Sparse Hydrodynamic Ocean Code V5199 User Manual

SHOC User Manual

12 Tests

A suite of tests has been collected for this model. These tests are used to validate the model against analytic solutions or other known, simply understood situations. They are particularly useful to check the correct operation of the model after modifications to the model code.

12.1 No forcing

Test 1 is an extremely simple, null case, test, where no forcing is applied to a closed model domain. The purpose is to demonstrate that no model variables deviate from their initial values. A rectangular grid is used with a horizontal grid of 5 by 10 cells, and 5 layers in the vertical having 1m vertical spacing. The bathymetry varies and the water is initially vertically stratified. Vertical diffusion of salt and heat is turned off in this case to avoid diffusive changes. Each run consists of a 1000 second integration with no externally applied forcing.

The initial variable values (zero elevation, zero velocity, salinity and temperature) should remain unchanged for the duration of the integration.

12.2 Ekman Spiral

A wind of constant stress and direction is blown over a homogeneous open ocean of constant depth. The model uses cyclic open boundaries reflect the open ocean condition, and utilizes constant vertical viscosity and linear bottom friction for simplicity. According to Kowalik and

Murty (1993, p27) the linear resistance coefficient is related to the bottom drag coefficient via:

r

=

ρ

C d v

12.2.1 where C d

is the drag coefficient, v is the bottom current speed and

ρ

is the density. Given a constant eddy viscosity, the Ekman depth, D

E

, is given by (Pond and Pickard, 1983, p108):

D

E

=

π

2

V z

/

f

12.2.2 where V z

is the eddy viscosity and f is the Coriolis parameter. The surface current speed, V o

, is then given by (Pond and Pickard, 1983, eqn 9.10):

V o

=

D

E

2

πτ

ρ

f s

12.2.3 where

τ s

is the wind stress, and the current speed, V b

, at the Ekman layer depth is (Pond and

Pickard, 1983, p108):

V b

=

V o

exp(

−

π

) ~ 0 .

04

V o

12.2.4

Therefore, using a wind stress of 0.01Nm

-2

on and f-plane with f = 1e

-4

, and V z

= 0.0507 m

2 the Ekman layer depth D

E

4.33x10

-3

ms

-1

and V b s

-1

,

= 100m. Furthermore, using 12.2.3 with

ρ

= 1025 kgm

-3

gives V o

=

= 1.87x10

-4

ms

-1

. Using the bottom velocity in 12.2.1 and a nominal drag coefficient of C d

= 0.003 gives a resistance coefficient of r = 0.00058. Linear friction is achieved by setting the parameter

UF

to a large value and

Z0

to a low value. The value of

Z0 below which bottom drag is set to the parameter

QBFC

is given by:

Z

0

<

0 .

5

∆

z bot

exp(

κ

.

sqrt

( 1 /

QBFC

)

−

1 )

12.2.5

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Using the value of

QBFC

quoted above,

Z0

< 7.7x10

-8

. Using these values with

UF

= 1.0 provides linear bottom friction with the required resistance coefficient. Using the above configuration, model results should show an Ekman spiral with velocities rotating clockwise with depth, surface current speed ~ 0.0043 ms

-1

and bottom current speed ~ 0.00019 ms

-1

.

Surface elevation should be equal to zero.

12.3 Constant wind stress – closed basin

This test examines the set-up due to a steady wind applied to a 1-layer (depth-averaged) model domain. If a constant wind is applied to a homogeneous closed basin of constant depth then depth averaged velocities are equal to zero in the steady state. For a linear model and constant wind stress in the x direction the surface slope should balance the applied wind stress, and the equations of motion reduce to an expression for the sea level gradient in the x direction:

∂

η

∂

x

= −

τ

s

ρ

gD

12.3.1 temperature 20 o basin is equal to 9.958x10

-6

-3

, then the slope in a 10m deep

, and the depth averaged velocities should be near zero. where D is the water depth. If a wind stress of 1Nm

-2

is applied to a homogeneous ocean of

C and salinity 35 psu so

ρ

= 1024.76 kgm

12.4 Constant wind stress – alongshore open channel

An analytical solution exists for a linear model of constant wind stress applied in a longshore direction along an infinitely long coast. Assuming cross-shelf transport and alongshore sea level gradient are small, then along shelf transport, U, is given by (Chapman, 1985, eqn 4.5):

U

=

D

τ

ρ

r s

(

1

− exp(

−

tr

/

D

)

)

12.4.1 with a steady state velocity (

t

= ∞

) given by:

U

=

D

τ

ρ

r s

12.4.2

Note U is the transport, hence velocity u = U/D.

The sea surface slope is given by (Chapman, 1985, eqn 4.6):

∂

η

∂

y

= −

fU gD

= −

f

τ

s

ρ

gr

(

1

− exp(

−

tr

/

D

)

)

12.4.3 with steady state sea level given by:

η

(

y

)

= −

f

τ

s

ρ

gr

(

L

2

−

y

)

12.4.4 where L is the width of the channel and it is assumed

η

=

0

at y = -L/2.

Cyclic open boundaries are used to represent an infinite coastline. Using a wind stress of 0.1

Nm

-2

in a channel 500km wide (the dimensions of this domain are the same as the test

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domain used by Palma and Matano (1998) except the Southern hemisphere is considered) with linear resistance coefficient r = 0.0005, Coriolis = -1.028e

-4

and

ρ

= 1024.76 kgm

-3

, the along-shore depth averaged velocity is 0.195 ms

-1

, the cross-shore depth averaged velocity is zero and elevation at the coast is 0.49 m (slope of 2.05x10

-6

. Note the first elevation cell center is found at y = 10km). This result assumes a linear depth averaged model is used, and a non-linear 3-D model with quadratic bottom friction is expected to give different results.

12.5 Constant wind stress – cross-shore open domain

A wind applied perpendicular to an infinitely long coastline (on the southern boundary in this case) will result in an elevation setup against the coast with zero depth averaged currents everywhere. The on-shore wind stress drives depth averaged flow to the west (east) in the northern (southern) hemisphere, and the sea level gradient resulting from setup at the coast drives this flow to the east (west). In a perfect situation the sea level gradient and wind stress forces balance resulting in no flow. Boundary effects and numerical error may make one of these forces dominate, leading to non-zero flow in the east (west) direction in the northern

(southern) hemisphere if the sea level pressure gradient dominates, and vice versa if wind stress dominates. Assuming a linear model with linear bottom friction, the analytical solution for sea level profile is given by (Chapman, 1985, eqn 4.13a):

∂

2

η

∂

y

2

= −

τ

s

ρ

gD

2

∂

D

∂

y

12.5.1

Using a domain with two cyclic cross-shelf open boundaries and one offshore boundary with elevation clamped to zero, and the linear depth profile used by Chapman (1985, eqn 4.1), then the sea level profile is given by Chapman (1985, eqn 4.14) and shown in Figure 12.5.1 and Table 12.5.1. The boundary conditions of eqn. 12.5.1. for this domain are:

∂

D

∂

y

= −

D o

at y

=

0

∴

∂

η

∂

y

=

τ

s

ρ

gD o

and

η

=

0 at y

=

L

12.5.2 where L is the distance to the offshore boundary.

Figure 12.5.1 : Sea level profile for onshore wind stress.

0.035

0.03

0.025

0.02

0.015

0.01

0.005

0

0 10 20 30 70 80 90 100 40 50 60

Offshore distance (km)

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Offshore distance (km) Surface elevation (m)

5

15

0.0301

0.0189

25

35

45

55

65

75

85

95

0.0137

0.0103

0.0077

0.0057

0.0039

0.0025

0.0012

0.0001

Table 12.5.1 : Sea level profile for onshore wind stress.

Again, a non-linear 3-D model with quadratic bottom friction is expected to give different results, and bottom friction is generally required to be increased for solutions to match theory.

Specifically, adequate solutions were obtained using the

CONSTANT

mixing scheme with background vertical viscosity

VZ0

= 0.0507 (see Test2) and minimum bottom drag coefficient of

QBFC

= 0.003 (

UF

=

Z0

=1e-8). Horizontal viscosity of

U1VH

=

U2VH

= 800 is also required for stability.

12.6 Propagation of a bore

This test simulates a wetting bore propagating along an initially dry channel. The model domain represents a channel 2km wide and 100km long with uniform (flat) bathymetry. A constant velocity of 1 ms

-1

is applied at one end of the initially dry channel. A bore propagates along the channel, with a parabolic shape (surface elevation profile) determined by a balance between the quadratic bottom friction and surface slope. The length of the bore is related to the depth at the inflow via:

L

=

gD

2

12.6.1

2

C U d

2

12.7 Wind stress curl – closed basin

Wind stress possessing curl applied to a closed basin with a gradient of f/D results in the formation of a gyre due to conservation of potential vorticity which is biased to the east if f/D <

0 and biased to the west if f/D > 0 (e.g. Herzfeld and Tomczak, 1999). The gradient of f/D may result from a gradient of f (

β

effect) or a change in topography. This test consists of a closed basin in the southern hemisphere with constant depth in the east – west direction, 50m depth at the southern coast and 100m depth at the northern coast. Wind stress in the e1 direction is applied, with 0.1 Nm

-2

at the southern boundary and 0 at the northern boundary, hence this stress possesses negative curl. The gradient of f/D is positive in this case, thus an anticyclonic gyre biased to the west is expected, generated by topographically induced conservation of potential vorticity. Theory predicts that:

•

A negative gradient of f/D (i.e.

CORIOLIS = 1.0e-4

) results in an eastward biased gyre.

•

A flat bottom (

BATHYMAX = 50

) results in an unbiased gyre.

•

Wind stress with positive curl (

WIND_SPEED SCALE = -1

) results in a cyclonic gyre with unaltered bias.

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