roms bed model tests
DESCRIPTION
ROMS Bed Model Tests. We consider three sediment classes We call them clay, silt, sand We can call them anything we want right now. 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 - PowerPoint PPT PresentationTRANSCRIPT
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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
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The 30-Box Test Grid
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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
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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.
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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.
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Test 2, Time Series of Mass in Bed
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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.
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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
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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.
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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
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Shear Stresses
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Clay in 30 boxes after 720 days
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Silt in 30 boxes after 720 days
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Sand in 30 boxes after 720 days
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Model cell 8, near bay mouth
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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
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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)
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ROMS Erosion Formula
• What does Eo mean?• It’s the erosion rate when Tau = 2 * Taucrit• In fact, erosion can take any value
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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