ROMS Bed Model Tests

• We consider three sediment classes

• We call them clay, silt, sand

• We can call them anything we want right now

Clay Silt Sand

D50, µm 3 250 250

ρ, kg/m3 1350 2000 2650

Wstl, µm/s 0.174 17.4 174

Taucrit, P 0.05 0.5 2

Erate, g/sq m/s 0.15 0.15 0.15

The 30-Box Test Grid

Test 1, Continuous Erosion

• Set Taucrit for all classes to 0.05 P

• Impose constant stress of 0.064 P

• Erosion rate for all classes 0.3 g/sq m/s

• No settling

• One third of bed, by volume, in each size class initially

• Thirty days duration

Conclusions/Findings

• The number of layers in the ROMS bed model is fixed. Layers do not appear or disappear.

• At present, no provision exists to reduce erosion as a function of depth in the bed or age of sediments.

• In the limiting case, the bed thickness will go to zero under constant erosion. Material is not entrained from deep sediments beneath the bottom of the bed

• Mass is conserved.

Test 2, Continuous Erosion

• Increase Taucrit for sand to 5 P.

• In this case, clay and silt will erode but not sand.

• All other conditions the same as Test 1.

Test 2, Time Series of Mass in Bed

Conclusions/Findings

• The total mass of sand in the bed does not change. The vertical distribution of sand does change.

• The thickness of the bed diminishes.

• The vertical distribution of clay and silt change.

• The bed does not appear to armor. Clay and silt erode continuously.

Test 3, Deposition

• Eliminate sediment erosion

• Activate settling in the water column

• One third of bed, by volume, in each size class initially.

• Thirty days duration

Conclusions/Findings

• Mass is conserved.

• When a new layer is added at the top, the bottom two layers are combined. Total thickness of the bed increases.

• No long-term burial out the bottom of the bed.

Test 4, Long-Term Behavior

• Activate the 30-box model of Chesapeake Bay.• Boundary conditions for clay and silt = 10 mg/L.

Boundary condition for sand = zero except at bottom of bay mouth.

• Erosion and settling active for all constituents.• Uniform initial conditions in bed. 1% clay, 49.5%

silt, 49.5% sand by volume.• 720 days duration

Shear Stresses

Clay in 30 boxes after 720 days

Silt in 30 boxes after 720 days

Sand in 30 boxes after 720 days

Model cell 8, near bay mouth

Conclusions/Findings

• Clay largely equilibrates with boundary conditions. After 720 days, however, erosion still occurs.

• A turbidity maximum consisting of silt forms.• The bed is erosional near the mouth,

depositional in the turbidity maximum.• Bed equilibrium is not attained after 720 days.• The bed exhibits similar properties to earlier

tests. E.g. the vertical distribution of sand changes

The ROMS Erosion Formula

11crit

EoE

E = erosion rate (kg m-2 s-1) calculated at shear stress τ Eo = Specified base erosion rate (kg m-2 s-1)φ = porosity (0 < φ < 1)τcrit = critical shear stress for erosion (P)

ROMS Erosion Formula

• What does Eo mean?• It’s the erosion rate when Tau = 2 * Taucrit• In fact, erosion can take any value

ROMS Erosion Formula

• We need some way of limiting the amount of sediment that can erode at any imposed stress

• The simplest approaches are patterned after Jerome Maa– Put a time constant on erosion rate so erosion

declines as a function of the time stress is imposed

– Erode only when current is accelerating