4
$\begingroup$

The 2-layer hydrostatic model is like thisenter image description here

And the pressure of the two layers and the top areenter image description here and it is easy to find the horizontal gradient of $p_1$ and $p_2$ \begin{equation} \begin{aligned} \nabla p_1 &= \nabla p_H\\ \nabla p_2 &= \nabla p_H + g(\rho_2 - \rho_1)\nabla h_2\\ \end{aligned} \end{equation} Now my problem is that how to get the result \begin{equation} p_2 - p_1 = g(\rho_2 - \rho_1)\eta \end{equation}

from the equations above, where$\eta = h_2 - H_2$. I find this result in the book"Fundamentals of Geophysical Fluid Dynamics" at page 164.

$\endgroup$
10
  • 1
    $\begingroup$ What have you tried? Is this the solution with the rigid layer (top) or with variable water level? $\endgroup$
    – arkaia
    Apr 14 at 13:31
  • 1
    $\begingroup$ Is that equation in your reference? I can't find it $\endgroup$ Apr 14 at 13:58
  • $\begingroup$ Just to help us verify... not a homework question, correct? $\endgroup$ Apr 15 at 1:40
  • 1
    $\begingroup$ @JeopardyTempest no…I find this problem when I was reading the book $\endgroup$
    –  Hou
    Apr 15 at 8:12
  • 1
    $\begingroup$ I'm with @J.Fregin... I do not see your final result equation in the book??? $\endgroup$ Apr 15 at 10:25

1 Answer 1

4
$\begingroup$

Let's say our goal is to find $\eta$ which is the displacement of the fluid relative to its resting position at $z = H_2$.

We find \begin{equation} \begin{split} p_2 - p_1 &= \require{cancel} \bcancel{p_H} + \rho_1 g(\require{cancel} \bcancel{H}-h_2) + \rho_2g(h_2 - z) \require{cancel} \bcancel{-p_H} - \rho_1g(\require{cancel} \bcancel{H}-z) \\ &= (\rho_2 - \rho_1)gh_2 - (\rho_2 - \rho_1)gz \\ & = (\rho_2 - \rho_1)g(H_2 + \eta) - (\rho_2 - \rho_1)gz \end{split} \end{equation}

Dividing by $(\rho_2 - \rho_1)g$ yields

\begin{equation} \eta + H_2 - z = \frac{p_2 - p_1}{(\rho_2 - \rho_1)g}. \end{equation} The equation above tells us the distance to the interface at some height $z$ (remember the coordinate origin is at the bottom of the domain). We are interested in the displacement relative to the mean interface height $z = H_2$, so we have \begin{equation} \eta = \frac{p_2 - p_1}{(\rho_2 - \rho_1)g}, \end{equation} which is the result you are looking for. However, in the document they say that they use four more equations to derive the result. Maybe it's a mistake or maybe I'm missing something.

I was a little confused that you said we can find the result in the document - so for anyone wondering:

Using $\phi_n = p_n / \rho_0$ and $g' = g(\rho_2 - \rho_1)/\rho_0$, we find \begin{equation} \eta = \frac{(p_2 - p_1)\rho_0}{(\rho_2 - \rho_1)g\rho_0} = -\frac{\phi_1 - \phi_2}{g'}, \end{equation} which corresponds to what's shown in the document.

$\endgroup$
5
  • $\begingroup$ I’m little confused about the coordinate $z$ here, I mean, to compute the pressure difference between two locations in two layers respectively, why you only used one $z$ in your answer not $z_1$(corresponding to the first layer $/rho_1$) and $z_2$(corresponding to $/rho_2$)? I think the pressure difference is a function of $z_1$ and $z_2$. $\endgroup$
    –  Hou
    Apr 17 at 4:22
  • $\begingroup$ Good question - I have two for you too: What happens if you just use $z= H_2$ ? Does this capture the qualitative behaviour of the displacement $\eta$? I think we can't interpret the equations as a difference between two locations. There is just one. $\endgroup$ Apr 17 at 8:16
  • $\begingroup$ But what is the physical meaning of the equation when you set $z=H_2$? $p_1$ and $p_2$ should be the pressure of the two layers respectively. $\endgroup$
    –  Hou
    Apr 17 at 9:10
  • $\begingroup$ The way I understand it is that you just assume that either layer 1 or layer 2 extends to $H_2$, which of course is unphysical in the case of one layer. I struggle to find another explanation for this using the simple rearrangement in my answer above. Notice that in practice you would solve the pde-system to find your unknowns instead of using this equation. Maybe someone else can help out here? $\endgroup$ Apr 17 at 11:11
  • $\begingroup$ Well, I have no ideas now, actually I think it is no need to compute the pressure difference, since we actually use the gradient of the pressure in the momentum equations $\endgroup$
    –  Hou
    Apr 17 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.