1. Aula Teórica 9&10 Equação de Evolução. Exemplos. Caso 1D

2. Equação de evolução Ou:

3. Diffusion plus advection Where is diffusive flux maximum?

4. 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)? x1 x2

5. 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.

6. Case of concentration Consider 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. y

7. 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?

8. 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. E se existisse decaimento?.

9. With first order decay How will the concentration evolve in each case? t0t0 t1t1 t2t2 t∞t∞ In case of decay, concentration are lower and tend to zero.

10. 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.

11. 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:

12. The Eulerian formulation 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:

13. Let’s split the channel into elementary volumes Cell i-1 Cell i Cell i+1 And apply the equation to each of them: If the property can be considered uniform inside each cell and along each surface: Where A is the area of the cross section between the elementary volumes, assumed constant in the academic example.

14. Numerical solution Dividing all the equation by the volume: 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?

15. 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:

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

17. Explicit calculation That can be solved explicitly, but has stability problems. If we had assumed: We would have got:

18. Implicit calculation That has less computation than the “average” approach, but still requires the resolution of a system of equations. If we had assumed: We would have got:

19. Results (D<0.5) D=0.25 DT=100 dx20 difusividate1 Timek=0 i-3i-2i-1ii+1i+2i+3 000000100000 1000000.250.50.25000 200000.060.250.380.250.0600 30000.020.090.230.310.230.090.020 40000.030.110.220.270.220.110.030 5000.010.030.120.210.250.210.120.040.01 The results evolve as expected.

20. Results (D=0.75) D=0.75 DT=300 dx20 difusividate1 Timek=0 i-3i-2i-1ii+1i+2i+3 000000100000 3000000.75-0.50.75000 600000.56-0.81.38-0.80.5600 90000.42-0.81.83-1.81.83-0.80.420 12000.32-0.82.11-2.93.65-2.92.11-0.80.32 1500-0.790.71-3.95.77-6.25.77-3.92.24-0.8 The results are strange. Concentration bigger than initial are obtained. Negative concentrations are also obtained…. The method is unstable

21. Stability The method became unstable because the parenthesis becomes negative when D>0.5; When the parenthesis changes sign, the effect of c i 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.

22. 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.