Converting depth to pressure

Typically, a constant density approximation has been made:

P(at depth) = Depth(m) x density x acceleration of gravity, (4)

where, the density of the overlying water column is assumed to be 1.035 and the acceleration of gravity is 9.8 m/s2. There are problems with this approach. Water is slightly compressible, so its density increases with depth. While small, the effect of this difference on pressure at depth is cumulative. The variation of gravity with latitude, makes this calculation site specific. These effects can be seen quite clearly in figure 1. Notice that below the seafloor, these effects work in reverse due to the geothermal gradient where pore-water temperature increases with depth. At higher temperatures, the density of the pore-water is greatly diminished and the cumulative pressure error can become negative.

Figure 1. Calculated pressure differences vs depth between the simplified pressure relation and a calculation including the compressibility of seawater and the variation of gravity (Fofonoff and Millard, 1982) as a function of latitude (solid line, 90°; dashed line, 60°; dot-dash line, 30°; and dotted line, 0°). A standard ocean model was used: seawater salinity was 35 at a constant temperature of 0°C; porewater salinity was 32.5 and the geothermal gradient in the sediments was 35°C/km. Note that one atmosphere equals 101.3 kPA, which is equivalent to about 10m in depth.

Pressure [kPa]

Figure 1. Calculated pressure differences vs depth between the simplified pressure relation and a calculation including the compressibility of seawater and the variation of gravity (Fofonoff and Millard, 1982) as a function of latitude (solid line, 90°; dashed line, 60°; dot-dash line, 30°; and dotted line, 0°). A standard ocean model was used: seawater salinity was 35 at a constant temperature of 0°C; porewater salinity was 32.5 and the geothermal gradient in the sediments was 35°C/km. Note that one atmosphere equals 101.3 kPA, which is equivalent to about 10m in depth.

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