Choose a western boundary current in the ECCO dataset available as preprocessed netcdf in fox-kemper.com/data (I've regridded to eliminate staggered grids and time-averaged for you). Study its volume transport (via u and v) and return flow (via the Sverdrup relation). Time mean fields are fine, but you might want to check out the full set.
Quantitative aspects: Does the volume budget across a latitude line close? Is the meridional flow at the surface or deep or throughout? Is there a northward heat flux implied by this flow?
That is, consider \int v dz. There are 5 ways to get at it. You should contrast some of these results:
Consider a western boundary current, perhaps one you have used. We will use the ECCO dataset, which I will preprocess for you to make it easy at fox-kemper.com/data
Choose one latitude, and make a section of some variables. Here are some questions you might ask:
Overall Objective: How can you understand aspects of the western boundary current: is it driven by surface height anomalies (eta), or thermal wind (rhoanom), or pressure (phihydid which has both eta and rhoanom effects), or the return flow of Sverdrup? What makes the most sense to you?
NOTES on rhoanom and phihydid******
Rhoanom is the *potential* density deviation from the background density (rhoanom=potlrho-rho_0). You can expect flow to generally be directed along surfaces of constant rhoanom. You can also understand the values at any depth as corresponding to one another, i.e., rhoanom=60kg/m^3 at 1000m would be denser if adiabatically rearranged to 500m than a parcel already at 500m with rhoanom=40kg/m^3. You can also expect horizontal gradients of rhoanom to correspond to roughtly to thermal wind dv/dz, except for thermobaric effects which should be small. The units are kg/m^3. Generally, you should expect rhoanom to increase with depth.
phihydid is the perturbation pressure away from hydrostatic (in class we called it p'), which includes both the effects of surface height elevation and the accumulation with depth due to density by the hydrostatic relation on the perturbation pressure (rho-rho_0). Thus, by looking at phihydid, you don't have to mess with calculating the accumulating pressures. The hydrostatic equation relevant for phihydid is:
d/dz(phihydid)= - (rho-rho_0) g/(rho_0)
Thus, while phihydid is proportional to pressure, it has units of m^2/s^2, so multiply by rho_0 to get to Pascals if you like. Perhaps best of all is: phihydid*rho_0/10^4 which is the pressure in decibars, roughly equivalent to sea surface height displacements. Thus, at the surface you expect
phihydid*rho_0/10^4 \approx eta