Aula Teórica 9&10

Equação de Evolução. Exemplos. Caso 1D

Equação de evolução

iojjj

j SSxxx

utdt

d

iojjj

j SSxxx

utdt

d

Ou:

• Diffusion plus advection

• Where is diffusive flux maximum?

Analysis of the evolution equation

• For the control volume:• What is the sign of ?• What does that mean physically (how does the

advective flux vary with x1)?• What is the relative value of diffusive flux in the lower

and upper faces of the control volume?• How does diffusive flux along x2 contribute to the

concentration inside the control volume? • If the flow is stationary and the material is

conservative what is the relation between advection and diffusion? what is the divergence of the total flux (advection + diffusion)?

iojjj

j SSxc

xxcu

tc

dtdc

x1

x2

11 xcu

Answers• The advective term is negative. The velocity is positive and • In other words, the advective flux decreases with x1, i.e., the quantity entering is bigger

than the quantity leaving (per unit of area).• The diffusive flux in the lower face is about zero. If located exactly over the symmetry line,

the flux will be exactly zero, because the gradient is null • On the upper face the diffusive flux is positive, i.e., the material is transported along the axis

x2 because the concentration is higher inside the control volume than above the control volume.

• The divergence of the vertical diffusive flux contributes to decrease the concentration inside the volume.

• The divergence of the horizontal diffusive flux is negative because the gradient on the left side of the volume is higher than on the right side. Horizontal advection is however the main mechanism to increase the concentration inside the control volume.

• If the flow is stationary the horizontal advection plus the horizontal diffusion balance the vertical diffusion. The divergence of both fluxes would be zero

• If the velocity was increased, the concentration inside the control volume would tend to increase because the quantity entering would increase and thus the quantity leaving would have to increase too.

01

xc

cu1

2xc

dif

0

jj

j xccu

x

Case of concentrationConsider two parallel plates and a property with a parabolic type distribution (blue) with maximum concentration at the center. Assume a stationary flow and a conservative property.a) Draw a control volume and indicate the fluxes (sense and relative magnitude). b) Where would the diffusive flux be maximum? Could this property be a

concentration? What kind of property could it be?c) What is the sign of the material time derivative?d) If the material is conservative (no sink or source), can the problem be

stationary?e) If it was a concentration without adsortion, what would be the final profile?

a) Diffusive flux increase with “r”. Advective flux depends on the longitudinal gradient, not yet known. b) Property gradient is maximum at the boundary. This means that there the diffusive flux is maximum and

consequently the property can pass through the boundary. The property can be a concentration only if there is adsortion at the boundary. It could be a temperature or momentum as well.

c) Property is being lost across the boundary and consequently there is a longitudinal negative gradient. The total derivative is negative (and so is the advective derivative)

d) yes the system could be stationary, but the fluid would loose property along the flow.e) If it is stationary and conservative what would be the sign of the longitudinal gradient component?f) If it was a concentration this profile could not maintain. The gradient should become null at the solid boundary (no

diffusive flux). The final profile would be uniform. Transversal diffusion would homogenize it.

yyxxx x

cxx

cxx

cu

y

Nitrogen Transport in the Baltic Sea• The figure shows the

distribution of nitrate discharged by the river Oder after 4, 8, 12 and 16 years of emission.

• In the region next to the river mouth the concentration becomes constant after 8 years. What does that mean?

• Is advection important in this system?

1D case• Vamos supor o caso de um reservatório com a forma de um

canal rectangular de 10 m de largura e 200 de comprimento, com velocidade nula. A concentração é elevada na zona central e nula na generalidade do canal. O material é conservativo.

• Escreva a equação que rege a evolução da concentração.

iojjj

j SSxc

xxcu

tc

dtdc

• E se existisse decaimento?.

With first order decay

• How will the concentration evolve in each case?

kcxc

xtc

dtdc

t0

t1

t2

t∞• In case of decay, concentration

are lower and tend to zero.

If there is a lateral discharge?

• Concentration will grow because of the discharge, will get homogenized because of diffusion and will decay because of decay.

• When discharge balances total decay, the system will reach equilibrium.

Solution of the problem

• The first case (pure diffusion) has analytical solution, but not the others.

• How to solve the problem numerically?• The Reynolds Theorem:

• The lagrangian formulation:

dAnvdVdtddV

t surfacesistemavc

.

dAncSiSodVdtd

surfacesistema

.)(

The Eulerian formulation )(.. SiSodAncdAnudV

t surfacesurfacevc

• This evolution equation states that “the rate of change inside a control volume balances the fluxes, plus sources minus sinks”.

• In differential form the equation becomes:

ioj

jj

iojjj

j

SSxccu

xtc

SSxc

xxcu

tc

)(.. SiSodAncnudVt surfacevc

Let’s split the channel into elementary volumes

Cell i-1 Cell i Cell i+1

And apply the equation to each of them:

dVolkcdAnccdVt vc

.

If the property can be considered uniform inside each cell and along each surface:

xAVol

kcVolxccA

xccA

tccVol iiiiti

tti

11

Where A is the area of the cross section between the elementary volumes, assumed constant in the academic example.

Numerical solution

• Dividing all the equation by the volume:

xAVol

kcVolxccA

xccA

tccVol iiiiti

tti

11

kcxcc

xcc

tcc iiiiti

tti

21

21

• In this equation we have two variables, , and an extra variable, c, which time we have not defined. Are the actual concentrations and are the concentrations that we want to calculate. The concentration c is the concentration used to calculate the fluxes and decay. What is it?

ttic

tictic

ttic

Flux calculation

• The flux equation gives quantity per unit of time. When we equation as:

• We are computing the amount that crossed the surface during a time period ( ) and dividing it by the length of the time. The most convenient time to allocate to c is the middle of the time interval:

kcxcc

xcc

tcc iiiiti

tti

21

21

t

2

tti

ti ccc

The equation becomes

21

21

21

21

22

222

xcc

xcc

cckxcc

xcc

tcc

tti

tti

tti

tti

ti

tti

ti

ti

ti

ti

ti

tti

222222 21

21

21

21

ti

ti

ti

ti

ti

tti

tti

tti

tti

tti

ti

tti kc

xcc

xcckc

xcc

xcc

tcc

ti

ti

ti

tti

tti

tti c

xtck

xtc

xtc

xtck

xtc

xt

1221212212 22221

222221

2

ti

ti

ti

tti

tti

tti cDckDcDcDckDcD 1111 22

1222

12

This equation has 3 unknowns. The all channel requires the resolution of a system of equations in each time step.

Explicit calculation

• That can be solved explicitly, but has stability problems.

• If we had assumed:ticc

• We would have got: t

iti

ti

tti DcckDDcc 11 21

Implicit calculation

• That has less computation than the “average” approach, but still requires the resolution of a system of equations.

• If we had assumed:tt

icc

• We would have got:

ti

ti

tti

tti cDcckDDc

11 21

Results (D<0.5)D= 0.25

DT= 100

dx 20

difusividate 1

Time k= 0

i-3 i-2 i-1 i i+1 i+2 i+3

0 0 0 0 0 0 1 0 0 0 0 0

100 0 0 0 0.25 0.5 0.25 0 0 0

200 0 0 0.06 0.25 0.38 0.25 0.06 0 0

300 0 0.02 0.09 0.23 0.31 0.23 0.09 0.02 0

400 0 0.03 0.11 0.22 0.27 0.22 0.11 0.03 0

500 0.01 0.03 0.12 0.21 0.25 0.21 0.12 0.04 0.01

The results evolve as expected.

Results (D=0.75)D= 0.75

DT= 300

dx 20

difusividate 1

Time k= 0

i-3 i-2 i-1 i i+1 i+2 i+3

0 0 0 0 0 0 1 0 0 0 0 0

300 0 0 0 0.75 -0.5 0.75 0 0 0

600 0 0 0.56 -0.8 1.38 -0.8 0.56 0 0

900 0 0.42 -0.8 1.83 -1.8 1.83 -0.8 0.42 0

1200 0.32 -0.8 2.11 -2.9 3.65 -2.9 2.11 -0.8 0.32

1500 -0.79 0.71 -3.9 5.77 -6.2 5.77 -3.9 2.24 -0.8

The results are strange. Concentration bigger than initial are obtained. Negative concentrations are also obtained….The method is unstable

Stability

• The method became unstable because the parenthesis becomes negative when D>0.5;

• When the parenthesis changes sign, the effect of ci at time t changes. If it is positive the larger is the initial concentration the larger is the subsequent concentration. If it is negative the larger is the initial concentration the smaller is the subsequent, which is physically impossible.

• This is the limitation of the explicit methods. They are conditionally stable. In pure diffusion problems the D (the diffusion number) must be smaller than 0.5. If other processes exist (e.g. k≠0) D must be samller.

ti

ti

ti

tti DcckDDcc 11 21

Implicit methods

• Implicit (as well as semi-implicit) methods are more difficult to program, but do not have stability limitations allowing larger values for D, meaning that we can combine small grid sizes with large time steps, while in explicit methods the time step is associated to the square of the spatial step.