# Volume Fluxes and Associated Velocities

1. Sep 28, 2014

### geojon

1. The problem statement, all variables and given/known data
Suppose that the volume transport associated with the wind-driven circulation's western boundary current (such as the Gulf Stream) is 50 x 10^6 m^3/second. Assume that this volume transport is carried in a current of uniform speed which is 50 km wide and 1 km thick. Calculate the average velocity of the current.

2. Relevant equations
Q = v * A
(volumetric flow rate = velocity*area)

3. The attempt at a solution
Reworking the equation Q = v*A to solve for velocity yields the equation, v = Q/A.
Area has been converted from km to meters.
So, velocity = (50 x 10^6 m^3/sec) / (50,000,000 m^2) = 1 meter/second.

So, the answer I've found is a velocity of one meter per second. This is a graduate level course, however, so I am not sure that this correct. The math is not wrong, though; could it be my approach?

2. Sep 29, 2014

### Staff: Mentor

Does the answer of 1m/s seem reasonable to you for the Gulf Stream current?

Chet

3. Sep 29, 2014

### pasmith

If the cross-sectional area is $50 \times 10^6\,\mathrm{m}^2$ and the volume flux is $50 \times 10^6\,\mathrm{m}^3\,\mathrm{s}^{-1}$ then the average velocity is indeed $1\,\mathrm{m}\,\mathrm{s}^{-1}$.

Wikipedia says that the maximum velocity in the Gulf Stream is about 2.5 ms-1, so 1 ms-1 is the right order of magnitude for an average.