9.3: Ekman Mass Transport

We may want to know the total mass transported in the layer. The Ekman mass transport \(M_{E}\) is defined as the integral of the Ekman velocity \(U_{E}\), \(V_{E}\) from the surface to a depth \(d\) below the Ekman layer. The two components of the transport are \(M_{Ex}, M_{Ey}\):

\[M_{Ex} = \int_{-d}^{0} \rho U_{E} \ dz, \quad\quad M_{Ey} = \int_{-d}^{0} \rho V_{E} \ dz \nonumber \]

The transport has units \(\text{kg/(m} \cdot \text{s)}\). It is the mass of water passing through a vertical plane one meter wide that is perpendicular to the transport and extending from the surface to depth \(-d\) (figure \(\PageIndex{1}\)).

Figure \(\PageIndex{1}\): Sketch for defining Left: mass transports, and Right: volume transports.

We calculate the Ekman mass transports by integrating \((8.3.4-5)\) in \((\PageIndex{1})\):

\[\begin{align} f \int_{-d}^{0} \rho V_{E} \ dz = f M_{Ey} &= -\int_{-d}^{0} dT_{xz} \nonumber \\ f M_{Ey} &= -T_{xz} \mid_{z=0} + T_{xz} \mid_{z = -d}\end{align} \nonumber \]

A few hundred meters below the surface the Ekman velocities approach zero, and the last term of \((\PageIndex{2})\) is zero. Thus mass transport is due only to wind stress at the sea surface \((z = 0)\). In a similar way, we can calculate the transport in the \(x\) direction to obtain the two components of the Ekman transport: \[\begin{align} f M_{Ey} &= -T_{xz}(0) \\ f M_{Ex} &= T_{yz}(0) \end{align} \nonumber \]

where \(T_{xz}(0), T_{yz}(0)\) are the two components of the stress at the sea surface.

Notice that the transport is perpendicular to the wind stress, and to the right of the wind in the northern hemisphere. If the wind is to the north in the positive \(y\) direction (a south wind), then \(T_{xz}(0) = 0\), \(M_{Ey} = 0\), and \(M_{Ex} = T_{yz}(0)/f\). In the northern hemisphere, \(f\) is positive, and the mass transport is in the \(x\) direction, to the east.

It may seem strange that the drag of the wind on the water leads to a current at right angles to the drag. The result follows from the assumption that friction is confined to a thin surface boundary layer, that it is zero in the interior of the ocean, and that the current is in equilibrium with the wind so that it is no longer accelerating.

Volume transport \(Q\) is the mass transport divided by the density of water and multiplied by the width perpendicular to the transport. \[Q_{x} = \frac{Y M_{x}}{\rho}, \quad\quad Q_{y} = \frac{X M_{y}}{\rho} \nonumber \]

where \(Y\) is the north-south distance across which the eastward transport \(Q_{x}\) is calculated, and \(X\) is the east-west distance across which the northward transport \(Q_{y}\) is calculated. Volume transport has dimensions of cubic meters per second. A convenient unit for volume transport in the ocean is a million cubic meters per second. This unit is called a Sverdrup, and it is abbreviated \(\text{Sv}\).

Recent observations of Ekman transport in the ocean agree with the theoretical values \((\PageIndex{1})\). Chereskin and Roemmich (1991) measured the Ekman volume transport across 11\(^{\circ}\)N in the Atlantic using an acoustic Doppler current profiler as described in Chapter 10. They calculated a transport of \(Q_{y} = 12.0 \ pm 5.5 \ \text{Sv}\) (northward) from direct measurements of current, \(Q_{y} = 8.8 \pm 1.9 \ \text{Sv}\) from measured winds using \((\PageIndex{1})\) and \((\PageIndex{2})\), and \(Q_{y} = 13.5 \pm 0.3 \ \text{Sv}\) from mean winds averaged over many years at 11\(^{\circ}\)N.

Use of Transports

Mass transports are widely used for two important reasons. First, the calculation is much more robust than calculations of velocities in the Ekman layer. By robust, I mean that the calculation is based on fewer assumptions, and that the results are more likely to be correct. Thus the calculated mass transport does not depend on knowing the distribution of velocity in the Ekman layer or the eddy viscosity.

Second, the variability of transport in space has important consequences. Let’s look at a few applications.