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Test Case 2


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This test case provides a fundamental check of the ability of a model to represent 1) mixing processes typical of estuarine conditions, 2) resuspension, advection, and deposition for suspended-sediment transport, 3) temporal dynamics of upper bed layer, and 4) interaction of suspended-sediment and the bed.

Figure showing the layout of test case 2, an estuary.


The domain is a long, narrow rectangular channel.
Length (east-west) l = 100,000 m
Width (north-south) w =1000 m
Depth h = 10 m at the western end, decreasing linearly to 5 m at the eastern end
Constant temperature, 10° C

Bottom Sediment

Single grain size on bottom:
Size (D50) = 0.15 mm (fine sand)
Density ρs = 2650 kg/m³
Settling velocity = 0.50 mm/s
Critical shear stress τc= 0.05 N/m²
Fractional bed concentration (1-porosity) = 0.90
Sediment layer thickness = 1 mm


No Coriolis
No wind
No heating/cooling

Initial Conditions

Salinity distribution from west end = 35 to east end = 0
(Vertically well mixed.)

Boundary Conditions

North and south sides = walls with no fluxes, no friction
Bottom roughness ks = 0.15 m (zo = Ks/30 = 0.005 m)
Sediment flux calculated by model
Surface = free surface, no fluxes

Flow, Western end, tidal
qwest = 4000 m³ /s * sin (ω t ) - qeast
ω = 2π/T
T = 12 * 3600, seconds
t = model time step, seconds
Flow, Eastern end, constant riverine (varying depth)
qeast = 400 m³ /s
Salinity, western end
Radiation condition with nudging, tnudge = 3 hours
Salinity at eastern end = 0

Output (ASCII files suitable for plotting)

At t = 10 days (20 tidal cycles):
Tidally averaged velocity field (average over last 2 tidal cycles)
Bed profile
Tidal-mean suspended sediment field

Physical Constants

Gravitational acceleration g = 9.81 m/s²
Von Karman's constant κ = 0.41
Dynamic viscosity (and minimum diffusivity) ν = 1e-6 m² /s


If a model incorporates physical constants that differ from these, and/or automatically calculates some values specified here, please specify the values used.


Solution to Test Case 2: Estuary

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