Column of water in the sea

Storyboard

In the case of the ocean, the density of water, depending on its temperature and salinity, varies with depth. For this reason the pressure cannot be calculated with the traditional pressure formula for the water column. It is necessary to consider the effect of the variation in density and calculate by integrating the mass along the column the pressure that occurs at the depth that we wish to estimate.

>Model

ID:(1598, 0)



Characterization of the ocean layers

Image

>Top


Ekman's transport causes the boundaries between the surface and deepest layers in the ocean to shift. These are characterized by sudden changes in parameters depending on the temperature. In particular there are changes in:

Temperature (thermocline)
Salinity (halocline)
Density (pycnocline)

ID:(11684, 0)



Column with variable density

Image

>Top


To calculate the pressure under the sea at a given depth, one must first estimate the mass of a volume element at a certain depth:

The problem in this case is that the density is not constant so the typical relationship of the pressure of the water column cannot be applied.

ID:(12008, 0)



Mass element

Equation

>Top, >Model


A water element of a height dz , section S and density \ rho has a mass:

$ dm = \rho S dz $

ID:(12010, 0)



Force element

Equation

>Top, >Model


Con la definición de la fuerza gravitacional

$ F_g = m_g g $



el aumento de la fuerza en función de la masa es

$ dF = g \, dm $

ID:(12012, 0)



Variation of force with depth

Equation

>Top, >Model


Con la variación de la masa

$ dm = \rho S dz $



y la variación de la fuerza en función de la masa

$ dF = g \, dm $



con lo que se obtiene

$ dF = \rho g S dz $

ID:(12009, 0)



Pressure element

Equation

>Top, >Model


Con la definición de la presión

$ p \equiv\displaystyle\frac{ F }{ S }$



la presión aumenta con la fuerza según

$ dp =\displaystyle\frac{ dF }{ S }$

en donde se asume que la sección no varia.

ID:(12013, 0)



Pressure increase ratio with depth

Equation

>Top, >Model


Con la definición de la presión

$ dF = \rho g S dz $



el aumento de la fuerza

$ dp =\displaystyle\frac{ dF }{ S }$



lleva a un aumento de la presión

$ dp = \rho g dz $

ID:(12011, 0)



Density modeling

Image

>Top


If you look at the curve of the density of ocean water as a function of depth, you see that it has the shape of an inverted exponential. In other words, the upper part is allowed to compress, reaching a limit where the weight of the column does not lead to greater compression:

ID:(12014, 0)



Seawater Density Model

Equation

>Top, >Model


If you observe the curve of density with depth you can model this with a value for surface density \ rho_0 \ sim 1,025 , g / cm ^ 3 and a very deep value of \ rho _ {\ infty} \ sim 1,028 , g / cm ^ 3 to which it converges exponentially with an average depth of (reaching 36% of the original value) of about 500 , m whose inverse value We will call \ lambda . In this way we have the model:

$ \rho = \rho_{\infty} - (\rho_{\infty}-\rho_0)e^{-\lambda z }$

ID:(11882, 0)



Depth pressure calculation

Equation

>Top, >Model


Con el incremento de la presión dp cuando se incrementa la profundidad dz

$ dp = \rho g dz $



se puede mediante integración calcular la presión para cualquier profundidad:

$ p = p_0 + g\displaystyle\int_0^z \rho\,du$

ID:(11881, 0)



Pressure as a function of depth

Equation

>Top, >Model


Si se emplea la función de la densidad

$ \rho = \rho_{\infty} - (\rho_{\infty}-\rho_0)e^{-\lambda z }$



en la ecuación de la presión

$ p = p_0 + g\displaystyle\int_0^z \rho\,du$



se obtiene

$ p = p_0 + \rho_{\infty} g z - \displaystyle\frac{( \rho_{\infty} - \rho_0 ) g }{ \lambda }(1- e^{- \lambda z })$

ID:(11883, 0)



0
Video

Video: Column of water in the sea