this change in surroundingsinfluence the mathematical model?


This problem concerns heat transfer in a solid. There are no heat generationsources so the appropriate governing equation reads

C p∂T

∂t = κ∇2T . (1.34)

We have used that the material is homogeneous with heat conduction co-efficient κ. Futhermore, is the isolating material’s density, and C p is thematerial’s heat capacity. We assume two-dimensional conditions, i.e., T =T (x,y,t) and ∂/∂z = 0.

The boundary conditions are more or less stated in the problem descrip-tion: cooling law at both boundaries. A complete initial-boundary value prob-lem can therefore be written as

C p∂T

∂t = κ∇2T, (1.35)

−κ ∂T ∂n

= hf (T − T m − A sin ωt) on inner boundary, (1.36)

−κ ∂T ∂n

= ha(T − T m − A sin ωt) on outer boundary, (1.37)T (x,y, 0) = f (x, y) . (1.38)

Here, we have introduced different heat transfer coefficients between the iso-lating material and the fluid (hf ) and between the material and the outer air(ha).

The boundary conditions do not depend on the position of the bound-ary, neither do the coefficients in the equations. The shape of the geometry


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will therefore govern symmetry lines. The geometry is seen to be symmetricabout four lines, dividing the domain into eight similar pieces. Figure 1.5

shows one of these eight minimum domain sizes that can be used for solvingthe problem. Along the symmetry lines 2 and 4 in Figure 1.5 we have the con-dition ∂T/∂n = 0, whereas the boundaries 1 and 3 correspond to the originalphysical boundaries, i.e., the outer and inner boundaries with cooling laws.

Any initial condition will fade out as t → ∞. The solution approachesa steady state oscillation driven by the temperature conditions on the outerboundary. The initial condition has no effect on the this steady state solution.Therefore, we can in a numerical simulation start with any initial conditionf (x, y) and obtain the desired steady state solution. However, if we choosef (x, y) to be significantly different from T f and T m, we see from the boundaryconditions that large temperature gradients are formed at the boundary att = 0. This can cause trouble when solving the problem numerically. Themost natural choice is therefore to let f (x, y) be a smooth function withvalues T f and T m at the inner and outer boundaries. Such a smooth functioncan be constructed by solving a Laplace problem:

∇2T = 0, (1.39)T = T f on inner boundary, (1.40)

T = T m on outer boundary . (1.41)

Physically, this f implies that that we have perfect conduction at the innerboundary and constant outer temperature for a long time period up to t = 0.Then we turn on cooling law conditions and an oscillating outer temperature.(A ”long time period” means that these conditions act for a sufficiently longtime such that time variations can be neglected. This is necessary for the

plain Laplace equation to be valid.)Considering the situation in Figure 1.4b, two things in the above solution

change: (i) the heat transfer coefficient is different in the air and soil parts,and (ii) the boundary conditions are symmetric in horizontal direction only.The latter fact implies that we only have symmetry along the vertical linethrough the center of the pipe. Regarding the heat transfer coefficient, it isnow natural to let ha model the transfer between the material and air, andintroduce a new coefficient hs for the transfer between the material and thesoil.

1.6 Heat Transfer in Pipeflow

A fluid is flowing in a straight pipe with a constant cross section depictedin Figure 1.6. We seek the temperature distribution in the fluid when it isknown that the temperature is constant at the walls of the pipe. The fluidvelocity is available as a function v = w(x,y,t)k, which we consider as knownwhen solving for the temperature. Heat is generated by viscous dissipation.


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1.6. Heat Transfer in Pipeflow 15

Derive a mathematical model for the temperature distribution T (x,y,t) forthe case of a Newtonian and a power-law generalized Newtonian fluid.


The governing equation for the temperature distribution in a fluid is theenergy equation

C p(∂T

∂t + v · ∇T ) = κ∇2T + 2µε̇ij ε̇ij,

where the last term models heat generation by viscous dissipation. This termis valid for all generalized Newtonian fluids, because the general form of theterm, σij ε̇ij , equals 2µε̇ij ε̇ij when σij = − pδ ij+2muε̇ij, regardless of whetherµ is constant or not.

First, we assume that all physical properties are constant along the pipe,which justifies the assumption ∂T/∂z = 0, i.e., T = T (x,y,t). The termv ·∇T = w∂T/∂z vanishes. The only non-vanishing components in the strain-rate tensor ε̇ij are ε̇xz and ε̇yz . Therefore,

2µε̇ij ε̇ij = 4µ(ε̇2xz + ε̇

2yz) .

We have that

ε̇xz = 1

2

∂w

∂x, ε̇yz =

1

2

∂w

∂y

and hence the dissipation term becomes

2µε̇ij ε̇ij = µ∂w∂x2

+∂w∂y2

= µ||∇w||2 .

The equation for T can now be written

C p∂T

∂t = κ

∂ 2T

∂x2 +

∂ 2T

∂y2

+ µ||∇w||2 .

This equation is to be solved in a domain Ω , which equals the cross sectionof the pipe, as depicted in Figure 1.6. The boundary conditions are T = T won the wall of the pipe. Moreover, we need to prescribe an initial conditionT (x,y, 0).

The time dependence of T must be due to a time-dependent velocity w,since , C p, κ, and the boundary conditions are time independent.

Symmetry can be utilized; the equation for T can be solved in the left (orright) part of the domain, with the symmetry condition ∂T/∂n = 0 alongthe symmetry line.

For Newtonian fluid, µ is constant. A power-law fluid has

µ = µ0γ̇ n−1, γ̇ =

2ε̇ij ε̇ij = ||∇w|| .


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The equation for T can therefore be expressed as follows for a power-lawgeneralized Newtonian fluid:

C p∂T

∂t = κ

∂ 2T

∂x2 +

∂ 2T

∂y2

+ µ0||∇w||n+1 .

1.7 Transient Heat Conduction in a 2D Geometry

Heat conduction problems are often described by two-dimensional mathe-matical models, but the conduction takes place in all three space directions.This problem addresses how one can reduce the three-dimensional physics to just two dimensions in a mathematical model.

In bodies with a large extent in the third direction (z) we may assumethat the temperature remains approximately constant in this direction. A

requirement is that the boundary conditions also remain constant in z di-rection and that there is no significant heat loss to the surroundings at the”end” surfaces (typically z = const). The simplification is then ∂T/∂z ≈ 0.Inserting this in the governing equation and boundary conditions yields amodel where T only depends on the spatial x and y coordinates and possiblyon time.

Consider now a thin plate where we want to omit calculations through thesmall thickness. Let z be orthogonal to the plate. We assume a homogeneousplate and stationary conditions. Set up the heat conduction equation onintegral form. Choose a volume with small extent ∆x and ∆y in the x and ydirections, respectively, and thickness h equal to the plate. Apply a coolinglaw on the plate surfaces (z = const) and let ∆x, ∆y → 0 to obtain thegoverning two-dimensional equation

∂ 2T

∂x2 +

∂ 2T

∂y2 = 2

hT h

(T − T s), (1.42)

with any relevant type of boundary conditions on the sides of the two-dimensional domain.

Alternatively, one can start with the governing equation ∇2T = 0 in threedimensions, apply this to the whole plate, integrate the equation through thethickness, and arrive at the (1.42). Carry out the details of this calculation.


The integral form of the heat conduction equation is readily found from theintegral form of the energy equation. In case of conduction only, only oneterm is relevant:

∂V

q · ndS = 0 . (1.43)


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1.7. Transient Heat Conduction in a 2D Geometry 17

Here, q is the heat flux, q = −k∇T , and n is the outward unit normal tothe surface ∂V of some arbitrary volume V . What we want is to ”integrate

away” the z direction and get a partial differential equation in the x and ydirections. The method for arriving at such a model from the integral form isto choose V to fill the thickness of the plate but have infinitesimal extensionsin the x and y directions. That is,

V = {(x,y,z) | x0 ≤ x ≤ x0 + ∆x,y0 ≤ y ≤ y0 + ∆y, 0 ≤ z ≤ h} .V is a cube with six sides so the surface integral gets six contributions:

∂V

q ·ndS = h∆yq|x0+∆x · i + h∆yq|x0 · (−i) +

h∆xq|y0+∆y · j + h∆xq|y0 · (− j) +∆x∆yq|h · k + ∆x∆yq|h · (−k) .

Applying the cooling law in the last line yields

∆x∆y(q|h·k+q|0·(−k)) = −∆x∆y (hT (T − T s)|h + hT (T − T s)) |0 ≈ −2∆x∆yhT (T −T s),if h is small. The first two lines form finite differences that tend to derivativesif we divide by the volume h∆x∆y and let ∆x, ∆y → ∞. Inserting thetemperature then results in

∂V

q ·ndS = k ∂ 2T

∂x2 + k

∂ 2T

∂y2 − 2 hT

h (T − T s) .

The alternative derivation consists in starting with ∇2T = 0 and inte-grating in the z direction:

h 0

k

∂ 2T

∂x2 + k

∂ 2T

∂y2

dz +

h 0

k∂ 2T

∂z2 dz = 0 .

We assume that the variation in z direction is sufficiently small in the firstintegrals so we can regard T is independent of z. In the other integral weperform integration once and insert the cooling law:

h 0

k∂ 2T

∂z2 dz = k

∂T

∂z

h

− k ∂T ∂z

0

= −hT (T − T s)|h − hT (T − T s)|0

≈ −2hT (T

−T s) .

Note that, e.g., k∂T/∂z at z = 0 equals −k∂T/∂n = hT (T − T s). Dividingby h in the integrated equation gives (1.42).

The heat loss in the third direction over a thin plate can with this tech-nique be modeled as an extra source term in a two-dimensional simulation of the temperature evolution.


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1.8 Cooling of an Object; PDE Models

Imagine that we bring an object, initially at temperature T = T 0, into sur-rounding medium at temperature T = T s. We are interested in the time ittakes to cool (or heat) the object to reach the steady state condition wherethe object’s temperature equals that of the surroundings (T s). As an exam-ple, think of taking a cold bottle out of the refrigerator or placing hot foodon a plate.

To simplify the problem we assume that the object is spherical and thatwe can utilize spherical symmetry in the problem, i.e., the spatial variationsdepend only the distance r from the center of the spherical object. We shallformulate three different models for the problem, one with a fixed tempera-ture T s at the surface of the object, one with a cooling law at the surface,and one with the object and its surroundings. The extent of the (hollow)spherical surrounding domain is taken as M diameters of the sphere, whereM can vary. In the cooling law we assume that the surrounding medium hasa constant temperature T s, while in the two-medium case we may imposeT = T s at the outer boundary of the surrounding medium.

Show that the simplified governing PDE for the first two formulations of this problem reads

C p∂T

∂t = k

1

r2∂

∂r

r2

∂T

∂r

, 0 < r < a, t > 0, (1.44)

where a is the radius of the object to be cooled or heated. Argue why theboundary condition at r = 0 is

∂T

∂r = 0 . (1.45)

For the third formulation of the problem, explain why the governing PDEtakes the form

C p∂T

∂t =

1

r2∂

∂r

k(r)r2

∂T

∂r

, 0 < r 0 . (1.46)

Find the associated boundary conditions in the different cases. Set up thethree complete mathematical models.

A common trick when dealing with Laplace terms in spherical coordinatesis to introduce a new variable:

v(r, t) = rT (r, t) .

This transformation leads to a standard one-dimensional heat equation, asin Cartesian coordinates, for the unknown v(r, t). Set up the complete math-ematical models for the first two models in this case. Show that the sim-plification does not apply to the third model with a heterogeneous medium


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1.8. Cooling of an Object; PDE Models 19

(k = k(r)). The great advantage with the transformation v = rT is that wecan analyze the present spherical problem using either a 1D diffusion equation

program or analytical solutions of the 1D diffusion equation.Introduce a suitable scaling.


Basic Equations. This is a 3D heat conduction problem. In the solid objectthe temperature T is governed by

C p∂T

∂t = ∇ · (k∇T ) .

The parameters , C p, and k are the object’s density, heat capacity, andheat conduction coefficient, respectively. This equation also applies to thesurrounding medium if we assume that there is no associated flow.

In air, for instance, the object will heat/cool the surrounding air, whichmay induce a small flow field because of buoyancy effects (hot air rises, cold airfalls). This small air flow may have a significant impact on transporting heat.However, analyzing the heat transfer problem in air under such conditionsrequires us to formulate a free thermal convection problem, i.e., a couplingof a heat transfer equation for fluids (with convective acceleration term) anda flow model with variable density (in the gravity term). This is beyond thescope of this exercise. When using a cooling law at the object’s surface, theheat transfer coefficient in this law often incorporates the effect free thermalconvection in the surrounding medium.

When it comes to boundary conditions, we need one condition at eachpoint at the surface of the object, if the domain of interest is the object. Atthe surface we either prescribe the temperature T s or we impose a coolinglaw

q ·n = hT (T − T s),where hT is a heat transfer coefficient and q is the heat flux, q = −k∇T .

In the two-medium case, the interface between the object and the sur-rounding medium is an internal boundary, hence no explicit boundary condi-tion is needed here when we work with a PDE model with variable coefficientsk(r), (r), and C p(r). The relevant boundary condition is T = T s at the outerboundary of the surrounding medium.

The initial condition reads T = T 0 in the object and T = T s in thesurrounding medium.

Simplifications. Since we have spherical symmetry, we introduce spherical

coordinates and assume that T only depends on the distance r from thecenter of the object (the origin) and time: T = T (r, t). The operator on theright-hand side of the governing equation then reduces to

1

r2∂

∂r

k(r)r2

∂T

∂r

.


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1. In the first case, we model the heat conduction inside the object only andassume that T is fixed at T s at the boundary. The complete mathematical

problems becomes

C p∂T

∂t = k

1

r2∂

∂r

r2

∂u

∂r

, 0 < r < a, t > 0, (1.47)

T (r, 0) = T 0, 0 ≤ r ≤ a, (1.48)T (a, t) = T s, t > 0, (1.49)

∂

∂rT (0, t) = 0, t > 0 . (1.50)

The last condition arises because the solution is symmetric about r = 0(the symmetry condition being vanishing normal derivative).

2. In the second problem we replace the boundary condition T (a, t) = T s

by a cooling law q · n = hT (T − T s).Since q = −k∇T , ∇T = ∂T

∂r ir, and n = ir, we get

−k ∂ ∂r

T (a, t) = hT (T (a, t) − T s) .

3. In the third problem we operate with two domains, the object and thesurrounding medium. We then introduce variable material properties:

ξ (r) =

ξ O, 0 ≤ r ≤ a,ξ S , a < r ≤ M a

where ξ can be either , C p, or k. The subscript O refers to the object,while S refers to the surrounding medium. The material properties ξ Oand ξ S are constant.

Since we deal with non-constant material properties, k must appear inside∇ · (k∇T ), and this term reduces to

1

r2∂

∂r

k(r)r2

∂T

∂r

under spherical symmetry. The complete mathematical problem is then

C p∂T

∂t =

1

r2∂

∂r k(r)r2

∂u

∂r , 0 < r 0,(1.51)

T (r, 0) =

T 0, 0 ≤ r ≤ a,T s, a < r ≤ M a (1.52)

T ((M + 1)a, t) = T s, t > 0, (1.53)

∂

∂rT (0, t) = 0, t > 0 . (1.54)


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1.8. Cooling of an Object; PDE Models 21

Transformation. The substitution v(r, t) = rT (r, t) implies

1r2 ∂ ∂r

k(r)r2 ∂T ∂r

= ∂ ∂r

k(r) ∂v∂r− vr2 ∂k∂r .

For k constant, we achieve a Laplace term of the same form as encountered if r were a Cartesian coordinate, modulo the factor 1/r. This factor is cancelledby the same factor on the left hand side,

∂T

∂t =

1

r

∂v

∂t .

That is, for the first two problems, where k = const, we get the governingPDE

C p∂v

∂t = k

∂ 2v

∂r2 . (1.55)

The boundary condition at r = 0 becomes

∂

∂rT (0, t) =

1

r

∂

∂rv(0, t) − 1

r2v(0, t) = 0 .

Multiplication by r2 and inserting r = 0 gives

0 · ∂ v∂r

− v(0, t) = 0 ⇒ v(0, t) = 0 .

The Dirichlet condition T (a, t) = T s simply becomes

v(a, t) = aT s,

while the cooling condition at r = a leads to

−k ∂v∂r

= 1

av(a, t) + hT (v(a, t) − aT s) .

Finally, the initial condition is v(r, 0) = rT 0.We may summarize the initial-boundary value problems for v(r, t):

C p∂v

∂t = k

∂ 2v

∂r2, 0 < r < a, t > 0, (1.56)

v(r, 0) = rT 0, 0 ≤ r ≤ a, (1.57)v(0, t) = 0, t > 0, (1.58)

v(a, t) = aT s, t > 0, case 1, (1.59)−k ∂v

∂r =

1

av(a, t) + hT (v(a, t) − aT s), t > 0, case 2 . (1.60)


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Scaling. The obvious space scale is a. The temperature variation is T s − T 0,assuming T s > T 0, and T 0 may then be taken as a reference value for T .

Dimensionless variables can be introduced by

r̄ = r

a, T̄ =

T − T 0T s − T 0 ,

t̄ = t

tc,

where tc is the time scale. Inserting r = ar̄ etc. in the governing equation,when k = const, leads to

∂ T̄

∂ ̄t = α

1

r̄2∂

∂ ̄r

r̄2

∂ T̄

∂ ̄r

, α =

tck

C pa2 . (1.61)

Taking α = 1, which implies that the two terms in the PDE are of orderunity, determines the time scale:

tc = C pa2/k .

The initial condition becomes

T̄ (r̄, 0) = 0 .

The boundary condition at r = a simply reads

T̄ (1, t) = 1,

whereas the cooling condition at r = a takes the form

− ∂ T̄

∂ ̄r = β ( T̄ − 1), r̄ = 1,

where

β = hT akis a dimensionless number.

Scaling in the heterogeneous case follows the same procedure, the onlydifference being that we introduce a dimensionless ξ̄ (r̄) function:

ξ̄ (r̄) =

1, 0 ≤ r̄ ≤ 1,ξ S /ξ 0, 1 < r̄ ≤ M

The problem for v can also be scaled, but since v = rT , there will only bea new factor r in the scaling, which cancels in the equations. Thus, no newinformation is provided. The scaled PDE for v̄ = r̄ T̄ is then

∂ ̄v

∂ ̄t = α

∂ 2v̄

∂ ̄r2

as a counterpart to (1.61). Boundary conditions are v̄ = 0 at r̄ = 0, for case1 v̄ = 1 at r̄ = 1, or for case 2

−∂ ̄v∂ ̄r

= v̄ + β (v̄ − 1) .


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1.9. Cooling of an Object; Averaged Model 23

Looking at the scaled problem, the PDE is independent of the physical in-put parameters in the model. In case 1, a single numerical solution is sufficient

to provide complete insight into the problem (i.e., there are no parametersto vary). In case 2, the boundary condition at r̄ = 1 ,

− ∂ T̄

∂ ̄r = β ( T̄ − 1), β = hT a

k ,

contains one parameter, β . The ratio hT a/k is therefore the only parameterthat influences the solution. In the two-medium case, the solution dependson

S c pS Oc pO

, kS

kO, M .

1.9 Cooling of an Object; Averaged Model

We study the same physical problem is in Problem 1.8. An object with initialtemperature T = T 0 is moved to a medium such that the object’s surround-ing temperature is T S . The purpose of the present problem is to derive asimple model for the temperature evolution in time inside the object, basedon integral equations rather than PDEs.

Find an integral formulation for stationary heat conduction without heatsources. (Hint: Start either with the general energy equation on integral formand remove irrelevant terms, or start with the PDE and “go backwards”in the derivation, i.e., integrate the PDE over a volume and use the diver-gence theorem if appropriate.) Let the integration volume V coincide withthe object to be cooled (or heated). The object’s geometry can now be taken

as arbitrary. At the outer surface we apply a cooling law. To simplify theintegral equation, we introduce two averaged temperature quantities:

T̄ V = 1

V

V

T dV, T̄ ∂V = 1

S

S

T dS . (1.62)

Here, V and S are the volume and surface area of the object, and ∂ V denotesthe surface. Derive from the integral equation the relation

d

dtT̄ V = −α1 T̄ ∂V + α2, (1.63)

where α1 and α2 are positive constants. Find, under the assumption T̄ V ≈T̄ ∂V , the time evolution of the object’s temperature. Provide expressions

for the special case of a spherical object, and discuss how the accuracy of the expressions can be evaluated by comparison with “exact” solutions fromProblem 1.8.


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The general integral form of the first law of thermodynamics reads

d

dt

1

2

V

v · vdV + V

udV

=

∂V

n · σ · vdS + V

b · vdV

− ∂V

q · ndS + V

hdV .

We know that the K term cancels with other terms by taking the equation of continuity and the equation of motion. Omitting heat generated by stressesand body forces, as well as internal heat sources, gives

d

dt

V udV

= −

∂V q ·ndS

Reynolds’ transport theorem transforms the first term to V

∂u

∂tdV +

∂V

uv · ndS .

The latter term vanishes in solids at rest. Further, in a solid we may use thethermodynamical relation

∂u

∂t ≈ C p ∂T

∂t .

Making use of Fourier’s law in the q term leads to the final integral form V

C p∂T ∂t

dV = ∂V

k ∂T ∂n

dS .

This equation corresponds to the PDE C pT ,t = ∇ · (k∇T ).The next step is to incorporate the cooling law

−k ∂T ∂n

= hT (T − T s),

in the surface integral term. This implies V

C p∂T

∂t dV = −

∂V

hT (T − T s)dS .

Assuming , C p, hT , and T s are constants in these integrations, and that V is a fixed volume in time, we get

C p ∂

∂t

V

T dV = −hT ∂V

T dS + hT T s

∂V

dS .


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1.9. Cooling of an Object; Averaged Model 25

Introducing the averaged quantities in the exercise, we may rewrite the lastexpression as

C pV ∂ T̄ V ∂t

= −hT S T̄ ∂V + hT T sS,or

d

dtT̄ V = −α1 T̄ ∂V + α2,

for

α1 = hT S

C pV , α2 = α1T s .

Assuming T̄ V ≈ T̄ ∂V ≡ T̄ (t), we havedT̄

dt = −α1 T̄ + α2 .

The solution of this differential equation is

T̄ = α2

α1+ Ce−α1t .

The initial condition reads T̄ = T 0, and α2/α1 = T s, resulting in the finalsolution

T̄ = T 0e−α1t + T s

1 − e−α1t , α1 = hT S

C pV .

We may scale the problem. Using the same scaling for T and t as inProblem 1.8,

T̄ = T − T 0T s − T 0 ,

t̄ = tC pka2

,

we get the differential equation

dT̄ dt̄

= γ β

(1 − T̄ ),

where

β = hT a

k , γ =

Sa

V are two dimensionless numbers; β occured in Problem 1.8, while γ is a newgeometry factor in the current approximation problem. The solution reads

T̄ = 1 − e−tγ/β . (1.64)For a sphere, γ = Sa/V = 3, so

T̄ = 1 − e−3t/β , (1.65)

showing that β is the main parameter influencing the solution, just as wefound from the scaling of case 2 in Problem 1.8. The solution of the currentproblem tells that the temperature increases exponentially to that of thesurroundings with a time scale hT a/k.

The accuracy of (1.65) can be assessed by comparing T (t) with the solu-tion T (a, t) in case 2 from Problem 1.8.


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1.10 Diffusion of Ink in a Water Tube

A tube of length L and radius a is filled with water. At one end we inject,at time t = 0, ink such that the concentration of ink is large in a small areaaround the center axis of the tube. The tube is fully closed so there is no flowof water.

Set up a mathematical model for predicting the distribution of ink in spaceand time. Use a suitable functional description of the ink concentration attime t = 0 (a localized Gaussian bell function, for instance). Present themodel both in three-dimensional Cartesian coordinates and in cylindricalcoordinates. In the latter case one can assume radial symmetry.

Under which conditions is it reasonable to work with a one-dimensionalmodel?


Basic Equations. The ink will spread out in water by diffusion. There is noconvective transport since there is no flow of water. Diffusion of a substancein a medium at rest is governed by diffusion equation

∂c

∂t = k∇2c + f, (1.66)

where c is the concentration of ink, k is the diffusion constant (from Fick’slaw), and f models injection or extraction. In the present case, we inject inkat t = 0, and model the result of this injection as an initial condition c(x, 0).The function f is therefore zero. The complete tube is closed. Therefore,there is no possibility for the ink to flow through the boundaries. This means

that the flux of ink is zero on the boundaries. Mathematically, the boundarycondition reads

∂c

∂n = 0 . (1.67)

Modeling of the Initial Condition. At t = 0 we specify c as a localized con-centration the x axis at x = 0, x being the coordinate along the axis of thetube. A Gaussian bell is a suitable model for localized functions:

c(x,y,z,t = 0) = c0 exp

− 1

σ2

x

σx

2+

y

σy

2+

z

σz

2. (1.68)

The parameters σx, σy, and σz describe the width of the Gaussian bell in the

x, y, and z directions. This means that σx, σy, σz must be quite small forthe initial concentration to be localized. The parameter c0 is the maximumconcentration at x = y = z = 0. We then place the origin on the tubeaxis at the left end of the tube. Outside the tube the initial c function is just truncated (i.e., ignored). The three-dimensional initial-boundary valueproblem consists of (1.66), (1.67), and (1.68).


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1.10. Diffusion of Ink in a Water Tube 27

Simplifications. If the initial value of c is radially symmetric, we assume thatthe further evolution of c exhibits cylindrical symmetry. The governing equa-

tions can therefore be written in cylindrical coordinates (r, x). This affectsonly the Laplace term k∇2 when written explicitly in terms of coordinates.In the initial condition we collect y2 + z2 as r2, under the assumption thatσy = σz ≡ σr .

The complete initial-boundary value problem for c(r,x,t) becomes

∂c

∂t = k

1

r

∂

∂r

r

∂c

∂r

+ k

∂ 2c

∂x2, (r, x) ∈ Ω, t > 0, (1.69)

c(r,x, 0) = c0 exp

−

r2

σ2r+

x2

σ2x

, (r, x) ∈ Ω, (1.70)

∂c

∂n = 0, r = a, x = 0, L, t > 0, (1.71)

whereΩ = {(r, x) | 0 ≤ r < a, 0 < x < L}

is the domain of the tube.

Scaling. It is natural to scale the problem:

c̄ = c

c0, x̄ =

x

L, r̄ =

r

L, t̄ =

t

tc.

Now, c̄ = 1 is pure ink and c̄ = 0 is pure water. Inserting the scaling in themodel and fitting tc such that the time-dependent term is of the same sizeas the Laplace term (i.e., tck/L2 = 1), and dropping bars as usual, leads to

∂c

∂t =

1

r

∂

∂r

r

∂c

∂r

+

∂ 2c

∂x2, (r, x) ∈ Ω, t > 0, (1.72)

c(r, x, 0) = exp

− 1

L2σ2r

r2 +

x2

σ2

, (r, x) ∈ Ω, (1.73)

∂c

∂n = 0, r = 1, x = 0, 1, t > 0, (1.74)

whereΩ = {(r, x) | 0 ≤ r


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may neglect the r dependence. The initial-boundary value problem then be-comes

∂c

∂t = k

∂ 2c

∂x2, 0 < x < L, t > 0, (1.75)

c(r, x, 0) = c0 exp

− x

2

σ2x

, 0 < x < L, (1.76)

∂c

∂x = 0, x = 0,L, t > 0 . (1.77)

In scaled form we have

∂c

∂t =

∂ 2c

∂x2, 0 < x < 1, t > 0, (1.78)

c(r,x, 0) = exp− x2

L2σ2x, 0 < x < 1, (1.79)

∂c

∂x = 0, x = 0, 1, t > 0 . (1.80)

Simulations. Figures 1.7 and 1.8 present some simulation of the two-dimensionalscaled model1. The purpose is to see whether a one-dimensional model is ad-equate or not. The ratio of L and a is 10. The results in Figure 1.7 wereproduced with an initial condition having σx = 0.6 and σr = 5. That is, theink fills the complete cross section at x = 0.

We then change the initial condition to be more localized by reducingσr to 0.5. Figure 1.8 shows the results. Initially, the concentration is clearlytwo-dimensional, but as time increases, the ink fills the whole cross sectionand the further development is well described as one-dimensional.

The relevance of a one-dimensional model can be investigated in moredetail by plotting the solution in Figure 1.8 along the tube axis in a curveplot, see Figure 1.9a. A corresponding one-dimensional model, solving (1.80)–(1.80), leads to the solution in Figure 1.9b. The difference is visible at earlytimes.

1 Actually, we have solved the PDE in two-dimensional Cartesian coordinates in-stead of radially symmetric cylindrical coordinates. The difference is assumed tobe very small when L = 10a as in the computational examples.


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(a)

air

soil

(b)

Fig.1.4. Sketch of the geometry of the heat conduction problem to be solved in

Problem 1.5. An isolated pipe with (a) : (a) Pipe surrounded by air. (b) Pipepartially digged into the ground and hence surrounded by air and soil.


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1

2

3

4

Fig.1.5. Reduced domain due symmetry in Problem 1.5.

Fig.1.6. Cross section of a tube. (Problem 1.6)


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0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

Fig.1.7. Diffusion of ink in a long and thin tube simulated with a two-dimensionalmathematical model. The top figure shows the initial concentration (dark is ink,white is water). The three figures below show the concentration of ink at (scaled)times t = 0.25, t = 0.5, t = 1, and t = 3, respectively. The evolution is clearlyone-dimensional.


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0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

0 10

0

1

Fig.1.8. Same diffusion problem and two-dimensional mathematical model as inFigure 1.7, but the initial concentration of ink at the left end does not fill the tubecross-section entirely, thus inducing some small initial two-dimensional effects. Howappropriate a one-dimensional model is for the present case becomes evident bycomparing the plots with those in Figure 1.7. See also Figure 1.9.


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0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

u(x,t=0)u(x,t=0.25)

u(x,t=0.5)u(x,t=1)u(x,t=3)

(a)

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

u(x,t=0)u(x,t=0.25)

u(x,t=0.5)u(x,t=1)u(x,t=3)

(b)

Fig.1.9. (a) Concentration along the tube axis in Figure 1.8; (b) concentrationcomputed by a corresponding one-dimensional model.


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Chapter 2

Fluid Flow Problems

2.1 Pressure in a Fluid at Rest

Dermine the pressure field in a fluid at rest in the gravity field, using thebasic equations of continuum mechanics and the only assumption that theconstitutive law is on the form

σ =

− pI + τ ,

where σ is the stress tensor, p is the pressure field, and τ denotes stressesthat vanish if the velocity is zero.


The governing equations for fluid motion are the Navier-Stokes equations(in some appropriate form) and the equation of continuity. If thermal effectsenter the problem, we also need an energy equation for the heat transport. Inthe present case, the fluid is a rest, implying that the velocity field v vanishes.There are no temperature effects in the present problem.

Only the equation of motion can then give us something non-trivial. This

equation reads

Dv

dt = ∇ · σ + b .

We have that σ = − pI , since τ = 0 when v = 0. Gravity is the only bodyforce, here written as b = −gk. The equation of motion then takes the form

0 = −∇ p − gk or ∇ p = −gk .This equation says that ∂p/∂x = ∂p/∂y = 0, that is, p = p(z), and integratingthe z component of the equation gives

p(z) = −gz + C,where C is an integration constant that must be determined from some con-

dition, e.g., that p(0) = p0. The p(z) function with this condition becomes

p(z) = p0 − gz.That is, in a fluid at rest, only the pressure gives contributions to stresses,and the pressure increases linearly with the depth.


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2.2. Pressure Force on a Box 35

2.2 Pressure Force on a Box

A tank with the shape of a box is submerged in deep water (at rest). Find thetotal pressure force on the tank from the water. Figure 2.1 depicts the prob-lem. Express the result in a form which shows that the result is in accordancewith Archimedes’ law.

symmetric

stress

p(x,y,z,t) = az + b x

z

W

H

x

z=D

z=D-H

Fig.2.1. Pressure on a box of size W x × W y × H . (Problem 2.2)


The total force S on a body, due to surface stress, can in general be writtenas

S (t) =

S

s(x, t)dA,

where s is the stress vector at the surface S of the body. The stress dueto pressure p is s = − pn, n being the outward unit normal to S . In caseof hydrostatic pressure, p varies linearly with coordinate along the directionof gravity, here such a variation is written compactly as p = az + b. The

exact expressions for a and b are developed in Problem 2.1: a = g andb = p0 + gz0, if g is the acceleration of gravity, is the density of water, and p = p0 for z = z0.

We then have

S = − S

(az + b)ndA .


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36 2. Fluid Flow Problems

The surface S is conveniently split into the six faces of a box:

S (t) =

6 faces

(az + b)(−n)dA

=

top

(az + b)z=D(−k)dxdy +

left side

(az + b)i dzdy +

bottom

(az + b)z=D−H k dxdy +

right side

(az + b)(−i)dzdy +

2 other sides

(az + b)(± j)dzdx

Further calculations give

S = W y

0 W x

0

(aD + b)dxdy(

−k) +

DD−H

W y0

(az + b)dydz i +

W y0

W x0

(a(D − H ) + b)dxdy k + DD−H

W y0

(az + b)dydz(−i)= [−(aD + b)W xW y + (a(D − H ) + b)W xW y ]k= −aW xW yH k = −aV k = M gk,

where V = W xW yH is the volume of the box and M = V is the mass of the water displaced by the box. The result S = M gk is in accordance withArchimedes’ law.

2.3 Pressure Force on a Cylinder

A tank with the shape of a cylinder is submerged in deep water (at rest).Find the total pressure force on the tank. Figure 2.2 depicts the problem.Express the result in a form which shows that the result is in accordancewith Archimedes’ law.


The total force S on a body, due to surface stress, can in general be writtenas

S (t) = S

s(x, t)dA,

where s is the stress vector at the surface S of the body. The stress dueto pressure p is s = − pn, n being the outward unit normal to S . In caseof hydrostatic pressure, p varies linearly with coordinate along the direction


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2.3. Pressure Force on a Cylinder 37

R

(p,q) p(z)=az+b L

Fig.2.2. Pressure on a cylinder with radius R and length L. (Problem 2.3)

of gravity, here such a variation is written compactly as p = az + b. Theexact expressions for a and b are developed in Problem 2.1: a = g andb = p0 + gz0, if g is the acceleration of gravity, is the density of water, and p = p0 for z = z0.

The relevant expression for S further work is then

S = − S

(az + b)ndA .

The surface S is naturally split into the two end surfaces S 1 and S 2 and thecurved, cylindrical surface S c. For analytical integration it is convenient touse cylindrical coordinates, defined relative to the centerline of the cylinder:

x = p + r cos θ,

z = q + r sin θ,

y = y .

The stress due to pressure on S 1 and S 2 is −(az + b)(± j), whereas on S cthe stress is −(az + b)ir. Since ir depends on the integration variables, itis natural to work with this dependence explicitly through the relation ir =i cos θ+k sin θ, which has constant unit vectors. Putting the elements togetherwe have for the S 1 surface that

S 1 = 2π

0 R

0

(a(q + r sin θ) + b)y=L jrdrdθ .

The corresponding expression for the S 2 surface reads

S 2 =

2π0

R0

(a(q + r sin θ) + b)y=0(− j)rdrdθ .


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We see that the contributions S 1 and S 2 cancel each other. On the S c surfacewe have the contribution

S c =

L0

2π0

(a(q + R sin θ) + b)[−(i cos θ + k sin θ)]Rdθdy

= L

2π0

aR sin2 θ(−k)Rdθ

= −aLπR2k = M gk,

with M = LπR2 being the mass of the water displaced by the cylinder. Theresult S = M gk is in accordance with Archimedes’ law.

2.4 Pressure Force on a Sphere

A tank with the shape of a sphere is submerged in deep water (at rest).Find the total pressure force on the tank. Figure 2.3 depicts the problem.Express the result in a form which shows that the result is in accordancewith Archimedes’ law.

p(z)=az+b R

(p,q)

Fig.2.3. Pressure on a sphere. (Problem 2.4)


The total force S on a body, due to surface stress, can in general be written

asS (t) =

S

s(x, t)dA,

where s is the stress vector at the surface S of the body. The stress dueto pressure p is s = − pn, n being the outward unit normal to S . In case


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2.5. Stationary Channel Flow; Newtonian Fluid 39

of hydrostatic pressure, p varies linearly with coordinate along the directionof gravity, here such a variation is written compactly as p = az + b. The

exact expressions for a and b are developed in Problem 2.1: a = g andb = p0 + gz0, if g is the acceleration of gravity, is the density of water, and p = p0 for z = z0.

The pressure force on a sphere can now be written

S (t) =

S

(az + b)(−n)dA .

Using spherical coordinates for convenience in analytical calculations, we have

x = p + r cos θ sin φ

z = q + r cos φ

y = 0 + r sin θ sin φ

andn = ir = i cos θ sin φ + k cos φ + j sin θ sin φ .

This results in the integral

S =

π0

2π0

(a(q + R cos φ) + b) ×

[−i cos θ sin φ − k cos φ − j sin θ sin φ]R2 sin φdθdφ= −2π

π0

R2(a(q + R cos φ) + b)cos φ sin φ dφk

= −a4

3 πR3

k = M gk,

where M = 43

πR3 is the mass of the water displaced by the sphere. Thisspecial form of the result shows that the end result is in accordance withArchimedes’ law.

2.5 Stationary Channel Flow; Newtonian Fluid

We consider incompressible fluid flow in a channel confined by two planewalls, z = 0 and z = H , see Figure 2.4. The flow is laminar and parallel to thewalls. The lower and upper walls move with velocity V 0 and V H , respectively.There migh be a prescribed pressure gradient in the direction of the flow.

The flow can also be driven by a component of gravity. The fluid is classifiedas Newtonian.

Develop a mathematical model for the fluid flow where it is assumed thatthe flow is stationary (i.e., the pressure gradient and the velocities of thewalls are constant). Introduce a scaling of the coordinates and the dependent


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z

x

V

V

H

0

pressure gradient

g H

Fig.2.4. Flow in a channel with plane (moving) walls. (Problem 2.5)

variables. Find the analytical solution for the velocity (v) and the pressure( p) fields. Thereafter, find a formula relating the total volume flux,

Q = H

0

v

·i dz,

to the pressure gradient. (i is a unit vector in the direction of the flow (andwalls)).


Incompressible Newtonian flow is governed by the Navier-Stokes equations,

∂ v

∂t + v · ∇v = − 1

∇ p + ν ∇2v + b,

∇ · v = 0 .

The body force b is here the acceleration of gravity, and we set b = g in thefollowing. On each wall, the velocity field v must equal the velocity of thewall.

We assume that the flow is parallel to the planes, in direction of the xaxis:

v = u(x , y , z , t)i

If the planes have infinite extension in the y direction, we can omit variationswith y. The equation of continuity requires ∇ · v = ∂u/∂x = 0. In thestationary case we also have ∂/∂t = 0 such that v = u(z)i. Inserting thisexpression in the Navier-Stokes equations and multiplying by gives

0 = −∇ p + µu(z)i + g .

Derivation of this equation with respect to x, y, or z shows that ∇ p is con-stant. In the x component of the equation, we introduce β = −∂p/∂x + gx(gx = b · i). The governing equation for u(z) then becomes

u(z) = −β µ


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2.5. Stationary Channel Flow; Newtonian Fluid 41

with boundary conditions

u(0) = V 0, u(H ) = V H .A possible scaling of this problem goes as follows:

z̄ = z

H , ū =

u − V 0uc

,

resulting in

d2ū

dz̄2 = −βH

2

µuc,

ū(0) = 0,

ū(1) = V H − V 0

uc.

Hereafter, we drop the bars. The choice of uc depends on whether the flowis driven mainly by the planes or by the pressure gradient. Aiming at u of order unity, one can assume that u is also of order unity, i.e.,

βH 2

µuc= 1 ⇒ uc = βH

2

µ .

The boundary condition at z = 1 then becomes

u(1) = µV H − V 0

βH 2 .

With β of the same order as V H − V 0, or larger, u(1) is of order unity andconsistent with the basic assumption. However, if V H − V 0 is much largerthan the pressure gradient, i.e., the flow is mainly driven by the movingwalls, u(1) 1 and the basic assumption of u(z) of order unity fails. Weshould in that case set uc = V H − V 0 resulting in the problem

u = − βH 2

µ(V H − V 0) ,u(0) = 0,

u(1) = 1 .

Again, if V H − V 0 is larger than β , u is of order unity and consistent withthe assumption, whereas the case V H − V 0 β might lead to a large u andthe scaling fails.

The analytical solution is easy to find as this is a matter of integratingtwice and applying the boundary conditions. Assuming a scaling that corre-sponds to pressure-driven flow, we get

ū(z̄) = 1

2z̄(1 − z̄) + µ V H − V 0

βH 2 z̄ .


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Transformed to unscaled variables, this reads

u(z) = βH

2µ z

1 − z

H

+ (V H − V 0) z

H + V 0 .

We can see that the velocity profile consists of the parabolic profile due tothe pressure gradient plus the linear Couette flow profile due to the movingplanes.

The volume flux,

Q =

H 0

v · i dz,

is then be computed to be

Q = H 3β

12µ +

1

2(V H − V 0)H .

The solution of the present problem can be viewed as a superposition(addition) of the solution of the classical Poiseulle problem, where a pressuregradient β drives the flow and the walls are at rest, and the classical Couetteproblem, where the flow is driven by moving walls and there is no pressuregradient. To see this, we can write up the two bou

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Continuum mechanics problems