# 11.13: Ekman Spiral of Ocean Currents

- Page ID
- 10214

Frictional drag between the atmosphere and ocean enables winds to drive ocean-surface currents. Coriolis force causes the surface current to be 45° to the right of the wind direction in the Northern Hemisphere. Drag between that surface-current and deeper water drives deeper currents that are slower, and which also are to the right of the current above. The result is an array of ocean-current vectors that trace a spiral (Fig. 11.61) called the **Ekman spiral**.

The equilibrium horizontal ocean-current components (U, V, for a rotated coordinate system having U aligned with the wind direction) as a function of depth (z) are:

\(\ \begin{align} U=\left[\frac{u_{*} \operatorname{water}^{2}}{\left(K \cdot f_{c}\right)^{1 / 2}}\right] \cdot\left[e^{z / D} \cdot \cos \left(\frac{z}{D}-\frac{\pi}{4}\right)\right]\tag{11.52a}\end{align}\)

\(\ \begin{align} V=\left[\frac{u_{*} w a t e r^{2}}{\left(K \cdot f_{c}\right)^{1 / 2}}\right] \cdot\left[e^{z / D} \cdot \sin \left(\frac{z}{D}-\frac{\pi}{4}\right)\right]\tag{11.52b}\end{align}\)

where z is negative below the ocean surface, f_{c} = Coriolis parameter, and u_{*water} is a **friction velocity** (m s^{–1}) for water. It can be found from

\(\ \begin{align} u_{*water}^{2}=\frac{\rho_{a i r}}{\rho_{\text {water}}} \cdot u_{* a i r}^{2}\tag{11.53}\end{align}\)

where the density ratio ρ_{air}/ρ_{water} ≈ 0.001195 for sea water. The friction velocity (m s–1) for air can be approximated using **Charnock’s relationship**:

\(\ \begin{align} u_{* a i r}^{2} \approx 0.00044 \cdot M^{2.55}\tag{11.54}\end{align}\)

where M is near-surface wind speed (at z = 10 m) in units of m s^{–1}.

The **Ekman-layer depth** scale is

\(\ \begin{align} D=\sqrt{\frac{2 \cdot K}{f_{c}}}\tag{11.55}\end{align}\)

where K is the ocean eddy viscosity (a measure of ability of ocean turbulence to mix momentum). One approximation is K ≈ 0.4 |z| u_{*water} . Although K varies with depth, for simplicity in this illustration I used constant K corresponding to its value at z = –0.2 m, which gave K ≈ 0.001 m^{2} s^{–1}.

The average water-mass transport by Ekman ocean processes is 90° to the right (left) of the near-surface wind in the N. (S.) Hemisphere (Fig. 11.61). This movement of water affects sea-level under hurricanes (see the Tropical Cyclone chapter).

**Sample Application**

For an east wind of 14 m/s at 30°N, graph the Ekman spiral.

**Find the Answer**

Given: M = 14 m·s^{–1}, ϕ = 30°N

Find: [U , V ] (m s^{–1}) vs. depth z (m)

Using a relationship from the Forces and Winds chapter, find the Coriolis parameter: f_{c} = 7.29x10^{–5} s^{–1} .

Next, use Charnock’s relationship. eq. (11.54): u_{*air}^{2} = 0.00044 (14 ^{2.55}) = 0.368 m^{2} s^{–2} .

Then use eq. (11.53):

u_{*water}^{2} = 0.001195 · (0.368 m^{2} s^{–2}) = 0.00044 m^{2} s^{–2}

Estimate K at depth 0.2 m in the ocean from

K ≈ 0.4 · (0.2 m) ·[0.00044 m^{2} s^{–2}] ^{1/2} = 0.00168 m^{2} s^{–1}

Use eq. (11.55):

D = [2 · (0.00168 m^{2} s^{–1}) / (7.29x10^{–5} s^{–1}) ]^{1/2} = 6.785 m

Use a spreadsheet to solve for (U, V) for a range of z, using eqs. (11.52). I used z = 0, –1, –2, –3 m etc.

Finally, rotate the graph 180° because the wind is from the East.

**Check:** Physics & units are reasonable.

**Exposition:** Net transport of ocean water is toward the north for this East wind case.