Numerical solution of the Graetz problem and related ... · PDF fileNumerical solution of the...

Numerical solution of the Graetz problem and

related parabolic problems in MEK4300

A rather simple task ?

G. Pedersen

Department of Mathematics, University of Oslo

MEK4300

G. Pedersen Numerical solution of the Graetz problem and related parabolic

Solution strategies in hydrodynamics

“Analytic” methods applied to simple problems.Sometimes useful, on other occations not worthwile.

Importance highly overrated by senior researches

Application of general computational tools; CFD(like Navier-Stokes solvers).Become increasingly important in science; are the future.

Incompetent use has yet miscredited this approach.

It is often difficult to assess quality of solution. Tradition in

this respect is frustratingly inadequate.

Specifically adapted/developed numerical techniques,dependent on problem.Much work, good controle of quality often enabled.

Still an indispensable tool in research.

Which one to choose ?Poor question! One should consider any method.


Plane transient Couette flow

Initial value problem

∂u

∂t= ν

∂2u

∂y2, 0 < y < h

with u(y , 0) = 0, u(0, t) = U, u(h, t) = 0.

May U be a function of t (for instance gradual start-up) ?

Dimensionlessu = Uu, x = hx , t = h2t/ν

∂u

∂ t=

∂2u

∂y2, 0 < y < 1

with u(y , 0) = 0, u(0, t) = 1, u(1, t) = 0.


Separation solution; omitting ˆ

Form AStationary profile - Fourier sum

un = 1− y −

n∑

k=1

2

πke−k2π2t sin(kπy)

Form BAlso stationary profile is expanded

un =n

∑

k=1

2

πk

(

1− e−k2π2t)

sin(kπy)

Sound convergence for t > 0 (save B and y = 1).t = 0 slow convergence due to changing signs of harmonic parts.


Form B, n = 20

Too coarse; not good.


Form B, n = 200

Still not impressive.


Form A, n = 20

Poor for small t.


Form A, n = 200

Still not good for t = 0. What about τ = µ∂u∂y

?


Solution near moving lid for t = 0.0025

Fourier (Fou) methodmarked with n.

Fou already good.

Finite difference (FD)marked with number ofpoints

FD presented later.

FD also challenged byjump at t = y = 0.


Separation solution for transient Couette flow

+ Simple explicit expression, easily worked out by hand

− Poor convergence for small t due to singularity at t = y = 0.

− Small time properties for ∂u∂y

even poorer

! FD solution also influenced by singularity.

+ Fourier series converge rapidly even for moderately small t

+ One term gives good approximations for large t

Graetz’ solution shares all the disadvantages and only the lastadvantage; Graetz functions must be analyzed and solvednumerically and coefficients must be found.This is more numerical work than direct application of a numericaltechnique.


Standard heat equation

Boundary value problem; dimensionless

∂T

∂t=

∂2T

∂y2, a ≤ y ≤ b, t ≥ 0

T (y , 0) = To(y), T (a, t) = Ta, T (b, t) = Tb = 0.

Fixed temperatures at the ends and initial condition.May for, instance, describe the evolution of temperature in aninsulated rod.


Standard heat equation, conservative form

Dimensionless energy density: E = T

Dimensionless energy flux: q = −∂T∂y

(Fourier’s law)

Heat equation conservative form

∂E

∂t= −

∂q

∂y, a ≤ y ≤ b, t ≥ 0.

It is immediately apparent that, for instance,

∂

∂t

b∫

a

Edy = q(a)− q(b)

Rate of energy change in rod = net energy rate provided at theends


Controle volume discretization

1 We define a spatial grid, yj , with a = y0 < y1 < ... < yn = b.Time discretized according to t(k) = k∆t.

2 Energy densities defined at grid points (yj , t(k)): E

(k)j = T

(k)j .

3 Spatial domain (rod) a ≤ y ≤ b divided into controle volums(or intervals, rather) centered at grid pointsIj = [yj− 1

2, yj+ 1

2] where yj− 1

2= 1

2(yj + yj+1)

Observe: two “half-volumes”, one at each end.

4 Energy in Ij at t(k): E

(k)j = ∆yjT

(k)j , with ∆yj = yj+ 1

2− yj− 1

2.

5 Energy flux from Ij to Ij+1 in t(k) < t < t(k+1): ∆t q(k+ 1

2)

j+ 12

.

Energy balance for each volume during one ∆t:Energy change in cell i = - net flux out of cell


Controle volume discretization, II

E(k+1)j − E

(k)j = −∆t

(

q(k+ 1

2)

j+ 12

− q(k+ 1

2)

j− 12

)

.

So far: completely based on physics, legible for people withoutknowledge of PDEs.Summation yields energy conservation

n−1∑

i=1

E(k+1)j −

n−1∑

i=1

E(k)j = −∆t

(

q(k+ 1

2)

n− 12

− q(k+ 1

2)

12

)

,

which is a crucial property.When available, controle volume methods are preferable fordeveloping finite diference schemes.We still need to relate the q’s to the T ’s


Controle volume discretization, III

Discrete Fourier’s law

q(k+ 1

2)

j+ 12

= −1

2

(

T(k+1)j+1 + T

(k)j+1 − T

(k+1)j − T

(k)j

)

/∆yj+ 12

where ∆yj+ 12= yj+1 − yj .

Interpretation: energy flux is proportional with energy difference inneighbouring cells.

The energy balance is now a discrete equation in T(k)j (next slide).

In explicit form it coresponds to a difference equation, where allderivatives of the heat equation is replaced by divided differences.This could have been formulated directly, but the controle volumeapproach is generally favourable.


Uniform resolution ∆yj = ∆yj+ 12= ∆y

InitiationSet T

(0)j = To(yj)

Time integration; for k = 0, 1, 2...

Advance from k∆t to (k + 1)∆t solving T(k+1)j from

(T(k+1)j − T

(k)j

∆t=

1

2

T(k+1)j+1 − 2T

(k+1)j + T

(k+1)j−1

∆y2+

T(k)j+1 − 2...

∆y2

.

for j = 1, ...n − 1, with T(k+1)0 = Ta and T

(k+1)n = Tb.

A linear, tri-diagonal equation solved by direct method.

Cranck-Nicholson’s method. Stable and efficient with one spatialdimension.


Generalized heat equation

PDE in terms of T ; dimensionless

F (y)∂T

∂t=

∂

∂y(G (y)

∂T

∂y) + S(y , t), a ≤ y ≤ b, t ≥ 0.

where G , F and S (source term) are given functions.

Conservative form

∂E

∂t= −

∂q

∂y+ S , a ≤ y ≤ b, t ≥ 0.

where E = FT and q = −G ∂T∂y

.Quantities may, or may not, have direct physical interpretations

Extension of numerical procedure to generalized problem isstraightforward. Examples: Graetz’ problem, transient Poiseulleflow, transient Couette flow.


The Graetz problem

The Graetz problem (dimensionless)

r(1− r2)∂T

∂x=

∂

∂r

(

r∂T

∂r

)

,

T (1, x) = 0, T (r , 0) = 1, r∂T (0, x)

∂r= 0.

The last one may seem trivial, but is needed in the numerics.There are (weak?) singularities at r = 0, 1 and a jump atr = x = 0 ⇒ affect numerical solution.


FDM

Grid: r0 = −12∆r ... rn = 1 = (n − 1

2)∆r

and x (k) = k∆x .First set u0j = 1, then step forward according to

rj(1− r2j )(

u(k+1)j − u

(k)j

)

=∆x

∆r(q

(k+ 12)

j+ 12

− q(k+ 1

2)

j− 12

),

where the discrete flux is given by

q(k+ 1

2)

j+ 12

=1

2∆r

(

u(k+1)j+1 + u

(k)j+1 − u

(k+1)j − u

(k)j

)

.

u(k+1)j−1 , u

(k+1)j , u

(k+1)j+1 do combine in each equation ⇒ tri-diagonal

set closed by u(k+1)n = 0 and u

(k+1)1 = u

(k+1)0 and solved by direct

method.Simple algorithm and simple code.


Early evolution

x = 0.0005 and x = 0.002.Rapid evolution near the wall for small x why ?Why must r → 1 and x → 0 be challenging ?


Later stages

x = 0.01, 0.05, 0.1, 0.2.Do we really have a good scaling for x ?By the way, ∆x/∆r is 0.01 or 0.1.


Transient Poiseulle flow

Scaling/transformation

r =r

r0, t =

ν t

r20, u = −

r204µ

dp

dx

(

1− r2 + u)

,

where tildes mark dimensional quantities.

The transformed problem (start from rest)

∂u

∂t=

1

r

∂

∂r

(

r∂u

∂r

)

,

u(1, t) = 0, u(r , 0) = r2 − 1, r∂u(0, t)

∂r= 0.

The last one may seem trivial, but is needed in the numerics.Same numerical technique as for the Graetz problem.


Early evolution

Half velocity profile w = 1− r2 + u for t = 0.05, 0.15, 0.25.Plug-like flow why ?


Later stages

t = 0.2, 0.6, 1.0, 1.4, 1.8.Approaching parabolic profile.


Plane transient Poiseulle flow

∂u

∂t= −

1

ρ

dp0

dx+ ν

∂2u

∂y2,

with u(y , 0) = 0, u(−h, t) = u(h, t) = 0.

Necessary to remove inhomogeneity ?May p0 be a function of t (for instance gradual start-up) ?We ought to make problem dimensionless.....


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