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Conductivity, temperature, depth sensors are used by oceanography to calculate the velocity of sound at varying depths.

These measurment values are used to comprise sound velocity profiles like the one below to increase the presition of deep water sonars.

Sound velocity profile

My question is, how is the actual calculation from conductivity, temperature and depth to sound velocity conducted?

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[Posted as answer instead of comment as suggested by David Zaslavsky]

While not able to give all details, I understand it like this: salinity=salinity(conductivity, temperature), density=density(temperature,salinity), stiffness=stiffness(temperature,pressure), and finally $c=\sqrt{\mbox{stiffness}/\mbox{density}}$). In the end, $c$ is a function of conductivity, temperature and pressure (depth).

There is an eq with references at wikipedia.

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  • $\begingroup$ Salinity isn't a direct function of temperature, it's far more complex. It's been measured directly, though, just by dragging it up. Similarly, temperature is influenced by many factors (e.g. currents, evaporation) so it's often measured instead of calculated. $\endgroup$
    – MSalters
    Commented Sep 12, 2011 at 10:07
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    $\begingroup$ @MSalters: I worked with the data OP said are provided by the sensors. I wrote computed salinity was function of conductivity and temperature. Temperature is measured directly, the sensor name reveals that. (.. evaporation?!) $\endgroup$
    – eudoxos
    Commented Sep 12, 2011 at 11:45
  • $\begingroup$ From what I found, yes. Of course it's not the only effect, but it's an interesting one as it affects both temperature and salinity. At tropical temperatures, heat loss through evaporation exceeds radiation. $\endgroup$
    – MSalters
    Commented Sep 12, 2011 at 13:03
  • $\begingroup$ @MSalters: evaporation will have only effect in the upmost layer of water. Since we talk about kilometers of depth, it is safe to ignore it even in tropical climate (unless the water is boiling). $\endgroup$
    – eudoxos
    Commented Sep 13, 2011 at 9:05
  • $\begingroup$ It's even a boundaray effect, but you need boundary conditions to solve your heat (temperature) equations. AFAICT tidal heating is quite minor, and the main heat exchange is therefore at the surface: solar radiation, radiative loss and evaporation. (Rain and river influx is also negligable) $\endgroup$
    – MSalters
    Commented Sep 13, 2011 at 9:20

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