SME
Oceanography · Problem type
Ekman transport and coastal upwelling
Use this for “A 10 m/s wind blows toward the equator along 1,000 km of coast at 40° N. How much water upwells?”
Diagram
Northern Hemisphere (φ = 40°): the surface current runs 45° and the net transport 90° to the right of the wind.
The wind blows toward the south, so the water is carried west.
M
=
τ
ρw f
M
=
0.156
Pa
1,025 × 9.374 ×
10
-5
s
-1
M = 1.624 m²/s
W
=
ML
=
1.624
m²/s
×
1 ×
10
6
m
W = 1.624 Sv
The transport carries surface water offshore, and deep, cold water upwells along the coast.
This is an example. Type your own values to start a new problem.
U
Wind speed
example · m/s
ρₐ
Air density
example · kg/m³
Cd
Drag coefficient
example
τ
Wind stress
calculated · Pa
φ
Latitude
example · °
f
Coriolis parameter
calculated · s⁻¹
M
Ekman transport
calculated · m²/s
L
Coast length
example
W
Upwelling
calculated
Formulas
τ
=
ρₐ
×
Cd
×
U
2
0.156
=
1.2
×
0.0013 ×
10
2
f
=
2
×
7.292 ×
10
-5
×
sin(
φ
)
9.3744 ×
10
-5
=
2
×
7.292 ×
10
-5
×
sin(40°)
M
=
τ
1,025 ×
f
1.6235
=
0.156
1,025 × (9.3744 ×
10
-5
)
W
=
M
×
L
1,623,516.0282 = 1.6235 × 1,000,000
Formulas are worked in m, m³/s. The step-by-step shows the conversions.
Assumptions
•
The wind’s stress drives a top layer whose net flow, the Ekman transport, runs 90° to the right of the wind in the north.
•
A wind toward the equator along a west coast pushes surface water offshore, and deep water rises to replace it.
•
Seawater is 1,025 kg/m³ and Earth’s spin rate Ω = 7.292 × 10⁻⁵ rad/s.
Step-by-step
WE KNOW
U = 10 m/s, ρₐ = 1.2 kg/m³, Cd = 0.0013, φ = 40°, L = 1,000 km
FIND
wind stress (τ), Coriolis parameter (f), Ekman transport (M), upwelling (W)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m, m³/s).
L = 1,000 km = 1,000,000 m (1 km = 1000 m)
Step 2 · Find wind stress (τ)
Use τ = ρₐ × Cd × U²
The wind’s drag on the sea grows with the square of its speed.
τ
=
1.2
×
0.0013 ×
10
2
τ
=
1.2
×
0.0013
×
100
τ
=
0.00156
×
100
τ
=
0.156
Pa
Step 3 · Find Coriolis parameter (f)
Use f = 2 × 7.292 × 10⁻⁵ × sin(φ)
The Coriolis parameter is twice Earth’s spin rate times the sine of the latitude.
f
=
2
×
7.292 ×
10
-5
×
sin(40°)
f
=
2
×
(7.292
×
10
-5
)
×
0.64279
f
=
0.00014584
×
0.64279
f
=
9.3744 ×
10
-5
s
-1
Step 4 · Find Ekman transport (M)
Use M = τ ÷ (1,025 × f)
The
Coriolis
force
on
the
moving
layer
balances
the
wind’s
stress,
so
divide
by
the
seawater
density
and
f
.
M
=
0.156
1,025 × (9.3744 ×
10
-5
)
M
=
0.156
0.096088
M
=
1.6235
m²/s
Step 5 · Find upwelling (W)
Use W = M × L
Each
meter
of
coast
loses
M
offshore,
and
the
same
flow
rises
from
below.
W
=
1.6235
×
1,000,000
W
=
1,623,516.0282
m³/s
Step 6 · Convert the answers
Change each answer back to the unit you chose.
W = 1,623,516.0282 m³/s = 1.6235 Sv (1 Sv = 10⁶ m³/s)
Check
Put the numbers back in. Both sides should match.
0.156
=
1.2
×
0.0013 ×
10
2
✓
9.3744 ×
10
-5
=
2
×
7.292 ×
10
-5
×
sin(40°)
✓
1.6235
=
0.156
1,025 × (9.3744 ×
10
-5
)
✓
1,623,516 = 1.623516 × 1,000,000 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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