MODELLING & ASSESSMENT (RESRAD) Why? Modelling Cant - PowerPoint PPT Presentation

MODELLING & ASSESSMENT (RESRAD)

Why? • Modelling – Can’t measure everything – Need to make predictions when designing new facilities • Assessment – Waste management and disposal – Compliance with regulatory requirements – Testing remediation strategies – Testing the design of new facilities

Problems • Internal dosimetry • Atmospheric dispersion • Tailings dams • General waste management strategies • Waste repositories/dumps • Landfill • Discharges to lakes, rivers, ocean • Legacy sites • Planning/designing of new facilities

Mathematical modelling • Mathematics is a scientific discipline in its own right • It is also an extremely useful tool for developing theories and models because it allows us to express ideas in very precise and concise terms, and because once the problem is formulated in mathematical terms all the power of the mathematics becomes available • Once the mathematical problem is solved the results have to be converted back into the language of the original problem

Scientific problem Conceptual model Formulate in mathematical terms, & solve the mathematical problem Data from Set up the (mathematical) model measurement program Modify the model VERIFY the model if necessary Make predictions VALIDATE the model Interpret the mathematical results by comparing predictions in terms of the original scientific problem against measurements

Conceptual model • Which processes to include (assumptions) • Which processes to exclude (assumptions) • Flow diagram • Each assumption places some restrictions on the use of the model or on the interpretation of the model predictions

Internal dosimetry injection wound skeleton adsorption skin muscle sweat inhalation blood liver respiratory tract exhalation ingestion GI tract urinary kidney tract faeces urine

Compartment models • For first-order, linear transfer between compartments, a compartment model for a single radionuclide can be described by the matrix-vector equation ∂ X = + A X P ∂ t • The general solution of this equation is ( ) ( ) P ( ) = + − e A e A t t X t X 0 I

Mathematical problem: Serial decay chain of length N • Chain (non-branching) with different biokinetics ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 0 0 0 A X X P 1 1 1 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ λ 0 0 0 I A X X P ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 1 2 2 2 ∂ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = λ + . 0 0 0 . . I A 2 dt ⎢ ⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎢ ⎥ . 0 . . . 0 . . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ λ 0 0 . ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ I A X X P − N N 1 N N N • This is actually the same equation as before, but written to show the relationship between the members of the decay chain

Waste burial/heap resuspension drinking inhalation gamma rain evaporation irrigation food radon run-off well cover (erosion) infiltration waste base leaching aquifer

Tailings dam evaporation precipitation inflow overflow water sediment (tailings) base aquifer

Fluid mechanics General conservation equation In any region the rate of change of a quantity (mass, momentum, angular momentum, energy) that can be considered to be conserved is given by an expression of the form Rate of change in region = + rate of flow into region – rate of flow out of region + rate of generation/loss within region by non-flow processes (chemistry, radioactive decay) This approach is valid for both microscopic situations (e.g. the equations of classical fluid mechanics) or macroscopic situations (e.g. estimating radionuclide concentrations inside large slabs of material)

• Flow equation for a one-constituent fluid (conservation of mass) ∂ C ( ) = −∇ • − λ U C C ∂ t

• Flow equation for a radioactive contaminant in a fluid ∂ C ( ) = −∇ • − λ a U C C ∂ a a a a t • Fick’s law (derived from experiment) states that ( ) − U = − ∇ U a C K C a a a • The conservation of mass equation now becomes ∂ C ( ) ( ) = −∇ • + ∇ • ∇ − λ a a U C K C C ∂ a a t • This is a form of the diffusion equation

• A more familiar form is ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎛ ∂ ⎞ ∂ ∂ ⎛ ⎞ ⎛ ⎞ C ( ) ( ) ( ) C C C + + + = + ⎜ ⎟ + − λ ⎜ ⎟ ⎜ ⎟ UC VC WC K K K C ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ x y z ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ t x y z x x y y z z • If the fluid is homogeneous, and the coordinate system is oriented so that the fluid is flowing in the x-direction, then ∂ ∂ ∂ ∂ ∂ 2 2 2 C C C C C + = + + − λ U K K K C ∂ ∂ ∂ ∂ ∂ x y z 2 2 2 t x x y z

• If the fluid is isotropic then ∂ ∂ ∂ ∂ ∂ 2 2 2 C C C C C + = + + − λ U K K K C ∂ ∂ ∂ ∂ ∂ 2 2 2 t x x y z • This equation can be used as the basis of models of atmospheric dispersion – Power stations – Ventilation shafts • It can also be used for area sources

Porous media ( ) ( ) ′ ′ = ε + ε − ε + − ε • Put C C C C 1 tot t s • where • ε ´ = total porosity (pore space/total space) • ε = effective porosity (connected pore space/total space) • C tot = total concentration of contaminant • C = concentration of contaminant in connected pores • C s = concentration of contaminant on pore surfaces • C t = concentration of contaminant in unconnected pores

Flow equation for a contaminant in a porous medium (the same balance approach as before) is Rate of increase in a small volume ∆ V = net rate at which flowing water brings contaminant into ∆ V + net rate at which contaminant diffuses into ∆ V - rate at which contaminant decays in ∆ V This leads to ( ) ( ) ( ) ( ) ∂ ∂ ε ∂ ε ∂ ε ∂ ε 2 2 2 C C C C C + = + + − λ tot U K K K C ∂ ∂ ∂ ∂ ∂ x y z tot 2 2 2 t x x y z which is the starting point for the discussion of groundwater transport of contaminants

• Substituting for C tot gives ( ( ) ( ) ) ( ) ′ ′ ∂ ε + ε − ε + − ε ∂ ε C C C 1 C + t s U ∂ ∂ t x ( ) ( ) ( ) ∂ ε ∂ ε ∂ ε 2 2 2 ( ( ) ( ) ) C C C ′ ′ = + + − λ ε + ε − ε + − ε K K K C C 1 C ∂ ∂ ∂ x y z t s 2 2 2 x y z • Assume that C t = C • This gives ( ( ) ) ( ) ′ ′ ∂ ε + − ε ∂ ε C C 1 C + s U ∂ ∂ t x ( ) ( ) ( ) ∂ ε ∂ ε ∂ ε 2 2 2 C C C ( ( ) ) ′ ′ = + + − λ ε + − ε K K K C 1 C ∂ ∂ ∂ x y z s 2 2 2 x y z

• Partition coefficient K d ( the ratio of the concentration of contaminant on the pore surfaces to the concentration of contaminant in solution) s = C K C d • The definition of K d assumes that absorption- desorption process are much faster than flow processes – confirmed by experiments ( ( ) ) ( ) ′ ′ ∂ ε + − ε ∂ ε C 1 K C C + d U ∂ ∂ t x ( ) ( ) ( ) ∂ ε ∂ ε ∂ ε 2 2 2 C C C ( ( ) ) ′ ′ = + + − λ ε + − ε K K K C 1 K C ∂ ∂ ∂ x y z d 2 2 2 x y z

• Final step – retardation factor • Put ( ) ′ ′ ε + − ε = ε 1 K d R • Then ∂ ∂ ∂ ∂ ∂ 2 K 2 2 K C U C C C K C + = + + − λ y x z C ∂ ∂ ∂ ∂ ∂ 2 2 2 t R x R x R y R z

∂ ∂ ∂ ∂ ∂ 2 K 2 2 K K C U C C C C + = + + − λ y x z C ∂ ∂ ∂ ∂ ∂ 2 2 2 t R x R x R y R z • This equation has exactly the same form as the atmospheric diffusion (fluid flow) equation - this means that the mathematical solutions of the porous medium equation have the same general form as those for the atmospheric diffusion equation • For most radionuclides K d >>1 and therefore R >> 1 which implies that the water moves through the porous medium much faster than the contaminant – again this is confirmed by measurement

ASSESSMENT • Internal dosimetry – Dose calculations – Bioassay interpretation – Hiroshima, Maralinga • Environmental impact assessment – Check on existing facilities – Design of new facilities – Checking waste management strategies – Checking remediation strategies for legacy sites – Hiroshima, Maralinga

Internal dosimetry injection wound skeleton adsorption skin muscle sweat inhalation blood liver respiratory tract exhalation ingestion GI tract urinary kidney tract faeces urine

Examples • Consumption of sea-food containing Po- 210 • Po-210 poisoning (London) • Pu fabrication plant accident

Context (Po-210) • The dose per unit intake for ingestion of Po-210 is approximate 1.2 µSv/Bq • To get a dose of 1 Sv would require an intake of approximately 1 MBq • The half-life of Po-210 is 138.4 days, so 1 MBq corresponds to 6 nano grams